Datasets:

Infi-MM
/

InfiMM-WebMath-40B

Dataset card Data Studio Files Files and versions Community

Split (1)

train · ~22.8M rows (showing the first 514k)

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.

URL stringlengths 15 1.68k	text_list listlengths 1 199	image_list listlengths 1 199	metadata stringlengths 1.19k 3.08k
https://getcalc.com/math-log-base3of69.htm	[ "# What is Log Base 3 of 69?", null, "Logarithm calculator to generate step by step calculation for how to find what is log base 3 of 69? Logarithm base 3 of 69 is 3.854.\n\nlog3(69) = 3.854\n\n## How to Find log3(69)? - Work with Steps\n\nThe below is the work with steps to find what is log base 3 of 69 shows how the input values are being used in the log base 3 functions. Formula:\nlogb(x) = y, if by = x\n\nInput:\nx = 69\nb = 3\n\nSolution:\ny = log3 69\n= log(3 x 23)\n= log3(3) + log3(23)\n= 1 + 2.854\n= 3.854\nlog3(69) = 3.854", null, "" ]	[ null, "https://getcalc.com/calculator-images/math/logarithm-antilog-calculator.png", null, "https://d7kr1t5cwstl7.cloudfront.net/graphics/getcalc-logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7867903,"math_prob":0.99962145,"size":348,"snap":"2021-43-2021-49","text_gpt3_token_len":137,"char_repetition_ratio":0.12209302,"word_repetition_ratio":0.0,"special_character_ratio":0.44827586,"punctuation_ratio":0.11363637,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999193,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-09T00:55:12Z\",\"WARC-Record-ID\":\"<urn:uuid:219c7515-dad7-471c-86e6-43522a8f0ebb>\",\"Content-Length\":\"15678\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:76772e2d-ef0c-485f-a86d-4446b1c1c556>\",\"WARC-Concurrent-To\":\"<urn:uuid:b70db1cf-8afb-4613-ae31-7f3c6aa10c89>\",\"WARC-IP-Address\":\"104.248.116.242\",\"WARC-Target-URI\":\"https://getcalc.com/math-log-base3of69.htm\",\"WARC-Payload-Digest\":\"sha1:MQ5RNY3ZM4YVVWBRGYIBZFDXLA2MGZWX\",\"WARC-Block-Digest\":\"sha1:ZGAIDDUTNFJLWLTDCRXKNAAKXS2KK5XJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363641.20_warc_CC-MAIN-20211209000407-20211209030407-00194.warc.gz\"}"}
https://ij-healthgeographics.biomedcentral.com/articles/10.1186/s12942-020-00251-z	[ "# Beyond standardized mortality ratios; some uses of smoothed age-specific mortality rates on small areas studies\n\n## Abstract\n\n### Background\n\nMost epidemiological risk indicators strongly depend on the age composition of populations, which makes the direct comparison of raw (unstandardized) indicators misleading because of the different age structures of the spatial units of study. Age-standardized rates (ASR) are a common solution for overcoming this confusing effect. The main drawback of ASRs is that they depend on age-specific rates which, when working with small areas, are often based on very few, or no, observed cases for most age groups. A similar effect occurs with life expectancy at birth and many more epidemiological indicators, which makes standardized mortality ratios (SMR) the omnipresent risk indicator for small areas epidemiologic studies.\n\n### Methods\n\nTo deal with this issue, a multivariate smoothing model, the M-model, is proposed in order to fit the age-specific probabilities of death (PoDs) for each spatial unit, which assumes dependence between closer age groups and spatial units. This age–space dependence structure enables information to be transferred between neighboring consecutive age groups and neighboring areas, at the same time, providing more reliable age-specific PoDs estimates.\n\n### Results\n\nThree case studies are presented to illustrate the wide range of applications that smoothed age specific PoDs have in practice . The first case study shows the application of the model to a geographical study of lung cancer mortality in women. This study illustrates the convenience of considering age–space interactions in geographical studies and to explore the different spatial risk patterns shown by the different age groups. Second, the model is also applied to the study of ischaemic heart disease mortality in women in two cities at the census tract level. Smoothed age-standardized rates are derived and compared for the census tracts of both cities, illustrating some advantages of this mortality indicator over traditional SMRs. In the latest case study, the model is applied to estimate smoothed life expectancy (LE), which is the most widely used synthetic indicator for characterizing overall mortality differences when (not so small) spatial units are considered.