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[ "# §9.2: Small subsets of the hypercube are noise-sensitive\n\nAn immediate consequence of the Bonami Lemma is that for any $f : \\{-1,1\\}^n \\to {\\mathbb R}$ and $k \\in {\\mathbb N}$, \\begin{equation} \\label{eqn:2-4-hypercon-k} \\|\\mathrm{T}_{1/\\sqrt{3}} f^{=k}\\|_4 = \\tfrac{1}{\\sqrt{3}^k} \\|f^{=k}\\|_4 \\leq \\|f^{=k}\\|_2. \\end{equation}\n\nThis is a special case of the $(2,4)$-Hypercontractivity Theorem (whose name will be explained shortly), which says that the assumption of degree-$k$ homogeneity is not necessary:\n\n$\\mathbf{(2,4)}$-Hypercontractivity Theorem Let $f : \\{-1,1\\}^n \\to {\\mathbb R}$. Then $\\|\\mathrm{T}_{1/\\sqrt{3}} f\\|_4 \\leq \\|f\\|_2.$\n\nIt almost looks as though you could prove this theorem simply by summing \\eqref{eqn:2-4-hypercon-k} over $k$. In fact that proof strategy can be made to work given a few extra tricks (see the exercises), but it’s just as easy to repeat the induction technique used for the Bonami Lemma.\n\nProof of the $(2,4)$-Hypercontractivity Theorem: We’ll prove $\\mathop{\\bf E}[\\mathrm{T}_{1/\\sqrt{3}} f({\\boldsymbol{x}})^4] \\leq \\mathop{\\bf E}[f({\\boldsymbol{x}})^2]^2$ using the same induction as in the Bonami Lemma. Retaining the notation $\\boldsymbol{d}$ and $\\boldsymbol{e}$, and using the shorthand $\\mathrm{T} = \\mathrm{T}_{1/\\sqrt{3}}$, we have $\\mathrm{T} \\boldsymbol{f} = {\\boldsymbol{x}}_n \\cdot \\tfrac{1}{\\sqrt{3}} \\mathrm{T} \\boldsymbol{d} + \\mathrm{T} \\boldsymbol{e}.$ Similar computations to those in the Bonami Lemma proof yield \\begin{align*} \\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{f})^4] &= \\bigl(\\tfrac{1}{\\sqrt{3}}\\bigr)^4\\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{d})^4] + 6 \\bigl(\\tfrac{1}{\\sqrt{3}}\\bigr)^2 \\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{d})^2(\\mathrm{T} \\boldsymbol{e})^2] + \\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{e})^4] \\\\ &\\leq \\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{d})^4] + 2 \\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{d})^2(\\mathrm{T} \\boldsymbol{e})^2] + \\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{e})^4] \\\\ &\\leq \\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{d})^4] + 2 \\sqrt{\\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{d})^4]}\\sqrt{\\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{e})^4]} + \\mathop{\\bf E}[(\\mathrm{T} \\boldsymbol{e})^4] \\\\ &\\leq \\mathop{\\bf E}[\\boldsymbol{d}^2]^2 + 2 \\mathop{\\bf E}[\\boldsymbol{d}^2]\\mathop{\\bf E}[\\boldsymbol{e}^2] + \\mathop{\\bf E}[\\boldsymbol{e}^2]^2 \\\\ &= \\bigl(\\mathop{\\bf E}[\\boldsymbol{d}^2] + \\mathop{\\bf E}[\\boldsymbol{e}^2]\\bigr)^2 = \\mathop{\\bf E}[\\boldsymbol{f}^2]^2, \\end{align*} where the second inequality is Cauchy–Schwarz, the third is induction, and the final equality is a simple computation analogous to equation (2) in the proof of the Bonami Lemma. $\\Box$\n\nThe name “hypercontractivity” in this theorem describes the fact that not only is $\\mathrm{T}_{1/\\sqrt{3}}$ a “contraction” on $L^2(\\{-1,1\\}^n)$ — meaning $\\|\\mathrm{T}_{1/\\sqrt{3}} f\\|_2 \\leq \\|f\\|_2$ for all $f$ — it’s even a contraction when viewed as an operator from $L^2(\\{-1,1\\}^n)$ to $L^4(\\{-1,1\\}^n)$. You should think of hypercontractivity theorems as quantifying the extent to which $\\mathrm{T}_\\rho$ is a “smoothing”, or “reasonable-izing” operator.\n\nUnfortunately the quantity $\\|\\mathrm{T}_{1/\\sqrt{3}}f\\|_4$ in the $(2,4)$-Hypercontractivity Theorem does not have an obvious combinatorial meaning. On the other hand, the quantity $\\|\\mathrm{T}_{1/\\sqrt{3}}f\\|_2 = \\sqrt{\\langle \\mathrm{T}_{1/\\sqrt{3}}f, \\mathrm{T}_{1/\\sqrt{3}}f \\rangle} = \\sqrt{\\langle f, \\mathrm{T}_{1/\\sqrt{3}} \\mathrm{T}_{1/\\sqrt{3}} f\\rangle} = \\sqrt{\\mathbf{Stab}_{1/3}[f]},$ does have a nice combinatorial meaning. And we can make this quantity appear in the Hypercontractivity Theorem via a simple trick from analysis, just using the fact that $\\mathrm{T}_{1/\\sqrt{3}}$ is a self-adjoint operator. We “flip the norms across $2$” using Hölder’s inequality:\n\n$\\mathbf{(4/3,2)}$-Hypercontractivity Theorem Let $f : \\{-1,1\\}^n \\to {\\mathbb R}$. Then $\\|\\mathrm{T}_{1/\\sqrt{3}} f\\|_2 \\leq \\|f\\|_{4/3};$ i.e., \\begin{equation} \\label{eqn:4/3-2} \\mathbf{Stab}_{1/3}[f] \\leq \\|f\\|_{4/3}^2. \\end{equation}\n\nProof: Writing $\\mathrm{T} = \\mathrm{T}_{1/\\sqrt{3}}$ for brevity we have \\begin{equation} \\label{eqn:hc-holder} \\|\\mathrm{T} f\\|_2^2 = \\langle \\mathrm{T} f, \\mathrm{T} f \\rangle = \\langle f, \\mathrm{T} \\mathrm{T} f \\rangle \\leq \\|f\\|_{4/3} \\|\\mathrm{T} \\mathrm{T} f\\|_4 \\leq \\|f\\|_{4/3} \\|\\mathrm{T} f\\|_2 \\end{equation} by Hölder’s inequality and the $(2,4)$-Hypercontractivity Theorem. Dividing through by $\\|\\mathrm{T} f\\|_2$ (which we may assume is nonzero) completes the proof. $\\Box$\n\nIn the inequality \\eqref{eqn:4/3-2} the left-hand side is a natural quantity. The right-hand side is just $1$ when $f : \\{-1,1\\}^n \\to \\{-1,1\\}$, which is not very interesting. But if we instead look at $f : \\{-1,1\\}^n \\to \\{0,1\\}$ we get something very interesting:\n\nCorollary 8 Let $A \\subseteq \\{-1,1\\}^n$ have volume $\\alpha$; i.e., let $1_A : \\{-1,1\\}^n \\to \\{0,1\\}$ satisfy $\\mathop{\\bf E}[1_A] = \\alpha$. Then $\\mathbf{Stab}_{1/3}[1_A] = \\mathop{\\bf Pr}_{\\substack{{\\boldsymbol{x}} \\sim \\{-1,1\\}^n \\\\ \\boldsymbol{y} \\sim N_{1/3}({\\boldsymbol{x}})}}[{\\boldsymbol{x}} \\in A, \\boldsymbol{y} \\in A] \\leq \\alpha^{3/2}.$ Equivalently (for $\\alpha > 0$), $\\mathop{\\bf Pr}_{\\substack{{\\boldsymbol{x}} \\sim A \\\\ \\boldsymbol{y} \\sim N_{1/3}({\\boldsymbol{x}})}}[\\boldsymbol{y} \\in A] \\leq \\alpha^{1/2}.$\n\nProof: This is immediate from inequality \\eqref{eqn:4/3-2}, since $\\|1_A\\|_{4/3}^2 = \\Bigr(\\mathop{\\bf E}_{{\\boldsymbol{x}}}[|1_A({\\boldsymbol{x}})|^{4/3}]^{3/4}\\Bigr)^2 = \\mathop{\\bf E}_{{\\boldsymbol{x}}}[1_A({\\boldsymbol{x}})]^{3/2} = \\alpha^{3/2}. \\qquad \\Box$\n\nSee Section 5 of this chapter for the generalization of this corollary to noise rates other than $1/3$.\n\nExample 9 Assume $\\alpha = 2^{-k}$, $k \\in {\\mathbb N}^+$, and $A$ is a subcube of codimension $k$; e.g., $1_A : {\\mathbb F}_2^n \\to \\{0,1\\}$ is the logical $\\mathrm{AND}$ function on the first $k$ coordinates. For every $x \\in A$, when we form $\\boldsymbol{y} \\sim N_{1/3}(x)$ we’ll have $\\boldsymbol{y} \\in A$ if and only if the first $k$ coordinates of $x$ do not change, which happens with probability $(2/3)^k = (2/3)^{\\log(1/\\alpha)} = \\alpha^{\\log(3/2)} \\approx \\alpha^{.585} \\leq \\alpha^{1/2}$. In fact, the bound $\\alpha^{1/2}$ in Corollary 8 is essentially sharp when $A$ is a Hamming ball; see the exercises.\n\nWe can phrase Corollary 8 in terms of the expansion in a certain graph:\n\nDefinition 10 For $n \\in {\\mathbb N}^+$ and $\\rho \\in [-1,1]$, the $n$-dimensional $\\rho$-stable hypercube graph is the edge-weighted, complete directed graph on vertex set $\\{-1,1\\}^n$ in which the weight on directed edge $(x,y) \\in \\{-1,1\\}^n \\times \\{-1,1\\}^n$ is equal to $\\mathop{\\bf Pr}[({\\boldsymbol{x}},\\boldsymbol{y}) = (x,y)]$ when $({\\boldsymbol{x}},\\boldsymbol{y})$ is a $\\rho$-correlated pair. If $\\rho = 1-2\\delta$ for $\\delta \\in [0,1]$, we also call this the $\\delta$-noisy hypercube graph. Here the weight on $(x,y)$ is $\\mathop{\\bf Pr}[({\\boldsymbol{x}}, \\boldsymbol{y}) = (x,y)]$ where ${\\boldsymbol{x}} \\sim \\{-1,1\\}^n$ is uniform and $\\boldsymbol{y}$ is formed from ${\\boldsymbol{x}}$ by negating each coordinate independently with probability $\\delta$.\n\nRemark 11 The edge weights in this graph are nonnegative and sum to $1$. The graph is also “regular” in the sense that for each $x \\in \\{-1,1\\}^n$ the sum of all the edge weight leaving (or entering) $x$ is $2^{-n}$. You can also consider the graph to be undirected, since the weight on $(x,y)$ is the same as the weight on $(y,x)$; in this viewpoint, the weight on the undirected edge $(x,y)$ would be $2^{1-n}\\delta^{\\Delta(x,y)}(1-\\delta)^{n-\\Delta(x,y)}$. In fact, the graph is perhaps best thought of as the discrete-time Markov chain on state space $\\{-1,1\\}^n$ in which a step from state $x \\in \\{-1,1\\}^n$ consists of moving to state $\\boldsymbol{y} \\sim N_\\rho(x)$. This is a reversible chain with the uniform stationary distribution. Each discrete step is equivalent to running the “usual” continuous-time Markov chain on the hypercube for time $t = \\ln(1/\\rho)$ (assuming $\\rho \\in [0,1]$).\n\nWith this definition in place, we can see Corollary 8 as saying that the $1/3$-stable (equivalently, $1/3$-noisy) hypercube graph is a “small-set expander”: given any small $\\alpha$-fraction of the vertices $A$, almost all of the edge weight touching $A$ is on its boundary. More precisely, if we choose a random vertex ${\\boldsymbol{x}} \\in A$ and take a random edge out of ${\\boldsymbol{x}}$ (with probability proportional to its edge weight), we end up outside $A$ with probability at least $1 – \\alpha^{1/2}$. You can compare this with the discussion surrounding the Level-$1$ Inequality in Chapter 5.4, which is the analogous statement for the $\\rho$-stable hypercube graph “in the limit $\\rho \\to 0^+$”. The appropriate statement for general $\\rho$ is appears in Section 5 of this chapter as the “Small-Set Expansion Theorem”.\n\nCorollary 8 would apply equally well if $1_A$ were replaced by a function $g : \\{-1,1\\}^n \\to \\{-1,0,1\\}$, with $\\alpha$ denoting $\\mathop{\\bf Pr}[g \\neq 0] = \\mathop{\\bf E}[|g|] = \\mathop{\\bf E}[g^2]$. This situation occurs naturally when $g = \\mathrm{D}_i f$ for some boolean-valued $f : \\{-1,1\\}^n \\to \\{-1,1\\}$. In this case $\\mathbf{Stab}_{1/3}[g] = \\mathbf{Inf}^{(1/3)}_i[f]$, the $1/3$-stable influence of $i$ on $f$. We conclude that for a boolean-valued function, if the influence of $i$ is small then its $1/3$-stable influence is much smaller:\n\nCorollary 12 Let $f : \\{-1,1\\}^n \\to \\{-1,1\\}$. Then $\\mathbf{Inf}_i^{(1/3)}[f] \\leq \\mathbf{Inf}_i[f]^{3/2}$ for all $i$.\n\nWe remark that the famous KKL Theorem (stated in Chapter 4.2) more or less follows by summing the above inequality over $i \\in [n]$; we could now prove it quite easily in a paragraph or so.\n\nLet’s take one more look at the “small-set expansion result”, Corollary 8. Since noise stability roughly measures how “low” a function’s Fourier weight is, this corollary implies that a function $f : \\{-1,1\\}^n \\to \\{0,1\\}$ with small mean $\\alpha$ cannot have much of its Fourier weight at low degree. More precisely, for any $k \\in {\\mathbb N}$ we have \\begin{equation} \\label{eqn:weak-level-k} \\alpha^{3/2} \\geq \\mathbf{Stab}_{1/3}[f] \\geq (1/3)^k \\mathbf{W}^{\\leq k}[f] \\quad\\Rightarrow\\quad \\mathbf{W}^{\\leq k}[f] \\leq 3^k \\alpha^{3/2}. \\end{equation} For $k = 1$ this gives $\\mathbf{W}^{\\leq 1}[f] \\leq 3\\alpha^{3/2}$, which is nontrivial but not as strong as the Level-1 Inequality from Chapter 5.4. But \\eqref{eqn:weak-level-k} also gives us “level-$k$ inequalities” for larger values of $k$. For example, $\\mathbf{W}^{\\leq .25 \\log(1/\\alpha)}[f] \\leq \\alpha^{-.25 \\log 3 + 3/2} \\leq \\alpha^{1.1} \\ll \\alpha = \\|f\\|_2^2;$ i.e., almost all of $f$’s Fourier weight is above degree $.25 \\log(1/\\alpha)$. We will give slightly improved versions of these level-$k$ inequalities in Section 5 of this chapter.\n\n### 2 comments to §9.2: Small subsets of the hypercube are noise-sensitive\n\n•", null, "Mitchell Johnston\n\nHi Ryan,\n\nFor the equations in the (2,4) theorem you leave the Ts in when you apply the induction hypothesis which gives you the wrong result in the end.\n\nBest,\nMitchell\n\n•", null, "Ryan O'Donnell\n\nThanks Mitchell — that was a big mistake! I think it’s fixed now." ]

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[ "Korean J Ophthalmol > Volume 28(6); 2014 > Article\nMoon, Lim, and Lee: Accuracy of Biometry for Intraocular Lens Implantation Using the New Partial Coherence Interferometer, AL-scan\n\n### Purpose\n\nTo compare the refractive results of cataract surgery measured by applanation ultrasound and the new partial coherence interferometer, AL-scan.\n\n### Methods\n\nMedical records of 76 patients and 104 eyes who underwent cataract surgery from January 2013 to June 2013 were retrospectively reviewed. Biometries were measured using ultrasound and AL-scan and intraocular lens power was calculated using the SRK-T formula. Automatic refraction examination was done 1 month after the operation, and differences between the ultrasound group and AL-scan group were compared and analyzed by mean absolute error.\n\n### Results\n\nMean axial length measured preoperatively by the ultrasound method was 23.53 ┬▒ 1.17 mm while the lengths measured using the AL-scan were 0.03 mm longer than that of the ultrasound group (23.56 ┬▒ 1.15 mm). However, there was not a significant difference in this finding (p = 0.638). Mean absolute error was 0.34 ┬▒ 0.27 diopters in the ultrasound group and 0.36 ┬▒ 0.31 diopters in AL-scan group, which showed no significant difference (p = 0.946) in precision of predicting postoperative refraction.\n\n### Conclusions\n\nAlthough the difference was not statistically significant, intraocular lens calculations done by the AL-scan were nearly similar in predicting postoperative refraction compared to those of applanation ultrasound, however more precise measurements may be obtained if the axial length is longer than 24.4 mm. Except in the case of opacity in the media, which makes obtaining measurements with the AL-scan difficult, AL-scan could be a useful biometry in cataract surgery.\n\nSatisfaction of patients undergoing cataract surgery is dependent on precise predictions of refractory outcomes. Over the years, development of biometry, phacoemulsification, and intraocular lens (IOL) calculation enabled precise prediction of postoperative refractory status.\nTo obtain accurate IOL power, a number of factors are needed. These biometries include axial length, corneal refractive power, and anterior chamber depth. Among these factors, Olsen reported that axial length plays a main role in determining postoperative refraction and is responsible for 54% of the actual refractive error. Axial length error of 100 ┬Ąm translates to a postoperative refraction error of 0.28 diopters (D).\nUntil recently, axial length was measured by using applanation ultrasound technique, which involves contact with the cornea and can result in corneal epithelial injury, infection, and patient discomfort. Error due to corneal indentation, which can lead to axial lengths 0.1 to 0.3 mm shorter than those measured by the immersion technique, is also a major disadvantage of the applanation ultrasound method .\nTo overcome this limitation, a partial coherence interferometer (PCI), which is based on the principle similar to that of optical coherence tomography, was introduced. Axial length measured by this method was comparable to that of other methods in precision and repeatability. Especially considering that the method is of the non-applanation type, it has the advantage of giving the patient less discomfort and has a low interobserver error [3,4,5,6].\nCommonly used PCIs in the clinical setting include the IOL Master (Carl Zeiss Meditiec, Jena, Germany) and Lenstar (Haag Steit AG, Koeniz, Switzerland). Recently, the new PCI AL-scan (Nidek, Gamagori, Japan) has been introduced and increases precision by using a 3-dimentional ocular tracking technique. By using PCI and scheimpflug imaging techniques, AL-scan made it possible to measure axial length, corneal refractive power, anterior chamber depth, central corneal thickness, white-to-white distance, and pupil size in a single sitting based on those values and automatically calculates the appropriate IOL power to be used in cataract surgery by onboard software.\nAccording to previous studies, it is known that the IOL Master and Lenstar show greater or similar accuracy compared to the conventional ultrasound techniques [3,7,8]. However, studies on the newly launched PCI, AL-scan, are lacking. Therefore, in this study we used AL-scan to measure axial length and refraction. With the obtained values, IOL power is calculated to look for the degree of error compared with the predicted value preoperatively and analyze the accuracy compared to values obtained from ultrasound.\n\n### Materials and Methods\n\nMedical records of 76 patients (104 eyes) who underwent phacoemulsification and posterior chamber IOL (AcrySof IQ SN60WF; Alcon, Forth Worth, TX, USA) implantation from January 2013 to June 2013 were retrospectively reviewed. The operation was the temporal clear corneal incision technique. Patients had follow up of more than 1 month.\nThose with factors known to inf luence visual acuity were excluded such as history of intraocular operation, inflammation, retinopathy, and others. Cases of posterior capsule rupture during cataract surgery and those requiring sutures at the corneal incision site were excluded. Furthermore, patients with axial length that could not be measured by the AL-scan were not included in the statistical process.\nIn the AL-scan group, axial length, anterior chamber depth, and corneal refractive power were measured by the AL-scan. In the ultrasound group, corneal refractive power was measured by auto kerato refractometer (KR-1; Topcon, Tokyo, Japan), and applanation ultrasound (Echoscan US-4000, Nidek) was used to obtain anterior chamber depth and axial length.\nIOL power was calculated using the SRK-T formula and the A-constant was maintained at 118.7, a value provided by the manufacturer. The predicted refraction value targeted emmetropia and myopia on patient request. Automatic refraction examination was done at the 1-month follow-up after the operation to obtain the actual postoperative refraction value.\nMean numerical error (MNE) was achieved based on the difference between actual refraction value and desired refraction value using the SRK-T formula. We assessed the result as hyperopic if the value was negative and myopic when positive. Also, with the absolute value of actual difference and by averaging it, mean absolute error (MAE) could be calculated to evaluate the precision of the IOL power calculation. Differences between the ultrasound group and AL-scan group were compared and analyzed by MNE and MAE.\nIt has been reported that biometry measurements by applanation ultrasound technique and by PCI technique vary by axial length. Generally the measurement attained from the PCI is higher [6,9] so if the reverse result is drawn in the actual clinic setting the reliability of the measurement must be questioned. Also, Lee et al. reported that postoperative refraction predicted by using the SRK-T formula is generally precise but refraction error had a relationship with the axial length. Based on the axial length of 24.4 mm, which is the value used to compensate axial length in the formula, axial lengths shorter than 24.4 mm tend to have hyperopic shift as eyes get shorter, and vice versa. Therefore, in this study we separated the group of patients with longer axial length measurements by ultrasound and AL-scan and analyzed the results drawn from those groups. Also, patients were divided into two groups by axial length of 24.4 mm, which was measured by ultrasound.\nDifferences of axial length measured by ultrasound and PCI were compared and analyzed using the paired t-test. Comparison of predictive error between those two methods was analyzed by absolute value using Wilcoxon rank test. A p-value of less than 0.05 indicated statistical significance.\n\n### Results\n\nA total of 75 patients and 104 eyes were included in this study. The mean age was 67.4 ┬▒ 9.6 years. A total of 6 eyes failed to have axial length measured by the AL-scan (5.77%) and were excluded from the statistic analysis.\nAxial length measurement by the two methods showed a statistically significant relationship (r = 0.976, p < 0.01) (Fig. 1). The dotted lines are the line of equality for the corresponding two methods. The Bland-Altman plot was examined to evaluate the agreement between the two methods (Fig. 2). Most values are along the dotted line (┬▒1.96 standard deviation [SD], -0.3919 to 0.3251), meaning the methods were comparable. Furthermore, the slope of the regression line being 0.012 represents low error between the two methods and high reliability.\nMean axial length measured preoperatively by ultrasound was 23.53 ┬▒ 1.17 mm, and AL-scan measurements were 0.03 mm longer than that of the ultrasound group (23.56 ┬▒ 1.15 mm). However, there was no significant difference in this finding (p = 0.638). Anterior chamber depth measured by the two methods was 3.17 ┬▒ 0.34 mm and 3.14 ┬▒ 0.39 mm, with the ultrasound group having the lower value, but the difference was not significant (p = 0.764) (Table 1).\nPost operative MNE in the ultrasound group was slightly hyperopic by -0.05 ┬▒ 0.43 D. In addition the AL-scan group had a tendency to be myopic by 0.13 ┬▒ 0.45 D, which was a statistically significant difference (p = 0.02). MAE was 0.34 ┬▒ 0.27 D in the ultrasound group and 0.36 ┬▒ 0.31 D in the AL-scan group, which showed no significant difference (p = 0.946) in precision in predicting post-operative refraction (Table 1). There was no statistically significant difference in corneal refractive power between the two groups as group in 73.5% of cases, and 75.5% of cases in the AL-scan group. Furthermore, MAE less than 1 D was 99% and 97% each, and less than 1.5 D was 100% and 99% each, respectively (Table 2).\nThe group with an axial length shorter than 24.4 mm included a total of 86 eyes, mean axial length measured by applanation ultrasound method and AL-scan was 23.18 ┬▒ 0.65 mm and 23.23 ┬▒ 0.68 mm, respectively and neither finding had a significant difference (p = 0.551) (Table 3). There were 12 eyes with an axial length longer than 24.4 mm and mean axial length measured by each method was 25.84 ┬▒ 1.21 mm (ultrasound group) and 25.95 ┬▒ 1.08 mm (AL-scan group), which also showed no significant difference (p = 0.751) (Table 4). However, the difference between the two methods was 0.11 mm, which was relatively higher than the group with an axial length of less than 24.4 mm.\nIn the case of an axial length less than 24.4 mm, MNE was -0.03 ┬▒ 0.41 D in the ultrasound group and MAE was 0.32 ┬▒ 0.25 D. In the AL-scan group, it was 0.12 ┬▒ 0.48 D, 0.38 ┬▒ 0.32 D each, which showed no significant difference in precision between two groups (p = 0.088, 0.416) (Table 3). In the case of an axial length longer than 24.4 mm, MNE was -0.14 ┬▒ 0.56 D in the ultrasound group and MAE was 0.44 ┬▒ 0.34 D. In the AL-scan group, it was 0.22 ┬▒ 0.22 D, 0.22 ┬▒ 0.21 D each, which has shown no significant difference in precision between two groups (p = 0.076, 0.064) (Table 4). However, comparing the results of the axial length shorter than 24.4 mm group (MAE, 0.06), MAE was relatively higher in axial length longer than 24.4 mm group by 0.22 D. Also, by looking at the result distribution of each method, when the axial length is shorter than 24.4 mm, the distribution between each method was even. On the other hand, in the group with an axial length longer than 24.4 mm, the percentage of patients with an MAE lower than 0.5 D was 83.3% in AL-scan group, which was higher than the ultrasound group (58.3%) (Table 2).\nThe axial length of 21 eyes measured by the ultrasound method was longer than that measured by the AL-scan. In this case, MAE values of the ultrasound and AL-scan groups were 0.49 ┬▒ 0.33 and 0.53 ┬▒ 0.38 D, respectively, and there was no significant difference in precision between the two groups (p = 0.876) (Table 5). However, both groups had higher MAE than any of the other cases, and also rated the MAE lower than 0.5 D was shown to be lower by 46.2% and 50%, respectively (Table 2).\n\n### Discussion\n\nSince PCI principle-based IOL Master was introduced in 1992, many studies have been launched to compare this new technology to the conventional applanation ultrasound method. Generally PCI is capable of measuring axial lengths 0.1 to 0.5 mm longer than applanation ultrasound [11,12]. In this study using AL-scan which is also based on the PCI principle, total patients had longer axial length than that measured by the ultrasound method. This study showed that there was no statistically significant difference between these two groups, but a strong relationship was identified. This result is thought to be from the different measurement technique. PCI and ultrasound have different refraction planes and also, in the case of applanation ultrasound, the degree of corneal indentation differs depending on the skill of the examiner [13,14].\nBy reviewing the previous studies regarding prediction of postoperative refraction, there are many studies reporting PCI benefit over the ultrasound method [12,15]. Conversely, there are studies reporting similar precision between those two methods . This study of the AL-scan also has shown no significant difference in MAE between the two groups using each method. However, the study revealed that MAE in the AL-scan group had the tendency of becoming statistically significantly myopic compared to the ultrasound group. We used the A constant offered by the manufacturer but studies revealed a need for adjusting the A-constant .\nFurthermore, to compare precision in predicting postoperative refraction by the axial length, patients were divided into two groups based on axial length above and below 24.4 mm. Analysis performed by these groups revealed no significant difference in precision between the two groups. However, in the axial length over 24.4 mm group, MAE difference between the ultrasound group and AL-scan group was larger and the percentage of patients showing a desired refraction of +0.5 D was higher in the AL-scan group. There is controversy regarding the relevance of refraction prediction error by axial length using PCI, so further studies will be necessary [9,16].\nThe axial length of 26 eyes was longer using the ultrasound method than the AL-scan method, and there was no significant difference between the two groups using each method. However, MAE was higher than other cases in both the ultrasound and AL-scan group. Postoperative target refraction of +0.5 D was less than 50% in both groups. Theoretically, ultrasound measures sound wave refracted from the internal limiting membrane while PCI measures the light refracted from the retinal pigment epithelium, which results in a difference of 130 ┬Ąm. When considering not only the theoretical background but also the results of this study, systemic error may have played an important role when axial length was longer with the ultrasound technique. Such systemic error includes examiner error, patient error, and possible error when using the AL-scan. So, in this case, remeasurement is required.\nPCI is a simple method and is more comfortable for the patient compared to the ultrasound method. It is also a noncontact method, and thus risk for infection is lower. However, in cases of severe cataracts, posterior capsular cataract, and difficult fixation, PCI cannot measure axial length with an accuracy greater than 10% to 20% . In our study, of the 104 eyes, six eyes (5.77%) could not be measured by the AL-scan. According to the manufacturer, AL-scan can measure eyes with dense cataracts, as advanced measurement algorithms enhance the signal-tonoise ratio by decreasing noise and boosting the signal. However, further study is warranted if AL-scan has higher success rates in measuring axial length.\nHere, we compared the precision in predicting postoperative refraction between AL-scan and applanation ultrasound. Eventually, IOL calculations made with the AL-scan were nearly similar in predicting postoperative refraction compared to that of using the applanation ultrasound, but it may be more precise in predicting postoperative refraction when the axial length is longer than 24.4 mm. Except in the case of opacity of the media, which makes it more difficult to obtain measurements by the AL-scan, this technique could be a useful biometry in cataract surgery.\n\n### REFERENCES\n\n1. Olsen T. Sources of error in intraocular lens power calculation. J Cataract Refract Surg 1992;18:125-129.", null, "", null, "2. Giers U, Epple C. Comparison of A-scan device accuracy. J Cataract Refract Surg 1990;16:235-242.", null, "", null, "3. Findl O, Drexler W, Menapace R, et al. Improved prediction of intraocular lens power using partial coherence interferometry. J Cataract Refract Surg 2001;27:861-867.", null, "", null, "4. Nemeth J, Fekete O, Pesztenlehrer N. Optical and ultrasound measurement of axial length and anterior chamber depth for intraocular lens power calculation. J Cataract Refract Surg 2003;29:85-88.", null, "", null, "5. Findl O, Drexler W, Menapace R, et al. High precision biometry of pseudophakic eyes using partial coherence interferometry. J Cataract Refract Surg 1998;24:1087-1093.", null, "", null, "6. Rajan MS, Keilhorn I, Bell JA. Partial coherence laser interferometry vs conventional ultrasound biometry in intraocular lens power calculations. Eye (Lond) 2002;16:552-556.", null, "", null, "7. Cruysberg LP, Doors M, Verbakel F, et al. Evaluation of the Lenstar LS 900 non-contact biometer. Br J Ophthalmol 2010;94:106-110.", null, "", null, "8. Buckhurst PJ, Wolffsohn JS, Shah S, et al. A new optical low coherence reflectometry device for ocular biometry in cataract patients. Br J Ophthalmol 2009;93:949-953.", null, "", null, "9. Hasemeyer S, Hugger P, Jonas JB. Preoperative biometry of cataractous eyes using partial coherence laser interferometry. Graefes Arch Clin Exp Ophthalmol 2003;241:251-252.", null, "", null, "10. Lee MK, Hwang KY, Kim MS. Effects of axial length and vitrectomy on refractive error after cataract surgery using SRK/T formula. J Korean Ophthalmol Soc 2013;54:257-264.", null, "11. Drexler W, Findl O, Menapace R, et al. Partial coherence interferometry: a novel approach to biometry in cataract surgery. Am J Ophthalmol 1998;126:524-534.", null, "", null, "12. Haigis W, Lege B, Miller N, Schneider B. Comparison of immersion ultrasound biometry and partial coherence interferometry for intraocular lens calculation according to Haigis. Graefes Arch Clin Exp Ophthalmol 2000;238:765-773.", null, "", null, "13. Tehrani M, Krummenauer F, Blom E, Dick HB. Evaluation of the practicality of optical biometry and applanation ultrasound in 253 eyes. J Cataract Refract Surg 2003;29:741-746.", null, "", null, "14. Packer M, Fine IH, Hoffman RS, et al. Immersion A-scan compared with partial coherence interferometry: outcomes analysis. J Cataract Refract Surg 2002;28:239-242.", null, "", null, "15. Kiss B, Findl O, Menapace R, et al. Refractive outcome of cataract surgery using partial coherence interferometry and ultrasound biometry: clinical feasibility study of a commercial prototype II. J Cataract Refract Surg 2002;28:230-234.", null, "", null, "16. Song BY, Yang KJ, Yoon KC. Accuracy of partial coherence interferometry in intraocular lens power calculation. J Korean Ophthalmol Soc 2005;46:775-780.\n17. Madge SN, Khong CH, Lamont M, et al. Optimization of biometry for intraocular lens implantation using the Zeiss IOLMaster. Acta Ophthalmol Scand 2005;83:436-438.", null, "##### Fig.┬Ā1\nScatter plots to compare means for axial length measured with ultrasound and AL-scan. Axial length measurement by two methods showed a statistically significant relationship. USAXL = axial length measured with applanation ultrasound; ALAXL = axial length measured with AL-scan. r = 0.976, p < 0.01.", null, "##### Fig.┬Ā2\nBland-Altman plots for assessing agreement of pairs of two methods. y = 0.012 X - 0.317; mean = -0.0334 ; mean +2 SD = 0.32508 ; mean -2 SD = -0.39188. The solid line represents the average mean difference (-0.0334) and the dotted line represents the 95 percentile confidence interval. The slope of regression line is 0.012 and represents low error between two methods and high reliability. In the scatter plot, most values are on the dotted line (┬▒1.96 SD, -0.3919 to 0.3251), suggesting good comparability. SD = standard deviation; USAXL = axial length measured with applanation ultrasound; ALAXL = axial length measured with AL-scan.", null, "##### Table┬Ā1\nRefractive results: comparison of the ultrasound group and the AL-scan group (all eyes, n = 98)", null, "Values are presented as mean ┬▒ standard deviation.\n\nD = diopter; K = corneal refractive power (keratometric diopter).\n\n##### Table┬Ā2\nPercentage of cases predicted to within ┬▒0.50 D, ┬▒1.00 D, and ┬▒1.50 D of each group", null, "Values are presented as %.\n\nD = diopter; US = ultrasound; USAXL = axial length measured with applanation ultrasound; ALAXL = axial length measured with AL-scan.\n\n##### Table┬Ā3\nRefractive results: comparison of the ultrasound group and the AL-scan group (axial length Ōēż24.4 mm, n = 86)", null, "Values are presented as mean ┬▒ standard deviation.\n\nD = diopter; K = corneal refractive power (keratometric diopter).\n\n##### Table┬Ā4\nRefractive results: comparison of the ultrasound group and the AL-scan group (axial length >24.4 mm, n = 12)", null, "Values are presented as mean ┬▒ standard deviation.\n\nD = diopter; K = corneal refractive power (keratometric diopter).\n\n##### Table┬Ā5\nRefractive results: comparison of the ultrasound group and the AL-scan group (ultrasound axial length > AL-scan axial length, n = 26)", null, "Values are presented as mean ┬▒ standard deviation.\n\nD = diopter; K = corneal refractive power (keratometric diopter).\n\nEditorial Office\nSKY 1004 Building #701\n50-1 Jungnim-ro, Jung-gu, Seoul 04508, Korea\nTel: +82-2-583-6520    Fax: +82-2-583-6521    E-mail: [email protected]", null, "", null, "", null, "", null, "", null, "" ]

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[ "Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nDear morphometricians,\n\nBelow you will find an update to our function for digitizing curves in 2d: curves2d(). This solves a problem with the function plotting landmarks and semilandmarks out of sequence. To use it, you can “source()” the code from a directory, or copy and paste it during digitizing.\n\nCheers,\n\nErik\n\n CODE: curves2d<-function(file, nsliders){lm<-readland.nts(spec.name<-basename(file))lm <- matrix(lm, ncol = dim(lm), byrow=T)spec.name<-unlist(strsplit(spec.name, \"\\\\.\"))plot(lm[,1],lm[,2],cex=1,pch=21,bg=”white”)text(lm[,1],lm[,2],label=paste(“LM”,1:dim(lm)),adj=.5,pos=1)selected<-matrix(NA,ncol=3,nrow=nsliders)select<-NULLfor(i in 1:nsliders){for(j in 1:3){select<-identify(lm,n=1,plot=FALSE,cex=5,pch=25)selected[i,j]<-selectif(j==2){points(lm[select,],lm[select,],cex=1.5,pch=19,col=”red”)arrows(lm[selected[i,j],],lm[selected[i,j],],lm[selected[i,j-1],],lm[selected[i,j-1],],col=”red”,lwd=2,length=.15)} else {points(lm[select,],lm[select,],cex=1.1,pch=19,col=”blue”)}if(j==3){arrows(lm[selected[i,j],],lm[selected[i,j],],lm[selected[i,j-1],],lm[selected[i,j-1],],col=”red”,lwd=2,length=.15,code=1)#lines(rbind(lm[selected[i,j],],lm[selected[i,j-1],]),col=”red”,lwd=2)} else { NA} }}output<-selectedcat(paste(‘”sliders’,sep=””),file=paste(“sliders.nts”,sep=””),sep=”\\n”)cat(paste(1,dim(output),3,0, “dim=3″),file=”sliders.nts”,sep=”\\n”,append=TRUE)write.table(output,file=”sliders.nts”,col.names = FALSE, row.names = FALSE,sep=” “,append=TRUE)return(list(sliders=output))}" ]

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[ "# In classless addressing, we know the first and the last address in the block. Can we find the prefix length? If the answer is yes, show the process and give an example.\n\n631 views", null, "+1 vote", null, "by\nselected by (user.guest)\n\nIf the first and the last addresses are known, the block is fully defined. We can first find the number of addresses in the block. We can then use the relation\n\nN = 232 − n → n = 32 − log2N\n\nto find the prefix length. For example, if the first address is 17.24.12.64 and the last address is 17.24.12.127, then the number of addresses in the block is 64. We can find the prefix length as\n\nn = 32 − log2N = 32 − log264 = 26\n\nThe block is then 72.24.12.64/26.\n\n+1 vote\n–1 vote\n+1 vote\n+1 vote\n+1 vote\n+1 vote\n–1 vote\n–1 vote\n+1 vote\n+1 vote\n+1 vote\n+1 vote\n+1 vote\n+1 vote" ]

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[ "# which of the analog sinusoidal frequency can not pass through the filter?", null, "First time I am encountering this type of question so i just tried but not getting whether my logic correct or not.\n\nFirst let the sinusoidal signal be $X(t)=\\cos(2\\pi ft)$.\n\nAfter sampling this signal I will get $X(nT_s)=\\cos(2\\pi fnT_s)$; now at sampling rate $f_s=18\\, \\text{kHz}$ it becomes $X(n)=\\cos(\\frac{n\\pi f}{9})$.\n\nNow I am checking optionS $(C)$ (given correct answer is also C) to see if the output of filter will zero or not at $f=12\\,\\text{kHz}$ and at 12kHz X(n) becomes $X(nT_s)=\\cos\\left(\\frac{4\\pi n}{3}\\right)$ .\n\n$H(z)=1+z^{-1}+z^{-2}$ can be written as $H(e^{jw})=1+e^{-jw}+e^{-2jw}$, so from eigenvalue concept the value of $H(e^{jw})$ at $w=\\frac{4\\pi}{3}$ is $H(e^{jw})=1+e^{\\frac{-j4\\pi }{3}}+e^{\\frac{-j8\\pi }{3}}=0$, so it's proven that output at 12 kHz will be zero.\n\nTherefore, 12 kHz frequency will not pass.\n\nI am doing by checking options, option B also satisfying this condition.\n\nIs there any alternate method to solve this?\n\n• That was barely readable. Seriously, I know that English might not be your native language, but you definitely understand how English sentences work (your English is actually pretty good). So add \".\" at the end of your sentences. I had to add eleven punctuation marks to make this readable. Also, there's exactly one correct way to write \"kHz\", it's \"kHz\", not \"Khz\" nor \"khz\". Luckily, you even have added a picture that contains the right spelling! – Marcus Müller Sep 3 '17 at 19:11\n• H of Hertz ! V of Volt B of Bell... these are derived from the names of people... and capitalized. – Fat32 Sep 3 '17 at 19:37\n• @MarcusMüller sorry sir :( and Thanks... next time i will try to improve it. – Rohit Sep 4 '17 at 6:12\n\nThe impulse response of the given transfer function $H(z)= 1 + z^{-1} + z^{-2}$ is $h[n] = \\delta[n] + \\delta[n-1] + \\delta[n-2]$.\n\nThis is a rectangular window with three taps and its Frequency response magnitude is given by:\n\n$$H(w) = \\frac{\\sin( 3 w/2 )}{\\sin(w/2)}$$\n\nIt'll have zeros (nulls) at the frequencies given by:\n\n$$\\sin(3w/2) = 0 \\longrightarrow 3 w_k / 2 = \\pi k \\longrightarrow w_k = 2\\pi k / 3 ~~~~, \\text{for} ~~~k = 0,1,2$$\n\nNote that the zero for $k=0$ is cancelled by the denominator and therefore there exists only two zeros $w_1 = 2\\pi/3$ and $w_2 = 4\\pi/3$\n\nNow note that those two zeros are at mirror locations about $w=\\pi$ which means that they will be the same, indistinguishable, discrete time frequencies.\n\nAlso you shall consider only the first zero , $w_1 = 2\\pi/3$ to be the frequency of the analog sinusoidal that will be nulled by the filter.\n\nThe given sampling frequency of $f_s = 18$ kHz maps $w_1= 2\\pi/3$ to the analog frequency of\n\n$$f_1 = \\frac{f_s w_1}{2\\pi} = \\frac{18k 2\\pi/3}{2\\pi} = 6 kHz$$\n\nThe answer therefore is $f_1 = 6$ kHz which cannot pass through your filter when sampled properly at the sampling rate of $f_s = 18$ kHz. This assumes that there is no aliasing during the sampling process which is prevented by an analogue anti-alasing filter before sampling. If this filter is omitted and sampling is performed then a sinusoid at the frequency of $f_2 = 12$ kHz will be aliased into a frequency of $f_1 = 6$ kHz and therefore it won't also pass through the filter. Note that there will be infinetely many new frequencies which will map into $f_1 = 6$ kHz and won't pass through the filter in such a case.\n\n• @Rohit Did you deleted your comment here? You're asking for some details. I didn't have time then. You can ask it again if you need... – Fat32 Sep 6 '17 at 22:26" ]

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[ "Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!\n\n#### Online Quiz (WorksheetABCD)\n\nQuestions Per Quiz = 2 4 6 8 10\n\n### Geometry4.21 The Area of a Circular Path - II\n\n Example: The radius of a circular path of garden is 60 metres. A circular path of width 3.5 metres is laid around and outside the garden. Find its area. Solution: We have given that, r = 60 m and width of the path = 3.5 m. So, the radius of the outer ring = 60 + 3.5 = 63.5 m. Hence the area of the path = p (R+r)(R-r) = 22/7 * (63.5 + 60)(3.5) = 22/7 * 123.5 * 3.5 = 1358.5 sq m. Directions: Read the above example carefully and answer the following questions. Also write at least 10 examples of your own.\n Q 1: The radius of a circular path of garden is 'a' m. A circular path of width 'b' m is laid around and outside the garden. Find its area.(a + b) psq unitsbpsq units(a + b)ap sq unitsabp sq units Q 2: The radius of a circular path of garden is (325/5) m. A circular path of width (3.5/5) metres is laid around and outside the garden. Find its area.2875.4 sq m28.754 sq m287.54 sq m287.45 sq m Q 3: A circular path of width 10.5 m is laid around and outside a garden. Find the radius of the circular path if its area is given by 3910.5 sq m.62 m58 m100 m54 m Q 4: The radius of a circular path of garden is 30 m. Find the width laid around and outside the garden if its area is given by 552.64 sq m.1.2 m3.1 m2.8 m9 m Q 5: The radius of a circular path of garden is 9 m. A circular path of width 3.5 m is laid around and outside the garden. Find its area.2014.5 sq m2016.5 sq m2012.5 sq m2018.5 sq m Q 6: The radius of a circular path of garden is 64 m. Find the width laid around and outside a garden if its area is given by 5758.53 sq m.15 m13 m10 m12 m Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!" ]

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[ "# Number system conversion", null, "## Decimal to number system conversion\n\nTo convert a decimal number system to a binary number system division method is used. The given Decimal number is divided by 2 until the quotient is less than 2. To convert the fraction number to binary multiply the fraction with the base (2) to get a new integer and a fraction. Remove the integer from the result and the multiplication with the new fraction is repeated till the result becomes zero.\n\nExample 1: (25)10 = () 2\n\n2510 = 110012\n\nExample 2: (34.45)10 = (?)2\n\nWhole number Part:\n\nDecimal Part:\n\n0.4510=0.011102\n\nThe bunary equivallent of  (34.45)10= (100010.01110)2\n\n## Decimal to Octal number system conversion\n\nTo convert a decimal number system to an Octal number system division method is used. The given\n\nDecimal number is divided by octal number system’s base 8 until the quotient is less than 8. To convert the fraction number to octal multiply the fraction with the base of octal number (8) to get a new integer and a fraction. Remove the integer from the result and the multiplication with the new fraction is repeated till the result becomes zero.\n\nExample 3: (144.25)10 = (?)8\n\nWhole number Part:\n\n14410 = 2208\n\nDecimal Part:\n\n0.2510=0.208\n\n(144.25)10= (220.20)8\n\n## Decimal to Hexadecimal number system conversion\n\nTo convert a decimal number system to a Hexadecimal number system division method is used. The given Decimal number is divided by 16 until the quotient is less than 16. To convert the fraction number to Hexadecimal multiply the fraction with the base of hexadecimal number (16) to get a new integer and a fraction. Remove the integer from the result and the multiplication with the new fraction is repeated till the result becomes zero.\n\nExample 4: (100)10 = (?)16\n\n10010 = 6416\n\n## Binary to Decimal number system conversion\n\nTo convert a Binary number system to Decimal number system multiplication method is used. The given binary number is multiplied by the increasing powers of 2 based on the bit position\n\nExample 6: (110110)2 = (?)10\n\nGraphical Method:\n\nEquation Method:\n\n= 0 × 20 + 1 × 21 + 1 × 22 + 0 × 23 + 1 × 24 + 1 × 25\n\n= 0 + 2 + 4 + 0 + 16 + 32\n\n= 54\n\n∴ (110110)2 = (54)10\n\nExample 7: (11110.0101)2 = (?)10\n\nGraphical Method:\n\nEquation Method:\n\n=1×2-4 + 0×2-3 + 1×2-2 + 0×2-1 + 0×20 + 1×21 +1×22 +1×23 +1×24\n\n=0.0625+0+0.250+0+0+2+4+8+16\n\n=54.3125\n\n∴ (11110.0101)2 = (54.3125)10\n\n## Binary to Octal number system conversion\n\nTo convert a Binary number system to octal number system the given binary number is split into three bit groups. For this group the equivalent octal values are written\n\nExample 8: (11110101.0101110)2 = (?)8\n\n## Binary to Hexadecimal number system conversion\n\nTo convert a Binary number system to Hexadecimal number system the given binary number is split into four bit groups. For this group the equivalent octal values are written\n\nExample 9: (11110101.0101110)2 = (?)16\n\n(11110101.0101110)2 = (F5.5C)16\n\n## Octal to Binary number system conversion\n\nTo convert the octal number system to binary system represent the values of each octal number as a three bit binary number. Example 10: (7405)8 = (?)2\n\n(7405)8 = (111100000101)2\n\nExample 11: (627.35)8 = (?)2\n\n(627.35)8 = (110010111011101)2\n\n## Hexadecimal to Binary number system conversion\n\nTo convert Hexadecimal number system to binary system represents the values of each Hexadecimal number as four bit binary number.\n\nExample 12: (F405)8 = (?)2\n\n(F405)8 = (1111010000000101)2\n\nExample 13: (ACF.E1)8 = (?)2\n\n(ACF.E1)16 = (101011001111.11100001)2\n\n## Octal to Hexadecimal number system conversion\n\nConvert the octal number to binary. Split the binary number into four bits and represent the equivalent hexadecimal system\n\nExample 14: (405)8 = (?)16\n\n(405)8 = (100000101)2\n\n(405)8 = (105)16\n\nWe will be happy to hear your thoughts", null, "" ]

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[ "IS-LM model/Tutorials", null, "", null, "Main Article Talk Related Articles  [?] Bibliography  [?] External Links  [?] Citable Version  [?] Tutorials [?] Addendum [?] Tutorials relating to the topic of IS-LM model.\n\nThe model\n\nEquilibriums in the real market : IS curve\n\n“IS” stands for “investment and saving”. The IS curve represents all equilibriums for which total spending (consumption and investment) equals total output (income). This equation is equivalent to the equality between investment and saving, which is a basic accounting equation in a closed economy.", null, "Investment is a decreasing function of interest rate. Indeed, the interest rate represents the opportunity cost of an investment. Higher is the interest rate, higher must be the return on investment. It implies that when the interest rate is high many investments whose profitability would be low are not undertaken.", null, "I is a decreasing function of i, and so is Y. Then IS is a decreasing curve.\n\nEquilibriums in the monetary market : LM curve\n\n“LM” refers to “Liquidity preference and money supply”. The LM curve represents all equilibriums where demand equals supply on the monetary market.\n\nAccording to Keynes, people hold liquidities for two reasons.\n\n• Citizens need cash for every day life transactions and because of uncertainty (they are cautious). Keynes calls those needs “transactions demand for money”. It is positively related to income (Y).\n• When people want to save they must choose between securities (financial markets) and money. Money can be considered as an asset among others and its price is the interest rate. Keynes suggests that when the interest rate seems too low, speculators expect it to rise and thus \"buy\" (by selling financial assets for example) or borrow money. On the contrary, when they think that the interest rate is too high, speculators expect it to drop and sell money (by buying financial assets for example).\n\nSo, according to Keynes, demand for money includes transactions demand (L1) and speculative demand (L2). In equilibrium we have the following relationship.", null, "And since", null, "then LM is a increasing curve.\n\nGeneral equilibrium\n\nThe general equilibrium is given by the intersection of IS and LM. The level of income and the interest rate can diverge from their equilibrium level only on a temporary basis. Indeed,\n\n• If the interest rate and the income are situated at point A, then it means that in the goods and services market, supply is insufficient to balance demand. Prices will rise and so will production. It also means that demand is insufficient to balance supply on the monetary market which will make the interest rate fall.\n• At point B, supply of goods and services is also insufficient and so is money supply. This combination implies a rise of both output and interest rate.\n• At point C, supply exceeds demand in the real market, and then companies must reduce their output. The lack of money make interest rate rise.\n• At point D, supply exceeds demand in both the real market and the monetary market. Both production and interest rate will drop.\n\nAnalysis of macroeconomic policies\n\nAccording to Keynes, the labour market is not a market among others. The demand of labour (that is to say the demand of corporations for workers) is not always determined by the level of wages, but also by the aggregate demand (C+I=Y). During a recession, the aggregate demand may be insufficient to require all production factors (capital and labour). In this case, the economy can be in equilibrium with persisting unemployment. In such a situation, the State can stimulate the economy with macroeconomic policies without creating inflationary tensions.\n\nIt can be noted that Keynes suggests implementing such measures in case of recession, while keynesian economists have then defended those policies as permanent regulation.\n\nSlope of the curves\n\nThe efficiency of macroeconomic policies is directly affected by the slope of LM and IS. For example, a liquidity trap, which implies a flat LM curve, makes the monetary policy inefficient.\n\nThe liquidity trap occurs when interest rates would be so low that everyone would expect them to rise. In this case, an infinite speculative demand would actually make them reach a more realistic level. Keynes concludes that interest rates cannot fall under a certain level whatever the income Y. In this situation, LM is horizontal.\n\nOn the contrary, there is a level of interest rate for which speculators have lent all their money. There is no more saving to finance any expansion of the output. In this case, LM is vertical.\n\nIn John Hicks views, the vertical part of LM corresponds to the general equilibrium of classical economics, while the horizontal part was purely keynesian. In fact, the IS-LM model was an extent of the Walrasian theory which had forgotten some particular cases described by Keynes.\n\nBetween those extremes, expansions of the output are possible, but they increase the transactions demand for money and make interest rates rise.\n\nThe slope of IS reflects the interest rate elasticity of investment, and thus the impact of the variations of the interest rate on the output.\n\nFiscal policy\n\nA more detailed model can be made in order to understand the effects of government spending. In this model, G represents governmental spending, T taxes, C consumption of private agents and I investment of private agents.", null, "If C is considered as a proportion of Y noted c, we obtain:", null, "", null, "", null, "", null, "The last equation implies that if the government increases its spending (G) without increasing tax revenues (T), by creating a deficit, then the effect on the national output will be 1/(1-c) times greater. For example, if c=0,8, then an increase of 1 of G implies an increase of 1/(1-0,8)=5 of Y. This mechanism is known as the spending multiplier.\n\nIn a few words, the government can change the level of Y by reducing or expanding the budget deficit.\n\nIn the case of a expansionary fiscal policy, the government increases its purchases or reduces taxes, which creates a budget deficit and makes Y rise as explained before. This deficit shifts the IS curve to the right (Y rises). It results in a new general equilibrium (E'). The output Y has become Y' while the interest rate has risen from i to i'. By the way of higher interest rates, such a policy is said to crowd-out private investment. According to keynesian economics, this negative effect can be exceeded by the positive effect of the spending multiplier at the following conditions :\n\n• The marginal propensity to consumption (c) is high. Higher is the marginal propensity to consumption, more important is the spending multiplier.\n• the slope of the LM curve is low (important speculative demand for money). The importance of the speculative demand for money reduces the rise of interest rates linked to the expansion, and thus the crowding-out effect.\n• and the slope of the IS curve is high (low interest rate elasticity of investment). Private investments are not very affected by the rise of interest rates.\n• prices must be rigid.\n• Some factors of production must be avaible.\n\nA contractionary fiscal policy would have the opposite effects.\n\nMonetary policy\n\nThe IS/LM model assumes that the monetary authority control the supply of money. If the supply of money is increased, then the price of money, i.e. the interest rate, will drop. A lower interest rate will stimulate investments and thus the output.\n\nOn the graph, an expansionary monetary policy makes LM shifts down (drop of the interest rate). Such a measure is particularly efficient under the following conditions:\n\n• A low interest elasticity of the demand for money. In this case, an important decrease of the interest rate would be necessary if demand is to equal supply.\n• A high interest elasticity of investment so that the fall of the interest rate has a large effect on the output.\n• An important multiplier effect.\n• prices must be rigid.\n• Some factors of production are avaible.\n\nA contractionary monetary policy would have the opposite effects." ]

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[ "videoGL.h -- glScalef32 not accepting 3 parameters\n\nRaid\nPosts: 2\nJoined: Tue Apr 20, 2010 4:43 am\n\nvideoGL.h -- glScalef32 not accepting 3 parameters\n\nLooking at the procedure 'glScalef' in videoGL.h, I see that there is 3 int32's being set to the MATRIX_SCALE address.\nThis is correct. This turns the floats from the parameters into int32's. The procedure tells me to use glScalef32 (with preconverted floats I assume of course).\n\nLooking at glScalef32 there is ONE parameter, called factor, and it's being sent 3 times to MATRIX_SCALE .. This scales in all axises with only one value instead of allowing me to input 3, like the previous glScalef method.\n\nThe diff from the old videoGL.h to the patched videoGL.h:\n\nCode: Select all\n\n@@ -554,14 +554,16 @@\n}\n\nGL_STATIC_INL\n-/*! \\fn void glScalef32(int32 factor)\n+/*! \\fn void glScalef32(int32 x, int32 y, int32 z)\n\\brief multiply the current matrix by a scale matrix<BR>\n-\\param factor the factor to scale by */\n- void glScalef32(int32 factor) {\n-\tMATRIX_SCALE = factor;\n-\tMATRIX_SCALE = factor;\n-\tMATRIX_SCALE = factor;\n+\\param x scaling on the x axis\n+\\param y scaling on the y axis\n+\\param z scaling on the z axis */\n+ void glScalef32(int32 x, int32 y, int32 z) {\n+\tMATRIX_SCALE = x;\n+\tMATRIX_SCALE = y;\n+\tMATRIX_SCALE = z;\n}\n\nGL_STATIC_INL\n\nWho is online\n\nUsers browsing this forum: Bing [Bot] and 1 guest" ]

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[ "# nLab Martin-Löf dependent type theory\n\nMartin-Lf dependent type theory\n\n# Martin-Löf dependent type theory\n\n## Idea\n\nPer Martin-Löf‘s dependent type theory, also known as intuitionistic type theory (Martin-Löf 75), or constructive type theory is a specific form of type theory developed to support constructive mathematics. (Note that both “dependent type theory” and “intuitionistic type theory” may refer more generally to type theories that contain dependent types or are intuitionistic, respectively.)\n\nMartin-Löf’s dependent type theory is notable for several reasons:\n\n• One can construct an interpretation of first-order intuitionistic logic by interpreting propositions as types (this is true of most any dependent type theory).\n\n• It has a version of a variant of the axiom of choice as a theorem (because of the properties of the above interpretation), see the discussion below.\n\n• In its intensional form, it has sufficient computational content to function as a programming language. At the same time, it then has identity types whose presence shows that one is really dealing with a form of homotopy type theory.\n\n## Syntax\n\nIf $t$ is a term, and $A$ a type, then $t : A$ is a typing declaration asserting that $t$ is a term of type $A$. We will define the forms of valid terms and types below; to begin with, we assume we have a stock of variables, atomic types, and parametrised types.\n\nA context is a finite ordered list of typing declarations, defined inductively as follows:\n\n• The empty list is a valid context.\n• If $\\Gamma$ is a valid context, $x$ is a variable not appearing in $\\Gamma$, and $A$ is a valid type in the context $\\Gamma$, then the context obtained by appending $x : A$ to $\\Gamma$ is also valid.\n\nNote that this must be considered a mutual inductive definition along with the definition of when a type is valid in a given context, which is to be given below in terms of type constructors.\n\nWe write $\\Gamma \\vdash t : A$ to assert that $t : A$ is a valid typing declaration in the context of $\\Gamma$, and by abuse of notation we may write $\\Gamma \\vdash A : \\mathrm{Type}$ to assert that $A$ is a valid type in the context of $\\Gamma$.\n\nFor any context $\\Gamma$, if $A$ is an atomic type, $\\Gamma \\vdash A : \\mathrm{Type}$.\n\nIf $x : A$ is a typing declaration in $\\Gamma$, then we have $\\Gamma \\vdash x : A$.\n\n### Binary product types\n\nIf $\\Gamma \\vdash A : \\mathrm{Type}$ and $\\Delta \\vdash B : \\mathrm{Type}$, then $\\Gamma, \\Delta \\vdash A \\times B : \\mathrm{Type}$.\n\nIf $\\Gamma \\vdash a : A$ and $\\Delta \\vdash b : B$ then $\\Gamma, \\Delta \\vdash \\langle a, b \\rangle : A \\times B$.\n\nIf $\\Gamma \\vdash c : A \\times B$, then $\\Gamma \\vdash \\pi_0(c) : A$ and $\\Gamma \\vdash \\pi_1(c) : B$.\n\n### Unit type\n\nThere is an atomic type called $\\top$.\n\nFor any context $\\Gamma$ we have $\\Gamma \\vdash \\mathrm{tt} : \\top$.\n\n### Dependent product types\n\nIf $\\Gamma, x : X \\vdash A(x) : \\mathrm{Type}$, then $\\Gamma \\vdash (\\Pi x : X) A(x) : \\mathrm{Type}$.\n\nIf $\\Gamma, x : X \\vdash a(x) : A(x)$, then $\\Gamma \\vdash (\\lambda x : X) a(x) : (\\Pi x : X) A(x)$.\n\nIf $\\Gamma \\vdash f : (\\Pi x : X) A(x)$ and $\\Delta \\vdash t : X$, then $\\Gamma, \\Delta \\vdash \\mathrm{apply}(f, t) : A(t)$.\n\n### Function types\n\nFunction types may be regarded as a special case of dependent product types, where $A(x)$ does not depend on $x$ at all. When we write out the rules for dependent products in this case, they become the following.\n\nFirstly, if $\\Gamma \\vdash X : \\mathrm{Type}$ and $\\Gamma \\vdash A : \\mathrm{Type}$, then $\\Gamma \\vdash X \\to A : \\mathrm{Type}$.\n\nIf $\\Gamma, x : X \\vdash a(x) : A$, then $\\Gamma \\vdash (\\lambda x : X) a(x) : X \\to A$.\n\nIf $\\Gamma \\vdash f : X \\to A$ and $\\Delta \\vdash t : X$, then $\\Gamma, \\Delta \\vdash \\mathrm{apply}(f, t) : A$.\n\n### Binary sum types\n\nIf $\\Gamma \\vdash A : \\mathrm{Type}$ and $\\Gamma \\vdash B : \\mathrm{Type}$, then $\\Gamma \\vdash A + B : \\mathrm{Type}$.\n\nIf $\\Gamma \\vdash a : A$ and $\\Gamma \\vdash A + B : \\mathrm{Type}$, then $\\Gamma \\vdash \\sigma_0 (a) : A + B$; and if $\\Gamma \\vdash b : B$ and $\\Gamma \\vdash A + B : \\mathrm{Type}$, then $\\Gamma \\vdash \\sigma_1(b) : A + B$.\n\nIf $\\Gamma \\vdash s : A + B$, $\\Delta, x : A \\vdash c(x) : C(\\sigma_0(x))$ and $E, y : B \\vdash c'(y) : C(\\sigma_1(y))$, then $\\Gamma, \\Delta, E \\vdash \\mathrm{cases}(s, (\\lambda x : A) c(x), (\\lambda y : B) c'(y)) : C(s)$.\n\n### Empty type\n\nThere is an atomic type called $\\bot$.\n\nIf $\\Gamma \\vdash A : \\mathrm{Type}$, then $\\Gamma, x : \\bot \\vdash \\mathrm{abort} : A$.\n\n### Dependent sum types\n\nIf $\\Gamma, x : X \\vdash A(x) : \\mathrm{Type}$, then $\\Gamma \\vdash (\\Sigma x : X) A(x) : \\mathrm{Type}$.\n\nIf $\\Gamma \\vdash t : X$, $\\Gamma \\vdash a : A(t)$ and $\\Gamma \\vdash (\\Sigma x : X) A(x) : \\mathrm{Type}$, then $\\Gamma \\vdash (t, a) : (\\Sigma x : X) A(x)$.\n\nIf $\\Gamma \\vdash s : (\\Sigma x : X) A(x)$ and $\\Delta, x : X, y : A(x) \\vdash b(x, y) : B((x, y))$, then $\\Gamma, \\Delta \\vdash \\mathrm{cases}(s, (\\lambda x : X)(\\lambda y : A(x)) b(x, y)) : B(s)$.\n\nNote that just as function types can be defined to be dependent products where $A(x)$ does not depend on $x$, binary product types (above) can be defined to be dependent sums where $A(x)$ does not depend on $x$.\n\n### Finite types\n\nMartin-Löf also includes in his type theory a type $\\mathbb{N}_n$ with exactly $n$ elements, for every external natural number $n$. The types $\\bot$ and $\\top$ can then be defined as $\\mathbb{N}_0$ and $\\mathbb{N}_1$, respectively. On the other hand, with $\\bot$ and $\\top$ given as above, one may define $\\mathbb{N}_2 = \\top + \\top$, $\\mathbb{N}_3 = \\top+\\top+\\top$, and so on (by recursion on the external natural number $n$).\n\nNote that we have not included any axiom of infinity, however.\n\n### Propositions as types\n\nUnder the Curry-Howard isomorphism, we may identify propositions with certain atomic types and predicates with certain parametrised types.\n\nAn inhabitant of such a proposition-as-a-type is interpreted as a ‘proof’ of that proposition.\n\nWe write $\\Gamma \\vdash A : \\mathrm{true}$ for the judgement that $A$ is inhabited: that is, if $\\Gamma \\vdash a : A$, then $\\Gamma \\vdash A : \\mathrm{true}$. The type-formation rules above are then seen to be the rules of inference for (a fragment of) intuitionistic first-order logic. Indeed, we have:\n\n• Conjunction introduction:\n$\\frac{\\Gamma \\vdash A : \\mathrm{true} ; \\Delta \\vdash B : \\mathrm{true}}{\\Gamma, \\Delta \\vdash A \\times B : \\mathrm{true}}$\n• Conjunction elimination:\n\\begin{aligned} \\frac{\\Gamma \\vdash A \\times B : \\mathrm{true}}{\\Gamma \\vdash A : \\mathrm{true}} \\,\\, & \\frac{\\Gamma \\vdash A \\times B : \\mathrm{true}}{\\Gamma \\vdash B : \\mathrm{true}} \\end{aligned}\n• Truth introduction\n$\\frac{}{\\Gamma \\vdash \\top : \\mathrm{true}}$\n• Universal generalisation:\n$\\frac{\\Gamma, x : X \\vdash A(x) : \\mathrm{true}}{\\Gamma \\vdash (\\Pi x : X) A(x) : \\mathrm{true}}$\n• Universal instantiation:\n$\\frac{\\Gamma \\vdash (\\Pi x : X) A(x) : \\mathrm{true} ; \\Delta \\vdash t : X}{\\Gamma, \\Delta \\vdash A(t) : \\mathrm{true}}$\n• Conditional proof (implication introduction):\n$\\frac{\\Gamma, X : \\mathrm{true} \\vdash A : \\mathrm{true}}{\\Gamma \\vdash X \\to A : \\mathrm{true}}$\n• Modus ponens (implication elimination):\n$\\frac{\\Gamma \\vdash X \\to A : \\mathrm{true} ; \\Delta \\vdash X : \\mathrm{true}}{\\Gamma, \\Delta \\vdash A : \\mathrm{true}}$\n• Disjunction introduction:\n\\begin{aligned} \\frac{\\Gamma \\vdash A : \\mathrm{true}}{\\Gamma \\vdash A + B : \\mathrm{true}} \\,\\, & \\frac{\\Gamma \\vdash B : \\mathrm{true}}{\\Gamma \\vdash A + B : \\mathrm{true}} \\end{aligned}\n• Disjunction elimination:\n$\\frac{\\Gamma \\vdash A + B : \\mathrm{true} ; \\Delta, A : \\mathrm{true} \\vdash C : \\mathrm{true} ; E, B : \\mathrm{true} \\vdash C : \\mathrm{true}}{\\Gamma, \\Delta, E \\vdash C : \\mathrm{true}}$\n• False elimination:\n$\\frac{\\Gamma \\vdash \\bot : \\mathrm{true}}{\\Gamma \\vdash A : \\mathrm{true}}$\n• Existential generalisation:\n$\\frac{\\Gamma \\vdash t : X ; \\Gamma \\vdash A(t) : \\mathrm{true}}{\\Gamma \\vdash (\\Sigma x : X) A(x) : \\mathrm{true}}$\n• Existential instantiation:\n$\\frac{\\Gamma \\vdash (\\Sigma x : X) A(x) : \\mathrm{true} ; \\Delta, x : X, A(x) : \\mathrm{true} \\vdash B : \\mathrm{true}}{\\Gamma, \\Delta \\vdash B : \\mathrm{true}}$\n\nThe only traditional rule of inference we are missing is the rule of excluded middle, so this logic is intuitionistic.\n\nThe above interpretation justifies the use of the symbols $\\forall$, $\\exists$, $\\wedge$, $\\vee$ instead of $\\Pi$, $\\Sigma$, $\\times$, $+$, when we are using types to represent propositions.\n\n### Equality types\n\nWe may introduce a family of equality types for each type and each pair of terms of that type: if $\\Gamma \\vdash a : A$ and $\\Delta \\vdash a' : A$, then $\\Gamma \\vdash a = a' : \\mathrm{Type}$.\n\nThese are not the intensional identity types which Martin-Löf later introduced as a particular sort of inductive family of types, but instead an extensional notion of equality defined by induction over the class of all types (regarded as inductively defined by the above type-formation clauses). Note that it is essential, for a purpose such as this, that the type-formation clauses be regarded as an inductive definition, rather than as “operations” defined on some pre-existing collection of types: in order to define equality in the way we are about to do, we have to be able to inspect a given type and decide uniquely which rule was used to construct it.\n\nThe equality type is reflexive, symmetric, and transitive. That means, for example, if $\\Gamma \\vdash a : A$, $\\Gamma \\vdash b : A$, $\\Gamma \\vdash c : A$, $\\Gamma \\vdash a = b : \\mathrm{true}$ and $\\Gamma \\vdash b = c : \\mathrm{true}$ then $\\Gamma \\vdash a = c : \\mathrm{true}$.\n\nMike Shulman: Is that (reflexivity, symmetry, and transitivity) a rule we are giving, or an assertion we are making about derivability? I’m a little unsure because the presentation here is slightly different from that in the reference: there Martin-Löf uses both a propositional equality and a judgmental one.\n\n### Finite product types\n\nIf $\\Gamma \\vdash a : A$, and $\\Gamma \\vdash b : B$, then $\\Gamma \\vdash a = \\pi_0 (\\langle a, b \\rangle) : \\mathrm{true}$ and $\\Gamma \\vdash \\pi_1 (\\langle a, b \\rangle) = b : \\mathrm{true}$.\n\nIf $\\Gamma \\vdash c : A \\times B$, then $c = \\langle \\pi_0 (c), \\pi_1 (c) \\rangle : \\mathrm{true}$.\n\n### Unit type\n\nIf $\\Gamma \\vdash a : \\top$, then $\\Gamma \\vdash a = \\mathrm{tt} : \\mathrm{true}$.\n\n### Dependent product types\n\nIf $\\Gamma, x : X \\vdash a(x) : A(x)$ and $\\Delta \\vdash t : X$, then $\\Gamma, \\Delta \\vdash \\mathrm{apply}((\\lambda x : X) a(x), t) = a(t) : \\mathrm{true}$.\n\nIf $\\Gamma \\vdash f : (\\Pi x : X) A(x)$, then $\\Gamma \\vdash f = (\\lambda x : X) \\mathrm{apply}(f, x) : \\mathrm{true}$.\n\n### Function types\n\nIf $\\Gamma, x : X \\vdash a(x) : A$ and $\\Delta \\vdash t : X$, then $\\Gamma, \\Delta \\vdash \\mathrm{apply}((\\lambda x : X) a(x), t) = a(t) : \\mathrm{true}$.\n\nIf $\\Gamma \\vdash f : X \\to A$, then $\\Gamma \\vdash f = (\\lambda x : X) \\mathrm{apply}(f, x) : \\mathrm{true}$.\n\n### Binary sum types\n\nIf $\\Gamma \\vdash a : A$ and $\\Gamma \\vdash A + B : \\mathrm{Type}$, $\\Delta, x : A \\vdash c(x) : C(\\sigma_0(x))$ and $E, y : B \\vdash c'(y) : C(\\sigma_1(y))$, then $\\Gamma, \\Delta, E \\vdash \\mathrm{cases}(a, (\\lambda x : A) c(x), (\\lambda y : B) c(y)) = c(a) : \\mathrm{true}$ and $\\Gamma, \\Delta, E \\vdash \\mathrm{cases}(b, (\\lambda x : A) c(x), (\\lambda y : B) c'(y)) = c'(b) : \\mathrm{true}$\n\n### Dependent sum types\n\nIf $\\Gamma \\vdash t : X$, $\\Gamma \\vdash a : A(t)$ and $\\Delta, x : X, y : A(x) \\vdash b(x, a) : B((x, a))$, then $\\Gamma, \\Delta \\vdash \\mathrm{cases}((t, a), (\\lambda x : X)(\\lambda y : A(x)) b(x, y)) = b(t, a) : \\mathrm{true}$.\n\nThe other equality rule for dependent sum types is derivable from the above.\n\n## Properties\n\n### Models and categorical semantics\n\nThe models of ML type theory depend crucially on whether one considers the variant of extensional type theory or that of intensional type theory.\n\nThe models of the extensional version are (just) locally cartesian closed categories.\n\nThe faithful models of the intensional version with identity types are however certain (∞,1)-categories, notably (∞,1)-toposes, presented by simplicial model categories (Warren). For this reason one speaks of homotopy type theory.\n\nFor a more detailed discussion of these matters see at relation between type theory and category theory.\n\n### Axiom of choice\n\nIn dependent type theory we can verify the following “logical form of the axiom of choice” (Bell, Tait), see also (Rijke, section 2.5.1).\n\n###### Theorem\n\n(ACL)\n\n$x : A, y : B(x) \\vdash C(x, y) : Type \\; \\vdash \\; acl : (\\forall x : A) (\\exists y : B(x)) C(x, y) \\to (\\exists f : (\\Pi x : A)B(x) )(\\forall x : A) C(x, \\mathrm{apply}(f, x)) \\,.$\n\nOne should note carefully that this “is” only “the axiom of choice” under the above propositions-as-types interpretation of the quantifiers $\\forall$ and $\\exists$.\n\n###### Remark\n\nIn the categorical semantics of this expression, using the propositions-as-types logic corresponds roughly to working with the subobject lattices in the ex/lex completion of a locally cartesian closed category; the ACL then follows since every object of the original category becomes projective in its ex/lex completion.\n\n###### Remark\n\nIf we use instead a different logic over the same type theory, such as bracket types to model the actual subobject lattices in an arbitrary lccc (not necessarily the ex/lex completion of anything), or the hProp logic in homotopy type theory, then the ACL in that logic will not be derivable.\n\nThe original articles:\n\nAs a programming language:\n\n• Per Martin-Löf, Constructive Mathematics and Computer Programming, Studies in Logic and the Foundations of Mathematics Volume 104, 1982, Pages 153-175 (doi:10.1016/S0049-237X(09)70189-2)\n\n• Bengt Nordström, Kent Petersson, Jan M. Smith, Programming in Martin-Löf’s Type Theory, Oxford University Press 1990 (webpage, pdf)\n\nA philosophical examination:\n\nAn introduction and survey (with an eye towards homotopy type theory) is in chapter 1 of\n\n• Michael Warren, Homotopy theoretic aspects of constructive type theory, PhD thesis (2008) (pdf)\n\nas well as\n\n• Egbert Rijke, Homotopy type theory (2012) (pdf)\n\nA discussion of ML dependent type theory as the internal language of locally cartesian closed categories is in\n\n• R. A. G. Seely, Locally cartesian closed categories and type theory, Math. Proc. Camb. Phil. Soc. (1984) 95 (pdf)\n\nDiscussion of the logical axiom of choice in dependent type theory is in\n\n• John Bell, The Axiom of choice in type theory, Stanford Encyclopedia of philosophy, 2008 (web)\n\n• Tait (1994)" ]

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[ "", null, "Education Bhaskar\nSymbol of Educational Excellence\n\nAdvertisement\n\nThe word hyperbola is a Greek word that means excessive. It is a group of all those points, the difference of whose distances from two fixed points is always same or constant. The fixed points are called as the foci(foci is plural for the word focus.) A hyperbola has two curves that are known as its arms or branches. These branches continue up to infinity.\n\n### Facts About Hyperbola:\n\n• So unlike an ellipse, a hyperbola is an open figure.\n• It may be symmetric about either of the x-axis or y-axis.\n• If a it is symmetric about x-axis, then its equation is of the form: x2/a2 – y2/b2 = 1\n• If it is symmetric about the y-axis, then its equation is of the form: y2/a2 – x2/b2 = 1\n• The axis about which the hyperbola is symmetric is called the major axis of it , and the other one is called the minor axis.\n• The major axis is also known as transverse axis and the minor axis as the conjugate axis.\n• Half of major axis is the semi-major axis or semi-transverse axis, and half of minor axis is semi-minor axis or semi-conjugate axis. In the equations of the hyperbola, a is the length of semi-major axis, and b is the length of the semi-minor axis. Therefore the length of the major or transverse axis is 2 x a, and the length of minor axis or conjugate axis is 2 x b. It has two focus. The difference of the distances of any point on the hyperbola from the two focus is always constant. The distance of focus from the centre of it is represented by c. It is known as semi-focal length.\n\nRelation between c(the distance of focus and center) , a(length of semi transverse axis) and b(length of semi conjugate axis) :c2 = a2 + b2.\n\nThe ratio of c(the distance of focus and centre) to a(length of semi-transverse axis) is constant and is called as eccentricity. It is represented by e. The value of e for a hyperbola is greater than 1. A line that passes through the focus and intersects it at two points is called as latus rectum. It is perpendicular to the major axis of it.\n\n### Some important formulae:\n\n•       Eccentricity = c/a.\n•       Relation between c , a and b is c2 = a2 + b2.\n•       Length of major axis = 2 x a.\n•       Length of minor axis = 2 x b.\n•       Length of latus rectum = 2b2/a.\n\nAs we know that the word hyperbola means excessive. This name is chosen because an ellipse has eccentricity less than 1,  it has an eccentricity greater than 1, and a parabola has an eccentricity equals to 1.\n\n### Hyperbola in Nature (Real Life):", null, "Gear transmission is the most practical example\n\n### The difference between a parabola, a hyperbola and a catenary curve Equations:\n\nThe equations of the four types of conic sections are as follows.\n\n• Circle- x2+y2=1\n• Ellipse- x2/a2+ y2/b2= 1\n• Parabola- y2=4ax\n• Hyperbola- x2/a2– y2/b2= 1\n\nIf you have any Question, you may write in comment section below\n\nAdvertisement\n\nYou might also like\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.\n\nThis website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Accept Read More" ]