\n\n### Conclusion\n\nOur age–space model is an appropriate and flexible proposal that provides more reliable estimates of the probabilities of death, which allow the calculation of enhanced epidemiological indicators (smoothed ASR, smoothed LE), thus providing alternatives to traditional SMR-based studies of small areas.\n\n## Background\n\nSpatial epidemiology deals with the description and analysis of geographically indexed health data with respect to demographic, environmental, behavioral, socioeconomic, genetic and infectious risk factors . Most of these risk factors, especially those related to mortality, strongly depend on age ; therefore, the comparison of raw (unstandardized) rates may be misleading due to differences in the population structure of the units of study, which may be confused with real differences in risk. Age-adjusted (standardized) rates were proposed in the mid-nineteenth century as an answer to this problem, although they are not a completely satisfactory solution.\n\nStandardization seeks to remove the effect of having different age distributions for the populations being compared. It produces a single summary index per population, easier to compare than a full set of age-specific rates. The most popular standardization method is the so-called indirect method, which would yield standardized mortality ratios (SMRs) if mortality was the event of interest. SMRs are calculated as the ratio of the number of deaths observed over a specific time interval in a population group, in our case and from now on “spatial units”, to the expected deaths in that population assuming that it had the same age-specific death rates as a reference population. This indicator is commonly used to compare the mortality for different geographical areas to that of the reference population, and its main advantage is that SMRs do not depend on the age-specific rates of each spatial unit, which may be completely unreliable when working with small areas. This property has led to SMRs being, by far, the most commonly used epidemiological indicator for small areas spatial studies. Nevertheless, SMRs do exhibit some particular problems. SMRs of different spatial units are not directly comparable, that is, even if the age-specific rates of two populations were all proportional, this would not mean that the same proportionality holds between their SMRs . In particular, areas with identical age-specific rates do not necessarily yield the same SMRs. Therefore, strictly speaking, SMRs are only valid for comparing the units of study against the standard population of the study, but not against one another.\n\nAlternatively, a second age-standardization method for rates, the direct method, produces age-standardized rates (ASRs), a weighted average of the age-specific rates of each unit of study also using a reference population to determine that common set of weights. When mortality is the outcome of study, ASRs have a direct interpretation as the hypothetical death rate that would have occurred in the corresponding unit of study if its age composition were the same as that of the reference population . Therefore, the standardization process of ASRs removes the effect that the age composition of the different population groups could have on raw rates. One advantage of direct over indirect standardization is that spatial units with proportional age-specific rates necessarily yield ASRs keeping that same proportion , and so they are also comparable to each other. Nevertheless, ASRs are not as popular as SMRs for small areas geographical analyses since ASRs are only reliable when trustworthy age-specific rates are available for each spatial unit. This usually requires having a considerable amount of observed health events, from now on “deaths”, for every combination of age group and spatial unit, which small areas do not usually have. As a consequence, when ASRs are pursued on small areas, statistical modeling becomes necessary.