[ null, "https://www.facebook.com/tr", null, "https://educationbhaskar.com/science/hyperbola/6239/None", null ]

[ "Optimizing AI and Machine Learning with eFPGAs\n\nWhy the performance and flexibility offered by eFPGA is turning out to be a game changer for anyone designing AI and machine learning and struggling to meet the compute demands.\n\nThe market for artificial intelligence (AI) and machine learning applications has been growing substantially over the last several years. Designers have a tough row to hoe when it comes to satisfying these applications’ seemingly insatiable compute hunger. They are finding that traditional Von Neumann processor architectures are not optimal solutions for the neural networks fundamental to AI and machine learning.\n\nWhen GPUs are used to train neural networks, they require floating pointing math that is very compute intensive. However, using integer math for inference, designers can speed computation by turning to FPGAs for neural network processing. Many companies are starting to recognize this, with Microsoft’s Project Brainwave, which uses FPGA chips to accelerate AI, as a perfect example.", null, "Figure 1: Neurons and the brain\n\nHow FPGAs Speed AI\nFPGAs have many advantages for AI. An FPGA’s many multiplier-accumulators (MACS) can accelerate AI computation with massive parallelism. FPGAs are also reconfigurable, which is critical in the AI and machine learning market because these applications are still at an early stage, and algorithms are changing very quickly.\n\n“How many MACs per second can I get?” is one of the first things chip designers start thinking about when they begin an AI project. The bulk of the computation that neural network inference requires is 8-bit (or sometimes 16-bit or even 8×16 or 16×8) integer matrix multiplies of very large matrices, driving the demand for MACs or GigaMACs, also referred to as GigaOps where one Op equals one multiply+accumulate. Thus, the GigaMACs/second determines throughput of the hardware.\n\nNeural networks are a digital approximation of the neuron structure in the human brain (Figure 1). Each neuron gets inputs from a large number of other neurons. The output or firing of a neuron occurs when the sum of the inputs exceeds a certain value determined by the neuron’s activation function. Figure 2 shows a simple neural network for computational purposes, which roughly approximates the function of a human brain.", null, "Figure 2:  A simple neural network for computational purposes\n\nEach layer of the neural network is a matrix multiply. The input layer is a vector, which is multiplied by a 2-dimensional matrix of weights (determined in a training phase), to generate the values for the next layer. Then each successive layer is another matrix multiply. After each matrix multiply, the new values go through an activation phase before becoming inputs to the next step.", null, "Figure 3:  Doing the matrix math demands the reading of a substantial amount of data.\n\nOn a side note, memory bandwidth is another, separate factor as a lot of data has to be read in order to do the matrix math (Figure 3). A full tutorial on the math can be viewed at this link http://www.flex-logix.com/eflx4k-ai/\n\neFPGAs Help Balance Performance and Configurability\nAt Flex Logix, we have customers such as Harvard already designing deep learning chips using eFPGAs. AI designers seeking to balance performance and reconfigurability are finding eFPGAs optimal for this purpose. FPGAs and eFPGAs are good for AI because of the large number of built-in MACs, originally used for digital signal processing (DSP).\n\nIn a typical Altera FPGA, the ratio of logic (Look up table (LUT)6) to MACs is about 200:1. Xilinx is about 100:1, whereas the Flex Logix EFLX4K DSP eFPGA core has a ratio of about 50:1. Thus, for a given array size, eFPGA delivers two-to-four times more MACs— key to AI performance.\n\nHowever, the MACs in all eFPGAs and FPGAs today are optimized for DSP. They have pre-adders, large multipliers (22×22 or 18×25, for example) and accumulators. AI prefers a MAC that has an 8×8 multiplier with accumulator with an option to configure as a 16×16, 16×8 or 8×16 MAC as well. Since an 8-bit MAC is smaller, it can run faster.\n\nFlex Logix has architected its eFPGA to be optimized for AI in many ways.  For example, it has optimized eFPGA for AI by using smaller 8×8 MACs, which are smaller (3 fit in the space of a 22×22 MAC) and run faster, and increasing the ratio of MACs to LUTs. The result is 441 8×8 MACs fit in an EFLX4K AI eFPGA core: more than 10 times the MACs in about the same area as the EFLX4K DSP core, which already had more MACs per square millimeter than any other FPGA/eFPGA. The 8×8 MACs can also be configured to do 16×16 multiplies if preferred. Customers with Verilog/VHDL code for neural network processing on FPGA will be able to use this new AI-optimized eFPGA without chaining code but achieving 10x the throughput.\n\nFigure 4 shows details of the Flex Logix EFLX4K-AI eFPGA core. Just like all EFLX cores, this is a complete FPGA with an input/output pins (the small yellow squares, >1000) circling the core. The core consists of two types of logic: MLUTS or memory-LUTs for local weight storage and DSP blocks which consist of 3 x 8-bit MACs each, pipelined in long rows for high-speed vector math. This core can be arrayed to >7×7 and mixed with other EFLX4K cores such as Logic and DSP. The EFLX4K AI core can be implemented in any CMOS process in ~6 months on customer demand. A smaller EFLX1K AI core is also available for 40nm-180nm applications.", null, "Figure 4: The Flex Logix EFLX4K-AI eFPGA core\n\nConclusion\nMany companies are starting to use FPGAs to implement AI and more specifically machine learning, deep learning, and neural networks as approaches to achieve AI. Foundational for AI are matrix multipliers, which consist of arrays of MACs. In existing FPGAs and eFPGAs, the MACs are optimized for DSPs with larger multipliers, pre-adders, and other logic—overkill for AI. For AI applications, smaller multipliers such as 16 bits or 8 bits, with the ability to support both modes with accumulators, allow more neural network processing per square millimeter.\n\nAI chip designers want more MACs/second and more MACs/square millimeter, but they also want the flexibility of eFPGA to reconfigure designs as AI algorithms are changing rapidly. eFPGAs enable them to switch between 8 -and 16-bit modes as needed and implement matrix multipliers of varying sizes to meet their applications’ performance and cost constraints.", null, "Cheng Wang is Senior VP, Architecture/Software/Silicon, Flex Logix, Inc. Cheng is originally from Shanghai, PRC. BSEECS, UC Berkeley. He spent two years as VLSI designer at Zoran. MSEE, EE PhD UCLA: designed five FPGA chips from 90nm to 40nm. 2013 Distinguished PhD Dissertation Award. 2014 ISSCC Lewis Winner Award for Outstanding Paper. Multiple patents at UCLA and Flex Logix.\n\nTags:" ]

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[ "\\begin{equation*} \\DeclareMathOperator\\Coim{Coim} \\DeclareMathOperator\\Coker{Coker} \\DeclareMathOperator\\Ext{Ext} \\DeclareMathOperator\\Hom{Hom} \\DeclareMathOperator\\Im{Im} \\DeclareMathOperator\\Ker{Ker} \\DeclareMathOperator\\Mor{Mor} \\DeclareMathOperator\\Ob{Ob} \\DeclareMathOperator\\Sh{Sh} \\DeclareMathOperator\\SheafExt{\\mathcal{E}\\mathit{xt}} \\DeclareMathOperator\\SheafHom{\\mathcal{H}\\mathit{om}} \\DeclareMathOperator\\Spec{Spec} \\newcommand\\colim{\\mathop{\\mathrm{colim}}\\nolimits} \\newcommand\\lim{\\mathop{\\mathrm{lim}}\\nolimits} \\newcommand\\Qcoh{\\mathit{Qcoh}} \\newcommand\\Sch{\\mathit{Sch}} \\newcommand\\QCohstack{\\mathcal{QC}\\!\\mathit{oh}} \\newcommand\\Cohstack{\\mathcal{C}\\!\\mathit{oh}} \\newcommand\\Spacesstack{\\mathcal{S}\\!\\mathit{paces}} \\newcommand\\Quotfunctor{\\mathrm{Quot}} \\newcommand\\Hilbfunctor{\\mathrm{Hilb}} \\newcommand\\Curvesstack{\\mathcal{C}\\!\\mathit{urves}} \\newcommand\\Polarizedstack{\\mathcal{P}\\!\\mathit{olarized}} \\newcommand\\Complexesstack{\\mathcal{C}\\!\\mathit{omplexes}} \\newcommand\\Pic{\\mathop{\\mathrm{Pic}}\\nolimits} \\newcommand\\Picardstack{\\mathcal{P}\\!\\mathit{ic}} \\newcommand\\Picardfunctor{\\mathrm{Pic}} \\newcommand\\Deformationcategory{\\mathcal{D}\\!\\mathit{ef}} \\end{equation*}\n\n## 15.18 Flattening over a closed subset of the base\n\nLet $R \\to S$ be a ring map. Let $I \\subset R$ be an ideal. Let $M$ be an $S$-module. In the following we will consider the following condition\n\n15.18.0.1\n\\begin{equation} \\label{more-algebra-equation-flat-at-primes-over} \\forall \\mathfrak q \\in V(IS) \\subset \\mathop{\\mathrm{Spec}}(S) : M_{\\mathfrak q}\\text{ is flat over }R. \\end{equation}\n\nGeometrically, this means that $M$ is flat over $R$ along the inverse image of $V(I)$ in $\\mathop{\\mathrm{Spec}}(S)$. If $R$ and $S$ are Noetherian rings and $M$ is a finite $S$-module, then (15.18.0.1) is equivalent to the condition that $M/I^ nM$ is flat over $R/I^ n$ for all $n \\geq 1$, see Algebra, Lemma 10.98.11.\n\nLemma 15.18.1. Let $R \\to S$ be a ring map. Let $I \\subset R$ be an ideal. Let $M$ be an $S$-module. Let $R \\to R'$ be a ring map and $IR' \\subset I' \\subset R'$ an ideal. If (15.18.0.1) holds for $(R \\to S, I, M)$, then (15.18.0.1) holds for $(R' \\to S \\otimes _ R R', I', M \\otimes _ R R')$.\n\nProof. Assume (15.18.0.1) holds for $(R \\to S, I \\subset R, M)$. Let $I'(S \\otimes _ R R') \\subset \\mathfrak q'$ be a prime of $S \\otimes _ R R'$. Let $\\mathfrak q \\subset S$ be the corresponding prime of $S$. Then $IS \\subset \\mathfrak q$. Note that $(M \\otimes _ R R')_{\\mathfrak q'}$ is a localization of the base change $M_{\\mathfrak q} \\otimes _ R R'$. Hence $(M \\otimes _ R R')_{\\mathfrak q'}$ is flat over $R'$ as a localization of a flat module, see Algebra, Lemmas 10.38.7 and 10.38.19. $\\square$\n\nLemma 15.18.2. Let $R \\to S$ be a ring map. Let $I \\subset R$ be an ideal. Let $M$ be an $S$-module. Let $R \\to R'$ be a ring map and $IR' \\subset I' \\subset R'$ an ideal such that\n\n1. the map $V(I') \\to V(I)$ induced by $\\mathop{\\mathrm{Spec}}(R') \\to \\mathop{\\mathrm{Spec}}(R)$ is surjective, and\n\n2. $R'_{\\mathfrak p'}$ is flat over $R$ for all primes $\\mathfrak p' \\in V(I')$.\n\nIf (15.18.0.1) holds for $(R' \\to S \\otimes _ R R', I', M \\otimes _ R R')$, then (15.18.0.1) holds for $(R \\to S, I, M)$.\n\nProof. Assume (15.18.0.1) holds for $(R' \\to S \\otimes _ R R', IR', M \\otimes _ R R')$. Pick a prime $IS \\subset \\mathfrak q \\subset S$. Let $I \\subset \\mathfrak p \\subset R$ be the corresponding prime of $R$. By assumption there exists a prime $\\mathfrak p' \\in V(I')$ of $R'$ lying over $\\mathfrak p$ and $R_{\\mathfrak p} \\to R'_{\\mathfrak p'}$ is flat. Choose a prime $\\overline{\\mathfrak q}' \\subset \\kappa (\\mathfrak q) \\otimes _{\\kappa (\\mathfrak p)} \\kappa (\\mathfrak p')$ which corresponds to a prime $\\mathfrak q' \\subset S \\otimes _ R R'$ which lies over $\\mathfrak q$ and over $\\mathfrak p'$. Note that $(S \\otimes _ R R')_{\\mathfrak q'}$ is a localization of $S_{\\mathfrak q} \\otimes _{R_{\\mathfrak p}} R'_{\\mathfrak p'}$. By assumption the module $(M \\otimes _ R R')_{\\mathfrak q'}$ is flat over $R'_{\\mathfrak p'}$. Hence Algebra, Lemma 10.99.1 implies that $M_{\\mathfrak q}$ is flat over $R_{\\mathfrak p}$ which is what we wanted to prove. $\\square$\n\nLemma 15.18.3. Let $R \\to S$ be a ring map of finite presentation. Let $M$ be an $S$-module of finite presentation. Let $R' = \\mathop{\\mathrm{colim}}\\nolimits _{\\lambda \\in \\Lambda } R_\\lambda$ be a directed colimit of $R$-algebras. Let $I_\\lambda \\subset R_\\lambda$ be ideals such that $I_\\lambda R_\\mu \\subset I_\\mu$ for all $\\mu \\geq \\lambda$ and set $I' = \\mathop{\\mathrm{colim}}\\nolimits _\\lambda I_\\lambda$. If (15.18.0.1) holds for $(R' \\to S \\otimes _ R R', I', M \\otimes _ R R')$, then there exists a $\\lambda \\in \\Lambda$ such that (15.18.0.1) holds for $(R_\\lambda \\to S \\otimes _ R R_\\lambda , I_\\lambda , M \\otimes _ R R_\\lambda )$.\n\nProof. We are going to write $S_\\lambda = S \\otimes _ R R_\\lambda$, $S' = S \\otimes _ R R'$, $M_\\lambda = M \\otimes _ R R_\\lambda$, and $M' = M \\otimes _ R R'$. The base change $S'$ is of finite presentation over $R'$ and $M'$ is of finite presentation over $S'$ and similarly for the versions with subscript $\\lambda$, see Algebra, Lemma 10.13.2. By Algebra, Theorem 10.128.4 the set\n\n$U' = \\{ \\mathfrak q' \\in \\mathop{\\mathrm{Spec}}(S') \\mid M'_{\\mathfrak q'}\\text{ is flat over }R'\\}$\n\nis open in $\\mathop{\\mathrm{Spec}}(S')$. Note that $V(I'S')$ is a quasi-compact space which is contained in $U'$ by assumption. Hence there exist finitely many $g'_ j \\in S'$, $j = 1, \\ldots , m$ such that $D(g'_ j) \\subset U'$ and such that $V(I'S') \\subset \\bigcup D(g'_ j)$. Note that in particular $(M')_{g'_ j}$ is a flat module over $R'$.\n\nWe are going to pick increasingly large elements $\\lambda \\in \\Lambda$. First we pick it large enough so that we can find $g_{j, \\lambda } \\in S_{\\lambda }$ mapping to $g'_ j$. The inclusion $V(I'S') \\subset \\bigcup D(g'_ j)$ means that $I'S' + (g'_1, \\ldots , g'_ m) = S'$ which can be expressed as $1 = \\sum z_ sh_ s + \\sum f_ jg'_ j$ for some $z_ s \\in I'$, $h_ s, f_ j \\in S'$. After increasing $\\lambda$ we may assume such an equation holds in $S_\\lambda$. Hence we may assume that $V(I_\\lambda S_\\lambda ) \\subset \\bigcup D(g_{j, \\lambda })$. By Algebra, Lemma 10.162.1 we see that for some sufficiently large $\\lambda$ the modules $(M_\\lambda )_{g_{j, \\lambda }}$ are flat over $R_\\lambda$. In particular the module $M_\\lambda$ is flat over $R_\\lambda$ at all the primes lying over the ideal $I_\\lambda$. $\\square$\n\nIn your comment you can use Markdown and LaTeX style mathematics (enclose it like $\\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar)." ]

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[ "# 2021 JMPSC Invitationals Problems/Problem 4\n\n## Problem\n\nLet", null, "$(x_n)_{n\\geq 0}$ and", null, "$(y_n)_{n\\geq 0}$ be sequences of real numbers such that", null, "$x_0 = 3$,", null, "$y_0 = 1$, and, for all positive integers", null, "$n$,", null, "$$x_{n+1}+y_{n+1} = 2x_n + 2y_n,$$", null, "$$x_{n+1}-y_{n+1}=3x_n-3y_n.$$ Find", null, "$x_5$.\n\n## Solution\n\nWe notice that", null, "$$x_5 + y_5 = 2(x_4 + y_4)$$", null, "$$= 2(2(x_3 + y_3))$$", null, "$$= 2(2(2(x_2 + y_2)))$$", null, "$$= 2(2(2(2(x_1 + y_1))))$$", null, "$$= 2(2(2(2(2(x_0 + y_0))))).$$ Since we are given that", null, "$x_0 = 3$ and", null, "$y_0 = 1$, we can plug these values in to get that", null, "$$x_5 + y_5 = 2(2(2(2(2(3 + 1))))) = 2(2(2(2(2(4))))) = 128 \\qquad (1).$$\n\nSimilarly, we conclude that", null, "$$x_5 - y_5 = 3(3(3(3(3(x_0 - y_0)))))= 3(3(3(3(3(3 - 1))))) = 3(3(3(3(3(2))))) = 486 \\qquad (2).$$\n\nAdding", null, "$(1)$ and", null, "$(2)$ gives us", null, "$2 \\cdot x_5 = 614.$ Dividing both sides by", null, "$2$ yields", null, "$x_5 = \\boxed{307}.$\n\n~mahaler\n\nThe problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.", null, "" ]

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[ "# Renaming parameters in a loop, Comparing Z-Scores\n\n2 views (last 30 days)\nA on 1 Aug 2011\nI have a 48x64x41 array where the dimensions are vertical x horizontal x time. I need to compare the z-score array values for times 1 to 41, ie:\nzscores(array(:,:,i)) for i=1:41\nand find the time of the array with the greatest values.\nI am attempting to use a for loop to assign the array at different times to a parameter:\nfor i=1:41 z=zscore(array(:,:,i)) end\nHowever, obviously like that z is reassigned each time the loop iterates. What I would like to do is have 41 new \"z\"'s, corresponding to each time.\ni.e\nat the end of the for loop there will be z1 through z41, corresponding respectively to zscore(array(:,:,1)) to zscore(array(:,:,41).\nI then plan to compare these 41 results to find the time at which the array values are the greatest.\nTherefore, my questions are: -How can I change the value of z to be z(i), that is z1 through z41 are the outputs at the end of the for loop?\n-Does anyone have a smarter suggestion as to how I can obtain the location time of the greatest array values?\n\nthe cyclist on 1 Aug 2011\nThis is better. [I had not realized that zscore() is a MATLAB function.]\narraySize = size(array);\nz = zeros(size(array));\nfor i = 1:41\nz(:,:,i) = zscore(array(:,:,i));\nend\n\nthe cyclist on 1 Aug 2011\nUse a cell array:\n% Preallocate\nz = cell(41,1);\n% Run loop\nfor i=1:41\nz{i} = zscore(array(:,:,i))\nend\nthe cyclist on 2 Aug 2011\nIn fairness to Amina, I did several quick edits to this answer, while I (too?) hastily posted answers, before I realized that each \"zi\" was not expected to be a scalar. In the first incarnation, I did not use cell arrays. [Regardless, my other answer, which was accepted, was better.]" ]

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[ "# Is there a faster algorithm for my graph problem?\n\nI'm working on a tool that can analyze sizes of individual functions in a compiled binary. For each one it calculates how much space would be saved if the function was removed.\n\nHowever, the current algorithm is quadratic in the number of functions and as such is quite slow. Is there a better one? The problem formulation follows.\n\nLet $G=(V, E)$ be a directed graph with labeling $f\\colon V\\to\\mathbb{N}$, and $R\\subseteq V$ be a set of root vertices. Here, $V$ represents the functions in the binary. The graph edges represent the dependencies between the functions: $(u, v)\\in E$ iff $u$ references $v$, thus forcing $v$ to be present in the binary. $f$ assigns to each function its size in bytes. Functions in $R$ (the main function in the executable or the set of exported functions in an .so) are the functions we want in the binary, the rest of them were pulled in directly or indirectly from $R$.\n\nFor a set of vertices $U\\subseteq V$ and edges $F\\subseteq E$, we define $r(U, F)$ as the set of vertices reachable from $U$ along the edged from $F$.\n\n$$r(U, F)=\\{v\\in V\\mid \\exists r\\in U. (r, v)\\in F^*\\}$$\n\nWe guarantee that $r(R, E)=V$, that is all vertices are reachable from $R$ in the original graph (not that it helps).\n\nWe define $s(U) = \\sum_{u\\in U} f(u)$ as the total size of all functions from $U$.\n\nFor each $u\\in V$, let $E_u\\subseteq E$ be the restriction of $E$ in which all edges adjacent to $u$ are removed. Then $r(R\\setminus\\{u\\}, E_u)$ is the set of functions that will be present in the binary if we remove $u$. Note that other functions besides $u$ may disappear as they will no longer be reachable from $R$.\n\nOur goal is to compute for each $u\\in V$ the size of the binary after removing $u$, i.e. to compute $g\\colon V\\rightarrow\\mathbb{N}$ at all points, where\n\n$$g(u)=s(r(R\\setminus\\{u\\}, E_u))$$\n\nI can easily compute $g(u)$ in $O(|V|+|E|)$ with a simple depth-first-search. Computing the entire $g$ thus yields $O(|V|(|V| + |E|))$. Is there a faster algorithm?\n\n• I wonder if algorithms for dynamic graph reachability would be useful here. They allow to update reachability information as you insert/delete edges, more efficiently than re-computing the reachability information. – D.W. Aug 28 '16 at 23:28\n\nIn practice, just as you can calculate g (u) for one particular u quite quickly, you can calculate g (u) for all $u \\in V$ in parallel, operating with bit vectors of $|V|$ bits instead of boolean values - this changes the number of operations only be a constant, but by a rather large constant. This is quite useful if you actually want the number of bytes saved for every u.\n\nIf you are only interested in one u, then you only want to know for each function: 1. Can it be reached from u? 2. Can it be reached from any function other than u? To decide that, just combine all the other functions in V into one single function which calls everything that any of these functions calls. You don't need to know all the individual call graphs, just what can be reached from any of the functions in V other than u.\n\nSecond comment: I thought that you wanted to know how many bytes are saved by removing one function. That took you O (|V| * (|V| + |U|) by calculating all g (u). Which looks quite bad and can be improved if you want to know the answer for one function. If you want to answer the question for all functions, then calculating all g (u) is fine. It still takes O (|V| * (|V| + |U|), but the work can be reused for every function.\n\nThat's a very common situation, where solving n problems can be done a lot quicker than n times solving one problem.\n\n• In your first paragraph, that constant is usually 4 or 8, right? I didn't fully get the part with bit vectors, is there some other trick that makes the constat higher? – avakar Aug 29 '16 at 16:07\n• I pretty much didn't get you second paragraph at all, can you elaborate? I'm interested in exactly the values of $g(u)$ for every $u\\in V$, in other words, I need exactly $|V|$ numbers. – avakar Aug 29 '16 at 16:08\n• Just noticed your edit. I might not have formulated the problem clearly. If I want to know how much is saved by removing one particular function, say $u$, which in my formulation is equivalent to computing $g(u)$, I can easily do it in $O(|V| + |E|)$ by computing the total size, then computing reachability on the graph with $u$ removed, and then subtracting. That's fine and I think it's optimal. However, I need to compute the savings for every function; I can trivially do it by computing $g(u)$ with the above algorithm for all $u$, yielding $O(|V|(|V| + |E|))$. I'm looking for a faster way. – avakar Aug 30 '16 at 19:49\n\nFor $v\\in V$, define the set of necessary vertices for $v$ as\n\n$$N(v)=\\{v' \\in V : \\text{for any path } u_0,\\dots, u_n \\text{ so that } u_0 \\in R\\text{ and } u_n = v, \\text{ there is an } i \\text{ so that }u_i = v'\\}$$\n\nThe idea here is that $v'\\in N(v)$ iff any path from $R$ to $v$ goes through $v'$ iff removing $v'$ makes $v$ unreachable.\n\nAssuming you have computed $N(v)$ for all $v$, then $$g(v)=\\sum_{v'\\in V}f(v')-\\sum_{\\substack{v'\\in V\\\\v\\not\\in N(v')}}f(v')=\\sum_{\\substack{v'\\in V\\\\v\\in N(v')}}f(v')$$\n\nFor each v\ng[v] := 0\nFor each v'\nFor each v in N(v')\ng[v] += f(v')\n\n\nComputing $g$ therefore costs $O(|V|^2)$ at most (but if you look closer, it's $\\displaystyle O\\left(\\sum_{v'\\in V}|N(v')|\\right)$ which should often be way smaller).\n\nLet's first try to compute $N$ in acyclic graphs:\n\nI'll call $\\operatorname{Pred}(v')$ the set of predecessors of $v'$ : $\\operatorname{Pred}(v'):=\\{v\\in V : v \\to v'\\}$.\n\nThen, I believe that $$N(v')=\\{v'\\}\\cup \\left(\\bigcap_{v\\in \\operatorname{Pred}(v')}N(v')\\right)$$\n\nFor v' in V in topological order\nsets = map((x -> N(x)), pred(v'))\nsmallest_set, remaining_sets := smallest_set(sets)\nS := smallest_set\nfor set in remaining_sets\nfor v in S\nif v not in set\nremove v from S\n\nThe cost is $O(|V|+|E|)$ for the topological sort, $O(|V|^2)$ to find the smallest set and then $O(|V|^3\\log |V|)$ to compute the intersection (but it's also $\\displaystyle O(\\sum_{v'\\in V}d^-(v')k_{v'}\\log K_{v'})$ with $k_{v'}$ the size of the smallest necessary set of the predecessors of $v'$ and $K_{v'}$ the size of the biggest, and $d^-(v')=|\\operatorname{Pred}(v')|$, and overall this shouldn't be too big because many predecessors implies many sets to intersect which implies small sets). You can therefore compute $N$ in $O(|V|^3\\log |V|)$." ]

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[ "C C++ JAVA PYTHON SQL HTML CSS DSA Robotics AWS CODING INTERVIEW PREPARATION\n\nMade with", null, "", null, "& Code\n\n# Addition of two matrices in C\n\nWritten By -\n\n## Program to add two matrices in c\n\n• Sum of matrices can be done when the matrices are compatible with each other.\n• The matrices are said to be compatible with each other when the number of rows and columns in both the matrices are equal.\n• Thus, if the first matrix has m rows and n columns second matrix should also have m rows and n columns.\n• Thus,  the elements can be added using the following formula: Ci,j = Ai,j + Bi,j   where i is the number of rows and j is the number of columns.\n\nApproach:\n\n• First we will take the number of rows and columns of each matrix as our input.\n• Next we validate if addition is possible, based upon number of rows and columns of both matrices are equal or not and accordingly either proceed with addition if valid and if invalid inform the user of it.\n• Next, using the above mentioned formula we calculate the sum of the matrices.\n\nCode:\n\n``````#include<stdio.h>\n\nint main()\n\n{\n\nint arr1, arr2, sum; //declaring array of predefined size 5 x 5\n\nint i, j, rows1,col1, rows2, col2;\n\nprintf (\"Enter the number of rows in the first matrix\\n\");\n\nscanf(\"%d\", &rows1);\n\nprintf (\"Enter the number of columns in the first matrix\\n\");\n\nscanf(\"%d\", &col1);\n\nprintf (\"Enter the number of rows in the second matrix\\n\");\n\nscanf(\"%d\", &rows2);\n\nprintf (\"Enter the number of columns in the second matrix\\n\");\n\nscanf(\"%d\", &col2);\n\nif ((rows1 != rows2) || (col1!=col2))\n\n{\n\nprintf(\"\\nThe matrices are not compatible. In order to perform sum of matrices number of rows and columns of the matrices should be equal.\\n\");\n\n}\n\nelse\n\n{\n\nprintf(\"Enter First Matrix Elements:\\n\");   //First Matrix\n\nfor(i = 0; i < rows1; i++)\n\n{\n\nfor(j = 0; j < col1; j++)\n\n{\n\nscanf(\"%d\", &arr1[i][j]);\n\n}\n\n}\n\nprintf(\"\\nEnter Second Matrix Elements:\\n\"); //Second Matrix\n\nfor(i = 0; i < rows2; i++)\n\n{\n\nfor(j = 0; j < col2; j++)\n\n{\n\nscanf(\"%d\", &arr2[i][j]);\n\n}\n\n}\n\nfor(i = 0; i < rows1; i++) //Performing addition of Matrix 1 and 2\n\n{\n\nfor(j = 0; j < col1; j++)\n\n{\n\nsum[i][j] = arr1[i][j] + arr2[i][j];\n\n}\n\n}\n\nprintf(\"\\nSum of matrices is as follows:\\n\");\n\nfor(i = 0; i < rows1; i++)\n\n{\n\nfor(j = 0; j < col1; j++)\n\n{\n\nprintf(\"%d\\t\", sum[i][j]);\n\n}\n\nprintf(\"\\n\");\n\n}\n\n}\n\nreturn 0;\n\n}``````\n\nOutput:\n\nCase 1: When number of rows and columns of both matrices do not match:\n\n``````Enter the number of rows in the first matrix\n\n3\n\nEnter the number of columns in the first matrix\n\n2\n\nEnter the number of rows in the second matrix\n\n1\n\nEnter the number of columns in the second matrix\n\n4``````\n\nThe matrices are not compatible. In order to perform sum of matrices number of rows and columns of the matrices should be equal.\n\nCase 2: When number of rows and columns of both matrices match:\n\n``````Enter the number of rows in the first matrix\n\n2\n\nEnter the number of columns in the first matrix\n\n3\n\nEnter the number of rows in the second matrix\n\n2\n\nEnter the number of columns in the second matrix\n\n3\n\nEnter First Matrix Elements:\n\n1\n\n2\n\n1\n\n4\n\n1\n\n5\n\nEnter Second Matrix Elements:\n\n2\n\n0\n\n3\n\n4\n\n1\n\n0\n\nSum of matrices is as follows:\n\n3 2 4\n\n8 2 5``````\n\nRelated Posts:\n\nOnline Compilers" ]

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[ "Show commands for: Magma / Pari/GP / SageMath\n\n## Minimal Weierstrass equation\n\nsage: E = EllipticCurve([1, -1, 0, -5639958, 5156776575]) # or\n\nsage: E = EllipticCurve(\"12789f2\")\n\ngp: E = ellinit([1, -1, 0, -5639958, 5156776575]) \\\\ or\n\ngp: E = ellinit(\"12789f2\")\n\nmagma: E := EllipticCurve([1, -1, 0, -5639958, 5156776575]); // or\n\nmagma: E := EllipticCurve(\"12789f2\");\n\n$$y^2 + x y = x^{3} - x^{2} - 5639958 x + 5156776575$$\n\n## Mordell-Weil group structure\n\n$$\\Z\\times \\Z/{2}\\Z \\times \\Z/{2}\\Z$$\n\n### Infinite order Mordell-Weil generator and height\n\nsage: E.gens()\n\nmagma: Generators(E);\n\n $$P$$ = $$\\left(\\frac{15079}{4}, \\frac{1533223}{8}\\right)$$ $$\\hat{h}(P)$$ ≈ $7.4721146276154595$\n\n## Torsion generators\n\nsage: E.torsion_subgroup().gens()\n\ngp: elltors(E)\n\nmagma: TorsionSubgroup(E);\n\n$$\\left(1374, -687\\right)$$, $$\\left(\\frac{5475}{4}, -\\frac{5475}{8}\\right)$$\n\n## Integral points\n\nsage: E.integral_points()\n\nmagma: IntegralPoints(E);\n\n$$\\left(-2742, 1371\\right)$$, $$\\left(1374, -687\\right)$$\n\n## Invariants\n\n sage: E.conductor().factor()  gp: ellglobalred(E)  magma: Conductor(E); Conductor: $$12789$$ = $$3^{2} \\cdot 7^{2} \\cdot 29$$ sage: E.discriminant().factor()  gp: E.disc  magma: Discriminant(E); Discriminant: $$126250019124003369$$ = $$3^{12} \\cdot 7^{10} \\cdot 29^{2}$$ sage: E.j_invariant().factor()  gp: E.j  magma: jInvariant(E); j-invariant: $$\\frac{231331938231569617}{1472026689}$$ = $$3^{-6} \\cdot 7^{-4} \\cdot 13^{3} \\cdot 29^{-2} \\cdot 47221^{3}$$ Endomorphism ring: $$\\Z$$ Geometric endomorphism ring: $$\\Z$$ (no potential complex multiplication) Sato-Tate group: $\\mathrm{SU}(2)$\n\n## BSD invariants\n\n sage: E.rank()  magma: Rank(E); Rank: $$1$$ sage: E.regulator()  magma: Regulator(E); Regulator: $$7.47211462762$$ sage: E.period_lattice().omega()  gp: E.omega  magma: RealPeriod(E); Real period: $$0.294262799202$$ sage: E.tamagawa_numbers()  gp: gr=ellglobalred(E); [[gr[i,1],gr[i]] | i<-[1..#gr[,1]]]  magma: TamagawaNumbers(E); Tamagawa product: $$32$$  = $$2^{2}\\cdot2^{2}\\cdot2$$ sage: E.torsion_order()  gp: elltors(E)  magma: Order(TorsionSubgroup(E)); Torsion order: $$4$$ sage: E.sha().an_numerical()  magma: MordellWeilShaInformation(E); Analytic order of Ш: $$1$$ (exact)\n\n## Modular invariants\n\nModular form 12789.2.a.k\n\nsage: E.q_eigenform(20)\n\ngp: xy = elltaniyama(E);\n\ngp: x*deriv(xy)/(2*xy+E.a1*xy+E.a3)\n\nmagma: ModularForm(E);\n\n$$q + q^{2} - q^{4} - 2q^{5} - 3q^{8} - 2q^{10} - 4q^{11} + 2q^{13} - q^{16} + 2q^{17} + 4q^{19} + O(q^{20})$$\n\n sage: E.modular_degree()  magma: ModularDegree(E); Modular degree: 294912 $$\\Gamma_0(N)$$-optimal: no Manin constant: 1\n\n#### Special L-value\n\nsage: r = E.rank();\n\nsage: E.lseries().dokchitser().derivative(1,r)/r.factorial()\n\ngp: ar = ellanalyticrank(E);\n\ngp: ar/factorial(ar)\n\nmagma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12);\n\n$$L'(E,1)$$ ≈ $$4.39753073255$$\n\n## Local data\n\nThis elliptic curve is not semistable.\n\nsage: E.local_data()\n\ngp: ellglobalred(E)\n\nmagma: [LocalInformation(E,p) : p in BadPrimes(E)];\n\nprime Tamagawa number Kodaira symbol Reduction type Root number ord($$N$$) ord($$\\Delta$$) ord$$(j)_{-}$$\n$$3$$ $$4$$ $$I_6^{*}$$ Additive -1 2 12 6\n$$7$$ $$4$$ $$I_4^{*}$$ Additive -1 2 10 4\n$$29$$ $$2$$ $$I_{2}$$ Non-split multiplicative 1 1 2 2\n\n## Galois representations\n\nThe image of the 2-adic representation attached to this elliptic curve is the subgroup of $\\GL(2,\\Z_2)$ with Rouse label X25k.\n\nThis subgroup is the pull-back of the subgroup of $\\GL(2,\\Z_2/2^3\\Z_2)$ generated by $\\left(\\begin{array}{rr} 3 & 0 \\\\ 0 & 5 \\end{array}\\right),\\left(\\begin{array}{rr} 3 & 0 \\\\ 4 & 1 \\end{array}\\right),\\left(\\begin{array}{rr} 5 & 0 \\\\ 0 & 1 \\end{array}\\right),\\left(\\begin{array}{rr} 1 & 2 \\\\ 0 & 5 \\end{array}\\right),\\left(\\begin{array}{rr} 3 & 0 \\\\ 0 & 3 \\end{array}\\right)$ and has index 24.\n\nsage: rho = E.galois_representation();\n\nsage: [rho.image_type(p) for p in rho.non_surjective()]\n\nmagma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];\n\nThe mod $$p$$ Galois representation has maximal image $$\\GL(2,\\F_p)$$ for all primes $$p$$ except those listed.\n\nprime Image of Galois representation\n$$2$$ Cs\n\n## $p$-adic data\n\n### $p$-adic regulators\n\nsage: [E.padic_regulator(p) for p in primes(3,20) if E.conductor().valuation(p)<2]\n\n$$p$$-adic regulators are not yet computed for curves that are not $$\\Gamma_0$$-optimal.\n\n## Iwasawa invariants\n\n $p$ Reduction type $\\lambda$-invariant(s) $\\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 ordinary add ordinary add ordinary ordinary ordinary ordinary ss nonsplit ordinary ordinary ordinary ordinary ordinary 5 - 1 - 1 1 1 1 1,1 1 1 1 1 1 1 0 - 0 - 0 0 0 0 0,0 0 0 0 0 0 0\n\nAn entry - indicates that the invariants are not computed because the reduction is additive.\n\n## Isogenies\n\nThis curve has non-trivial cyclic isogenies of degree $$d$$ for $$d=$$ 2 and 4.\nIts isogeny class 12789.k consists of 6 curves linked by isogenies of degrees dividing 8.\n\n## Growth of torsion in number fields\n\nThe number fields $K$ of degree up to 7 such that $E(K)_{\\rm tors}$ is strictly larger than $E(\\Q)_{\\rm tors}$ $\\cong \\Z/{2}\\Z \\times \\Z/{2}\\Z$ are as follows:\n\n$[K:\\Q]$ $K$ $E(K)_{\\rm tors}$ Base change curve\n$2$ $$\\Q(\\sqrt{21})$$ $$\\Z/2\\Z \\times \\Z/4\\Z$$ Not in database\n$4$ $$\\Q(\\sqrt{-3}, \\sqrt{-7})$$ $$\\Z/2\\Z \\times \\Z/8\\Z$$ Not in database\n$4$ $$\\Q(\\sqrt{-21}, \\sqrt{-87})$$ $$\\Z/2\\Z \\times \\Z/4\\Z$$ Not in database\n$4$ $$\\Q(\\sqrt{-21}, \\sqrt{87})$$ $$\\Z/2\\Z \\times \\Z/4\\Z$$ Not in database\n\nWe only show fields where the torsion growth is primitive. For each field $K$ we either show its label, or a defining polynomial when $K$ is not in the database." ]

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[ "# Probability and Prime Numbers\n\nWhat is the probability that a positive divisor of 8748 million is the product of exactly 20 non-distinct primes?\n\nHint: $8748$ million $= 8748 \\times 10^6 = 2^8 \\cdot 3^7 \\cdot 5^6$, which has $8+7+6 = 21$ non-distinct prime factors. Now, simply count the number of ways to remove just one prime factor, count the number of divisors, and divide." ]

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[ "/* 获取文章的评论人数 by zwwooooo | zww.me */\n\\$i=0; \\$j=0; \\$commentusers=array();\n++\\$i;\nif (\\$i==1) { \\$commentusers[] = \\$comment->comment_author_email; ++\\$j; }\nif ( !in_array(\\$comment->comment_author_email, \\$commentusers) ) {\n\\$commentusers[] = \\$comment->comment_author_email;\n++\\$j;\n}\n}\n\\$output = array(\\$j,\\$i);\n\\$which = (\\$which == 0) ? 0 : 1;\nreturn \\$output[\\$which]; //返回评论人数\n}\nreturn 0; //没有评论返回0\n}\n\n1\n\n1", null, "" ]

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[ "# Redshift Academy", null, "", null, "", null, "Wolfram Alpha:", null, "Search by keyword:", null, "Astronomy\n\n-\n-\n-\n-", null, "Chemistry\n\n-\n-\n-\n-", null, "Classical Mechanics\n\n-", null, "Classical Physics\n\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-", null, "Climate Change\n\n-", null, "Cosmology\n\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-", null, "Finance and Accounting\n\n-\n-\n-\n-\n-\n-\n-\n-\n-", null, "Game Theory\n\n-", null, "General Relativity\n\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-", null, "Group Theory\n\n-\n-\n-\n-\n-\n-", null, "Lagrangian and Hamiltonian Mechanics\n\n-\n-\n-\n-\n-\n-", null, "Macroeconomics\n\n-\n-\n-", null, "Mathematics\n\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-", null, "Mathjax\n\n-", null, "Microeconomics\n\n-", null, "Nuclear Physics\n\n-\n-", null, "Particle Physics\n\n-\n-\n-\n-\n-\n-\n-", null, "Probability and Statistics\n\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-", null, "Programming and Computer Science\n\n-\n-\n-\n-\n-\n-", null, "-", null, "Quantum Computing\n\n-\n-\n-", null, "Quantum Field Theory\n\n-\n-\n-\n-\n-", null, "Quantum Mechanics\n\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-", null, "Semiconductor Reliability\n\n-", null, "Solid State Electronics\n\n-\n-\n-\n-\n-", null, "Special Relativity\n\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-", null, "Statistical Mechanics\n\n-\n-\n-", null, "String Theory\n\n-\n-\n-\n-\n-\n-", null, "Superconductivity\n\n-\n-\n-\n-\n-\n-", null, "Supersymmetry (SUSY) and Grand Unified Theory (GUT)\n\n-\n-\n-\n-\n-", null, "The Standard Model\n\n-\n-\n-\n-\n-\n-\n-\n-\n-\n-", null, "Topology\n\n-", null, "Units, Constants and Useful Formulas\n\n-", null, "Monte-Carlo Methods\n-------------------\n\nThe Monte Carlo method is just one of many methods for analyzing\nuncertainty propagation, where the goal is to determine how random\nvariation, lack of knowledge, or error affects the sensitivity,\nperformance, or reliability of the system that is being modeled.\n\nMonte Carlo simulation is categorized as a sampling method because\nthe inputs are randomly generated from probability distributions to\nsimulate the process of sampling from an actual population. So, we try\nto choose a distribution for the inputs that most closely matches data\nwe already have, or best represents our current state of knowledge.\nThe data generated from the simulation can be represented as probability\ndistributions (or histograms) or converted to error bars, reliability\npredictions, tolerance zones, and confidence intervals.\n\n------\n| |\n---x1--| |---y1\n---x2--| f(x) |\n---x3--| |---y2\n| |\n------\n\n1. Create a parametric model y = f(x1,x2,...xn)\n2. Generate a random set of inputs x1,x1,...xn\n3. Evaluate the model and store the results as yn\n4. Repeat steps 2 and 3 for i = 1 to n\n5. Analyze the results using histograms, summary statistics,\nconfidence intervals, etc.\n\nExample: Compute the value of π\n\nInscribe circle inside square with sides x = 1, y = 1. Consider\n\ny\n|\n|.\n| .\n1 | .\n| .\n| .\n------------ x\n1\n\nArea of square = 4\n\nGenerate random combination of x and y between 0 and 1. From\nPythagoras:\n\nif √(x2 + y2) <= 1 then count as a 'hit'. if √(x2 + y2) > 1 then\ncount as a 'miss'.\n\n4 * hits\nπ = ---------------\n(hits + misses)\n\nIt is simple to write a simple javascript program to do this:\n\nvar hits = 0;\nvar i = 0;\nvar pi = 0;\nvar x = 0;\nvar y = 0;\nvar hyp = 0;\nfor(i=0; i<=1000000; i++)\n{\nx = Math.random();\ny = Math.random();\nhyp = Math.sqrt((x*x)+(y*y));\n\nif(hyp <= 1)\n{\nhits++;\n}\n}\n\npi = (4*hits)/i;\n\nfunction displaypi(){\nvar hits = 0;\nvar i = 0;\nvar pi = 0;\nvar x = 0;\nvar y = 0;\nvar hyp = 0;\nfor(i=0; i<=1000000; i++)\n{\nx = Math.random();\ny = Math.random();\nhyp = Math.sqrt((x*x)+(y*y));\n\nif(hyp <= 1)\n{\nhits++;\n}\n}\n\npi = (4*hits)/i;", null, "" ]

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[ "# #01202f Color Information\n\nIn a RGB color space, hex #01202f is composed of 0.4% red, 12.5% green and 18.4% blue. Whereas in a CMYK color space, it is composed of 97.9% cyan, 31.9% magenta, 0% yellow and 81.6% black. It has a hue angle of 199.6 degrees, a saturation of 95.8% and a lightness of 9.4%. #01202f color hex could be obtained by blending #02405e with #000000. Closest websafe color is: #003333.\n\n• R 0\n• G 13\n• B 18\nRGB color chart\n• C 98\n• M 32\n• Y 0\n• K 82\nCMYK color chart\n\n#01202f color description : Very dark (mostly black) blue.\n\n# #01202f Color Conversion\n\nThe hexadecimal color #01202f has RGB values of R:1, G:32, B:47 and CMYK values of C:0.98, M:0.32, Y:0, K:0.82. Its decimal value is 73775.\n\nHex triplet RGB Decimal 01202f `#01202f` 1, 32, 47 `rgb(1,32,47)` 0.4, 12.5, 18.4 `rgb(0.4%,12.5%,18.4%)` 98, 32, 0, 82 199.6°, 95.8, 9.4 `hsl(199.6,95.8%,9.4%)` 199.6°, 97.9, 18.4 003333 `#003333`\nCIE-LAB 10.882, -4.799, -13.203 1.042, 1.245, 2.874 0.202, 0.241, 1.245 10.882, 14.048, 250.025 10.882, -7.179, -10.328 11.156, -2.852, -7.467 00000001, 00100000, 00101111\n\n# Color Schemes with #01202f\n\n• #01202f\n``#01202f` `rgb(1,32,47)``\n• #2f1001\n``#2f1001` `rgb(47,16,1)``\nComplementary Color\n• #012f27\n``#012f27` `rgb(1,47,39)``\n• #01202f\n``#01202f` `rgb(1,32,47)``\n• #01092f\n``#01092f` `rgb(1,9,47)``\nAnalogous Color\n• #2f2701\n``#2f2701` `rgb(47,39,1)``\n• #01202f\n``#01202f` `rgb(1,32,47)``\n• #2f0109\n``#2f0109` `rgb(47,1,9)``\nSplit Complementary Color\n• #202f01\n``#202f01` `rgb(32,47,1)``\n• #01202f\n``#01202f` `rgb(1,32,47)``\n• #2f0120\n``#2f0120` `rgb(47,1,32)``\n• #012f10\n``#012f10` `rgb(1,47,16)``\n• #01202f\n``#01202f` `rgb(1,32,47)``\n• #2f0120\n``#2f0120` `rgb(47,1,32)``\n• #2f1001\n``#2f1001` `rgb(47,16,1)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000000\n``#000000` `rgb(0,0,0)``\n• #000f16\n``#000f16` `rgb(0,15,22)``\n• #01202f\n``#01202f` `rgb(1,32,47)``\n• #023148\n``#023148` `rgb(2,49,72)``\n• #024261\n``#024261` `rgb(2,66,97)``\n• #03537a\n``#03537a` `rgb(3,83,122)``\nMonochromatic Color\n\n# Alternatives to #01202f\n\nBelow, you can see some colors close to #01202f. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #012c2f\n``#012c2f` `rgb(1,44,47)``\n• #01282f\n``#01282f` `rgb(1,40,47)``\n• #01242f\n``#01242f` `rgb(1,36,47)``\n• #01202f\n``#01202f` `rgb(1,32,47)``\n• #011c2f\n``#011c2f` `rgb(1,28,47)``\n• #01182f\n``#01182f` `rgb(1,24,47)``\n• #01152f\n``#01152f` `rgb(1,21,47)``\nSimilar Colors\n\n# #01202f Preview\n\nThis text has a font color of #01202f.\n\n``<span style=\"color:#01202f;\">Text here</span>``\n#01202f background color\n\nThis paragraph has a background color of #01202f.\n\n``<p style=\"background-color:#01202f;\">Content here</p>``\n#01202f border color\n\nThis element has a border color of #01202f.\n\n``<div style=\"border:1px solid #01202f;\">Content here</div>``\nCSS codes\n``.text {color:#01202f;}``\n``.background {background-color:#01202f;}``\n``.border {border:1px solid #01202f;}``\n\n# Shades and Tints of #01202f\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000609 is the darkest color, while #f4fbff is the lightest one.\n\n• #000609\n``#000609` `rgb(0,6,9)``\n• #01131c\n``#01131c` `rgb(1,19,28)``\n• #01202f\n``#01202f` `rgb(1,32,47)``\n• #012d42\n``#012d42` `rgb(1,45,66)``\n• #023a55\n``#023a55` `rgb(2,58,85)``\n• #024769\n``#024769` `rgb(2,71,105)``\n• #03547c\n``#03547c` `rgb(3,84,124)``\n• #03618f\n``#03618f` `rgb(3,97,143)``\n• #036ea2\n``#036ea2` `rgb(3,110,162)``\n• #047cb5\n``#047cb5` `rgb(4,124,181)``\n• #0489c9\n``#0489c9` `rgb(4,137,201)``\n• #0596dc\n``#0596dc` `rgb(5,150,220)``\n• #05a3ef\n``#05a3ef` `rgb(5,163,239)``\n``#0eadfa` `rgb(14,173,250)``\n• #21b3fa\n``#21b3fa` `rgb(33,179,250)``\n• #34bafb\n``#34bafb` `rgb(52,186,251)``\n• #48c1fb\n``#48c1fb` `rgb(72,193,251)``\n• #5bc7fc\n``#5bc7fc` `rgb(91,199,252)``\n• #6ecefc\n``#6ecefc` `rgb(110,206,252)``\n• #81d4fc\n``#81d4fc` `rgb(129,212,252)``\n• #94dbfd\n``#94dbfd` `rgb(148,219,253)``\n• #a8e1fd\n``#a8e1fd` `rgb(168,225,253)``\n• #bbe8fe\n``#bbe8fe` `rgb(187,232,254)``\n• #ceeefe\n``#ceeefe` `rgb(206,238,254)``\n• #e1f5fe\n``#e1f5fe` `rgb(225,245,254)``\n• #f4fbff\n``#f4fbff` `rgb(244,251,255)``\nTint Color Variation\n\n# Tones of #01202f\n\nA tone is produced by adding gray to any pure hue. In this case, #171819 is the less saturated color, while #01202f is the most saturated one.\n\n• #171819\n``#171819` `rgb(23,24,25)``\n• #15191b\n``#15191b` `rgb(21,25,27)``\n• #131a1d\n``#131a1d` `rgb(19,26,29)``\n• #121a1e\n``#121a1e` `rgb(18,26,30)``\n• #101b20\n``#101b20` `rgb(16,27,32)``\n• #0e1c22\n``#0e1c22` `rgb(14,28,34)``\n• #0c1c24\n``#0c1c24` `rgb(12,28,36)``\n• #0a1d26\n``#0a1d26` `rgb(10,29,38)``\n• #081d28\n``#081d28` `rgb(8,29,40)``\n• #071e29\n``#071e29` `rgb(7,30,41)``\n• #051f2b\n``#051f2b` `rgb(5,31,43)``\n• #031f2d\n``#031f2d` `rgb(3,31,45)``\n• #01202f\n``#01202f` `rgb(1,32,47)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #01202f is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "#71a783 Color Information\n\nIn a RGB color space, hex #71a783 is composed of 44.3% red, 65.5% green and 51.4% blue. Whereas in a CMYK color space, it is composed of 32.3% cyan, 0% magenta, 21.6% yellow and 34.5% black. It has a hue angle of 140 degrees, a saturation of 23.5% and a lightness of 54.9%. #71a783 color hex could be obtained by blending #e2ffff with #004f07. Closest websafe color is: #669999.\n\n• R 44\n• G 65\n• B 51\nRGB color chart\n• C 32\n• M 0\n• Y 22\n• K 35\nCMYK color chart\n\n#71a783 color description : Mostly desaturated dark cyan - lime green.\n\n#71a783 Color Conversion\n\nThe hexadecimal color #71a783 has RGB values of R:113, G:167, B:131 and CMYK values of C:0.32, M:0, Y:0.22, K:0.35. Its decimal value is 7448451.\n\nHex triplet RGB Decimal 71a783 `#71a783` 113, 167, 131 `rgb(113,167,131)` 44.3, 65.5, 51.4 `rgb(44.3%,65.5%,51.4%)` 32, 0, 22, 35 140°, 23.5, 54.9 `hsl(140,23.5%,54.9%)` 140°, 32.3, 65.5 669999 `#669999`\nCIE-LAB 63.987, -25.595, 13.043 24.724, 32.786, 26.497 0.294, 0.39, 32.786 63.987, 28.727, 152.996 63.987, -26.54, 22.251 57.259, -23.127, 12.644 01110001, 10100111, 10000011\n\nColor Schemes with #71a783\n\n• #71a783\n``#71a783` `rgb(113,167,131)``\n• #a77195\n``#a77195` `rgb(167,113,149)``\nComplementary Color\n• #7aa771\n``#7aa771` `rgb(122,167,113)``\n• #71a783\n``#71a783` `rgb(113,167,131)``\n• #71a79e\n``#71a79e` `rgb(113,167,158)``\nAnalogous Color\n• #a7717a\n``#a7717a` `rgb(167,113,122)``\n• #71a783\n``#71a783` `rgb(113,167,131)``\n• #9e71a7\n``#9e71a7` `rgb(158,113,167)``\nSplit Complementary Color\n• #a78371\n``#a78371` `rgb(167,131,113)``\n• #71a783\n``#71a783` `rgb(113,167,131)``\n• #8371a7\n``#8371a7` `rgb(131,113,167)``\n• #95a771\n``#95a771` `rgb(149,167,113)``\n• #71a783\n``#71a783` `rgb(113,167,131)``\n• #8371a7\n``#8371a7` `rgb(131,113,167)``\n• #a77195\n``#a77195` `rgb(167,113,149)``\n• #4e7e5e\n``#4e7e5e` `rgb(78,126,94)``\n• #588d6a\n``#588d6a` `rgb(88,141,106)``\n• #619d75\n``#619d75` `rgb(97,157,117)``\n• #71a783\n``#71a783` `rgb(113,167,131)``\n• #81b191\n``#81b191` `rgb(129,177,145)``\n• #90bb9e\n``#90bb9e` `rgb(144,187,158)``\n• #a0c4ac\n``#a0c4ac` `rgb(160,196,172)``\nMonochromatic Color\n\nAlternatives to #71a783\n\nBelow, you can see some colors close to #71a783. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #71a776\n``#71a776` `rgb(113,167,118)``\n• #71a77a\n``#71a77a` `rgb(113,167,122)``\n• #71a77f\n``#71a77f` `rgb(113,167,127)``\n• #71a783\n``#71a783` `rgb(113,167,131)``\n• #71a788\n``#71a788` `rgb(113,167,136)``\n• #71a78c\n``#71a78c` `rgb(113,167,140)``\n• #71a791\n``#71a791` `rgb(113,167,145)``\nSimilar Colors\n\n#71a783 Preview\n\nThis text has a font color of #71a783.\n\n``<span style=\"color:#71a783;\">Text here</span>``\n#71a783 background color\n\nThis paragraph has a background color of #71a783.\n\n``<p style=\"background-color:#71a783;\">Content here</p>``\n#71a783 border color\n\nThis element has a border color of #71a783.\n\n``<div style=\"border:1px solid #71a783;\">Content here</div>``\nCSS codes\n``.text {color:#71a783;}``\n``.background {background-color:#71a783;}``\n``.border {border:1px solid #71a783;}``\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #020302 is the darkest color, while #f6faf7 is the lightest one.\n\n• #020302\n``#020302` `rgb(2,3,2)``\n• #0a0f0c\n``#0a0f0c` `rgb(10,15,12)``\n• #111c15\n``#111c15` `rgb(17,28,21)``\n• #19281e\n``#19281e` `rgb(25,40,30)``\n• #203427\n``#203427` `rgb(32,52,39)``\n• #284030\n``#284030` `rgb(40,64,48)``\n• #2f4c39\n``#2f4c39` `rgb(47,76,57)``\n• #375842\n``#375842` `rgb(55,88,66)``\n• #3e644b\n``#3e644b` `rgb(62,100,75)``\n• #467054\n``#467054` `rgb(70,112,84)``\n• #4d7c5d\n``#4d7c5d` `rgb(77,124,93)``\n• #558966\n``#558966` `rgb(85,137,102)``\n• #5c956f\n``#5c956f` `rgb(92,149,111)``\n• #659f78\n``#659f78` `rgb(101,159,120)``\n• #71a783\n``#71a783` `rgb(113,167,131)``\n• #7daf8e\n``#7daf8e` `rgb(125,175,142)``\n• #89b698\n``#89b698` `rgb(137,182,152)``\n• #95bea3\n``#95bea3` `rgb(149,190,163)``\n``#a1c5ad` `rgb(161,197,173)``\n• #aecdb8\n``#aecdb8` `rgb(174,205,184)``\n``#bad4c2` `rgb(186,212,194)``\n• #c6dccd\n``#c6dccd` `rgb(198,220,205)``\n• #d2e3d8\n``#d2e3d8` `rgb(210,227,216)``\n• #deebe2\n``#deebe2` `rgb(222,235,226)``\n• #eaf2ed\n``#eaf2ed` `rgb(234,242,237)``\n• #f6faf7\n``#f6faf7` `rgb(246,250,247)``\nTint Color Variation\n\nTones of #71a783\n\nA tone is produced by adding gray to any pure hue. In this case, #8c8c8c is the less saturated color, while #21f768 is the most saturated one.\n\n• #8c8c8c\n``#8c8c8c` `rgb(140,140,140)``\n• #839589\n``#839589` `rgb(131,149,137)``\n• #7a9e86\n``#7a9e86` `rgb(122,158,134)``\n• #71a783\n``#71a783` `rgb(113,167,131)``\n• #68b080\n``#68b080` `rgb(104,176,128)``\n• #5fb97d\n``#5fb97d` `rgb(95,185,125)``\n• #56c27a\n``#56c27a` `rgb(86,194,122)``\n• #4eca77\n``#4eca77` `rgb(78,202,119)``\n• #45d374\n``#45d374` `rgb(69,211,116)``\n• #3cdc71\n``#3cdc71` `rgb(60,220,113)``\n• #33e56e\n``#33e56e` `rgb(51,229,110)``\n• #2aee6b\n``#2aee6b` `rgb(42,238,107)``\n• #21f768\n``#21f768` `rgb(33,247,104)``\nTone Color Variation\n\nColor Blindness Simulator\n\nBelow, you can see how #71a783 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

[ null ]

[ "# Subgrouping elements of list based on their values\n\nIn a list of elements I want to group the elements which are in consecutive order. For eg.\n\nlist = {2,3,4,5,6,9,12,13,14,16,17}\nresult = {{2,3,4,5,6},{9},{12,13,14},{16,17}}\n\n\nI have written the code below for it and it works fine. I wanted to know if I can use any in built Function to do it in a better way.\n\n For[\ni = 1; res = {},\ni <= Length[list],\ni++,\ntemp = {list[[i]]} ; While[If[i + 1 > Length[list], Break[]];list[[i + 1]] - list[[i]] == 1,AppendTo[temp, list[[i + 1]]]; i++]; AppendTo[res, temp]\n];\n\n• – kglr\nJul 4, 2017 at 15:10\n\nSplit is the built-in function you need:\n\nSplit[list, #2 == # + 1 &]\n\n\n{{2, 3, 4, 5, 6}, {9}, {12, 13, 14}, {16, 17}}\n\n• Looks neat. Thanks. Jul 4, 2017 at 15:12\n• @mrkbtr, my pleasure. Thank you for the accept.\n– kglr\nJul 4, 2017 at 15:13\n• kglr, I think this question reasonably already has an answer in (23607); do you disagree? Jul 5, 2017 at 15:38\n• @Mr.Wizard, the linked question(s) asks more than what Split gives. One of the answers does use Split as a step, but imho it would be difficult for an average user to use that answer to identify an answer for the current question.\n– kglr\nJul 5, 2017 at 19:15\n\nThis question is arguably answered in Find subsequences of consecutive integers inside a list.\n\nNevertheless as it remains open it is beneficial to apply similar methods here.\n\nUsing intervals from my answer there this is solved with:\n\nRange @@@ intervals[list]\n\n{{2, 3, 4, 5, 6}, {9}, {12, 13, 14}, {16, 17}}\n\n\nThis can be much faster than Split:\n\nintervals[a_List] :=\n{a[[Prepend[# + 1, 1]]], a[[Append[#, -1]]]}\\[Transpose] & @\n\na = Delete[#, List /@ RandomSample[#, 15000]] &@Range@1*^7;\n\n(r1 = Range @@@ intervals[a]) // RepeatedTiming // First\n\n(r2 = Split[a, #2 == # + 1 &]) // RepeatedTiming // First\n\nr1 === r2\n\n0.13\n\n6.20\n\nTrue" ]

[ null ]