\n\nThere have been various approaches to the smooth estimation of ratios, such as mortality rates. In this regard, Kafadar proposed a linear smoother for geographically-defined data that are in form of ratios. Although it shows that the smoothed rate has less variability than the age-adjusted unsmoothed rate, it uses parametric weights and does not consider the dependence between age groups. Ezzati et al. used information on the number of deaths and county-level sociodemographic characteristics to estimate mortality rates and probabilities of death, but merged the counties, pooled the data on deaths and population over 5 years and also did not consider the dependence between age groups.\n\nMany more epidemiologic measures show similar problems produced by the age composition of the units of study. For example, life expectancy at birth (LE) is an alternative mortality summary which shows similar problems. This indicator has the advantage of being very intuitive and easier to understand for the majority of people, therefore its use is increasingly demanded from health authorities. Its calculation also depends on the age distribution of mortality for each unit of study. Thus, although directly comparable between spatial units, it also poses the small area estimation problems arising from the age-specific risk estimates for each spatial unit . Several Bayesian hierarchical approaches have been previously proposed to estimate smoothed LEs in small areas studies. Congdon considers the smoothing of LEs but his model does not contain a specific age–space interaction term, so age–space risk variations are just a simple sum of separate age and space components. In other works [10,11,12,13,14] model area and age specific mortality rates in order to subsequently calculate LEs; nevertheless, their approaches do not include a fully structured (dependent) age–space interaction term jointly considering the variation of these two sources of dependence. These works do not use the dependence of contiguous age groups and locations for the age–space interaction term. As a consequence, these models fit the underlying overdispersion but could possibly fail to appropriately describe the space–age group structured interaction of the data. Consequently, the technical difficulties of structuring the dependence structure in age–space terms for LEs have not yet been properly addressed in the literature.\n\nIn this paper, an autoregressive proposal, based on the original model developed by Botella-Rocamora et al. , is made in order to model the age-specific probabilities of death (PoDs) for each spatial unit, which assumes stronger dependence between closer age groups and spatial units. This age–space dependence structure enables information to be transferred between neighboring (consecutive) age groups and neighboring areas, at the same time, providing more reliable age-specific PoDs estimates. Moreover, the model proposed can be easily generalized to the analysis of age–space–time datasets producing spatio-temporal age-specific rates, for example, or include any additional factor, such as sex, which could be of interest to be studied. The smoothed age-specific PoDs from this model could be later used to derive enhanced indicators, such as smoothed ASRs or LEs, of high epidemiological value.\n\nThis article is organized as follows. \"Methods\" section shows our modeling proposal for estimating age-specific PoDs in small areas geographical studies and its application to the construction of additional smoothed epidemiological indicators such as ASRs and LEs. The following section shows three case studies illustrating some real uses that smoothed age-specific PoDs could have, enhancing therefore traditional SMRs-based small areas studies. The first case study shows the application of the model introduced in \"Methods\" section to a real setting, a geographical study of lung cancer mortality in women. Its goal is to illustrate the convenience of considering age–space interaction in geographical studies. In the second case, the model is applied to study mortality from ischaemic heart disease for women in two cities at the census tract level. Smoothed ASRs are derived and compared for the census tracts of both cities, illustrating some advantages of this mortality indicator. In the last case study, the model is applied to estimate smoothed LEs, which is the most widely used synthetic indicator for characterizing overall mortality differences throughout areas. Finally, \"Discussion\" section summarizes the main contributions of our work.\n\n## Methods\n\n### Spatial smoothing of age-specific PoDs\n\nMartinez-Beneito proposed a unifying framework for multivariate disease mapping, i.e., for jointly mapping several diseases taking into account the spatial correlation that they might show. A second reformulation of this proposal, the M-model, was proposed in Botella-Rocamora et al. which enables the joint smoothing of tens of diseases in a single study. This approach, with important computational benefits, allows researchers to determine and take advantage of the correlations among diseases, revealing the existence of common risk factors or unknown associations among some diseases. In this model, dependence between spatial units is introduced in a structured way, while the relationship between diseases is introduced in an unstructured way, no prior dependence structure is assumed between diseases. However, this model allows considering also structured relationships between diseases, or geographical patterns in general, (such as an autoregressive structure) if this was considered convenient. Thus, the relationship between geographical patterns of different age groups should be possibly considered as a structured factor, reproducing its ordinal character. This modeling proposal could also be used in our case to model several risk patterns, one per age group, instead of several diseases, although assuming stronger correlations for contiguous age groups. This could be achieved, for example, by assuming spatially dependent first-order autoregressive dependency structures for age within each spatial unit.\n\nOur model proposes to model PoDs instead of age-specific rates. Age-specific rates are the relation between observed deaths and the person-years at risk available. Those person-years are not so easily available for small areas studies, which is why we propose to model PoDs instead of rates. Nevertheless, if mortality rates were really our goal, the person-years in the study could be reasonably approximated by the population at risk for each area multiplied by the years in the period of study, as we will see later. Additionally, LEs depend directly on PoDs instead of rates so the modeling of PoDs will be also suitable for deriving these indicators.\n\nThe autoregressive adaptation of the M-model for PoDs was made as follows. We changed the typical Poisson likelihood made for most disease mapping studies by a binomial distribution since the population of each (small) spatial unit is further divided into several age groups, which makes the denominator of many age specific PoDs quite low. As a consequence, the binomial assumption seems substantially more appropriate in this context as, in contrast to the Poisson case, it implicitly assumes that the number of deaths for each spatial unit and age group may not be higher than the corresponding population. Thus, let $$Y_{sa}$$, $$N_{sa}$$ and $$P^_{sa}$$ be the number of observed deaths, population and (unknown) PoDs, respectively, for the s-th spatial unit and a-th age group, $$s=1,\\ldots,S$$; $$a=1,\\ldots,A$$. We use the asterisk in $$P^_{sa}$$ in order to distinguish the smoothed age-specific PoDs that we are modeling from the raw age-specific PoDs, $$P_{sa}=Y_{sa}/N_{sa}$$. We assume that $$Y_{sa}$$ follows the binomial distribution:\n\n$$Y_{sa} \\sim \\text {Binomial}(N_{sa}, P^_{sa}), \\quad s=1,\\ldots,S, a=1,\\ldots,A,$$\n\nand model the logit of the PoDs as:\n\n\\begin{aligned} logit(P^_{sa}) = \\mu _{a} + \\theta _{sa} \\end{aligned}\n(1)\n\nwhere $$\\mu _{a}$$ represents the intercept of the a-th age group and $$\\theta _{sa}$$ models the PoDs variability between age groups and spatial units. Note that $$P^_{sa}$$ in this model, and in general the PoDs that we are referring to, are the conditional probabilities of death at the age interval a given that you have survived the previous age intervals.