[ "Intensive Care Medicine Experimental (2016-09-01)\n\n# ESICM LIVES 2016: part one\n\n• L. Bos,\n• L. Schouten,\n• L. van Vught,\n• M. Wiewel,\n• D. Ong,\n• O. Cremer,\n• A. Artigas,\n• I. Martin-Loeches,\n• A. Hoogendijk,\n• T. van der Poll,\n• J. Horn,\n• N. Juffermans,\n• M. Schultz,\n• N. de Prost,\n• T. Pham,\n• G. Carteaux,\n• A. Mekontso Dessap,\n• C. Brun-Buisson,\n• E. Fan,\n• G. Bellani,\n• J. Laffey,\n• A. Mercat,\n• L. Brochard,\n• B. Maitre,\n• LUNG SAFE investigators and the ESICM study group,\n• P. A. Howells,\n• D. R. Thickett,\n• C. Knox,\n• D. P. Park,\n• F. Gao,\n• O. Tucker,\n• T. Whitehouse,\n• D. F. McAuley,\n• G. D. Perkins,\n• T. Pham,\n• J. Laffey,\n• G. Bellani,\n• E. Fan,\n• LUNG SAFE Investigators and the ESICM Trials Group,\n• L. Pisani,\n• J. P. Roozeman,\n• F. D. Simonis,\n• A. Giangregorio,\n• L. R. Schouten,\n• S. M. Van der Hoeven,\n• J. Horn,\n• A. Serpa Neto,\n• E. Festic,\n• A. M. Dondorp,\n• S. Grasso,\n• L. D. Bos,\n• M. J. Schultz,\n• M. Koster-Brouwer,\n• D. Verboom,\n• B. Scicluna,\n• K. van de Groep,\n• J. Frencken,\n• M. Schultz,\n• T. van der Poll,\n• M. Bonten,\n• O. Cremer,\n• J. I. Ko,\n• K. S. Kim,\n• G. J. Suh,\n• W. Y. Kwon,\n• K. Kim,\n• J. H. Shin,\n• O. T. Ranzani,\n• E. Prina,\n• R. Menendez,\n• A. Ceccato,\n• R. Mendez,\n• C. Cilloniz,\n• A. Gabarrus,\n• M. Ferrer,\n• A. Torres,\n• A. Urbano,\n• L. A. Zhang,\n• D. Swigon,\n• F. Pike,\n• R. S. Parker,\n• G. Clermont,\n• C. Scheer,\n• S. O. Kuhn,\n• A. Modler,\n• M. Vollmer,\n• C. Fuchs,\n• K. Hahnenkamp,\n• S. Rehberg,\n• M. Gründling,\n• A. Taggu,\n• N. Darang,\n• N. Öveges,\n• I. László,\n• K. Tánczos,\n• M. Németh,\n• G. Lebák,\n• B. Tudor,\n• D. Érces,\n• J. Kaszaki,\n• W. Huber,\n• D. Trásy,\n• Z. Molnár,\n• G. Ferrara,\n• V. S. Kanoore Edul,\n• H. S. Canales,\n• E. Martins,\n• C. Canullán,\n• G. Murias,\n• M. O. Pozo,\n• J. F. Caminos Eguillor,\n• M. G. Buscetti,\n• C. Ince,\n• A. Dubin,\n• H. D. Aya,\n• A. Rhodes,\n• N. Fletcher,\n• R. M. Grounds,\n• M. Cecconi,\n• M. Jacquet-Lagrèze,\n• M. Riche,\n• R. Schweizer,\n• P. Portran,\n• W. Fornier,\n• M. Lilot,\n• J. Neidecker,\n• J. L. Fellahi,\n• A. Escoresca-Ortega,\n• A. Gutiérrez-Pizarraya,\n• L. Charris-Castro,\n• Y. Corcia-Palomo,\n• J. Garnacho-Montero,\n• C. Roger,\n• L. Muller,\n• L. Elotmani,\n• J. Lipman,\n• J. Y. Lefrant,\n• J. A. Roberts,\n• R. Muñoz-Bermúdez,\n• M. Samper,\n• C. Climent,\n• F. Vasco,\n• V. Sara,\n• S. Luque,\n• N. Campillo,\n• S. Grau Cerrato,\n• J. R. Masclans,\n• F. Alvarez-Lerma,\n• S. Carvalho Brugger,\n• G. Jimenez Jimenez,\n• M. Miralbés Torner,\n• J. Trujillano Cabello,\n• B. Balsera Garrido,\n• X. Nuvials Casals,\n• F. Barcenilla Gaite,\n• M. Vallverdú Vidal,\n• M. Palomar Martínez,\n• V. Gusarov,\n• D. Shilkin,\n• M. Dementienko,\n• E. Nesterova,\n• N. Lashenkova,\n• A. Kuzovlev,\n• M. Zamyatin,\n• A. Demoule,\n• S. Carreira,\n• S. Lavault,\n• O. Palancca,\n• E. Morawiec,\n• J. Mayaux,\n• I. Arnulf,\n• T. Similowski,\n• B. S. Rasmussen,\n• R. G. Maltesen,\n• M. Hanifa,\n• S. Pedersen,\n• S. R. Kristensen,\n• R. Wimmer,\n• G. Li Bassi,\n• O. T. Ranzani,\n• T. Kolobow,\n• A. Zanella,\n• M. Cressoni,\n• L. Berra,\n• V. Parrini,\n• H. Kandil,\n• G. Salati,\n• S. Livigni,\n• A. Amatu,\n• A. Andreotti,\n• F. Tagliaferri,\n• G. Moise,\n• G. Mercurio,\n• A. Costa,\n• A. Vezzani,\n• S. Lindau,\n• J. Babel,\n• M. Cavana,\n• D. Consonni,\n• A. Pesenti,\n• L. Gattinoni,\n• A. Torres,\n• for the GRAVITY-VAP TRIAL NETWORK,\n• P. Mansouri,\n• F. Zand,\n• L. Zahed,\n• M. Bahrani,\n• M. Ghorbani,\n• B. Cambiaghi,\n• O. Moerer,\n• T. Mauri,\n• N. Kunze-Szikszay,\n• C. Ritter,\n• A. Pesenti,\n• M. Quintel,\n• L. M. Vilander,\n• M. A. Kaunisto,\n• S. T. Vaara,\n• V. Pettilä,\n• FINNAKI Study Group,\n• J. L. G. Haitsma Mulier,\n• S. Rozemeijer,\n• A. M. E. Spoelstra-de Man,\n• P. E. Elbers,\n• P. R. Tuinman,\n• M. C. de Waard,\n• H. M. Oudemans-van Straaten,\n• A. M. A. Liberatore,\n• R. B. Souza,\n• A. M. C. R. P. F. Martins,\n• J. C. F. Vieira,\n• I. H. J. Koh,\n• M. Galindo Martínez,\n• R. Jiménez Sánchez,\n• L. Martínez Gascón,\n• M. D. Rodríguez Mulero,\n• A. Ortín Freire,\n• S. Rebollo Acebes,\n• Á. Fernández Martínez,\n• S. Moreno Aliaga,\n• L. Herrera Para,\n• J. Murcia Payá,\n• F. Rodríguez Mulero,\n• P. Guerci,\n• Y. Ince,\n• P. Heeman,\n• B. Ergin,\n• C. Ince,\n• Z. Uz,\n• M. Massey,\n• Y. Ince,\n• R. Papatella,\n• E. Bulent,\n• P. Guerci,\n• F. Toraman,\n• C. Ince,\n• E. R. Longbottom,\n• H. D. Torrance,\n• H. C. Owen,\n• C. J. Hinds,\n• R. M. Pearse,\n• M. J. O’Dywer,\n• Z. Trogrlic,\n• M. van der Jagt,\n• H. Lingsma,\n• H. H. Ponssen,\n• J. F. Schoonderbeek,\n• F. Schreiner,\n• S. J. Verbrugge,\n• S. Duran,\n• T. van Achterberg,\n• J. Bakker,\n• D. A. M. P. J. Gommers,\n• E. Ista,\n• A. Krajčová,\n• P. Waldauf,\n• F. Duška,\n• A. Shah,\n• N. Roy,\n• S. McKechnie,\n• C. Doree,\n• S. Fisher,\n• S. J. Stanworth,\n• J. F. Jensen,\n• D. Overgaard,\n• M. H. Bestle,\n• D. F. Christensen,\n• I. Egerod,\n• The RAPIT Group,\n• A. Pivkina,\n• V. Gusarov,\n• I. Zhivotneva,\n• M. Zamyatin,\n• J. F. Jensen,\n• I. Egerod,\n• M. H. Bestle,\n• D. F. Christensen,\n• A. Alklit,\n• R. L. Hansen,\n• H. Knudsen,\n• L. B. Grode,\n• D. Overgaard,\n• The RAPIT group,\n• M. Hravnak,\n• L. Chen,\n• A. Dubrawski,\n• G. Clermont,\n• M. R. Pinsky,\n• S. M. Parry,\n• L. D. Knight,\n• B. C. Connolly,\n• C. E. Baldwin,\n• Z. A. Puthucheary,\n• L. Denehy,\n• N. Hart,\n• P. E. Morris,\n• J. Mortimore,\n• C. L. Granger,\n• H. I. Jensen,\n• R. Piers,\n• B. Van den Bulcke,\n• J. Malmgren,\n• V. Metaxa,\n• A. K. Reyners,\n• M. Darmon,\n• K. Rusinova,\n• D. Talmor,\n• A. P. Meert,\n• L. Cancelliere,\n• L. Zubek,\n• P. Maia,\n• A. Michalsen,\n• J. Decruyenaere,\n• E. Kompanje,\n• S. Vanheule,\n• E. Azoulay,\n• S. Vansteelandt,\n• D. Benoit,\n• B. Van den Bulcke,\n• R. Piers,\n• H. I. Jensen,\n• J. Malmgren,\n• V. Metaxa,\n• A. K. Reyners,\n• M. Darmon,\n• K. Rusinova,\n• D. Talmor,\n• A. P. Meert,\n• L. Cancelliere,\n• L. Zubek,\n• P. Maia,\n• A. Michalsen,\n• J. Decruyenaere,\n• E. Kompanje,\n• S. Vanheule,\n• E. Azoulay,\n• S. Vansteelandt,\n• D. Benoit,\n• C. Ryan,\n• D. Dawson,\n• J. Ball,\n• K. Noone,\n• B. Aisling,\n• S. Prudden,\n• A. Ntantana,\n• D. Matamis,\n• S. Savvidou,\n• M. Giannakou,\n• M. Gouva,\n• G. Nakos,\n• V. Koulouras,\n• J. Aron,\n• G. Lumley,\n• D. Milliken,\n• B. A. McGrath,\n• S. J. Lynch,\n• B. Bovento,\n• G. Sharpe,\n• E. Grainger,\n• S. Pieri-Davies,\n• S. Wallace,\n• B. McGrath,\n• S. J. Lynch,\n• B. Bovento,\n• E. Grainger,\n• S. Pieri-Davies,\n• G. Sharpe,\n• S. Wallace,\n• M. Jung,\n• J. Cho,\n• H. Park,\n• G. Suh,\n• O. Kousha,\n• L. Gamrin Gripenberg,\n• M. Sundström Rehal,\n• J. Wernerman,\n• O. Rooyackers,\n• H. J. de Grooth,\n• W. P. Choo,\n• A. M. Spoelstra-de Man,\n• E. L. Swart,\n• H. M. Oudemans-van Straaten,\n• L. Talan,\n• G. Güven,\n• N. D. Altıntas,\n• G. Uusvel,\n• L. Starkopf,\n• J. Starkopf,\n• A. Reintam Blaser,\n• M. S. Kalaiselvan,\n• A. S. Arunkumar,\n• M. K. Renuka,\n• R. L. Shivkumar,\n• M. Volbeda,\n• D. ten Kate,\n• M. Hoekstra,\n• J. M. van der Maaten,\n• M. W. Nijsten,\n• A. Komaromi,\n• O. Rooyackers,\n• J. Wernerman,\n• Å. Norberg,\n• M. Smedberg,\n• M. Mori,\n• Å. Norberg,\n• O. Rooyackers,\n• J. Wernerman,\n• M. Theodorakopoulou,\n• T. Christodoulopoulou,\n• A. Diamantakis,\n• F. Frantzeskaki,\n• M. Kontogiorgi,\n• E. Chrysanthopoulou,\n• M. Lygnos,\n• C. Diakaki,\n• A. Armaganidis,\n• K. Gundogan,\n• E. Dogan,\n• R. Coskun,\n• S. Muhtaroglu,\n• M. Sungur,\n• T. Ziegler,\n• M. Guven,\n• A. Kleyman,\n• W. Khaliq,\n• D. Andreas,\n• M. Singer,\n• R. Meierhans,\n• R. Schuepbach,\n• I. De Brito-Ashurst,\n• F. Zand,\n• G. Sabetian,\n• R. Nikandish,\n• F. Hagar,\n• M. Masjedi,\n• B. Maghsudi,\n• A. Vazin,\n• M. Ghorbani,\n• K. C. Kao,\n• L. C. Chiu,\n• C. Y. Hung,\n• C. H. Chang,\n• S. H. Li,\n• H. C. Hu,\n• S. El Maraghi,\n• M. Ali,\n• D. Rageb,\n• M. Helmy,\n• J. Marin-Corral,\n• C. Vilà,\n• J. R. Masclans,\n• A. Vàzquez,\n• I. Martín-Loeches,\n• E. Díaz,\n• J. C. Yébenes,\n• A. Rodriguez,\n• F. Álvarez-Lerma,\n• H1N1 SEMICYUC/GETGAG Working Group,\n• N. Varga,\n• A. Cortina-Gutiérrez,\n• L. Dono,\n• M. Martínez-Martínez,\n• E. Papiol,\n• M. Pérez-Carrasco,\n• R. Ferrer,\n• K. Nweze,\n• B. Morton,\n• I. Welters,\n• M. Houard,\n• B. Voisin,\n• G. Ledoux,\n• S. Six,\n• E. Jaillette,\n• S. Nseir,\n• S. Romdhani,\n• R. Bouneb,\n• D. Loghmari,\n• N. Ben Aicha,\n• J. Ayachi,\n• K. Meddeb,\n• I. Chouchène,\n• A. Khedher,\n• M. Boussarsar,\n• K. S. Chan,\n• W. L. Yu,\n• J. Marin-Corral,\n• C. Vilà,\n• J. R. Masclans,\n• J. Nolla,\n• L. Vidaur,\n• J. Bonastre,\n• B. Suberbiola,\n• J. E. Guerrero,\n• A. Rodriguez,\n• H1N1 SEMICYUC/GETGAG working group,\n• N. Ramon Coll,\n• G. Jiménez Jiménez,\n• S. Carvalho Brugger,\n• J. Codina Calero,\n• B. Balsera Garrido,\n• M. García,\n• M. Palomar Martínez,\n• M. Vallverdú Vidal,\n• M. C. de la Torre,\n• E. Vendrell,\n• E. Palomera,\n• E. Güell,\n• J. C. Yébenes,\n• M. Serra-Prat,\n• J. F. Bermejo-Martín,\n• J. Almirall,\n• E. Tomas,\n• A. Escoval,\n• F. Froe,\n• M. H. Vitoria Pereira,\n• N. Velez,\n• E. Viegas,\n• E. Filipe,\n• C. Groves,\n• M. Reay,\n• L. C. Chiu,\n• H. C. Hu,\n• C. Y. Hung,\n• C. H. Chang,\n• S. H. Li,\n• K. C. Kao,\n• A. Ballin,\n• F. Facchin,\n• G. Sartori,\n• F. Zarantonello,\n• E. Campello,\n• S. Rossi,\n• C. Ori,\n• P. Simioni,\n• N. Umei,\n• I. Shingo,\n• A. C. Santos,\n• C. Candeias,\n• I. Moniz,\n• R. Marçal,\n• Z. Costa e Silva,\n• J. M. Ribeiro,\n• J. F. Georger,\n• J. P. Ponthus,\n• M. Tchir,\n• V. Amilien,\n• M. Ayoub,\n• E. Barsam,\n• G. Martucci,\n• G. Panarello,\n• F. Tuzzolino,\n• G. Capitanio,\n• V. Ferrazza,\n• T. Carollo,\n• L. Giovanni,\n• M. López Sánchez,\n• M. A. González-Gay,\n• F. J. Llorca Díaz,\n• M. I. Rubio López,\n• E. Zogheib,\n• L. Villeret,\n• M. Bernasinski,\n• P. Besserve,\n• T. Caus,\n• H. Dupont,\n• P. Morimont,\n• S. Habran,\n• R. Hubert,\n• T. Desaive,\n• F. Blaffart,\n• N. Janssen,\n• J. Guiot,\n• A. Pironet,\n• P. Dauby,\n• B. Lambermont,\n• F. Zarantonello,\n• A. Ballin,\n• F. Facchin,\n• G. Sartori,\n• E. Campello,\n• T. Pettenuzzo,\n• G. Citton,\n• S. Rossi,\n• P. Simioni,\n• C. Ori,\n• C. Kirakli,\n• O. Ediboglu,\n• S. Ataman,\n• M. Yarici,\n• F. Tuksavul,\n• S. Keating,\n• A. Gibson,\n• M. Gilles,\n• M. Dunn,\n• G. Price,\n• N. Young,\n• P. Remeta,\n• P. Bishop,\n• M. D. Fernández Zamora,\n• J. Muñoz-Bono,\n• E. Curiel-Balsera,\n• E. Aguilar-Alonso,\n• R. Hinojosa,\n• A. Gordillo-Brenes,\n• J. A. Arboleda-Sánchez,\n• ARIAM-CARDIAC SURGERY PROJECT AUTHORS,\n• I. Skorniakov,\n• D. Vikulova,\n• C. Whiteley,\n• O. Shaikh,\n• A. Jones,\n• M. Ostermann,\n• L. Forni,\n• M. Scott,\n• J. Sahatjian,\n• W. Linde-Zwirble,\n• D. Hansell,\n• P. Laoveeravat,\n• N. Srisawat,\n• M. Kongwibulwut,\n• S. Peerapornrattana,\n• N. Suwachittanont,\n• T. O. Wirotwan,\n• P. Chatkaew,\n• P. Saeyub,\n• K. Latthaprecha,\n• K. Tiranathanagul,\n• S. Eiam-ong,\n• J. A. Kellum,\n• R. E. Berthelsen,\n• A. Perner,\n• A. E. K. Jensen,\n• J. U. Jensen,\n• M. H. Bestle,\n• D. J. Gebhard,\n• J. Price,\n• C. E. Kennedy,\n• A. Akcan-Arikan,\n• A. M. A. Liberatore,\n• R. B. Souza,\n• A. M. C. R. P. F. Martins,\n• J. C. F. Vieira,\n• Y. R. Kang,\n• M. N. Nakamae,\n• I. H. J. Koh,\n• K. Hamed,\n• M. M. Khaled,\n• R. Aly Soliman,\n• M. Sherif Mokhtar,\n• G. Seller-Pérez,\n• D. Arias-Verdú,\n• E. Llopar-Valdor,\n• I. De-Diós-Chacón,\n• M. E. Herrera-Gutierrez,\n• R. Hafes,\n• G. Carroll,\n• P. Doherty,\n• C. Wright,\n• I. G. Guerra Vera,\n• M. Ralston,\n• M. L. Gemmell,\n• A. MacKay,\n• E. Black,\n• C. Wright,\n• R. I. Docking,\n• R. Appleton,\n• M. R. Ralston,\n• L. Gemmell,\n• R. Appleton,\n• C. Wright,\n• R. I. Docking,\n• E. Black,\n• A. Mackay,\n• S. Rozemeijer,\n• J. L. G. Haitsma Mulier,\n• J. G. Röttgering,\n• P. W. G. Elbers,\n• A. M. E. Spoelstra-de Man,\n• P. R. Tuinman,\n• M. C. de Waard,\n• H. M. Oudemans-van Straaten,\n• N. Mejeni,\n• J. Nsiala,\n• A. Kilembe,\n• P. Akilimali,\n• G. Thomas,\n• I. Egerod,\n• A. M. Fagerdahl,\n• V. Knudsen,\n• P-INFECT,\n• K. Meddeb,\n• A. Ben Cheikh,\n• Y. Hamdaoui,\n• J. Ayachi,\n• A. Guiga,\n• N. Fraj,\n• S. Romdhani,\n• N. Sma,\n• R. Bouneb,\n• I. Chouchene,\n• A. Khedher,\n• N. Bouafia,\n• M. Boussarsar,\n• A. Amirian,\n• B. Ziaian,\n• M. Masjedi,\n• C. Fleischmann,\n• D. O. Thomas-Rueddel,\n• A. Schettler,\n• D. Schwarzkopf,\n• A. Stacke,\n• K. Reinhart,\n• E. Filipe,\n• A. Escoval,\n• A. Martins,\n• P. Sousa,\n• N. Velez,\n• E. Viegas,\n• E. Tomas,\n• G. Snell,\n• R. Matsa,\n• T. T. S. Paary,\n• M. S. Kalaiselvan,\n• A. M. Cavalheiro,\n• L. L. Rocha,\n• C. S. Vallone,\n• A. Tonilo,\n• M. D. S. Lobato,\n• D. T. Malheiro,\n• G. Sussumo,\n• N. M. Lucino,\n• F. Zand,\n• V. D. Rosenthal,\n• M. Masjedi,\n• G. Sabetian,\n• B. Maghsudi,\n• M. Ghorbani,\n• A. Sanaei Dashti,\n• A. Yousefipour,\n• J. R. Goodall,\n• M. Williamson,\n• E. Tant,\n• N. Thomas,\n• C. Balci,\n• C. Gonen,\n• E. Haftacı,\n• H. Gurarda,\n• E. Karaca,\n• B. Paldusová,\n• I. Zýková,\n• D. Šímová,\n• S. Houston,\n• L. D’Antona,\n• J. Lloyd,\n• V. Garnelo-Rey,\n• M. Sosic,\n• V. Sotosek-Tokmazic,\n• J. Kuharic,\n• I. Antoncic,\n• S. Dunatov,\n• A. Sustic,\n• C. T. Chong,\n• M. Sim,\n• T. Lyovarin,\n• F. M. Acosta Díaz,\n• S. Narbona Galdó,\n• M. Muñoz Garach,\n• O. Moreno Romero,\n• A. M. Pérez Bailón,\n• A. Carranza Pinel,\n• M. Colmenero,\n• A. Gritsan,\n• A. Gazenkampf,\n• E. Korchagin,\n• N. Dovbish,\n• R. M. Lee,\n• M. P. P. Lim,\n• C. T. Chong,\n• B. C. L. Lim,\n• J. J. See,\n• R. Assis,\n• F. Filipe,\n• N. Lopes,\n• L. Pessoa,\n• T. Pereira,\n• N. Catorze,\n• M. S. Aydogan,\n• C. Aldasoro,\n• P. Marchio,\n• A. Jorda,\n• M. D. Mauricio,\n• S. Guerra-Ojeda,\n• M. Gimeno-Raga,\n• M. Colque-Cano,\n• A. Bertomeu-Artecero,\n• M. Aldasoro,\n• S. L. Valles,\n• D. Tonon,\n• T. Triglia,\n• J. C. Martin,\n• M. C. Alessi,\n• N. Bruder,\n• P. Garrigue,\n• L. Velly,\n• S. Spina,\n• V. Scaravilli,\n• C. Marzorati,\n• E. Colombo,\n• D. Savo,\n• A. Vargiolu,\n• G. Cavenaghi,\n• G. Citerio,\n• P. Bulgarelli,\n• J. A. P. Araujo,\n• V. Gonzalez,\n• V. A. Souza,\n• A. Costa,\n• C. Massant,\n• C. A. C. Abreu Filho,\n• R. A. Morbeck,\n• L. E. Burgo,\n• R. van Groenendael,\n• L. T. van Eijk,\n• G. P. Leijte,\n• B. Koeneman,\n• M. Kox,\n• P. Pickkers,\n• A. García-de la Torre,\n• A. Fernández-Porcel,\n• C. Rueda-Molina,\n• P. Nuevo-Ortega,\n• T. Tsvetanova-Spasova,\n• E. Cámara-Sola,\n• A. García-Alcántara,\n• L. Salido-Díaz,\n• X. Liao,\n• T. Feng,\n• J. Zhang,\n• X. Cao,\n• Q. Wu,\n• Z. Xie,\n• H. Li,\n• Y. Kang,\n• M. S. Winkler,\n• A. Nierhaus,\n• E. Mudersbach,\n• A. Bauer,\n• L. Robbe,\n• C. Zahrte,\n• E. Schwedhelm,\n• S. Kluge,\n• C. Zöllner,\n• B. Morton,\n• E. Mitsi,\n• S. H. Pennington,\n• J. Reine,\n• A. D. Wright,\n• R. Parker,\n• I. D. Welters,\n• J. D. Blakey,\n• G. Rajam,\n• D. M. Ferreira,\n• D. Wang,\n• S. B. Gordon,\n• R. Koch,\n• M. Kox,\n• J. Rahamat-Langedoen,\n• J. Schloesser,\n• M. de Jonge,\n• P. Pickkers,\n• J. Bringue,\n• R. Guillamat-Prats,\n• E. Torrents,\n• M. L. Martinez,\n• M. Camprubí-Rimblas,\n• A. Artigas,\n• L. Blanch,\n• S. Y. Park,\n• Y. B. Park,\n• D. K. Song,\n• S. Shrestha,\n• S. H. Park,\n• Y. Koh,\n• M. J. Park,\n• C. W. Hong,\n• O. Lesur,\n• D. Coquerel,\n• X. Sainsily,\n• J. Cote,\n• A. Murza,\n• L. Dumont,\n• R. Dumaine,\n• M. Grandbois,\n• P. Sarret,\n• E. Marsault,\n• D. Salvail,\n• M. Auger-Messier,\n• F. Chagnon,\n• Apelin Group,\n• M. P. Lauretta,\n• E. Greco,\n• A. Dyson,\n• M. Singer,\n• S. Preau,\n• M. Ambler,\n• A. Sigurta,\n• S. Saeed,\n• M. Singer,\n• L. Topcu Sarıca,\n• N. Zibandeh,\n• D. Genc,\n• F. Gul,\n• T. Akkoc,\n• E. Kombak,\n• L. Cinel,\n• T. Akkoc,\n• I. Cinel,\n• S. J. Pollen,\n• N. Arulkumaran,\n• M. Singer,\n• H. D. Torrance,\n• E. R. Longbottom,\n• G. Warnes,\n• C. J. Hinds,\n• D. J. Pennington,\n• K. Brohi,\n• M. J. O’Dwyer,\n• H. Y. Kim,\n• S. Na,\n• J. Kim,\n• Y. F. Chang,\n• A. Chao,\n• P. Y. Shih,\n• C. T. Lee,\n• Y. C. Yeh,\n• L. W. Chen,\n• Z. Trogrlic,\n• E. Ista,\n• H. Lingsma,\n• W. Rietdijk,\n• H. H. Ponssen,\n• J. F. Schoonderbeek,\n• F. Schreiner,\n• S. J. Verbrugge,\n• S. Duran,\n• D. A. M. P. J. Gommers,\n• M. van der Jagt,\n• S. Funcke,\n• S. Sauerlaender,\n• B. Saugel,\n• H. Pinnschmidt,\n• D. A. Reuter,\n• R. Nitzschke,\n• S. Perbet,\n• C. Biboulet,\n• A. Lenoire,\n• D. Bourdeaux,\n• B. Pereira,\n• B. Plaud,\n• J. E. Bazin,\n• V. Sautou,\n• A. Mebazaa,\n• J. M. Constantin,\n• M. Legrand,\n• Y. Boyko,\n• P. Jennum,\n• M. Nikolic,\n• H. Oerding,\n• R. Holst,\n• P. Toft,\n• H. K. Nedergaard,\n• T. Haberlandt,\n• H. I. Jensen,\n• P. Toft,\n• S. Park,\n• S. Kim,\n• Y. J. Cho,\n• Y. J. Lim,\n• A. Chan,\n• S. Tang,\n• S. L. Nunes,\n• S. Forsberg,\n• H. Blomqvist,\n• L. Berggren,\n• M. Sörberg,\n• T. Sarapohja,\n• C. J. Wickerts,\n• J. G. M. Hofhuis,\n• L. Rose,\n• B. Blackwood,\n• E. Akerman,\n• J. Mcgaughey,\n• I. Egerod,\n• M. Fossum,\n• H. Foss,\n• E. Georgiou,\n• H. J. Graff,\n• M. Kalafati,\n• R. Sperlinga,\n• A. Schafer,\n• A. G. Wojnicka,\n• P. E. Spronk,\n• F. Zand,\n• F. Khalili,\n• R. Afshari,\n• G. Sabetian,\n• M. Masjedi,\n• B. Maghsudi,\n• P. Petramfar,\n• S. Nasimi,\n• A. Vazin,\n• B. Ziaian,\n• H. Tabei,\n• A. Gunther,\n• J. O. Hansen,\n• P. Sackey,\n• H. Storm,\n• J. Bernhardsson,\n• Ø. Sundin,\n• A. Bjärtå,\n• A. Bienert,\n• P. Smuszkiewicz,\n• P. Wiczling,\n• K. Przybylowski,\n• A. Borsuk,\n• I. Trojanowska,\n• J. Matysiak,\n• Z. Kokot,\n• M. Paterska,\n• E. Grzeskowiak,\n• A. Messina,\n• E. Bonicolini,\n• D. Colombo,\n• G. Moro,\n• S. Romagnoli,\n• A. R. De Gaudio,\n• F. Della Corte,\n• S. M. Romano,\n• J. A. Silversides,\n• E. Major,\n• E. E. Mann,\n• A. J. Ferguson,\n• D. F. Mcauley,\n• J. C. Marshall,\n• B. Blackwood,\n• E. Fan,\n• J. A. Diaz-Rodriguez,\n• R. Silva-Medina,\n• E. Gomez-Sandoval,\n• N. Gomez-Gonzalez,\n• R. Soriano-Orozco,\n• P. L. Gonzalez-Carrillo,\n• M. Hernández-Flores,\n• K. Pilarczyk,\n• J. Lubarksi,\n• D. Wendt,\n• F. Dusse,\n• J. Günter,\n• B. Huschens,\n• E. Demircioglu,\n• H. Jakob,\n• A. Palmaccio,\n• A. M. Dell’Anna,\n• D. L. Grieco,\n• F. Torrini,\n• C. Iaquaniello,\n• F. Bongiovanni,\n• M. Antonelli,\n• L. Toscani,\n• D. Antonakaki,\n• D. Bastoni,\n• H. D. Aya,\n• A. Rhodes,\n• M. Cecconi,\n• M. Jozwiak,\n• F. Depret,\n• J. L. Teboul,\n• J. Alphonsine,\n• C. Lai,\n• C. Richard,\n• X. Monnet,\n• I. László,\n• G. Demeter,\n• N. Öveges,\n• K. Tánczos,\n• M. Németh,\n• D. Trásy,\n• I. Kertmegi,\n• D. Érces,\n• B. Tudor,\n• J. Kaszaki,\n• Z. Molnár,\n• A. Hasanin,\n• A. Lotfy,\n• H. Nassar,\n• S. Mahmoud,\n• A. Abougabal,\n• A. Mukhtar,\n• F. Quinty,\n• S. Habchi,\n• A. Luzi,\n• E. Antok,\n• G. Hernandez,\n• B. Lara,\n• L. Enberg,\n• M. Ortega,\n• P. Leon,\n• C. Kripper,\n• P. Aguilera,\n• E. Kattan,\n• J. Bakker,\n• W. Huber,\n• M. Lehmann,\n• S. Sakka,\n• B. Bein,\n• R. M. Schmid,\n• J. Preti,\n• J. Creteur,\n• A. Herpain,\n• J. Marc,\n• E. Zogheib,\n• F. Trojette,\n• S. Bar,\n• L. Kontar,\n• D. Titeca,\n• J. Richecoeur,\n• B. Gelee,\n• N. Verrier,\n• R. Mercier,\n• E. Lorne,\n• J. Maizel,\n• H. Dupont,\n• M. Slama,\n• M. E. Abdelfattah,\n• M. A. Ali Elsayed,\n• A. Mukhtar,\n• A. Pedraza Montenegro,\n• E. Monares Zepeda,\n• J. Franco Granillo,\n• J. S. Aguirre Sánchez,\n• G. Camarena Alejo,\n• A. Rugerio Cabrera,\n• A. A. Tanaka Montoya,\n• C. Lee,\n• F. Hatib,\n• M. Cannesson,\n• P. Theerawit,\n• T. Morasert,\n• Y. Sutherasan,\n• G. Zani,\n• S. Mescolini,\n• M. Diamanti,\n• R. Righetti,\n• A. Scaramuzza,\n• M. Papetti,\n• M. Terenzoni,\n• C. Gecele,\n• M. Fusari,\n• K. A. Hakim,\n• A. Chaari,\n• M. Ismail,\n• A. H. Elsaka,\n• T. M. Mahmoud,\n• K. Bousselmi,\n• V. Kauts,\n• W. F. Casey,\n• S. D. Hutchings,\n• D. Naumann,\n• J. Wendon,\n• S. Watts,\n• E. Kirkman,\n• Z. Jian,\n• S. Buddi,\n• C. Lee,\n• J. Settels,\n• F. Hatib,\n• M. R. Pinsky,\n• P. Bertini,\n• F. Guarracino,\n• C. Trepte,\n• P. Richter,\n• S. A. Haas,\n• V. Eichhorn,\n• J. C. Kubitz,\n• D. A. Reuter,\n• M. S. Soliman,\n• W. I. Hamimy,\n• A. M. Mukhtar,\n• M. Charlton,\n• L. Tonks,\n• L. Mclelland,\n• T. J. Coats,\n• J. P. Thompson,\n• M. R. Sims,\n• D. Williams,\n• D. Z. Roushdy,\n• R. A. Soliman,\n• R. A. Nahas,\n• M. Y. Arafa,\n• W. T. Hung,\n• C. C. Chiang,\n• W. C. Huang,\n• K. C. Lin,\n• S. C. Lin,\n• C. C. Cheng,\n• P. L. Kang,\n• S. R. Wann,\n• G. Y. Mar,\n• C. P. Liu,\n• M. Lopez Carranza,\n• H. Sancho Fernandez,\n• J. A. Sanchez Roman,\n• F. Lucena,\n• A. Campanario Garcia,\n• A. Loza Vazquez,\n• A. Lesmes Serrano,\n• ARIAM-SEMICYUC Registry Investigators,\n• L. Sayagues Moreira,\n• R. Vidal-Perez,\n• U. Anido Herranz,\n• J. M. Garcia Acuna,\n• C. Pena Gil,\n• J. L. Garcia Allut,\n• C. Martin Lopez,\n• E. Saborido Paz,\n• C. Galban Rodriguez,\n• J. R. Gonzalez-Juanatey,\n• A. Vallejo-Baez,\n• M. V. de la Torre-Prados,\n• P. Nuevo-Ortega,\n• A. Fernández-Porcel,\n• E. Cámara-Sola,\n• T. Tsvetanova-Spasova,\n• C. Rueda-Molina,\n• L. Salido-Díaz,\n• A. García-Alcántara,\n• ARIAM Group,\n• J. Aron,\n• R. Marharaj,\n• K. Gervasio,\n• M. Bottiroli,\n• M. Mondino,\n• D. De Caria,\n• A. Calini,\n• E. Montrasio,\n• F. Milazzo,\n• M. P. Gagliardone,\n• A. Vallejo-Báez,\n• M. V. de la Torre-Prados,\n• P. Nuevo-Ortega,\n• A. Fernández-Porcel,\n• E. Cámara-Sola,\n• T. Tsvetanova-Spasova,\n• C. Rueda-Molina,\n• L. Salido-Díaz,\n• A. García-Alcántara,\n• ARIAM group,\n• L. Sayagues Moreira,\n• R. Vidal-Perez,\n• U. Anido,\n• C. Pena Gil,\n• J. M. Garcia Acuna,\n• C. Martin Lopez,\n• E. Saborido Paz,\n• J. L. Garcia Allut,\n• C. Galban Rodriguez,\n• J. R. Gonzalez-Juanatey,\n• Y. Hamdaoui,\n• A. Khedher,\n• M. Cheikh-Bouhlel,\n• J. Ayachi,\n• K. Meddeb,\n• N. Sma,\n• N. Fraj,\n• N. Ben Aicha,\n• S. Romdhani,\n• R. Bouneb,\n• I. Chouchene,\n• M. Boussarsar,\n• M. P. R. D. L. Dela Cruz,\n• J. M. Bernardo,\n• F. Galfo,\n• A. Dyson,\n• M. Singer,\n• A. Marino,\n• A. Dyson,\n• M. Singer,\n• C. C. Chao,\n• P. Hou,\n• W. C. Huang,\n• C. C. Hung,\n• C. H. Chiang,\n• W. T. Hung,\n• K. C. Lin,\n• S. C. Lin,\n• Y. J. Liou,\n• S. M. Hung,\n• Y. S. Lin,\n• C. C. Cheng,\n• F. Y. Kuo,\n• K. R. Chiou,\n• C. J. Chen,\n• L. S. Yan,\n• C. Y. Liu,\n• H. H. Wang,\n• P. L. Kang,\n• H. L. Chen,\n• C. K. Ho,\n• G. Y. Mar,\n• C. P. Liu,\n• S. Grewal,\n• S. Gopal,\n• C. Corbett,\n• A. Wilson,\n• J. Capps,\n• W. Ayoub,\n• A. Lomas,\n• S. Ghani,\n• J. Moore,\n• D. Atkinson,\n• M. Sharman,\n• W. Swinnen,\n• J. Pauwels,\n• K. Mignolet,\n• E. Pannier,\n• A. Koch,\n• T. Sarens,\n• W. Temmerman,\n• A. M. Elmenshawy,\n• A. M. Fayed,\n• M. Elboriuny,\n• E. Hamdy,\n• E. Zakaria,\n• A. C. Falk,\n• A. Petosic,\n• K. Olafsen,\n• H. Wøien,\n• H. Flaatten,\n• K. Sunde,\n• J. J. Cáceres Agra,\n• J. L. Santana Cabrera,\n• J. D. Martín Santana,\n• L. Melián Alzola,\n• H. Rodríguez Pérez,\n• T. Castro Pires,\n• H. Calderón,\n• A. Pereira,\n• S. Castro,\n• C. Granja,\n• I. Norkiene,\n• I. Urbanaviciute,\n• G. Kezyte,\n• D. Ringaitiene,\n• T. Jovaisa,\n• G. Vogel,\n• U. B. Johansson,\n• A. Sandgren,\n• C. Svensen,\n• E. Joelsson-Alm,\n• M. A. Leite,\n• L. D. Murbach,\n• E. F. Osaku,\n• C. R. L. M. Costa,\n• M. Pelenz,\n• N. M. Neitzke,\n• M. M. Moraes,\n• M. M. M. Silva,\n• R. S. Zaponi,\n• L. R. L. Abentroth,\n• S. M. Ogasawara,\n• A. C. Jorge,\n• P. A. D. Duarte,\n• L. D. Murbach,\n• M. A. Leite,\n• E. F. Osaku,\n• J. Barreto,\n• S. T. Duarte,\n• S. Taba,\n• D. Miglioranza,\n• D. P. Gund,\n• C. F. Lordani,\n• C. R. L. M. Costa,\n• S. M. Ogasawara,\n• A. C. Jorge,\n• P. A. D. Duarte,\n• H. Vollmer,\n• M. Gager,\n• C. Waldmann,\n• A. T. Mazzeo,\n• R. Tesio,\n• C. Filippini,\n• M. E. Vallero,\n• C. Giolitti,\n• S. Caccia,\n• M. Medugno,\n• T. Tenaglia,\n• R. Rosato,\n• I. Mastromauro,\n• L. Brazzi,\n• P. P. Terragni,\n• R. Urbino,\n• V. Fanelli,\n• V. M. Ranieri,\n• L. Mascia,\n• J. Ballantyne,\n• L. Paton,\n• A. Mackay,\n• P. Perez-Teran,\n• O. Roca,\n• J. C. Ruiz-Rodriguez,\n• A. Zapatero,\n• J. Serra,\n• J. R. Masclans,\n• S. Bianzina,\n• P. Cornara,\n• G. Rodi,\n• G. Tavazzi,\n• M. Pozzi,\n• G. A. Iotti,\n• F. Mojoli,\n• A. Braschi,\n• A. Vishnu,\n• D. Buche,\n• R. Pande,\n• D. L. J. Moolenaar,\n• F. Bakhshi-Raiez,\n• D. A. Dongelmans,\n• N. F. de Keizer,\n• D. W. de Lange,\n• I. Fuentes Fernández,\n• D. Martínez Baño,\n• J. L. Buendía Moreno,\n• R. Jara Rubio,\n• J. Scott,\n• D. Phelan,\n• D. Morely,\n• J. O’Flynn,\n• P. Stapleton,\n• M. Lynch,\n• B. Marsh,\n• E. Carton,\n• C. O’Loughlin,\n• K. C. Cheng,\n• M. I. Sung,\n• M. O. Elghonemi,\n• M. H. Saleh,\n• T. S. Meyhoff,\n• M. Krag,\n• P. B. Hjortrup,\n• A. Perner,\n• M. H. Møller,\n• T. Öhman,\n• T. Sigmundsson,\n• E. Redondo,\n• M. Hallbäck,\n• F. Suarez-Sipmann,\n• H. Björne,\n• C. Hällsjö Sander,\n• KARISMA,\n• M. Cressoni,\n• D. Chiumello,\n• C. Chiurazzi,\n• M. Brioni,\n• I. Algieri,\n• M. Guanziroli,\n• G. Vergani,\n• T. Tonetti,\n• I. Tomic,\n• A. Colombo,\n• F. Crimella,\n• E. Carlesso,\n• A. Colombo,\n• V. Gasparovic,\n• L. Gattinoni,\n• R. El-Sherif,\n• M. Abd Al-Basser,\n• A. Raafat,\n• A. El-Sherif,\n• F. D. Simonis,\n• L. R. A. Schouten,\n• O. L. Cremer,\n• D. S. Y. Ong,\n• G. Amoruso,\n• G. Cinnella,\n• M. J. Schultz,\n• L. D. J. Bos,\n• W. Huber,\n• P. Schmidle,\n• M. Findeisen,\n• P. Hoppmann,\n• J. Jaitner,\n• F. Brettner,\n• R. M. Schmid,\n• T. Lahmer,\n• EXODUS-investigators,\n• E. Festic,\n• G. Rajagopalan,\n• V. Bansal,\n• R. Frank,\n• R. Hinds,\n• J. Levitt,\n• United States Critical Illness and Injury Trials Group/LIPS-B investigators,\n• S. Siddiqui,\n• SICM NICER Group,\n• J. P. Gilbert,\n• K. Sim,\n• C. H. Wang,\n• H. C. Hu,\n• I. J. Li,\n• W. R. Tang,\n• K. C. Kao,\n• P. Persona,\n• A. De Cassai,\n• M. Franco,\n• F. Facchin,\n• C. Ori,\n• S. Rossi,\n• A. Goffi,\n• S. H. Li,\n• H. C. Hu,\n• L. C. Chiu,\n• C. Y. Hung,\n• C. H. Chang,\n• K. C. Kao,\n• B. Llorente Ruiz,\n• J. Lujan Varas,\n• R. Molina Montero,\n• O. Navarrete,\n• M. Vazquez Mezquita,\n• E. Alonso Peces,\n• M. A. M. Nakamura,\n• L. A. Hajjar,\n• F. R. B. G. Galas,\n• T. A. Ortiz,\n• M. B. P. Amato,\n• L. Bitker,\n• N. Costes,\n• D. Le Bars,\n• F. Lavenne,\n• D. Mojgan,\n• J. C. Richard,\n• C. Chiurazzi,\n• M. Cressoni,\n• D. Massari,\n• M. Guanziroli,\n• G. Vergani,\n• M. Gotti,\n• M. Brioni,\n• I. Algieri,\n• T. Tonetti,\n• D. Chiumello,\n• L. Gattinoni,\n• A. Zerman,\n• M. Türkoğlu,\n• G. Arık,\n• F. Yıldırım,\n• Z. Güllü,\n• I. Kara,\n• N. Boyacı,\n• B. Basarık Aydoğan,\n• Ü. Gaygısız,\n• K. Gönderen,\n• G. Aygencel,\n• M. Aydoğdu,\n• Z. Ülger,\n• G. Gürsel,\n• J. Riera,\n• C. Mazo,\n• M. Martínez,\n• J. Baldirà,\n• L. Lagunes,\n• A. Roman,\n• M. Deu,\n• J. Rello,\n• D. J. Levine,\n• R. M. Mohus,\n• J. Paulsen,\n• A. Mehl,\n• A. T. Dewan,\n• J. K. Damås,\n• E. Solligård,\n• B. O. Åsvold,\n• Mid-Norway Sepsis Research Center,\n• J. Paulsen,\n• R. M. Mohus,\n• A. Mehl,\n• A. DeWan,\n• E. Solligård,\n• J. K. Damås,\n• B. O. Åsvold,\n• O. Aktepe,\n• A. Kara,\n• H. Yeter,\n• A. Topeli,\n• M. Norrenberg,\n• M. Devroey,\n• J. C. Preiser,\n• Z. Tang,\n• C. Qiu,\n• L. Tong,\n• C. Cai,\n• M. Theodorakopoulou,\n• A. Diamantakis,\n• M. Kontogiorgi,\n• E. Chrysanthopoulou,\n• T. Christodoulopoulou,\n• F. Frantzeskaki,\n• M. Lygnos,\n• O. Apostolopoulou,\n• A. Armaganidis,\n• J. Y. Moon,\n• M. R. Park,\n• I. S. Kwon,\n• G. R. Chon,\n• J. Y. Ahn,\n• S. J. Kwon,\n• Y. J. Chang,\n• J. Y. Lee,\n• S. Y. Yoon,\n• J. W. Lee,\n• The Korean Chungcheong Critical Care Research Group,\n• M. Kostalas,\n• J. Mckinlay,\n• G. Kooner,\n• G. Dudas,\n• A. Horton,\n• C. Kerr,\n• N. Karanjia,\n• B. Creagh-Brown,\n• N. D. Altintas,\n• S. Izdes,\n• O. Keremoglu,\n• A. Alkan,\n• S. Neselioglu,\n• O. Erel,\n• N. Tardif,\n• T. Gustafsson,\n• O. Rooyackers,\n• K. N. MacEachern,\n• M. Traille,\n• I. Bromberg,\n• S. E. Lapinsky,\n• M. J. Moore,\n• Z. Tang,\n• C. Cai,\n• L. Tong,\n• J. L. García-Garmendia,\n• F. Villarrasa-Clemente,\n• F. Maroto-Monserrat,\n• O. Rufo-Tejeiro,\n• V. Jorge-Amigo,\n• M. Sánchez-Santamaría,\n• C. Colón-Pallarés,\n• A. Barrero-Almodóvar,\n• S. Gallego-Lara,\n• C. T. Anthon,\n• R. B. Müller,\n• N. Haase,\n• K. Møller,\n• P. B. Hjortrup,\n• J. Wetterslev,\n• A. Perner,\n• M. Nakanishi,\n• A. Kuriyama,\n• T. Fukuoka,\n• M. A. Abd el Halim,\n• M. H. Elsaid hafez,\n• A. M. Moktar,\n• H. M. Elazizy,\n• K. Abdel Hakim,\n• A. Chaari,\n• M. Elbahr,\n• M. Ismail,\n• T. Mahmoud,\n• V. Kauts,\n• K. Bousselmi,\n• E. Khalil,\n• W. Casey,\n• S. H. Zaky,\n• A. Rizk,\n• M. O. Elghonemi,\n• R. Ahmed,\n• J. C. F. Vieira,\n• R. B. Souza,\n• A. M. A. Liberatore,\n• I. H. J. Koh,\n• G. A. Ospina-Tascón,\n• A. F. Garcia Marin,\n• G. J. Echeverry,\n• W. F. Bermudez,\n• J. D. Valencia,\n• E. Quiñonez,\n• A. Marulanda,\n• C. A. Arango-Dávila,\n• A. Bruhn,\n• G. Hernandez,\n• D. De Backer,\n• D. Orbegozo Cortes,\n• F. Su,\n• J. L. Vincent,\n• J. Creteur,\n• L. Tullo,\n• L. Mirabella,\n• P. Di Molfetta,\n• G. Cinnella,\n• M. Dambrosio,\n• C. Villavicencio Lujan,\n• J. Leache irigoyen,\n• M. Cartanya ferré,\n• R. Carbonell García,\n• A. Mukhtar,\n• M. Ahmed,\n• M. El Ayashi,\n• A. Hasanin,\n• E. Ayman,\n• M. Salem,\n• S. Fathy,\n• H. Nassar,\n• A. Zaghlol,\n• M. F. Aguilar Arzapalo,\n• Å. Valsø,\n• K. Sunde,\n• T. Rustøen,\n• I. Schou-Bredal,\n• K. Tøien,\n• Y. Palmeiro,\n• W. Egbaria,\n• R. Kigli,\n• B. Maertens,\n• K. Blot,\n• S. Blot,\n• E. Santana-Santos,\n• E. R. dos Santos,\n• R. E. D. L. Ferretti-Rebustini,\n• R. D. C. C. D. O. dos Santos,\n• R. G. S. Verardino,\n• L. A. Bortolotto,\n• A. M. Doyle,\n• I. Naldrett,\n• J. Tillman,\n• S. Price,\n• S. Shrestha,\n• P. Pearson,\n• J. Greaves,\n• D. Goodall,\n• A. Berry,\n• A. Richardson,\n• G. O. Odundo,\n• P. Omengo,\n• P. Obonyo,\n• N. M. Chanzu,\n• R. Kleinpell,\n• S. J. Sarris,\n• P. Nedved,\n• M. Heitschmidt,\n• H. Ben-Ghezala,\n• S. Snouda,\n• S. Djobbi,\n• H. Ben-Ghezala,\n• S. Snouda,\n• L. Rose,\n• D. Leasa,\n• D. Fergusson,\n• D. A. Mckim,\n• J. Weblin,\n• O. Tucker,\n• D. McWilliams,\n• F. Doesburg,\n• F. Cnossen,\n• W. Dieperink,\n• W. Bult,\n• M. W. N. Nijsten,\n• G. A. Galvez-Blanco,\n• E. Monares Zepeda,\n• C. I. Olvera Guzman,\n• J. S. Aguirre Sánchez,\n• J. Franco Granillo,\n• J. Santos Stroud,\n• R. Thomson,\n• A. Lobo-Civico,\n• M. Pi-Guerrero,\n• I. Blanco-Sanchez,\n• A. Piñol-Tena,\n• C. Paños-Espinosa,\n• Y. Alabart-Segura,\n• B. Coloma-Gomez,\n• A. Fernandez-Blanco,\n• F. Braga-Dias,\n• M. Treso-Geira,\n• A. Valeiras-Valero,\n• L. Martinez-Reyes,\n• A. Sandiumenge,\n• M. F. Jimenez-Herrera,\n• CAPCRI Study,\n• P. Juárez,\n• R. Argandoña,\n• J. J. Díaz,\n• C. Sánchez Ramirez,\n• P. Saavedra,\n• S. Ruiz Santana,\n• O. Obukhova,\n• S. Kashiya,\n• I. A. Kurmukov,\n• A. M. Pronina,\n• P. Simeone,\n• L. Puybasset,\n• G. Auzias,\n• O. Coulon,\n• B. Lesimple,\n• G. Torkomian,\n• L. Velly,\n• A. Bienert,\n• A. Bartkowska-Sniatkowska,\n• P. Wiczling,\n• O. Szerkus,\n• D. Siluk,\n• J. Bartkowiak-Wieczorek,\n• J. Warzybok,\n• A. Borsuk,\n• R. Kaliszan,\n• E. Grzeskowiak,\n• C. Hernandez Caballero,\n• S. Roberts,\n• G. Isgro,\n• D. Hall,\n• G. Guillaume,\n• O. Passouant,\n• F. Dumas,\n• W. Bougouin,\n• B. Champigneulle,\n• M. Arnaout,\n• J. Chelly,\n• J. D. Chiche,\n• O. Varenne,\n• J. P. Mira,\n• E. Marijon,\n• A. Cariou,\n• M. Beerepoot,\n• H. R. Touw,\n• K. Parlevliet,\n• C. Boer,\n• P. W. Elbers,\n• P. R. Tuinman,\n• Á. J. Roldán Reina,\n• Y. Corcia Palomo,\n• R. Martín Bermúdez,\n• L. Martín Villén,\n• I. Palacios García,\n• J. R. Naranjo Izurieta,\n• J. B. Pérez Bernal,\n• F. J. Jiménez Jiménez,\n• Cardiac Arrest Group HUVR,\n• M. V. de la Torre-Prados,\n• A. Fernández-Porcel,\n• P. Nuevo-Ortega,\n• E. Cámara-Sola,\n• T. Tsvetanova-Spasova,\n• C. Rueda-Molina,\n• L. Salido-Díaz,\n• A. García-Alcántara,\n• T. Kaneko,\n• H. Tanaka,\n• M. Kamikawa,\n• R. Karashima,\n• S. Iwashita,\n• H. Irie,\n• S. Kasaoka,\n• O. Arola,\n• R. Laitio,\n• A. Saraste,\n• J. Airaksinen,\n• M. Pietilä,\n• M. Hynninen,\n• J. Wennervirta,\n• M. Bäcklund,\n• E. Ylikoski,\n• P. Silvasti,\n• E. Nukarinen,\n• J. Grönlund,\n• V. P. Harjola,\n• J. Niiranen,\n• K. Korpi,\n• M. Varpula,\n• R. O. Roine,\n• T. Laitio,\n• for the Xe-HYPOTHECA study group,\n• S. Salah,\n• B. G. Hassen,\n• A. Mohamed Fehmi,\n• S. Kim,\n• Y. C. Hsu,\n• J. Barea-Mendoza,\n• C. García-Fuentes,\n• M. Castillo-Jaramillo,\n• R. Viejo-Moreno,\n• L. Terceros-Almanza,\n• S. Bermejo Aznárez,\n• C. Mudarra-Reche,\n• W. Xu,\n• M. Chico-Fernández,\n• J. C. Montejo-González,\n• K. Crewdson,\n• M. Thomas,\n• M. Merghani,\n• L. Fenner,\n• P. Morgan,\n• D. Lockey,\n• E. J. van Lieshout,\n• B. Oomen,\n• D. A. Dongelmans,\n• R. J. de Haan,\n• N. P. Juffermans,\n• M. B. Vroom,\n• R. Algarte,\n• L. Martínez,\n• B. Sánchez,\n• I. Romero,\n• F. Martínez,\n• S. Quintana,\n• O. Sheikh,\n• D. Pogson,\n• R. Clinton,\n• F. Riccio,\n• L. Gemmell,\n• A. MacKay,\n• A. Arthur,\n• L. Young,\n• A. Sinclair,\n• D. Markopoulou,\n• K. Venetsanou,\n• L. Filippou,\n• E. Salla,\n• S. Stratouli,\n• I. Alamanos,\n• A. H. Guirgis,\n• R. Gutiérrez Rodriguez,\n• M. J. Furones Lorente,\n• I. Macias Guarasa,\n• A. Ukere,\n• S. Meisner,\n• G. Greiwe,\n• B. Opitz,\n• D. Benten,\n• B. Nashan,\n• L. Fischer,\n• C. J. C. Trepte,\n• D. A. Reuter,\n• S. A. Haas,\n• C. R. Behem,\n• G. Tavazzi,\n• B. Ana,\n• A. Vazir,\n• D. Gibson,\n• S. Price,\n• M. Masjedi,\n• M. Riahi alam,\n• M. R. Sasani,\n• N. Parenti,\n• F. Agrusta,\n• C. Palazzi,\n• B. Pifferi,\n• R. Sganzerla,\n• F. Tagliazucchi,\n• A. Luciani,\n• M. Möller,\n• J. Müller-Engelmann,\n• G. Montag,\n• C. Lange,\n• J. Neuzner,\n• K. H. Wodack,\n• F. Thürk,\n• A. D. Waldmann,\n• M. F. Grässler,\n• S. Nishimoto,\n• S. H. Böhm,\n• E. Kaniusas,\n• D. A. Reuter,\n• C. J. Trepte,\n• T. Sigmundsson,\n• T. Öhman,\n• E. Redondo,\n• M. Hallbäck,\n• M. Wallin,\n• F. Suarez Sipman,\n• A. Oldner,\n• C. Hällsjö Sander,\n• H. Björne,\n• L. Colinas,\n• G. Hernandez,\n• R. Vicho,\n• M. Serna,\n• R. Cuena,\n• A. Canabal,\n• ECOCRITIC group,\n• A. Chaari,\n• K. Abdel Hakim,\n• M. Etman,\n• M. El Bahr,\n• A. El Sakka,\n• K. Bousselmi,\n• A. Arali,\n• V. Kauts,\n• W. F. Casey,\n• O. Bond,\n• P. De Santis,\n• E. Iesu,\n• F. Franchi,\n• J. L. Vincent,\n• J. Creteur,\n• S. Scolletta,\n• F. S. Taccone,\n• Z. Marutyan,\n• L. Hamidova,\n• A. Shakotko,\n• V. Movsisyan,\n• I. Uysupova,\n• A. Evdokimov,\n• S. Petrikov,\n• C. Gonen,\n• E. Haftacı,\n• C. Balci,\n• F. J. Redondo Calvo,\n• N. Bejarano,\n• R. Villazala,\n• J. Redondo,\n• P. Villarejo,\n• A. Akcan-Arikan,\n• C. E. Kennedy,\n• M. F. Aguilar Arzapalo,\n• C. Gomez-Gonzalez,\n• S. Mas-Font,\n• A. Puppo-Moreno,\n• M. Herrera-Gutierrez,\n• M. Garcia-Garcia,\n• S. Aldunate-Calvo,\n• NEFROCON Investigators,\n• E. P. Plata-Menchaca,\n• X. L. Pérez-Fernández,\n• M. Estruch,\n• A. Betbese-Roig,\n• P. Cárdenas Campos,\n• M. Rojas Lora,\n• N. D. Toapanta Gaibor,\n• R. S. Contreras Medina,\n• V. D. Gumucio Sanguino,\n• E. J. Casanova,\n• J. Sabater Riera,\n• SIRAKI group,\n• K. Kritmetapak,\n• S. Peerapornratana,\n• P. Kittiskulnam,\n• T. Dissayabutra,\n• K. Tiranathanagul,\n• P. Susantithapong,\n• K. Tungsanga,\n• S. Eiam-Ong,\n• N. Srisawat,\n• T. Winkelmann,\n• T. Busch,\n• J. Meixensberger,\n• S. Bercker,\n• E. M. Flores Cabeza,\n• M. Sánchez Sánchez,\n• N. Cáceres Giménez,\n• C. Gutierrez Melón,\n• E. Herrero de Lucas,\n• P. Millán Estañ,\n• M. Hernández Bernal,\n• A. Garcia de Lorenzo y Mateos,\n• B. Ergin,\n• P. Guerci,\n• P. A. C. Specht,\n• Y. Ince,\n• C. Ince,\n• M. Balik,\n• M. Zakharchenko,\n• F. Los,\n• H. Brodska,\n• C. de Tymowski,\n• P. Augustin,\n• M. Desmard,\n• P. Montravers,\n• S. N. Stapel,\n• R. de Boer,\n• H. M. Oudemans,\n• A. Hollinger,\n• T. Schweingruber,\n• F. Jockers,\n• M. Dickenmann,\n• M. Siegemund,\n• Clinical Intensive Care Research Basel,\n• N. Runciman,\n• M. Ralston,\n• R. Appleton,\n• T. Mauri,\n• L. Alban,\n• C. Turrini,\n• T. Sasso,\n• T. Langer,\n• P. Taccone,\n• E. Carlesso,\n• C. Marenghi,\n• G. Grasselli,\n• A. Pesenti,\n• P. Wibart,\n• T. Reginault,\n• M. Garcia,\n• B. Barbrel,\n• A. Benard,\n• F. Vargas,\n• H. N. Bui,\n• G. Hilbert,\n• J. M. Serrano Simón,\n• P. Carmona Sánchez,\n• F. Ruiz Ferrón,\n• M. García de Acilu,\n• J. Marin,\n• V. Antonia,\n• L. Ruano,\n• M. Monica,\n• R. Ferrer,\n• J. R. Masclans,\n• O. Roca,\n• G. Hong,\n• D. H. Kim,\n• Y. S. Kim,\n• J. S. Park,\n• Y. K. Jee,\n• Z. Yu xiang,\n• W. Jia-xing,\n• W. Xiao dan,\n• N. Wen long,\n• W. Yu,\n• Z. Yan,\n• X. Cheng,\n• T. Kobayashi,\n• Y. Onodera,\n• R. Akimoto,\n• A. Sugiura,\n• H. Suzuki,\n• M. Iwabuchi,\n• M. Nakane,\n• K. Kawamae,\n• P. Carmona Sanchez,\n• M. D. Bautista Rodriguez,\n• V. Martínez de Pinillos Sánchez,\n• A. Mula Gómez,\n• J. M. Serrano Simón,\n• P. Beuret,\n• C. Fortes,\n• M. Lauer,\n• M. Reboul,\n• J. C. Chakarian,\n• X. Fabre,\n• B. Philippon-Jouve,\n• S. Devillez,\n• M. Clerc,\n• N. Rittayamai,\n• M. Sklar,\n• M. Dres,\n• M. Rauseo,\n• C. Campbell,\n• B. West,\n• D. E. Tullis,\n• L. Brochard,\n• Y. Onodera,\n• R. Akimoto,\n• H. Suzuki,\n• M. Nakane,\n• K. Kawamae,\n• M. Wood,\n• A. Glossop,\n• J. Higuera Lucas,\n• A. Blandino Ortiz,\n• D. Cabestrero Alonso,\n• R. De Pablo Sánchez,\n• L. Rey González,\n• R. Costa,\n• G. Spinazzola,\n• A. Pizza,\n• G. Ferrone,\n• M. Rossi,\n• M. Antonelli,\n• G. Conti,\n• H. Ribeiro,\n• J. Alves,\n• M. Sousa,\n• P. Reis,\n• C. S. Socolovsky,\n• R. P. Cauley,\n• J. E. Frankel,\n• A. L. Beam,\n• K. O. Olaniran,\n• F. K. Gibbons,\n• K. B. Christopher,\n• J. Pennington,\n• P. Zolfaghari,\n• H. S. King,\n• H. H. Y. Kong,\n• H. P. Shum,\n• W. W. Yan,\n• C. Kaymak,\n• N. Okumus,\n• A. Sari,\n• B. Erdogdu,\n• S. Aksun,\n• H. Basar,\n• A. Ozcan,\n• N. Ozcan,\n• D. Oztuna,\n• J. A. Malmgren,\n• S. Lundin,\n• K. Torén,\n• M. Eckerström,\n• A. Wallin,\n• A. C. Waldenström,\n• for the Section on Ethics of the ESICM,\n• F. C. Riccio,\n• D. Pogson,\n• A. C. P. Antonio,\n• A. F. Leivas,\n• F. Kenji,\n• E. James,\n• P. Morgan,\n• G. Carroll,\n• L. Gemmell,\n• A. MacKay,\n• C. Wright,\n• J. Ballantyne,\n• C. S. Gerrard,\n• N. Jones,\n• J. D. Salciccioli,\n• D. C. Marshall,\n• M. Komorowski,\n• A. Hartley,\n• M. C. Sykes,\n• R. Goodson,\n• J. Shalhoub,\n• J. R. Fernández Villanueva,\n• R. Fernández Garda,\n• A. M. López Lago,\n• E. Rodríguez Ruiz,\n• R. Hernández Vaquero,\n• C. Galbán Rodríguez,\n• E. Varo Pérez,\n• C. Hilasque,\n• I. Oliva,\n• G. Sirgo,\n• M. C. Martin,\n• M. Olona,\n• M. C. Gilavert,\n• M. Bodí,\n• C. Ebm,\n• G. Aggarwal,\n• S. Huddart,\n• N. Quiney,\n• M. Cecconi,\n• S. M. Fernandes,\n• J. Santos Silva,\n• J. Gouveia,\n• D. Silva,\n• R. Marques,\n• H. Bento,\n• A. Alvarez,\n• Z. Costa Silva,\n• D. Díaz Diaz,\n• M. Villanova Martínez,\n• E. Palencia Herrejon,\n• A. Martinez de la Gandara,\n• G. Gonzalo,\n• M. A. Lopez,\n• P. Ruíz de Gopegui Miguelena,\n• C. I. Bernal Matilla,\n• P. Sánchez Chueca,\n• M. D. C. Rodríguez Longares,\n• R. Ramos Abril,\n• A. L. Ruíz Aguilar,\n• R. Garrido López de Murillas,\n• R. Fernández Fernández,\n• P. Morales Laborías,\n• M. A. Díaz Castellanos,\n• M. E. Morales Laborías,\n• J. Cho,\n• J. Kim,\n• J. Park,\n• S. Woo,\n• T. West,\n• E. Powell,\n• A. Rimmer,\n• C. Orford,\n• N. Jones,\n• J. Williams,\n• C. I. Bernal Matilla,\n• P. Ruiz de Gopegui Miguelena,\n• P. Sánchez Chueca,\n• R. Ramos Abril,\n• M. D. C. Rodríguez Longares,\n• A. L. Ruíz Aguilar,\n• R. Garrido López de Murillas,\n• R. S. Bourne,\n• R. Shulman,\n• M. Tomlin,\n• G. H. Mills,\n• M. Borthwick,\n• W. Berry,\n• D. García Huertas,\n• F. Manzano,\n• F. Villagrán-Ramírez,\n• A. Ruiz-Perea,\n• C. Rodríguez-Mejías,\n• F. Santiago-Ruiz,\n• M. Colmenero-Ruiz,\n• C. König,\n• B. Matt,\n• A. Kortgen,\n• C. S. Hartog,\n• A. Wong,\n• C. Balan,\n• G. Barker,\n• N. Srisawat,\n• S. Peerapornratana,\n• P. Laoveeravat,\n• S. Tachaboon,\n• S. Eiam-ong,\n• J. Paratz,\n• G. Kayambu,\n• R. Boots,\n• M. F. Aguilar Arzapalo,\n• R. Vlasenko,\n• E. Gromova,\n• M. Kiselevskiy,\n• Y. Dolgikova,\n• K. B. Tang,\n• C. M. Chau,\n• K. N. Lam,\n• E. Gil,\n• G. Y. Suh,\n• C. M. Park,\n• J. Park,\n• C. R. Chung,\n• C. T. Lee,\n• A. Chao,\n• P. Y. Shih,\n• Y. F. Chang,\n• C. H. Lai,\n• Y. C. Hsu,\n• Y. C. Yeh,\n• Y. J. Cheng,\n• V. Colella,\n• N. Zarrillo,\n• M. D’Amico,\n• F. Forfori,\n• B. Pezza,\n• V. Beltramelli,\n• M. L. Pizzaballa,\n• A. Doronzio,\n• B. Balicco,\n• D. Kiers,\n• W. van der Heijden,\n• J. Gerretsen,\n• Q. de Mast,\n• S. el Messaoudi,\n• G. Rongen,\n• M. Gomes,\n• M. Kox,\n• P. Pickkers,\n• N. P. Riksen,\n• Y. Kashiwagi,\n• K. Hayashi,\n• Y. Inagaki,\n• S. Fujita,\n• M. N. Nakamae,\n• Y. R. Kang,\n• R. B. Souza,\n• A. M. A. Liberatore,\n• I. H. J. Koh,\n• A. Blet," ]

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[ "Schering bridge is an AC bridge which is used for the measurement of unknown capacitance. This is one of the most commonly used AC bridge. Schering bridge is shown in figure 1 under balance condition.\n\nContents\n\n## Introduction\n\nFrom figure 1, there are four arms as\n\narm 1 = ab\n\narm 3 = bc\n\narm 4 = dc\n\nwhere\n\nC1 is an unknown value of capacitance to be measured.\nr1 is a series resistance representing loss in C1.\nC2 is a standard capacitor. It is an air or gas capacitor and hence loss-less capacitor.\nR3 is a non–inductive resistance.\nC4 is a variable capacitor.\nR4 is a variable non-inductive resistance.\n\nTherefore, the impedances of arms 1, 2, 3 and 4 are respectively,", null, "At balance condition,\n\nZ1Z4 = Z2Z3", null, "Equating real and imaginary parts on both sides,", null, "", null, "The independent variables in r1 and C1 are C4, R4 respectively. Therefore, Balancing is done by adjustment of R4, C4 with R3, C2 fixed.\n\nDissipation factor (D) is given by", null, "1. The balanced equation obtained is independent of frequency terms.\n\n2. By using fixed values of C2, R4, the dial of R3 may be calibrated to read the capacitance (C1) directly.\n\n3. In case of fixed frequency, the dial of capacitor C4 can be calibrated to read dissipation factor directly as\n\nD= ωC4R4.\n\nThere is a difficulty in obtaining balance as R3 appears in both equations.\n\n### Application of Schering Bridge\n\nThis bridge is used for the measurement of the relative permittivity of dielectric materials.\n\nThis site uses Akismet to reduce spam. Learn how your comment data is processed.\n\nerror: Content is protected !!" ]

[ null, "https://electricalvoice.com/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://electricalvoice.com/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://electricalvoice.com/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://electricalvoice.com/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://electricalvoice.com/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null ]

[ "#### Start\n\n2019-10-27 08:00 AKDT\n\n## Problem Set 5\n\n#### End\n\n2019-11-11 07:00 AKST\nThe end is near!\nSession is over.\nNot yet started.\nSession is starting in -285 days 13:22:25\n\n360:00:00\n\n0:00:00\n\n# Problem IGinger Candy\n\nMr. G is one of the most excellent students in North River High School for Gifted Students. Despite having impressive performance in a programming competition and making it to the next round, he was not totally happy since his best friend did not get such a great achievement. In order to appease the poor girl, Mr. G has to deal with a very hard challenge: Buy her some ginger candies!\n\nThe road system in North River province consists of $N$ junctions and $M$ bidirectional roads. Junctions are numbered from $1$ to $N$, and roads are numbered from $1$ to $M$, inclusive. On the $i^{th}$ road, which connects two junctions $u_ i$ and $v_ i$, there is a shop in which Mr. G can buy $c_ i$ ginger candies. No two roads have the same number of candies. Mr. G wants to meet his friend at some junction, travel through several roads without visiting the same road twice, buy all candies on those roads, and finish at the same junction where he starts.\n\nUsing his humble knowledge in Physics, Mr. G calculates the amount of energy he needs to spend as follow: Let $L$ be the maximum number of candies he buys in one road, and $K$ be the number of roads he passes through. The amount of energy he needs to spend is $L^2+\\alpha K$, where $\\alpha$ is some constant he has already known.\n\nHelp him to satisfy his friend with the minimum amount of energy.\n\n## Input\n\n• The first line contains three integers $N$, $M$, $\\alpha$, the number of junctions, the number of roads and the predefined constant Mr. G uses to calculate the amount of energy, respectively ($1 \\leq N \\leq 10^5$, $1 \\leq M \\leq 2 \\times 10^5$, $1 \\leq \\alpha \\leq 20$).\n\n• In the next $M$ lines, each contains three integers $u$, $v$, $c$ ($1 \\leq u \\leq N$, $1 \\leq v \\leq N$, $10^6 \\leq c \\leq 10^9$), meaning that there is a road connecting two junctions $u$ and $v$, which sells $c$ ginger candies.\n\nIt is guaranteed that all $c$ in the above $M$ lines are distinct.\n\n## Output\n\nWrite one integer denoting the minimum amount of energy Mr. G has to spend. If there is no route satisfying the condition, output Poor girl instead.\n\nSample Input 1 Sample Output 1\n7 7 10\n1 2 1000000\n2 3 2000000\n3 4 3000000\n4 5 4000000\n5 6 5000000\n6 7 6000000\n7 1 7000000\n\n49000000000070\n\nSample Input 2 Sample Output 2\n6 6 7\n1 3 1000000\n3 5 3000000\n5 1 5000000\n2 4 2000000\n4 6 4000000\n6 2 6000000\n\n25000000000021\n\nSample Input 3 Sample Output 3\n5 4 1\n1 2 22081999\n1 3 28021999\n2 4 19992208\n2 5 19992802\n\nPoor girl" ]

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[ "# GATE | GATE CS 2018 | Question 54\n\n• Difficulty Level : Expert\n• Last Updated : 09 Mar, 2018\n\nConsider the first-order logic sentence\n\nφ ≡ ∃s∃t∃u∀v∀w∀x∀y ψ(s, t, u, v, w, x, y)\n\nwhere ψ(s, t, u, v, w, x, y) is a quantifier-free first-order logic formula using only predicate symbols, and possibly equality, but no function symbols. Suppose φ has a model with a universe containing 7 elements.\n\nWhich one of the following statements is necessarily true?\n\n(A) There exists at least one model of φ with universe of size less than or equal to 3\n(B) There exists no model of φ with universe of size less than or equal to 3\n(C) There exists no model of φ with universe size of greater than 7\n(D) Every model of φ has a universe of size equal to 7" ]

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[ "Cody\n\n# Problem 105. How to find the position of an element in a vector without using the find function\n\nSolution 2025291\n\nSubmitted on 17 Nov 2019 by Le Huu Hai\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\nx = [1 3 5 4 2]; y = 2 posX_correct = 5; assert(isequal(findPosition(x,y),posX_correct))\n\ny = 2\n\n2   Pass\nx = [1 5 8 6 7 6 0]; y = 8 posX_correct = 3; assert(isequal(findPosition(x,y),posX_correct))\n\ny = 8" ]

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[ "# Problem: Assume that an exhaled breath of air consists of 74.7 % N2, 15.4 % O2, 3.8 % CO2, and 6.1 % water vapor.If the total pressure of the gases is 0.990 atm, calculate the partial pressure of N2.\n\n###### FREE Expert Solution\n\nRecall that the partial pressure of a gas (PGas) in a mixture is given by:\n\n$\\overline{){{\\mathbf{P}}}_{{\\mathbf{gas}}}{\\mathbf{=}}{{\\mathbf{\\chi }}}_{{\\mathbf{gas}}}{{\\mathbf{P}}}_{{\\mathbf{total}}}}$\n\nwhere:\n\nχGas = mole fraction of the gas\n\nPtotal = total pressure of the gas mixture", null, "###### Problem Details\n\nAssume that an exhaled breath of air consists of 74.7 % N2, 15.4 % O2, 3.8 % CO2, and 6.1 % water vapor.\n\nIf the total pressure of the gases is 0.990 atm, calculate the partial pressure of N2." ]

[ null, "https://cdn.clutchprep.com/assets/button-view-text-solution.png", null ]

[ "# 查看主机名\n\n``````//查看一下当前主机名的情况，查看全部三种主机名\nhostnamectl\n\n//或者，查看全部三种主机名\nhostnamectl status\n\n//只查看静态、瞬态或灵活主机名，分别使用--static，--transient或--pretty选项\n[[email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-yjshash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-yjsemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ ~]# hostnamectl --static\n\nxh00\n[[email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-yjshash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-yjsemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ ~]# hostnamectl --transient\n\nxh01\n[[email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-yjshash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-yjsemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ ~]# hostnamectl --pretty\n\n//或者，查看到的是瞬态的（Tansient hostname）\nhostname\n\n//或者查看主机名配置文件，查看到的是静态的（Static hostname）\ncat /etc/hostname\n``````\n\n## 查看当前Linux操作系统相关信息（内核版本号、硬件架构、主机名称和操作系统类型等）:\n\nuname -a //查看到的是瞬态的（Tansient hostname）\ncat /etc/redhat-release //查看操作系统环境\n\n# 修改主机名\n\n## 临时有效\n\nhostname 主机名 //只能临时修改的主机名，当重启机器后，主机名称又变回来了。\n\n``````hostname hx01\n``````\n\n## 永久生效\n\n``````hostnamectl set-hostname hx01\n``````\n\nvim /etc/hostname\n\n## 重启命令\n\n``````reboot init 6 shutdown -r now\n``````\n\n## 关机\n\n``````poweroff init 0 shoutdown -h now\n``````\n\n## 安装图形\n\n``````yum groupinstall \"GNOME Desktop\" \"Graphical Administration Tools\"\n\ninit 5\n``````\n\n### 您可能感兴趣的文章", null, "" ]

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[ "# Overview\n\nOverview\n\nA current transformers (CT) is a device that can detect a current through a wire passing through its center. This wire\nconstitutes a one-turn primary for the transformer, and the winding on the CT constitutes the secondary of the transformer. If the secondary winding is terminated to a suitable resistor, then a current will be generated in the secondary winding which is  proportional to the current in the one-turn primary, as well as to the ratio of the transformer. This current in turn generates a voltage VL across the terminating resistor which is  proportional to the value of the terminating resistor. Thus a desired output voltage can be obtained for a specific input current by a suitable selection of the turns ratio and the terminating resistor.\n\nTheory of Operation\n\nA current Ip through the CT will generate a current Is in the winding of the transformer according to the relation\n\nIs = Ip / N\n\nWhere N is the ratio of the CT.\n\nWhen the secondary is terminated to a load resistor RL, the voltage drop across RL is given by\n\nVL = Is X RL\n\n= Ip X RL / N\n\nHence the voltage VL is a linear indicator of the line current Ip within the linear range of the CT.\n\nThe limits to this range are set by the saturation point of the core material at the high end, and by its ability to respond to low magnetic fields at the low end.\n\nBy selecting a suitable value for RL, a desired voltage range corresponding to the input current range can be obtained.\n\nFor example, if it is desired to generate 2V at 20A line current, and if transformer ratio is 1000:1, then\n\nRL = VL X N / Ip\n\n= 2 X 1000 / 20\n\n= 100 ohms.\n\nThen the current to be measured will be given by\n\nIp = N X VL / RL\n\n= 1000 X VL / 100\n\n= 10 X VL  Amps.\n\nThe resistor rating is given by\n\nW = Is^2 X RL\n\n= Ip^2 X RL / N^2 watts.\n\n=  0.04 W for the example above." ]

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[ "Solutions by everydaycalculation.com\n\n## Reduce 113/1260 to lowest terms\n\n113/1260 is already in the simplest form. It can be written as 0.089683 in decimal form (rounded to 6 decimal places).\n\n#### Steps to simplifying fractions\n\n1. Find the GCD (or HCF) of numerator and denominator\nGCD of 113 and 1260 is 1\n2. Divide both the numerator and denominator by the GCD\n113 ÷ 1/1260 ÷ 1\n3. Reduced fraction: 113/1260\nTherefore, 113/1260 simplified to lowest terms is 113/1260.\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]

[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]

[ "# Controlling a seven segment display\n\n(Difference between revisions)\n\n## Overview\n\nIn this example, we will be using PIC 18F4520 to control a double digit 7-segment display LTD-4708JS. A few notes about using this display is that it is only designed to display one digit at a time, so when trying to use both digits, you have to use a method to switch between the two digits fast enough to create the illusion that both are on simultaneously. We've made a simple loop that switches between the two digits every 5ms, but an ISR could also work.\n\nNow, since there are 7 segments, one would assume that you would need 7 outputs from the PIC to display 1 digit. But it is very wasteful to use outputs from the PIC for just displaying, so we will be using a chip (HEF 4543B) that will take 4-bits and convert it to the 7-bit output for the 7-segment display.\n\nIf you want to display a decimal, this particular 7-segment display will allow you to do so. However, it will take up another output from your PIC.\n\nFinally, if you are planning on displaying more than one digit, you will need an extra bit for every digit that you want to display.\n\n### This Example\n\nThe example that you will see here will simply display one digit on the 7-segment display. We will use the PIC to output a value between 0 and 9 using output pins D0-D3. The 4-bits will then be passed to the HEF 4543B chip which will convert it into 7-bits that will control the display. We did not use the decimal point in our example and we only used one of the digits, so we've hard wired the pins that control them to +5V or ground where appropriate.\n\n### Capabilities\n\nTo display a 2-digit number where a decimal point is needed, you will need a total of 7 outputs from the PIC. 4 for the value, 1 for the decimal point, 2 for each of the digits. However, since you will only be displaying 1 of the 2 digits at any given moment, you can use a logic inverter gate (74LS05) to reduce 1 output needed for selecting which digit to display. A logic inverter gate is a chip that will take an input and return a high if the input is low or low if the input is high. Finally, you will need to make a loop with a delay of 5ms to switch between the digits, or use an ISR if you will be doing other computations with the PIC.\n\n## Circuit\n\nThere are three primary elements required for a circuit involving a 7-segment display.\n\nThe elements are:\n\n1) the 7-segment display\n\n2) the controller (in this case the PIC)\n\n3) the decoder\n\nThe 7-segment display used in this demonstration was the LTD-4708JS which can display 2 digits with trailing decimal points. The controller was the PIC 18F4520. The decoder used was the HEF 4543B which takes one 4-bit binary input and converts it to the appropriate 7 logic outputs to drive the 7-segment display.\n\n### 7-Segment Display\n\nThe LTD-4708JS 7-segment display has 10 pins. Seven pins correspond to the seven LED segments, one pin corresponds to the decimal point, and two pins select which digit is being activated. Both digits can be activated simultaneously, however they would not be able to display independent digits in such a manner. To achieve multi-digit display, the digits must alternate back and forth at a rate preferably greater than 45 Hz for a complete cycle of display (approximately the flicker fusion frequency for the human eye, allowing it to appear that all displayed digits are on continuously).\n\nThe LTD-4708JS has 10 pins and they are designated as follows:\n\nPin 1: Segment \"C\"\n\nPin 2: Decimal Point\n\nPin 3: Segment \"E\"\n\nPin 4: Select pin for the second digit\n\nPin 5: Segment \"D\"\n\nPin 6: Segment \"F\"\n\nPin 7: Segment \"G\"\n\nPin 8: Segment \"B\"\n\nPin 9: Select pin for the first digit\n\nPin 10: Segment \"A\"\n\n### Controller\n\nThe controller used in this demonstration was the PIC 18F4520. The demonstration outputs were the digital output pins D0-D3 representing the output number in binary form. The PIC was not used to control which digit was used by the display (the display was hardwired instead to display only one digit). For a multiple-digit display a multiplexer could be used to decode a binary output from the PIC specifying the desired active display digit while the PIC outputs the appropriate digit on the data pins, although there is an upper limit to the number of digits which can be output while retaining a display frequency of greater than 45 Hz. This limit is dependent on the time between data outputs (likely to be calls of an ISR) used to display a digit. At 50 Hz each digit must be displayed every 20ms, so if the PIC outputs to a new digit every 4ms, then at most 5 digits may be displayed.\n\n### Decoder\n\nThe decoder used in this demonstration was the HEF 4543B. It receives a 4-bit binary input as an input and outputs seven logic pins which correspond to the respective segments of a 7-segment display unit. Some display units require that the inputs be logic low to set a segment active. To achieve this, the chip has a phase input pin which will invert the outputs when the pin is set high. There is also a Blanking pin which will inhibit outputs, ensuring a blank display when set high. For more information regarding this pin, click on the chip name to access the data sheet or click here.\n\n### Circuit Wiring\n\nFor the simplest of cases (which is what this module will demonstrate), the decoder needs to receive the number to be displayed in 4-bit binary form. In addition, the 7-segment display will need the seven independent segment outputs from the decoder as well as one decimal point input and one digit select input for each digit being used (all others can be set to ground to ensure they are not active). In the simplest case (displaying one digit only) the PIC need not have an output to control which digit is active. For more complex displays which have multiple digits, up to three digits could reasonably be controlled by the PIC directly, or a decoder could be used for four or more digits so that the PIC need only output the binary number of the desired digit and the decoder would activate the appropriate digit on the display.\n\nThe yellow wires in the circuit are the four digital output bits from the PIC. In order from LSB to MSB they connect to pins 5, 3, 2 and 4 of the HEF 4543B. If this wiring is not correct, cycling through the numbers from 0-9 will show numbers out of sequence and even missing numbers (binary numbers higher than nine have no output from the decoder).\n\nThe red and black wires track the +5V and ground from the PIC board respectively. Be careful to keep all of the circuit on the same reference node so that the +5V is not floating for one or more of the elements as this may prevent you from getting an appropriate output.\n\nThe blue and green wires represent the data lines from the decoder to the 7-segment display module. They are two colors to assist in visualization. The wires cross over and travel to both sides of the display module, so it may be easier to track which outputs go to which display module pins using the circuit diagram. On the circuit diagram, crossing lines only connect when a connector dot is present.\n\n## Code\n\nThe code below simply loops through 0-9 with a 1 second delay and outputs them to D0-D3.\n\n```/*\nj.yeung, a.care, e.nickel 2008-01-31\n\nThis code will count from 0-9 continuously and output the binary value\non D0-D3 for a HEF 4543B chip to control a 7-segment display.\n*/\n\n#include <18f4520.h>\n#fuses HS, NOLVP, NOWDT, NOPROTECT\n#use delay(clock=20000000) // 20 MHz crystal on PCB\n\nvoid main() {\n\nint i=0;\n\nwhile (TRUE) {\nif (i>9) i=0; // if i is 9, set it to 0\noutput_d(i & 15); // display nibble on pins RD0...RD3\ni++;\ndelay_ms(1000);\n}\n}\n```\n\nThe code below loops through 00-99 with a 1 second delay and outputs them to D0-D3 and using simple delays for digit control on D4 and D5.\n\n```/*\nj.yeung, a.care, e.nickel 2008-01-31\n\nThis code will count from 00-99 continuously and output the binary value\non D0-D3 for a HEF 4543B chip to control a 7-segment display. Output D4\ncontrols the one's digit and D5 controls the ten's digit.\n*/\n\n#include <18f4520.h>\n#fuses HS, NOLVP, NOWDT, NOPROTECT\n#use delay(clock=20000000) // 20 MHz crystal on PCB\n\nvoid main() {\n\nint i=0, j=0, count=0;\n\nwhile (TRUE) {\nif (i>9){\ni=0; // if i is greater than 9, set it to 0\nj++; // since i counted to 10, increment j\n}\nif (j>9) j=0; // if j is greater than 9, set it to 0\nfor(count=0; count<100; count++){\noutput_d(i & 15); // display nibble on pins RD0...RD3\noutput_d4(1); // enable one's digit\ndelay_ms(5);\noutput_d(j & 15); // display nibble on pins RD0...RD3\noutput_d5(1); // enable ten's digit\ndelay_ms(5);\n}\ni++;\n}\n}\n```" ]

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[ "Current Conditions Temp", null, "1.0 C RH", null, "26 % Wind", null, "NE 10 mph Road", null, "Open (4x4)\nMauna Kea Weather Center Meteorology Glossary/Dictionary\n\nA", null, "B", null, "C", null, "D", null, "E", null, "F", null, "G", null, "H", null, "I", null, "J", null, "K", null, "L", null, "M", null, "N", null, "O", null, "P", null, "Q", null, "R", null, "S", null, "T", null, "U", null, "V", null, "W", null, "X", null, "Y", null, "Z\n Search for: Search Definitions:\n\n• G system - Same as grid navigation.", null, "• gain - An increase or amplification. There are two general uses of the term in radar meteorology: 1) antenna gain (or gain factor), which is the ratio of the power transmitted along the beam axis to that of an isotropic radiator transmitting the same total power; and 2) receiver gain (or video gain), which is the amplification given a signal by the receiver.", null, "• gating - (Or range gating.) The use of digital or electrical methods in radar to eliminate or reject the target signals from all targets that are outside certain range limits. Such methods make it possible to measure properties of the echoes from particular targets without interference from the signals returned from closer or more distant targets.", null, "• general forecast - See weather forecast.", null, "• global model - See general circulation model.", null, "• GOES - Abbreviation for Geostationary Operational Environmental Satellite.", null, "• gradient - 1. The space rate of decrease of a function. The gradient of a function in three space dimensions is the vector normal to surfaces of constant value of the function and directed toward decreasing values, with magnitude equal to the rate of decrease of the function in this direction. The gradient of a function f is denoted by -f (without the minus sign in the older literature) and is itself a function of both space and time. The ascendent is the negative of the gradient. In Cartesian coordinates, the expression for the gradient is -f = i + j + k. For expressions in other coordinate systems, see Berry et al. (1945). 2. Often loosely used to denote the magnitude of the gradient or ascendent (i.e., without regard to sign) of a horizontal pressure field.", null, "• grading - The degree of mixing of particle size classes in a sediment. Well-graded sediments are those with a more or less uniform distribution of sizes; poorly graded implies uniformity in size or lack of a continuous distribution.", null, "• gravity - (Or force of gravity.) The force imparted by the earth to a mass that is at rest relative to the earth. Since the earth is rotating, the force observed as gravity is the resultant of the force of gravitation and the centrifugal force arising from this rotation. It is directed normal to sea level and to its geopotential surfaces. The magnitude of the force of gravity at sea level decreases from the poles, where the centrifugal force is zero, to the equator, where the centrifugal force is a maximum but directed opposite to the force of gravitation. This difference is accentuated by the shape of the earth, which is nearly that of an oblate spheroid of revolution slightly depressed at the poles. Also, because of the asymmetric distribution of the mass of the earth, the force of gravity is not directed precisely toward the earth's center. The magnitude of the force of gravity per unit mass (acceleration of gravity) g may be determined at any latitude f[&phgr;] and at any geometric height z (meters) above sea level in the free air from the following empirical formula: g = gf[&phgr;] - (3.085462 ´[×] 10-4 + 2.27 ´[×] 10-7cos 2f[&phgr;])z + (7.254 ´[×] 10-11 + 1.0 ´[×] 10-13cos 2f[&phgr;])z2 - (1.517 ´[×] 10-17 + 6 ´[×] 10-20cos 2f[&phgr;])z3 (cm s-2), where gf[&phgr;] = 980.6160 (1 - 0.0026373 cos 2f[&phgr;] + 0.0000059 cos2 2f[&phgr;]) is the sea level value of gravity (cm s-2) at latitude f[&phgr;]. This formula as applied near the earth indicates that gravity changes very little with height or latitude, so that for rough calculations a constant value of 980 cm s-2 may be used. Besides these variations in the magnitude of the force of gravity, there are more localized variations controlled by the topography of the earth's surface, and the distribution of mass beneath. The magnitude of the force of gravity is usually called either gravity, acceleration of gravity, or apparent gravity. See virtual gravity, geopotential height, standard gravity.", null, "• gravity wave - (Also called gravitational wave.) A wave disturbance in which buoyancy (or reduced gravity) acts as the restoring force on parcels displaced from hydrostatic equilibrium. There is a direct oscillatory conversion between potential and kinetic energy in the wave motion. Pure gravity waves are stable for fluid systems that have static stability. This static stability may be 1) concentrated in an interface or 2) continuously distributed along the axis of gravity. The following remarks apply to the two types, respectively. 1) A wave generated at an interface is similar to a surface wave, having maximum amplitude at the interface. A plane gravity wave is characteristically composed of a pair of waves, the two moving in opposite directions with equal speed relative to the fluid itself. In the case where the upper fluid has zero density, the interface is a free surface and the two gravity waves move with speeds where U is the current speed of fluid, g the acceleration of gravity, L the wavelength, and H the depth of the fluid. For deep-water waves (or Stokesian waves or short waves), H >> L and the wave speed reduces to For shallow-water waves (or Lagrangian waves or long waves), H << L, and c = U ±[±] (gH)½. All waves of consequence on the ocean surface or interfaces are gravity waves, for the surface tension of the water becomes negligible at wavelengths of greater than a few centimeters (see capillary wave). 2) Heterogeneous fluids, such as the atmosphere, have static stability arising from a stratification in which the environmental lapse rate is less than the process lapse rate. The atmosphere can support short internal gravity waves and long external gravity waves. The short waves (of the order of 10 km) have been associated, for example, with lee waves and billow waves. Such waves have vertical accelerations that cannot be neglected in the vertical equation of perturbation motion. The long gravity waves, moving relative to the atmosphere with speed ±[±](gH)½, where H is the height of the corresponding homogeneous atmosphere, have small vertical accelerations and are therefore consistent with the quasi-hydrostatic approximation. In neither type of gravity wave, however, is the horizontal divergence negligible. For meteorological purposes in which neither type is desired as a solution, for example, numerical forecasting, they may be eliminated by some restriction on the magnitude of the horizontal divergence. The above discussion is based upon the method of small perturbations. In certain special cases of water waves, for example, the Gerstner wave or the solitary wave, a theory of finite-amplitude disturbances exists. See shear-gravity wave.", null, "• grid - A set of points arranged in an orderly fashion on which specified variables are analyzed or predicted. Various forms of horizontal and vertical grids, each with particular characteristics, have been devised for use in numerical weather prediction.", null, "• gust - 1. A sudden, brief increase in the speed of the wind. It is of a more transient character than a squall and is followed by a lull or slackening in the wind speed. Generally, winds are least gusty over large water surfaces and most gusty over rough land and near high buildings. According to U.S. weather observing practice, gusts are reported when the peak wind speed reaches at least 16 knots and the variation in wind speed between the peaks and lulls is at least 9 knots. The duration of a gust is usually less than 20 s. 2. With respect to aircraft turbulence, a sharp change in wind speed relative to the aircraft; a sudden increase in airspeed due to fluctuations in the airflow, resulting in increased structural stresses upon the aircraft. 3. (Rare.) Same as cloudburst.", null, "" ]

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[ "# How can a constant secondary side current cause a primary current to flow in a transformer?\n\nThis is a question which was originally asked by my power electronics professor but left unanswered.\n\nSuppose you are given the following system (kindly borrowed from this question)\n\nSuppose that:\n\n1. The inductance L is so big that the load current is practically constant (remove Vx and think of the load as an ideal current source).\n2. The diodes are ideal.\n3. The voltage source is a secondary side of an ideal transformer (no leakage reactance no resistance and no magnetizing current).\n\nAs you can see, the secondary current is a square wave. How can a square wave (i.e. a constant current on a half period basis) cause a primary current in the transformer to flow?\n\nWe can see the secondary current as a Fourier series, that is as an infinite sum of sinewaves. This means that, at least on the transformer part of the circuit which is linear, we can apply the superposition principle and therefore obtain a lot of AC circuits to be solved and then combined. It's the fundamental frequency and the harmonics which are causing primary current to flow. If it were pure DC it wouldn't work.\n\nIn reality of course, the transformer has a leakage reactance which will smooth out the edges of the square wave, providing a (perhaps more evident) time varying current on both its sides.\n\n• Would you maybe expect to rectify secondary Vi into coil then Idc will continue to rise with primary current as a staircase result or if open a staircase voltage with zero crossing BEMF from current rectifiers ? ( with ideal no Bsat or BDV on insulation) Mar 30 '18 at 14:39\n• I'm not sure I understand your comment. Could you clarify? Mar 30 '18 at 20:28", null, "Here a rectifed +-100Vp sine 50Hz drives a full bridge to 1H 1:10 step-up\nSee the coil current rise with DC steps ( shaped by sine ). This is what I meant by staircase rise in current. Meanwhile AC current increases as well. Then diode drop increases slightly.", null, "" ]

[ null, "https://i.stack.imgur.com/gSBRO.jpg", null, "https://i.stack.imgur.com/qaGqK.jpg", null ]

[ "## Sunday, February 28, 2016\n\n### Tweetable Brainfuck Interpreter in C\n\nBackground\n\nBrainfuck is a turing complete esoteric programming language created by Urban Müller in 1993. The language is made up of only eight instructions that easily translate into C. A Hello World program could look something like:\n\n--[>--->->->++>-<<<<<-------]>--.>---------.>\n--..+++.>----.>+++++++++.<<.+++.------.<-.>>+.\n\nUrban’s primary goal was to create a language that could be compiled with the smallest possible compiler. Several small compilers has been developed over the years. The smallest compiler is less than 100 bytes of binary code.\n\nIn addition to create small compiler binaries, it has become a sport to write interpreters in as few bytes of source code as possible, often referred to as code golfing. This article describes a tweetable implementation written in the C language.\n\nHowever, it may be good to tell up front, that it is only a tweetable method. Although fully functional in 140 characters of code, it needs to be called by a separate main method. Extending the method to a fully compilable program requires 151 characters of code which places the implementation among the shortest presented to date.\n\nLanguage Overview\n\nA brainfuck program is a sequence of commands and an instruction pointer keeps track of the instruction to execute next. The program starts at the first instruction and terminates when the last instruction is executed. The language uses an array of memory cells that can be modified by the program. In addition the language supports reading data from standard in and writing data to standard out.\n\nProgram environment:\nchar cells[infinitely large] = { 0 };\nchar* cell_ptr = cells;\n\nBrainfuck Command     C equivalent code\n> ++cell_ptr;\n< --cell_ptr;\n+ ++*cell_ptr;\n- --*cell_ptr;\n. putchar(*cell_ptr);\n, *cell_ptr = getchar();\n[ while(*cell_ptr) {\n] }\n\nGolfing the Interpreter\n\nTo start off, let's begin with a fairly standard brainfuck interpreter implementation:\n\nint m, *d=m, t;\nvoid g(char *p) {\nfor (; *p != 0 ; p++) {\nswitch (*p) {\ncase '>': ++d; break;\ncase '<': --d; break;\ncase '+': ++*d; break;\ncase '-': --*d; break;\ncase '.': putchar(*d); break;\ncase ',': *d = getchar(); break;\ncase '[':\nif (*d == 0) {\nfor (t = 1; t != 0;) {\n++p;\nif (*p == '[') ++t;\nelse if (*p == ']') --t;\n}\n}\nbreak;\ncase ']':\nfor (t = 1; t != 0;) {\n--p;\nif (*p == '[') --t;\nelse if (*p == ']') ++t;\n}\n--p;\nbreak;\n}\n}\n}\n\nThere are multiple approaches to reduce the number of code characters. Some implementations use recursive calls to the method, some tries to eliminate / combine for loops, some use modulus on the ascii value of the instructions and look at bit patterns, and some use the question mark operator and implicit conversions from boolean expressions to integers.\n\nAll these techniques are great for code golfing, although the implementation described here uses almost none of these. There are a few places where these techniques actually do help, but the main code size reduction is done in a different way.\n\nThere are four groups of brainfuck commands, memory cell increment/decrement, cell pointer increment/decrement, input/output, and loops. An interesting observation is that the ascii codes of the two commands in each group differs by 2. The input and output needs to be handled by special casing, but we could implement the increment/decrement the same way, for example:\n\nif (*p == ‘>’) d++; else if (*p == ‘<’) d--;\nif (*p == 62) d++; else if (*p == 60) d--;\nd += (*p == 62); *d -= (*p ==60);\nd += (*p==62)-(*p==60);\nd += *p-60?*p==62:-1;\nd += 1/(61-*p);\n\nThe last one has a caveat, and that is if *p == ‘=’. All characters that aren’t brainfuck commands are supposed to be ignored, but with ‘=’ we do get a division by zero. It is relatively easy to avoid this, but for the tweetable version we will just accept that programs containing ‘=’ characters won’t run.\n\nThe interesting thing with the last example is that we can save some extra characters by assigning *p to a temporary variable and subtract 44 which is the ascii value of the input command ‘,’. In addition to making the check for an input command short, it will make one of our increment/decrement operations sort as well. Although simple, it turns out that these savings on the non loop commands are larger than any advanced modulus or bitmasking.\n\nAnother saving can be made by combining the two inner for loops. This is actually quite straightforward. One for loop increments the cell pointer, and the other one decrements it. As we have an efficient way to compute increment/decrement, we can use it to determine the step to advance the cell pointer in the inner loop. If assign s to the cell pointer step and t to the nesting count we get something like:\n\nfor(t=s=1/(t-48);*d?t>0:t;t+=1/(*p-92))p-=s;\n\nNote that we initialize the nesting counter t to -1 if we are searching forward, and 1 if we are searching backward. This is to allow us to use the short increment/decrement statement when updating the nesting count. The only caveat is the additional decrement at the end of one of the loops. But we can avoid that decrement by not incrementing the data pointer at the end of the outer loop in this case:\n\np+=s<1\n\nA few characters can also be saved by using read() and write() instead of getchar() and putchar(). The saving comes from the fact that we can read or write 1 or 0 bytes using boolean to integer conversion in the method call, and also that we can embed a d+=... in one of the calls instead of having it as a separate statement.\n\nFrom a golfing perspective the C language evolution hasn’t been all that great. If we limit ourselves to C89, we can get rid of type declarations and instantly save some code characters. The types default to int. The brainfuck language doesn’t dictate the size of each cell, so we could without violating the language accept that the cell size is int.\n\nLastly, but quite unfortunately, the code after all golfing isn’t tweetable with 30,000 memory cells, so the only thing left is to limit the number of data cells. Small brainfuck programs run fine with only 10 cells, so the last thing we’ll do is to reduce the number of cells from the recommended 30,000 to 10 and we are tweetable!\n\nm,*d=m,s,t;g(char*p){for(;t=*p;p+=s<1)\nfor(t-=44,t?*d-=1/t,write(1,d+=1/(t-17),t==2),\n\nCompilable Program\n\nThe easiest way to make a compilable program with the tweetable method is to implement a main method that calls the brainfuck method, e.g.:\n\nmain(int a,char**b){g(b);}\n\nBut this adds a whopping 29 characters of code. If we limit ourselves to 32 bit computer systems we can merge the main method into the brainfuck method. By using implicit integer to pointer conversions we can get something reasonable compact:\n\nm,*d;main(s,t){char*p=1[d=t];for(d=m;t=*p;p+=s<1)\nfor(t-=44,t?*d-=1/t,write(1,d+=1/(t-17),t==2),t=s=1/\n\nThe most interesting part here is probably the assignment of p. The arguments of main defaults to int types in C89, but we know the second argument of main is a pointer to a character array, where the first array holds the executable name, and the second the argument to the program. So by temporarily assigning d to the pointer value stored in t, we can index the second character array which holds the code. then we assign p using an implicit cast from int* to char*.\n\nIt is of course also possible to remove all compiler and platform specific features and write a fully portable interpreter adhering to the initial language specification (including ignoring all non command characters). Here is one 171 character long version:\n\nchar m,*d=m,s;main(int t,char**p){for(++p;t=**p;\n*p+=s<1)for(t-=44,t?*d-=1/t,write(1,d+=t-17?0:1/(t-17)\n*p-=s;}\n\nConclusion\n\nAs mentioned in the introduction the article do present a tweetable brainfuck method written in C, but falls short of being a complete program. Despite that it may be interesting reading for any code golf enthusiasts and perhaps it gives some ideas to someone to get all the way to a complete tweetable brainfuck interpreter program written in C.\n\nAnd finally, if you feel the urge to tweet, copy paste the code below and send it off to your friends:\n\nm,*d=m,s,t;g(char*p){for(;t=*p;p+=s<1)\nfor(t-=44,t?*d-=1/t,write(1,d+=1/(t-17),t==2),\n\n## Tuesday, January 20, 2015\n\n### ColecoVision Driving Module\n\nIntroduction\n\nColecoVision became one of the major game consoles when released in 1982 and sold over two million units, mainly in North America. However, sales and popularity of the console dropped quickly in the big video game crash of 1983 and was withdrawn from the video game market in 1985. ColecoVision carried titles such as Donkey Kong, Smurf, and Zaxxon.\n\nThe console was powered by an 8 bit Z80 processor running at 3.58MHz Z80 and had a whopping 1 kB RAM. The video processor had 16kB video ram providing 256 x 192 pixel resolution in 16 predefined colors. The sound was powered by a three channel square wave sound generator with an additional noise channel. The games were released on cartridges with up to 32 kB ROM.\n\nOne of the more interesting aspects of the console was the hardware expandability through the Expansion Module interface. The first expansion module made the ColecoVision compatible with Atari 2600, enabling the biggest game library at the time. The second expansion module was a driving controller with a steering wheel and gas pedal, and other expansion modules followed.\n\nThis article focuses on the driving controller, and the challenges using it in modern homebrew development projects. The steering wheel was first introduced with the game Turbo, and it was later used in a handful of games. Typically for the games in the 80’s is that these where one person games, and we’ll go into details on why and what actually can be done to allow two steering wheels to be used at the same time in a game.\n\nMotivation\n\nThe main reason for me learning about and solving the problems with the driving module came when me and Vincent van Dam decided to develop a new racing game for the ColecoVision with graphics help from Luc Miron. Our goal was to do something never done before on ColecoVision; a variation of the microcars game, including multi directional scrollers, real physics engines to model the cars, advanced music players, and of course support for the driving module allowing two players to compete. Many of the features we wanted had never been done on ColecoVision, so it was a quite ambitious goal. The hardware constraints of the console made the development hard and the game took over 2,000 hours to develop. The finished game was a game published by CollectorVision in 2011:\n\nDriving Module Overview\n\nThe steering wheel has a 2 bit rotary encoder with two concentric rings with contact surfaces and openings offset by 90 degrees as the picture below illustrates.\n\nThe rings move as a person turns the steering wheel, metal plates at fixed position generates an ‘on’ signal when they touch the metal surface of the rings, and an ‘off’ signal otherwise. These signals can be read by software on I/O port \\$FC for the first steering wheel and I/O port \\$FF for the second steering wheel, allowing the software to read the angle of the rotary wheel as a 2 bit value:\n\nContact 1 Contact 2\nAngle\nOff Off\n0° to 90°\nOff On\n90° to 180°\nOn On\n180° to 270°\nOn Off\n270° to 360°\n\nOn the ColecoVision, contact 1 is mapped to bit 4 and contact 2 to bit 5.\n\nDriving games typically want to know how much the user moved the steering wheel. A convenient measurement is a single integer value (called steeringCunter in the algorithms below) where a negative value means steering to the left, and a positive value steering to the right. A small value means steering a little bit and a large value steering a lot. We’ll discuss two approaches, one pure polling based, and one that is using interrupts that calculates the steeringCunter value.\n\nPolling\n\nA straightforward way of maintaining the steeringCounter value is to initialize it to 0, then read the values of contact 1 and contact 2 at a periodic interval. The contact bits are Gray coded, which means that it is a binary counter where two adjacent codes differ by only one bit position. The algorithm for a polling function increases or decreases the steeringCounter value by one depending on the previous values of the contact bits and which contact changed since last read. If both contact bits has changed we don’t do anything as we can’t say if the wheel turned left or right.\n\nThe code snippet below illustrates how the polling method could be implemented in a fairly efficient way utilizing the Gray coded binary counter.\n\n#define CONTACT_1 0x10\n#define CONTACT_2 0x20\n\nvoid pollSteeringWheel() {\nint wheelCounter = 0;\nunsigned char lastState = 0;\n\nvoid pollSteeringWheel() {\n/* Read the current state of the steering wheel contacts */\n/* and compute the bits that changed since last call. */\n\nunsigned char bitsChanged = lastState ^ newState;\n/* Mirror bitsChanged if the high bit is set in lastState\n/* to allow a single test for increment and decrement. */\n\nif (lastState & CONTACT_2) {\n}\n\n/* Increase or decrease wheelCounter if only one */\n/* bit has changed. */\n\nif (bitsChanged == CONTACT_1) {\nwheelCounter++;\n}\nif (bitsChanged == CONTACT_2) {\nwheelCounter--;\n}\nlastState = newState;\n}\n\nThe code can easily be extended to support two steering wheel controllers. Just duplicate the code for the second steering wheel, reading from port \\$FF instead of port \\$FC.\n\nThe drawback of using a polling method like the one above, is that it needs to be called at a frequency that is higher than the rotary wheel moves 90 degrees as the user moves the steering wheel. Although the code is not complicated it uses a fair bit of CPU on a ColecoVision system, and interleaving calls to the polling method is sometimes not that easy.\n\nInterrupt Driven Approach\n\nTo allow software to register steering wheel movements in real time, the first contact is connected to a non maskable interrupt, which means that when the first contact changes from ‘off’ to ‘on’ the cpu is interrupted and the software can read the bits at that time. We can use the interrupt and the state of the rotary encoder to implement a counter that measures how much the wheel has been rotated to the left or to the right.\n\nThe first contact can turn from ‘off’ to ‘on’ in two places, either if the wheel is turned clockwise from ‘off’ to ‘on’ or counter clockwise from ‘off’ to ‘on’. By looking at the state of the second contact we can see if the wheel was turned clockwise or counter clockwise. The second contact is ‘on’ when the transition from ‘off’ to ‘on’ occurs on the first contact when we rotate clockwise. Similarly, the second contact is ‘off’ if we rotate counter clockwise.\n\nThere are however two quite big problems with using interrupts on the steering wheel controllers.\n1. Both steering wheels share the same nonmaskable interrupt, so when supporting two steering wheels, it is not possible to tell which of them that caused the interrupt to be fired.\n2. The mechanical contacts can cause jitter if the steering wheel is positioned just at the edge of the connecting metal plates, causing multiple interrupts to be fired even though the wheel is not spinning.\n\nAlgorithm using Interrupts\n\nTo address both these issues, we’ll implement a clock pulse for each steering wheel that is set in the interrupt service routine, and reset in a background polling task. When the interrupt is triggered, the program checks if the logical clock pulse. If the clock pulse is high, no action is taken. If the clock pulse is low, we know that the steering wheel has moved, and we update the counter based on the state of the second contact. The pseudo code below implements the algorithm for one of the steering wheel controller. The real code linked at the end of the article duplicates the code for the second controller both in the polling method and the interrupt service routine.\n\n#define CONTACT_1 0x10\n#define CONTACT_2 0x20\n\nunsigned char clockPulse = CONTACT_1;\nint wheelCounter = 0;\n\nvoid pollSteeringWheel() {\n/* Clear clock pulse if contact 1 is off. */\n}\n\nvoid interruptServiceRoutine() {\n/* Update counter for steering wheel if clock pulse */\n/* was reset in the polling routine. */\nif (clockPulse == 0) {\n/* Read contact 2 to check direction. */\nwheelCounter++;\n} else {\nwheelCounter--;\n}\n/* Set clock pulse. */\nclockPulse = CONTACT_1;\n}\n}\n\nUsing one clock pulse for each steering wheel (setting in the interrupt service routine and clearing it in the background) serves two purposes:\n1. It allows for two steering wheels to share a single interrupt\n2. It provides a pseudo low-pass filter that significantly reduces any jitter caused by the mechanical contacts.\n\nAlthough this approach also require polling, it has some advantages. First the polling interval can be twice as long as the pure polling based version. Second, the polling method is rather short compared to the one in the polling only version. Together this gives a solution that require much less CPU time.\n\nTuning\n\nTo get a responsive input from the controllers, the background task needs to run around once every one millisecond. Running it too often causes more jitter, and running it less often reduces the responsiveness of the steering wheel. The algorithm is not that sensitive for some variance in when the background task is run, and occasional large delays between runs does not cause any large impact to the user. An easy way during development to see that the background task is called at the desired interval is to add a code snippet that changes the border color every time it is been called.\n\nAlso, a game may want to downshift the wheelCounter values to get the desired sensitivity of the steering wheel.\n\nSample Code\n\nA Hitech C implementation (assembly code using c calling conventions) of the interrupt based algorithm can be downloaded here.\n\n## Tuesday, July 9, 2013\n\n### Rewinding CRC - Calculating CRC backwards\n\nBackground\n\nIn two earlier articles (Calculating Reverse CRC and Reverse CRC Patch with Readable Characters) I've described algorithms to find a sequence of bytes and ASCII characters that will append to a buffer to change the CRC to a desired CRC.\nSome readers have said that it would be nice to be able to modify the data and apply a patch when the beginning of the buffer is unknown. This article describes a method to do just that, although it does require the final CRC of the original buffer to be known.\n\nProblem\n\nThe algorithms described in earlier articles require that the CRC at the beginning of the buffer is known. But suppose you have a message with an unknown beginning and you want to change a sequence in the buffer and apply a patch to make the mutated buffer end up with the same CRC as the original. For example changing:\n\n{unknown data} + {known buffer} -> Known CRC\n\nto\n\n{same unknown data} + {modified buffer} + {patch} -> Same Known CRC\n\nIn order for the previously described patch algorithms to work, you would need to know the CRC at the beginning of the known buffer to use the algorithms. This article describes an algorithm that starts at the end of a known buffer and calculates the CRC at an earlier point in the buffer.\n\nSolution\n\nThe solution is not that complicated but has a few caveats that we'll go into. The algorithm is basically the reverse of the normal CRC algorithm that looks like:\n\nnewCrc = (oldCrc >> 8) ^ crc32Table[value ^ (oldCrc & 0xff)];\n\nAssume that the CRC at the current location in a buffer contains the octets [An Bn Cn Dn] where An is the high eight bits and Dn is the low eight bits. And the CRC before the last byte Vn was added to the CRC was [Ao Bo Co Do]. If we put these two CRCs into the equation and performs the shifts, we'll get:\n\n[An Bn Cn Dn] = [0 Ao Bo Co] ^ crcTable[Vn ^ Do]\n\nNote the 0 in the high byte of the old CRC. Looking at this it is quite clear that the high byte of the new/current CRC is equivalent to the high byte in the CRC table lookup. Knowing this we can easily look at all 256 CRC table indices and find the ones that matches the equation:\n\nAn = crcTable[i] >> 24\n\nOnce we know a table index that matches the equation, we can calculate the old octet Do:\n\nDo = i ^ Vn\n\nwhere Vn was the byte that we added to the old CRC to get the new CRC. This is however where the caveat comes in. There may be more than one table entry that satisfy the equation, which means that there will be multiple possible previous CRCs. The good news is that the standard CRC with the polynomial 04c11db7h and most other polynomials has exactly one table entry that do, so if using the standard CRC polynomial there is a single solution to the equation and thus only a single possible previous CRC.\n\nLet us continue. Once Do is calculated, it is easy to calculate the other three octets in the previous CRC, by inserting the found Dn into the equation:\n\n[0 Ao Bo Co] = [An Bn Cn Dn] ^ crcTable[Vn ^ Do]\n\nSo to address the original problem where we wanted to find the CRC X number of bytes before the end, we'll apply this algorithm recursively on the known buffer starting at the last byte and the known CRC and end when we reach the beginning of the known buffer:\n\nfor (int i = len(buffer) - 1; i >= 0; --i) {\nfor (int j = 0; j < 256; ++j) {\nif ((crc32Table[j] >> 24) == (crc >> 24)) {\ncrc = ((crc ^ crc32Table[j]) << 8) | (j ^ buffer[i]);\nbreak;\n}\n}\n}\n\nThis gives us the CRC at the beginning of the known buffer, and we can apply the patch algorithms described in earlier articles.\n\nAlthough not that common, it may be worth looking at a solution to address the case where the polynomial doesn't have unique high octets. In this case we need to do a bit more work to find all solutions. We can use a recursive function call, but this may not scale very well, so the best option is to use a stack where we keep track of each found match in the CRC table.\nThere are two pieces of information that needs to be pushed on the stack for each match; the current location in the buffer and the CRC at that location. Then we need to modify the program slightly to use the stack. The CRCs we found at the beginning of the known buffer is then stored in a list.\n\nstd::list<UInt32> startCrcs;\nstd::stack<match> stack;\n\n// Push last octet in the buffer to seed the loop.\nstack.push(new Match(size, INV(endCrc)));\n\nwhile (!stack.empty()) {\nMatch* node = stack.top();\nstack.pop();\nif (node->offset == 0) {\n// Reached the beginning of the buffer; save the CRC.\nstartCrcs.push_back(INV(node->crc));\n}\nelse {\n// Find previous CRCs and push nodes to stack.\nfor (int i = 0; i < 256; ++i) {\nif ((crc32Table[i] >> 24) == (node->crc >> 24)) {\nUInt32 prevCrc = ((node->crc ^ crc32Table[i]) << 8) |\n(i ^ buffer[node->offset - 1]);\nstack.push(new Node(node->offset - 1, prevCrc));\n}\n}\n}\ndelete node;\n}\n\nUsage\n\nAs mentioned, this algorithm may be useful if you want to back calculate CRC in a stream of data or in a frame when you know the final CRC in order to modify the data. There are some limitations though, one is that there may be multiple solutions depending on polynomial. The bigger limitation is that the algorithm requires that the CRC at the end (or at least at some point later than where the to calculate CRC at) which may be unknown.\n\nSource\n\nI have added an implementation of the Rewind algorithm to the previous CRC package with some examples and test code. The source code can be found here.\n\n## Sunday, May 13, 2012\n\n### How to win (or not to win) IOCCC\n\nBackground\n\nThe International Obfuscated C Code Contest, IOCCC, is a programming contest where the goal is to write the most obscure/obfuscated C program that shows the importance of programming style, stress the compilers and illustrates some of the subtleties of the C language. The competition started in 1984 and is now the longest running contest on the Internet. Through the years, the competition has produced amazing small programs. Man y of them are unique and fun, but there are also programs that solve the same problems as previous winners. One of the most popular is to generate small primes.\n\nThis article talks about different ways to make a small primes program and reasons for why they didn’t win, followed by what eventually led to the 7th winning small primes program.\n\nSimple Small Primes Program\n\nOne of the entries I made to the competition is what may be the shortest small primes program with its 52 characters:\n\nl;main(O){for(;2-l||printf(\"%8d\",O);l=O%--l?l:++O);}\n\nThis program has some features of winning entries:\n• It is short and compact\n• uses hard to read variable names\n• Uses question mark operator\n• implicit type definitions\n• depends on order of Boolean operators\nThis is a fairly good list of features that makes up a good obfuscated program. But it has one big flaw (apart from calculating primes), which is that the algorithm is quite easy to spot in the code. Anything that looks for modulus of one variable by a decreasing or increasing second variable is most likely a prime number generator. To win over the previous winners the program really need to be a bit more creative in hiding what it does, or do it in a more interesting way than easy to spot brute force.\n\nDisguising the Algorithm\n\nA simple modification to the program is to make the brute force approach a bit less revealing. One way is to look at what makes a prime. A prime number p is a number greater than 1 that satisfies the equation with only integer numbers:\n\na * b = p    iff:   a = 1 or b = 1\n\nA simple substitution of a and b where\n\na = i + j\nb = i – j\n\ngives us the following equation:\n\n(i + j) * (i – j) = i2 – j2 = p    iff:   i – j = 1\n\nWith this equation we can be a bit more creative. We still need to find an integer solution to the equation and one approach is to iterate over x and y until we find two numbers x and y that satisfy the equation above. Knowing that the derivate of n2 is 2n, we can find the squares iteratively where\n\n12 = 1\nd1 = 1\ndn = dn - 1 + 2\nn2 = (n-1)2 + dn\n\nThe nice thing with this iteration is that we can use only additions and subtractions to find whether a number is a prime. We need two derivate variables in our iteration, say x and y to test the potential prime number. An interesting observation from the initial algorithm is that we actually don’t need to keep track of both squared values as we are only interested in iterating until we find two values of x and y that gives us an integer solution to the equation. Once the integer solution is found, we can check if x – y = 2, which means that p must be a prime.\n\nWe still need one variable to hold the accumulated value of the squares, so we introduce a variable z. Each step in the iteration for finding will increase z by x or decrease z by y depending on whether z is greater or less than 0. We have an integer solution to the problem once z is 0. The iterative algorithm is implemented in the following obfuscated program:\n\nx,y,z;main(p){for(;;z+=!z?x-y-2?p:printf(\n\"%8d\",p),x=y=1,p+=2:z>0?y-=2:-(x-=2));}\n\nThis program has many of the characteristics of the first implementation, but it also hides the obvious modulus that killed the first version. For most people it would be pretty hard to see that this version calculates prime numbers, but for anyone skilled in hyperbolic functions, the program still gives enough hints on what it’s doing, and since the program is trying to compete in a category where the IOCCC judges explicitly mention that it is much harder to win with a small primes program, its simply not good enough.\n\nBe Creative in the Implementation\n\nAlthough the last program may have been too trivial, it does open up for several interesting implementations as it the problem is solved by simple additions and subtractions. One interesting feature of C is the pre-processor. A prime number generator using the pre-processor has already won IOCCC, so it may be a long shot going that route. That particular entry did hide the math in several pre-processor macros, where defined variable holds some value needed for the calculation. In fact it is very easy to find the actual test for a prime number as it is limited to one line doing a modulus test.\n\nSince we have a simple algorithm, we could use it to do something less obvious using the pre-processor. What if we didn’t assign any values to the variables we use? We could let a defined variable represent a 1 bit, and if the variable is not defined, it is a 0. This way we can use 10 variable names to represent a 10 bit number.\n\nWe could also use recursive includes, where we have a state variable indicating what one pass of the program code does. For example we could exercise part of the program to do an addition, and a second to do subtraction etc.\n\nIn order to do an addition of two numbers with these constraints, we need to have a one bit adder, that generates a carry. This is really not that hard. Assume we use x and y as the bits, c, as carry, and place the output back in x and c, we can do this in the pre-processor as:\n\n#ifdef x\n#ifdef y\n#ifndef c\n#define c\n#endif\n#else\n#ifdef c\n#undef x\n#endif\n#endif\n#else\n#ifdef y\n#ifndef c\n#define x\n#endif\n#else\n#ifdef c\n#undef c\n#endif\n#endif\n#endif\n\nThen we can use the one bit adder to add our 10 bit values A0-A9 and B0-B9 by first defining x and y for each of the bits in the two values and include the program recursively to exercise the code above, like:\n\n#ifdef A0\n#define x\n#else\n#undef x\n#endif\n#ifdef B0\n#define y\n#else\n#undef y\n#endif\n#include \"program.c\"\n\nAnd then do the same for for A1-A9 and B1-B9. We can do similar for subtractions. Furthermore, we can save the value of the last carry and use it as an overflow bit. Then we can leverage that for testing if a value is greater than another. Finally we need some similar logic to run the actual algorithm by including and exercising the right portions of the code. The whole program can be found at the IOCCC who won page.\n\nWhat makes this a winner compared to the other examples is probably that this really pushes the compilers to the limit. Not a single compiler was able to calculate prime numbers up to 1024 at the time of the competition. Linus Torvalds did work hard to fix one compiler within 12 hours after the source code was released to the public, but that is really an exception. Most compilers still don’t handle the program. The number of recursive includes are over 6.8 million, and the amount of memory required by the compiler is significant. It is also extremely slow; it takes hours to compute just a few hundred prime numbers. Furthermore, the actual pre-processed code is many megabytes and the only actual code, is one print statement for each prime number.\n\nConclusion\n\nTo win IOCCC with something that is not unique or new is a challenge, but it is possible to find a new approach that is interesting enough, even with something as common as a prime number generator. In general though, it is much easier to win with something new, but even then, some fairly clever and interesting techniques need to be present. The good thing is that it is possible to improve an idea and reach something worthy by looking at alternatives and some out of the box thinking, just as these examples show. And the improvements can be introduces iterative and doesn’t need to be thought out in advance.\n\nHappy coding.\n\n## Sunday, January 29, 2012\n\n### Finding Reverse CRC Patch with Readable Characters\n\nBackground\n\nIn the article Calculating Reverse CRC I described an algorithm to create a four byte patch sequence that can be inserted into a message to give the message a desired CRC. A reader asked if and how the algorithm could be adopted to return a patch sequence with only ASCII characters. This article details how to extend the original algorithm to achieve this.\n\nProblem\n\nAs described in the original article, one of the properties of CRC is that you can find and append a four byte sequence to any message and get any desired resulting CRC hash.\n\nFor a given input CRC and desired CRC, we can easily check if the four patch bytes are ASCII characters. The problem is that for many combinations of input CRC and desired CRC the byte sequence includes non-ASCII characters. The problem is how we address the case when we don’t get ASCII characters in our patch sequence.\n\nObservations\n\nBefore we go into the solution, we can make some observations that may be beneficial. The binary patch sequence is four bytes, or 32 bits. As we want to limit the patch sequence to ASCII characters, each patch byte need to be limited to characters 0-9, A-Z, and a-z. The number of bits we can encode with these characters are roughly 6 (actually 62 different character values, but close enough to 6 bits for our needs). Using four characters would only give us 24 bits of information and we can’t represent all possible patch bytes. By adding two additional characters to our patch sequence, we will get 36 bits which would be sufficient.\n\nSolution\n\nThe problem is that the CRC algorithm and the reverse CRC algorithm are working on 8 bit values, and can’t be changed. So we need to find a solution around how we could add the extra two patch characters without changing the nature of the original CRC algorithm.\n\nThis can fortunately be done relatively easy. Assume that our patch sequence is six characters represented by the names A-F:\n\nABCDEF\n\nWe know the initial CRC (before the patch bytes are added to the CRC) as well as the desired CRC. We also know that the reverse algorithm operates on four characters. We can use this knowledge to break the problem into two by introducing a temporary CRC.\n\nIf we pick two characters for the substring AB and then append these characters to the original CRC we get a new CRC value. The new CRC value can then be used in our original algorithm to get the patch bytes CDEF. If we pick the two characters AB carefully, we can get patch bytes CDEF that are only ASCII.\n\nSo the problem is now reduced to finding two characters AB that satisfy the requirements that the patch sequence CDEF only contains ASCII characters.\n\nThis problem can be tough to solve, but as we are only trying to find two characters which means that there are only 62 * 62 = 3844 possible solutions. This is a very small number for trial and error on a computer, and the best approach is to do a brute force trial to find these characters, as shown in the following pseudo code:\n\nfor (char a in ValidCharset) {\nfor (char b in ValidCharset) {\nCrc32 crc32(msgCrc); // Temporary CRC object\ncrc32.append(a);\ncrc32.append(b);\nif (ContainsValidChars(crc32.findReverse(desiredCrc)) {\n// Return the found patch bytes....\n}\n}\n}\n\nUsage\n\nFinding a patch sequence with a limited character set that generates a desired CRC can be useful to for example finding weak passwords in systems that use CRC for encryption. Other uses may include cases where you want to CRC any message but for some reason have constraints on what the generated CRC can be and the only option for patching is to use printable characters.\n\nSource Code\n\nThe complete source code of a CRC implementation that includes a method for finding a six readable character patch sequence for the reverse CRC, as well as some test cases can be found here." ]

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[ "Scilab Home page | Wiki | Bug tracker | Forge | Mailing list archives | ATOMS | File exchange\nChange language to: Français - Português - 日本語\n\nSee the recommended documentation of this function\n\n# lu\n\nLU factorization with pivoting\n\n### Calling Sequence\n\n```[L,U]= lu(A)\n[L,U,E]= lu(A)```\n\n### Arguments\n\nA\n\nreal or complex matrix (m x n).\n\nL\n\nreal or complex matrices (m x min(m,n)).\n\nU\n\nreal or complex matrices (min(m,n) x n ).\n\nE\n\na (n x n) permutation matrix.\n\n### Description\n\n`[L,U]= lu(A)` produces two matrices `L` and `U` such that `A = L*U` with `U` upper triangular and `L` a general matrix without any particular structure. In fact, the matrix `A` is factored as `E*A=B*U` where the matrix `B` is lower triangular and the matrix `L` is computed from `L=E'*B`.\n\nIf `A` has rank `k`, rows `k+1` to `n` of `U` are zero.\n\n`[L,U,E]= lu(A)` produces three matrices `L`, `U` and `E` such that `E*A = L*U` with `U` upper triangular and `E*L` lower triangular for a permutation matrix `E`.\n\nIf `A` is a real matrix, using the function `lufact` and `luget` it is possible to obtain the permutation matrices and also when `A` is not full rank the column compression of the matrix `L`.\n\n### Example #1\n\nIn the following example, we create the Hilbert matrix of size 4 and factor it with A=LU. Notice that the matrix L is not lower triangular. To get a lower triangular L matrix, we should have given the output argument E to Scilab.\n\n```a = testmatrix(\"hilb\",4);\n[l,u]=lu(a)\nnorm(l*u-a)```\n\n### Example #2\n\nIn the following example, we create the Hilbert matrix of size 4 and factor it with EA=LU. Notice that the matrix L is lower triangular.\n\n```a = testmatrix(\"hilb\",4);\n[l,u,e]=lu(a)\nnorm(l*u-e*a)```\n\n### Example #3\n\nThe following example shows how to use the lufact and luget functions.\n\n```a=rand(4,4);\n[l,u]=lu(a)\nnorm(l*u-a)\n\n[h,rk]=lufact(sparse(a))\n[P,L,U,Q]=luget(h);\nludel(h)\nP=full(P);\nL=full(L);\nU=full(U);\nQ=full(Q);\nnorm(P*L*U*Q-a)```\n\n### Used Functions\n\nlu decompositions are based on the Lapack routines DGETRF for real matrices and ZGETRF for the complex case." ]

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[ "# How do you factor completely 15a^2y-30ay?\n\n$15 {a}^{2} y - 30 a y = 15 a y \\left(a - 2\\right)$\nBoth of the terms are divisible by $15$, $a$ and $y$, so by $15 a y$.\n$15 {a}^{2} y - 30 a y = 15 a y \\cdot a - 15 a y \\cdot 2 = 15 a y \\left(a - 2\\right)$" ]

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[ "```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`\n\n```\n\n```\n`\t`" ]

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[ "", null, "# Bharathiar University PG Calculation of GPA CGPA and Division\n\n(Last Updated On: 05/11/2019)\n\nBharathiar University PG Calculation of GPA/CGPA\n\nGrade Point Average (GPA) and Cumulative Grade Point Average (CGPA) are calculated as given here.\n\nContents\n\n## Bharathiar University PG Grade Point Average (GPA)\n\nCalculation for a semester\n\nGPA = Σ(iCi×GPi)/ΣiCi", null, "where iCi = credit assigned for a course\n\nGPi = grade point for that course.\n\nThe failed courses in the semester are also accounted for in the summation.\n\nCі = Credits earned for course i in any semester\n\nGi = Grade Point obtained for course i in any semester\n\nn = refers to the semester in which such course were credited\n\n## Bharathiar University PG Cumulative Grade Point Average (CGPA)\n\nCalculation for entire programme\n\nCGPA =ΣnΣi Cni GPni / Σn Σi Cni\n\nwhere iCi = credit assigned for a course\n\nGPni = grade point for that course. The summation is done for all courses registered by the student during all the semesters for which the CGPA is needed. Bharathiar University PG CGPA for the UG programme is arrived at by considering all course credits that are needed for the degree and their respective grade points.", null, "## Bharathiar University Division", null, "Bharathiar University, Coimbatore, Tamil Nadu  https://www.b-u.ac.in/\n\n(Visited 4,524 times, 1 visits today)\n\n### 1 thought on “Bharathiar University PG Calculation of GPA CGPA and Division”\n\n1.", null, "How to calculate Bharathiar University PG result percentage from CGPA ?" ]

[ null, "https://indorestudents.com/wp-content/uploads/2019/11/Bharathiar-University.png", null, "https://indorestudents.com/wp-content/uploads/2019/11/bharathiar-university-gpa.png", null, "https://indorestudents.com/wp-content/uploads/2019/11/bharathiar-university-cgpa.png", null, "https://indorestudents.com/wp-content/uploads/2019/11/Bharathiar-University-Division.png", null, "https://secure.gravatar.com/avatar/59b938d096da6e601ed7900078601fea", null ]

[ "# C API\n\nThe API can be used to extend CVXOPT with interfaces to external C routines and libraries. A C program that creates or manipulates the dense or sparse matrix objects defined in CVXOPT must include the cvxopt.h header file in the src directory of the distribution.\n\nBefore the C API can be used in an extension module it must be initialized by calling the macro import_cvxopt. As an example we show the module initialization from the cvxopt.blas module, which itself uses the API:\n\n#if PY_MAJOR_VERSION >= 3\n\nstatic PyModuleDef blas_module = {\n\"blas\",\nblas__doc__,\n-1,\nblas_functions,\nNULL, NULL, NULL, NULL\n};\n\nPyMODINIT_FUNC PyInit_blas(void)\n{\nPyObject *m;\nif (!(m = PyModule_Create(&blas_module))) return NULL;\nif (import_cvxopt() < 0) return NULL;\nreturn m;\n}\n\n#else\n\nPyMODINIT_FUNC initblas(void)\n{\nPyObject *m;\nm = Py_InitModule3(\"cvxopt.blas\", blas_functions, blas__doc__);\nif (import_cvxopt() < 0) return ;\n}\n\n#endif\n\n\n## Dense Matrices\n\nAs can be seen from the header file cvxopt.h, a matrix is essentially a structure with four fields. The fields nrows and ncols are two integers that specify the dimensions. The id field controls the type of the matrix and can have values DOUBLE, INT, and COMPLEX. The buffer field is an array that contains the matrix elements stored contiguously in column-major order.\n\nThe following C functions can be used to create matrices.\n\nmatrix *Matrix_New(int nrows, int ncols, int id)\n\nReturns a matrix object of type id with nrows rows and ncols columns. The elements of the matrix are uninitialized.\n\nmatrix *Matrix_NewFromMatrix(matrix *src, int id)\n\nReturns a copy of the matrix src converted to type id. The following type conversions are allowed: 'i' to 'd', 'i' to 'z', and 'd' to 'z'.\n\nmatrix *Matrix_NewFromSequence(PyListObject *x, int id)\n\nCreates a matrix of type id from the Python sequence type x. The returned matrix has size (len(x), 1). The size can be changed by modifying the nrows and ncols fields of the returned matrix.\n\nTo illustrate the creation and manipulation of dense matrices (as well as the Python C API), we show the code for the cvxopt.uniform function described in the section Randomly Generated Matrices.\n\nPyObject * uniform(PyObject *self, PyObject *args, PyObject *kwrds)\n{\nmatrix *obj;\nint i, nrows, ncols = 1;\ndouble a = 0, b = 1;\nchar *kwlist[] = {\"nrows\", \"ncols\", \"a\", \"b\", NULL};\n\nif (!PyArg_ParseTupleAndKeywords(args, kwrds, \"i|idd\", kwlist,\n&nrows, &ncols, &a, &b)) return NULL;\n\nif ((nrows<0) || (ncols<0)) {\nPyErr_SetString(PyExc_TypeError, \"dimensions must be non-negative\");\nreturn NULL;\n}\n\nif (!(obj = Matrix_New(nrows, ncols, DOUBLE)))\nreturn PyErr_NoMemory();\n\nfor (i = 0; i < nrows*ncols; i++)\nMAT_BUFD(obj)[i] = Uniform(a,b);\n\nreturn (PyObject *)obj;\n}\n\n\n## Sparse Matrices\n\nSparse matrices are stored in compressed column storage (CCS) format. For a general nrows by ncols sparse matrix with nnz nonzero entries this means the following. The sparsity pattern and the nonzero values are stored in three fields:\n\nvalues\n\nAn array of floating-point numbers of length nnz with the nonzero entries of the matrix stored columnwise.\n\nrowind\n\nAn array of integers of length nnz containing the row indices of the nonzero entries, stored in the same order as values.\n\ncolptr\n\nAn array of integers of length ncols + 1 with for each column of the matrix the index of the first element in values from that column. More precisely, colptr is 0, and for k = 0, 1, …, ncols - 1, colptr[k+1] is equal to colptr[k] plus the number of nonzeros in column k of the matrix. Thus, colptr[ncols] is equal to nnz, the number of nonzero entries.\n\nFor example, for the matrix", null, "the elements of values, rowind, and colptr are:\n\nvalues:\n\n1.0, 2.0, 3.0, 4.0, 5.0, 6.0\n\nrowind:\n\n0, 1,3, 1, 0, 2\n\ncolptr:\n\n0, 3, 3, 4, 6.\n\nIt is crucial that for each column the row indices in rowind are sorted; the equivalent representation\n\nvalues:\n\n3.0, 2.0, 1.0, 4.0, 5.0, 6.0\n\nrowind:\n\n3, 1, 0, 1, 0, 2\n\ncolptr:\n\n0, 3, 3, 4, 6\n\nis not allowed (and will likely cause the program to crash).\n\nThe nzmax field specifies the number of non-zero elements the matrix can store. It is equal to the length of rowind and values; this number can be larger that colptr[nrows], but never less. This field makes it possible to preallocate a certain amount of memory to avoid reallocations if the matrix is constructed sequentially by filling in elements. In general the nzmax field can safely be ignored, however, since it will always be adjusted automatically as the number of non-zero elements grows.\n\nThe id field controls the type of the matrix and can have values DOUBLE and COMPLEX.\n\nSparse matrices are created using the following functions from the API.\n\nspmatrix *SpMatrix_New(int_t nrows, int_t ncols, int_t nzmax, int id)\n\nReturns a sparse zero matrix with nrows rows and ncols columns. nzmax is the number of elements that will be allocated (the length of the values and rowind fields).\n\nspmatrix *SpMatrix_NewFromMatrix(spmatrix *src, int id)\n\nReturns a copy the sparse matrix var{src}.\n\nspmatrix *SpMatrix_NewFromIJV(matrix *I, matrix *J, matrix *V, int_t nrows, int_t ncols, int id)\n\nCreates a sparse matrix with nrows rows and ncols columns from a triplet description. I and J must be integer matrices and V either a double or complex matrix, or NULL. If V is NULL the values of the entries in the matrix are undefined, otherwise they are specified by V. Repeated entries in V are summed. The number of allocated elements is given by nzmax, which is adjusted if it is smaller than the required amount.\n\nWe illustrate use of the sparse matrix class by listing the source code for the real method, which returns the real part of a sparse matrix:\n\nstatic PyObject * spmatrix_real(spmatrix *self) {\n\nif (SP_ID(self) != COMPLEX)\nreturn (PyObject *)SpMatrix_NewFromMatrix(self, 0, SP_ID(self));\n\nspmatrix *ret = SpMatrix_New(SP_NROWS(self), SP_NCOLS(self),\nSP_NNZ(self), DOUBLE);\nif (!ret) return PyErr_NoMemory();\n\nint i;\nfor (i=0; i < SP_NNZ(self); i++)\nSP_VALD(ret)[i] = creal(SP_VALZ(self)[i]);\n\nmemcpy(SP_COL(ret), SP_COL(self), (SP_NCOLS(self)+1)*sizeof(int_t));\nmemcpy(SP_ROW(ret), SP_ROW(self), SP_NNZ(self)*sizeof(int_t));\nreturn (PyObject *)ret;\n}" ]

[ null, "http://cvxopt.org/userguide/_images/math/9173aacc17760942bd0ce61e5ad70788a84297d5.png", null ]

[ "### Question\n\nHow is the seasonality of a commodity price measured?\n\n### Seasonality Tool\n\nUses monthly price data of a commodity to measure the seasonal pattern in prices. Seasonality is expressed as 12 indexes that represent the ratio of the price each month to the average annual price. The indexes are often graphed to show the seasonal pattern. This spreadsheet provides an example.\n\n### Data Requirements\n\n• Monthly price data over multiple years\n\n### Background\n\nAgricultural prices often follow a seasonal pattern because production is seasonal and storage is costly. In some cases, seasonal demand (such as holiday consumption) may also contribute to seasonality in agricultural prices. The table below shows the degree of seasonality depending on outcome under different conditions. If production and consumption are not seasonal, then prices will not be seasonal. This is the case for most non-agricultural commodities. If production or consumption are seasonal, but storage is inexpensive, then prices will not be strongly seasonal because of low storage costs from harvest period to off-season. This is the case for some cereals, such as sorghum. If production or consumption are seasonal, storage is expensive, and demand is price-elastic (consumers are not willing to pay higher prices), then the commodity will be available only during part of the year, because consumers are not willing to pay the cost of storage. This is the case of many fruits in developing countries. If production or consumption is seasonal, storage is expensive, and demand is price-inelastic (consumers are willing to pay higher prices), then the commodity will be available all year but the price will be strongly seasonal.\n\nSeasonality in production Storage cost Price elasticity of demand Seasonality in supply\nLow Not relevant Not relevant Little seasonality in prices due to steady supply, little storage\nHigh Low Not relevant Little seasonality in prices due to inexpensive storage\nHigh High Low Availability is seasonal, no supply in offseason\nHigh High High High seasonality in prices. Although storage is costly, consumers are willing to pay to have it in the offseason.\n\nThe seasonality of prices can be measured by comparing the prices in a given month to the prices over the whole year. There are two methods:\n\nThe calendar-year seasonal index compares the price in each month to the average over the same calendar year and then averages this ratio over multiple years. In other words,", null, "Variables\nY number of years of data\nPm Price in month m\ny index for the year\ni index for the month in year y\n\nThe term in parentheses is the ratio of the price in month m to the average price over that calendar year. This is ratio is summed over the Y years and divided by Y to obtain the average.\n\nThe moving-average seasonal index is similar except that it compares the price in each month to the average over the 13-month period centered on that month.", null, "Each method has its strengths and weaknesses: The calendar-year price index is easier to understand and to explain. However, the moving-average price index has the advantage that it is less affected by an upward or downward trend in the prices. For example, an upward trend in the prices will cause the January-February indexes in the calendar-year method to be low and the November-December indexes to be high. The moving-average seasonal index is also easier to calculate with a spreadsheet.\n\nIn practice, the two methods tend to give fairly similar results.\n\n### Example\n\nThis spreadsheet shows the calculation of these two types of seasonal price indexes. Column A has the names of the month, which are used in the graph of the seasonal index. Column B has the month and year of the price data . Column C contains the price data for each month.\n\nColumns D to F calculate the calendar-year seasonal index. Column D calculates the calendar-year average price, which is the denominator inside parentheses in the first equation above. Column E calculates the ratio of the price each month to the annual average (column D). This is ratio in parentheses in the first equation above. Column F calculates the average of all the January ratios, February ratios, and so on. This is the calendar-year seasonal index.\n\nColumns G to I calculate the moving average seasonal index. Column G calculates the 13-month moving average, meaning the average price from six months before to six months after. This is the denominator in the ratio in parentheses in the second equation. Column H calculates the ratio of the price each month to the 13-month moving average (column G). This is the ratio in parentheses in the second equation above. Column I calculates the average of all the January ratios, February ratios, and so on. This is the moving-average seasonal index.\n\nThe first graph plots the two seasonal indexes. It is apparent that the two methods give almost the same results: prices climb in the first half of the year and peak in July and August. Price then fall to their lowest point in December. This suggests that the main part of the harvest is in the last three months of the year.\n\nThe moving-average index in July is 1.15, meaning that July prices are generally 15% higher than the annual average. In contrast, the moving-average index in December is 0.81, meaning that December prices are 19% below the annual average. In other words, the prices in the peak month are 42% higher than in the lowest month (1.15/0.81 = 1.42).\nThe second graph shows the 13-month moving average from column G in red. It also shows the moving average with seasonality added. This is obtained by multiplying the moving average (column G) and the moving-average seasonal index (column I).\n\nThe third graph shows the original price (blue line) and the price with seasonality removed (green line). This is calculated by dividing the original price by the seasonal index. If the main variation in prices was seasonal, then removing the seasonal component would result in a flat line. The fact that the green line is volatile indicates that seasonality is not the main explanation for the variation in this price.\n\nPreviewAttachmentSize\nSeasonality_in_prices.xls93.5 KB" ]

[ null, "http://www.foodsecurityportal.org/sites/default/files/seasonality_calyear.jpg", null, "http://www.foodsecurityportal.org/sites/default/files/seasonality_movavg.jpg", null ]

[ "18.02 | Fall 2007 | Undergraduate\n\n# Multivariable Calculus\n\n## Calendar\n\nLEC # TOPICS KEY DATES\nI. Vectors and matrices\n0 Vectors\n1 Dot product\n2 Determinants; cross product\n3 Matrices; inverse matrices\n4 Square systems; equations of planes Problem set 1 due\n5 Parametric equations for lines and curves\n6\n\nVelocity, acceleration\n\nKepler’s second law\n\n7 Review Problem set 2 due\nExam 1 (covering lectures 1-7)\nII. Partial derivatives\n8 Level curves; partial derivatives; tangent plane approximation\n9 Max-min problems; least squares Problem set 3 due\n10 Second derivative test; boundaries and infinity\n11 Differentials; chain rule\n12 Gradient; directional derivative; tangent plane Problem set 4 due\n13 Lagrange multipliers\n14 Non-independent variables\n15 Partial differential equations; review Problem set 5 due\nExam 2 (covering lectures 8-15)\nIII. Double integrals and line integrals in the plane\n16 Double integrals Problem set 6 due\n17 Double integrals in polar coordinates; applications\n18 Change of variables\n19 Vector fields and line integrals in the plane Problem set 7 due\n20 Path independence and conservative fields\n21 Gradient fields and potential functions\n22 Green’s theorem Problem set 8 due\n23 Flux; normal form of Green’s theorem\n24 Simply connected regions; review\nExam 3 (covering lectures 16-24) Problem set 9 due\nIV. Triple integrals and surface integrals in 3-space\n25 Triple integrals in rectangular and cylindrical coordinates\n26 Spherical coordinates; surface area\n27 Vector fields in 3D; surface integrals and flux Problem set 10 due\n28 Divergence theorem\n29 Divergence theorem (cont.): applications and proof\n30 Line integrals in space, curl, exactness and potentials\n31 Stokes’ theorem Problem set 11 due\n32 Stokes’ theorem (cont.); review\nExam 4 (covering lectures 25-32)\n33\n\nTopological considerations\n\nMaxwell’s equations\n\nProblem set 12 due\n34 Final review\n35 Final review (cont.)\n36 Final exam\n\n## Course Info\n\nFall 2007\n##### Learning Resource Types\nLecture Videos\nProblem Sets\nExams with Solutions\nLecture Notes" ]

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[ "# How to turn arbitrary function to polinomial series in Mathematica?\n\nCan I turn any multivariate function into polinomial series in Mathematica?\n\nSuppose I have a function\n\nFwd[x_, α_] := x (1/Sin[Pi x/2])^α\n\n\nand wish to express it with a formula, consisting only of multiplications and summations. Is it possible?\n\nIf I apply\n\nNormal[Series[Fwd[x, α], {x, 0, 3}, {α, 0, 3}]]\n\n\nI get complex formula", null, "which contains Log function. Why?\n\nFwd[x_,\\[Alpha]_]:=x (1/Sin[Pi x/2])^\\[Alpha] = x* Exp[\\[Alpha]*Log[1/Sin[Pi x/2]]]" ]

[ null, "https://i.stack.imgur.com/GQ0gj.png", null ]

[ "##### Personal tools\nYou are here: Home / PhD Defence: Symbolic Solutions of First-Order Algebraic Differential Equations\n\n# PhD Defence: Symbolic Solutions of First-Order Algebraic Differential Equations\n\nFiled under:\nDI Georg Grasegger (Doctoral Program, Johannes Kepler University Linz), 16 July 2015, 10:00 a.m., HF 9905\nWhen Jul 16, 2015 from 10:00 AM to 11:30 AM HF 9905", null, "vCal", null, "iCal\n\nSymbolic Solutions of First-Order Algebraic Differential Equations\n\nDifferential equations have been intensively studied for a long time. Various exact solution methods have been proposed for specific cases. Nevertheless, there is no general algorithm for finding explicit exact solutions. The main aim of this thesis is to develop and investigate new methods for computing explicit exact solutions of algebraic differential equations. For this purpose, the differential problem is transformed into an algebraic geometric one by considering the differential equation to be an algebraic equation. Such an equation defines an algebraic variety and hence, tools from algebraic geometry can be applied. In particular, parametrizations of algebraic varieties are intrinsically used to solve the problem and prove properties of the obtained solutions. A general idea for solving first-order autonomous algebraic differential equations is presented.\n\nThe main results of the thesis are applications of this general idea to ordinary and partial differential equations. The idea is introduced for first-order autonomous algebraic ordinary differential equations. The presented method is a generalization of an existing algorithm for computing rational solutions. It admits an extension to the computation of radical solutions. Moreover, it allows a further generalization to higher-order algebraic ordinary differential equations.\n\nA second focus lies on the application of the general idea to partial differential equations in arbitrary many variables. The presented method reduces the problem to another one for which solution methods exist. Various well-known differential equations are solved by this method. Furthermore, classes of differential equations with rational, radical or algebraic solutions are presented. With the help of linear transformations a solution method for certain non-autonomous differential equations is achieved.\n\nThe procedures are constructed in such a way that the obtained solutions thereof satisfy certain requirements. It is shown that algebraic solutions of ordinary differential equations are general solutions. Rational solutions of partial differential equations are proven to be proper and complete." ]

[ null, "https://www.dk-compmath.jku.at/icon_export_vcal.png", null, "https://www.dk-compmath.jku.at/icon_export_ical.png", null ]

[ "# What is really M-theory? (non-pertubatively)\n\nI don´t really understand what M-theory is supposed to be. Going beyond the dualities relating different string theories (for example the common $11-D$ limit of IIA and $E_{8}\\times E_{8}$) I don't understand what is really M-theory and why is supposed to exist such a theory. I mean, what is incomplete the theory?\n\nThe motivation of this question arises in the context of string compactification and string landscape. The current paradigm includes such an enormous range of combinations (choice of: compactification manifold, version of the string theory, bundle/brane configuration, flux, critical point of $V_{eff}$...) which fixes the vacua in an approximate way that there would be interesting to know what does it lack in the final formulation of string/M-theory (in this case M-theory).\n\nIn this context I do not understand the interest of Matrix theory (I would be thankful if someone could explain to me its physical motivation). Although some of the calculations using that formalism are coherent with perturbative M-theory ones; in which sense it is expected to be a generalization of M-theory perturbative limit/11-D SUGRA? (Having in mind it is only formulated in Minkowski spacetime and toroidal compactifications).\n\nThe other path for \"complete\" M-theory is even more obscure for me: AdS/CFT dualities does not seem to suggest (for me) a physical idea/motivation for a different theory beyond the dualities between certain theories.\n\nEDIT 1: Regarding the branes allowed in M-theory: how it can be a generalization of the five string theories if lots of the branes allowed in them are not present in M-theory (it only allows M2-branes and M5-branes (respectively) electrically and magnetically coupled to the four-field strength of the theory). What happens for example with D7-branes and D3-branes which are a fundamental ingredient to allow realistic string compactifications in IIB-theory? If M-theory is true, they just don't have physical existance?\n\n• While you wait: encrypted.google.com/search?q=M-theory+non-pertubative - does that provide no answers? – Rob Feb 8 '18 at 0:55\n• @Chequez I edited my answer, you might want to have a look. Cheers!!! – DiSp0sablE_H3r0 Feb 17 '18 at 16:40\n• @Konstantinos Yes, I understood the way that M-branes include D-branes by reading the link mentioned in the previous comment (arxiv.org/abs/hep-th/9906108) and I have confirmed those explanations with your last edit. Thanks! – Chequez Feb 18 '18 at 19:26\n• Ignatios is a great physicist and a great person. You might also want to look the following link arxiv.org/abs/hep-th/9512062. Cheers!!! – DiSp0sablE_H3r0 Feb 18 '18 at 19:28\n\nI don't think that the questions have definite answers yet, but I think they are great for general discussion.\n\nI am addressing the second question related to the AdS/CFT, and I am quoting -copying and pasting- some known facts from a recent paper. In case one wants to study all the details of the paper, find it below.\n\nhttps://arxiv.org/abs/1712.08570\n\nand the references therein of course; in particular the eight first ones.\n\nPeople are trying to calculate beyond the SUGRA corrections to M-theory, in order to get some first lessons on this theory\n\nM-theory arises in the strong coupling limit of type IIA string theory.\n\nIt is known that M-theory reduces to $11$d SUGRA at low energies, and its fundamental degrees of freedom are $2$d and $5$d objects known as M$2$-branes and M$5$-branes, respectively.\n\nThe worldvolume theory for M$2$-branes (on an orbifold) was found to be a $3$d superconformal Chern-Simons matter theory with classical $\\mathcal{N} = 6$ supersymmetry.\n\nThe M$5$-brane worldvolume theory is expected to be a $6$d superconformal field theory with $OSp(8_{*}|4)$ symmetry; this is usually called the $6$d $(2, 0)$ theory.\n\nIf one considers a single M$5$-brane, one can formulate the theory in terms of an Abelian $(2, 0)$ tensor multiplet, consisting of a self-dual $2$-form gauge field, $5$ scalars, and $8$ fermions, but it is not known how to generalize this construction to describe multiple M$5$-branes.\n\nThis is where the AdS/CFT enters the game to give a hint. The statement is that the worldvolume theory for a stack of N M$5$-branes is dual to M-theory on $AdS_7 × S^4$ with N units of flux through the $4$-sphere, which reduces to $11$d SUGRA on this background in the limit large N limit; $N \\rightarrow \\infty$.\n\nThe above limit is the SUGRA approximation.\n\nRecently, people have started thinking ways of going beyond the SUGRA approximation, by using the conformal bootstrap programme. This is how people hope to gain new insight into M-theory beyond the SUGRA approximation.\n\nIt is elusive yet how one can formulate the $6$d $(2, 0)$ theory, and this is why people are using the superconformal and crossing symmetry to deduce the structure of the $4$-point correlators in this theory.\n\nThe implications of the aforementioned symmetries for the $4$-point correlators of stress tensor multiplets in the $6$d $(2, 0)$, adapting the holographic arguments, have showed that these solutions scale in the large-N limit. The spin-$0$ solution scales like $N^{−5}$, all higher-spin solutions are suppressed by at least $N^{−19/3}$. As a result, the spin-$0$ solution corresponds to the leading correction to the supergravity prediction for the $4$-point correlator (which scales like $N^{−3}$) and all higher-spin solutions are subleading compared to the $1$-loop supergravity prediction (which scales like $N^{−6}$ but has not been computed yet).\n\nThe lessons are about the low energy effective action of M-theory on $AdS^7 \\times S_4$. In particular, noting that the anomalous dimensions of the spin-$0$ solution scale like $n^6$ compared to those of the SUGRA prediction (where n is the twist), this suggests that the corresponding terms in the effective action contain six more derivatives than the SUGRA Lagrangian, and are subsequently of the form $\\mathcal{R}^4$, where $\\mathcal{R}$ is the Riemann tensor.\n\nEdit: M-branes and D-branes.\n\nPunchline: The M-branes realize all D-branes, and this is why D-branes are consistent objects that you don't need to throw away.\n\nEleven-dimensional supergravity has only an anti-symmetric tensor field $A_{(3)}$ -the rank of the tensor is 3- and the realisation of branes is restricted.\n\nThe consistent supergravity solutions are known as M$2$ and M$5$. The are further solutions to classical supergravity known as F$1$ -the fundamental string- and its magnetic dual the NS$5$-brane.\n\nYou can classify the theories based on the type of supergravity theory you get in the classical limit.\n\nThe way that M-theory sees D-branes is via the net of dualities.\n\nAll of the D-branes and the NS$5$ brane are solutions to type II theories, both A and B.\n\nWhen you reduce the M-theory on a circle, you get back to TypeIIA. TypeIIA is the same as TypeIIB when you perform a T-duality. The M$2$ and M$5$ branes reduce to the various D-branes and S-dualizing the D$5$ brane you get the NS$5$. I don't remember the actual dualities you need to apply, but you can go to the seminal papers and look there for more detailed answers.\n\n• Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. – Qmechanic Feb 9 '18 at 10:05\n• Sorry about that. I thought it would be ok, since it is an arXiv file. I will keep that in mind. Thanks – DiSp0sablE_H3r0 Feb 9 '18 at 10:20\n• Thanks for the information, it is very useful. Some questions, though: regarding the branes allowed in M-theory: how it can be a generalization of the five string theories if lots of the branes allowed in them are not present in M-theory (it only allows M2-branes and M5-branes (respectively) electrically and magnetically coupled to the four-field strength of the theory). What happens for example with D7-branes and D3-branes which are a fundamental ingredient to allow realistic string compactifications in IIB-theory? If M-theory is true, they just don't have physical existance ? – Chequez Feb 9 '18 at 12:35\n• No worries. Glad you liked it. I will try to answer to your question on M and D branes when I get back home, as I am swamped right now. Sorry for that. By the way, I think -I am not quite sure- that it is one of the rules that each question will have a different topic, so you might want to ask a new one about branes. Cheers!!! – DiSp0sablE_H3r0 Feb 9 '18 at 13:02\n• Sorry, I have not replied yet, but I am working on a project that consumes most of my time and energy; I saw the edit thoug. I will get back to this question as soon as possible. Thanks for the understanding. Cheers!!! – DiSp0sablE_H3r0 Feb 10 '18 at 21:35" ]

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[ "# What is the difference between an Ordinary Demand equation and an Engel curve equation?\n\nI mean, an Ordinary Demand Equation measures changes in quantity of X due to changes in price Px. Engel Curve measures changes in quantity of X due to changes in income M. But ODE has M in it's equation, and is simply just an algebraic variant of the Engel curve!\n\n• Engel curve assumes the price is constant, and demand curve assumes income constant. There is only demand function, no engel curve equation or demand equation. – Kun Feb 14 '16 at 2:43\n• You are very welcome – Kun Feb 14 '16 at 21:54" ]

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[ "# 数学代写|线性代数代写linear algebra代考|MAST10007", null, "#### Doug I. Jones\n\nLorem ipsum dolor sit amet, cons the all tetur adiscing elit\n\ncouryes-lab™ 为您的留学生涯保驾护航 在代写线性代数linear algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写线性代数linear algebra代写方面经验极为丰富，各种代写线性代数linear algebra相关的作业也就用不着说。\n\n• Statistical Inference 统计推断\n• Statistical Computing 统计计算\n• (Generalized) Linear Models 广义线性模型\n• Statistical Machine Learning 统计机器学习\n• Longitudinal Data Analysis 纵向数据分析\n• Foundations of Data Science 数据科学基础\ncouryes™为您提供可以保分的包课服务\n\n## 数学代写|线性代数代写linear algebra代考|Big Data Considerations and the Texas Sharpshooter Fallacy\n\nOne final thought before we close this chapter. As we have discussed, randomness plays such a role in validating our findings and we always have to account for its possibility. Unfortunately with big data, machine learning, and other data-mining tools, the scientific method has suddenly become a practice done backward. This can be precarious; allow me to demonstrate why, adapting an example from Gary Smith’s book Standard Deviations (Overlook Press).\n\nLet’s pretend I draw four playing cards from a fair deck. There’s no game or objective here other than to draw four cards and observe them. I get two $10 \\mathrm{~s}$, a 3 , and a 2 . “This is interesting,” I say. “I got two 10s, a 3, and a 2. Is this meaningful? Are the next four cards I draw also going to be two consecutive numbers and a pair? What’s the underlying model here?”\n\nSee what I did there? I took something that was completely random and I not only looked for patterns, but I tried to make a predictive model out of them. What has subtly happened here is I never made it my objective to get these four cards with these particular patterns. I observed them after they occurred.\n\nThis is exactly what data mining falls victim to every day: finding coincidental patterns in random events. With huge amounts of data and fast algorithms looking for patterns, it’s easy to find things that look meaningful but actually are just random coincidences.\n\nThis is also analogous to me firing a gun at a wall. I then draw a target around the hole and bring my friends over to show off my amazing marksmanship. Silly, right? Well, many people in data science figuratively do this every day and it is known as the Texas Sharpshooter Fallacy. They set out to act without an objective, stumble on something rare, and then point out that what they found somehow creates predictive value.\n\n## 数学代写|线性代数代写linear algebra代考|Matrix Vector Multiplication\n\nThis brings us to our next big idea in linear algebra. This concept of tracking where $\\hat{i}$ and $\\hat{j}$ land after a transformation is important because it allows us not just to create vectors but also to transform existing vectors. If you want true linear algebra enlightenment, think why creating vectors and transforming vectors are actually the same thing. It’s all a matter of relativity given your basis vectors being a starting point before and after a transformation.\n\nThe formula to transform a vector $\\vec{v}$ given basis vectors $\\hat{i}$ and $\\hat{j}$ packaged as a matrix is: \\begin{aligned} &{\\left[\\begin{array}{l} x_{n e w} \\ y_{n e w} \\end{array}\\right]=\\left[\\begin{array}{ll} a & b \\ c & d \\end{array}\\right]\\left[\\begin{array}{l} x \\ y \\end{array}\\right]} \\ &{\\left[\\begin{array}{l} x_{n e w} \\ y_{\\text {new }} \\end{array}\\right]=\\left[\\begin{array}{l} a x+b y \\ c x+d y \\end{array}\\right]} \\end{aligned}\n$\\hat{i}$ is the first column $[a, c]$ and $\\hat{j}$ is the column $[b, d]$. We package both of these basis vectors as a matrix, which again is a collection of vectors expressed as a grid of numbers in two or more dimensions. This transformation of a vector by applying basis vectors is known as matrix vector multiplication. This may seem contrived at first, but this formula is a shortcut for scaling and adding $\\hat{i}$ and $\\hat{j}$ just like we did earlier adding two vectors, and applying the transformation to any vector $\\vec{v}$.\nSo in effect, a matrix really is a transformation expressed as basis vectors.\nTo execute this transformation in Python using NumPy, we will need to declare our basis vectors as a matrix and then apply it to vector $\\vec{v}$ using the dot() operator (Example 4-7). The dot () operator will perform this scaling and addition between our matrix and vector as we just described. This is known as the dot product, and we will explore it throughout this chapter.\n\n# 线性代数代考\n\n## 数学代写|线性代数代写linear algebra代考| Matrix Vector Multiplication\n\n$\\left[x_{n e w} y_{\\text {new }}\\right]=\\left[\\begin{array}{lll}a & b c & d\\end{array}\\right][x y] \\quad\\left[x_{\\text {new }} y_{\\text {new }}\\right]=[a x+b y c x+d y]$\n$\\hat{i}$ 是第一列 $[a, c]$ 和 $\\hat{j}$ 是列 $[b, d]$.㧴们将这两个基向量打包为一个矩阵，该矩阵又是一个向\n\n## 有限元方法代写\n\ntatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。\n\n## MATLAB代写\n\nMATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。\n\nDays\nHours\nMinutes\nSeconds\n\n# 15% OFF\n\n## On All Tickets\n\nDon’t hesitate and buy tickets today – All tickets are at a special price until 15.08.2021. Hope to see you there :)" ]

[ null, "https://couryes.com/wp-content/uploads/2023/08/speaker-2-297x300.png", null ]

[ "# How to Calculate the Annual Growth Rate for Real GDP\n\nShare It\n\nThe Gross Domestic Product (GDP) for a country is a total market value of all domestically produced goods and services. The GDP growth rate indicates the current growth trend of the economy. When calculating GDP growth rates, the U.S. Bureau of Economic Analysis uses real GDP, which equalizes the actual figures to filter out the effects of inflation. Using real GDP allows you to compare previous years without inflation affecting the results.\n\nLook up the real GDP for two consecutive years. These figures are found on the U.S. Department of Commerce Bureau of Economic Analysis' website.\n\nSubtract the first year's real GDP from the second year's GDP. As an example, the real GDP in the U.S. for 2009 and 2010 were \\$12.7 trillion and \\$13.1 trillion, respectively. Subtracting the 2009 figure from the 2010 figure results in a difference of \\$384.9 billion.\n\nDivide this difference by the first year's read GDP. In the example, you would divide \\$354.9 billion by \\$12.7 trillion, which gives you an annual growth rate of 0.030, or 3 percent." ]

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[ "# An intuitive derivation of Taylor polynomial coefficients\n\nI'd like to introduce Taylor polynomials by generalizing the linear approximation of a function $f(x)$ to a quadratic approximation.\n\nThe linear approximation formula $L(x)=f(a)+f'(a)(x-a)$ of $f(x)$ near $a$ can be seen to be the point-slope formula of a line passing through the point $(a,f(a))$ with slope $f'(a)$.\n\nNow, I'd like to derive the formula for a quadratic approximation of $f$ near $a$. Let this quadratic approximation be denoted by $T_2(x)$, and require that it satisfy the following $3$ conditions:\n\n1. $T_2(a)=f(a)$\n2. $T_2'(a)=f'(a)$\n3. $T_2''(a)=f''(a)$\n\nWhat is a nice way to show that $T_2(x)$ is what we know it to be without just writing it down and showing it has the desired properties? How can I motivate the coefficients? In other words, we don't really have a well-known point-slope-concavity\" formula for parabolas that generalizes the point-slope formula we used to write down the linear approximation. I know I could start with a general form of $T_2(x)=Ax^2+Bx+C$, and use the above conditions to compute the coefficients, but for $a\\ne 0$ this will be messy and unintuitive. If I could somehow motivate the switch to consider the form $T_2(x)=A(x-a)^2+B(x-a)+C$, that would be better, but I'm unsure how to convince the students that this is a natural form to consider.\n\nThe best I've been able to do is start off assuming that $T_2(x)$ is of the form $$T_2(x)=f(a)+f'(a)(x-a)+c(x-a)^2$$ for some constant $c$ but I think this is too much assuming too soon. Sure we can test that this form will satisfy $1$ and $2$ above, and that $c$ gives us the flexibility to specify $T_2''(a)$ so as to satisfy $3$, but this is still unmotivating to me. Why did I think this would work? It seems I've just written down what I know will work, and have only left the challenge of finding the right $c$ (which is just a computation and gives no intuition).\n\nHow can I motivate the form for $T_2(x)$ I've started with? Or is there another way to do this? I'd like to avoid viewing Taylor polynomials as truncated Taylor series, as I'm hoping to introduce Taylor polynomials before discussing infinite and power series.\n\nOf course, I'm hoping that if I can intuitively motivate the derivation of $T_2(x)$, then a generalization to $T_n(x)$ will be clear.\n\n• It seems to me that the simplest approach would be to assume the form you want and go from there (without worrying about rigor, since I assume you're looking for an algebraic \"permanence of form\" argument). That is, start with $f(x) = a_0 + a_1x + a_{2}x^2 + \\cdots$ First, plug in $x=0$ to both sides to get $a_0.$ Then differentiate both sides and plug in $x=0$ to both sides to get $a_1.$ Then differentiate both sides again and plug in $x=0$ to both sides to get $a_2.$ Continue in this manner . . . – Dave L Renfro Feb 20 '15 at 17:26\n• matheducators.stackexchange.com/questions/6135/… Is this in any way useful? – Shai Feb 25 '15 at 8:09\n• My teacher used physics, and talked about acceleration. I don't remember the details, but it made me understand perfectly! – geodude Mar 4 '15 at 8:47\n\nIt seems to me that the goal here is to exemplify that a higher degree polynomial is capable of providing a higher fidelity approximation. I think this is a good goal, but I don't think that there's a lot of value in pursuing a 'clean' process of producing the parabola. Your thoughts on that line are good, but will be difficult for most students to follow and more trouble than it's worth to instruct given that the learned technique is essentially a one-off for exploratory purposes.\n\nOn the other hand, this is an opportunity for a bit of cross pollination. Students are likely to be able to understand the motivation behind the three constraints $T(a) = f(a)$, $T'(a) = f'(a),$ and $T''(a) = f''(a)$, especially when compared with the constraints that produce the linear approximation at a point. They're also (very) likely to be familiar with the general $Ax^2 + Bx + C$ form of a parabola. I'd suggest working directly from here by solving the three-variable, three-equation system. This gives an understandable, workable method for producing 2nd order approximations of continuous functions at a point. An in-class worked example or two comparing the linear and quadratic approximation, paired with a homework question or two doing the same, is likely to meet your goals, eg, providing a bridge to understanding Taylor series.\n\n• I'm not sure if this was clear, but the great side effect here is that students get to exercise the skill of solving a system in a completely legitimate, non-contrived context. Its use is systemically required to achieve a reasonable, understandable goal. Much better than 'solve these systems' on a homework sheet. – NiloCK Feb 21 '15 at 19:00\n\nMotivation is a slippery goal. What does your audience know? What do they appreciate? These questions are central. I advocate the view that your audience probably knows more about algebra than geometry. What is the geometry of $f^{(n)}(x)$? Will connecting to that help? Perhaps, I give a constructive approach a bit further down in this answer.\n\nI tend to think your approach has more motivation than you give it credit. The key idea is that the coefficient of the expansion has to do with a single power of the derivative provided we're talking about the center point.\n\nContrast, $$y = mx+b \\qquad \\& \\qquad y=y_o+m(x-a)$$ the point-slope formula is explicitly connected to the point whereas the slope-intercept is implicitly connected to the intercept $(0,b)$. Next, you prove a simple lemma to show $f(x)=c_o+c_1(x-a)+c_2(x-a)^2+ \\cdots$ once more connects the formula of the function to data $f'(a),f''(a)$ etc.. about the center point. The advantage of the expansion in $(x-a)$ verses $x$ is that the expansion in $(x-a)$ is localized at $a$. Of course, the expansion in $x$ is localized at $x=0$. So, why the $1/n!$ factors?\n\nHere I might try a boot-strap approach. Let me explain the first step. For the linearization, a nice alternative is the theorem of Caratheodory: $f'(a)$ exists iff there is a continuous function $\\phi$ for which $$f(x) = f(a)+\\phi(x)(x-a)$$ and $\\phi(a)=f'(a)$ (I assume $f$ is defined on $\\mathbb{R}$ for brevity). If $f$ is twice differentiable then $\\phi$ must be differentiable at $a$ hence: \\begin{align} f''(x) &= \\frac{d}{dx}\\frac{d}{dx} \\left[f(a)+\\phi(x)(x-a) \\right]\\\\ &= \\frac{d}{dx} \\left[\\phi'(x)(x-a)+\\phi(x)\\right] \\\\ &= \\phi''(x)(x-a)+2\\phi'(x). \\end{align} Setting $x=a$ in the above reveals $f''(a)=2\\phi'(a)$. Interesting. On the flip-side, we can apply Caratheodory's Theorem to $\\phi$ at $a$ to find $\\psi$ such that $$\\phi(x)=\\phi(a)+\\psi(x)(x-a) = f'(a)+\\psi(x)(x-a)$$ where $\\phi'(a)=\\psi(a)$ and so $f''(a)=2\\psi(a)$. Likewise, if triply differentiable there exists $\\lambda$ such that $\\psi(x)= \\psi(a)+\\lambda(x)(x-a) = \\frac{1}{2}f''(a)+\\lambda(x)(x-a)$. Nesting these gives: \\begin{align} f(x) &= f(a)+\\phi(x)(x-a) \\\\ &= f(a)+\\left[f'(a)+\\psi(x)(x-a)\\right](x-a) \\\\ &= f(a)+ \\left[f'(a)+\\left(\\frac{1}{2}f''(a)+\\lambda(x)(x-a)\\right)(x-a)\\right](x-a) \\\\ &= f(a)+f'(a)(x-a)+\\frac{1}{2}f''(a)(x-a)^2+\\lambda(x)(x-a)^3. \\end{align} If there is interest, I could attempt to show inductively how this explains the $1/n!$ in the $n$-th coefficient. That said, I think your approach is more appropriate for the second-semester calculus course. I do love these formulas for their exactness. All the nasty remainder is tied up in the function $\\lambda(x)$ rather than some fuzzy $+\\cdots$. I am no expert, but I think this has connection to Morse's Lemma.\n\nTo motivate the writing of $T_2$ under the form $A+B(x-a)+C(x-a)^2$ (note the change of order), you can explain that the formula you seek is an approximation formula at $a$, and that $(x-a)$ is precisely the gap between the central value $a$ and the considered point $x$.\n\nThe zeroth approximation, given that $f$ is continuous, is to simply approximate $f(x)$ by a constant $A$ when $x$ is near $a$; we want the approximation to be perfect if $x$ is equal to $a$, so that we pick $A=f(a)$ (at this point you can discuss the merits of continuity if needed).\n\nThe first-order approximation naturally takes the form $A+B(x-a)$ because we want a linear approximation, and we want to compare the error made in the zeroth approximation with the gap between $x$ and $a$. (For example, $a$ can be the imperfect measurement of a physical quantity $x$: can we estimate the error $f(x)-f(a)$ in terms of $x-a$?). It also backs up nicely with the formula for the derivative. At this point it is easy to see that when $f$ is derivable, the best choice for $A$ is still $f(a)$ and the best choice for $B$ is $f'(a)$.\n\nThen, if one has to take this process one step further, using $(x-a)$ as a gap, it makes much sense to try to approximate $f$ near $a$ by a second-order polynomial in the gap (x-a). This happens to be a second-order polynomial in $x$ also, but this is mostly a happy coincidence.\n\nOne benefit of this approach is that you have a good opportunity to stress that when $(x-a)$ is small (in particular less than $1$ in absolute value), $(x-a)^2$ is even smaller. One difficulty in this business is to remove this idea that squares are necessarily bigger.\n\nWhy not just pose the problem of how best to approximate functions locally by polynomial functions? (In more general terms, that of asymptotic expansions.)\n\nThat is, get $f(x_{0}+h)=\\sum_{0}^{\\infty}A_{i}(x_{0})h^{i}$ and just truncate at $k$ with the remainder then being o[$h^{k}$]. Then, you can say that $f$ is $k$-differentiable iff $f$ has a best polynomial approximation of degree $k$ and define $f^{(k)}$ to be the function which for $x_{0}$ returns $A_{k}(x_{0})k!$\n\nThe reason for $k!$ is to make derivatives recursively computable: $f^{(k+1)}=f^{(k)\\prime}$.\n\nThis is in fact what Lagrange did. (But much misunderstood by people who confuse series and asymptotic expansions.)\n\n• But a function as $x\\mapsto x^{12} \\sin(x^{-11})$ (extended by $0$ at $0$) has a Taylor series of order $11$ at each point, but is not even $C^1$. – Benoît Kloeckner Mar 4 '15 at 13:29\n• @Benoit Kloeckner The existence of an osculating polynomial of degree $>1$ does not of course ensure the existence of any derivative of order $>1$. E.g., for $f(x)=x^{3}\\sin\\frac{1}{x}, x\\neq0, 0$ otherwise, $f(h)= 0+0h+0h^{2}+o[h^{2}]$ but $f$ does not have a second derivative at $0$. On the other hand, given the OP, that the asymptotic expansion gives the curvature near $0$ seems rather to the point. – schremmer Mar 6 '15 at 15:42\n• But then how can you suggest to define higher differentiability by best polynomial approximation? – Benoît Kloeckner Mar 6 '15 at 15:51\n• $f$ is $n$-differentiable at $x_{{0}}$ iff $f$ has a Best Polynomial Approximation of degree $n$ near $x_{{0}}$. I am not sure what the issue is. – schremmer Mar 6 '15 at 16:19\n• so you call $n$-differentiable a function that does not admit a $n$-th derivative? I confess I'm lost. – Benoît Kloeckner Mar 6 '15 at 18:10\n\nGiven $f\\colon I\\to \\mathbb R$ and $x\\in\\mathrm{int}\\, I$, you could construct the unique second degree polynomial $P_\\varepsilon$ that passes by the points $(x-\\varepsilon, f(x-\\varepsilon))$, $(x, f(x))$ and $(x+\\varepsilon, f(x+\\varepsilon))$ (Lagrange interpolation) and then let $\\varepsilon \\to 0$. This seems the most natural approach generalizing the idea of a tangent line and if you argue $f''(x)$ exists, it is not hard to see that $\\displaystyle\\lim_{\\varepsilon\\to 0}P_\\varepsilon=T_2$.\n\nAnother interesting thing to say is that a function may be well-approximated by a second degree polynomial in the above sense but not be twice differentiable; check $1+x+x^2+x^3\\sin(1/x)$ out around zero.\n\nYou may extend the interpolation idea to higher degrees with more work, but the computations get messy fast. The idea of constructing the second degree polynomial that passes by the points $(x+\\delta, f(x+\\delta))$, $(x, f(x))$ and $(x+\\varepsilon, f(x+\\varepsilon))$ and then letting $(\\delta,\\varepsilon)\\to (0,0)$ is also probably too cumbersome.\n\n• Just want to comment that in a typical (large, American at least) University, using $\\varepsilon$ approximations, although clear to us, will be over the heads of many calculus students who see this for the first time. But overall, interesting approach. – Chris C Mar 4 '15 at 22:03" ]

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[ "# Algorithmic Methods in Combinatorial Algebra\n\nUniversity dissertation from Centre for Mathematical Sciences, Box 118, SE-221 00 Lund\n\nAbstract: This thesis consists of a collection of articles all using and/or developing algorithmic methods for the investigation of different algebraic structures. Part A concerns orthogonal decompositions of simple Lie algebras. The main result of this part is that the symplectic Lie algebra C3 has no orthogonal decomposition of so called monomial type. This was achieved by developing an algorithm for finding all monomial orthogonal decompositions and implementing it in Maple. In part B we study subalgebras on two generators of the univariate polynomial ring and the semigroups of degrees associated to such subalgebras. The generators f and g of the subalgebra constitute a so called SAGBI basis if and only if the semigroup of degrees is generated by deg(f) and deg(g). We show that this occurs exactly when both the generators are contained in a subalgebra k[h] for some polynomial h of degree equal to the greatest common divisor of the degrees of f and g. In particular this is the case whenever the degrees of f and g are relatively prime. There is an algorithmic test to check if a set polynomials constitute a SAGBI basis and our proof is by showing that the condition of this test is satisfied. We present two ways of proving this. The first one uses the fact that g is integral over k[f] and therefore satisfies a polynomial equation over k[f], while the second one gives this equation explicitly as a resultant related to f and g. Part C of the thesis is about maximal symmetry groups of hyperbolic three-manifolds. Those are groups of orientation preserving isometries of three-dimensional hyperbolic manifolds that are of maximal order in relation to the volume of the manifold. One can show that maximal symmetry groups are the quotients by normal torsion free subgroups of a certain finitely presented group. We use different computational methods to find such quotients. Our main results are the following: PGL(2,9) is the smallest maximal symmetry group, and for each prime p there is some prime power q=pk such that either PSL(2,q) or PGL(2,q) is a maximal symmetry group, and all but finitely many alternating and symmetric groups are maximal symmetry groups.", null, "This dissertation MIGHT be available in PDF-format. Check this page to see if it is available for download.", null, "" ]

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[ "## 255334000000001\n\n255,334,000,000,001 (two hundred fifty-five trillion three hundred thirty-four billion one) is an odd fifteen-digits composite number following 255334000000000 and preceding 255334000000002. In scientific notation, it is written as 2.55334000000001 × 1014. The sum of its digits is 23. It has a total of 2 prime factors and 4 positive divisors. There are 235,692,923,076,912 positive integers (up to 255334000000001) that are relatively prime to 255334000000001.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 15\n• Sum of Digits 23\n• Digital Root 5\n\n## Name\n\nShort name 255 trillion 334 billion 1 two hundred fifty-five trillion three hundred thirty-four billion one\n\n## Notation\n\nScientific notation 2.55334000000001 × 1014 255.334000000001 × 1012\n\n## Prime Factorization of 255334000000001\n\nPrime Factorization 13 × 19641076923077\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 255334000000001 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 255,334,000,000,001 is 13 × 19641076923077. Since it has a total of 2 prime factors, 255,334,000,000,001 is a composite number.\n\n## Divisors of 255334000000001\n\n4 divisors\n\n Even divisors 0 4 4 0\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 2.74975e+14 Sum of all the positive divisors of n s(n) 1.96411e+13 Sum of the proper positive divisors of n A(n) 6.87438e+13 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.59792e+07 Returns the nth root of the product of n divisors H(n) 3.71429 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 255,334,000,000,001 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 255,334,000,000,001) is 274,975,076,923,092, the average is 68,743,769,230,773.\n\n## Other Arithmetic Functions (n = 255334000000001)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 235692923076912 Total number of positive integers not greater than n that are coprime to n λ(n) 58923230769228 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7942530460867 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares\n\nThere are 235,692,923,076,912 positive integers (less than 255,334,000,000,001) that are coprime with 255,334,000,000,001. And there are approximately 7,942,530,460,867 prime numbers less than or equal to 255,334,000,000,001.\n\n## Divisibility of 255334000000001\n\n m n mod m 2 3 4 5 6 7 8 9 1 2 1 1 5 6 1 5\n\n255,334,000,000,001 is not divisible by any number less than or equal to 9.\n\n## Classification of 255334000000001\n\n• Arithmetic\n• Semiprime\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n\n## Base conversion (255334000000001)\n\nBase System Value\n2 Binary 111010000011100110010100010111100111110000000001\n3 Ternary 1020111001200200110211102200212\n4 Quaternary 322003212110113213300001\n5 Quinary 231431343013000000001\n6 Senary 2303014453102323505\n8 Octal 7203462427476001\n10 Decimal 255334000000001\n12 Duodecimal 24779533b89595\n20 Vigesimal 14idjdf00001\n36 Base36 2iiatj2kn5\n\n## Basic calculations (n = 255334000000001)\n\n### Multiplication\n\nn×i\n n×2 510668000000002 766002000000003 1021336000000004 1276670000000005\n\n### Division\n\nni\n n⁄2 1.27667e+14 8.51113e+13 6.38335e+13 5.10668e+13\n\n### Exponentiation\n\nni\n n2 65195451556000510668000000001 16646615427599899586354668000766002000000001 4250446903590809407597710399207172709336001021336000000001 1085283609681459979726457377880571834275997691954515560001276670000000001\n\n### Nth Root\n\ni√n\n 2√n 1.59792e+07 63440.9 3997.4 761.065\n\n## 255334000000001 as geometric shapes\n\n### Circle\n\n Diameter 5.10668e+14 1.60431e+15 2.04818e+29\n\n### Sphere\n\n Volume 6.97292e+43 8.1927e+29 1.60431e+15\n\n### Square\n\nLength = n\n Perimeter 1.02134e+15 6.51955e+28 3.61097e+14\n\n### Cube\n\nLength = n\n Surface area 3.91173e+29 1.66466e+43 4.42251e+14\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 7.66002e+14 2.82305e+28 2.21126e+14\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.12922e+29 1.96182e+42 2.08479e+14\n\n## Cryptographic Hash Functions\n\nmd5 1400e82b1a98f5d79a47ce80f5652150 4df576b1d1e7da98fb95b46712262e57937fa2f7 1731cbc0aa8475714f6294209fcee6faa53f0178d94a9b17d9f46e578fe328da 2679a69df0d92a5a97995b3fe63919a3b108af2a9339b31528957c0870717b541e6285db6233c3d13db565c2bfd6c2dff44b2f96a7db5c88fadb8959522a76dc f41fb1e0813f49e13c772bce95a46d3b08c5f690" ]

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[ "# Does dilution of a buffer affect pH?\n\nThe Henderson-Hasselbalch equation for the $\\mathrm{pH}$ of a buffer solution of the monoprotic acid $\\ce{HA}$ is given by $$\\mathrm{pH}=\\mathrm pK_\\mathrm a+\\log{\\frac{[\\ce{A-}]}{[\\ce{HA}]}}$$ Since concentration appears in both the numerator and denominator of the fraction $\\frac{[\\ce{A-}]}{[\\ce{HA}]}$ and $\\mathrm pK_\\mathrm a$ is constant (at a fixed temperature), it appears that dilution of the solution with pure $\\ce{H2O}$ would not change the $\\mathrm{pH}$. However, since $$\\mathrm{pH}=-\\log{[\\ce{H+}]}$$ the amount of substance of $\\ce{H+}$ must increase in order for $\\mathrm{pH}$ to stay constant upon dilution.\n\nWhere is this additional $\\ce{H+}$ coming from? I know that diluting an acid causes it to dissociate to a greater extent. But at the same time, you would be diluting its conjugate base and causing it to associate more, cancelling the dissociation of the acid.\n\n• Your intuition is right. Strictly speaking, dilution does affect the pH of the buffer because it affects the position of the equilibrium $\\ce{HA + H2O <=> A- + H3O+}$. However the effect is really very small which is why it is commonly said that the pH is unchanged. – orthocresol Sep 7 '16 at 7:48\n• – JM97 Sep 7 '16 at 11:53\n\n## 1 Answer\n\nIn the Henderson-Hasselbalch equation, $K_\\mathrm{a}$ is a product of concentrations and considered a constant.\n\nIn reality, $K_\\mathrm{a}$, when defined as a product of concentrations, is not a constant:", null, "Upon dilution (decrease in ionic strength) the $\\mathrm{p}K_\\mathrm{a}$ will change, and therefore the pH of the solution will change.\n\nIn addition to the above reason, pH will always approach 7 at extreme dilution as it approaches being pure water.\n\n• So if I did dilute a buffer solution with water, in a 10 to 1 ratio, would that affect the pH or not? I'm still unable to understand... My book says that the pH will remain unchanged, but as per your answer, the pH should change, and approach 7 – AbhigyanC Sep 16 '17 at 1:24\n• @Abhihyan it won't charge much, but it could change on the order of 0.1 pH units – DavePhD Sep 16 '17 at 1:29\n• So basically, there is a pH change, but the change is negligible... That's what I understand... Is that right? (Also, it's Abhigyan :) ) – AbhigyanC Sep 17 '17 at 6:30\n• @Abhigyan if you only dilute by a factor of 10, and you consider changes on the order of 0.1 units negligible, then it's negligible. – DavePhD Sep 17 '17 at 13:43\n• thanks so much... that really clears things up for me, and hope it does the same for many people to come... – AbhigyanC Sep 17 '17 at 13:57" ]

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[ "# How to calculate a pump's operating point for a fully open control valve\n\n## Summary:\n\nHow to calculate the flow rate and pressure of a centrifugal pump through a fully open valve.\n\n## Main Question or Discussion Point\n\nHow do you calculate the flow rate and differential pressure (the operating point on the pump curve) for a centrifugal pump if all of the flow is through a single control valve with known $C_V$, discharging to atmosphere?\n\nClearly the flow rate and differential pressure of the pump will be some point on the pump curve to match the flow and pressure drop across the valve. As the $C_V$ increases, that point will move further down the curve i.e. more flow at less pressure drop. It is easy enough to calculate the flow rate through a valve for a known pressure drop and $C_V$, but if only the $C_V$ is known I struggle to work this out.\n\nThanks\n\nRelated Materials and Chemical Engineering News on Phys.org\nThe traditional way to do this is graphically. You have a pump curve from the manufacturer and can plot a 'system curve' for your valve (at any single fixed Cv). Where they intersect is your operating point. It is possible to 'calculate' the intersection, but you'll need to come up with a function to describe the pump curve. Don't forget any other significant (plumbing, filters, etc) losses in your system curve.\n\n•", null, "fonz, ChemAir and russ_watters\nThanks for the reply that is very helpful. On most system curves I have seen the friction factor is used or K factor for fittings etc. Do I just add the head loss due to Cv directly from the flow rate equation for a valve or do I need to convert to a K factor first?" ]

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[ "# Operations on COO arrays¶\n\n## Operators¶\n\nCOO objects support a number of operations. They interact with scalars, Numpy arrays, other COO objects, and scipy.sparse.spmatrix objects, all following standard Python and Numpy conventions.\n\nFor example, the following Numpy expression produces equivalent results for both Numpy arrays, COO arrays, or a mix of the two:\n\nnp.log(X.dot(beta.T) + 1)\n\n\nHowever some operations are not supported, like operations that implicitly cause dense structures, or numpy functions that are not yet implemented for sparse arrays.\n\nnp.svd(x) # sparse svd not implemented\n\n\n### elemwise¶\n\nThis function allows you to apply any arbitrary broadcasting function to any number of arguments where the arguments can be SparseArray objects or scipy.sparse.spmatrix objects. For example, the following will add two arrays:\n\nsparse.elemwise(np.add, x, y)\n\n\nWarning\n\nPreviously, elemwise was a method of the COO class. Now, it has been moved to the sparse module.\n\n### Auto-Densification¶\n\nOperations that would result in dense matrices, such as operations with Numpy arrays raises a ValueError. For example, the following will raise a ValueError if x is a numpy.ndarray:\n\nx + y\n\n\nHowever, all of the following are valid operations.\n\nx + 0\nx != y\nx + y\nx == 5\n5 * x\nx / 7.3\nx != 0\nx == 0\n~x\nx + 5\n\n\nWe also support operations with a nonzero fill value. These are operations that map zero values to nonzero values, such as x + 1 or ~x. In these cases, they will produce an output with a fill value of 1 or True, assuming the original array has a fill value of 0 or False respectively.\n\nIf densification is needed, it must be explicit. In other words, you must call COO.todense on the COO object. If both operands are COO, both must be densified.\n\n### Operations with NumPy arrays¶\n\nIn certain situations, operations with NumPy arrays are also supported. For example, the following will work if x is COO and y is a NumPy array:\n\nx * y\n\n\nThe following conditions must be met when performing element-wise operations with NumPy arrays:\n\n• The operation must produce a consistent fill-values. In other words, the resulting array must also be sparse.\n• Operating on the NumPy arrays must not increase the size when broadcasting the arrays.\n\n## Operations with scipy.sparse.spmatrix¶\n\nCertain operations with scipy.sparse.spmatrix are also supported. For example, the following are all allowed if y is a scipy.sparse.spmatrix:\n\nx + y\nx - y\nx * y\nx > y\nx < y\n\n\nIn general, if operating on a scipy.sparse.spmatrix is the same as operating on COO, as long as it is to the right of the operator.\n\nNote\n\nResults are not guaranteed if x is a scipy.sparse.spmatrix. For this reason, we recommend that all Scipy sparse matrices should be explicitly converted to COO before any operations.\n\nAll binary operators support broadcasting. This means that (under certain conditions) you can perform binary operations on arrays with unequal shape. Namely, when the shape is missing a dimension, or when a dimension is 1. For example, performing a binary operation on two COO arrays with shapes (4,) and (5, 1) yields an object of shape (5, 4). The same happens with arrays of shape (1, 4) and (5, 1). However, (4, 1) and (5, 1) will raise a ValueError.\n\n## Element-wise Operations¶\n\nCOO arrays support a variety of element-wise operations. However, as with operators, operations that map zero to a nonzero value are not supported.\n\nTo illustrate, the following are all possible, and will produce another COO array:\n\nnp.abs(x)\nnp.sin(x)\nnp.sqrt(x)\nnp.conj(x)\nnp.expm1(x)\nnp.log1p(x)\nnp.exp(x)\nnp.cos(x)\nnp.log(x)\n\n\nAs above, in the last three cases, an array with a nonzero fill value will be produced.\n\nNotice that you can apply any unary or binary numpy.ufunc to COO arrays, and numpy.ndarray objects and scalars and it will work so long as the result is not dense. When applying to numpy.ndarray objects, we check that operating on the array with zero would always produce a zero.\n\n## Reductions¶\n\nCOO objects support a number of reductions. However, not all important reductions are currently implemented (help welcome!) All of the following currently work:\n\nx.sum(axis=1)\nnp.max(x)\nnp.min(x, axis=(0, 2))\nx.prod()\n\n\nNote\n\nIf you are performing multiple reductions along the same axes, it may be beneficial to call COO.enable_caching.\n\n### COO.reduce¶\n\nThis method can take an arbitrary numpy.ufunc and performs a reduction using that method. For example, the following will perform a sum:\n\nx.reduce(np.add, axis=1)\n\n\nNote\n\nThis library currently performs reductions by grouping together all coordinates along the supplied axes and reducing those. Then, if the number in a group is deficient, it reduces an extra time with zero. As a result, if reductions can change by adding multiple zeros to it, this method won’t be accurate. However, it works in most cases.\n\n### Partial List of Supported Reductions¶\n\nAlthough any binary numpy.ufunc should work for reductions, when calling in the form x.reduction(), the following reductions are supported:\n\n## Indexing¶\n\nCOO arrays can be indexed just like regular numpy.ndarray objects. They support integer, slice and boolean indexing. However, currently, numpy advanced indexing is not properly supported. This means that all of the following work like in Numpy, except that they will produce COO arrays rather than numpy.ndarray objects, and will produce scalars where expected. Assume that z.shape is (5, 6, 7)\n\nz\nz[1, 3]\nz[1, 4, 3]\nz[:3, :2, 3]\nz[::-1, 1, 3]\nz[-1]\n\n\nAll of the following will raise an IndexError, like in Numpy 1.13 and later.\n\nz\nz[3, 6]\nz[1, 4, 8]\nz[-6]\n\n\nAdvanced indexing (indexing arrays with other arrays) is supported, but only for indexing with a single array. Indexing a single array with multiple arrays is not supported at this time. As above, if z.shape is (5, 6, 7), all of the following will work like NumPy:\n\nz[[0, 1, 2]]\nz[1, ]\nz[1, 4, [3, 6]]\nz[:3, :2, [1, 5]]\n\n\n## Package Configuration¶\n\nBy default, when performing something like np.array(COO), we allow the array to be converted into a dense one. To prevent this and raise a RuntimeError instead, set the environment variable SPARSE_AUTO_DENSIFY to 0.\n\nIf it is desired to raise a warning if creating a sparse array that takes no less memory than an equivalent desne array, set the environment variable SPARSE_WARN_ON_TOO_DENSE to 1.\n\n## Other Operations¶\n\nCOO arrays support a number of other common operations. Among them are dot, tensordot, concatenate and stack, transpose and reshape. You can view the full list on the API reference page.\n\nNote\n\nSome operations require zero fill-values (such as nonzero) and others (such as concatenate) require that all inputs have consistent fill-values. For details, check the API reference." ]

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[ "# #00ac81 Color Information\n\nIn a RGB color space, hex #00ac81 is composed of 0% red, 67.5% green and 50.6% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 25% yellow and 32.5% black. It has a hue angle of 165 degrees, a saturation of 100% and a lightness of 33.7%. #00ac81 color hex could be obtained by blending #00ffff with #005903. Closest websafe color is: #009999.\n\n• R 0\n• G 67\n• B 51\nRGB color chart\n• C 100\n• M 0\n• Y 25\n• K 33\nCMYK color chart\n\n#00ac81 color description : Dark cyan.\n\n# #00ac81 Color Conversion\n\nThe hexadecimal color #00ac81 has RGB values of R:0, G:172, B:129 and CMYK values of C:1, M:0, Y:0.25, K:0.33. Its decimal value is 44161.\n\nHex triplet RGB Decimal 00ac81 `#00ac81` 0, 172, 129 `rgb(0,172,129)` 0, 67.5, 50.6 `rgb(0%,67.5%,50.6%)` 100, 0, 25, 33 165°, 100, 33.7 `hsl(165,100%,33.7%)` 165°, 100, 67.5 009999 `#009999`\nCIE-LAB 62.582, -47.839, 11.754 18.713, 31.088, 25.782 0.248, 0.411, 31.088 62.582, 49.262, 166.196 62.582, -52.669, 23.739 55.757, -37.665, 11.614 00000000, 10101100, 10000001\n\n# Color Schemes with #00ac81\n\n• #00ac81\n``#00ac81` `rgb(0,172,129)``\n• #ac002b\n``#ac002b` `rgb(172,0,43)``\nComplementary Color\n• #00ac2b\n``#00ac2b` `rgb(0,172,43)``\n• #00ac81\n``#00ac81` `rgb(0,172,129)``\n• #0081ac\n``#0081ac` `rgb(0,129,172)``\nAnalogous Color\n• #ac2b00\n``#ac2b00` `rgb(172,43,0)``\n• #00ac81\n``#00ac81` `rgb(0,172,129)``\n• #ac0081\n``#ac0081` `rgb(172,0,129)``\nSplit Complementary Color\n• #ac8100\n``#ac8100` `rgb(172,129,0)``\n• #00ac81\n``#00ac81` `rgb(0,172,129)``\n• #8100ac\n``#8100ac` `rgb(129,0,172)``\nTriadic Color\n• #2bac00\n``#2bac00` `rgb(43,172,0)``\n• #00ac81\n``#00ac81` `rgb(0,172,129)``\n• #8100ac\n``#8100ac` `rgb(129,0,172)``\n• #ac002b\n``#ac002b` `rgb(172,0,43)``\nTetradic Color\n• #006048\n``#006048` `rgb(0,96,72)``\n• #00795b\n``#00795b` `rgb(0,121,91)``\n• #00936e\n``#00936e` `rgb(0,147,110)``\n• #00ac81\n``#00ac81` `rgb(0,172,129)``\n• #00c694\n``#00c694` `rgb(0,198,148)``\n• #00dfa7\n``#00dfa7` `rgb(0,223,167)``\n• #00f9ba\n``#00f9ba` `rgb(0,249,186)``\nMonochromatic Color\n\n# Alternatives to #00ac81\n\nBelow, you can see some colors close to #00ac81. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #00ac56\n``#00ac56` `rgb(0,172,86)``\n• #00ac64\n``#00ac64` `rgb(0,172,100)``\n• #00ac73\n``#00ac73` `rgb(0,172,115)``\n• #00ac81\n``#00ac81` `rgb(0,172,129)``\n• #00ac8f\n``#00ac8f` `rgb(0,172,143)``\n• #00ac9e\n``#00ac9e` `rgb(0,172,158)``\n• #00acac\n``#00acac` `rgb(0,172,172)``\nSimilar Colors\n\n# #00ac81 Preview\n\nText with hexadecimal color #00ac81\n\nThis text has a font color of #00ac81.\n\n``<span style=\"color:#00ac81;\">Text here</span>``\n#00ac81 background color\n\nThis paragraph has a background color of #00ac81.\n\n``<p style=\"background-color:#00ac81;\">Content here</p>``\n#00ac81 border color\n\nThis element has a border color of #00ac81.\n\n``<div style=\"border:1px solid #00ac81;\">Content here</div>``\nCSS codes\n``.text {color:#00ac81;}``\n``.background {background-color:#00ac81;}``\n``.border {border:1px solid #00ac81;}``\n\n# Shades and Tints of #00ac81\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000f0b is the darkest color, while #fafffe is the lightest one.\n\n• #000f0b\n``#000f0b` `rgb(0,15,11)``\n• #00231a\n``#00231a` `rgb(0,35,26)``\n• #003629\n``#003629` `rgb(0,54,41)``\n• #004a37\n``#004a37` `rgb(0,74,55)``\n• #005e46\n``#005e46` `rgb(0,94,70)``\n• #007155\n``#007155` `rgb(0,113,85)``\n• #008564\n``#008564` `rgb(0,133,100)``\n• #009872\n``#009872` `rgb(0,152,114)``\n• #00ac81\n``#00ac81` `rgb(0,172,129)``\n• #00c090\n``#00c090` `rgb(0,192,144)``\n• #00d39e\n``#00d39e` `rgb(0,211,158)``\n• #00e7ad\n``#00e7ad` `rgb(0,231,173)``\n• #00fabc\n``#00fabc` `rgb(0,250,188)``\nShade Color Variation\n• #0fffc3\n``#0fffc3` `rgb(15,255,195)``\n• #23ffc8\n``#23ffc8` `rgb(35,255,200)``\n• #36ffcd\n``#36ffcd` `rgb(54,255,205)``\n• #4affd2\n``#4affd2` `rgb(74,255,210)``\n• #5effd7\n``#5effd7` `rgb(94,255,215)``\n• #71ffdc\n``#71ffdc` `rgb(113,255,220)``\n• #85ffe0\n``#85ffe0` `rgb(133,255,224)``\n• #98ffe5\n``#98ffe5` `rgb(152,255,229)``\n• #acffea\n``#acffea` `rgb(172,255,234)``\n• #c0ffef\n``#c0ffef` `rgb(192,255,239)``\n• #d3fff4\n``#d3fff4` `rgb(211,255,244)``\n• #e7fff9\n``#e7fff9` `rgb(231,255,249)``\n• #fafffe\n``#fafffe` `rgb(250,255,254)``\nTint Color Variation\n\n# Tones of #00ac81\n\nA tone is produced by adding gray to any pure hue. In this case, #4f5d59 is the less saturated color, while #00ac81 is the most saturated one.\n\n• #4f5d59\n``#4f5d59` `rgb(79,93,89)``\n• #49635d\n``#49635d` `rgb(73,99,93)``\n• #426a60\n``#426a60` `rgb(66,106,96)``\n• #3c7063\n``#3c7063` `rgb(60,112,99)``\n• #357767\n``#357767` `rgb(53,119,103)``\n• #2e7e6a\n``#2e7e6a` `rgb(46,126,106)``\n• #28846d\n``#28846d` `rgb(40,132,109)``\n• #218b70\n``#218b70` `rgb(33,139,112)``\n• #1a9274\n``#1a9274` `rgb(26,146,116)``\n• #149877\n``#149877` `rgb(20,152,119)``\n• #0d9f7a\n``#0d9f7a` `rgb(13,159,122)``\n• #07a57e\n``#07a57e` `rgb(7,165,126)``\n• #00ac81\n``#00ac81` `rgb(0,172,129)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00ac81 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ "# TriangleWave\n\nTriangleWave[x]\n\ngives a triangle wave that varies between", null, "and", null, "with unit period.\n\nTriangleWave[{min,max},x]\n\ngives a triangle wave that varies between min and max with unit period.\n\n# Details", null, "• TriangleWave[x] is symmetric about", null, ", and has value", null, "at", null, ".\n\n# Examples\n\nopen allclose all\n\n## Basic Examples(3)\n\nEvaluate numerically:\n\nPlot over a subset of the reals:\n\nTriangleWave is a piecewise function over finite domains:\n\n## Scope(33)\n\n### Numerical Evaluation(4)\n\nEvaluate numerically:\n\nEvaluate to high precision:\n\nThe precision of the output tracks the precision of the input:\n\nEvaluate efficiently at high precision:\n\nTriangleWave threads over lists in the last argument:\n\n### Specific Values(4)\n\nValue at zero:\n\nValues of TriangleWave at fixed points:\n\nEvaluate symbolically:\n\nFind a value of", null, "for which the TriangleWave[x]=0.5:\n\n### Visualization(4)\n\nPlot the TriangleWave function:\n\nVisualize scaled TriangleWave functions:\n\nVisualize TriangleWave functions with different maximum and minimum values:\n\nPlot TriangleWave in three dimensions:\n\n### Function Properties(11)\n\nFunction domain of TriangleWave:\n\nIt is restricted to real inputs:\n\nFunction range of TriangleWave[x]:\n\nTriangleWave is periodic with period 1:\n\nTriangleWave is an odd function:\n\nThe area under one period is zero:\n\nTriangleWave is not an analytic function because it is singular at the half-integers:\n\nHowever, it is continuous:\n\nTriangleWave[x] is neither nondecreasing nor nonincreasing:\n\nTriangleWave is not injective:\n\nTriangleWave[x] is not surjective:\n\nTriangleWave[x] is neither non-negative nor non-positive:\n\nTriangleWave is neither convex nor concave:\n\n### Differentiation and Integration(5)\n\nFirst derivative with respect to", null, ":\n\nDerivative of the two-argument form with respect to", null, ":\n\nThe second (and higher) derivatives are zero except at points where the derivative does not exist:\n\nIf a==b, TriangleWave[{a,b},x] is constant and its derivatives are zero everywhere:\n\nIntegrals over finite domains:\n\n### Series Expansions(5)\n\nSince TriangleWave is odd, FourierTrigSeries gives a simpler result:\n\nThe two results are equivalent:\n\nFourierCosSeries of a scaled TriangleWave:\n\nMaclaurin series:\n\nSeries expansion at a singular point:\n\nTaylor expansion at a generic point:\n\n## Applications(2)\n\nCoefficients of Fourier series:\n\nExplicit Fourier series approximant:\n\nPlot the residual term:\n\nTriangle wave sound sample:\n\n## Properties & Relations(3)\n\nUse FunctionExpand to expand TriangleWave in terms of elementary functions:\n\nUse PiecewiseExpand to obtain piecewise representation on an interval:\n\nTriangleWave[x] is both upper and lower semicontinuous, and thus continuous, at the origin:\n\nThis is different from SquareWave[x], which is only upper semicontinuous:\n\nAs well as SawtoothWave[x], which is only lower semicontinuous:\n\nVisualize the three functions:\n\n## Possible Issues(1)\n\nTriangleWave is undefined for complex numbers:" ]

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[ "# The wave-induced solute flux from submerged sediment\n\n## Abstract\n\nThe issue of the transport of dissolved nutrients and contaminants between the sediment in the bottom of a lake or reservoir and the body of water above it is an important one for many reasons. In particular the biological and chemical condition of the body of water is intricately linked to these mass transport processes. As the review by Boudreau (Rev Geophys 38(3):389–416, 2000) clearly demonstrates those transport processes are very complex involving mechanisms as diverse as the wave-induced flux between the sediment and the overlying water and the effect of burrowing animals on the transport within the sediment as well as basic diffusion mechanisms. The present paper focuses on one facet of these transport processes; we re-examine the balance of diffusion and wave-induced advection and demonstrate that the wave-induced flux of a solute from submerged sediment is not necessarily purely diffusive as suggested by Harrison et al. (J Geophys Res 88:7617–7622, 1983) but can be dominated by a mean or time-averaged flux induced by the advective fluid motion into and out of the sediment caused by the fluctuating pressure waves associated with wave motion. Indeed along the subtidal shoreline where the fluctuating bottom pressures are greatest, wave-induced advection will dominate the mean, time-averaged transport of solute into or out of the sediment as suggested in the work of Riedl et al. (Mar Biol 13:210–221, 1972). However, the present calculations also indicate that this advective flux decreases rapidly with increasing depth so that further away from the shoreline the advective flux becomes negligible relative to the diffusive flux and therefore the latter dominates in deeper water.\n\n## Introduction\n\nThis paper presents a re-evaluation of the wave-induced solute transport between lake- or ocean-bottom sediment and the overlying body of water. The issue is of particular importance to the biological and chemical condition of the overlying water because of the reactions that occur within the sediment and the associated transport of solute to and from the water. For example, nutrients generated by the decomposition of deposited organic material are recycled by transportation back up into the water. Also contaminants within the sediment may be similarly mixed up into the lake or ocean. For these reasons, significant attention has been given in the literature to these transport processes. In this paper we focus on the transport to and from the sediment that occurs in the subtidal zone. The transport that occurs along the beaches and in the tidal zone is much more complex involving factors such as breaking wave transport, tidal effects and the influence of freshwater runoff (see, for example, ).\n\nIn subtidal waters, the possible transport mechanisms are fewer though still diverse: in addition to porous-media molecular diffusion they include wave-induced advection, tide-induced advection, the transport enhancement due to wave-induced ripples on the sediment surface, transport due to animal burrowing (see [2, 3]) and that due to gas bubbles (or hot water) percolating up through the sediment . Of these, this paper will be concerned only with wave-induced advection. Enhancement of transport by tide-induced advection has been detected in some North Sea sediments by (purely diffusive transport was measured in other areas). The ripple-induced enhancement that has been measured in the laboratory by and analysed by [7, 8]. It probably pertains in some locations and is clearly related to the detailed investigations carried out by and [12, 13] on the solute flux between a stream and migrating sediment bedforms. Also site specific are the transport due to animal burrowing and percolating bubbles.\n\nThe focus of the present paper is restricted to the possible enhancement of the solute transport in the subtidal sediment caused by wave-induced pressures and flows. In particular, in the absence of the other possible transport mechanisms listed above, is there significant wave-induced advective enhancement of the solute transport or is porous-media diffusion the dominant process? This question has received significant attention in the literature, for example in the reviews by and . Transport in excess of that by molecular diffusion has certainly been observed by a number of researchers. Some have ascribed that excess to an increase in the effective diffusivity in the sediment due to tortuosity or to turbulent mixing . However experimental measurements of solute transport such as those of off the coast of Georgia and Florida suggest that advective enhancement is important in some locations.\n\nThe literature contains a surprising diversity of analyses reflecting a significant divergence in what is considered the primary resistance to solute transport. Key analyses have been published by and by who present two different understandings and models of the wave-induced transport processes. The model of contains no purely advective time-averaged transport due to flows into and out of the sediment that might be induced by the pressure fluctuations caused by surface or internal waves. Indeed the only effect that is included in their model is purely diffusive transport and the sole effect of the sediment is the modification of the porous medium diffusivity caused by the tortuosity of the diffusion pathway within the medium and by possible elevated levels of the diffusivity at higher Peclet numbers. In an alternative approach, imply that the primary wave effect is due to advection into and out of the sediment. They associate the exchange of nutrients or contaminants between the sediment and the overlying water with the total volume of liquid that passes into and out of the sediment during the passage of one wavelength. Their analysis focuses on the total volume “filtered through the sediment” by adding the volume flowing into and out of the sediment (per unit surface area per unit time averaged over one cycle of the wave motion). focus on this total advective transport volume; no diffusive contribution is included in their analysis. In this paper we will attempt to resolve this conflict and to identify those circumstances in which either of the two mechanisms, diffusion or advection, might dominate the time-averaged solute transport into or out of the sediment.\n\n## Transport equation\n\nThe basic equations for solute transport in a granular bed have been carefully developed and presented by and and need only be quoted here. The transport equation for the concentration, $$c$$, of a solute (which could be a contaminant or a nutrient dissolved in the liquid) is:\n\n\\begin{aligned} \\nabla ( D . \\nabla c ) - \\mathbf{u} . \\nabla c + q^\\prime (c_\\infty -c) - \\frac{\\partial c}{\\partial t} = 0 \\end{aligned}\n(1)\n\nwhere $$t$$ is time, $$u$$ is the pore water velocity field, and $$D$$ is the diffusivity tensor. address the circumstances in which the solute is being absorbed in the sediment, but only a minor modification is required to handle cases in which solute is produced rather than absorbed. To do so the term $$-qc$$ in their version of the transport equation (where $$qc$$ is the mass rate of absorption of solute per unit volume of the sediment per unit time) has been replaced in Eq. 1 by a mass rate of production per unit sediment volume per unit time of $$+q^\\prime (c_\\infty -c)$$ where $$+q^{\\prime }$$ is the solute production factor and $$c_\\infty$$ is the concentration deep in the bed far from the sediment surface. We note that this form that includes the factor $$(c_\\infty -c)$$ is necessary in order for a steady state solute concentration to exist.\n\nWe also note as observe that even in homogeneous isotropic sediments, the diffusivity in Eq. 1 must be represented by a tensor because of anisotropy caused by the direction of the pore water velocity vector, u. Experiments show (e.g. ) and analyses confirm that in isotropic sediments the dispersion tensor in a coordinate system oriented with its first axis parallel to the pore water velocity vector u, has the following diagonal elements (the off-diagonal elements are zero): along this axis the “longitudinal” dispersion coefficient, denoted by $$D_L$$, may be much larger than the “transverse” dispersion coefficient, $$D_T$$, perpendicular to u. Both dispersion coefficients approach the porous media molecular diffusivity, $$D_P$$, when the Peclet number, $$Pe = \\vert \\mathbf{u} \\vert d /D_M \\ll 1$$ ($$d$$ is some measure of the grain size, most conveniently the equivalent spherical diameter). Note also that $$D_P$$ is somewhat less than the pure liquid molecular diffusivity, $$D_M$$, due to the more tortuous path for diffusion within the porous medium (usually $$D_P \\approx 0.67 D_M$$). More notably for $$Pe \\gg 1, D_M > D_T$$, usually by an order of magnitude. It is the combined effect on the diffusivity of the tortuosity and of the high Peclet number that explore.\n\nWe define a coordinate system, $$(x, y, z)$$, in which $$x$$ and $$z$$ are horizontal coordinates, $$y$$ is vertical down and the pore velocity in the $$z$$ direction is zero. Defining $$\\theta$$ to be the angle of the pore velocity, u, relative to a horizontal plane identify the following form of the diffusion term in the transport equation:\n\n\\begin{aligned} \\nabla (D.\\nabla c)&= (\\alpha c_x)_x + (\\alpha c_y)_y + D_T (c_z)_z + ( \\beta \\cos {2\\theta } \\ c_x - \\beta \\sin {2\\theta } \\ c_y )_x \\nonumber \\\\&\\quad + ( - \\beta \\sin {2\\theta } \\ c_x - \\beta \\cos {2\\theta } \\ c_y )_y \\end{aligned}\n(2)\n\nwhere the subscripts $$x,y$$ refer to partial differentiation with respect to those variables and $$\\alpha$$ is the mean dispersion coefficient given by\n\n\\begin{aligned} \\alpha = (D_L+D_T)/2 \\quad \\text {and} \\quad \\beta = (D_L-D_T)/2 \\end{aligned}\n(3)\n\nNote that when $$Pe \\ll 1, \\alpha \\rightarrow D_P$$ and $$\\beta \\rightarrow 0$$.\n\n## Pore water velocity field\n\nThe pore water velocity may be evaluated using the incompressibility condition and Darcy’s law for incompressible flow in a porous medium . We focus here on the flows and solute fluxes induced by waves in the overlying liquid; both traveling surface waves and internal waves will be considered and it is convenient to consider an individual wavenumber, $$k$$, in the $$x$$ direction associated with a radian frequency, $$\\omega$$, so that the pore velocity field, ($$u, v, w$$), can be represented by\n\n\\begin{aligned} u = Ue^{-ky} \\cos {(kx - \\omega t)};\\quad v = -Ue^{-ky} \\sin {(kx - \\omega t)}; \\quad w = 0, \\end{aligned}\n(4)\n\nwhere $$t$$ is time and $$U$$ is the velocity fluctuation amplitude at the sediment surface. Note that the angle, $$\\theta$$, which the velocity vector makes with the $$x$$ axis is given by $$\\theta = (kx-\\omega t)$$.\n\nAccording to the simplest form of Darcy’s law, the sediment surface pressure variation, $$p(x,t)$$, associated with the above velocity field is\n\n\\begin{aligned} p = - \\frac{\\rho g U}{\\kappa k} \\sin {(kx - \\omega t)}, \\end{aligned}\n(5)\n\nwhere $$\\rho$$ is the liquid density, $$g$$ is the acceleration due to gravity and $$\\kappa$$ is the hydraulic conductivity of the sediment divided by the porosity. The velocity amplitude $$U$$ is thus given by\n\n\\begin{aligned} U = \\frac{\\kappa k P}{\\rho g}, \\end{aligned}\n(6)\n\nwhere $$P$$ is the amplitude of the pressure fluctuations acting on the sediment surface as a result of the waves in the overlying water. The above applies to all traveling wave motions of small amplitude, whether surface waves or internal waves. In the particular case of linear surface waves of amplitude $$a$$ on a homogeneous ocean of depth, $$H$$:\n\n\\begin{aligned} P = \\frac{\\rho g a}{\\cosh {kH}}\\quad \\text {and}\\quad \\omega = (gk \\tanh {(kH)})^\\frac{1}{2} \\end{aligned}\n(7)\n\nbut other forms of $$P$$ and $$\\omega$$ are readily incorporated into the system of governing equations including the effect of sediment surface slope [18, 21]. The validity of such expressions for the fluid velocity within the sediment have been well tested by the extensive experimental measurements of and others. make note of additional limitations in the simple form of Darcy’s law used in the above system of equations, including the neglect of fluid inertial terms (see also [22, 23]).\n\nFinally, we note that the advective term in Eq. 1 can be written in terms of the above velocity field as\n\n\\begin{aligned} \\mathbf{u} . \\nabla c = U e^{-ky} ( c_x \\cos {\\theta } - c_y \\sin {\\theta } ) \\end{aligned}\n(8)\n\nand the corresponding form of the solute transport Eq. 1 becomes\n\n\\begin{aligned}&(\\alpha c_x)_x + (\\alpha c_y)_y + D_T (c_z)_z + ( \\beta c_x \\cos {2\\theta } - \\beta c_y \\sin {2\\theta } )_x \\nonumber \\\\&\\quad \\quad - U e^{-ky} ( c_x \\cos {\\theta } - c_y \\sin {\\theta } ) + q^{\\prime } (c_\\infty - c) - c_t = 0 \\end{aligned}\n(9)\n\nTo investigate the solute flux due to the above wave motion, we seek a solution of this equation which is purely periodic in $$x, t$$ and therefore $$\\theta$$. Consequently,\n\n\\begin{aligned} \\frac{\\partial }{\\partial x} \\equiv k \\frac{\\partial }{\\partial \\theta } \\quad \\text {and} \\quad \\frac{\\partial }{\\partial t} \\equiv - \\omega \\frac{\\partial }{\\partial \\theta } \\end{aligned}\n(10)\n\nand, if we assume that $$\\alpha$$ and $$\\beta$$ are constant and uniform and that there is no change in the $$z$$ direction, then Eq. 9 becomes\n\n\\begin{aligned}&k^2 \\alpha c_{\\theta \\theta } + \\alpha c_{yy} + \\beta k (\\cos {(2\\theta )} \\ k c_\\theta - \\sin {(2\\theta )} \\ c_y )_\\theta \\nonumber \\\\&\\quad - \\beta ( \\sin {(2\\theta )} \\ k c_\\theta + \\cos {(2\\theta )} \\ c_y )_y - U e^{-ky} ( k \\cos {\\theta } \\ c_\\theta - \\sin {\\theta } \\ c_y) \\nonumber \\\\&\\quad + q^{\\prime } (c_\\infty - c) + \\omega c_\\theta = 0 \\end{aligned}\n(11)\n\nThough this appears complicated it has the advantage of being linear and homogeneous in the concentration, $$c$$. It is useful to rewrite it using a non-dimensional $$y$$ coordinate, $$s$$, defined by\n\n\\begin{aligned} s = (q^\\prime / \\alpha )^\\frac{1}{2} y \\quad \\text {so that} \\quad \\frac{\\partial }{\\partial y} \\equiv \\left( \\frac{q^\\prime }{\\alpha } \\right) ^\\frac{1}{2} \\frac{\\partial }{\\partial s} \\end{aligned}\n(12)\n\nThen, using the dimensionless parameters\n\n\\begin{aligned} \\xi = q^\\prime / \\omega \\quad ; \\quad \\eta =\\beta / 2 \\alpha \\quad ; \\quad \\tau = U / ( q^\\prime \\alpha )^\\frac{1}{2} \\quad ; \\quad \\Gamma = k (\\alpha / q^\\prime )^\\frac{1}{2} \\end{aligned}\n(13)\n\nthe governing Eq. 11 becomes\n\n\\begin{aligned}&c_{ss}+(c_\\infty - c) + c_\\theta / \\xi - 2 \\eta \\cos {(2\\theta )} \\ c_{ss} + \\tau \\sin {\\theta } \\ e^{- \\Gamma s} \\ c_s \\nonumber \\\\&\\quad - \\Gamma [ 4 \\eta ( \\cos {(2\\theta )} \\ c_s + \\sin {(2\\theta )} \\ c_{s\\theta }) + \\tau \\cos {\\theta } \\ e^{-\\Gamma s} \\ c_\\theta ]\\nonumber \\\\&\\quad + 2 \\eta \\Gamma ^2 [ \\cos {(2\\theta )} \\ c_{\\theta \\theta } - 2 \\sin {(2\\theta )} \\ c_\\theta ] = 0 \\end{aligned}\n(14)\n\nEquation 14 is the governing equation to which we seek solutions. It can be greatly simplified by considering the typical magnitude of the various parameters as we will do in the sections which follow and in the Appendix. Once a solution is obtained we will evaluate the solute flux at the sediment surface in the downward direction (into the sediment), $$F(\\theta )$$, from\n\n\\begin{aligned} F(\\theta ) = \\left( v c - \\alpha \\frac{\\partial c}{\\partial y} \\right) _{y=0} = \\left( v c - ( q^\\prime \\alpha )^\\frac{1}{2}\\frac{\\partial c}{\\partial s} \\right) _{s=0} \\end{aligned}\n(15)\n\nand we will be most interested in the flux, $$\\bar{F}$$, averaged over one period of $$\\theta$$.\n\n## In the absence of waves\n\nFirst, however, we note that in the absence of waves ($$k \\rightarrow 0$$ and $$\\omega \\rightarrow 0$$) the governing equation reduces to the simple diffusion equation\n\n\\begin{aligned} c_{ss}+(c_\\infty - c) = 0 \\end{aligned}\n(16)\n\nto which the steady state solution is\n\n\\begin{aligned} c = c_\\infty - (c_\\infty - c_{y=0}) e^{-s}, \\end{aligned}\n(17)\n\nwhere $$c_{y=0}$$ is the concentration at the sediment surface which would be uniform and constant in the absence of waves. It follows that the thickness of the “diffusion-reaction sublayer” just underneath the sediment surface (in which most of the solute production and diffusion occurs) is given by\n\n\\begin{aligned} \\Delta s = 1 \\quad \\text {or} \\quad \\Delta y = (\\alpha / q^\\prime )^\\frac{1}{2} \\end{aligned}\n(18)\n\nIn this elementary case in which $$v_{y=0} = 0$$, the flux, $$\\bar{F}$$, simply becomes\n\n\\begin{aligned} \\bar{F} \\ = - ( q^\\prime \\alpha )^\\frac{1}{2} (c_\\infty - c_{y=0}) \\end{aligned}\n(19)\n\nNote that the absence of a contribution to $$F(\\theta )$$ (Eq. 15) from the advective term is due to the imposed constant and uniform value of $$c_{y=0}$$. Equation 15 is also the result that obtain when waves are present because they impose the condition that $$c$$ is constant on the sediment surface. As a consequence, they are left only with the effects on the diffusivity of tortuosity and of high Peclet number. Thus, in the absence of waves and in the Harrison et al. wave effect analysis, the flux, $$\\bar{F}$$, divided by the flux, $$\\bar{F}_M$$ through a pure, non-moving liquid “sediment” with diffusivity $$D_\\mathrm{{M}}$$, becomes\n\n\\begin{aligned} \\frac{\\bar{F}}{\\bar{F}_M} = \\left( \\frac{\\alpha }{D_M} \\right) ^\\frac{1}{2} \\end{aligned}\n(20)\n\n## Simplification\n\nWe proceed with the evaluation of the effect of waves by examining some simplifications to the governing Eq. 14. The first step in this process is to gauge the magnitude of the parameter, $$\\Gamma$$. From the definition 13 and Eq. 18, $$\\Gamma$$ is readily seen to be the ratio of the typical diffusion-reaction sublayer thickness, $$(\\alpha / q^\\prime )^\\frac{1}{2}$$, to the wavelength divided by $$2 \\pi , 1/k$$. The diffusion-reaction sublayer thickness is usually of the order of millimeters to several centimeters [24, 25] compared with wavelengths of the order of meters. Though much larger diffusion-reaction sublayer thicknesses are possible we will focus here on thicknesses that are smaller than meters. The parameter $$\\Gamma$$ is therefore much less unity in many cases of practical interest and we will confine our attention to this range. Since $$\\eta$$ is always less than one, it follows that when $$\\Gamma \\ll 1$$ we need only be concerned with the sediment layer in which $$s<\\Gamma ^{-1}$$, and the second and third lines of Eq. 14 become negligible ($$\\tau \\Gamma$$ must also be small and this is addressed in Appendix). Equation 14 then takes the manageable, large wavelength form\n\n\\begin{aligned} c_{ss}+(c_\\infty - c) + c_\\theta /\\xi - 2 \\eta \\cos {(2\\theta )} \\ c_{ss} + \\tau \\sin {\\theta } \\ c_s = 0 \\end{aligned}\n(21)\n\nA detailed solution to this governing equation is delineated in Appendix. In that appendix we demonstrate that within some reasonable parametric limitations, an appropriate, approximate solution to the governing Eq. 21 which is consistent with the desired form of the sediment surface boundary condition (Eq. 24) leads to a time-averaged flux into the sediment, $$\\bar{F}$$, given by\n\n\\begin{aligned} \\bar{F} = - (q^\\prime \\alpha )^\\frac{1}{2} (c_\\infty - \\bar{c}_0) - \\tilde{c} \\ U / 2 \\end{aligned}\n(22)\n\nwhere $$\\bar{c}_0$$ and $$\\tilde{c}$$ are the time-averaged solute concentration and the fluctuating solute concentration at the sediment surface. Not surprisingly therefore, the boundary condition at the sediment surface plays a crucial role in determining the magnitude of the solute flux and we turn to that issue in the sections which follow. (We note that Eq. 22 could, of course, have been constructed heuristically without any of the complicated algebra presented in Appendix. However, the authors believe it is still useful to delineate the parametric limitations and assumptions behind that result.)\n\nThe above expression, Eq. 22, displays an additional advective contribution, $$- \\tilde{c} \\ U / 2$$, over and above that in the absence of waves or in the analysis of . Consequently a key issue is the relative importance of diffusive and advective contributions to the time-averaged solute flux. From Eq. 22 the ratio of the advective flux to the diffusive flux is given by\n\n\\begin{aligned} \\frac{\\text {Advective Flux}}{\\text {Diffusive Flux}} = \\frac{U}{ (q^\\prime \\alpha )^\\frac{1}{2}} \\frac{\\tilde{c}}{2 (c_\\infty - \\bar{c}_0)} = \\frac{\\tau \\tilde{c}}{2 (c_\\infty - \\bar{c}_0)} \\end{aligned}\n(23)\n\nTherefore the issue depends on the product of two key quantities, the parameter $$\\tau$$ and the ratio of the amplitude of the sediment surface concentration fluctuation to the difference between the deep sediment concentration and the time-averaged sediment surface concentration. We examine each of these quantities in the sections which follow, beginning with an estimate of $$\\tau$$ followed by an examination of the sediment surface boundary condition and the quantity $$\\tilde{c}/2 (c_\\infty - \\bar{c}_0)$$.\n\n## Parameter estimation\n\nMeasurements and estimates by [18, 25] and others allow estimates of some of the controlling parameters, particularly the key parameter, $$\\tau =U/(q^\\prime \\alpha )^\\frac{1}{2}$$. Beginning with the velocity $$U$$ in the sediment, we will make use of the extensive measurements by of the volume of liquid $$Q$$ (per unit surface area per unit time averaged over a cycle of the wave motion) that passes into and out of the sediment during the passage of a wave. Note that $$Q = U/2 \\pi ^2$$ in the present notation. Reidl et al. made measurements and estimates of $$Q$$ for the sediment of the Atlantic Continental Shelf and concluded that $$Q$$ varied from values of the order of $$Q=2 \\times 10^{-6}$$ m/s ($$U=4 \\times 10^{-5}$$ m/s) within a few miles of the coast to values smaller by an order of magnitude or more further out on the Continental Shelf. This decrease is primarily due to the increasing depth and its effect on the wave-induced bottom pressures. We will call this data for $$U$$ and the associated dimensionless parameters, “coastal” data having $$U$$ values of the order of $$4 \\times 10^{-5}$$ m/s and decreasing with increasing depth.\n\nHowever, sample calculations readily demonstrate that much larger values of $$U$$ will pertain along the shoreline and, in the discussion, we will refer to these as “shoreline” data. For example, using Eq. 6 a $$1$$ m amplitude wave of wavelength $$10$$ m in shallow water with $$P/\\rho g = 1$$ m will produce $$U$$ values of $$10^{-4}$$ for the $$130\\,\\upmu$$m Oklahoma 90 sand of (which has a $$\\kappa = 0.00011$$ m/s) and $$U= 10^{-3}$$ m/s for Elliott’s 470 $$\\upmu$$m Ottawa sand (which has a $$\\kappa = 0.0014$$ m/s). These much larger values are consistent with the measurements by of the pore water velocities in the breaker zone of a laboratory beach. They imply very different solute transport along the shoreline.\n\nNext we note that a typical value for the molecular diffusivity $$D_P$$ is $$10^{-9}\\,{\\text {m}}^2\\,{\\text {s}}^{-1}$$ for fine sand, corresponding to a value of $$\\alpha \\approx 4 \\times 10^{-9}\\,{\\text {m}}^2\\,{\\text {s}}^{-1}$$. Moreover $$(\\alpha / q^\\prime )^\\frac{1}{2}$$ is the typical diffusion-reaction sublayer thickness in the sediment and this thickness is estimated and measured to be about $$0.03\\,{\\text {m}}$$ for fine sand [18, 25]. It follows that with coastal $$U$$ values of $$4 \\times 10^{-5}\\, {\\text {m/s}}$$ and less and with fine sand the pertinent values of $$\\tau$$ are $$300$$ and less. On the other hand the above shoreline values of $$U$$ lead to $$\\tau$$ values of about $$10^3$$ for the Oklahoma 90 sand and $$10^4$$ for the Ottawa sand. In summary, the key parameter $$\\tau =U/(q^\\prime \\alpha )^\\frac{1}{2}$$ appears to decrease from values of the order of $$10^4$$ for coarse sand (less for fine sand) at the subtidal shoreline to values of the order of 300 for fine sand in deeper coastal waters.\n\nAs a postscript we also note that other parameters of interest include $$\\Gamma$$, the ratio of the typical diffusion-reaction sublayer thickness to the wavelength (divided by $$2 \\pi$$), which we have earlier concluded is much less than unity in almost all cases of practical interest and the parameter $$\\eta$$ which is clearly of order unity or smaller. We also note that, by definition, $$(q^\\prime )^{-1}$$ is a measure of the time required for the source type used to deplete the concentration in the sediment by diffusion and since $$(\\omega )^{-1}$$ is typical of the period of the wave motions we might reasonably expect that in almost all circumstances of interest the former time is much longer than the latter so that $$\\xi \\ll 1$$ and $$\\xi \\eta \\ll 1$$.\n\n## Sediment surface boundary\n\nClearly the time-averaged solute transport may depend crucially on the boundary condition(s) at the sediment surface. If the conditions are such that the solute concentration remains constant at the sediment surface (as assume) then, since $$\\tilde{c} \\approx 0$$, the transport is purely diffusive; there would be no advective transport and the only significant effect on the time-averaged transport would be the modification of the porous medium diffusivity caused by the tortuosity of the diffusion pathway within the sediment as suggested by . This is because when $$\\tilde{c}=0$$ the fluid entering and leaving the bed has the same concentration at all times and the net advective flux averaged over one cycle of the wave motion would be zero. In their alternative approach, associate the exchange of nutrients or contaminants between the sediment and the overlying water with the total volume of liquid $$Q$$ (per unit surface area per cycle of the wave motion), that passes into and out of the sediment during the passage of a wave (in the present notation $$Q = U/2 \\pi ^2$$); no mention is made of any diffusive contribution. Moreover, the relation between that volume flux and the actual solute transport is not clarified.\n\nIn taking a closer look at the sediment surface concentration, we shall not assume a priori that $$c_{y=0}$$ is constant but rather that it may change depending on whether the instantaneous advective flow is into or out of the sediment. We shall heuristically assume that $$c_{y=0}$$ is correlated with the fluid velocity normal to the surface, namely $$v_{y=0}$$ (positive downward) which we previously denoted by $$v_{y=0}=-U \\sin {\\theta }$$ (note that for $$0<\\theta <\\pi$$ the flow is up out of the sediment whereas for $$\\pi <\\theta <2\\pi$$ the flow is into the sediment) and that\n\n\\begin{aligned} c_{y=0} = \\bar{c}_0 + \\tilde{c} \\ \\sin {\\theta } \\end{aligned}\n(24)\n\n(Note that when $$\\tilde{c}$$ is positive the surface concentration is greater when the flow is emerging from the sediment and less when the flow is into the sediment).\n\nConsider first the half-cycle during which flow and solute are emerging from the sediment into the overlying water in a lake, reservoir or ocean. After emerging it will be mixed by the turbulence in the benthic boundary layer . During that half-period a layer of thickness, $$U/\\omega$$ would be ejected from the sediment. Therefore, the extent to which the surface concentration will change during that half-period will depend on the ratio of $$U/\\omega$$ to the thickness of the potential diffusion-reaction sublayer in the sediment, $$(\\alpha / q^\\prime )^\\frac{1}{2}$$. In other words it will depend on the parameter, $$U/\\omega (\\alpha / q^\\prime )^\\frac{1}{2}$$. If that parameter is much larger than unity then the change in the surface concentration will be comparable with the overall concentration difference, $$(c_\\infty - \\bar{c}_0)$$ since the diffusion-reaction sublayer in the sediment would be rapidly ejected and the surface concentration would approach $$c_\\infty$$. On the other hand, if the parameter is much smaller than unity the surface concentration fluctuation will be much less. Using the values for $$U$$ and $$(\\alpha / q^\\prime )^\\frac{1}{2}$$ from the preceding section and radian frequencies for the waves of $$\\omega =0.6 \\rightarrow 6.0\\,{\\text {s}}^{-1}$$ leads to maximum coastal values of $$U/\\omega (\\alpha / q^\\prime )^\\frac{1}{2}$$ of $$2 \\times 10^{-4} \\rightarrow 2 \\times 10^{-3}$$ and to shoreline values of $$U/\\omega (\\alpha / q^\\prime )^\\frac{1}{2}$$ up to $$0.5$$ for the coarse sand. We conclude that during the emerging flow half-cycle the surface concentration in the coastal zone and beyond will change very little but that the surface concentration in the shoreline zone will change significantly.\n\nNow consider the other half-period when the flow is into the sediment. Clearly a key consideration will be the mixing in the benthic boundary layer next to the bottom of oceans, lakes and reservoirs for that will determine the solute concentration entering the sediment. This mixing has been the subject of much research and debate (see, for example, [14, 2730]). It can be crudely modelled using a turbulent diffusivity, $$D_t$$, in the waters above the sediment. This allows an estimate of the potential diffusion layer thickness in the water in the absence of advection of $$(D_t/\\omega )^\\frac{1}{2}$$ (averaged over an integer number of wave periods). We seek the ratio of this to the thickness of layer ingested to the sediment during the same period, namely $$U/\\omega$$. If that ratio, $$(D_t \\omega )^\\frac{1}{2}/U$$, is small then the diffusive benthic layer would be rapidly ingested and the surface concentration would approach the concentration in the water above that layer. If, on the other hand, that ratio is large the surface concentration fluctuation will be much less. Typical values of $$D_t$$ are the $$3 \\times 10^{-6}\\,{\\text {m}}^2\\,{\\text {s}}^{-1}$$ measured by in Lake Kinnaret in Israel (see also ) and the $$10^{-5}\\,{\\text {m}}^2\\,{\\text {s}}^{-1}$$ measured by in the Baltic Sea. Assuming $$D_t \\approx 10^{-5}\\,{\\text {m}}^2\\,{\\text {s}}^{-1}$$ it follows that for the coastal zone with $$U$$ values of $$4 \\times 10^{-5}\\,{\\text {m}}\\,{\\text {s}}^{-1}$$ (and smaller) and a radian wave frequency of $$\\omega =0.6 \\rightarrow 6.0{\\text {s}^{-1}}$$, the ratio, $$(D_t \\omega )^\\frac{1}{2}/U$$, is in the range $$60 \\rightarrow 200$$ though with much larger values for smaller $$U$$. On the other hand with subtidal shoreline values of $$U$$ and with $$D_t \\approx 10^{-5}\\, {\\text {m}}^2\\,{\\text {s}^{-1}}$$ the parameter $$(D_t \\omega )^\\frac{1}{2}/U$$ could be as small as $$0.2 \\rightarrow 8$$; however since the effective “turbulent” diffusivity in the shoreline zone is likely to be much greater than $$10^{-5}\\, {\\text {m}}^2\\, {\\text {s}^{-1}}$$ we estimate that the effective value for $$(D_t \\omega )^\\frac{1}{2}/U$$ at the shoreline is also likely to be very much greater than unity. These values suggests a mimimal change in the surface concentration during the sediment inflow half-period in both the coastal and shoreline zones.\n\nWe conclude from the preceding two paragraphs that the sediment surface concentration varies by very little in the coastal zone and beyond but may change significantly in the shoreline zone. Therefore, factoring in the respective $$\\tau$$ values, we conclude that the ratio of the advective flux to the diffusive flux, given by Eq. 23 as $$\\tau \\tilde{c}/2 (c_\\infty - \\bar{c}_0)$$, will be large in the shoreline zone (perhaps as much as $$10^3 \\rightarrow 10^4$$ and therefore that solute transport at the shoreline will be dominated by wave-induced advection. On the other hand, the wave-induced advective transport in the coastal zone may be of the same order as the diffusive flux closest to the coast but will rapidly decrease with water depth so that, in deeper waters, the diffusive transport will dominate.\n\nParenthetically we note that, for simplicity, we have tacitly assumed similar grain sizes in the coastal and shoreline zones. When these differ substantially, the different grain sizes need to be included in evaluating the zonal sediment hydraulic conductivities and diffusivities and the relevant zonal parameters in which they appear.\n\n## Concluding remarks\n\nWe have demonstrated that the wave-induced flux of a solute from submerged sediment is not necessarily purely diffusive as suggested by but can be dominated by a mean or time-averaged flux induced by the advective fluid motion into and out of the sediment caused by the fluctuating pressures associated with wave motion. For convenience we include the flowchart, Fig. 1, which illustrates the dominant processes and steps involved in evaluating and comparing the diffusive and advective contributions to the flux out of the sediment.\n\nAlong the shoreline where the fluctuating bottom pressures are greatest, wave-induced advection will dominate the mean, time-averaged transport of solute into or out of the sediment. In this zone, the time-averaged transport will therefore be dominated by the mechanism suggested in the work of . However, the present calculations also indicate that this advective flux decreases rapidly with increasing water depth so that further away from the shoreline the advective flux becomes negligible relative to the diffusive flux and, consequently, the diffusive flux dominates in deeper water where the calculations of are therefore relevant.\n\nWhether one can therefore assume that most of the solute transport occurs in the immediate vicinity of the subtidal shoreline or whether it includes significant contributions from deeper waters depends on the relative magnitudes of the concentration differences in the two zones. Clearly, models of the mixing processes in the entire body of the lake, reservoir or ocean and of the production/adsorption processes in the sediments at various depths will be needed to determine the prevailing concentration differences that drive mass transport. It is unlikely that these concentration differences would be similar in the subtidal shoreline and coastal zones. The present paper indicates the appropriate mechanism for transport in the different zones.\n\nAs a footnote, we observe that the present analyses do not apply in the tidal or beach zone where water is thrown above the mean waterline and mass transport occurs during backwash, where tidal fluctuations also generate flow through the sediment and where freshwater runoff may play a significant role (see, for example, [1, 26, 32]). Further work is needed to determine the relation between the magnitudes of the mass transport in the beach zone and that in the subtidal zone and in deeper waters.\n\n## Abbreviations\n\n$$a$$ :\n\nSurface wave amplitude ($$\\mathrm{m}$$)\n\n$$c$$ :\n\nConcentration of solute ($$\\mathrm{{kg}}\\,\\mathrm{m}^{-3}$$)\n\n$$c_{\\infty }$$ :\n\nConcentration as y tends to infinity ($$\\mathrm{{kg}}\\,\\mathrm{m}^{-3}$$)\n\n$$\\bar{c}_0$$ :\n\nTime-averaged concentration at the sediment surface ($$\\mathrm{{kg}}\\,\\mathrm{m}^{-3}$$)\n\n$$\\tilde{c}$$ :\n\nAmplitude of surface concentration fluctuation ($$\\mathrm{{kg}}\\,\\mathrm{m}^{-3}$$)\n\n$$d$$ :\n\nGrain size ($$\\mathrm{{m}}$$)\n\n$$D$$ :\n\nDiffusivity tensor ($$\\mathrm{m}^2 \\, \\mathrm{s}^{-1}$$)\n\n$$D_L ,D_T$$ :\n\nLongitudinal and transverse diffusivities ($$\\mathrm{m}^2 \\, \\mathrm{s}^{-1}$$)\n\n$$D_M$$ :\n\nPure liquid molecular diffusivity ($$\\mathrm{m}^2 \\, \\mathrm{s}^{-1}$$)\n\n$$D_P$$ :\n\nPorous medium molecular diffusivity ($$\\mathrm{m}^2 \\, \\mathrm{s}^{-1}$$)\n\n$$D_t$$ :\n\nTurbulent diffusivity in overlying water ($$\\mathrm{m}^2 \\, \\mathrm{s}^{-1}$$)\n\n$$F$$ :\n\nFlux of solute ($$\\mathrm{{kg}} \\ \\mathrm{{m}}^{-2} \\, \\mathrm{{s}}^{-1}$$)\n\n$$\\bar{F}$$ :\n\nTime-averaged flux of solute ($$\\mathrm{{kg}} \\ \\mathrm{{m}}^{-2} \\, \\mathrm{{s}}^{-1}$$)\n\n$$g$$ :\n\nAcceleration due to gravity ($$\\mathrm{{m}} \\, \\mathrm{{s}}^{-2}$$)\n\n$$H$$ :\n\nWater depth ($$\\mathrm{m}$$)\n\n$$i$$ :\n\n$$(-1)^\\frac{1}{2}$$\n\n$$k$$ :\n\nWavenumber ($$\\mathrm{{m}}^{-1}$$)\n\n$$P\\mathrm{{e}}$$ :\n\nPeclet number $$\\vert \\mathbf{u} \\vert d /D_M$$\n\n$$q$$ :\n\nSink strength factor ($$\\mathrm{{s}}^{-1}$$)\n\n$$q^{\\prime }$$ :\n\nSource strength factor ($$\\mathrm{{s}}^{-1}$$)\n\n$$Q$$ :\n\nFiltered water volume per unit area time averaged over wave period ($$\\mathrm{{m}} \\, \\mathrm{{s}}^{-1}$$)\n\n$$t$$ :\n\nTime ($$\\mathrm{s}$$)\n\nu :\n\nPore water velocity vector ($$\\mathrm{{m}} \\, \\mathrm{{s}}^{-1}$$)\n\n$$U$$ :\n\nMaximum wave-induced pore water velocity ($$\\mathrm{{m}} \\, \\mathrm{{s}}^{-1}$$)\n\n$$u,v,w$$ :\n\nPore water velocities in $$x,y,z$$ directions ($$\\mathrm{{m}} \\, \\mathrm{{s}}^{-1}$$)\n\n$$x$$ :\n\nHorizontal coordinate in direction of wave motion ($$\\mathrm{{m}}$$)\n\n$$y$$ :\n\nCoordinate measured vertically downward ($$\\mathrm{{m}}$$)\n\n$$z$$ :\n\nHorizontal coordinate perpendicular to $$x$$ and $$y$$ ($$\\mathrm{{m}}$$)\n\n$$\\alpha$$ :\n\n$$(D_L+D_T)/2$$ ($$\\mathrm{{m}} \\, \\mathrm{{s}}^{-1}$$)\n\n$$\\beta$$ :\n\n$$(D_L-D_T)/2$$ ($$\\mathrm{{m}} \\, \\mathrm{{s}}^{-1}$$)\n\n$$\\gamma , \\delta$$ :\n\nConstants\n\n$$\\xi , \\eta , \\Gamma$$ :\n\nDimensionless parameters\n\n$$\\tau$$ :\n\nTransport parameter, $$U / ( q^\\prime \\alpha )^\\frac{1}{2}$$\n\n$$\\kappa$$ :\n\nRatio of hydraulic conductivity to the porosity ($$\\mathrm{{m}} \\, \\mathrm{{s}}^{-1}$$)\n\n$$\\rho$$ :\n\nWater density ($$\\mathrm{{kg}} \\, \\mathrm{{m}}^{-3}$$)\n\n$$\\theta$$ :\n\n$$kx -\\omega t$$\n\n$$\\omega$$ :\n\nRadian frequency of wave motion ($$\\mathrm{{rad}} \\, \\mathrm{{s}}^{-1}$$)\n\n## References\n\n1. 1.\n\nBoufadel MC, Suidan MT, Rauch CH, Ahn C-H, Venosa AD (1999) Nutrient transport in beaches subject to freshwater input and tides. Proc Int Oil Spill Conf 1:471–476\n\n2. 2.\n\nBoudreau BP (2000) The mathematics of early diagenesis: from worms to waves. Rev Geophys 38(3):389–416\n\n3. 3.\n\nAllanson BR, Skinner D, Imberger J (1992) Flow in prawn burrows. Estuar Coast Shelf Sci 35:253–266\n\n4. 4.\n\nMeier JA, Jewell JS, Brennen CE, Imberger J (2010) Bubbles emerging from a submerged granular bed. J Fluid Mech 666:189–203\n\n5. 5.\n\nLohse L, Epping EHG, Helder W, van Raaphorst W (1996) Oxygen pore water profiles in continental shelf sediments of the North Sea: turbulent versus molecular diffusion. Mar Ecol Prog Ser 145:63–75\n\n6. 6.\n\nPrecht E, Huettel M (2003) Advective pore-water exchange driven by surface gravity waves and its ecological implications. Limnol Oceanogr 48(4):1674–1684\n\n7. 7.\n\nShum KT (1992) Wave-induced advective transport below a rippled water-sediment interface. J Geophys Res 97:789–808\n\n8. 8.\n\nShum KT (1993) The effects of wave-induced pore water circulation on the transport of reactive solutes below a rippled sediment bed. J Geophys Res 98:10289–10301\n\n9. 9.\n\nElliott AH (1991) Transfer of solutes into and out of streambeds. Ph.D. Thesis, California Institute of Technology\n\n10. 10.\n\nPackman A, Salehin M, Zaramella M (2004) Hyporheic exchange with gravel beds: basic hydrodynamic interactions and bedform-induced advective flows. ASCE J Hydr Eng 130(7):647–656\n\n11. 11.\n\nPackman AI (1997) Exchange of colloidal kaolinite between stream and sand bed in a laboratory flume. in a laboratory flume. Ph.D. Thesis, California Institute of Technology\n\n12. 12.\n\nEylers H (1994) Transport of adsorbing metal ions between stream water and sediment bed in a laboratory flume. Ph.D. Thesis, California Institute of Technology\n\n13. 13.\n\nEylers H, Brooks NH, Morgan JJ (1995) Transport of adsorbing metals from stream water to a stationary sand-bed in a laboratory flume. Mar Freshw Res 46(1):209–214\n\n14. 14.\n\nWüest A, Lorke A (2003) Small-scale hydrodynamics in lakes. Ann Rev Fluid Mech 35:373–412\n\n15. 15.\n\nHarrison WD, Musgrave D, Reeburgh WS (1983) A wave-induced transport process in marine sediments. J Geophys Res 88:7617–7622\n\n16. 16.\n\nRutgers van der Loeff MM (1981) Wave effects on sediment water exchange in a submerged sand bed. Neth J Sea Res 15(1):100–112\n\n17. 17.\n\nMarinelli RL, Jahnke RA, Craven DB, Nelson JR, Eckman JE (1998) Sediment nutrient dynamics on the South Atlantic Bight continental shelf. Limnol Oceanogr 43:1305–1320\n\n18. 18.\n\nRiedl RJ, Huang N, Machan R (1972) The subtidal pump: a mechanism of interstitial water exchange by wave action. Mar Biol 13:210–221\n\n19. 19.\n\nFried JJ, Combarnous MA (1971) Dispersion in porous media. Adv Hydrosci 7:169–282\n\n20. 20.\n\nKoch DL, Brady JF (1985) Dispersion in fixed beds. J Fluid Mech 154:399–427\n\n21. 21.\n\nPhillips OM (1966) The dynamics of the upper ocean. Cambridge Univ. Press, Cambridge\n\n22. 22.\n\nReid RO, Kajiura K (1957) On the damping of sea waves over a permeable sea bed. Eos Trans AGU 38(5):662–666\n\n23. 23.\n\nYamamoto T, Koning HL, Sellmeijer H, van Hijum E (1978) On the response of a poro-elastic bed to water waves. J Fluid Mech 87:193–206\n\n24. 24.\n\nJorgensen BB, Revsbech NP (1983) Colorless sulphur bacteria, Beggiatoa spp. and Thiovulum spp., in $$O_2$$ and $$H_2S$$ microgradients. Appl Environ Microbiol 45:1261–1270\n\n25. 25.\n\nHoltappels M, Kuypers MMM, Schlüter M, Brüchert V (2011) Measurement and interpretation of solute concentration gradients in the benthic boundary layer. Limnol Oceanogr 9:113\n\n26. 26.\n\nKakinuma T, Ohishi S, Nakamura K (2010) Seepage Flow in a breaker zone. Proceedings of the International Conference on Coastal Engineering, No 32\n\n27. 27.\n\nLorke AW, Umlauf L, Jonas T, Wüest A (2002) Dynamics of turbulence in low-speed oscillating bottom-boundary layers of stratified basins. Environ Fluid Mech 2:291313\n\n28. 28.\n\nLorke AW, Müller B, Maerki M, Wüest A (2003) Breathing sediments: the control of diffusive transport across the sediment–water interface by periodic boundary-layer turbulence. Limnol Oceanogr 48(6):2077–2085\n\n29. 29.\n\nLemckert C, Antenucci J, Saggio A, Imberger J (2004) Physical properties of turbulent benthic boundary layers generated by internal waves. ASCE J Hydr Eng 130(1):58–69\n\n30. 30.\n\nHoltappels M, Lorke A (2011) Estimating turbulent diffusion in a benthic boundary layer. Limnol Oceanogr 9:2941\n\n31. 31.\n\nGloor M, Wüest A, Imboden DM (2000) Dynamics of mixed bottom boundary layers and its implication for diapycnal transport in a stratified, natural water basin. J Geophys Res 105(C4):86298646\n\n32. 32.\n\nHorn D, Li L (2006) Measurement and modelling of gravel beach groundwater response to wave run-up: effect of beach profile changes. J Coast Res 22(5):1241–1249\n\n## Acknowledgments\n\nThe lead author would like to acknowledge the support received from CWR during many visits to the Centre for Water Research, University of Western Australia.\n\n## Author information\n\nAuthors\n\n### Corresponding author\n\nCorrespondence to Christopher Earls Brennen.\n\n## Appendix: Detailed solution of transport problem\n\n### Appendix: Detailed solution of transport problem\n\nThe linear form of the governing Eq. 21 allows us to construct the following seperable solution which allows flexibility in applying an appropriate boundary condition at the sediment surface:\n\n\\begin{aligned} c - c_\\infty = ( \\bar{c}_0 - \\ c_\\infty ) e^{-s} + \\ Re \\{ G (\\theta ) e^{- (\\gamma + i \\delta ) s } \\} \\end{aligned}\n(25)\n\nin which $$i=(-1)^{1/2}$$ and $$Re\\{ \\}$$ refers to the real part. The quantity $$\\bar{c}_0$$ will be seen to be the time-averaged solute concentration at the sediment surface, and the real constants $$\\gamma$$ and $$\\delta$$ as well as the potentially complex function $$G(\\theta )$$ remain to be determined. Substituting the expression 25 into the governing Eq. 21 and integrating yields\n\n\\begin{aligned} G(\\theta )&= exp \\{ \\xi (1-(\\gamma +i\\delta )^2) \\theta - (\\gamma +i\\delta ) \\xi \\tau \\cos {\\theta } +\\xi \\eta (\\gamma +i\\delta )^2 \\sin {( 2 \\theta )}\\nonumber \\\\&+ \\Phi + i \\Psi \\} \\end{aligned}\n(26)\n\nwhere $$(\\Phi + i \\Psi )$$ is a complex integration constant. In the present problem we are only interested in solutions which are periodic in $$\\theta$$. It therefore follows that\n\n\\begin{aligned} \\gamma ^2 - \\delta ^2 = 1 \\end{aligned}\n(27)\n\nTherefore the desired solution must be of the form\n\n\\begin{aligned} G(\\theta )&= exp \\{- \\gamma \\xi \\tau \\cos {\\theta } + \\eta \\xi \\sin {( 2 \\theta )} + \\Phi \\nonumber \\\\&+\\,i [- 2 \\xi \\gamma \\delta \\theta - \\delta \\xi \\tau \\cos {\\theta } + 2 \\gamma \\delta \\xi \\eta \\sin {(2\\theta )} + \\Psi ]\\} \\end{aligned}\n(28)\n\nTo use this solution to explore the flux of solute through the sediment surface we next need to consider the boundary condition at that surface which we assume takes the form of Eq. 24. To conform with the required $$\\sin {\\theta }$$ fluctuation in Eq. 24 we need to choose\n\n\\begin{aligned} \\Psi = \\pm \\pi /2 \\quad \\text {and} \\quad 2 \\gamma \\delta \\xi = 1 \\end{aligned}\n(29)\n\nso that, in combination with the relation 27, it follows that\n\n\\begin{aligned} \\gamma = [\\{ \\xi +(\\xi ^2 + 1)^{1/2} \\} / 2 \\xi ]^{1/2} \\quad \\text {and} \\quad \\delta = [ 2 \\xi \\{ \\xi +(\\xi ^2 + 1)^{1/2} \\} ]^{-1/2} \\end{aligned}\n(30)\n\nThen\n\n\\begin{aligned} G = \\pm \\ i \\ exp \\{ \\ - i \\theta ^* + \\xi \\eta \\sin {(2\\theta )} - \\gamma \\xi \\tau \\cos {\\theta } + \\Phi \\}, \\end{aligned}\n(31)\n\nwhere\n\n\\begin{aligned} \\theta ^* = \\theta - \\eta \\sin {(2\\theta )} + \\delta \\xi \\tau \\cos {\\theta } \\end{aligned}\n(32)\n\nand\n\n\\begin{aligned} c - c_\\infty = ( \\bar{c}_0 - \\ c_\\infty ) e^{-s} \\pm \\ e^{(\\Phi - \\gamma s)} \\ e^{ ( \\xi \\eta \\sin {(2\\theta )} - \\xi \\tau \\gamma \\cos {\\theta } ) } \\sin { (\\theta ^* + \\delta s) } \\end{aligned}\n(33)\n\nwhich yields a surface concentration of\n\n\\begin{aligned} c_{y=0} = \\bar{c}_0 \\pm \\ e^{\\Phi } \\ e^{ ( \\xi \\eta \\sin {(2\\theta )} - \\gamma \\xi \\tau \\cos {\\theta } ) } \\sin { \\theta ^* } \\end{aligned}\n(34)\n\n\\begin{aligned} \\left( \\frac{\\partial c}{\\partial s} \\right) _{s=0} = (c_\\infty - \\bar{c}_0) \\mp e^{\\Phi } \\ e^{ ( \\xi \\eta \\sin {(2\\theta )} - \\gamma \\xi \\tau \\cos {\\theta } ) } ( \\gamma \\sin { \\theta ^*} - \\delta \\cos { \\theta ^*} ) \\end{aligned}\n(35)\n\nThese lead to the following expression for the flux at the sediment surface:\n\n\\begin{aligned} F (\\theta )&= - U \\bar{c}_0 \\sin {\\theta } - (q^\\prime \\alpha )^{1/2} (c_\\infty - \\bar{c}_0) \\nonumber \\\\&\\mp e^{\\Phi } \\ e^{ ( \\xi \\eta \\sin {(2\\theta )} - \\gamma \\xi \\tau \\cos {\\theta } ) } [ U \\sin {\\theta } \\sin { \\theta ^*} - (q^\\prime \\alpha )^{1/2} (\\gamma \\sin { \\theta ^*} - \\delta \\cos { \\theta ^*}) ]\\qquad \\end{aligned}\n(36)\n\nSince $$\\xi \\ll 1$$ and $$\\xi \\eta \\ll 1$$ (see Sect. 8), it follows from the definitions 30 that\n\n\\begin{aligned} \\gamma \\approx ( 1 / 2 \\xi )^{1/2} \\quad \\text {and} \\quad \\delta \\approx ( 1 / 2 \\xi )^{1/2} \\end{aligned}\n(37)\n\nand that $$\\xi \\gamma \\ll 1$$ and $$\\xi \\delta \\ll 1$$. Moreover, using the values of $$\\tau$$ estimated in Sect. 8 it also follows that $$\\xi \\delta \\tau \\ll 1$$ and $$\\xi \\gamma \\tau \\ll 1$$. Then from Eq. 32, $$\\theta ^* \\approx \\theta$$ and the second exponential function in Eqs. 33, 34, 35, and 36, is approximately unity. With these approximations the amplitude of the surface concentration oscillation, $$\\tilde{c} = e^{\\Phi }$$ and therefore\n\n\\begin{aligned} F (\\theta )&= - U \\bar{c}_0 \\sin {\\theta } - (q^\\prime \\alpha )^{1/2} (c_\\infty - \\bar{c}_0) \\nonumber \\\\&- \\tilde{c} \\ [ U \\sin {\\theta } \\sin { \\theta ^*} - (q^\\prime \\alpha )^{1/2} (\\gamma \\sin { \\theta ^*} - \\delta \\cos { \\theta ^*}) ] \\end{aligned}\n(38)\n\nso that the time-averaged flux into the sediment, $$\\bar{F}$$, is\n\n\\begin{aligned} \\bar{F} = - (q^\\prime \\alpha )^{1/2} (c_\\infty - \\bar{c}_0) - \\tilde{c} \\ U / 2 \\end{aligned}\n(39)\n\nThis includes the additional advective solute flux, $$- \\tilde{c} U / 2$$, that is discussed in the main text.\n\n## Rights and permissions\n\nReprints and Permissions\n\nBrennen, C.E., Imberger, J. The wave-induced solute flux from submerged sediment. Environ Fluid Mech 14, 221–234 (2014). https://doi.org/10.1007/s10652-013-9307-2" ]

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[ "# Year 10 Term 3\n\n*Functions Continued* (also Additional)", null, "Introduction to function notation:\n\nComposite functions:\n\nInverse functions:\n\nFinding if a graph has an inverse:\n\nFinding domains of functions:\n\nFinding vertical asymptotes:\n\nCompleting the square and sketching the resultant graph:\n\nSketching quadratic graphs from equations:\n\nUsing the discriminate to find the number of solutions to a quadratic:\n\nAn introduction to transforming graphs (translations):\n\nAn introduction to transforming graphs (stretches):\n\nSolve absolute value equations:\n\nGraphing absolute values of functions 1:\n\nGraphing absolute values of functions 2:\n\nProbability", null, "Tree diagrams with replacement:\n\nTree diagrams (including without replacement):\n\nSample space diagram:\n\nVenn Diagram probabilities:" ]

[ null, "https://bispmaths.files.wordpress.com/2015/08/year10-14.jpg", null, "https://bispmaths.files.wordpress.com/2015/08/year10-16.jpg", null ]

[ "### [Back]\n\nContributions to Books:\n\nJ. Melenk, D. Praetorius, B. Wohlmuth:\n\"Simultaneous quasi-optimal convergence in FEM-BEM coupling\";\nin: \"ASC Report 13/2014\", issued by: Institute of Applied Mathematics and Numerical Analysis; Vienna University of Technology, Wien, 2014, ISBN: 978-3-902627-07-0, 1 - 21.\n\nEnglish abstract:\nWe consider the symmetric FEM-BEM coupling that connects two linear\nelliptic second order partial differential equations posed in a\nbounded domain Ω and its complement, where the exterior problem is\nrestated by an integral equation on the coupling boundary Gamma. We\nassume that the corresponding transmission problem admits a shift\ntheorem by more than 1/2. We analyze the discretization by piecewise\npolynomials of degree k for the domain variable and piecewise\npolynomials of degree k-1 for the flux variable on the coupling\nboundary. Given sufficient regularity we show that (up to logarithmic\nfactors) the optimal convergence order k+1/2 in the H^{−1/2}-norm\nis obtained for the flux variable, while classical arguments by\nCea-type quasi-optimality and standard approximation results provide\nonly convergence order k for the overall error in the natural\nproduct norm.\n\nKeywords:\nFEM-BEM coupling, a priori convergence analysis, transmission problem\n\nElectronic version of the publication:\n\nCreated from the Publication Database of the Vienna University of Technology." ]

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[ "# Are there winding-number vacua in Weinberg-Salam (Or are they a gauge artifact)?\n\nIn pure SU(2) Yang-Mills the vacua van be grouped in homotopy classes labeled by their winding number. Instantons connect these giving rise to the theta-vacuum.\n\nI’m studying the SU(2) sphaleron in the Weinberg-Salam theory without fermions (Original paper by Manton). Which is the same SU(2) Yang-Mills theory with the addition of an SSB Higgs doublet $$\\Phi$$. The theory looses its instantons (Derricks theorem) and obtains a sphaleron. (The theory has no soliton monopoles.)\n\nIn the paper Manton constructs a non-contractible loop starting and ending at the same vacuum. It is important to note that using his gauge fixing $$A_r = 0$$ (as well as fixing a base point on $$\\Phi^\\infty$$) there is only a single vacuum left. Once the construction is over he claims that by letting go of $$A_r = 0$$ and imposing some additional new gauge constraints the loop can be turned into a path connecting topologically distinct vacuums (very reminiscent of how the instanton connected vacuums with different winding numbers in the pure Yang-Mills theory).\n\nHas the inclusion of the Higgs somehow gotten rid of the pure Yang-Mills theory’s multiple vacuum structure? Or is it that these multiple vacua are somehow gauge artifacts only observable in some gauges and not others ($$A_r$$ = 0)? This appears strange to me since these many vacuums give rise to real physical effects.\n\nedit Added the remark that I’m studying the model in absence of fermions.\n\nThere are indeed interesting winding number instanton configurations in the $$SU(2)$$ weak interaction sector of the standard model. There is no analogue of the QCD vacuum angle however, as the instantons have exact fermion zero modes even when the fermions gain mass via the Higgs field. Here are some of the key references:\n\nN.V.Krasnikov, V.A. Rubakov, V.F.Tokarev, \"Zero-fermion modes in models with spontaneous symmetry- breaking\". J. Phys A 12 (1979) L343-346.\n\nA.A.Anselm, A.A.Johansen, \"Baryon nonconservation in standard model and Yukawa interaction\", Nuclear Physics B 407 (1993) 313-327\n\nA.A.Anselm, A.A.Johansen, \"Can the electroweak $$\\theta$$-term be observable?\", Nuclear Physics B 412 (1994)553-573.\n\n• Thank you for the response, I have not heard about these fermion zero modes. However, I’m studying Weinberg-Salam without fermions for the moment (so only with Higgs) and I’m trying to understand the vacuum structure of that particular theory (and whether the theory has true winding-number vacuums rather than those being a gauge artifact). Are you sure these modes are relevant? I’ll look into them if that is the case. Oct 14, 2019 at 8:09\n• Ah I see the point of Derrick's theorem. There may not be exact classical solutions, but the topology still alows winding number changing fluctuations and so, I suppose, theta vacua. I'm not an expert on this though. I expect, from the condensed-matter analogues that the Higgs expectation would be reduced in the core of the instanton. The actual configuration would be similar to that in the Rubakov paper. Oct 14, 2019 at 12:08\n\nI've also been puzzling over this question and I believe I've found the answer --- it is the latter, that the vacua the sphaleron path connects are topologically distinct only in certain gauges.\n\nIt's clear that for the incontractible loop detailed in Manton's paper that at $$\\mu = \\pi$$ the vacuum is topologically trivial. However, the path itself is topologically non-trivial, as it cannot be contracted to a point --- more specifically there is a non-zero value of $$\\int d^4 x \\, \\mathrm{Tr}\\, F_{\\mu \\nu} \\tilde{F}^{\\mu \\nu}$$ over the path if $$\\mu$$ is promoted to a function of time.\n\nIn Klinkhamer and Manton's follow-up paper where they study the sphaleron configuration itself, they calculate the sphaleron Chern-Simons number and note just above Equation (39) that the sphaleron can be considered as connecting two topologically distinct vacua in a gauge where the the gauge field vanishes faster than $$1/r$$ at spatial infinity. Note that this is more stringent than the condition $$A_r = 0$$.\n\nThere's a more detailed calculation in sections 2 and 3 of this paper where they show that to recover the Chern-Simons number of the field configuration along Manton's path you either need to introduce time-dependence or transform to a different gauge. With either of these choices the Chern-Simons number is\n\n$$N_\\mathrm{CS} = \\frac{2\\mu - \\sin 2\\mu}{2 \\pi}.$$\n\nFor vacuum configurations $$\\mu = \\pi n$$ for integer $$n$$, so $$N_\\mathrm{CS} \\in \\mathbb{Z}$$.\n\n• I’ve recently discovered an essay called “Demystifying the QCD Vacuum“ by J. Schwichtenberg. He discusses this precise problem and its unclear history. As far as I have been able to tell, your answers definitely goed in the right direction. The vacuum structure appears to be gauge dependent, but the conclusion (That Yang-Mills has a vacuum angle) does not change. Dec 20, 2019 at 11:38" ]

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[ "Categories: Physics, Quantum mechanics.\n\n# Imaginary time\n\nLet $$\\hat{A}_S$$ and $$\\hat{B}_S$$ be time-independent in the Schrödinger picture. Then, in the Heisenberg picture, consider the following expectation value with respect to thermodynamic equilibium (as found in Green’s functions for example):\n\n\\begin{aligned} \\expval*{\\hat{A}_H(t) \\hat{B}_H(t')} &= \\frac{1}{Z} \\Tr\\!\\Big( \\exp\\!(-\\beta \\hat{H}_{0,S}(t)) \\: \\hat{A}_H(t) \\: \\hat{B}_H(t') \\Big) \\end{aligned}\n\nWhere the “simple” Hamiltonian $$\\hat{H}_{0,S}$$ is time-independent. Suppose a (maybe time-dependent) “difficult” $$\\hat{H}_{1,S}$$ is added, so that the total Hamiltonian is $$\\hat{H}_S = \\hat{H}_{0,S} + \\hat{H}_{1,S}$$. Then it is easier to consider the expectation value in the interaction picture:\n\n\\begin{aligned} \\expval*{\\hat{A}_H(t) \\hat{B}_H(t')} &= \\frac{1}{Z} \\Tr\\!\\Big( \\exp\\!(-\\beta \\hat{H}_S(t)) \\: \\hat{K}_I(0, t) \\hat{A}_I(t) \\hat{K}_I(t, t') \\hat{B}_I(t') \\hat{K}_I(t', 0) \\Big) \\end{aligned}\n\nWhere $$\\hat{K}_I(t, t_0)$$ is the time evolution operator of $$\\hat{H}_{1,S}$$. In front, we have $$\\exp\\!(-\\beta \\hat{H}_S(t))$$, while $$\\hat{K}_I$$ is an exponential of an integral of $$\\hat{H}_{1,I}$$, so we are stuck. Keep in mind that exponentials of operators cannot just be factorized, i.e. in general $$\\exp\\!(\\hat{A} \\!+\\! \\hat{B}) \\neq \\exp\\!(\\hat{A}) \\exp\\!(\\hat{B})$$\n\nTo get around this, a useful mathematical trick is to use an imaginary time variable $$\\tau$$ instead of the real time $$t$$. Fixing a $$t$$, we “redefine” the interaction picture along the imaginary axis:\n\n\\begin{aligned} \\boxed{ \\hat{A}_I(\\tau) \\equiv \\exp\\!\\bigg(\\frac{\\tau \\hat{H}_{0,S}}{\\hbar}\\bigg) \\: \\hat{A}_S \\: \\exp\\!\\bigg( \\!-\\! \\frac{\\tau \\hat{H}_{0,S}}{\\hbar}\\bigg) } \\end{aligned}\n\nIronically, $$\\tau$$ is real; the point is that this formula comes from the real-time definition by replacing $$t \\to -i \\tau$$. The Heisenberg and Schrödinger pictures can be redefined in the same way.\n\nIn fact, by substituting $$t \\to -i \\tau$$, all the key results of the interaction picture can be updated, for example the Schrödinger equation for $$\\ket{\\psi_S(\\tau)}$$ becomes:\n\n\\begin{aligned} \\hbar \\dv{t} \\ket{\\psi_S(\\tau)} = - \\hat{H}_S \\ket{\\psi_S(\\tau)} \\quad \\implies \\quad \\ket{\\psi_S(\\tau)} = \\exp\\!\\bigg( \\!-\\! \\frac{\\tau \\hat{H}_S}{\\hbar} \\bigg) \\ket{\\psi_H} \\end{aligned}\n\nAnd the interaction picture’s time evolution operator $$\\hat{K}_I$$ turns out to be given by:\n\n\\begin{aligned} \\boxed{ \\hat{K}_I(\\tau, \\tau_0) = \\mathcal{T} \\bigg\\{ \\exp\\!\\bigg( \\!-\\! \\frac{1}{\\hbar} \\int_{\\tau_0}^\\tau \\hat{H}_{1,I}(\\tau') \\dd{\\tau'} \\bigg) \\bigg\\} } \\end{aligned}\n\nWhere $$\\mathcal{T}$$ is the time-ordered product with respect to $$\\tau$$. This operator works as expected:\n\n\\begin{aligned} \\ket{\\psi_I(\\tau)} = \\hat{K}_I(\\tau, \\tau_0) \\ket{\\psi_I(\\tau_0)} \\end{aligned}\n\nWhere $$\\ket{\\psi_I(\\tau)}$$ is related to the Schrödinger and Heisenberg pictures as follows:\n\n\\begin{aligned} \\ket{\\psi_I(\\tau)} \\equiv \\exp\\!\\bigg(\\frac{\\tau \\hat{H}_{0,S}}{\\hbar}\\bigg) \\ket{\\psi_S(\\tau)} = \\exp\\!\\bigg(\\frac{\\tau \\hat{H}_{0,S}}{\\hbar}\\bigg) \\exp\\!\\bigg( \\!-\\! \\frac{\\tau \\hat{H}_S}{\\hbar}\\bigg) \\ket{\\psi_H} \\end{aligned}\n\nIt is interesting to combine this definition with the action of time evolution $$\\hat{K}_I(\\tau, \\tau_0)$$:\n\n\\begin{aligned} \\ket{\\psi_I(\\tau)} &= \\hat{K}_I(\\tau, \\tau_0) \\ket{\\psi_I(\\tau_0)} \\\\ \\exp\\!\\bigg(\\frac{\\tau \\hat{H}_{0,S}}{\\hbar}\\bigg) \\exp\\!\\bigg( \\!-\\! \\frac{\\tau \\hat{H}_S}{\\hbar}\\bigg) \\ket{\\psi_H} &= \\hat{K}_I(\\tau, \\tau_0) \\exp\\!\\bigg(\\frac{\\tau_0 \\hat{H}_{0,S}}{\\hbar}\\bigg) \\exp\\!\\bigg( \\!-\\! \\frac{\\tau_0 \\hat{H}_S}{\\hbar}\\bigg) \\ket{\\psi_H} \\end{aligned}\n\nRearranging this leads to the following useful alternative expression for $$\\hat{K}_I(\\tau, \\tau_0)$$:\n\n\\begin{aligned} \\boxed{ \\hat{K}_I(\\tau, \\tau_0) = \\exp\\!\\bigg(\\frac{\\tau \\hat{H}_{0,S}}{\\hbar}\\bigg) \\exp\\!\\bigg(\\!-\\! \\frac{(\\tau \\!-\\! \\tau_0) \\hat{H}_{S}}{\\hbar}\\bigg) \\exp\\!\\bigg(\\!-\\! \\frac{\\tau_0 \\hat{H}_{0,S}}{\\hbar}\\bigg) } \\end{aligned}\n\nReturning to our initial example, we can set $$\\tau = \\hbar \\beta$$ and $$\\tau_0 = 0$$, so $$\\hat{K}_I(\\tau, \\tau_0)$$ becomes:\n\n\\begin{aligned} \\hat{K}_I(\\hbar \\beta, 0) &= \\exp\\!\\big(\\beta \\hat{H}_{0,S}\\big) \\exp\\!\\big(\\!-\\! \\beta \\hat{H}_{S}\\big) \\\\ \\implies \\quad \\exp\\!\\big(\\!-\\! \\beta \\hat{H}_{S}\\big) &= \\exp\\!\\big(\\!-\\! \\beta \\hat{H}_{0,S}\\big) \\hat{K}_I(\\hbar \\beta, 0) \\end{aligned}\n\nUsing the easily-shown fact that $$\\hat{K}_I(\\hbar \\beta, 0) \\hat{K}_I(0, \\tau) = \\hat{K}_I(\\hbar \\beta, \\tau)$$, we can therefore rewrite the thermodynamic expectation value like so:\n\n\\begin{aligned} \\expval*{\\hat{A}_H(\\tau) \\hat{B}_H(\\tau')} &= \\frac{1}{Z} \\Tr\\!\\Big(\\! \\exp\\!(-\\beta \\hat{H}_{0,S}) \\hat{K}_I(\\hbar \\beta, \\tau) \\hat{A}_I(\\tau) \\hat{K}_I(\\tau, \\tau') \\hat{B}_I(\\tau') \\hat{K}_I(\\tau', 0) \\!\\Big) \\end{aligned}\n\nWe now introduce a time-ordering $$\\mathcal{T}$$, letting us reorder the (bosonic) $$\\hat{K}_I$$-operators inside, and thereby reduce the expression considerably:\n\n\\begin{aligned} \\expval{\\mathcal{T}\\Big\\{\\hat{A}_H \\hat{B}_H\\Big\\}} &= \\frac{1}{Z} \\Tr\\!\\Big( \\mathcal{T} \\Big\\{ \\hat{K}_I(\\hbar \\beta, \\tau) \\hat{K}_I(\\tau, \\tau') \\hat{K}_I(\\tau', 0) \\hat{A}_I(\\tau) \\hat{B}_I(\\tau') \\Big\\} \\exp\\!(-\\beta \\hat{H}_{0,S}) \\Big) \\\\ &= \\frac{1}{Z} \\Tr\\!\\Big( \\mathcal{T}\\Big\\{ \\hat{K}_I(\\hbar \\beta, 0) \\hat{A}_I(\\tau) \\hat{B}_I(\\tau') \\Big\\} \\exp\\!(-\\beta \\hat{H}_{0,S}) \\Big) \\end{aligned}\n\nWhere $$Z = \\Tr\\!\\big(\\exp\\!(-\\beta \\hat{H}_S)\\big) = \\Tr\\!\\big(\\hat{K}_I(\\hbar \\beta, 0) \\exp\\!(-\\beta \\hat{H}_{0,S})\\big)$$. If we now define $$\\expval{}_0$$ as the expectation value with respect to the unperturbed equilibrium involving only $$\\hat{H}_{0,S}$$, we arrive at the following way of writing this time-ordered expectation:\n\n\\begin{aligned} \\boxed{ \\expval{\\mathcal{T}\\Big\\{\\hat{A}_H \\hat{B}_H\\Big\\}} = \\frac{\\expval{\\mathcal{T}\\Big\\{ \\hat{K}_I(\\hbar \\beta, 0) \\hat{A}_I(\\tau) \\hat{B}_I(\\tau') \\Big\\}}_0}{\\expval{\\hat{K}_I(\\hbar \\beta, 0)}_0} } \\end{aligned}\n\nFor another application of imaginary time, see e.g. the Matsubara Green’s function.\n\n1. H. Bruus, K. Flensberg, Many-body quantum theory in condensed matter physics, 2016, Oxford.\n\n© Marcus R.A. Newman, a.k.a. \"Prefetch\". Available under CC BY-SA 4.0." ]

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[ "## пятница, 10 января 2014 г.\n\n### Codility. Train. Genomic-range-query ★★★\n\nA non-empty zero-indexed string S is given. String S consists of N characters from the set of upper-case English letters A, C, G, T.\nThis string actually represents a DNA sequence, and the upper-case letters represent single nucleotides.\nYou are also given non-empty zero-indexed arrays P and Q consisting of M integers. These arrays represent queries about minimal nucleotides. We represent the letters of string S as integers 1, 2, 3, 4 in arrays P and Q, where A = 1, C = 2, G = 3, T = 4, and we assume that A < C < G < T.\nQuery K requires you to find the minimal nucleotide from the range (P[K], Q[K]), 0 ≤ P[i] ≤ Q[i] < N.\nFor example, consider string S = GACACCATA and arrays P, Q such that:\n``` P = 0 Q = 8\nP = 0 Q = 2\nP = 4 Q = 5\nP = 7 Q = 7```\nThe minimal nucleotides from these ranges are as follows:\n• (0, 8) is A identified by 1,\n• (0, 2) is A identified by 1,\n• (4, 5) is C identified by 2,\n• (7, 7) is T identified by 4.\nWrite a function:\ndef solution(S, P, Q)\nthat, given a non-empty zero-indexed string S consisting of N characters and two non-empty zero-indexed arrays P and Q consisting of M integers, returns an array consisting of M characters specifying the consecutive answers to all queries.\nThe sequence should be returned as:\n• a Results structure (in C), or\n• a vector of integers (in C++), or\n• a Results record (in Pascal), or\n• an array of integers (in any other programming language).\nFor example, given the string S = GACACCATA and arrays P, Q such that:\n``` P = 0 Q = 8\nP = 0 Q = 2\nP = 4 Q = 5\nP = 7 Q = 7```\nthe function should return the values [1, 1, 2, 4], as explained above.\nAssume that:\n• N is an integer within the range [1..100,000];\n• M is an integer within the range [1..50,000];\n• each element of array P, Q is an integer within the range [0..N − 1];\n• P[i] ≤ Q[i];\n• string S consists only of upper-case English letters A, C, G, T.\nComplexity:\n• expected worst-case time complexity is O(N+M);\n• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).\nElements of input arrays can be modified.\n\nSolution:\n```def solution(S, P, Q):\nvMap = dict(A = 1, C = 2, G = 3, T = 4)\nvLst = dict(A =[], C =[], G =[], T =[])\nfor i in xrange(len(S)):\nvLst[S[i]].append(i)\nfor i in xrange(len(P)):\nfor j in sorted(vLst):\niF, iL = 0, len(vLst[j])\nif iL == iF or \\\nvLst[j][iF] > Q[i] or \\\nvLst[j][iL - 1] < P[i]:\ncontinue\nwhile iF < iL:\niM = iF + (iL - iF) / 2\nif P[i] <= vLst[j][iM]:\nif vLst[j][iM] <= Q[i]:\nP[i] = vMap[j]\nbreak\niL = iM\nelse:\niF = iM + 1\nelse:\ncontinue\nbreak\nreturn P```" ]

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[ "# Python frozenset()\n\nFiled Under: Python\n\nPython frozenset is an unordered collection of distinct hashable objects. Frozenset is an immutable set, so its contents cannot be modified after it’s created.\n\n## Python frozenset()\n\nPython frozenset() function is used to create frozenset object. Its syntax is:\n\n``````\nclass frozenset([iterable])\n``````\n\nIf the input iterable argument is provided, then forzenset is created from the iterable elements. If no argument is provided, then an empty frozenset object is returned.\n\n## Python frozenset() example\n\nLet’s see how to use frozenset() function to create frozenset object.\n\n``````\nfs = frozenset()\nprint(fs)\nprint(type(fs))\n\n# frozenset from iterable\nlist_vowels = ['A', 'E', 'I', 'O', 'U']\nfs = frozenset(list_vowels)\nprint(fs)\n\ntuple_numbers = (1, 2, 3, 4, 5, 4, 3)\nfs = frozenset(tuple_numbers)\nprint(fs)\n``````\n\nOutput:\n\n``````\nfrozenset()\n<class 'frozenset'>\nfrozenset({'E', 'U', 'I', 'O', 'A'})\nfrozenset({1, 2, 3, 4, 5})\n``````\n\n## Iterating frozenset elements\n\nWe can use for loop to iterate through frozen set elements.\n\n``````\nfs = frozenset([1, 2, 3, 4, 5, 4, 3])\nfor x in fs:\nprint(x)\n``````\n\nOutput:\n\n``````\n1\n2\n3\n4\n5\n``````\n\nNote that the duplicate elements are removed while frozenset is being created.\n\n## Python frozenset functions\n\nSince frozenset is immutable, there are no methods available to alter its elements. So add(), remove(), update(), pop() etc. functions are not defined for frozenset.\n\nLet’s look at some of the methods available for the frozenset object.\n\n• `len(fs)`: returns the number of elements in the frozenset.\n• `x in fs`: returns True if x is present in fs, else returns False.\n• `x not in fs`: returns True if x is not present in fs, else returns False.\n• `isdisjoint(other)`: returns True if the frozenset has no elements in common with other. Two sets are disjoint if and only if their intersection is the empty set.\n• `issubset(other)`: returns True if every element of the set is present in the other set, else returns False.\n• `issuperset(other)`: returns True if every element in other is present in the set, else returns False.\n• `union(*others)`: returns a new frozenset object with elements from the frozenset and other sets.\n• `intersection(*others)`: returns a new frozenset with elements from this set and all the other sets.\n• `difference(*others)`: returns a new frozenset with elements in the frozenset that are not in the other sets.\n• `symmetric_difference(other)`: return a new frozenset with elements in either the frozenset or other but not both.\n``````\nfs = frozenset([1, 2, 3, 4, 5])\n\nsize = len(fs)\nprint('frozenset size =', size)\n\ncontains_item = 5 in fs\nprint('fs contains 5 =', contains_item)\n\nnot_contains_item = 6 not in fs\nprint('fs not contains 6 =', not_contains_item)\n\nis_disjoint = fs.isdisjoint(frozenset([1, 2]))\nprint(is_disjoint)\n\nis_disjoint = fs.isdisjoint(frozenset([10, 20]))\nprint(is_disjoint)\n\nis_subset = fs.issubset(set([1, 2]))\nprint(is_subset)\n\nis_subset = fs.issubset(set([1, 2, 3, 4, 5, 6, 7]))\nprint(is_subset)\n\nis_superset = fs.issuperset(frozenset([1, 2]))\nprint(is_superset)\n\nis_superset = fs.issuperset(frozenset([1, 2, 10]))\nprint(is_superset)\n\nfs1 = fs.union(frozenset([1, 2, 10]), set([99, 98]))\nprint(fs1)\n\nfs1 = fs.intersection(set((1, 2, 10, 20)))\nprint(fs1)\n\nfs1 = fs.difference(frozenset([1, 2, 3]))\nprint(fs1)\n\nfs1 = fs.symmetric_difference(frozenset([1, 2, 10, 20]))\nprint(fs1)\n\nfs1 = fs.copy()\nprint(fs1)\n``````\n\nOutput:\n\n``````\nfrozenset size = 5\nfs contains 5 = True\nfs not contains 6 = True\nFalse\nTrue\nFalse\nTrue\nTrue\nFalse\nfrozenset({1, 2, 3, 4, 5, 98, 99, 10})\nfrozenset({1, 2})\nfrozenset({4, 5})\nfrozenset({3, 20, 4, 5, 10})\nfrozenset({1, 2, 3, 4, 5})\n``````\n\n## Python frozenset to list, tuple\n\nWe can easily create list and tuple from frozenset object using built-in functions.\n\n``````\nfs = frozenset([1, 2, 3, 4, 5])\nl1 = list(fs)\nprint(l1)\n\nt1 = tuple(fs)\nprint(t1)\n``````\n\nOutput:\n\n``````\n[1, 2, 3, 4, 5]\n(1, 2, 3, 4, 5)\n``````\n\nThat’s all for python frozenset object and frozenset() built-in function.\n\nYou can checkout complete python script and more Python examples from our GitHub Repository.\n\nReference: Official Documentation\n\nclose\nGeneric selectors\nExact matches only\nSearch in title\nSearch in content\nSearch in posts\nSearch in pages" ]

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[ "# Document (#19553)\n\nAuthor\nHawkins, D.T.\nTitle\nWeb-based training for online retrieval : some examples\nSource\nOnline. 21(1997) no.5, S.73-75\nYear\n1997\nAbstract\nDescribes the efforts of Knight-Ridder Information (KRI) and Dun & Bradstreet (D&B) to provide Web based training systems. The KRI tutorial covers the DIALOG Web service, describes the technology, the tutorial and evaluates its usefulness. D&B's web based training trains users on its online system. It is more advanced than the KRI tutorial. Describes its training modules, and its testing of methods and technologies. The University of Texas as Austin has developed 2 tutorials on searching for patents and trademarks\nTheme\nInternet\nComputer Based Training\nObject\nKnight-Ridder\n\n## Similar documents (author)\n\n1. 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Hawkins, D.T.: Growth trends in the electronic information services market (1994) 5.39\n```5.3864803 = sum of:\n5.3864803 = weight(author_txt:hawkins in 7018) [ClassicSimilarity], result of:\n5.3864803 = fieldWeight in 7018, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.618368 = idf(docFreq=20, maxDocs=42740)\n0.625 = fieldNorm(doc=7018)\n```\n\n## Similar documents (content)\n\n1. 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Balas, J.: Online training resources (1998) 0.23\n```0.23324707 = sum of:\n0.23324707 = product of:\n0.8330253 = sum of:\n0.01946066 = weight(abstract_txt:technology in 2822) [ClassicSimilarity], result of:\n0.01946066 = score(doc=2822,freq=1.0), product of:\n0.057985857 = queryWeight, product of:\n1.006568 = boost\n4.2958136 = idf(docFreq=1582, maxDocs=42740)\n0.013410147 = queryNorm\n0.33561045 = fieldWeight in 2822, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.2958136 = idf(docFreq=1582, maxDocs=42740)\n0.078125 = fieldNorm(doc=2822)\n0.04223126 = weight(abstract_txt:online in 2822) [ClassicSimilarity], result of:\n0.04223126 = score(doc=2822,freq=3.0), product of:\n0.08490684 = queryWeight, product of:\n1.7225363 = boost\n3.6757057 = idf(docFreq=2942, maxDocs=42740)\n0.013410147 = queryNorm\n0.4973835 = fieldWeight in 2822, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n3.6757057 = idf(docFreq=2942, maxDocs=42740)\n0.078125 = fieldNorm(doc=2822)\n0.12565562 = weight(abstract_txt:tutorials in 2822) [ClassicSimilarity], result of:\n0.12565562 = score(doc=2822,freq=1.0), product of:\n0.20106584 = queryWeight, product of:\n1.8743522 = boost\n7.999329 = idf(docFreq=38, maxDocs=42740)\n0.013410147 = queryNorm\n0.6249476 = fieldWeight in 2822, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.999329 = idf(docFreq=38, maxDocs=42740)\n0.078125 = fieldNorm(doc=2822)\n0.024332728 = weight(abstract_txt:based in 2822) [ClassicSimilarity], result of:\n0.024332728 = score(doc=2822,freq=1.0), product of:\n0.09706247 = queryWeight, product of:\n2.2556326 = boost\n3.2088501 = idf(docFreq=4693, maxDocs=42740)\n0.013410147 = queryNorm\n0.2506914 = fieldWeight in 2822, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.2088501 = idf(docFreq=4693, maxDocs=42740)\n0.078125 = fieldNorm(doc=2822)\n0.040459838 = weight(abstract_txt:describes in 2822) [ClassicSimilarity], result of:\n0.040459838 = score(doc=2822,freq=1.0), product of:\n0.13623025 = queryWeight, product of:\n2.6722646 = boost\n3.8015487 = idf(docFreq=2594, maxDocs=42740)\n0.013410147 = queryNorm\n0.296996 = fieldWeight in 2822, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.8015487 = idf(docFreq=2594, maxDocs=42740)\n0.078125 = fieldNorm(doc=2822)\n0.22968635 = weight(abstract_txt:training in 2822) [ClassicSimilarity], result of:\n0.22968635 = score(doc=2822,freq=3.0), product of:\n0.33084157 = queryWeight, product of:\n4.808638 = boost\n5.130556 = idf(docFreq=686, maxDocs=42740)\n0.013410147 = queryNorm\n0.69424874 = fieldWeight in 2822, product of:\n1.7320508 = tf(freq=3.0), with freq of:\n3.0 = termFreq=3.0\n5.130556 = idf(docFreq=686, maxDocs=42740)\n0.078125 = fieldNorm(doc=2822)\n0.35119882 = weight(abstract_txt:tutorial in 2822) [ClassicSimilarity], result of:\n0.35119882 = score(doc=2822,freq=1.0), product of:\n0.5753863 = queryWeight, product of:\n5.4918976 = boost\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.013410147 = queryNorm\n0.6103705 = fieldWeight in 2822, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n7.8127427 = idf(docFreq=46, maxDocs=42740)\n0.078125 = fieldNorm(doc=2822)\n0.28 = coord(7/25)\n```\n3. 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Pack, T.: Greater than the sum of their parts : vendors and services taking the hybrid lead (1996) 0.18\n```0.17857878 = sum of:\n0.17857878 = product of:\n0.89289385 = sum of:\n0.06202847 = weight(abstract_txt:technologies in 4798) [ClassicSimilarity], result of:\n0.06202847 = score(doc=4798,freq=1.0), product of:\n0.0791148 = queryWeight, product of:\n1.175739 = boost\n5.0177994 = idf(docFreq=768, maxDocs=42740)\n0.013410147 = queryNorm\n0.78403115 = fieldWeight in 4798, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.0177994 = idf(docFreq=768, maxDocs=42740)\n0.15625 = fieldNorm(doc=4798)\n0.04876446 = weight(abstract_txt:online in 4798) [ClassicSimilarity], result of:\n0.04876446 = score(doc=4798,freq=1.0), product of:\n0.08490684 = queryWeight, product of:\n1.7225363 = boost\n3.6757057 = idf(docFreq=2942, maxDocs=42740)\n0.013410147 = queryNorm\n0.574329 = fieldWeight in 4798, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.6757057 = idf(docFreq=2942, maxDocs=42740)\n0.15625 = fieldNorm(doc=4798)\n0.33145466 = weight(abstract_txt:knight in 4798) [ClassicSimilarity], result of:\n0.33145466 = score(doc=4798,freq=1.0), product of:\n0.2418131 = queryWeight, product of:\n2.0555212 = boost\n8.772519 = idf(docFreq=17, maxDocs=42740)\n0.013410147 = queryNorm\n1.3707061 = fieldWeight in 4798, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.772519 = idf(docFreq=17, maxDocs=42740)\n0.15625 = fieldNorm(doc=4798)\n0.36972657 = weight(abstract_txt:ridder in 4798) [ClassicSimilarity], result of:\n0.36972657 = score(doc=4798,freq=1.0), product of:\n0.2600863 = queryWeight, product of:\n2.1317723 = boost\n9.097941 = idf(docFreq=12, maxDocs=42740)\n0.013410147 = queryNorm\n1.4215534 = fieldWeight in 4798, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.097941 = idf(docFreq=12, maxDocs=42740)\n0.15625 = fieldNorm(doc=4798)\n0.080919676 = weight(abstract_txt:describes in 4798) [ClassicSimilarity], result of:\n0.080919676 = score(doc=4798,freq=1.0), product of:\n0.13623025 = queryWeight, product of:\n2.6722646 = boost\n3.8015487 = idf(docFreq=2594, maxDocs=42740)\n0.013410147 = queryNorm\n0.593992 = fieldWeight in 4798, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n3.8015487 = idf(docFreq=2594, maxDocs=42740)\n0.15625 = fieldNorm(doc=4798)\n0.2 = coord(5/25)\n```\n5. Guenette, D.R.: ¬The CD-ROM-online connection (1996) 0.16\n```0.1628775 = sum of:\n0.1628775 = product of:\n0.58170533 = sum of:\n0.01946066 = weight(abstract_txt:technology in 4796) [ClassicSimilarity], result of:\n0.01946066 = score(doc=4796,freq=1.0), product of:\n0.057985857 = queryWeight, product of:\n1.006568 = boost\n4.2958136 = idf(docFreq=1582, maxDocs=42740)\n0.013410147 = queryNorm\n0.33561045 = fieldWeight in 4796, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.2958136 = idf(docFreq=1582, maxDocs=42740)\n0.078125 = fieldNorm(doc=4796)\n0.030587196 = weight(abstract_txt:examples in 4796) [ClassicSimilarity], result of:\n0.030587196 = score(doc=4796,freq=1.0), product of:\n0.0783869 = queryWeight, product of:\n1.1703178 = boost\n4.9946623 = idf(docFreq=786, maxDocs=42740)\n0.013410147 = queryNorm\n0.390208 = fieldWeight in 4796, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n4.9946623 = idf(docFreq=786, maxDocs=42740)\n0.078125 = fieldNorm(doc=4796)\n0.031014236 = weight(abstract_txt:technologies in 4796) [ClassicSimilarity], result of:\n0.031014236 = score(doc=4796,freq=1.0), product of:\n0.0791148 = queryWeight, product of:\n1.175739 = boost\n5.0177994 = idf(docFreq=768, maxDocs=42740)\n0.013410147 = queryNorm\n0.39201558 = fieldWeight in 4796, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n5.0177994 = idf(docFreq=768, maxDocs=42740)\n0.078125 = fieldNorm(doc=4796)\n0.101288155 = score(doc=4796,freq=2.0), product of:\n0.13822274 = queryWeight, product of:\n1.5540744 = boost\n6.6324525 = idf(docFreq=152, maxDocs=42740)\n0.013410147 = queryNorm\n0.7327894 = fieldWeight in 4796, product of:\n1.4142135 = tf(freq=2.0), with freq of:\n2.0 = termFreq=2.0\n6.6324525 = idf(docFreq=152, maxDocs=42740)\n0.078125 = fieldNorm(doc=4796)\n0.04876446 = weight(abstract_txt:online in 4796) [ClassicSimilarity], result of:\n0.04876446 = score(doc=4796,freq=4.0), product of:\n0.08490684 = queryWeight, product of:\n1.7225363 = boost\n3.6757057 = idf(docFreq=2942, maxDocs=42740)\n0.013410147 = queryNorm\n0.574329 = fieldWeight in 4796, product of:\n2.0 = tf(freq=4.0), with freq of:\n4.0 = termFreq=4.0\n3.6757057 = idf(docFreq=2942, maxDocs=42740)\n0.078125 = fieldNorm(doc=4796)\n0.16572733 = weight(abstract_txt:knight in 4796) [ClassicSimilarity], result of:\n0.16572733 = score(doc=4796,freq=1.0), product of:\n0.2418131 = queryWeight, product of:\n2.0555212 = boost\n8.772519 = idf(docFreq=17, maxDocs=42740)\n0.013410147 = queryNorm\n0.68535304 = fieldWeight in 4796, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n8.772519 = idf(docFreq=17, maxDocs=42740)\n0.078125 = fieldNorm(doc=4796)\n0.18486328 = weight(abstract_txt:ridder in 4796) [ClassicSimilarity], result of:\n0.18486328 = score(doc=4796,freq=1.0), product of:\n0.2600863 = queryWeight, product of:\n2.1317723 = boost\n9.097941 = idf(docFreq=12, maxDocs=42740)\n0.013410147 = queryNorm\n0.7107767 = fieldWeight in 4796, product of:\n1.0 = tf(freq=1.0), with freq of:\n1.0 = termFreq=1.0\n9.097941 = idf(docFreq=12, maxDocs=42740)\n0.078125 = fieldNorm(doc=4796)\n0.28 = coord(7/25)\n```" ]

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[ "# The density of rational points and invariants of genus one curves Doctoral thesis, 2019\n\nThe present thesis contains three papers dealing with two arithmetic problems on curves of genus one, which are closely related to elliptic curves.\n\nThe first problem is to study the density of rational points presented in Papers I and II. We give uniform upper bounds for the number of rational points of bounded height on smooth curves of genus one given by ternary cubics or complete intersections of two quadratic surfaces. The main tools used in these two papers are descent on elliptic curves and determinant methods. While working with the rational points counting problem, one need to deal with the smoothness of geometric objects and the bad reduction of polynomials. To characterize these properties, there is a classical object called the discriminant which naturally appears.\n\nThe above discriminant gives an inspiration to the study of the next problem in the thesis concerning invariants of models of genus one curves presented in Paper III. Here an invariant of a genus one curve is a polynomial in the coefficients of the model defining the curve that is stable under certain linear transformations. The discriminant is a classical example of an invariant. Besides that, there are two more important invariants which generate the ring of invariants of genus one models over a field. Fisher considered these invariants over the field of rational numbers and normalized them such that they are moreover defined over the integers. We provide an alternative way to express these normalized invariants using a natural connection to modular forms. In the case of the discriminant of ternary cubics over the complex numbers, we also present another approach using determinantal representations. This latter approach produces a natural connection to theta functions.\n\nThe common idea in the thesis is to link a smooth genus one curve to a Weierstrass form, which is a more well-understood object.\n\ngenus one\n\ndiscriminant\n\nmodular form\n\nElliptic curve\n\nheight\n\ndescent\n\ndeterminantal representation\n\ninvariant\n\nrational point\n\ntheta function.\n\ndeterminant method\n\nPascal, Hörsalsvägen 1\nOpponent: Professor Fabien Pazuki, University of Copenhagen, Denmark.\n\n## Author\n\n#### Manh Hung Tran\n\nChalmers, Mathematical Sciences, Algebra and geometry\n\n#### Counting rational points on smooth cubic curves\n\nJournal of Number Theory,; Vol. 189(2018)p. 138-146\n\nJournal article\n\n#### Manh Hung Tran. Invariants of models of genus one curves via and modular forms and determinantal representations, Submitted, arXiv:1911.01350.\n\nAlgebraic equations represent an important subject in mathematics. The solution sets of equations are viewed as geometric objects such as curves, surfaces, etc. Among them, there are classical objects called elliptic curves. These types of curves can be defined by cubic polynomials. Elliptic curves appear in many fields of mathematics such as number theory, algebraic geometry, complex analysis, etc.\n\nThis thesis studies two important arithmetic problems on curves of genus one, which are natural generalizations of elliptic curves. The first problem is to study the density of points over the rational numbers on such curves. If there are infinitely many rational points on a genus one curve, one might quantify the infinite by considering the density of points in a bounded domain. We provide upper bounds for the number of rational points in a box of given size on smooth curves of genus one defined by ternary cubics or intersections of two quadric surfaces, which are two typical ways of representing curves of genus one. The implicit constants in these bounds are moreover uniform which are independent of the equations defining the curves.\n\nWhen considering working with the density of rational points on genus one curves problem, there is a classical notion called the discriminant which naturally appears to characterize the smoothness of the curves. This discriminant is a classical example of what is called an invariant. Here an invariant of a genus one curve is a polynomial in the coefficients of the polynomials defining the curve, which is unchanged under certain natural transformations. The discriminant led the author to work on the second problem in this thesis which concerns invariants of genus one curves.\n\nAmong all invariants of genus one curves over a field, there are two important ones which can be used to describe all the others. These two invariants were normalized by Fisher in such a way that they are primitive polynomials with integer coefficients. In this thesis, we provide an alternative way to express these normalized invariants by naturally relating to modular forms. These modular forms are functions on the upperhalf plane satisfying a certain transformation and can be expressed in terms of Taylor series. If all coefficients of the Taylor series are integral, the corresponding modular form is defined over the integers and so is the associated invariant. The behavior of invariants will then be predicted by the transformation property of modular forms. The special case of the discriminant of ternary cubics over the complex numbers is also treated by an another approach.\n\nMANH HUNG TRAN\n\nMathematics\n\nGeometry\n\n#### ISBN\n\n978-91-7905-204-1\n\nDoktorsavhandlingar vid Chalmers tekniska högskola. Ny serie: 4671\n\n#### Publisher\n\nChalmers University of Technology\n\nPascal, Hörsalsvägen 1\n\nOpponent: Professor Fabien Pazuki, University of Copenhagen, Denmark." ]

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[ "# 10719850\n\n## 10,719,850 is an even composite number composed of four prime numbers multiplied together.\n\nWhat does the number 10719850 look like?\n\nThis visualization shows the relationship between its 4 prime factors (large circles) and 24 divisors.\n\n10719850 is an even composite number. It is composed of four distinct prime numbers multiplied together. It has a total of twenty-four divisors.\n\n## Prime factorization of 10719850:\n\n### 2 × 52 × 29 × 7393\n\n(2 × 5 × 5 × 29 × 7393)\n\nSee below for interesting mathematical facts about the number 10719850 from the Numbermatics database.\n\n### Names of 10719850\n\n• Cardinal: 10719850 can be written as Ten million, seven hundred nineteen thousand, eight hundred fifty.\n\n### Scientific notation\n\n• Scientific notation: 1.071985 × 107\n\n### Factors of 10719850\n\n• Number of distinct prime factors ω(n): 4\n• Total number of prime factors Ω(n): 5\n• Sum of prime factors: 7429\n\n### Divisors of 10719850\n\n• Number of divisors d(n): 24\n• Complete list of divisors:\n• Sum of all divisors σ(n): 20629260\n• Sum of proper divisors (its aliquot sum) s(n): 9909410\n• 10719850 is a deficient number, because the sum of its proper divisors (9909410) is less than itself. Its deficiency is 810440\n\n### Bases of 10719850\n\n• Binary: 1010001110010010011010102\n• Hexadecimal: 0xA3926A\n• Base-36: 6DRHM\n\n### Squares and roots of 10719850\n\n• 10719850 squared (107198502) is 114915184022500\n• 10719850 cubed (107198503) is 1231873535443596625000\n• The square root of 10719850 is 3274.1182018981\n• The cube root of 10719850 is 220.4937258581\n\n### Scales and comparisons\n\nHow big is 10719850?\n• 10,719,850 seconds is equal to 17 weeks, 5 days, 1 hour, 44 minutes, 10 seconds.\n• To count from 1 to 10,719,850 would take you about twenty-six weeks!\n\nThis is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)\n\n• A cube with a volume of 10719850 cubic inches would be around 18.4 feet tall.\n\n### Recreational maths with 10719850\n\n• 10719850 backwards is 05891701\n• The number of decimal digits it has is: 8\n• The sum of 10719850's digits is 31\n• More coming soon!\n\n## Link to this page\n\nHTML: To link to this page, just copy and paste the link below into your blog, web page or email.\n\nBBCODE: To link to this page in a forum post or comment box, just copy and paste the link code below:\n\n## Cite this page\n\nMLA style:\n\"Number 10719850 - Facts about the integer\". Numbermatics.com. 2021. Web. 20 June 2021.\n\nAPA style:\nNumbermatics. (2021). Number 10719850 - Facts about the integer. Retrieved 20 June 2021, from https://numbermatics.com/n/10719850/\n\nChicago style:\nNumbermatics. 2021. \"Number 10719850 - Facts about the integer\". https://numbermatics.com/n/10719850/\n\nThe information we have on file for 10719850 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!\n\nKeywords: Divisors of 10719850, math, Factors of 10719850, curriculum, school, college, exams, university, Prime factorization of 10719850, STEM, science, technology, engineering, physics, economics, calculator, ten million, seven hundred nineteen thousand, eight hundred fifty." ]

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[ "#", null, "Relevance Vector Machine (RapidMiner Studio Core)\n\n## Synopsis\n\nThis operator is an implementation of Relevance Vector Machine (RVM) which is a probabilistic method both for classification and regression.\n\n## Description\n\nThe Relevance Vector Machine operator is a probabilistic method both for classification and regression. The implementation of the relevance vector machine is based on the original algorithm described by 'Tipping/2001'. The fast version of the marginal likelihood maximization ('Tipping/Faul/2003') is also available if the rvm type parameter is set to 'Constructive-Regression-RVM'.\n\nA Relevance Vector Machine (RVM) is a machine learning technique that uses Bayesian inference to obtain parsimonious solutions for regression and classification. The RVM has an identical functional form to the support vector machine, but provides probabilistic classification. It is actually equivalent to a Gaussian process model with a certain covariance function. Compared to that of support vector machines (SVM), the Bayesian formulation of the RVM avoids the set of free parameters of the SVM (that usually require cross-validation-based post-optimizations). However RVMs use an expectation maximization (EM)-like learning method and are therefore at risk of local minima. This is unlike the standard sequential minimal optimization(SMO)-based algorithms employed by SVMs, which are guaranteed to find a global optimum.\n\n## Input\n\n•", null, "training set (IOObject)\n\nThis input port expects an ExampleSet. This operator cannot handle nominal attributes; it can be applied on data sets with numeric attributes. Thus often you may have to use the Nominal to Numerical operator before the application of this operator.\n\n## Output\n\n•", null, "model (Model)\n\nThe RVM is applied and the resultant model is delivered from this output port. This model can now be applied on unseen data sets.\n\n•", null, "example set (IOObject)\n\nThe ExampleSet that was given as input is passed without changing to the output through this port. This is usually used to reuse the same ExampleSet in further operators or to view the ExampleSet in the Results Workspace.\n\n## Parameters\n\n• rvm_typeThis parameter specifies the type of RVM Regression. The following options are available: Regression-RVM, Classification-RVM and Constructive-Regression-RVM. Range: selection\n• kernel_typeThe type of the kernel function is selected through this parameter. Following kernel types are supported: rbf, cauchy, laplace, poly, sigmoid, Epanechnikov, gaussian combination, multiquadric Range: selection\n• kernel_lengthscaleThis parameter specifies the lengthscale to be used in all kernels. Range: real\n• kernel_degreeThis is the kernel parameter degree. This is only available when the kernel type parameter is set to polynomial or epachnenikov. Range: real\n• kernel_biasThis parameter specifies the bias to be used in the poly kernel. Range: real\n• kernel_sigma1This is the kernel parameter sigma1. This is only available when the kernel type parameter is set to epachnenikov, gaussian combination or multiquadric. Range: real\n• kernel_sigma2This is the kernel parameter sigma2. This is only available when the kernel type parameter is set to gaussian combination. Range: real\n• kernel_sigma3This is the kernel parameter sigma3. This is only available when the kernel type parameter is set to gaussian combination. Range: real\n• kernel_shiftThis is the kernel parameter shift. This is only available when the kernel type parameter is set to multiquadric. Range: real\n• kernel_aThis is the kernel parameter a. This is only available when the kernel type parameter is set to sigmoid Range: real\n• kernel_bThis is the kernel parameter b. This is only available when the kernel type parameter is set to sigmoid Range: real\n• max_iterationThis parameter specifies the maximum number of iterations to be used. Range: integer\n• min_delta_log_alphaThe iteration is aborted if the largest log alpha change is smaller than min delta log alpha. Range: real\n• alpha_maxThe basis function is pruned if its alpha is larger than the alpha max. Range: real\n• use_local_random_seedThis parameter indicates if a local random seed should be used for randomization. Using the same value of local random seed will produce the same randomization. Range: boolean\n• local_random_seedThis parameter specifies the local random seed. This parameter is only available if the use local random seed parameter is set to true. Range: integer\n\n## Tutorial Processes\n\n### Introduction to the RVM operator\n\nThe 'Polynomial' data set is loaded using the Retrieve operator. The Split Validation operator is applied on it for training and testing a regression model. The Relevance Vector Machine operator is applied in the training subprocess of the Split Validation operator. All parameters are used with default values. The Relevance Vector Machine operator generates a regression model. The Apply Model operator is used in the testing subprocess to apply this model on the testing data set. The resultant labeled ExampleSet is used by the Performance operator for measuring the performance of the model. The regression model and its performance vector are connected to the output and it can be seen in the Results Workspace.", null, "" ]

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[ "", null, "# COMPARISON OF SYSTEM BETS AND EXPRESS BETS\n\n## Systems of bigger dimensions\n\nLet’s see what other system sizes are available and what their breakeven thresholds are.\n\nOf particular interest are the four-element systems. There is already a manoeuvre here. It is possible to make a dimension “2-4” or “3-4”.\n\nFor the size “2-4” the break-even point would be the coefficient of 1.42. At this variant from four outcomes 6 doubles are formed.\n\nIf a parlay were made and all 4 outcomes played, a profit would be made:\n\n300 * 1.42 * 1.42 * 1.42 * 1.42 = 1 220\n\nIn the system, \\$300 would be divided among six elements, \\$50 each If all played, we would get:\n\n6 * (50 * 1.42 * 1.42) = 605\n\nThat’s half the profit of the Express.\n\nIf one bet loses, 3 doubles lose and 3 win:\n\n3 * (50 * 1.42 * 1.42) = 303\n\nWe’re left with ours.\n\nOption “3-4” represents 4 trebles, it already has a break-even point at 1.59. Higher, than in the “2-4”, which is logical.\n\nAn express of four events at 1.59 would give a profit:\n\n300 * 1.59 * 1.59 * 1.59 * 1.59 = 1 917\n\nA fully winning 3-4 system would give:\n\n4 * (75 * 1.59 * 1.59 * 1.59) = 1 206\n\nFor the sake of comparison let’s calculate the system with the size “2-4”, but with the odds 1.59.\n\nIf all 4 bets are played, we get:\n\n6 * (50 * 1.59 * 1.59) = 758\n\nIf one loses, and 3 win:\n\n3 * (50 * 1.59 * 1.59) = 379\n\nAs you can see the total winning profit here is almost 2.5 times lower than in the Express, and by 37% in comparison with the winning systems of size “3-4”. But there is an option for winning just two positions out of four. In this case both the Express and the 3-4 will be lost.\n\n50 * 1.59 * 1.59 = 126\n\nThat is not all the money will be lost, but about 42% of the played money will be saved. This can be perceived as some kind of additional bonus.\n\n## Ratio of risks and profitability of expresses and systems\n\nWhen multiplying the odds and risks are also multiplied. Here we take the ideal case where the bookmaker’s quotes reflect the real probability minus the margin. This, of course, is not always the case. The bettor’s task is to avoid outcomes that are clearly undervalued or loaded with a bunch of “pop-ups”. It is necessary to take only adequately priced events, or with the presence of a certain preponderance in favor of the player. But these are nuances from the topic of other articles. Here we will assume that odds are adequate and margin is equal. The task is to compare the probability of winning the express and the system. By comparing the size of these winnings, we will be able to make important conclusions about the appropriateness of these designs.\n\nFor comparison, let’s take the same variants we started with: a treble and a “2-3” system with quotes of 1.74.\n\nLet’s assume that the pair of outcomes is priced as: 1.74 to 2.21. The margin is 2.72%. At a leverage of 1.74 it would be 1.52%. The real probability without margin would be 55.95%, or as a fraction: 0.5595. To find out the probability of winning the express, multiply the probabilities:\n\n0.5595 3 = 0.175\n\nOr 17.5%.\n\nThe probability of winning the whole system is identical, because all three pluses are needed.\n\nThe probability of winning a double in the system and getting a return:\n\n0.5595 2 = 0.313\n\nOr 31.3%.\n\nLet’s compare the potential profit and probabilities. Let’s take a distance of 3000 bets. From these, 1000 combinations of 3 will be formed, which will create the expresses and systems. Given the probability, only 175 out of the 1000 expresses will win.\n\nLet’s assume we still bet \\$300 on each Parlay. This would represent a total of \\$300,000\n\n175 * (300 * 1.74 * 1.74 * 1.74) = 276 571\n\nThe loss would be:\n\n276 571 – 300 000 = -23 429\n\nThe reason is the margin contained in each quote. Due to the multiplication within an express, this drawdown is obtained. Let’s take a look at the systems.\n\nGiven the calculated probabilities, at the same distance 175 systems will also win and another 313 will give a return.\n\n175 * [ 3 * (1.74 * 1.74 * 100) ] = 158 949\n\n313 * (1.74 * 1.74 * 100) = 94 784\n\n158 949 + 94 784 = 253 733\n\nThe loss will be:\n\n253 733 – 300 000 = -46 267\n\nWe have clearly proved that expresses are more profitable than systems over the distance. Curtain." ]

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[ "", null, "# Embedding GIF animations in IPython notebook\n\nA couple of days ago I was thinking about the possibility of generating GIF animations using matplotlib and visualizing this animations directly in IPython notebooks. So I did a quick search to find possible solutions, and found one solution to embed videos, and other to convert matplotlib animations to gif, so I combined both solution in a third solution converting the animation to GIF using imagemagick (you will need to have it installed in your computer), enconding the resulting file in a base64 string and embedding this in a img tag. Using IPython.display HTML class this resulting string will be displayed in the notebook (and saved with the notebook).\n\nI will briefly explain the elements of my solution.\n\nFirst, import HTML class from IPython.display module. This will enable results which are valid html string to be interpreted and displayed by the notebook. We will return a string with an inline image using DataUri\n\n```from IPython.display import HTML\nIMG_TAG = &amp;quot;&amp;quot;&amp;quot;&amp;amp;lt;img src=&amp;quot;data:image/gif;base64,{0}&amp;quot; alt=&amp;quot;some_text&amp;quot;&amp;amp;gt;&amp;quot;&amp;quot;&amp;quot;\n\n```\n\nThe other important piece of the script is the function that takes a matplotlib animation object, convert and save the GIF to a temporary file using imagemagick, and finally read the file and convert to a base64 string to be inserted in the inline image string.\n\n```def anim_to_gif(anim):\ndata=&amp;quot;0&amp;quot;\nwith NamedTemporaryFile(suffix='.gif') as f:\nanim.save(f.name, writer='imagemagick', fps=10);\ndata = data.encode(&amp;quot;base64&amp;quot;)\nreturn IMG_TAG.format(data)\n\ndef display_animation(anim):\nplt.close(anim._fig)\nreturn HTML(anim_to_gif(anim))\n```\n\nAfter that you just need to generate a matplotlib animation and pass it to the display_animation function. This method could be generalize to other libraries, as long as it has some away to convert animations to GIF. Once the conversion is completed that file can be loaded, converted to inline image and displayed in a IPython notebook.\n\nFull code:\n\n```from tempfile import NamedTemporaryFile\nimport matplotlib.pyplot as plt\nfrom matplotlib import animation\nfrom IPython.display import HTML\nimport numpy as np\n\nIMG_TAG = &amp;quot;&amp;quot;&amp;quot;&amp;amp;lt;img src=&amp;quot;data:image/gif;base64,{0}&amp;quot; alt=&amp;quot;some_text&amp;quot;&amp;amp;gt;&amp;quot;&amp;quot;&amp;quot;\n\ndef anim_to_gif(anim):\ndata=&amp;quot;0&amp;quot;\nwith NamedTemporaryFile(suffix='.gif') as f:\nanim.save(f.name, writer='imagemagick', fps=10);\ndata = data.encode(&amp;quot;base64&amp;quot;)\nreturn IMG_TAG.format(data)\n\ndef display_animation(anim):\nplt.close(anim._fig)\nreturn HTML(anim_to_gif(anim))\n\nfig = plt.figure()\nfig.set_dpi(100)\nfig.set_size_inches(7, 6.5)\n\nax = plt.axes(xlim=(0, 10), ylim=(0, 10))\npatch = plt.Circle((5, -5), 0.75, fc='y')\npatch2 = plt.Circle((5, -5), 0.1, fc='b')\nc= np.array([5,5])\n\ndef init():\npatch.center = (5, 5)\nreturn patch,\n\ndef animate(i):\nx = 5 + 3 * np.sin(np.radians(i))\ny = 5 + 3 * np.cos(np.radians(i))\npatch.center = (x, y)", null, "" ]

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[ "浙江省2014届理科数学专题复习试题精选1：集合(教师版)\n\n= {( x, y ) | 4 x ? y ? 6} , B ? {( x, y ) | 3 x ? 2 y ? 7}, 则 A ? B ? A． {x ? 1或y ? 2}\n【答案】B 2\n\n（ C． {1, 2}\n\nB． {(1, 2)}\n\nD． (1, 2)\n\n． 2013 年 普 通 高 等 学 校 招 生 统 一 考 试 浙 江 数 学 （ 理 ） 试 题 （ 纯 WORD 版 ） 设 集 合 （ ）\n\nS ? {x | x ? ?2}, T ? {x | x 2 ? 3x ? 4 ? 0},则 (CR S ) ?T ?\nA． (?2,1]\n【答案】C 3\n\n（ D． [1,??)\n\nB． (??,?4]\n\nC． (??,1]\n\n．（ 浙 江 省 五 校 2013 届 高 三 上 学 期 第 一 次 联 考 数 学 （ 理 ） 试 题 ） 若 全 集 U=R, 集 合\n\nM= x x ? 4 ,N= ? x\n2\n\n?\n\n?\n\n? 3? x ? ? 0 ? ,则 M ? ? ?U N ? 等 于 ? x ?1 ?\nB． {x x ? ?2或x ? 3} B．\n\nA． {x x ? ?2}\n【答案】\n\nC． {x x ? 3} D． {x ?2 ? x ? 3}\n\n4 ． （浙江省五校联盟 2013 届高三下学期第二次联考数学（理）试题）设集合 A ? { y |\n\ny ? sin x, x ? R} ,集合\n（ ）\n\nB ? {x | y ? lg x} ,则 (CR A) ? B ?\nA． (??, ?1) ? (1, ??)\n【答案】C\n\n， B． [?11]\n\nC． (1, ??)\n\nD． [1, ??)\n\n5 ． 浙 江 省 绍 兴 一 中 2013 届 高 三 下 学 期 回 头 考 理 科 数 学 试 卷 ） 设 集 合 A ? { y | y ? lg x } , 集 合 （\n\nB ? {x y ? 1 ? x } ,则 A ? B ?\nA．[0,1]\n【答案】D 6\n\n（ C． [0， ?） ? D． ? ?，] （ 1\n\nB．(0,1]\n\n．（ 浙 江 省 杭 州 市 2013 届 高 三 第 二 次 教 学 质 检 检 测 数 学 （ 理 ） 试 题 ） 已 知 集 合\n\nA = { k ? Z| s i n k q = ) p( p (0, )}, 则(?z A) ? B = 2\nA． {k | k = 2n, n C． {k | k = 4n, n\n【答案】A\n\np q iq ? , s n 2\n\n( 0B ,\n\n) }?, Z { k\n\npk | qc = s ( q q ) o\n\ncos ,\n\nZ} Z}\n\nB． {k | k = 2n - 1, n D． {k | k = 4n - 1, n\n\nZ} Z}\n\n7\n\n．（ 浙 江 省 杭 州 高 中\n\n2013\n\nU ? R, A ? {x | ? x 2 ? 3x ? 0}, B ? {x | x ? ?1} , 则图中阴影部分表示的集合为\n\n(第 1 题) A． {x | x ? 0}\n【答案】B\n\n（ B． {x | ?3 ? x ? ?1}\n\nC． {x | ?3 ? x ? 0} D． {x | x ? ?1}\n\n8 ． （浙江省温岭中学 2013 届高三高考提优冲刺考试（三）数学（理）试题 ）设 A ? ?1, 4, 2x? , B ? 1, x 2 ,若\n\n?\n\n?\n\nB ? A ,则 x ?\nA．0\n【答案】C\n\n（ B． ?2 C．0 或 ?2 D．0 或 ?2\n\n9 ． （浙江省杭州二中 2013 届高三 6 月适应性考试数学 （理） 试题） 已知集合 A ? {x | x ? 1} , B ? {x | x ? m} ,\n\n【答案】D 10 ． 浙 江 省 稽 阳 联 谊 学 校 2013 届 高 三 4 月 联 考 数 学 ( 理 ) 试 题 （ word 版 ） （\n\n）若集合\n\n2\n\n【答案】D 11 ．（ 浙 江 省 一 级 重 点 中 学 （ 六 校 ） 2013 届 高 三 第 一 次 联 考 数 学 （ 理 ） 试 题 ） 设 集 合\n\nP ? {y | y ? k , k ? R}, Q ? {y | y ? a x ? 1, a ? 0且a ? 1, x ? R} ,若集合 P ? Q 只有\n\n【答案】B 12 ． 浙 江 省 宁 波 市 金 兰 合 作 组 织 2013 届 高 三 上 学 期 期 中 联 考 数 学 （ 理 ） 试 题 ） 已 知 集 合 （\n\n)[来源:学&科&网] C． (1,??) D．[1,??)\n\nB． (??,1]\n\nA ? {1, 2,3, 4,5} , B ? {( x, y ) x ? A, y ? A, x ? y ? A} ;,则 B 中所含元 素的个数为\nA． 3\n【答案】D\n\nB． 6\n\nC． ?\n\nD． ??\n\n13． （浙江省十校联合体 2013 届高三上学期期初联考数学 （理） 试题） 已知集合 A={x|a-3<x<a+3},B={x|x≤―3\n\n14． （浙江省宁波市十校 2013 届高三下学期能力测试联考数学（理）试题）设集合 S\n\n? ?1,2,3,4,5,6,7,8,9? ,\n\n?\n\n【答案】B\n\n（ B．83 C．78 D．76\n\n15．（浙江省嘉兴市第一中学 2013 届高三一模数学（理）试题）已知函数f(x)=x2+bx+c,(b,c∈R),集合A =\n\n{x丨f(x)=0}, 是 A． b ? 0\n【答案】B\n\nB．= {x|f(f(x)))= 0},若 A ? B ? （ ） C． 0 ? b ? 4\n\nB．b<0或 b ? 4\n\nD． b ? 4或b ? 4\n\nM ? {1, a ? 5} , M ? U , ?U M ? ?5,7? ,则实数 a 的值为______.\n【答案】 8 17 ． 浙 江 省 宁 波 市 金 兰 合 作 组 织 2013 届 高 三 上 学 期 期 中 联 考 数 学 （ 理 ） 试 题 ） 已 知 集 合 （\n\nA ? {x ? R | x ? 2 ? 3},\n\nB ? {x ? R | ( x ? m)( x ? 2) ? 0},\n\nA ? B ? (?1, n),\n\nm ? __________, n ? __________.\n【答案】 m ? ?1, n ? 1 18 ．（ 浙 江 省 温 岭 中 学 2013 届 高 三 高 考 提 优 冲 刺 考 试 （ 三 ） 数 学 （ 理 ） 试 题 ）集合\n\nA ? { x | m2x ? x 0 ,为常数 , } ? m B ? {x | 2mx2 ? 2m(1 ? m) x ? 1 ? 0, m为常数},若 A ? B ? R ,则实数 m 的取值范围是:_____.\n【答案】 [0,2]\n\n1 } ,对于集合 B ,考虑: m\n\n? ? [2m(1 ? m)]2 ? 8m ? 4m(m2 ? 1)(m ? 2)\n①若 ? ? 0 ,即 0 ? m ? 2 时, B ? R ,满足 A ? B ? R .故 0 ? m ? 2 符合\n2 ②若 ? ? 0 ,即 m ? 2 时,考虑函数 f ( x) ? 2mx ? 2m(1 ? m) x ? 1 ,由于其对称轴\n\n1? m ? 0 ,结合图像可知: A ? B ? R 不可能成立.故 m ? 2 舍去. 2 1 2 (3)当 m ? 0 , A ? {x | x ? 或x ? 0} ,考虑函数 f ( x) ? 2mx ? 2m(1 ? m) x ? 1 ,结合图像可知:要使 m 1 1 2 A ? B ? R 成立,则必有 f (0) ? 0 且 f ( ) ? 0 ,但是由于 f ( ) ? ? 2m ? 1 ? 0 ,矛盾!故 m ? 0 舍 m m m x0 ?\n\n19．浙江省十校联合体 2013 届高三上学期期初联考数学 （ （理） 试题） 设关于 x 的不等式 x( x ? a ? 1) ? 0(a ? R )\n\n【答案】" ]

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[ "# FE2 multi-scale framework for the two-equation model of transient heat conduction in two-phase media\n\nMingzhao Zhuo*\n\n*Corresponding author for this work\n\nResearch output: Contribution to journalArticleScientificpeer-review\n\n1 Citation (Scopus)\n\n## Abstract\n\nIn the study of transient heat conduction in heterogeneous two-phase media, the local thermal non-equilibrium condition calls for the use of a two-equation model to appropriately describe different temperatures in the two phases. We propose for the two-equation model an FE2 multi-scale framework that is capable of addressing nonlinear conduction problems. The FE2 framework consists of volume-averaged macroscale equations, well-defined microscale problems, and the information exchange between the two scales. Compared to a traditional FE2 method for the one-equation model, the proposed FE2 framework introduces an additional source term at the macroscale that is upscaled from the microscale interfacial heat transfer. At variance with the tangent matrices (i.e., effective conductivity) of the heat flux, the tangent matrices of the interfacial heat transfer depend on the microscopic length scale. The proposed FE2 framework is validated against single-scale direct numerical simulations, and some numerical examples are employed to demonstrate its potential.\n\nOriginal language English 121683 16 International Journal of Heat and Mass Transfer 179 https://doi.org/10.1016/j.ijheatmasstransfer.2021.121683 Published - 2021\n\n## Keywords\n\n• computational homogenization\n• FE method\n• interfacial heat transfer\n• transient heat conduction\n• two-equation model" ]

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[ "# #5a7548 Color Information\n\nIn a RGB color space, hex #5a7548 is composed of 35.3% red, 45.9% green and 28.2% blue. Whereas in a CMYK color space, it is composed of 23.1% cyan, 0% magenta, 38.5% yellow and 54.1% black. It has a hue angle of 96 degrees, a saturation of 23.8% and a lightness of 37.1%. #5a7548 color hex could be obtained by blending #b4ea90 with #000000. Closest websafe color is: #666633.\n\n• R 35\n• G 46\n• B 28\nRGB color chart\n• C 23\n• M 0\n• Y 38\n• K 54\nCMYK color chart\n\n#5a7548 color description : Mostly desaturated dark green.\n\n# #5a7548 Color Conversion\n\nThe hexadecimal color #5a7548 has RGB values of R:90, G:117, B:72 and CMYK values of C:0.23, M:0, Y:0.38, K:0.54. Its decimal value is 5928264.\n\nHex triplet RGB Decimal 5a7548 `#5a7548` 90, 117, 72 `rgb(90,117,72)` 35.3, 45.9, 28.2 `rgb(35.3%,45.9%,28.2%)` 23, 0, 38, 54 96°, 23.8, 37.1 `hsl(96,23.8%,37.1%)` 96°, 38.5, 45.9 666633 `#666633`\nCIE-LAB 46.129, -18.738, 21.718 11.747, 15.364, 8.477 0.33, 0.432, 15.364 46.129, 28.684, 130.787 46.129, -13.356, 28.973 39.197, -15.099, 14.615 01011010, 01110101, 01001000\n\n# Color Schemes with #5a7548\n\n• #5a7548\n``#5a7548` `rgb(90,117,72)``\n• #634875\n``#634875` `rgb(99,72,117)``\nComplementary Color\n• #717548\n``#717548` `rgb(113,117,72)``\n• #5a7548\n``#5a7548` `rgb(90,117,72)``\n• #48754d\n``#48754d` `rgb(72,117,77)``\nAnalogous Color\n• #754871\n``#754871` `rgb(117,72,113)``\n• #5a7548\n``#5a7548` `rgb(90,117,72)``\n• #4d4875\n``#4d4875` `rgb(77,72,117)``\nSplit Complementary Color\n• #75485a\n``#75485a` `rgb(117,72,90)``\n• #5a7548\n``#5a7548` `rgb(90,117,72)``\n• #485a75\n``#485a75` `rgb(72,90,117)``\n• #756348\n``#756348` `rgb(117,99,72)``\n• #5a7548\n``#5a7548` `rgb(90,117,72)``\n• #485a75\n``#485a75` `rgb(72,90,117)``\n• #634875\n``#634875` `rgb(99,72,117)``\n• #36462b\n``#36462b` `rgb(54,70,43)``\n• #425535\n``#425535` `rgb(66,85,53)``\n• #4e653e\n``#4e653e` `rgb(78,101,62)``\n• #5a7548\n``#5a7548` `rgb(90,117,72)``\n• #668552\n``#668552` `rgb(102,133,82)``\n• #72955b\n``#72955b` `rgb(114,149,91)``\n• #7fa268\n``#7fa268` `rgb(127,162,104)``\nMonochromatic Color\n\n# Alternatives to #5a7548\n\nBelow, you can see some colors close to #5a7548. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #657548\n``#657548` `rgb(101,117,72)``\n• #627548\n``#627548` `rgb(98,117,72)``\n• #5e7548\n``#5e7548` `rgb(94,117,72)``\n• #5a7548\n``#5a7548` `rgb(90,117,72)``\n• #567548\n``#567548` `rgb(86,117,72)``\n• #537548\n``#537548` `rgb(83,117,72)``\n• #4f7548\n``#4f7548` `rgb(79,117,72)``\nSimilar Colors\n\n# #5a7548 Preview\n\nThis text has a font color of #5a7548.\n\n``<span style=\"color:#5a7548;\">Text here</span>``\n#5a7548 background color\n\nThis paragraph has a background color of #5a7548.\n\n``<p style=\"background-color:#5a7548;\">Content here</p>``\n#5a7548 border color\n\nThis element has a border color of #5a7548.\n\n``<div style=\"border:1px solid #5a7548;\">Content here</div>``\nCSS codes\n``.text {color:#5a7548;}``\n``.background {background-color:#5a7548;}``\n``.border {border:1px solid #5a7548;}``\n\n# Shades and Tints of #5a7548\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #060805 is the darkest color, while #fbfcfb is the lightest one.\n\n• #060805\n``#060805` `rgb(6,8,5)``\n• #0f140c\n``#0f140c` `rgb(15,20,12)``\n• #192014\n``#192014` `rgb(25,32,20)``\n• #222c1b\n``#222c1b` `rgb(34,44,27)``\n• #2b3823\n``#2b3823` `rgb(43,56,35)``\n• #35442a\n``#35442a` `rgb(53,68,42)``\n• #3e5132\n``#3e5132` `rgb(62,81,50)``\n• #475d39\n``#475d39` `rgb(71,93,57)``\n• #516941\n``#516941` `rgb(81,105,65)``\n• #5a7548\n``#5a7548` `rgb(90,117,72)``\n• #63814f\n``#63814f` `rgb(99,129,79)``\n• #6d8d57\n``#6d8d57` `rgb(109,141,87)``\n• #76995e\n``#76995e` `rgb(118,153,94)``\n• #80a369\n``#80a369` `rgb(128,163,105)``\n• #8aaa75\n``#8aaa75` `rgb(138,170,117)``\n• #95b281\n``#95b281` `rgb(149,178,129)``\n• #9fb98d\n``#9fb98d` `rgb(159,185,141)``\n• #a9c099\n``#a9c099` `rgb(169,192,153)``\n• #b3c8a6\n``#b3c8a6` `rgb(179,200,166)``\n• #becfb2\n``#becfb2` `rgb(190,207,178)``\n• #c8d7be\n``#c8d7be` `rgb(200,215,190)``\n• #d2deca\n``#d2deca` `rgb(210,222,202)``\n• #dce6d6\n``#dce6d6` `rgb(220,230,214)``\n• #e7ede2\n``#e7ede2` `rgb(231,237,226)``\n• #f1f5ee\n``#f1f5ee` `rgb(241,245,238)``\n• #fbfcfb\n``#fbfcfb` `rgb(251,252,251)``\nTint Color Variation\n\n# Tones of #5a7548\n\nA tone is produced by adding gray to any pure hue. In this case, #5e5f5e is the less saturated color, while #4db607 is the most saturated one.\n\n• #5e5f5e\n``#5e5f5e` `rgb(94,95,94)``\n• #5d6657\n``#5d6657` `rgb(93,102,87)``\n• #5b6e4f\n``#5b6e4f` `rgb(91,110,79)``\n• #5a7548\n``#5a7548` `rgb(90,117,72)``\n• #597c41\n``#597c41` `rgb(89,124,65)``\n• #578439\n``#578439` `rgb(87,132,57)``\n• #568b32\n``#568b32` `rgb(86,139,50)``\n• #54922b\n``#54922b` `rgb(84,146,43)``\n• #539924\n``#539924` `rgb(83,153,36)``\n• #51a11c\n``#51a11c` `rgb(81,161,28)``\n• #50a815\n``#50a815` `rgb(80,168,21)``\n• #4eaf0e\n``#4eaf0e` `rgb(78,175,14)``\n• #4db607\n``#4db607` `rgb(77,182,7)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #5a7548 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]

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[ " 离散半正边值问题正解的存在性及多解性\n\n# 离散半正边值问题正解的存在性及多解性Existence and Multiplicity of Semipositone Discrete Boundary Value Problems\n\nAbstract: By using the Guo-Krasnosel’skii fixed point theorem, a Dirichlet boundary value problem with sign-changing nonlinearity is discussed and some results of existence and multiplicity of positive solutions are established.\n\n Berger, H. (2008) Existence of Nontrivial Solutions of a Two-Point Boundary Value Problem for a 2nth-Order Nonlinear Difference Equation. Advances in Dynamical Systems and Applications, 3, 131-146.\n\n Anderson, D. (2003) Discrete Third-Order Three-Point Right-Focal Boundary Value Problems. Computers & Mathematics with Applications, 45, 861-871.\nhttp://dx.doi.org/10.1016/S0898-1221(03)80157-8\n\n Anderson, D. and Avery, R. (2001) Multiple Positive Solutions to a Third-Order Discrete Focal Boundary Value Problem. Computers & Mathematics with Applications, 42, 333-340.\nhttp://dx.doi.org/10.1016/S0898-1221(01)00158-4\n\n Aykut, N. (2004) Existence of Positive Solutions for Boundary Value Problems of Second Order Functional Difference Equations. Computers & Ma-thematics with Applications, 48, 517-527.\nhttp://dx.doi.org/10.1016/j.camwa.2003.10.007\n\n Bai, D. (2013) A Global Result for Discrete ϕ-Laplacian Ei-genvalue Problems. Advances in Difference Equations, 264, 10 p.\n\n Bai, D. and Xu, Y. (2007) Nontrivial Solutions of Boundary Value Problems of Second Order Difference Equations. Journal of Mathematical Analysis and Applications, 326, 297-302.\nhttp://dx.doi.org/10.1016/j.jmaa.2006.02.091\n\n Bai, D. and Xu, X. (2013) Existence and Multiplicity of Difference ϕ-Laplacian Boundary Value Problems. Advances in Difference Equations, 267, 13 p.\n\n Ji, D. and Ge, W. (2008) Existence of Multiple Positive Solutions for Sturm-Liouville-Like Four-Point Boundary Value Problem with p-Laplacian. Nonlinear Analysis: Theory, Methods & Applications, 68, 2638-2646.\nhttp://dx.doi.org/10.1016/j.na.2007.02.010\n\n Bai, D. and Xu, Y. (2008) Positive Solutions for Semipositone BVPs of Second-Order Difference Equations. Indian Journal of Pure and Applied Mathematics, 39, 59-68.\n\n Anuradha, V. and Shivaji, R. (1994) A Quadrature Method for Classes of Multi-Parameter Two Point Boundary Value Problems. Applicable Analysis, 54, 263-281.\nhttp://dx.doi.org/10.1080/00036819408840282\n\n Anuradha, A. and Shivaji, R. (1996) Existence Results for Superlinear Semipositone BVP’s. Proceedings of the American Mathematical Society, 124, 757-763.\nhttp://dx.doi.org/10.1090/S0002-9939-96-03256-X\n\n Feng, H. and Bai, D. (2011) Existence of Positive Solu-tions for Semipositone Multi-Point Boundary Value Problem. Mathematical and Computer Modelling, 54, 2287-2292.\nhttp://dx.doi.org/10.1016/j.mcm.2011.05.037\n\n Anuradha, V. and Shivaji, R. (1999) Positive Solutions for a Class of Nonlinear Boundary Value Problems with Neumann-Robin Boundary Conditions. Journal of Mathematical Analysis and Applications, 236, 94-124.\nhttp://dx.doi.org/10.1006/jmaa.1999.6439\n\n Bai, D. and Xu, Y. (2005) Existence of Positive Solutions for Boundary Value Problems of Second-Order Delay Differential Equations. Applied Mathematics Letters, 18, 621-630.\nhttp://dx.doi.org/10.1016/j.aml.2004.07.022\n\n Bai, D. and Xu, Y. (2005) Positive Solutions of Second-Order Two-Delay Differential Systems with Twin-Parameter. Nonlinear Analysis, 63, 601-617.\nhttp://dx.doi.org/10.1016/j.na.2005.05.021\n\n Castro, A. and Shivaji, R. (2000) Evolution of Positive Solution Curves in Semipositone Problems with Concave Nonlinearities. Journal of Mathematical Analysis and Applications, 245, 282-293.\nhttp://dx.doi.org/10.1006/jmaa.2000.6787\n\n Hai, D. and Shivaji, R. (1998) Positive Solutions of Quasilinear Boundary Value Problems. Journal of Mathematical Analysis and Applications, 217, 672-686.\nhttp://dx.doi.org/10.1006/jmaa.1997.5762\n\n Hai, D. and Shivaji, R. (2004) An Existence Result on Positive Solutions for a Class of p-Laplacian Systems. Nonlinear Analysis, 56, 1007-1010.\nhttp://dx.doi.org/10.1016/j.na.2003.10.024\n\n Ma, R. (2003) Multiple Positive Solutions for a Semipositone Fourth-Order Boundary Value Problem. Hiroshima Mathematical Journal, 33, 217-227.\n\n Maya, C. and Shivaji, R. (1999) Multiple Positive Solutions for a Class of Semilinear Elliptic Boundary Value Problems. Nonlinear Analysis, 38, 497-504.\nhttp://dx.doi.org/10.1016/S0362-546X(98)00211-9\n\n Sun, J. and Wei, J. (2008) Existence of Positive Solutions of Positive Solution for Semi-Positone Second-Order Three- Point Boundary-Value Problems. Electronic Journal of Differential Equations, 41, 1-7.\n\n Bai, D., Henderson, J. and Zeng, Y. (2015) Positive Solutions of Dis-crete Neumann Boundary Value Problems with Sign-Changing Nonlinearities. Boundary Value Problems, 1, 231.\nhttp://dx.doi.org/10.1186/s13661-015-0500-8\n\nTop" ]

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[ "# [OLD] Osmotic coefficients (again)\n\n## Recommended Posts\n\nFrom: James Cleverley\n\nSubject: osmotic coefficients (again)\n\nI am having trouble locating the exact reference source for the equation used to calculate the osmotic coefficient for NaCl salt solutions and used in GWB to calculate water activity. The equations and some regression coefficients are reproduced in Bethke (1996) as eq. 7.8, 7.9 and the table on page 114, however where in the literature are these defined and reproduced? The reason I ask is that I am having trouble reproducing the graph in Figure 7.5 of Bethke (1996) using the data on page 114 and equation 7.8, 7.9 and 7.7 (plus the A parameter from the GWB datafile) I am calculating much lower water activities at high Is. Any pointers to the original source of the regression data and those equations would be really appreciated.\n\nFrom: Craig Bethke\n\nSubject: Re: osmotic coefficients (again)\n\nBill Bourcier and Tom Wolery provide the following information: The oiginal literature reference is Helgeson, Brown, Nigrini, and Jones (1970) GCA 34:569. The actual parameters in the regression were taken from the PATH code, developed in the 1960s; it may (or may not) take some digging in the literature to find precisely how the parameter values were derived.\n\n## Join the conversation\n\nYou can post now and register later. If you have an account, sign in now to post with your account.", null, "Reply to this topic...\n\n×   Pasted as rich text.   Paste as plain text instead\n\nOnly 75 emoji are allowed.\n\n×   Your previous content has been restored.   Clear editor\n\n×   You cannot paste images directly. Upload or insert images from URL." ]

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[ "# Python - Normalized cross-correlation to measure similarites in 2 images\n\nI'm trying to measure per-pixel similarities in two images (same array shape and type) using Python.\n\nIn many scientific papers (like this one), normalized cross-correlation is used. Here's an image from the ict paper showing the wanted result:", null, "(b) and (c) are the 2 input images, and (d) is the per-pixel confidence.\n\nI only used OpenCV before to do template matching with normalized cross correlation using cv2.matchTemplate function, but in this case it seems to be a really different use of cross correlation.\n\nDo you know if I can approch this result using Python and image processing libraries (numpy, openCV, sciPy etc...), and the logic behind this use of cross correlation ?\n\nThanks.\n\n• It's not really too clear exactly what cross-correlation function you're trying to compute. Rather than give the link to the paper, could you write down the function here? OpenCV, numpy and scipy may not have a built-in method to do this, but I'm certain you can write a program using these tools to do what you need. – Praveen Jan 9 '16 at 6:46\n• Thanks for reply Praveen, the problem is that I don't know this use of cross correlation at all, and there is no formula in the paper(s), they just say ; italic We then compute normalized cross correlation between the static image (b) and the warped dynamic image (c) to produce the per-pixel confidence shown in (d). – Paul Parneix Jan 9 '16 at 14:38\n• stackoverflow.com/q/6991471/1461210 – ali_m Jan 10 '16 at 1:23\n\nI guess you can compute for each pixel the correlation coefficient between patches centered on this pixel in the two images of interest. Here is an example where I downloaded the figure attached here and tried to compute the correlation in such a way. The output looks different from the one of the article, but it was to be expected since the resolution is very different.\n\nfrom skimage import io, feature\nfrom scipy import ndimage\nimport numpy as np\n\ndef correlation_coefficient(patch1, patch2):\nproduct = np.mean((patch1 - patch1.mean()) * (patch2 - patch2.mean()))\nstds = patch1.std() * patch2.std()\nif stds == 0:\nreturn 0\nelse:\nproduct /= stds\nreturn product\n\nim1 = im[16:263, 4:146]\nsh_row, sh_col = im1.shape\nim2 = im[16:263, 155:155+sh_col]\n\n# Registration of the two images\ntranslation = feature.register_translation(im1, im2, upsample_factor=10)\nim2_register = ndimage.shift(im2, translation)\n\nd = 1\n\ncorrelation = np.zeros_like(im1)\n\nfor i in range(d, sh_row - (d + 1)):\nfor j in range(d, sh_col - (d + 1)):\ncorrelation[i, j] = correlation_coefficient(im1[i - d: i + d + 1,\nj - d: j + d + 1],\nim2[i - d: i + d + 1,\nj - d: j + d + 1])\n\nio.imshow(correlation, cmap='gray')\nio.show()", null, "The code above is a naive and slow implementation of the correlation, as the two for loops are very slow. For faster execution, you could for example port the script to Cython.\n\nIn the article, I think the idea is to measure whether face expressions look similar or not. If a pixel has a large correlation index between two images, it means that the region of the face where this pixel is located does not change much between the images.\n\nIf you use this method on good-resolution images, you should increase the patch size for more accurate results (d=2 or 3).\n\n• Merci Emmanuelle ! That's a lot more clear now :). Thanks for this very complete answer, I'll have a look at Cython, it seems to be quite cool and powerfull. – Paul Parneix Jan 11 '16 at 6:18\n• Thank you very much for this question and clarifying answer of Emmanuelle. In the code above, there are two calculations which I think are related to the co-registration of the Master(b) and Slave(c) images. However in the calculation of NCC map you used the same original images (im1 and im2). Should not one use im1 and im2_register for calculation of the NCC map? With regards Sina – Sinooshka Mar 14 '19 at 10:19\n• what is the purpose of two lines with translation and im2_register calculated ? How did you come up with numbers in im2 = im[16:263, 155:155+sh_col] ? – Färid Alijani Mar 6 at 14:56\n\nThis isn't a complete answer but nowadays Python's skimage module has a bunch of tools for template matching and feature detection:" ]

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[ "Inches To Mm\n\n# 4829 in to mm4829 Inch to Millimeters\n\nin\n=\nmm\n\n## How to convert 4829 inch to millimeters?\n\n 4829 in * 25.4 mm = 122656.6 mm 1 in\nA common question is How many inch in 4829 millimeter? And the answer is 190.118110236 in in 4829 mm. Likewise the question how many millimeter in 4829 inch has the answer of 122656.6 mm in 4829 in.\n\n## How much are 4829 inches in millimeters?\n\n4829 inches equal 122656.6 millimeters (4829in = 122656.6mm). Converting 4829 in to mm is easy. Simply use our calculator above, or apply the formula to change the length 4829 in to mm.\n\n## Convert 4829 in to common lengths\n\nUnitUnit of length\nNanometer1.226566e+11 nm\nMicrometer122656600.0 µm\nMillimeter122656.6 mm\nCentimeter12265.66 cm\nInch4829.0 in\nFoot402.416666667 ft\nYard134.138888889 yd\nMeter122.6566 m\nKilometer0.1226566 km\nMile0.0762152778 mi\nNautical mile0.0662292657 nmi\n\n## What is 4829 inches in mm?\n\nTo convert 4829 in to mm multiply the length in inches by 25.4. The 4829 in in mm formula is [mm] = 4829 * 25.4. Thus, for 4829 inches in millimeter we get 122656.6 mm.\n\n## 4829 Inch Conversion Table", null, "## Alternative spelling\n\n4829 Inch in Millimeters, 4829 Inches to Millimeter, 4829 Inches in mm, 4829 Inch to mm, 4829 Inch in mm, 4829 Inch to Millimeter, 4829 Inches to Millimeters, 4829 Inches in Millimeters, 4829 in in Millimeters," ]

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[ "# Five Math Puzzles (pattern recognition): Can You Solve Them?\n\n## MATH PUZZLES\n\nHere is a math puzzle challenge.\n\nHint: Each of the 5 patterns below has something in common.\n\nDirections: See if you can figure out which numbers go in the blanks.\n\n• 1, 2, 4, 6, 10, 12, 16, 18, 22, _, _\n• 4, 6, 10, 14, 22, 26, 34, 38, _, _\n• 3, 7, 13, 19, 29, 37, _, _\n• 4, 9, 25, 49, 121, 169, 289, _, _\n• 5, 8, 12, 18, 24, 30, 36, 42, 52, 60, 68, _, _\n\nIf you need help, you can find hints below.\n\nBut don’t scroll too far or you’ll run into the answers and explanations.\n\n## PUZZLE HINT\n\nEach pattern above has something in common.\n\nThey all involve prime numbers.\n\nA prime number is only evenly divisible by two integers: 1 and itself.\n\nFor example, 7 is a prime number because the only integers that can multiply together to make 7 are 1 and 7.\n\nIn contrast, 6 isn’t a prime number because 2 x 3 = 6 (in addition to 1 x 6).\n\nHere are the first several prime numbers.\n\n2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37\n\nEach of the puzzles above relates to these prime numbers.\n\nHere are the answers and explanations to the math puzzles:\n\n• 28, 30. Explanation: Subtract 1 from each prime number: 2 – 1 = 1, 3 – 1 = 2, 5 – 1 = 4, 7 – 1 = 6, 11 – 1 = 10, etc.\n• 46, 58. Explanation: Double each prime number: 2 x 2 = 4, 3 x 2 = 6, 5 x 2 = 10, 7 x 2 = 14, 11 x 2 = 22, etc.\n• 43, 53. Explanation: Every other prime number: 3 (skip 5) 7 (skip 11) 13 (skip 17) 19 (skip 23) 29 etc.\n• 361, 529. Explanation: Square each prime number: 2² = 4, 3² = 9, 5² = 25, 7² = 49, 11² = 121, etc.\n• 78, 84. Explanation: Add consecutive prime numbers together: 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34, etc.\n\n## WANT MORE MATH PUZZLES?\n\nOne way is to follow my blog. I will post occasional math puzzles in the future.\n\nAnother way is to check out my newest book, 300+ Mathematical Pattern Puzzles.\n\nIt starts out easy and the level of challenge grows progressively so that puzzlers of all abilities can find many puzzles to enjoy.\n\nA wide variety of topics are covered, including:\n\n• visual patterns\n• arithmetic\n• repeating patterns\n• Roman numerals\n• Fibonacci sequence\n• prime numbers\n• arrays\n• analogies\n• and much more", null, "The cover was designed by Melissa Stevens at www.theillustratedauthor.net.\n\n## CHRIS MCMULLEN, PH.D.\n\n• 300+ Mathematical Pattern Puzzles\n• Basic Linear Graphing Skills Practice Workbook\n• Systems of Equations: Simultaneous, Substitution, Cramer’s Rule\n\nImprove Your Math Fluency. Build fluency in:\n\n• arithmetic\n• long division\n• fractions\n• algebra\n• trigonometry\n• graphing\n1.", null, "Sheldon Glover\n•", null, "chrismcmullen" ]

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[ "# What edges are not in any MST?\n\nThis is a homework question. I do not want the solution - I'm offering the solution I've been thinking of and wish to know whether is it good or why is it flawed.\n\nConsider a weighted undirected graph. What edges of it are not a part of any minimum spanning tree (MST)? This problem only makes sense when several edges have the same weight, otherwise the MST is unique.\n\nMy idea comes from Prim's Algorithm with a slight change. Let $S$ and $T$ be two sets of vertices the algorithm works with. Instead of adding the minimum edge from $S$ to $T$ on every step, look for the minimum edge and more edges of the same value going from $S$ to the vertex the minimum edge goes to. By doing that, (so I suppose) we will obtain a graph containing all the edges which appear in any MST. If this is right, I can simply XOR the edges list with the original graph edges list to find what edges are not in any MST.\n\n• I think the question is probably meant to be more conceptual and less algorithmic. Here's a hint: What if I gave you a graph with $n$ edges (that is, a spanning tree with one edge that doesn't belong in it). How would you know which is the edge that doesn't belong? – Joe Dec 10 '12 at 23:19\n• Thanks Joe, I obviously meant a weighted graph (fixed that). About your hint: for example a graph with 3 vertex and 3 edges, fully connected with every edge weighing 2. Every edge is in at least one MST. – The-Q Dec 11 '12 at 5:46\n• How is the question exactly phrased. I was expecting something like \"the unique maximal edge is not in any MST\" as an answer. – A.Schulz Dec 11 '12 at 7:31\n• This sounds like one of those questions that only make sense if you know what the questioner is thinking. If the question was \"what is a case where an edge can never be part of a MST\", I can think of a good answer. (Hint: Think of a triangle.) – usul Dec 11 '12 at 7:51\n• Hey and thank you for your comments. The original question is \"Find an algorithm which returns what edges are not in any possible MST for a given undirected, weighted graph\". The algorithm I stated in the original question has already been found wrong. – The-Q Dec 11 '12 at 22:29\n\nA Google search for \"edges not in MST\" leads me to this question. The answer included in the question has already been found wrong, as OP said in the last comment. For future references, here is my solution to the question.\n\nLet $$G$$ be a weighted undirected (finite) graph. An edge $$e$$ is called superheavy if it is the unique heaviest edge in some cycle. Here is a characterization of the edges that are not in any MST.\n\n## An edge is not in any MST if and only if it is a superheavy edge\n\nThe well-known \"if\" part, a superheavy edge is not in any MST, is proved in the cycle property of MST.\n\nThe \"only if\" part, any edge $$e$$ that is not superheavy is in some MST, is harder to prove. Let me introduce an algorithm and a lemma first.\n\nThe delete-heaviest-edge algorithm (DHE) on $$G$$ removes a heaviest edge in any cycle until no cycle remains. Just to be extra clear, there might be multiple heaviest edges in one cycle. The edge removed can be chosen to be any one of them.\n\nLemma: DHE is an MST algorithm. That is, the final graph at the end of DHE is an MST of the original $$G$$.\n\nOne proof of the lemma can be developed by adapting the proof at reverse-delete algorithm.\n\nFor another proof of the lemma, reader can check my post at the delete-heaviest-edge algorithm on graphs with edges of distinct weights, where I proved the special case where all weights are distinct. For the general case of a given run of DHE where all weights of $$G$$ are not necessarily distinct, we can perturb the weights of all edges of $$G$$ so that all weights are distinct and that each edge removed is the unique heaviest edge in the corresponding cycle. We can then repeat the given run of DHE on the perturbed $$G$$, resulting in the same spanning tree, which is also the MST of the perturbed $$G$$, since this is the proven special case of distinct weights. By making the perturbation arbitrarily small, we can see the total weight of the resulting MST can be arbitrarily near to the weight of a (or any) MST of the original $$G$$. Once we take the limit of perturbation going to $$0$$, we can see that original given run of the DHE must produce a spanning tree with the minimal weight. The lemma is proved.\n\nNow let us prove the \"only if\" part. Suppose edge $$e$$ is not a superheavy edge. Let us start a particular run of DHE. We will adjust that run whenever necessary to avoid deleting $$e$$. Suppose $$e$$, as a heaviest edge in some cycle is removed in some step. Because $$e$$ is not superheavy, there is another heaviest edge, say $$f$$, in that cycle. Let us modify that step so that it removes $$f$$ instead of $$e$$. Continue to run the algorithm, alway avoiding removing $$e$$ as we just did. At the end of the algorithm, we must arrive at an MST thanks to the lemma, which contains $$e$$. Proof of the \"only if\" part is done.\n\nSince an edge not in any MST means it is superheavy, any algorithm that finds all the superheavy edges will also finds all edges that are not in any MST.\n\n## Algorithms to find all superheavy edges\n\nHere is a naive algorithm. Iterate over all edges. For each edge $$e$$, pick arbitrarily one of its two vertices. Starting from this chosen vertex, make a DFS or BFS along edges with weight lower than that of $$e$$, checking whether we have reached the other vertex of $$e$$ all along. Once we have, $$e$$ is superheavy. Otherwise we cannot at the end of the search, implying $$e$$ is not superheavy.\n\nIt often happens that there is a unique MST, for example when all edge-weights are distinct. We can use various algorithms such as Kruskal's algorithm or Prim's algorithm to compute that unique MST. The superheavy edges are the edges that are not in that unique MST.\n\nHere is the efficient algorithm to find all superheavy edges in general cases. Its time-complexity is about the time-complexity to sort the edges by weights, or $$O(m\\log m + n)$$, where $$n$$ is the number of vertices and $$m$$ is the number of edges. Its space-complexity is about $$O(m+n)$$.\n\n1. Sort all edges in groups of increasing weights so that we have $$E=E_1\\cup E_2\\cup \\cdots \\cup E_r$$ and $$w_1 where $$E$$ is the edge set of $$G$$ and every edge in $$E_i$$ has weight $$w_i$$.\n2. Let $$H$$ be a graph that has the same vertices as $$G$$ but without any edges. Track the connected components of $$H$$ by a disjoint-set data structure.\n3. For $$i$$ from 2 to $$r$$ do the following.\n1. Add all edges in $$E_{i-1}$$ to $$H$$.\n2. For each edge in $$E_i$$, check whether its endpoints are in the same connected component of $$H$$. If yes, it is a superheavy edge." ]

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[ "# Excel: Create Easier-to-Understand Formulas with Named Ranges\n\nThis page is an advertiser-supported excerpt of the book, Power Excel 2010-2013 from MrExcel - 567 Excel Mysteries Solved. If you like this topic, please consider buying the entire e-book.\n\nProblem: How can I create easier-to-understand formulas?", null, "1. This formula is not very intuitive.\n\nStrategy: It would be easier to understand the results if each component of every formula were named for what it represented and not just for the cell it came from. You can therefore use named ranges to make formulas easier to understand:\n\n1. Select cell B3. In the Name box (the area to the left of the formula bar), type Revenue and press Enter.\n2. Select cell B4. Click in the Name box, type COGS, and press Enter.", null, "2. Type a name in the Name Box and press Enter.\n1. Clear the formula in B6. Reenter the formula and use the mouse to select the cells. Type =. Using the mouse, touch B3. Type -. Using the mouse, touch B4. Excel will enter the formula as =Revenue-COGS. This is easier to understand than a typical formula.\n\nGotcha: You need a lot of foresight to use this technique. In order to have this work automatically, you are supposed to be smart enough to create the range names before you enter the formula.", null, "3. This formula is easier to understand.\n\nHowever, most people create a formula first and then decide to make the worksheet easier to understand. To assign range names after creating formulas, follow these steps:\n\n1. Select Formulas, Define Name dropdown, Apply Names. Gotcha: Don't click on the words Define Name; click on the dropdown icon to the right of Define Name.\n2. Select all the names you want to apply and click OK.", null, "4. Apply Names is hidden in the Define Name dropdown.\n\nResults: A formula like =B6-B11 will be updated to =GrossProfit-Expenses.\n\nAdditional Details: One advantage of named ranges: they are always treated as an absolute reference. You don't need to add dollar signs to have the formula always point to that cell." ]

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[ "", null, "PoissonProcess - Maple Help\n\nFinance\n\n PoissonProcess\n create new Poisson process", null, "Calling Sequence PoissonProcess(lambda) PoissonProcess(lambda, X)", null, "Parameters\n\n lambda - algebraic expression; intensity parameter X - algebraic expression; jump size distribution", null, "Description\n\n • A Poisson process with intensity parameter $0<\\mathrm{\\lambda }\\left(t\\right)$, where $\\mathrm{\\lambda }\\left(t\\right)$ is a deterministic function of time, is a stochastic process $N$ with independent increments such that $N\\left(0\\right)=0$ and\n\n$\\mathrm{Pr}\\left(N\\left(t+h\\right)-N\\left(t\\right)=1\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}|\\phantom{\\rule[-0.0ex]{0.3em}{0.0ex}}N\\left(t\\right)\\right)=\\mathrm{lambda}\\left(t\\right)h+o\\left(h\\right)$\n\n for all $0\\le t$. If the intensity parameter $\\mathrm{\\lambda }\\left(t\\right)$ itself is stochastic, the corresponding process is called a doubly stochastic Poisson process or Cox process.\n • A compound Poisson process is a stochastic process $J\\left(t\\right)$ of the form $J\\left(t\\right)={\\sum }_{i=1}^{N\\left(t\\right)}{Y}_{i}$, where $N\\left(t\\right)$ is a standard Poisson process and ${Y}_{i}$ are independent and identically distributed random variables. A compound Cox process is defined in a similar way.\n • The parameter lambda is the intensity. It can be constant or time-dependent. It can also be a function of other stochastic variables, in which case the so-called doubly stochastic Poisson process (or Cox process) will be created.\n • The parameter X is the jump size distribution. The value of X can be a distribution, a random variable or any algebraic expression involving random variables.\n • If called with one parameter, the PoissonProcess command creates a standard Poisson or Cox process with the specified intensity parameter.", null, "Examples\n\n > $\\mathrm{with}\\left(\\mathrm{Finance}\\right):$\n > $J≔\\mathrm{PoissonProcess}\\left(1.0\\right):$\n > $\\mathrm{PathPlot}\\left(J\\left(t\\right),t=0..3,\\mathrm{timesteps}=50,\\mathrm{replications}=20,\\mathrm{thickness}=3,\\mathrm{color}=\\mathrm{red}..\\mathrm{blue},\\mathrm{axes}=\\mathrm{BOXED},\\mathrm{gridlines}=\\mathrm{true},\\mathrm{markers}=\\mathrm{false}\\right)$", null, "Create a subordinated Wiener process with $J$ as a subordinator.\n\n > $W≔\\mathrm{WienerProcess}\\left(J\\right):$\n > $\\mathrm{PathPlot}\\left(W\\left(t\\right),t=0..3,\\mathrm{timesteps}=20,\\mathrm{replications}=10,\\mathrm{markers}=\\mathrm{false},\\mathrm{color}=\\mathrm{red}..\\mathrm{blue},\\mathrm{thickness}=3,\\mathrm{gridlines}=\\mathrm{true},\\mathrm{axes}=\\mathrm{BOXED}\\right)$", null, "Next define a compound Poisson process.\n\n > $Y≔{\\mathrm{Statistics}}_{\\mathrm{RandomVariable}}\\left(\\mathrm{Normal}\\left(0.3,0.5\\right)\\right):$\n > $\\mathrm{λ}≔0.5$\n ${\\mathrm{\\lambda }}{≔}{0.5}$ (1)\n > $X≔\\mathrm{PoissonProcess}\\left(\\mathrm{λ},Y\\right):$\n > $\\mathrm{PathPlot}\\left(X\\left(t\\right),t=0..3,\\mathrm{timesteps}=20,\\mathrm{replications}=10,\\mathrm{markers}=\\mathrm{false},\\mathrm{color}=\\mathrm{red}..\\mathrm{blue},\\mathrm{thickness}=3,\\mathrm{gridlines}=\\mathrm{true},\\mathrm{axes}=\\mathrm{BOXED}\\right)$", null, "Compute the expected value of $X\\left(T\\right)$ for $T=3$ and verify that this is approximately equal to $\\mathrm{\\lambda }T$ times the expected value of $Y$.\n\n > $T≔3$\n ${T}{≔}{3}$ (2)\n > $\\mathrm{ExpectedValue}\\left(X\\left(T\\right),\\mathrm{replications}={10}^{4},\\mathrm{timesteps}=100\\right)$\n $\\left[{\\mathrm{value}}{=}{0.4435146732}{,}{\\mathrm{standarderror}}{=}{0.007164725012}\\right]$ (3)\n > $\\mathrm{λ}T{\\mathrm{Statistics}}_{\\mathrm{ExpectedValue}}\\left(Y\\right)$\n ${0.45}$ (4)\n\nHere is an example of a doubly stochastic Poisson process for which the intensity parameter evolves as a square-root diffusion.\n\n > $\\mathrm{κ}≔0.354201$\n ${\\mathrm{\\kappa }}{≔}{0.354201}$ (5)\n > $\\mathrm{μ}≔1.21853$\n ${\\mathrm{\\mu }}{≔}{1.21853}$ (6)\n > $\\mathrm{ν}≔0.538186$\n ${\\mathrm{\\nu }}{≔}{0.538186}$ (7)\n > $\\mathrm{y0}≔1.81$\n ${\\mathrm{y0}}{≔}{1.81}$ (8)\n > $y≔\\mathrm{SquareRootDiffusion}\\left(\\mathrm{y0},\\mathrm{κ},\\mathrm{μ},\\mathrm{ν}\\right):$\n > $J≔\\mathrm{PoissonProcess}\\left(y\\left(t\\right)\\right):$\n > $\\mathrm{PathPlot}\\left(y\\left(t\\right),t=0..3,\\mathrm{timesteps}=100,\\mathrm{replications}=10,\\mathrm{thickness}=3,\\mathrm{color}=\\mathrm{red}..\\mathrm{blue},\\mathrm{axes}=\\mathrm{BOXED},\\mathrm{gridlines}=\\mathrm{true}\\right)$", null, "> $\\mathrm{PathPlot}\\left(J\\left(t\\right),t=0..3,\\mathrm{timesteps}=100,\\mathrm{replications}=10,\\mathrm{thickness}=3,\\mathrm{color}=\\mathrm{red}..\\mathrm{blue},\\mathrm{axes}=\\mathrm{BOXED},\\mathrm{gridlines}=\\mathrm{true}\\right)$", null, "", null, "References\n\n Glasserman, P., Monte Carlo Methods in Financial Engineering. New York: Springer-Verlag, 2004.", null, "Compatibility\n\n • The Finance[PoissonProcess] command was introduced in Maple 15." ]

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[ "# 20 Machine Learning, Data Science job interview questions you must know\n\nIf you are a Machine learning professional looking to ace an interview, you are at the right place. Here are top 20 ML job interview questions that you should be prepared to answer.", null, "Machine learning is a part of two of the most important technologies of our time, artificial intelligence and data science. Data science is one of the hottest tech professions right now. If you are a Machine learning professional looking to ace an interview, you are at the right place.\n\nHere is a list of top 20 ML job interview questions that you must know the answer to:\n\n1. What is the similarity between Hadoop and K?\n2. If a linear regression model shows a 90% confidence interval, what does that mean?\n3. A single-layer perceptron or a 2-layer decision tree, which one is superior in terms of expressiveness?\n4. How can a neural network be used for dimensionality?\n5. Name two utilities of the intercept term in linear regression?\n6. Why do a majority of machine learning algorithms involve some kind of matrix manipulation?\n7. Is time series really a simple linear regression problem with one response variable predictor?\n8. Can it be mathematically proven that finding the optimal decision trees for a classification problem among all decision trees is hard?\n9. Which is easier, a deep neural network or a decision tree model?\n10. Apart from back-propagation, what are some of the other alternative techniques to train a neural network?\n11. How can one tackle the impact of correlation among predictors on principal component analysis?\n12. Is there a way to work beyond the 99% accuracy mark on a classification model?\n13. How can one capture the correlation between continuous and categorical variables?\n14. Does k-fold cross-validation work well with a time-series model?\n15. Why can’t simple random sampling of training data set and validation set work for a classification problem?\n16. What should be a priority, a model accuracy or model performance?\n17. What is your preferred approach for multiple CPU cores, boosted tree algorithm, or random forest?\n18. What algorithm works best for tiny storage, logistic regression, or k-nearest neighbor?\n19. What are the criteria to choose the right ML algorithm?\n20. Why can’t logistic regression use more than 2 classes?\n\n#### Trending Videos", null, "" ]

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[ "# What algorithm should I use to find the shortest path in this graph?\n\nI have a graph with about a billion vertices, each of which is connected to about 100 other vertices at random.\n\nI want to find the length of the shortest path between two points. I don't care about the actual path used.\n\nNotes:\n\n• Sometimes edges will be severed or added. This happens about 500 times less often than lookups. It's also ok to batch up edge-changes if it lets you get better performance.\n• I can pre-process the graph.\n• If it takes more than 6 steps, you can just come back with infinity.\n• It's acceptable to be wrong 0.01% of the time, but only in returning a length that's too long.\n• All edges have a length of 1.\n• All edges are bidirectional.\n\nI'm looking for an algorithm. Psuedocode, english descriptions, and actual code are all great.\n\nI could use A*, but that seems optimized for pathfinding.\nI thought about using Dijkstra's algorithm, but it has a step which requires setting the shortest-path-found attribute of every vertice to infinity\n\n(If you're wondering about the use-case, it's for the Underhanded C Contest.)\n\n• Dijkstra's algo is A* with h=0, wrt to the 'pathfinding' do you mean that you have no way to estimate a better minimum cost?\n– jk.\nApr 3, 2013 at 7:20\n• Setting the shortest-path-found attribute of every vertex doesn't mean you have to write 'infinity' one billion times. You just need a function that returns 'infinity' when no value is set. Apr 3, 2013 at 22:16\n\nBasic Algorithm\n\nMaintain two sets of the nodes you can reach from the start and end node. In alternating fashion, go three steps from both sides. Each time replacing your set with nodes you can reach through one more step. After each step you check the two sets for common nodes.\n\nOptimizations\n\nMake sure you can iterate the sets as sorted so that you can search for common nodes in a single sweep: an O(n+m) operation. Lists will be up to a million nodes each.\n\nTo extend a set with one step, you have query all connections of the nodes in the original set and merge them into a new sorted set. Merging 2 sorted lists can again be done in a single sweep. So you also want to make sure that you can query the connections of a node as sorted. (This could be preprocessed).\n\nIn the last two steps each new set is the result of merging up to 10000 of these query results. It is best to do this merge adaptive (merging equally sized chunks). In that way, the sorted set data structure can be a simple linked list.\n\nThat way the whole algorithm becomes O(6*n + 6*n*log n) where n is max. 1,000,000.\n\n• How do you check for membership in a linked list in less than O(n)? That seems like a big problem. Apr 3, 2013 at 17:07\n• The trick is to run over both lists simultaneously. That way you can check for any doubles in a single sweep. Apr 3, 2013 at 21:02\n• Won't insertion get expensive using a sorted list? A tree or hash would work better. Apr 3, 2013 at 22:21\n• A tree might indeed be better. But either way, the insertion is optimized by having all connections of a node pre-sorted. Apr 4, 2013 at 21:59\n• Or rather... Single linked list is perfectly fine. See edits of the answer. Apr 4, 2013 at 23:07\n\nJust use breath first search (no need for Dijkstra's alg. because all edges have uniform length) (and as Kris Van Bael said, run it from both sides)\n\n`\"All edges have a length of 1\"` This is a best-case scenario making Dijkstra's Algorithm a perfect greedy algorithm choice. Even using the Floyd-Warshall algorithm involving fast matrix multiplication would work well.\n\n• I'm confused. From the link you gave, it looks like the Floyd-Warshall algorithm is aimed towards solving the all-pairs shortest path problem. Is it also best way to find distance between a single pair? Apr 3, 2013 at 17:24\n• @NickODell Floyd-Warshall is a greedy algorithm for finding shortest paths in a graph. It is but one, including Dijkstra which is quite popular, for finding shortest paths from the start node to all other nodes. Remember, you are finding the shortest path from a designated node to all other nodes and not just two points. Apr 3, 2013 at 17:33" ]

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[ "Location, Seattle WA USA\nSe Habla Espanol\[email protected]\n\n# Unbalance Force\n\nVibration Experts", null, "# Unbalance Force\n\nUnbalance force in rotors lead to premature machine failure. Excess unbalance forces cause bearings and seals to fail prematurely. Excessive vibration can cause other problems. It is therefore important to precision balance rotors.\n\nWe can define the unbalance force in two separate ways. First, Force (F) equals the unbalance mass (m) times the radius of the mass from the center of rotation (r) times the rotational speed (w) squared.\n\n## F = mrw2\n\nBecause the unbalance force is proportional to the rotational speed squared, it is especially important to precision balance high speed rotors.\n\nWhen diagnosing unbalance with vibration, we are essentially saying that if the unbalance force goes up, the vibration will also go up. It is important to note however that both an increase in unbalance mass AND and an increase in rotational speed will cause the vibration to increase in amplitude. This is why it is so important to always test machines as close to the same running speed as possible and trend the data. Otherwise it is difficult to know if increases in vibration are unbalance related or just due to a higher shaft speed.\n\n## Unbalance Force and Eccentricity\n\nAnother way to calculate the unbalance force (F) is to multiply the rotor mass (M) by the rotor eccentricity (e) times the rotational speed (w) squared.\n\n## F=Mew2\n\nWhat is happening here is the rotor wants to rotate around it’s center of mass, not its geometrical center. To visualize this, think about throwing a uniform wooden pole through the air so it flips end over end. It rotates around its geometric center. Now imagine throwing a hammer through the air the same way. Think about the motion it will create – it will not rotate around the geometrical center but around the head, eq the heavy part.\n\nThe rotor eccentricity (e) is the distance between the geometric center and the center of mass. This also explains why the rotor vibrates. If it was perfectly balanced and rotated around its geometric center, it wouldn’t vibrate!\n\nSince we have two equations for the unbalance force, we can say they are equal.\n\n## e=mr/M\n\nWe are talking here about “rigid rotors” or machines that are running well below their first critical speed. Once we go over the first critical speed, things change.\n\nThe formulas above will be useful to consider when we discuss balance standards and balance quality (in a future post.) If you are interested in this topic and would like to learn more, the best way is to sign up for one of my vibration courses: https://www.zencovibrations.com/events/ and get certified in accordance with ISO 18436-2" ]

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[ "# Hamilton Paths in Complete graph $K_n$\n\nIn complete graph $$K_n$$, is it true that we can have at least $$2*n$$ Hamilton paths?\n\n## 1 Answer\n\nSince in a complete graph $$K_n$$ each two vertices are adjacent, the set of Hamiltonian paths of $$K_n$$ is exactly a set $$S_n$$ of permutations of its vertices. But $$|S_n|=n!$$, which is at least $$2n$$ when $$n\\ge 3$$, but it is less than $$2n$$ for $$n\\le 2$$.\n\n• I know it’s stupid, but you have to divide by two since the reverse of a permutation gives the same Hamiltonian path. – Bob Krueger Dec 19 '18 at 8:50\n• @BobKrueger I was aware of this point, so I have checked it (in “Chromatic Graph Theory” by Chartrand and Zhang). A path is defined as a sequence of vertices such that... . So, for instance, since the sequences $(1,2,3)$ and $(3,2,1)$ are distinct, these paths are distinct too. – Alex Ravsky Dec 21 '18 at 7:02\n• Fair enough. Then with that, the OP can prove their claim for any graph with a Hamiltonian cycle. – Bob Krueger Dec 21 '18 at 9:05" ]

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[ "# Fetching a Row\n\nWe can fetch a row three different ways. Essentially each way retrieves a single record, the one being pointed to, and returns it in a unique structure.\n\n##### PHP\n` MySQL_fetch_row(Result); `\n\nThis returns the data in a simple array. Each column's value is put in a separate element of the array. The array is zero-based . The range is 0 “ Number_of_columns “ 1.\n\n##### PHP\n` MySQL_fetch_array(Result); `\n\nThis returns an array. But this array is different from the array that we just fetched in the previous command. This array is an associative array. The key to access any value is the name of the column that the datum was stored in. In addition we can use the PHP array-accessing features (show, list, each) to move through this array, which, since a meaning is attached to each variable, can then be used easily.\n\n##### PHP\n` MySQL_fetch_object(Result); `\n\nThere is only one major difference between this command and the one preceding it. This one returns an object. Therefore, all the data is stored as a property with the name of the column being the name of the property. Also, unlike the associated array that was fetched before, this data cannot be accessed by referring to the offset of data. And there is no way to discover the column names if you do not already know them.\n\nAlthough we did not create it and cannot directly access it, we can easily move the pointer to any row we choose. There are two functions we can use to do this. The MySQL_result command returns the value of any single field of the result. The Row is the number row of the result. The Column parameter is the zero-based column.\n\n##### PHP\n` MySQL_result(Result, Row, Column); `\n\nThis command moves the internal pointer within the result object to the row after Row. Any of the above commands that fetches a row affect the row immediately after the row that the retrieved field was in.\n\nA second command to move the internal pointer expressly does solely that.\n\n##### PHP\n` MySQL_data_seek(Result, NewRow); `\n\nThis command moves the pointer to NewRow. Since all fetch commands retrieve rows, no column definition is neccessary.\n\n#### Measuring Results\n\nWe can count how many rows were returned by MySQL_query .\n\n##### PHP\n` MySQL_num_rows(Result); `\n\nThis command returns the number of rows in the result ”very useful information for setting up loop exit conditions.\n\n#### Releasing Memory\n\nDepending on how much information a result is storing and depending on how many results we have, we may start having memory problems. We can free up some of the results we are not using anymore for garbage collection. Garbage collection is when the program disassociates chunks of RAM with variable names so that RAM can be conserved.\n\n##### PHP\n` MySQL_free_result(Result); `\n\nElusive, erratic, and evil bugs may be the result of running out of memory. Fortunately, on today's modern machines, it is not as big a worry as used to be.\n\nWe have to be careful, because once a result is freed, it can never be accessed again. Once something is deleted it is gone, after all. To attempt to use the results will cause errors to occur.\n\nThe changes we make to the codebase in this chapter are based on the work of Chapter 10, where the following code appeared:\n\n##### PHP\n` <? class Question { var \\$idkey; var \\$Q; var \\$A= array(); `\n\nFakedata exists only until the MySQL database is working. It is, so it will disappear in our new code.\n\n##### PHP\n` var \\$fakedata= array ( . . . lots and lots of hardcoded data . . . ); `\n\nThe \"database\" is queried during this object constructor, which is replaced later on in this chapter.\n\n##### PHP\n` function Question( \\$idkey ) { if( !\\$this->fakedata[\\$idkey][Q]) return null; \\$this->Q = \\$this->fakedata[\\$idkey][Q]; while ( list(\\$key, \\$choice) = each(\\$this->fakedata[\\$idkey][A] ) ) array_push( \\$this->A, \\$choice); \\$this->idkey = \\$idkey; return \\$this; } `\n\nThis function creates all the necessary XML output. It remains unchanged.\n\n##### PHP\n` function qzml(){ if( !\\$beenhere++ ){ echo ( '<?xml version=\"1.0\"?>' .\"\\ r\\ n\"); echo ( '<Question UIN = \"'.\\$this->idkey.'\" >' .\"\\ r\\ n\"); echo ( '<Q>'.\\$this->Q. '</Q>' .\"\\ r\\ n\"); while ( list( \\$n, \\$answer ) = each(\\$this->A) ) echo( '<A>'.\\$answer.'</A>' .\"\\ r\\ n\"); echo ( '</Question>' .\"\\ r\\ n\"); } } } srand(microtime()*1000000); // seed random number generator \\$q= new Question( rand(0,5) ); // create a question \\$q->qzml(); // display as xml ?> `\n\nNow we replace the object's constructor with one that queries our database. Most of it is straightforward code. A large conceptual problem, however, is choosing which question we want from the database.\n\nThere are several ways that we could use to decide which question we will withdraw from the Question table. One would be to plod through the questions sequentially, using a persistent variable to keep counting through all the QuID s. We could even store this counter in its own MySQL table. If we wanted to be even fancier, we could keep a separate persistent variable for each of the users. But let's keep it simple and just select a random question regardless of past question history.", null, "", null, "Flash and XML[c] A Developer[ap]s Guide\nISBN: 201729202\nEAN: N/A\nYear: 2005\nPages: 160" ]

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[ "General Question", null, "What mathematical sequence includes these two numbers?\n\nAsked by weeveeship (4660", null, ") December 15th, 2010\n\nThe numbers are:\n\n130812782.6503 (28th number in the sequence)\n523251130.6012 (52nd number in the sequence)\n\nP.S. I think the sequence is mathematical because it involves numbers but feel free to go beyond math sequences. Thanks a bunch.\n\nObserving members: 0", null, "Composing members: 0", null, "", null, "The ratio of 52nd to 28th term is exactly 4. This is the only real clue here.\n\nThis suggests a geometric series, in which case the terms quadruple every 52 – 28 = 24 positions, or double every 12. This is exactly the case with the well-tempered piano scale, where frequencies double each octave of 12 semi-tones. The ratio between successive terms is the twelfth root of two, which is 2^(1/12) = 1.059… (Each note is about 6 percent higher in frequency than its predecessor).\n\nWhat puzzles me is that the numbers in your sequence are on the order of a million times bigger than piano frequencies in cycles per second (Hertz).\n\ngasman (11315", null, ")“Great Answer” (12", null, ") Flag as…", null, "", null, "@gasman\n\nHaving frequencies double every 12 semi-tones is not unique to well temperament. The feature of well temperament is the relative frequencies within the octave. I’m quite sure nearly all Western tuning methods have a 2:1 frequency ratio every 12 semi-tones, otherwise you would not be able to sustain a scale over a wide range of notes.\n\nchocolatechip (3004", null, ")“Great Answer” (1", null, ") Flag as…", null, "", null, "@chocolatechip I understand—forgive me for not explaining it well. I meant that in the context of geometric / exponential series, where the ratio of each term to its predecessor is constant, it seems more than a coincidence to have exact doubling after exactly 12 steps.\n\nEarlier natural musical scales (where a perfect fifth is a 3:2 frequency ratio, etc.) still feature octave doubling, as you point out, but then the tones do not form a perfectly geometric series. My answer was based on assuming a geometric series. This is more or less a leap of faith, given practically no other information to work with.\n\ngasman (11315", null, ")“Great Answer” (1", null, ") Flag as…", null, "or" ]

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[ "$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n# 4.5E: Exercises for Section 4.5\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\| #1 \\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$\n\n1) If $$c$$ is a critical point of $$f(x)$$, when is there no local maximum or minimum at $$c$$? Explain.\n\n2) For the function $$y=x^3$$, is $$x=0$$ both an inflection point and a local maximum/minimum?\n\nIt is not a local maximum/minimum because $$f'$$ does not change sign\n\n3) For the function $$y=x^3$$, is $$x=0$$ an inflection point?\n\n4) Is it possible for a point $$c$$ to be both an inflection point and a local extremum of a twice differentiable function?\n\nNo\n\n5) Why do you need continuity for the first derivative test? Come up with an example.\n\n6) Explain whether a concave-down function has to cross $$y=0$$ for some value of $$x$$.\n\nFalse; for example, $$y=\\sqrt{x}$$.\n\n7) Explain whether a polynomial of degree $$2$$ can have an inflection point.\n\nIn exercises 8 - 12, analyze the graphs of $$f'$$, then list all intervals where $$f$$ is increasing or decreasing.\n\n8)", null, "Increasing for $$−2<x<−1$$ and $$x>2$$;\nDecreasing for $$x<−2$$ and $$−1<x<2$$\n\n9)", null, "10)", null, "Decreasing for $$x<1$$,\nIncreasing for $$x>1$$\n\n11)", null, "12)", null, "Decreasing for $$−2<x<−1$$ and $$1<x<2$$;\nIncreasing for $$−1<x<1$$ and $$x<−2$$ and $$x>2$$\n\nIn exercises 13 - 17, analyze the graphs of $$f',$$ then list all intervals where\n\na. $$f$$ is increasing and decreasing and\n\nb. the minima and maxima are located.\n\n13)", null, "14)", null, "a. Increasing over $$−2<x<−1,\\;0<x<1,x>2$$, Decreasing over $$x<−2, \\;−1<x<0, \\;1<x<2;$$\nb. Maxima at $$x=−1$$ and $$x=1$$, Minima at $$x=−2$$ and $$x=0$$ and $$x=2$$\n\n15)", null, "16)", null, "a. Increasing over $$x>0$$, Decreasing over $$x<0;$$\nb. Minimum at $$x=0$$\n\n17)", null, "In exercises 18 - 22, analyze the graphs of $$f'$$, then list all inflection points and intervals $$f$$ that are concave up and concave down.\n\n18)", null, "Concave up for all $$x$$,\nNo inflection points\n\n19)", null, "20)", null, "Concave up for all $$x$$,\nNo inflection points\n\n21)", null, "22)", null, "Concave up for $$x<0$$ and $$x>1$$,\nConcave down for $$0<x<1$$,\nInflection points at $$x=0$$ and $$x=1$$\n\nFor exercises 23 - 27, draw a graph that satisfies the given specifications for the domain $$x=[−3,3].$$ The function does not have to be continuous or differentiable.\n\n23) $$f(x)>0,\\;f'(x)>0$$ over $$x>1,\\;−3<x<0,\\;f'(x)=0$$ over $$0<x<1$$\n\n24) $$f'(x)>0$$ over $$x>2,\\;−3<x<−1,\\;f'(x)<0$$ over $$−1<x<2,\\;f''(x)<0$$ for all $$x$$\n\n25) $$f''(x)<0$$ over $$−1<x<1,\\;f''(x)>0,\\;−3<x<−1,\\;1<x<3,$$ local maximum at $$x=0,$$ local minima at $$x=±2$$\n\n26) There is a local maximum at $$x=2,$$ local minimum at $$x=1,$$ and the graph is neither concave up nor concave down.\n\n27) There are local maxima at $$x=±1,$$ the function is concave up for all $$x$$, and the function remains positive for all $$x.$$\n\nFor the following exercises, determine\n\na. intervals where $$f$$ is increasing or decreasing and\n\nb. local minima and maxima of $$f$$.\n\n28) $$f(x)=\\sin x+\\sin^3x$$ over $$−π<x<π$$\n\na. Increasing over $$−\\frac{π}{2}<x<\\frac{π}{2},$$ decreasing over $$x<−\\frac{π}{2},\\; x>\\frac{π}{2}$$\n\nb. Local maximum at $$x=\\frac{π}{2}$$; local minimum at $$x=−\\frac{π}{2}$$\n\n29) $$f(x)=x^2+\\cos x$$\n\nFor exercise 30, determine\n\na. intervals where $$f$$ is concave up or concave down, and\n\nb. the inflection points of $$f$$.\n\n30) $$f(x)=x^3−4x^2+x+2$$\n\na. Concave up for $$x>\\frac{4}{3},$$ concave down for $$x<\\frac{4}{3}$$\n\nb. Inflection point at $$x=\\frac{4}{3}$$\n\nFor exercises 31 - 37, determine\n\na. intervals where $$f$$ is increasing or decreasing,\n\nb. local minima and maxima of $$f$$,\n\nc. intervals where $$f$$ is concave up and concave down, and\n\nd. the inflection points of $$f.$$\n\n31) $$f(x)=x^2−6x$$\n\n32) $$f(x)=x^3−6x^2$$\n\na. Increasing over $$x<0$$ and $$x>4,$$ decreasing over $$0<x<4$$\nb. Maximum at $$x=0$$, minimum at $$x=4$$\nc. Concave up for $$x>2$$, concave down for $$x<2$$\nd. Inflection point at $$x=2$$\n\n33) $$f(x)=x^4−6x^3$$\n\n34) $$f(x)=x^{11}−6x^{10}$$\n\na. Increasing over $$x<0$$ and $$x>\\frac{60}{11}$$, decreasing over $$0<x<\\frac{60}{11}$$\nb. Maximum at $$x=0$$, minimum at $$x=\\frac{60}{11}$$\nc. Concave down for $$x<\\frac{54}{11}$$, concave up for $$x>\\frac{54}{11}$$\nd. Inflection point at $$x=\\frac{54}{11}$$\n\n35) $$f(x)=x+x^2−x^3$$\n\n36) $$f(x)=x^2+x+1$$\n\na. Increasing over $$x>−\\frac{1}{2}$$, decreasing over $$x<−\\frac{1}{2}$$\nb. Minimum at $$x=−\\frac{1}{2}$$\nc. Concave up for all $$x$$\nd. No inflection points\n\n37) $$f(x)=x^3+x^4$$\n\nFor exercises 38 - 47, determine\n\na. intervals where $$f$$ is increasing or decreasing,\n\nb. local minima and maxima of $$f,$$\n\nc. intervals where $$f$$ is concave up and concave down, and\n\nd. the inflection points of $$f.$$ Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator.\n\n38) [T] $$f(x)=\\sin(πx)−\\cos(πx)$$ over $$x=[−1,1]$$\n\na. Increases over $$−\\frac{1}{4}<x<\\frac{3}{4},$$ decreases over $$x>\\frac{3}{4}$$ and $$x<−\\frac{1}{4}$$\nb. Minimum at $$x=−\\frac{1}{4}$$, maximum at $$x=\\frac{3}{4}$$\nc. Concave up for $$−\\frac{3}{4}<x<\\frac{1}{4}$$, concave down for $$x<−\\frac{3}{4}$$ and $$x>\\frac{1}{4}$$\nd. Inflection points at $$x=−\\frac{3}{4},\\;x=\\frac{1}{4}$$\n\n39) [T] $$f(x)=x+\\sin(2x)$$ over $$x=[−\\frac{π}{2},\\frac{π}{2}]$$\n\n40) [T] $$f(x)=\\sin x+\\tan x$$ over $$(−\\frac{π}{2},\\frac{π}{2})$$\n\na. Increasing for all $$x$$\nb. No local minimum or maximum\nc. Concave up for $$x>0$$, concave down for $$x<0$$\nd. Inflection point at $$x=0$$\n\n41) [T] $$f(x)=(x−2)^2(x−4)^2$$\n\n42) [T] $$f(x)=\\dfrac{1}{1−x},\\quad x≠1$$\n\na. Increasing for all $$x$$ where defined\nb. No local minima or maxima\nc. Concave up for $$x<1$$; concave down for $$x>1$$\nd. No inflection points in domain\n\n43) [T] $$f(x)=\\dfrac{\\sin x}{x}$$ over $$x=[-2π,0)∪(0,2π]$$\n\n44) $$f(x)=\\sin(x)e^x$$ over $$x=[−π,π]$$\n\na. Increasing over $$−\\frac{π}{4}<x<\\frac{3π}{4}$$, decreasing over $$x>\\frac{3π}{4},\\;x<−\\frac{π}{4}$$\nb. Minimum at $$x=−\\frac{π}{4}$$, maximum at $$x=\\frac{3π}{4}$$\nc. Concave up for $$−\\frac{π}{2}<x<\\frac{π}{2}$$, concave down for $$x<−\\frac{π}{2},\\;x>\\frac{π}{2}$$\nd. Inflection points at $$x=±\\frac{π}{2}$$\n\n45) $$f(x)=\\ln x\\sqrt{x},\\quad x>0$$\n\n46) $$f(x)=\\frac{1}{4}\\sqrt{x}+\\frac{1}{x},\\quad x>0$$\n\na. Increasing over $$x>4,$$ decreasing over $$0<x<4$$\nb. Minimum at $$x=4$$\nc. Concave up for $$0<x<8\\sqrt{2}$$, concave down for $$x>8\\sqrt{2}$$\nd. Inflection point at $$x=8\\sqrt{2}$$\n\n47) $$f(x)=\\dfrac{e^x}{x},\\quad x≠0$$\n\nIn exercises 48 - 52, interpret the sentences in terms of $$f,\\;f',$$ and $$f''.$$\n\n48) The population is growing more slowly. Here $$f$$ is the population.\n\n$$f>0,\\;f'>0,\\;f''<0$$\n\n49) A bike accelerates faster, but a car goes faster. Here $$f=$$ Bike’s position minus Car’s position.\n\n50) The airplane lands smoothly. Here $$f$$ is the plane’s altitude.\n\n$$f>0,\\;f'<0,\\;f''>0$$\n\n51) Stock prices are at their peak. Here $$f$$is the stock price.\n\n52) The economy is picking up speed. Here $$f$$ is a measure of the economy, such as GDP.\n\n$$f>0,\\;f'>0,\\;f''>0$$\n\nFor exercises 53 - 57, consider a third-degree polynomial $$f(x),$$ which has the properties $$f'(1)=0$$ and $$f'(3)=0$$.\n\nDetermine whether the following statements are true or false. Justify your answer.\n\n53) $$f(x)=0$$ for some $$1≤x≤3$$.\n\n54) $$f''(x)=0$$ for some $$1≤x≤3$$.\n\nTrue, by the Mean Value Theorem\n\n55) There is no absolute maximum at $$x=3$$.\n\n56) If $$f(x)$$ has three roots, then it has $$1$$ inflection point.\n\n57) If $$f(x)$$ has one inflection point, then it has three real roots." ]

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[ "# Antoine's coefficients conversion\n\n Input : Formula scale", null, "log (base 10) ln (base e) Pressure Unit", null, "mmHg Bar mBar Pa kPa psia Antoine A Antoine B Antoine C Temperature min", null, "Temperature max", null, "Temperature Unit", null, "Kelvin Celsius Fahrenheit Rankine Output : Formula scale", null, "log (base 10) ln (base e) Pressure Unit", null, "mmHg Bar mBar Pa kPa psia Temperature Unit", null, "Kelvin Celsius Fahrenheit Rankine Results : Antoine A Antoine B Antoine C\n\nYou may use this tools for free once you have registered here in less than 2 minutes.\n\n# How to use this conversion tool\n\nThis tool converts Antoine's coefficients from formula log10(Pv)=A-B/(C+T) (log base 10) to formula ln(Pv)=A-B/(C+T) (natural log) and the other way round.\n\nAntoine's coefficients to fill in are those corresponding to the chosen input formula and that give a result using input unit.\n\nTemperature range must be a valid range for Antoine's coefficients filled in for input. The default range (283-303K) can be used when input Antoine's coefficients are valid for ambient temperature.\n\nOutput parameters (Formula scale and Pressure unit) describe the format of the formula to be used with calculated Antoine's coefficients and the unit of the result of this calculation.\n\nExample :\n\nAntoine's coefficients for ethanol with formula log10(P)=A-B/(C+T) (P in bar, T in kelvin) are : A=5.37229, B=1670.409, C=-40.191 for the temperature range 273 to 351 K.\nEquivalent coefficients for formula ln(P)=A-B/(C+T) (P in mmHg, T in kelvin) are : A=18.99031, B=3846.25886, C=-40.191.\nFor ethanol at 283K, P = 10(5.37229-1670.409/(-40.191+283)) = 0.0311 bar which corresponds to P = e(18.99031-3846.25886/(-40.191+283)) = 23.3275 mmHg.\n\nEnvironmental Models" ]

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[ "# How to limit optimization program to choose only natural integers numbers (not float) if my objective function is not linear?\n\n4 views (last 30 days)\nRA on 18 Oct 2021\nAnswered: Alan Weiss on 19 Oct 2021\nHello,\nI would like to ask you about my optimization problem on matlab:\nIs there a function to compute the minimum of a non-linear function by giving us naturals integers values for x belongs to N (x should be an integer which represent a discrete mode 1 or 2 or 10 or anything else, the important is that x is an natural integer and f is reel R).\n\nAlan Weiss on 19 Oct 2021\nIf you have Global Optimization Toolbox you can use Mixed Integer ga Optimization.\nAlan Weiss\nMATLAB mathematical toolbox documentation" ]

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[ "# Local outlier factor\n\nAlgorithm\n\nThe local outlier factor algorithm is used to detect (and usually remove) outliers, or anomalous data items, from a training set that is going to be used for some other regression or classification algorithm.\n\nWhether or not an item is an outlier depends on its average distance to its k nearest neighbours, but weighted according to the average distance of those neighbours to their own neighbours. This ensures that an average distance that would mark a data item out as an outlier in a very densely populated area of the vector space does not mark a data item out as an outlier in an area that is more sparsely populated.\n\nOutlier detection should only be used if there is a clear theoretical basis for believing that outliers represent errors, otherwise removing outliers might hide important insights from the regression or classification that is then performed on the artificially “cleaned up” training data.\n\nalias\nsubtype\nhas functional building block\nFBB_Classification\nhas input data type\nIDT_Vector of quantitative variables\nhas internal model\nhas output data type\nODT_Vector of quantitative variables\nhas learning style\nLST_Unsupervised\nhas parametricity\nPRM_Nonparametric with hyperparameter(s)\nhas relevance\nREL_Relevant\nuses\nALG_Nearest Neighbour\nsometimes supports\nALG_Least Squares Regression ALG_Logistic regression ALG_Nearest Neighbour\nmathematically similar to" ]

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[ "Write a complete C++ program for WorldParking Sdn Bhd. to perform the following:\n\na. Write a return-value function named calcCharges() to calculate and return the parking charges for the customers. The company charges a RM1.00 minimum fee to park for up to one hour. An additional RM0.50 will be charged for each hour exceeding the first one hour. The maximum charge for any given 24-hour period is RM10.00. Assume that no car parks for longer than 24 hours at a time.\n\nb. Write a void function named calcTotal() to calculate the total charges for all the customers.\n\nc. Write the main program that allows the user to input number of customers and the hours parked for the customers. The program should use the function calcCharges()above to calculate the parking charges for each customer and function calcTotal() above to calculate the total charges for all the customers.\n\nd. Finally, your program should output the parking charges for each customer and the total charges for all the customers. Use the appropriate parameters to pass values in and out of functions.\n\nFatur_1 commented: i got the answer for the question, can email me for the answer. [email protected] +0\n\n## All 2 Replies\n\nDo provide evidence of having done some work yourself if posting questions from school or work assignments\n\nWith that out of the way, keep the question(s) to what you need help with. That is, that line of code that fails so we can narrow it down to that one thing that stopped you.\n\n``````#include <iostream>\n#include <cmath>\nusing namespace std;\n\ndouble calcCharges(int hours){\n\ndouble totalcharges;\n\nif(hours<=19.0){\ntotalcharges = (hours - 1) * 0.5 + 1;\n}else{\ntotalcharges = 10.0;\n}\n\n}\n\nvoid calctotal(double sum, float all[], int arraysize){\n\nint a;\n\nsystem (\"CLS\");\n\nfor( a = 0; a < arraysize; a++){\nif(all[a] != 0){\ncout<<\" Customer \" <<a+1 <<\" : RM\" <<all[a] <<endl;\n}\n}\n\ncout<< \"Below Is the the total charges for all the customers: \\n\";\n// cout<< \"RM \" <<sum;\n\ncout<< \"\\n TOTAL : \" <<sum;\n}\n\nint main()\n{\nint hour, choice, arraysize, continue1, i = 0;\ndouble total, sum;\nfloat all[] = { 0.0 };\n\ndo{\ncout<<\" Welcome to World Parking Sdn Bhd. Calculator System \\n\";\ncout<< \" Please Select the action below: \\n\";\ncout<< \" 1 : Add customer \\n\";\ncout<< \" 2 : View Total \\n\";\ncin>>choice;\n\nswitch(choice) {\ncase 1:\ncout<<\" Please enter the total our hour: \\n\";\ncin>>hour;\n\ntotal = calcCharges(hour);\nall[i] = total;\ni++;\n\ncout<< \" Total charges: RM\" <<total;\ncout<< \"\\n Enter 1 to continue, 2 to exit: \\n\";\ncin>>continue1;\n\nchoice = 0;\n\n//system (\"CLS\");\n\nbreak;\n\ncase 2:\n\narraysize = sizeof(all);\nfor(int a = 0; a < arraysize; a++){\nsum += all[a];\n}\ncalctotal(sum, all, arraysize);\ncout<<\"\\n\\n THANK YOU BYEBYE\";\n\nbreak;\n\ndefault:\ncout<< \" Please Select the action below: (Please type 1 or 2)\\n\";\ncout<< \" 1 : Add customer \\n\";\ncout<< \" 2 : View Total \\n\";\n}\n\nsystem (\"CLS\");\n\n}\nwhile (continue1 == 1);\n\narraysize = sizeof(all);\nfor(int a = 0; a < arraysize; a++){\nsum += all[a];\n}\ncalctotal(sum, all, arraysize);\ncout<<\"\\n\\n THANK YOU BYEBYE\";\n\n}``````\ncommented: I thought it was buh-bye? +15\nBe a part of the DaniWeb community\n\nWe're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, learning, and sharing knowledge." ]

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[ "You are on page 1of 24\n\nCapital Budgeting\n\nTest ID: 7694293\n\nQuestion #1 of 57\n\nQuestion ID: 414743\n\nPolington Aircraft Co. just announced a sale of 30 aircraft to Cuba, a project with a net present value of \\$10 million. Investors did\nnot anticipate the sale because government approval to sell to Cuba had never before been granted. The share price of\nPolington should:\nA) increase by the NPV (1 - corporate tax rate) divided by the number of common shares\noutstanding.\nB) not necessarily change because new contract announcements are made all the time.\nC) increase by the project NPV divided by the number of common shares outstanding.\nExplanation\nSince the sale was not anticipated by the market, the share price should rise by the NPV of the project per common share. NPV\nis already calculated using after-tax cash flows.\n\nQuestion #2 of 57\n\nOne of the basic principles of capital budgeting is that:\n\nA) decisions are based on cash flows, not accounting income.\nB) opportunity costs should be excluded from the analysis of a project.\nC) cash flows should be analyzed on a pre-tax basis.\nExplanation\nThe five key principles of the capital budgeting process are:\n1. Decisions are based on cash flows, not accounting income.\n2. Cash flows are based on opportunity costs.\n3. The timing of cash flows is important.\n4. Cash flows are analyzed on an after-tax basis.\n5. Financing costs are reflected in the project's required rate of return.\n\nQuestion #3 of 57\n\nQuestion ID: 414742\n\nThe effect of a company announcement that they have begun a project with a current cost of \\$10 million that will generate future\ncash flows with a present value of \\$20 million is most likely to:\nA) increase value of the firm's common shares by \\$10 million.\nB) increase the value of the firm's common shares by \\$20 million.\nC) only affect value of the firm's common shares if the project was unexpected.\nExplanation\nStock prices reflect investor expectations for future investment and growth. A new positive-NPV project will increase stock price\n\n1 of 24\n\nonly if it was not previously anticipated by investors.\n\nQuestion #4 of 57\n\nQuestion ID: 434325\n\nAn analyst has gathered the following data about a company with a 12% cost of capital:\nProject P\n\nProject Q\n\nCost\n\n\\$15,000\n\n\\$25,000\n\nLife\n\n5 years\n\n5 years\n\nCash inflows\n\n\\$5,000/year\n\n\\$7,500/year\n\nIf Projects P and Q are mutually exclusive, what should the company do?\n\nA) Accept Project P and reject Project Q.\n\nB) Reject both Project P and Project Q.\nC) Accept Project Q and reject Project P.\nExplanation\nProject P:\nN = 5; PMT = 5,000; FV = 0; I/Y = 12; CPT PV = 18,024\nNPV for Project A = 18,024 15,000 = 3,024.\nProject Q:\nN = 5; PMT = 7,500; FV = 0; I/Y = 12; CPT PV = 27,036\nNPV for Project B = 27,036 25,000 = 2,036.\nFor mutually exclusive projects, accept the project with the highest positive NPV. In this example the NPV for Project P (3,024) is\nhigher than the NPV of Project Q (2,036). Therefore accept Project P.\n\nQuestion #5 of 57\n\nQuestion ID: 414739\n\nThe NPV profile is a graphical representation of the change in net present value relative to a change in the:\nA) internal rate of return.\nB) discount rate.\nC) prime rate.\nExplanation\nAs discount rates change the net present values change. The NPV profile is a graphic illustration of how sensitive net present\nvalues are to different discount rates. By comparison, every project has a single internal rate of return and payback period\nbecause the values are determined solely by the investment's expected cash flows.\n\nQuestion #6 of 57\n\nWhich of the following statements about independent projects is least accurate?\n\n2 of 24\n\nA) The internal rate of return and net present value methods can yield different accept/reject decisions\nfor independent projects.\n\nB) If the internal rate of return is less than the cost of capital, reject the project.\nC) The net present value indicates how much the value of the firm will change if the project is accepted.\nExplanation\nFor independent projects the IRR and NPV give the same accept/reject decision. For mutually exclusive projects the IRR and NPV\ntechniques can yield different accept/reject decisions.\n\nQuestion #7 of 57\n\nQuestion ID: 414707\n\nThe process of evaluating and selecting profitable long-term investments consistent with the firm's goal of shareholder wealth\nmaximization is known as:\nA) financial restructuring.\nB) capital budgeting.\nC) monitoring.\nExplanation\nIn the process of capital budgeting, a manager is making decisions about a firm's earning assets, which provide the basis for the\nfirm's profit and value. Capital budgeting refers to investments expected to produce benefits for a period of time greater than one\nyear. Financial restructuring is done as a result of bankruptcy and monitoring is a critical assessment aspect of capital budgeting.\n\nQuestion #8 of 57\n\nQuestion ID: 414711\n\nWhich of the following statements about the payback period is NOT correct?\nA) The payback period is the number of years it takes to recover the original cost of the\ninvestment.\nB) The payback method considers all cash flows throughout the entire life of a project.\nC) The payback period provides a rough measure of a project's liquidity and risk.\nExplanation\nThe payback period does not take any cash flows after the payback point into consideration.\n\nQuestion #9 of 57\n\nQuestion ID: 460660\n\nIn a net present value (NPV) profile, the internal rate of return is represented as the:\nA) intersection of the NPV profile with the vertical axis.\nB) intersection of the NPV profile with the horizontal axis.\nC) point where two NPV profiles intersect.\nExplanation\n\n3 of 24\n\nThe internal rate of return is the rate of discount at which the NPV of a project is zero. On an NPV profile, this is the point where\nthe profile intersects the horizontal axis.\n\nQuestion #10 of 57\n\nQuestion ID: 414704\n\nThe Chief Financial Officer of Large Closeouts Inc. (LCI) determines that the firm must engage in capital rationing for its capital\nbudgeting projects. Which of the following describes the most likely reason for LCI to use capital rationing? LCI:\nA) must choose between projects that compete with one another.\nB) would like to arrange projects so that investing in a project today provides the option to accept or\nreject certain future projects.\nC) has a limited amount of funds to invest.\nExplanation\nCapital rationing exists when a company has a fixed (maximum) amount of funds to invest. If profitable project opportunities\nexceed the amount of funds available, the firm must ration, or prioritize its funds to achieve the maximum value for shareholders\ngiven its capital limitations.\n\nQuestion #11 of 57\n\nQuestion ID: 414701\n\nThe CFO of Axis Manufacturing is evaluating the introduction of a new product. The costs of a recently completed marketing\nstudy for the new product and the possible increase in the sales of a related product made by Axis are best described\n(respectively) as:\n\nA) opportunity cost; externality.\n\nB) externality; cannibalization.\nC) sunk cost; externality.\nExplanation\nThe study is a sunk cost, and the possible increase in sales of a related product is an example of a positive externality.\n\nQuestion #12 of 57\n\nQuestion ID: 414731\n\nTwo projects being considered by a firm are mutually exclusive and have the following projected cash flows:\nYear\n\nProject 2 Cash Flow\n\n\\$4.0\n\n\\$3.0\n\n\\$1.7\n\n\\$5.0\n\n\\$3.2\n\n\\$2.0\n\n\\$5.8\n\nThe crossover rate of the two projects' NPV profiles is 9%. What is the initial cash flow for Project 2?\n\nA) \\$4.22.\n\n4 of 24\n\nB) \\$5.70.\nC) \\$4.51.\nExplanation\nThe crossover rate is the rate at which the NPV for two projects is the same. That is, it is the rate at which the two NPV profiles\ncross. At a discount rate of 9%, the NPV of Project 1 is: CF0 = -4; CF1 = 3; CF2 = 5; CF3 = 2; I = 9%; CPT NPV = \\$4.51. Now\nperform the same calculations except that we need to set the unknown CF0 = 0. The remaining entries are: CF1 = 1.7; CF2 = 3.2;\nCF3 = 5.8; I = 9%; CPT NPV = \\$8.73. Since by definition the crossover rate produces the same NPV for both projects, we\nknow that both projects should have an NPV = \\$4.51. Since the NPV of Project 2 (with CF0 = 0) is \\$8.73, the unknown cash flow\nmust be a large enough negative amount to reduce the NPV for Project 2 from \\$8.73 to \\$4.51. Thus the unknown initial cash flow\nfor Project 2 is determined as \\$4.51 = \\$8.73 + CF0, or CF0 = \\$4.22.\n\nQuestion #13 of 57\n\nFinancing costs for a capital project are:\n\nA) subtracted from estimates of a project's future cash flows.\nB) captured in the project's required rate of return.\nC) subtracted from the net present value of a project.\nExplanation\nFinancing costs are reflected in a project's required rate of return. Project specific financing costs should not be included as\nproject cash flows. The firm's overall weighted average cost of capital, adjusted for project risk, should be used to discount\nexpected project cash flows.\n\nQuestion #14 of 57\n\nQuestion ID: 414720\n\nEdelman Enginenering is considering including an overhead pulley system in this year's capital budget. The cash outlay for the\npully system is \\$22,430. The firm's cost of capital is 14%. After-tax cash flows, including depreciation are \\$7,500 for each of the\nnext 5 years.\nCalculate the internal rate of return (IRR) and the net present value (NPV) for the project, and indicate the correct accept/reject\ndecision.\nNPV\n\nIRR\n\nAccept/Reject\n\nA) \\$15,070\n\n14%\n\nReject\n\nB) \\$15,070\n\n14%\n\nAccept\n\nC) \\$3,318\n\n20%\n\nAccept\n\nExplanation\nUsing the cash flow keys:\nCF0 = -22,430; CFj = 7,500; Nj = 5; Calculate IRR = 20%\nI/Y = 14%; Calculate NPV = 3,318\nBecause the NPV is positive, the firm should accept the project.\n\n5 of 24\n\nQuestion #15 of 57\n\nQuestion ID: 460659\n\nA single independent project with a negative net present value has an initial cost of \\$2.5 million and would generate cash inflows\nof \\$1 million in each of the next three years. The discount rate the company used when evaluating this project is closest to:\nA) 10%.\nB) 9%.\nC) 8%.\nExplanation\nGiven that the NPV is negative, the discount rate used by the company evaluating the project must be greater than the IRR (the\ndiscount rate for which the NPV equals zero). On a financial calculator: CF0 = -2.5; CFj = 1; Nj = 3; CPT IRR = 9.7%. Since the\ndiscount rate used for this project is greater than 9.7%, it must be closer to 10% than to either of the other answer choices.\n\nQuestion #16 of 57\n\nQuestion ID: 414718\n\nA firm is reviewing an investment opportunity that requires an initial cash outlay of \\$336,875 and promises to return the following\nirregular payments:\nYear 1: \\$100,000\nYear 2: \\$82,000\nYear 3: \\$76,000\nYear 4: \\$111,000\nYear 5: \\$142,000\nIf the required rate of return for the firm is 8%, what is the net present value of the investment? (You'll need to use your financial\ncalculator.)\n\nA) \\$86,133.\nB) \\$99,860.\nC) \\$64,582.\nExplanation\nIn order to determine the net present value of the investment, given the required rate of return; we can discount each cash flow to\nits present value, sum the present value, and subtract the required investment.\nYear\n\nCash Flow\n\nPV of Cash flow at 8%\n\n-336,875.00\n\n-336,875.00\n\n100,000.00\n\n92,592.59\n\n82,000.00\n\n70,301.78\n\n76,000.00\n\n60,331.25\n\n111,000.00\n\n81,588.31\n\n142,000.00\n\n96,642.81\n\nNet Present Value\n\n64,581.74\n\n6 of 24\n\nQuestion #17 of 57\n\nLane Industries has a project with the following cash flows:\n\nYear\n\nCash Flow\n\n\\$200,000\n\n60,000\n\n80,000\n\n70,000\n\n60,000\n\n50,000\n\nThe project's cost of capital is 12%. The discounted payback period is closest to:\n\nA) 3.9 years.\nB) 2.9 years.\nC) 3.4 years.\nExplanation\nThe discounted payback period method discounts the estimated cash flows by the project's cost of capital and then calculates\nthe time needed to recover the investment.\n\nYear\n\nCash Flow\n\nDiscounted\nCash Flow\n\n\\$200,000 \\$200,000.00\n\nCumulative\nDiscounted\nCash Flow\n\\$200,000.00\n\n60,000\n\n53,571.43\n\n146,428.57\n\n80,000\n\n63,775.51\n\n82,653.06\n\n70,000\n\n49,824.62\n\n32,828.44\n\n60,000\n\n38,131.08\n\n5,302.64\n\n50,000\n\n28,371.30\n\n33,673.98\n\ndiscounted payback period =number of years until the year before full recovery +\n\nQuestion #18 of 57\n\nQuestion ID: 414719\n\nA company is considering the purchase of a copier that costs \\$5,000. Assume a cost of capital of 10 percent and the following\ncash flow schedule:\nYear 1: \\$3,000\nYear 2: \\$2,000\nYear 3: \\$2,000\nDetermine the project's payback period and discounted payback period.\nPayback Period\nA) 2.4 years\n\nDiscounted Payback Period\n\n1.6 years\n\n7 of 24\n\nB) 2.0 years\n\n2.4 years\n\nC) 2.0 years\n\n1.6 years\n\nExplanation\nRegarding the regular payback period, after 1 year, the amount to recover is \\$2,000 (\\$5,000 - \\$3,000). After the second year, the\namount is fully recovered.\nThe discounted payback period is found by first calculating the present values of each future cash flow. These present values of\nfuture cash flows are then used to determine the payback time period.\n\n3,000 / (1 + .10)1 = 2,727\n\n2,000 / (1 + .10)2 = 1,653\n2,000 / (1 + .10)3 = 1,503.\nThen:\n\n5,000 - (2,727 + 1,653) = 620\n\n620 / 1,503 = .4.\nSo, 2 + 0.4 = 2.4.\n\nQuestion #19 of 57\n\nQuestion ID: 414715\n\nWhich of the following statements about NPV and IRR is least accurate?\nA) For independent projects if the IRR is > the cost of capital accept the project.\nB) The NPV method assumes that all cash flows are reinvested at the cost of capital.\nC) For mutually exclusive projects you should use the IRR to rank and select projects.\nExplanation\nFor mutually exclusive projects you should use NPV to rank and select projects.\n\nQuestion #20 of 57\n\nQuestion ID: 414741\n\nGarner Corporation is investing \\$30 million in new capital equipment. The present value of future after-tax cash flows generated\nby the equipment is estimated to be \\$50 million. Currently, Garner has a stock price of \\$28.00 per share with 8 million shares\noutstanding. Assuming that this project represents new information and is independent of other expectations about the company,\nwhat should the effect of the project be on the firm's stock price?\nA) The stock price will remain unchanged.\nB) The stock price will increase to \\$30.50.\nC) The stock price will increase to \\$34.25.\nExplanation\nIn theory, a positive NPV project should provide an increase in the value of a firm's shares.\nNPV of new capital equipment = \\$50 million - \\$30 million = \\$20 million\n\n8 of 24\n\nValue of company prior to equipment purchase = 8,000,000 \\$28.00 = \\$224,000,000\n\nValue of company after new equipment project = \\$224 million + \\$20 million = \\$244 million\nPrice per share after new equipment project = \\$244 million / 8 million = \\$30.50\nNote that in reality, changes in stock prices result from changes in expectations more than changes in NPV.\n\nQuestion #21 of 57\n\nQuestion ID: 460661\n\nA firm is evaluating two mutually exclusive projects of the same risk class, Project X and Project Y. Both have the same initial\ncash outlay and both have positive NPVs. Which of the following is a sufficient reason to choose Project X over Project Y?\nA) Project X has both a shorter payback period and a shorter discounted payback period\ncompared to Project Y.\nB) Project Y has a lower profitability index than Project X.\nC) Project Y has a lower internal rate of return than Project X.\nExplanation\nThe correct method of choosing between two mutually exclusive projects is to choose the one with the higher NPV. The\nprofitability index is calculated as the present value of the future cash flows divided by the initial outlay for the project. Because\nboth projects have the same initial cash outlay, the one with the higher profitability index has both higher present value of future\ncash flows and the higher NPV. Ranking projects on their payback periods or their internal rates of return can lead to incorrect\nranking.\n\nQuestion #22 of 57\n\nQuestion ID: 414738\n\nWhich of the following projects would most likely have multiple internal rates of return (IRRs)? The cost of capital for all projects\nis 10.0%.\n\nCash Flows\n\nSouth\n\nEast\n\nWest\n\nCF0\n\n-15,000\n\n-12,000\n\n-8,000\n\nCF1\n\n10,000\n\n7,000\n\n4,000\n\nCF2\n\n-1,000\n\n2,000\n\nCF3\n\n15,000\n\n2,000\n\n6,000\n\nA) Projects East and West.\n\nB) Project South only.\nC) Projects South and West.\nExplanation\nThe multiple IRR problem occurs if a project has an unconventional cash flow pattern, that is, the sign of the cash flows changes more than\nonce (from negative to positive to negative, or vice-versa). Only Project South has this cash flow pattern. Neither the zero cash flow for\nProject West nor the likely negative net present value for Project East would result in multiple IRRs.\n\n9 of 24\n\nQuestion #23 of 57\n\nQuestion ID: 414700\n\nMason Webb makes the following statements to his boss, Laine DeWalt about the principles of capital budgeting.\nStatement 1: Opportunity costs are not true cash outflows and should not be considered in a capital budgeting analysis.\nStatement 2: Cash flows should be analyzed on an after-tax basis.\nShould DeWalt agree or disagree with Webb's statements?\nStatement 1\n\nStatement 2\n\nA) Disagree\n\nDisagree\n\nB) Agree\n\nAgree\n\nC) Disagree\n\nAgree\n\nExplanation\nDeWalt should disagree with Webb's first statement. Cash flows are based on opportunity costs. Any cash flows that the firm\ngives up because a project is undertaken should be charged to the project. DeWalt should agree with Webb's second statement.\nThe impact of taxes must be considered when analyzing capital budgeting projects.\n\nQuestion #24 of 57\n\nQuestion ID: 414729\n\nApple Industries, a firm with unlimited funds, is evaluating five projects. Projects A and B are independent and Projects C, D, and\nE are mutually exclusive. The projects are listed with their rate of return and NPV. Assume that the applicable discount rate is\n10%.\n\nProject\n\nStatus\n\nRate of Return Net Present Value\n\nIndependent\n\n14%\n\n\\$10,500\n\nIndependent\n\n12%\n\n\\$13,400\n\nMutually Exclusive\n\n11%\n\n\\$16,000\n\nMutually Exclusive\n\n15%\n\n\\$14,000\n\nMutually Exclusive\n\n12%\n\n\\$11,500\n\nA) Project A, Project B, and Project C.\n\nB) Project A, Project B, and Project D.\nC) All projects should be selected.\nExplanation\nWhen it comes to independent projects, financial managers should select all with positive NPVs, resulting in inclusion of Project\nA and Project B. Remember that projects with positive NPVs will increase the value of the firm. Among mutually exclusive\nprojects, financial managers would select the one with the highest NPV, in this case Project C. Although all projects have positive\nNPVs, only one of the latter three can be chosen. If the selection were based upon the internal rate of return, Project D would be\nchosen instead of Project C. This shows why NPV is the superior decision criteria because Project C is the investment that will\n\n10 of 24\n\ncause the greatest increase to the value of the firm.\n\nQuestion #25 of 57\n\nWhen using net present value (NPV) profiles:\n\nA) the NPV profile's intersection with the vertical y-axis identifies the project's internal rate of\nreturn.\nB) one should accept all independent projects with positive NPVs.\nC) one should accept all mutually exclusive projects with positive NPVs.\nExplanation\nWhere the NPV intersects the vertical y-axis you have the value of the cash inflows less the cash outflows, assuming an absence\nof money having a time value (i.e., the discount rate is zero). Where the NPV intersects the horizontal x-axis you have the\nproject's internal rate of return. At this cost of financing, the cash inflows and cash outflows offset each other. The NPV profile is a\ntool that graphically plots the project's NPV as calculated using different discount rates. Assuming an appropriate discount rate,\none should accept all projects with positive net present values, if the projects are independent. If projects are mutually exclusive\nselect the one with the higher NPV at any given level of the cost of capital.\n\nQuestion #26 of 57\n\nQuestion ID: 414709\n\nWhich of the following statements about the discounted payback period is least accurate? The discounted payback:\nA) period is generally shorter than the regular payback.\nB) method can give conflicting results with the NPV.\nC) frequently ignores terminal values.\nExplanation\nThe discounted payback period calculates the present value of the future cash flows. Because these present values will be less\nthan the actual cash flows it will take a longer time period to recover the original investment amount.\n\nQuestion #27 of 57\n\nQuestion ID: 414726\n\nWhich of the following statements about NPV and IRR is NOT correct?\nA) The NPV will be positive if the IRR is less than the cost of capital.\nB) The IRR can be positive even if the NPV is negative.\nC) When the IRR is equal to the cost of capital, the NPV equals zero.\nExplanation\nThis statement should read, \"The NPV will be positive if the IRR is greater than the cost of capital. The other statements are\ncorrect. The IRR can be positive (>0), but less than the cost of capital, thus resulting in a negative NPV. One definition of the IRR\nis the rate of return for which the NPV of a project is zero.\n\n11 of 24\n\nQuestion #28 of 57\n\nProject sequencing is best described as:\n\nA) an investment in a project today that creates the opportunity to invest in other projects in the\nfuture.\nB) arranging projects in an order such that cash flows from the first project fund subsequent projects.\nC) prioritizing funds to achieve the maximum value for shareholders, given capital limitations.\nExplanation\nProjects are often sequenced through time so that investing in a project today may create the opportunity to invest in other\nprojects in the future. Note that funding from the first project is not a requirement for project sequencing.\n\nQuestion #29 of 57\n\nQuestion ID: 414735\n\nWhich of the following statements regarding the internal rate of return (IRR) is most accurate? The IRR:\nA) can lead to multiple IRR rates if the cash flows extend past the payback period.\nB) assumes that the reinvestment rate of the cash flows is the cost of capital.\nC) and the net present value (NPV) method lead to the same accept/reject decision for independent\nprojects.\nExplanation\nNPV and IRR lead to the same decision for independent projects, not necessarily for mutually exclusive projects. IRR assumes\nthat cash flows are reinvested at the IRR rate. IRR does not ignore time value of money (the payback period does), and the\ninvestor may find multiple IRRs if there are sign changes after time zero (i.e., negative cash flows after time zero).\n\nQuestion #30 of 57\n\nQuestion ID: 434326\n\nAn analyst has gathered the following data about a company with a 12% cost of capital:\nProject P\n\nProject Q\n\nCost\n\n\\$15,000\n\n\\$25,000\n\nLife\n\n5 years\n\n5 years\n\nCash inflows\n\n\\$5,000/year\n\n\\$7,500/year\n\nA) Accept both Project P and Project Q.\n\nB) Reject both Project P and Project Q.\nC) Accept Project P and reject Project Q.\nExplanation\nProject P: N = 5; PMT = 5,000; FV = 0; I/Y = 12; CPT PV = 18,024; NPV for Project A = 18,024 15,000 = 3,024.\nProject Q: N = 5; PMT = 7,500; FV = 0; I/Y = 12; CPT PV = 27,036; NPV for Project B = 27,036 25,000 = 2,036.\nFor independent projects the NPV decision rule is to accept all projects with a positive NPV. Therefore, accept both projects.\n\n12 of 24\n\nQuestion #31 of 57\n\nA company is considering a \\$10,000 project that will last 5 years.\n\nAnnual after tax cash flows are expected to be \\$3,000\nTarget debt/equity ratio is 0.4\nCost of equity is 12%\nCost of debt is 6%\nTax rate 34%\nWhat is the project's net present value (NPV)?\n\nA) +\\$1,460.\nB) -\\$1,460.\nC) \\$+1,245\nExplanation\nFirst, calculate the weights for debt and equity\n\nwd + we = 1\nwe = 1 wd\nwd / we = 0.40\nwd = 0.40 (1 wd)\nwd = 0.40 0.40wd\n1.40wd = 0.40\nwd = 0.286, we = 0.714\nSecond, calculate WACC\n\nWACC = (wd kd) (1 t) + (we ke) = (0.286 0.06 0.66) + (0.714 0.12) = 0.0113 + 0.0857 = 0.0970\nThird, calculate the PV of the project cash flows\n\nN = 5, PMT = -3,000, FV = 0, I/Y = 9.7, CPT PV = 11,460\n\nAnd finally, calculate the project NPV by subtracting out the initial cash flow\n\nNPV = \\$11,460 \\$10,000 = \\$1,460\n\nQuestion #32 of 57\n\nQuestion ID: 414712\n\nA company is considering the purchase of a copier that costs \\$5,000. Assume a cost of capital of 10 percent and the following\ncash flow schedule:\nYear 1: \\$3,000\nYear 2: \\$2,000\nYear 3: \\$2,000\n\n13 of 24\n\nDetermine the project's NPV and IRR.\n\nNPV\n\nIRR\n\nA) \\$243\n\n20%\n\nB) \\$883\n\n20%\n\nC) \\$883\n\n15%\n\nExplanation\nTo determine the NPV, enter the following:\nPV of \\$3,000 in year 1 = \\$2,727, PV of \\$2,000 in year 2 = \\$1,653, PV of \\$2,000 in year 3 = \\$1,503. NPV = (\\$2,727 + \\$1,653 +\n\\$1,503) \\$5,000 = 883.\n\nYou know the NPV is positive, so the IRR must be greater than 10%. You only have two choices, 15% and 20%. Pick one and\nsolve the NPV. If it is not close to zero, then you guessed wrong; select the other one.\n[3000 (1 + 0.2)1 + 2000 (1 + 0.2)2 + 2000 (1 + 0.2)3] 5000 = 46 This result is closer to zero (approximation) than the \\$436\nresult at 15%. Therefore, the approximate IRR is 20%.\n\nQuestion #33 of 57\n\nQuestion ID: 414695\n\nWhich of the following steps is least likely to be an administrative step in the capital budgeting process?\nA) Forecasting cash flows and analyzing project profitability.\nB) Arranging financing for capital projects.\nC) Conducting a post-audit to identify errors in the forecasting process.\nExplanation\nArranging financing is not one of the administrative steps in the capital budgeting process. The four administrative steps in the\ncapital budgeting process are:\n1. Idea generation\n2. Analyzing project proposals\n3. Creating the firm-wide capital budget\n4. Monitoring decisions and conducting a post-audit\n\nQuestion #34 of 57\n\nQuestion ID: 414727\n\nThe underlying cause of ranking conflicts between the net present value (NPV) and internal rate of return (IRR) methods is the\nunderlying assumption related to the:\n\nA) reinvestment rate.\nB) cash flow timing.\nC) initial cost.\nExplanation\n\n14 of 24\n\nThe IRR method assumes all future cash flows can be reinvested at the IRR. This may not be feasible because the IRR is not\nbased on market rates. The NPV method uses the weighted average cost of capital (WACC) as the appropriate discount rate.\n\nQuestion #35 of 57\n\nQuestion ID: 414713\n\nA firm is considering a \\$200,000 project that will last 3 years and has the following financial data:\nAnnual after-tax cash flows are expected to be \\$90,000.\nTarget debt/equity ratio is 0.4.\nCost of equity is 14%.\nCost of debt is 7%.\nTax rate is 34%.\nDetermine the project's payback period and net present value (NPV).\nPayback Period NPV\nA) 2.22 years\n\n\\$18,716\n\nB) 2.22 years\n\n\\$21,872\n\nC) 2.43 years\n\n\\$18,716\n\nExplanation\nPayback Period\n\\$200,000 / \\$90,000 = 2.22 years\nNPV Method\nFirst, calculate the weights for debt and equity\nwd + we = 1\nwe = 1 wd\nwd / we = 0.40\nwd = 0.40 (1 wd)\nwd = 0.40 0.40wd\n1.40wd = 0.40\nwd = 0.286, we = 0.714\nSecond, calculate WACC\nWACC = (wd kd) (1 t) + (we ke) = (0.286 0.07 0.66) + (0.714 0.14) = 0.0132 + 0.100 = 0.1132\nThird, calculate the PV of the project cash flows\n90 / (1 + 0.1132)1 + 90 / (1 + 0.1132)2 + 90 / (1 + 0.1132)3 = \\$218,716\n\nAnd finally, calculate the project NPV by subtracting out the initial cash flow\nNPV = \\$218,716 \\$200,000 = \\$18,716\n\n15 of 24\n\nQuestion #36 of 57\n\nQuestion ID: 460657\n\nThe greatest amount of detailed capital budgeting analysis is typically required when deciding whether to:\nA) introduce a new product or develop a new market.\nB) replace a functioning machine with a newer model to reduce costs.\nC) expand production capacity.\nExplanation\nIntroducing a new product or entering a new market involves sales and expense projections that can be highly uncertain.\nExpanding capacity or replacing old machinery involves less uncertainty and analysis.\n\nQuestion #37 of 57\n\nIf two projects are mutually exclusive, a company:\n\nA) can accept one of the projects, both projects, or neither project.\nB) can accept either project, but not both projects.\nC) must accept both projects or reject both projects.\nExplanation\nMutually exclusive means that out of the set of possible projects, only one project can be selected. Given two mutually exclusive\nprojects, the company can accept one of the projects or reject both projects, but cannot accept both projects.\n\nQuestion #38 of 57\n\nQuestion ID: 460658\n\nThe effects that the acceptance of a project may have on other firm cash flows are best described as:\nA) pure plays.\nB) opportunity costs.\nC) externalities.\nExplanation\nExternalities refer to the effects that the acceptance of a project may have on other firm cash flows. Cannibalization is one\nexample of an externality.\n\nQuestion #39 of 57\n\nQuestion ID: 414710\n\nLanden, Inc. uses several methods to evaluate capital projects. An appropriate decision rule for Landen would be to invest in a\nproject if it has a positive:\nA) internal rate of return (IRR).\nB) net present value (NPV).\nC) profitability index (PI).\nExplanation\n\n16 of 24\n\nThe decision rules for net present value, profitability index, and internal rate of return are to invest in a project if NPV > 0, IRR >\nrequired rate of return, or PI > 1.\n\nQuestion #40 of 57\n\nQuestion ID: 414722\n\nA firm is considering a \\$5,000 project that will generate an annual cash flow of \\$1,000 for the next 8 years. The firm has the\nfollowing financial data:\nDebt/equity ratio is 50%.\nCost of equity capital is 15%.\nCost of new debt is 9%.\nTax rate is 33%.\nDetermine the project's net present value (NPV) and whether or not to accept it.\nNPV\n\nAccept / Reject\n\nA) +\\$4,968\n\nAccept\n\nB) -\\$33\n\nReject\n\nC) +\\$33\n\nAccept\n\nExplanation\nFirst, calculate the weights for debt and equity\n\nwd + we = 1\nwd = 0.50We\n0.5We + We = 1\nwd = 0.333, we = 0.667\nSecond, calculate WACC\n\nWACC = (wd kd) (1 t) + (we ke) = (0.333 0.09 0.67) + (0.667 0.15) = 0.020 + 0.100 = 0.120\n\nThird, calculate the PV of the project cash flows\n\nN = 8, PMT = -1,000, FV = 0, I/Y = 12, CPT PV = 4,967\n\nAnd finally, calculate the project NPV by subtracting out the initial cash flow\n\nNPV = \\$4,967 \\$5,000 = -\\$33\n\nQuestion #41 of 57\n\nQuestion ID: 414728\n\nWhich of the following statements regarding the net present value (NPV) and internal rate of return (IRR) is least accurate?\nA) For independent projects, the internal rate of return IRR and the NPV methods always yield\nthe same accept/reject decisions.\nB) The NPV tells how much the value of the firm will increase if you accept the project.\n\n17 of 24\n\nC) For mutually exclusive projects, you must accept the project with the highest NPV regardless of the\nsign of the NPV calculation.\nExplanation\nIf the NPV for two mutually exclusive projects is negative, both should be rejected.\n\nQuestion #42 of 57\n\nQuestion ID: 414698\n\nAshlyn Lutz makes the following statements to her supervisor, Paul Ulring, regarding the basic principles of capital budgeting:\nStatement 1: The timing of expected cash flows is crucial for determining the profitability of a capital budgeting project.\nStatement 2: Capital budgeting decisions should be based on the after-tax net income produced by the capital project.\nWhich of the following regarding Lutz's statements is most accurate?\nStatement 1\n\nStatement 2\n\nA) Correct\n\nCorrect\n\nB) Correct\n\nIncorrect\n\nC) Incorrect\n\nCorrect\n\nExplanation\nLutz's first statement is correct. The timing of cash flows is important for making correct capital budgeting decisions. Capital\nbudgeting decisions account for the time value of money. Lutz's second statement is incorrect. Capital budgeting decisions\nshould be based on incremental after-tax cash flows, not net (accounting) income.\n\nQuestion #43 of 57\n\nQuestion ID: 414737\n\nWhich of the following statements about the internal rate of return (IRR) for a project with the following cash flow pattern is\nCORRECT?\nYear 0: -\\$ 2,000\nYear 1: \\$10,000\nYear 2: -\\$ 10,000\n\nA) It has a single IRR of approximately 38%.\n\nB) It has two IRRs of approximately 38% and 260%.\nC) No IRRs can be calculated.\nExplanation\nThe number of IRRs equals the number of changes in the sign of the cash flow. In this case, from negative to positive and then\nback to negative. Although 38% seems appropriate, one should not automatically discount the value of 260%.\n10,000 1.38 - 10,000 1.382 = 1995.38\nAnd:\n\n18 of 24\n\n10,000 3.6 - 10,000 3.62 = 2006.17\n\nBoth discount rates give NPVs of approximately zero and thus, are IRRs.\n\nQuestion #44 of 57\n\nQuestion ID: 414730\n\nWhich of the following statements about the internal rate of return (IRR) and net present value (NPV) is least accurate?\n\nA) The IRR is the discount rate that equates the present value of the cash inflows with the present value\nof the outflows.\n\nB) For mutually exclusive projects, if the NPV rankings and the IRR rankings give conflicting signals, you\nshould select the project with the higher IRR.\n\nC) The discount rate that causes the project's NPV to be equal to zero is the project's IRR.\nExplanation\nThe NPV method is always preferred over the IRR, because the NPV method assumes cash flows are reinvested at the cost of capital.\nConversely, the IRR assumes cash flows can be reinvested at the IRR. The IRR is not an actual market rate.\n\nQuestion #45 of 57\n\nQuestion ID: 414717\n\nTapley Acquisition, Inc., is considering the purchase of Tangent Company. The acquisition would require an initial investment of\n\\$190,000, but Tapley's after-tax net cash flows would increase by \\$30,000 per year and remain at this new level forever. Assume\na cost of capital of 15%. Should Tapley buy Tangent?\n\nA) Yes, because the NPV = \\$30,000.\n\nB) No, because k > IRR.\nC) Yes, because the NPV = \\$10,000.\nExplanation\nThis is a perpetuity.\nPV = PMT / I = 30,000 / 0.15 = 200,000\n200,000 190,000 = 10,000\n\nQuestion #46 of 57\n\nQuestion ID: 414703\n\nRosalie Woischke is an executive with ColaCo, a nationally known beverage company. Woischke is trying to determine the firm's\noptimal capital budget. First, Woischke is analyzing projects Sparkle and Fizz. She has determined that both Sparkle and Fizz\nare profitable and is planning on having ColaCo accept both projects. Woischke is particularly excited about Sparkle because if\nSparkle is profitable over the next year, ColaCo will have the opportunity to decide whether or not to invest in a third project,\nBubble. Which of the following terms best describes the type of projects represented by Sparkle and Fizz as well as the\nopportunity to invest in Bubble?\nSparkle and Fizz\nA) Independent projects\n\nOpportunity to invest in Bubble\n\nProject sequencing\n\n19 of 24\n\nB) Mutually exclusive projects Project sequencing\n\nC) Independent projects\n\nExplanation\nIndependent projects are projects for which the cash flows are independent from one another and can be evaluated based on\neach project's individual profitability. Since Woischke is accepting both projects, the projects must be independent. If the projects\nwere mutually exclusive, only one of the two projects could be accepted. The opportunity to invest in Bubble is a result of project\nsequencing, which means that investing in a project today creates the opportunity to decide to invest in a related project in the\nfuture.\n\nQuestion #47 of 57\n\nQuestion ID: 485785\n\nA company is considering two mutually exclusive investment projects. The firm's cost of capital is 12%. Each project costs \\$7\nmillion and the after-tax cash flows for each are as follows:\nProject One\n\nProject Two\n\nYear 1\n\n\\$6.6 million\n\n\\$3.0 million\n\nYear 2\n\n\\$1.5 million\n\n\\$3.0 million\n\nYear 3\n\n\\$0.1 million\n\n\\$3.0 million\n\nIndicate which project should be accepted and whether the IRR and NPV methods would lead to the same decision.\nProject accepted? Same decision?\n\nA) Project Two\n\nNo\n\nB) Project One\n\nNo\n\nC) Project Two\n\nYes\n\nExplanation\nThe NPVs for Project One and Project Two are \\$0.160 million and \\$0.206 million, respectively, thus, Project Two should be\nselected. The IRRs for Projects One and Project Two are 14.2% and 13.7%, respectively. NPV is considered a superior method\nfor ranking mutually exclusive projects.\n\nQuestion #48 of 57\n\nQuestion ID: 414725\n\nWhich of the following projects would have multiple internal rates of return (IRRs)? The cost of capital for all projects is 9.75%.\n\nT0\n\nT1\n\n10,000\n\n7,000\n\n4,000\n\nT2\n\n15,000\n\n2,000\n\nT3\n\n-10,000\n\n2,000\n\n6,000\n\n20 of 24\n\nA) Projects Roulette and Keno.\n\nB) Projects Blackjack and Keno.\nC) Project Blackjack only.\nExplanation\nThe multiple IRR problem occurs if a project has non-normal cash flows, that is, the sign of the net cash flows changes from\nnegative to positive to negative, or vice versa. For the exam, a shortcut to look for is the project cash flows changing signs more\nthan once. Only Project Blackjack has this cash flow pattern. The 0 net cash flow in T2 for Project Keno and likely negative net\npresent value (NPV) for Project Roulette would not necessarily result in multiple IRRs.\n\nQuestion #49 of 57\n\nQuestion ID: 414736\n\nIf a project has a negative cash flow during its life or at the end of its life, the project most likely has:\nA) a negative internal rate of return.\nB) multiple net present values.\nC) more than one internal rate of return.\nExplanation\nProjects with unconventional cash flows (where the sign of the cash flow changes from minus to plus to back to minus) will have\nmultiple internal rates of return. However, one will still be able to calculate a single net present value for the cash flow pattern.\n\nQuestion #50 of 57\n\nQuestion ID: 414716\n\nAs the director of capital budgeting for Denver Corporation, an analyst is evaluating two mutually exclusive projects with the\nfollowing net cash flows:\nYear Project X Project Z\n0\n\n-\\$100,000 -\\$100,000\n\n\\$50,000\n\n\\$10,000\n\n\\$40,000\n\n\\$30,000\n\n\\$30,000\n\n\\$40,000\n\n\\$10,000\n\n\\$60,000\n\nA) Project X, since it has the higher net present value (NPV).\n\nB) Project X, since it has the higher IRR.\nC) Neither project.\nExplanation\nNPV for Project X = -100,000 + 50,000 / (1.15)1 + 40,000 / (1.15)2 + 30,000 / (1.15)3 + 10,000 / (1.15)4\n= -100,000 + 43,478 + 30,246 + 19,725 + 5,718 = -833\n\n21 of 24\n\nNPV for Project Z = -100,000 + 10,000 / (1.15)1 + 30,000 / (1.15)2 + 40,000 / (1.15)3 + 60,000 / (1.15)4\n= -100,000 + 8,696 + 22,684 + 26,301 + 34,305 = -8,014\nReject both projects because neither has a positive NPV.\n\nQuestion #51 of 57\n\nQuestion ID: 414706\n\nThe Seattle Corporation has been presented with an investment opportunity which will yield cash flows of \\$30,000 per year in\nyears 1 through 4, \\$35,000 per year in years 5 through 9, and \\$40,000 in year 10. This investment will cost the firm \\$150,000\ntoday, and the firm's cost of capital is 10%. The payback period for this investment is closest to:\nA) 5.23 years.\nB) 4.86 years.\nC) 6.12 years.\nExplanation\n\nYears\n\nCash Flows -\\$150,000 \\$30,000 \\$30,000 \\$30,000 \\$30,000 \\$35,000\n\n\\$150,000\n120,000 (4 years)(30,000/year)\n\\$30,000\nWith \\$30,000 unrecovered cost in year 5, and \\$35,000 cash flow in year 5; \\$30,000 / \\$35,000 = 0.86 years\n4 + 0.86 = 4.86 years\n\nQuestion #52 of 57\n\nQuestion ID: 414734\n\nIf the calculated net present value (NPV) is negative, which of the following must be CORRECT. The discount rate used is:\nA) greater than the internal rate of return (IRR).\nB) equal to the internal rate of return (IRR).\nC) less than the internal rate of return (IRR).\nExplanation\nWhen the NPV = 0, this means the discount rate used is equal to the IRR. If a discount rate is used that is higher than the IRR,\nthe NPV will be negative. Conversely, if a discount rate is used that is lower than the IRR, the NPV will be positive.\n\nQuestion #53 of 57\n\nQuestion ID: 414708\n\nLincoln Coal is planning a new coal mine, which will cost \\$430,000 to build, with the expenditure occurring next year. The mine\nwill bring cash inflows of \\$200,000 annually over the subsequent seven years. It will then cost \\$170,000 to close down the mine\nover the following year. Assume all cash flows occur at the end of the year. Alternatively, Lincoln Coal may choose to sell the site\n\n22 of 24\n\ntoday. What minimum price should Lincoln set on the property, given a 16% required rate of return?\nA) \\$325,859.\nB) \\$376,872.\nC) \\$280,913.\nExplanation\nThe key to this problem is identifying this as a NPV problem even though the first cash flow will not occur until the following year.\nNext, the year of each cash flow must be property identified; specifically: CF0 = \\$0; CF1 = -430,000; CF2-8 = +\\$200,000; CF9 =\n-\\$170,000. One simply has to discount all of the cash flows to today at a 16% rate. NPV = \\$280,913.\n\nQuestion #54 of 57\n\nQuestion ID: 414732\n\nFor a project with cash outflows during its life, the least preferred capital budgeting tool would be:\nA) internal rate of return.\nB) profitability index.\nC) net present value.\nExplanation\nThe IRR encounters difficulties when cash outflows occur throughout the life of the project. These projects may have multiple\nIRRs, or no IRR at all. Neither the NPV nor the PI suffer from these limitations.\n\nQuestion #55 of 57\n\nQuestion ID: 414696\n\nWhich of the following types of capital budgeting projects are most likely to generate little to no revenue?\nA) New product or market development.\nB) Regulatory projects.\nC) Replacement projects to maintain the business.\nExplanation\nMandatory regulatory or environmental projects may be required by a governmental agency or insurance company and typically\ninvolve safety-related or environmental concerns. The projects typically generate little to no revenue, but they accompany other\nnew revenue producing projects and are accepted by the company in order to continue operating.\n\nQuestion #56 of 57\n\nQuestion ID: 414723\n\nWhich of the following is the most appropriate decision rule for mutually exclusive projects?\n\n23 of 24\n\nA) Accept both projects if their internal rates of return exceed the firm's hurdle rate.\nB) Accept the project with the highest net present value, subject to the condition that its net present\nvalue is greater than zero.\nC) If the net present value method and the internal rate of return method give conflicting signals, select\nthe project with the highest internal rate of return.\nExplanation\nThe project that maximizes the firm's value is the one that has the highest positive NPV.\n\nQuestion #57 of 57\n\nQuestion ID: 414733\n\nWhen a company is evaluating two mutually exclusive projects that are both profitable but have conflicting NPV and IRR project\nrankings, the company should:\nA) use a third method of evaluation such as discounted payback period.\nB) accept the project with the higher net present value.\nC) accept the project with the higher internal rate of return.\nExplanation\nNet present value is the preferred criterion when ranking projects because it measures the firm's expected increase in wealth\nfrom undertaking a project.\n\n24 of 24" ]

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[ "# MCQ Questions Chapter 6 Thermodynamics Class 11 Chemistry\n\nPlease refer to MCQ Questions Chapter 6 Thermodynamics Class 11 Chemistry with answers provided below. These multiple-choice questions have been developed based on the latest NCERT book for class 11 Chemistry issued for the current academic year. We have provided MCQ Questions for Class 11 Chemistry for all chapters on our website. Students should learn the objective based questions for Chapter 6 Thermodynamics in Class 11 Chemistry provided below to get more marks in exams.\n\n## Chapter 6 Thermodynamics MCQ Questions\n\nPlease refer to the following Chapter 6 Thermodynamics MCQ Questions Class 11 Chemistry with solutions for all important topics in the chapter.\n\nMCQ Questions Answers for Chapter 6 Thermodynamics Class 11 Chemistry\n\nQuestion. For the reaction, C (graphite) + 1/2 O2 (g) → CO(g) at 2 298 K and 1 atm, ΔH = – 26. 4 kcal. What is ΔE, if the molar volwne of graphite is 0.0053 L?\n[R = 0. 002 kcal mol-1 K-1]\n(a) – 26.7 kcal\n(b) + 26.7 kcal\n(c) – 52.4 kcal\n(d) + 52.4 kcal\n\nA\n\nQuestion. The value of ~E for combustion of 16 g of CH4 is – 885389 J at 298 K. The ΔH combustion for CH4 in J mol-1 at this temperature will be\n(Given that,R = 8. 314 JK-1 mol-1)\n(a) – 55337\n(b) – 880430\n(c) – 885389\n(d) – 890348\n\nB\n\nQuestion. If the bond dissociation energies of XY, X 2 and Y2 (all diatomic molecules) are in the ratio of 1: 1: 0.5 and ΔHf I for the formation of XY is -200 kJ mol-1 . The bond dissociation energy of X2 will be\n(a) 400 kJ mol-1\n(b) 300 kJ mol-1\n(c) 20 kJ mol-1\n(d) None of these\n\nD\n\nQuestion. For the reaction (at 1240 Kand 1 atm.)\nCaCO3(s) → CaO(s) + CO2 (g)\nΔH = 176 kJ/mol; ΔE will be\n(a) 160 kJ\n(b) 165.6 kJ\n(c) 186.4 kJ\n(d) 180 kJ\n\nB\n\nQuestion. Which is correct for an endothermic reaction ?\n(a) ΔH is positive\n(b) ΔH is negative\n(c) ΔE isnegative\n(d) ΔH = 0\n\nA\n\nQuestion. In an isothermal process\n(a) q = 0 and ΔE = 0\n(b) q ≠ 0 and ΔE = 0\n(c) q = 0 and ΔE ≠ 0\n(d) q ≠ 0 and ΔE ≠ 0\n\nB\n\nQuestion. Heat of neutralisation will be minimum for which of the following combination ?\n(a) NaOH + H2SO4\n(b) NH4OH + CH3COOH\n(c) NaOH + HCl\n(d) NaOH + CH3COOH\n\nB\n\nQuestion. Which of the following indicates the heat ofreaction equal to heat of formation ?\n(a) C(graphite)  + O2  (latm)  → CO2 (latm)\n(b) C(diamond) + O2 (I atm) → CO2 (2 atm)\n(c) C(graphite)  + O2  (latm) → CO2 (2atm)\n(d) C(diamond) + O2 (l atrn) → CO2 (1 atm)\n\nA\n\nQuestion. For CaCO3 (s) ⇔ CaO(s)+ CO2(g) at 927° C ΔH = 176 kJ mol; then ΔE is\n(a) 180 kJ\n(b) 186.4 kJ\n(c) 166.0 kJ\n(d) 160 kJ\n\nC\n\nQuestion. The H—H bond energy is 430 kJ mol-1 and Cl—Cl bond energy is 240 kJ mol-1 , AH for HCI is – 90 kJ. The H—Cl bond energy is about\n(a) 180 kJ mol-1\n(b) 360kJ mol-1\n(c) 213 kJ mol-1\n(d) 425 kJ mol-1\n\nD\n\nQuestion. The ΔH°for CO2 (g), CO(g) and H2O(g) are – 393.5, – 110.5 and – 241.8 kJ/mol respectively. The standard enthalpy change (in kJ) for the reaction\nCO2(g)+H2(g) → CO(g)+H2O(g)is\n(a) 524.1\n(c) – 262.5\n(b) 41.2\n(d) – 4 1.2\n\nB\n\nQuestion. For the following two reactions,\n(i) CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O\nΔH = – 890.4 kJ\n(ii) 2HgO(s) → 2Hg(/) + O2 (g ); ΔH + 181.6 kJ\nWhich one of the following statements is correct ?\n(a) Both of them are exothermic\n(b) Both of them are endothermic\n(c) (i) is exothermic and (ii) is endothermic\n(d) (i) is endothem1ic and (ii) is exothermic\n\nC\n\nQuestion. The first law of thermodynan1ics is expressed as\n(a) q – W = ΔE\n(b) ΔE = q – W\n(c) q = M – W\n(d) W = q + ΔE\n(e) None of the above\n\nC\n\nQuestion. Hess’s law is used to calculate\n(a) enthalpy ofreaction\n(b) entropy of reaction\n(c) work done in reaction\n(d) All of these\n\nA\n\nQuestion. The pair of isochoric processes among the transformation of states is\n(a) K to L and L to M\n(b) L to M and N to K\n(c) L to M and M to N\n(d) M to N and N to K\n\nC\n\nQuestion. The intensive property among these quantities is\n(a) enthalpy\n(b) mass/volume\n(c) mass\n(d) volwne\n\nB\n\nQuestion. A gas can expand from 100 mL to 250 mL under a constant pressure of 2 attn. The work done by the gas is\n(a) 30.48 J\n(b) 25 J\n(c) 5 kJ\n(d) 16 J\n\nA\n\nQuestion. Which one of the following is not a state function ?\n(a) Enthalpy\n(b) Entropy\n(c) Work\n(d) Free energy\n\nC\n\nQuestion. We believe in the laws of thermodynamics because they are\n(a) theoretical\n(b) derived based on mathematical analysis\n(c) empirical and nobody disproved\n(d) mere statements\n\nD\n\nQuestion. The sublimation energy of I2 (s) is 57 .3 kJ/mol and the enthalpy of fusion is 15.5 kJ/mol. The enthalpy of vaporisation of I2 is\n(a) 41.8 kJ/mol\n(b) –41.8 k:J/ mol\n(c) 72.8 kJ/mol\n(d) –72.8 k:J/ mol\n\nA\n\nQuestion. The temperature of the system decreases in an\n(b) isothermal compression\n(c) isothermal expansion\n\nD\n\nQuestion. On complete combustion, 0.246 g of an organic compound gave 0.198 g of CO2 and 0.1014 g of H2O. The ratio of carbon and hydrogen atoms in the compound is\n(a) 1 : 3\n(b) 1 : 2\n(c) 2: 5\n(d) 2: 7\n\nC\n\nQuestion. When ice melts into water, the entropy\n(a) becomes zero\n(b) remains same\n(c) decreases\n(d) increases\n\nD\n\nQuestion. Which one of the following is correct ?\n(a) – ΔG = ΔH – TΔS\n(b) ΔH = [ΔG – TΔS]\n(c) ΔS = 1/T [ΔG – ΔH]\n(d) ΔS = 1/T [ΔH – ΔG]\n\nD\n\nQuestion. What is the relation between E° and K ?\n(a) E° = 0.0591/n logK\n(b) E° = n/0.0591 logK\n(c) E° = 2.303R/nF log K\n(d) E° = 2.303 RT/n log K\n\nA\n\nQuestion. One mole of ice is converted into water at 273 K.. The entropies of H2O(s) and H2O(/) are 38.20 and 60.01 J mol-1 K-1 respectively. The enthalpy change for the conversion is\n(a) 3 kJ mol-1\n(c) 5 kJ mol-1\n(b) 4 kJ mol-1\n(d) 6 kJ mol-1\n\nD\n\nQuestion. Which of the following statements is true?\n(a) The total entropy of the universe is continuously decreasing\n(b) The total energy of the universe is continuously decreasing\n(c) The total energy of the universe remains constant\n(d) The total entropy of the universe remains constant\n\nC\n\nQuestion. For the reaction H2O(/) ⇌ H2O(g) at 373 K and I atm pressure\n(a) ΔH = 0\n(b) ΔE = 0\n(c) ΔH = TΔS\n(d) ΔH = ΔE\n\nC\n\nQuestion. Given that dE = TdS – pdV and H = E + pV. Which one of the following relations is true?\n(a) dH = TdS +Vdp\n(b) dH = SdT + V dp\n(c) dH = – SdT+Vdp\n(d) dH = dE + pdV\n(e) dH = dE – TdS\n\nA\n\nQuestion. The entropy change involved in the isothermal reversible expansion of 2 moles ofan ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27°C is\n(a) 38.3 J mol-1 K-1\n(b) 35.8J mol-1 K-1\n(c) 32.3 J mol-1 K-1\n(d) 42.3 J mol-1 K-1\n\nA\n\nQuestion. The entropy of a perfectly crystalline material is zero at 0°C. This is statement of\n(a) first law of thermodynamics\n(b) second law of thermodynamics\n(c) third law of thermodynamics\n(d) law of conservation of energy\n\nC\n\nQuestion. In view of the signs of ΔrG° for the following reactions\nPbO2 + Pb → 2PbO, ΔrG° < 0,\nSnO2 + Sn → 2SnO, ΔrG° > 0,\nWhich oxidation states are more characteristic for lead and tin ?\n(a) For lead + 4, for tin + 2\n(b) For lead + 2, for tin + 2\n(c) For lead + 4, for tin + 4\n(d) For lead + 2, for tin + 4\n\nB\n\nQuestion. For the reversible reaction, A(s) + B(g) → C(g) + D(g); ΔG° =- 350kJ, which one of the following statements is true ?\n(a) The reaction is thermodynamically non-feasible\n(b) The entropy change is negative\n(c) Equilibrium constant is greater than one\n(d) The reaction should be instantaneous\n\nC\n\nQuestion. In any chemical reaction, a quantity that decrease to a minimum is\n(a) free energy\n(b) entropy\n(c) temperature\n(d) enthalpy\n\nA\n\nQuestion. Carbon cannot reduce Fe2O3 to Feat a temperature below 983 K because\n(a) free energy change for the formation of CO is more negative than that of Fe2O3\n(b) CO is thermodynamically more stable than Fe2O3\n(c) carbon has higher affinity towards oxygen than iron\n(d) iron has higher affinity towards oxygen than carbon\n\nC\n\nQuestion. In evaporation of water ΔH and ΔS are\n(a) + ,+\n(b) +, –\n(c) – , –\n(d) – , +\n\nA\n\nQuestion. ΔG for the reaction Ag2O ➔ 2Ag + 1/2 O2 at a certain 2 temperature is found to be -10.0kJ moI– 1 , which one of the following statements is correct at this temperature ?\n(a) Silver oxide decomposes to give silver and oxygen\n(b) Silver and oxygen combine to form silver oxide\n(c) The reaction is in equilibrium\n(d) The reaction can neither occur in the forward direction nor in the backward direction\n\nA\n\nQuestion. The heat of neutralisation is highest for the reaction between\n(a) NH4OH—CH3COOH\n(b) HNO3—NH4OH\n(c) NaOH—CH3COOH\n(d) HCI—NaOH\n\nD\n\nQuestion. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K, due to the combustion process. Given, that the heat capacity of the calorimeter is 2.5 kJ K-1 , the numerical value for the enthalpy of combustion of the gas in kJ mmol-1 is\n(a) 7\n(b) 9\n(c) 0\n(d) 8\n\nB\n\nQuestion. For which of the following equations is the change in enthalpy at 25°C and 1 atm equal to ΔH°f of CH2O(/)\n(a) C(g) + H2 (g)+(V2) O2 (g) → CH2O(/)\n(b) C(s) + H2 (g) + (l/2)O2 (g) → CH2O(/)\n(c) C(g)+ 2H2 (g)+ O(g) → CH2O(/)\n(d) CO(g)+ H2 (g) → CH2O(/)\n\nB\n\nQuestion. The standard enthalpies of formation of CO2 (g ), H2O(/) and glucose (s) at 25°C are – 400 kJ/ mol, –300 kJ/ mol and –1300 kJ/ mol, respectively. The standard enthalpy of combustion per gram of glucose at 25°C is\n(a) + 2900 kJ\n(b) –2900 kJ\n(c) –16.11 kJ\n(d) + 16.11 kJ\n\nC\n\nQuestion. The standard enthalpy of formation (ΔH°) at 298 K for methane, CH4 (g), is – 78.4 kJ mol-1. The additional information required to determine the average energy for C—H bond formation would be\n(a) the dissociation energy of H2 and enthalpy of sublimation of carbon\n(b) latent heat of vaporisation of methane\n(c) the first four ionisation energies of carbon and electron gain enthalpy of hydrogen\n(d) the dissociation energy of hydrogen molecule, H2\n\nA\n\nQuestion. The enthalpy changes for the following processes are listed below.\nCl2 (g) = 2Cl(g ), 242.3 kJ mol-1\nI2 (g)= 2I(g), 151.0 kJ mol-1\nICl (g)= I(g)+ Cl(g),211.3 kJ mol-1\nI2 (s) = I2 (g), 62.76 kJ mol-1\nGiven that the standard states for iodine and chlorine are I2 (s) and Cl2 (g), the standard enthalpy of formation of ICl (g) is\n(a) – 14.6kJ mol-1\n(b) – 16.8kJ mol-1\n(c) + 16.8kJ mol-1\n(d) + 244.8kJ mol-1\n\nC\n\nQuestion. A monoatornic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equals to 1. What is the molar heat capacity of the gas?\n(a) 4R/2\n(b) 3R/2\n(c) 5R\n(d) O\n\nA\n\nQuestion. (ΔH – ΔU) for the fommtion of carbon monoxide (CO) from its elements at 298 K is (R = 8.314 JK-1 mol-1 )\n(a) – 1238.78 J mol-1\n(b) 1238.78 J mol-1\n(c) – 2477.57 mol-1\n(d) 2477.57 J mol-1\n\nB\n\nQuestion. If heat of neutralisation of HCN with NaOH is — 12.1 kJ and the heat of neutralisation of HCI by NaOH is —55.9 kJ/mol, then the energy of dissociation of HCN is\n(a) 43.8 kJ\n(b) – 43.8 kJ\n(c) – 68 kJ\n(d) 68 kJ\n\nA\n\nQuestion. ln an isochoric process, ΔH for a system is equal to\n(a) p . ΔV\n(b) pV\n(c) E + p.ΔV\n(d) ΔE\n\nD\n\nQuestion. For the reaction, A (g) + 2B(g) → 2C (g) + 3D (g) the change of enthalpy at 27° C is 19 kcal. The value of M is\n(a) 21.2 kcal\n(b) 17.8 kcal\n(c) 18.4 kcal\n(d) 20.6 kcal\n\nB\n\nQuestion. When 50cm3 of 0.2 N H2SO4 is mixed with 50cm3 of I N KOH, the heat liberated is\n(a) 11.46 kJ\n(b) 157.3 kJ\n(c) 573 kJ\n(d) 573 J\n\nD\n\nQuestion. The enthalpy of reaction,\nH2(g) + 1/2 O2(g) → H2O(g) is ΔH1 and that of\nH2(g) + 1/2 O2(g) → H2O(/) is ΔH2.\nThen\n(a) ΔH1 < ΔH2\n(b) ΔH+ ΔH= 0\n(c) ΔH> ΔH2\n(d) ΔH= ΔH2\n\nA\n\nQuestion. Identify the reaction for which ΔH ≠ ΔE\n(a) S (rhombic) + O2 (g) → SO2 (g)\n(b) N2(g) + O2(g) → 2NO(g)\n(c) H2(g) + Cl2(g) → 2HCl(g)\n(d) CO(g) + 1/2 O2(g) → CO2(g)\n\nD\n\nQuestion. EnthalpyofformationofHFandHCl are -16l kJ and -92 kJ respectively. Which of the following statements is incorrect ?\n(a) HCl is more stable than HF\n(b) HF and HCl are exothermic compounds\n(c) The affinity of fluorine to hydrogen is greater than the affinity of chlorine to hydrogen\n(d) HF is more stable than HCl\n\nA\n\nQuestion. A mixture of two moles of carbon monoxide and one mole of oxygen, in a closed vessel is ignited to convert the carbon monoxide to carbon dioxide. If ΔH is the enthalpy change and ΔE is the change in internal energy, then\n(a) ΔH > ΔE\n(b) ΔH < ΔE\n(c) ΔH = ΔE\n(d) the relationship depends on the capacity of the vessel\n\nB\n\nQuestion. The cooling in refrigerator is due to\n(a) reaction of the refrigerator gas\n(b) expansion of ice\n(c) the expansion of the gas in the refrigerator\n(d) the work of the compressor\n\nC\n\nQuestion. What would be the heat released when an aqueous solution containing 0.5 mole of HNO3 is mixed with 0.3 mole of OH (enthalpy ofneutralisation is -57.1 kJ)?\n(a) 28.5 kJ\n(b) 17.1 kJ\n(c) 45.7 kJ\n(d) 1.7 kJ\n\nB\n\nQuestion. ΔE° of combustion of iso-butylene is – X kJ mol-1 . The value of ΔH°  is\n(a) = ΔE°\n(b) > MΔE°\n(c) = 0\n(d) < ΔE°\n\nD\n\nQuestion. Calculate ΔH (in Joules) for, C (graphite) → C ( diamond), from the following data\nC (graphite) + O2 (g) → CO2 (g ); ΔH = – 393.5 kJ\nC (diamond) + O2 (g) → CO2 (g ); ΔH= – 395.4 kJ\n(a) 1900\n(b) – 788.9x 103\n(c) 190000\n(d) + 788.9x 103\n\nA\n\nQuestion. In which of the following reactions, the heat liberated is known as “heat of combustion” ?\n(a) H+ (aq) + OH (aq)➔ H2O(l)+ heat\n(b) C (graphite) + 1/2 O2 (g) ➔ CO(g) + heat\n(c) CH4 (g)+2O2 (g) ➔ CO2(g)+2H2O(/)+heat\n(d) H2SO4 (aq)+ H2O (l)➔ H2SO4 (aq)+ heat\n\nC\n\nQuestion. Heat of formation of SO2 is -298 kJ. What is the heat of combustion of 4 g of S ?\n(a) + 37 kJ\n(b) – 37.25 kJ\n(c) + 298kJ\n(d) 18.6 kJ\n\nB\n\nQuestion. Among the following which is true forl mole of liquid ?\n(a) Cp ≈ Cv\n(b) Cp – Cv = R\n(c) Cp – Cv > R\n(d) Cp < Cv\n\nA\n\nQuestion. The heats of combustion of carbon monoxide at constant pressw-e and at constant volume at 27° C will differ from one another by\n(a) 27 cal\n(b) 54 cal\n(c) 300 cal\n(d) 600 cal\n\nC\n\nQuestion. Minimum work is obtained when 1 kg of…gas expanded under 500 kPa to 200 kPa pressure at 0°C.\n(a) chlorine\n(b) oxygen\n(c) nitrogen\n(d) methane\n\nA\n\nQuestion. Which of the following is not correct?\n(a) Dissolution of NH4Cl in excess of water is an endothermic process\n(b) Neutralisation process is always exothermic\n(c) The absolute value of enthalpy (H) can be determined experimentally\n(d) The heat ofreaction at constant volume is denoted by ΔE\n\nC\n\nQuestion. Heat liberated with 100 mL of l N NaOH is neutralised by 300 mL of IN HCI\n(a) 11.56kJ\n(b) 5.73kJ\n(c) 22.92kJ\n(d) 17.19kJ\n\nB\n\nQuestion. Given, 2Fe + 3/2 O2 → Fe2O3; ΔH =- 193.4 kJ;\nMg+ 1/2 O2 → MgO; ΔH= – 140.2kJ;\nWhat is the t’1H of the reaction ?\n3Mg + Fe2O→ 3MgO + 2Fe\n(a) – 227.2 kJ\n(b) – 272.3 kJ\n(c) 227.2 kJ\n(d) 272.2 kJ" ]

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[ "# Creating scatter plots using ggplot2\n\nNovember 6, 2009\nBy\n\n[This article was first published on Software for Exploratory Data Analysis and Statistical Modelling, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nThe ggplot2 package can be used as an alternative to lattice for producing high quality graphics in R. The package provides a framework and hopefully simple interface to producing graphs and is inspired by the grammar of graphics.\n\nThe main function for producing graphs in this package is qplot, which stands for quick plot. The first two arguments to the function are the name of objects that contain the x and y variables for the plot that is being created. Like many functions in R there is a data argument that can be used to specify a data frame to look in for the variables.\n\nAs a first example we create a scatterplot of age and circumference for the data set in R that has measurements of the growth of Orange trees. The code to produce this graph is very simple and is shown below:\n\n``` qplot(age, circumference, data = Orange) ```\n\nThis produces the following graph:\n\nThe main thing with this graph is that we are ignoring the different trees and looking at the overall trend. If we want to distinguish between the growth for the trees separately we can use different colours for the plotting symbols and add a legend to indicate which colour corresponds to a given tree. The colour argument is used to specify a variable and qplot will automatically created a legend based on the levels of this categorical variable. We adjust our code to be:\n\n``` qplot(age, circumference, data = Orange, colour = Tree) ```\n\nand the graph now looks like:\n\nThat is a nice improvement on the initial graph as we can visually compare the growth trends for the five trees.\n\nWe can build additional elements into our graph, such as adding a smoother to show a trend, by making use of the geom argument which is used to specify what type of display is being created. The package has a nice feat that allows us to specify a vector with multiple elements to build up additional elements to the graph. We can add a smoother to the original plot with the code below:\n\n``` qplot(age, circumference, data = Orange, geom = c(\"point\", \"smooth\")) ```\n\nThis produces the following graph:\n\nAn alternative would be to change from plotting with symbols to joining the points with lines. This change again makes use of the geom argument as follows:\n\n``` qplot(age, circumference, data = Orange, colour = Tree, geom = \"line\") ```\n\nThe graph now looks like:\n\nwith a separate coloured line for each tree.\n\nR-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nTags: , , ," ]

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[ "Free Decimal Place Value Worksheets I Love How It Is Color Coded And Decimals 4th Grade Pdf", null, "free decimal place value worksheets i love how it is color coded and decimals 4th grade pdf.\n\ndecimal worksheets 5th grade place value free math 6th pdf,decimal place value worksheets with answers 5th grade pdf free kindergarten of decimals worksheet on angles,free decimal place value worksheets for 4th grade chart 5th lessons and,decimal place value worksheets with answers for kids 5th grade pdf of whole numbers and decimals,sample decimal place value charts templates free worksheets 5th grade 4th pdf,decimal place value worksheets with answers chart worksheet the best image 5th grade pdf decimals grid,place value worksheets grade free decimal decimals 5th grid,free decimal place value worksheets for 4th grade decimals 5th pdf,free decimal place value worksheets 6th grade pdf worksheet to thousandths with answers,decimal place value worksheets with answer key decimals answers." ]

[ null, "http://deffufa.info/wp-content/uploads/2017/12/free-decimal-place-value-worksheets-i-love-how-it-is-color-coded-and-decimals-4th-grade-pdf.jpg", null ]

[ "Cody\n\n# Problem 716. Arrange vector in ascending order\n\nSolution 166331\n\nSubmitted on 21 Nov 2012 by Sourav Mondal\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1   Pass\n%% x = [4 5 1 2 9]; y_correct = [1 2 4 5 9]; assert(isequal(ascending_order(x),y_correct))\n\n2   Pass\n%% x = [10 8 -4 3 1 -1 0]; y_correct = [-4 -1 0 1 3 8 10]; assert(isequal(ascending_order(x),y_correct))\n\n3   Pass\n%% x = 0; y_correct = 0; assert(isequal(ascending_order(x),y_correct))" ]

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[ "", null, "# Communications of the ACM\n\nResearch highlights\n\n# Compressed Linear Algebra for Declarative Large-Scale Machine Learning\n\nView as: Print Mobile App ACM Digital Library Full Text (PDF) In the Digital Edition Share: Send by email Share on reddit Share on StumbleUpon Share on Hacker News Share on Tweeter Share on Facebook\n\nLarge-scale Machine Learning (ML) algorithms are often iterative, using repeated read-only data access and I/O-bound matrix-vector multiplications. Hence, it is crucial for performance to fit the data into single-node or distributed main memory to enable fast matrix-vector operations. General-purpose compression struggles to achieve both good compression ratios and fast decompression for block-wise uncompressed operations. Therefore, we introduce Compressed Linear Algebra (CLA) for lossless matrix compression. CLA encodes matrices with lightweight, value-based compression techniques and executes linear algebra operations directly on the compressed representations. We contribute effective column compression schemes, cache-conscious operations, and an efficient sampling-based compression algorithm. Our experiments show good compression ratios and operations performance close to the uncompressed case, which enables fitting larger datasets into available memory. We thereby obtain significant end-to-end performance improvements.\n\n### 1. Introduction\n\nLarge-scale ML leverages large data collections to find interesting patterns or build robust predictive models.7 Applications range from traditional regression, classification, and clustering to user recommendations and deep learning for unstructured data. The labeled data required to train these ML models is now abundant, thanks to feedback loops in data products and weak supervision techniques. Many ML systems exploit data-parallel frameworks such as Spark20 or Flink2 for parallel model training and scoring on commodity hardware. It remains challenging, however, to train ML models on massive labeled data sets in a cost-effective manner. We provide compression-based methods for accelerating the linear algebra operations that are central to training. The key ideas are to perform these operations directly on the compressed data, and to automatically determine the best lossless compression scheme, as required by declarative ML systems.\n\nDeclarative ML. State-of-the-art, large-scale ML systems provide high-level languages to express ML algorithms by means of linear algebra such as matrix multiplications, aggregations, element-wise and statistical operations. Examples at different abstraction levels are SystemML,4 Mahout Samsara,17 Spark MLlib,19 and TensorFlow.1 The high-level specification allows data scientists to create or customize ML algorithms without worrying about data and cluster characteristics, data representations (e.g., sparse or dense formats), and execution-plan generation.\n\nData-intensive ML algorithms. Many ML algorithms are iterative, with repeated read-only data access. These algorithms often rely on matrix-vector multiplications, which require one complete scan of the matrix with only two floating point operations per matrix element. This low operational intensity renders matrix-vector multiplication, even in-memory, I/O bound.18 Despite the adoption of flash-and NVM-based SSDs, disk bandwidth is usually 10x-100x slower than memory bandwidth, which is in turn 10x-40x slower than peak floating point performance. Hence, it is crucial for performance to fit the matrix into available memory without sacrificing operations performance. This challenge applies to single-node in-memory computations, data-parallel frameworks with distributed caching like Spark,20 and accelerators like GPUs with limited device memory. Even in the face of emerging memory and link technologies, the challenge persists due to increasing data sizes, different access costs in the memory hierarchy, and monetary costs.\n\nLossy versus lossless compression. Recently, lossy compression has received a lot of attention in ML. Many algorithms can tolerate a loss in accuracy because these algorithms are approximate in nature, and because compression introduces noise that can even improve the generalization of the model. Common techniques are (1) low- and ultra-low-precision storage and operations, (2) sparsification (which reduces the number of non-zero values), and (3) quantization (which reduces the value domain). However, these techniques require careful, manual application because they affect the accuracy in a data-and algorithm-specific manner. In contrast, declarative ML aims at physical data independence. Accordingly, we focus on lossless compression because it guarantees exact results and thus, it allows for automatic compression to fit large datasets in memory when needed.\n\nBaseline solutions. The use of general-purpose compression techniques with block-wise decompression per operation is a common baseline solution. However, heavyweight techniques like Gzip are not applicable because decompression is too slow, while lightweight methods like Snappy or LZ4 achieve only modest compression ratios. Existing compressed matrix formats with good performance like CSR-VI15 similarly show only moderate compression ratios. In contrast, our approach builds upon research on lightweight database compression, such as compressed bitmaps and dictionary coding, as well as sparse matrix representations.\n\nContributions. We introduce value-based Compressed Linear Algebra (CLA),9,10 in which lightweight compression techniques are applied to matrices and then linear algebra operations are executed directly on the compressed representations. Figure 1 shows the goals of this approach: we want to widen the sweet spot for compression by achieving both (1) performance close to uncompressed in-memory operations, and (2) good compression ratios to fit larger datasets into memory. Our contributions include:", null, "Figure 1. Goals of compressed linear algebra.\n\n• Adapted column-oriented compression schemes for numeric matrices, and cache-conscious linear algebra operations over these compressed matrices (Section 3).\n• A sampling-based algorithm for selecting a good compression plan, including techniques for compressed-size estimation and column grouping (Section 4).\n\nOur CLA framework is available open source in Apache SystemML, where CLA is enabled by default for matrices that are larger than aggregate cluster memory.\n\n### 2. Background and Motivation\n\nAfter giving an overview of SystemML as a representative ML system, we discuss common workload characteristics that directly motivate the design of our CLA framework.\n\nSystemML compiler and runtime. In SystemML,4 ML algorithms are expressed in a high-level language with R-like syntax for linear algebra and statistical operations. These scripts are automatically compiled into hybrid runtime plans that combine single-node, in-memory operations and distributed operations on MapReduce or Spark. During this compilation step, the system also applies optimizations such as common subexpression elimination, optimization of matrix-multiplication chains, algebraic simplifications, physical operator selection, and rewrites for dataflow properties like caching and partitioning. Matrices are represented in a binary block matrix format with fixed-size blocks, where individual blocks can be in dense, sparse, or ultra-sparse formats. For single-node operations, the entire matrix is represented as a block, which ensures consistency without unnecessary overheads. CLA can be seamlessly integrated by adding a new derived block representation and operations.\n\nCommon operation characteristics. Two important classes of ML algorithms are (1) iterative algorithms with matrix-vector multiplications (or matrix-matrix with a small second matrix), and (2) closed-form algorithms with transpose-self matrix multiplication. For both classes, few matrix operations dominate the overall algorithm runtime, apart from the costs for the initial read from distributed file system or object storage. This is especially true with hybrid runtime plans, where operations over small data are executed in the driver and thus, incur no latency for distributed computation. Examples for class (1) are linear regression via a conjugate gradient method (LinregCG), L2-regularized support vector machines (L2SVM), multinomial logistic regression (MLogreg), Generalized Linear Models (GLM), and Kmeans, while examples for class (2) are linear regression via a direct solve method (LinregDS) and Principal Component Analysis (PCA). Besides matrix-vector multiplication, we have vector-matrix multiplication, which is often caused by the rewrite Xv (vX) to avoid transposing X because computing X is expensive, whereas computing v involves only a metadata update. Many systems also implement physical operators for matrix-vector chains with optional element-wise weighting X(w(Xv) ), and transpose-self matrix multiplication (`tsmm`) XX.4,17 Most of these operations are I/O-bound, except for `tsmm` with m 1 features because its compute workload grows as O(m2). Other common operations over X are `cbind`, unary aggregates like `colSums`, and matrix-scalar operations.\n\nCommon data characteristics. The inputs to these algorithm classes often exhibit common data characteristics:\n\n• Tall and skinny matrices: Matrices usually have significantly more rows (observations) than columns (features), especially in enterprise ML, where data often originates from data warehouses (see Table 1).\n• Non-uniform sparsity: Sparse datasets usually have many features, often created via pre-processing such as dummy coding. Sparsity, however, is rarely uniform, but varies among features. For example, Figure 2 shows the skew of the Covtype and Mnist8m datasets.\n• Low column cardinalities: Many datasets exhibit features with few distinct values, for example, encoded categorical, binned or dummy-coded features. For example, Figure 3 shows the ratio of column cardinality to the number of rows of the Higgs and Census datasets.\n• Column correlations: Correlation among features is also very common and typically originates from natural data correlation, the use of composite features, or again pre-processing techniques like dummy coding. For example, exploiting column correlations improved the compression ratio for Census from 12.8x to 35.7x.", null, "Table 1. Compression ratios of real datasets.", null, "Figure 3. Cardinality ratios.\n\nThese data characteristics directly motivate the use of column-oriented compression schemes as well as heterogeneous encoding schemes and column co-coding.\n\n### 3. Compression Schemes\n\nWe now describe the overall CLA compression framework, encoding formats for compressed column groups, and cache-conscious operations over compressed matrices.", null, "3.1. Matrix compression framework\n\nCLA compresses matrices column-wise to exploit two key characteristics: few distinct values per column and high cross-column correlations. Taking advantage of few distinct values, we encode a column as a dictionary of distinct values, and a list of offsets per value or value references. Offsets represent row indexes where a given value appears, while references encode values by their positions in the dictionary.\n\nColumn co-coding. We further exploit column correlation by partitioning columns into groups such that columns within each group are highly correlated. Each column group is then encoded as a single unit. Conceptually, each row of a column group comprising m columns is an m-tuple t of floating-point values that represent reals or integers.\n\nColumn encoding formats. The lists of offsets and references are then stored in a compressed representation. Inspired by database compression techniques and sparse matrix formats, we adapt four effective encoding formats:\n\n• Offset-List Encoding (OLE) encodes the offset lists per value tuple as an ordered list of row indexes.\n• Run-Length Encoding (RLE) encodes the offset lists as sequence of runs of begin row index and run length.\n• Dense Dictionary Coding (DDC) stores tuple references to the dictionary including zeros.\n• Uncompressed Columns (UC) is a fallback for incompressible columns, stored as a sparse or dense block.\n\nEncoding may be heterogeneous, with different formats for different column groups. The decisions on co-coding and encoding formats are strongly data-dependent and thus, require automatic compression planning (Section 4).\n\nExample compressed matrix. Figure 4 shows an example compressed matrix block in its logical representation. The 10 × 5 input matrix is encoded as four column groups, where we use 1-based indexes. Columns 2, 4, and 5 are represented as single-column groups and encoded via RLE, DDC, and UC, respectively. For Column 2 in RLE, we have two distinct non-zero values and hence two associated offset lists encoded as runs. Column 4 in DDC has three distinct values (including zero) and encodes the data as tuple references, whereas Column 5 is a UC group in dense format. Finally, there is a co-coded OLE column group for the correlated Columns 1 and 3, which encodes offset lists for all three distinct non-zero value-pairs as lists of row indexes.", null, "Figure 4. Example compressed matrix block.\n\nNotation. For the ith column group, denote by Ti = {ti1, ti2, ..., tidi} the set of di distinct tuples, by\n\nNo entries found" ]

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[ "# Compound interest math formula\n\nCompound interest math formula keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website\n\n## Keyword Suggestions\n\nCompound interest math formula\n\nCompound interest formula for math\n\nRecurring compound interest formula math\n\nMath formula for compound interest monthly\n\nCompounded interest math formula\n\n( Please select at least 2 keywords )\n\n#### Most Searched Keywords\n\n## Websites Listing\n\nWe found at least 10 Websites Listing below when search with compound interest math formula on Search Engine\n\n### What is Compound Interest? Calculate Compound Interest\n\nThere are two types of interest, simple and compound. Compound interest is interest calculated on the initial principal and also on the accumulated interest of previous periods of a deposit or loan. Learn more about compound interest, the math formula for calculating it on your own, and how a worksheet can help you practice the concept.\n\nhttps://www.thoughtco.com/compound-interest-formula-2312228\n\nDA: 17 PA: 34 MOZ Rank: 51\n\n### Compound interest calculator - Free math calculators ...\n\nCompound Interest is calculated on the initial payment and also on the interest of previous periods. Example: Suppose you give \\\\$100 to a bank which pays you 10% compound interest at the end of every year. After one year you will have \\\\$100 + 10% = \\\\$110, and after two years you will have \\\\$110 + 10% = \\\\$121.\n\nhttps://www.mathportal.org/calculators/financial-calculators/compound-interest-calculator.php\n\nDA: 18 PA: 50 MOZ Rank: 85\n\n### Calculate Compound Interest: Formula with examples and ...\n\nCompound interest is when a bank pays interest on both the principal (the original amount of money)and the interest an account has already earned. To calculate compound interest use the formula below. In the formula, A represents the final amount in the account after t years compounded 'n' times at interest rate 'r' with starting amount 'p' .\n\nhttps://www.mathwarehouse.com/compound-interest/formula-calculate.php\n\nDA: 21 PA: 40 MOZ Rank: 61\n\n### Compound Interest Definition, Formula, Derivation, Examples\n\nCompound Interest Formula. The compound interest formula is given below: Compound Interest = Amount – Principal. Where the amount is given by: Where, A= amount. P= principal. R= rate of interest. n= number of years. It is to be noted that the above formula is the general formula for the number of times the principal is compounded in an year.\n\nhttps://byjus.com/maths/compound-interest/\n\nDA: 9 PA: 25 MOZ Rank: 34\n\n### What Is Compound Interest? Formula, Definition and Examples\n\nConfused? It may help to examine a graph of how compound interest works. Say you start with \\$1000 and a 10% interest rate. If you were paying simple interest, you'd pay \\$1000 + 10%, which is another \\$100, for a total of \\$1100, if you paid at the end of the first year. At the end of 5 years, the total with simple interest would be \\$1500.\n\nhttps://www.thoughtco.com/what-is-compound-interest-3863068\n\nDA: 17 PA: 34 MOZ Rank: 51\n\n### Compound Interest Formula- Explained - Basic-mathematics.com\n\nCompound interest formula. A simpler version of the compound interest formula is B = P( 1 + r) n where B is the final balance, P is the principal, r is the interest rate for 1 or each interest period, and n is the number of payment periods. The principal is the amount of money you deposit that you expect will grow over time.\n\nhttps://www.basic-mathematics.com/compound-interest-formula.html\n\nDA: 25 PA: 31 MOZ Rank: 56\n\n### Compound interest formula and examples - MathBootCamps\n\nCompound interest is when interest is earned not only on the initial amount invested, but also on any interest. In other words, interest is earned on top of interest and thus “compounds”. The compound interest formula can be used to calculate the value of such an investment after a given amount of time, or to calculate things like the ...\n\nhttps://www.mathbootcamps.com/compound-interest-formula/\n\nDA: 21 PA: 27 MOZ Rank: 48\n\n### Compound Interest - Math is Fun - Maths Resources\n\nAnd by rearranging that formula (see Compound Interest Formula Derivation) we can find any value when we know the other three: PV = FV(1+r) n. Finds the Present Value when you know a Future Value, the Interest Rate and number of Periods. r = (FV/PV) (1/n) − 1. Finds the Interest Rate when you know the Present Value, Future Value and number of ...\n\nhttps://www.mathsisfun.com/money/compound-interest.html\n\nDA: 18 PA: 29 MOZ Rank: 47\n\n### Interest Formulas For Simple and Compound Interests With ...\n\nInterest formulas mainly refer to the formulas of simple and compound interests. The simple interest (SI) is a type of interest that is applied to the amount borrowed or invested for the entire duration of the loan, without taking any other factors into account, such as past interest (paid or charged) or any other financial considerations. Simple interest is generally applied to short-term ...\n\nhttps://byjus.com/interest-formula/\n\nDA: 9 PA: 18 MOZ Rank: 27\n\n### Interest and Exponential Growth Table - math\n\nLoan Balance Situation: A person initially borrows an amount A and in return agrees to make n repayments per year, each of an amount P.While the person is repaying the loan, interest is accumulating at an annual percentage rate of r, and this interest is compounded n times a year (along with each payment).). Therefore, the person must continue paying these installments of amount P until the ...\n\nhttp://www.math.com/tables/general/interest.htm\n\nDA: 12 PA: 28 MOZ Rank: 40" ]

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