\n\nBy modeling $$\\theta _{sa}$$ we will induce dependence between age groups and spatial units at the same time. The ideas of the M-model will be used for this end. Specifically, Botella-Rocamora et al. show that it is possible to induce both spatial and multivariate dependence on $$\\varvec{\\Theta }=(\\theta _{sa})$$ as the product of matrices:\n\n\\begin{aligned} \\varvec{\\Theta } = \\varvec{\\Phi } \\mathbf{M } \\end{aligned}\n(2)\n\nwhere $$\\varvec{\\Phi }$$ is an $$S\\times A$$ matrix whose a-th column $$\\varvec{\\Phi }_{\\cdot a}$$ follows a spatially correlated distribution. In the $$\\varvec{\\Phi }$$ matrix, the columns are independent of each other, and therefore they do not consider dependence between groups. The second term in this expression M, is an $$A\\times A$$ matrix which induces in $$\\varvec{\\Theta }$$ the dependence between age groups that we pursue.\n\nThe j-th column of $$\\varvec{\\Theta }$$, say $$\\varvec{\\Theta }_{\\cdot j}$$, contains the spatially referenced logit probabilities for the j-th age group. One can, therefore, interpret that Eq. (2) defines the spatial patterns in $$\\varvec{\\Theta }$$ as a linear combination of underlying latent variables whose coefficients correspond to the j-th column of M. Thus,\n\n\\begin{aligned} \\varvec{\\Theta }_{\\cdot j}= \\varvec{\\Phi }_{\\cdot 1} m_{1j} + \\ldots + \\varvec{\\Phi }_{\\cdot A} m_{Aj} \\end{aligned}\n(3)\n\nwhere $$m_{ij}$$ is the (i, j)-th entry in M. Matrix M combines, through linear combinations, the spatial patterns of $$\\varvec{\\Phi }$$ yielding therefore dependent spatial patterns.\n\nIn particular, we model the set of columns as independent Proper Conditional Auto-Regressive (PCAR) distributions, that is:\n\n\\begin{aligned} \\varvec{\\Phi }_{\\cdot a} \\sim N(0, \\sigma ^2 ({\\mathbf {D}}- \\gamma _a {\\mathbf {W}})^{-1}), \\end{aligned}\n\nwhere $${\\mathbf {W}}$$ and $${\\mathbf {D}}$$ are the adjacency and diagonal matrices typically summarizing the geographical structure in PCAR distributions (see for example).\n\nAccording to Botella-Rocamora et al. , for the separable case with $$\\gamma _1=\\cdots=\\gamma _A$$, the variance covariance matrix between age groups $$\\varvec{\\Sigma }$$ is equal to $$\\mathbf{M }^T \\mathbf{M }$$. Therefore, if we want the correlation between age groups to decrease as a function of their distance, M should be modelled in order to reproduce that effect. We will assume for simplicity a first-order autoregressive dependence structure between age groups. In that case, the covariance matrix between age groups would take the form:\n\n\\begin{aligned} \\Sigma _{ij} = \\rho ^{\|i-j\|}, \\end{aligned}\n\nwhere $$\\rho$$ is an autoregressive dependence parameter controlling the strength of the dependence between age groups. Note that the variance parameter of $$\\varvec{\\Sigma }$$ has been removed from this latest expression since the overall variance of $$\\varvec{\\Theta } = \\varvec{\\Phi } \\mathbf{M }$$ is already controlled by the term $$\\varvec{\\Phi }$$ by means of the variance parameter $$\\sigma ^2$$ of its columns. An M matrix inducing the mentioned autoregressive dependence between age groups could be, for instance, the upper-triangular Cholesky matrix of $$\\varvec{\\Sigma }$$, which has this simple expression:\n\n\\begin{aligned} \\mathbf{M }= \\left( \\begin{array}{ccccc} 1 &{} \\rho &{} \\rho ^2 &{} \\cdots &{} \\rho ^{A-1}\\\\ 0 &{} (1-\\rho ^2)^{1/2} &{} \\rho (1-\\rho ^2)^{1/2}&{} \\cdots &{} \\rho ^{A-2} (1-\\rho ^2)^{1/2}\\\\ 0 &{} 0 &{} (1-\\rho ^2)^{1/2} &{} \\cdots &{} \\rho ^{A-3} (1-\\rho ^2)^{1/2}\\\\ \\vdots &{} \\vdots &{} \\vdots &{} \\ddots &{} \\vdots \\\\ 0 &{} 0 &{} 0 &{} \\cdots &{} (1-\\rho ^2)^{1/2} \\end{array} \\right) . \\end{aligned}\n\nTo complete the model specification, an improper flat uniform prior is proposed for $$\\mu _a$$ and a uniform prior distribution on the interval $$[-1,1]$$ for $$\\rho$$. A vague uniform prior on [0, C] for a high enough (non-informative) value of C is also proposed for the standard deviation $$\\sigma$$ (see Gelman or pages 164–171 of Martinez-Beneito and Botella-Rocamora for more details). Finally, a uniform prior distribution on $$]\\lambda _{(1)}^{-1},\\lambda _{(I)}^{-1}[$$, where $$\\lambda _{(1)}$$ and $$\\lambda _{(I)}$$ are the lowest and highest eigenvalues of $${\\mathbf {D}}^{-1/2}{\\mathbf {W}}{\\mathbf {D}}^{-1/2}$$ respectively, is proposed for $$\\gamma _1,\\ldots,\\gamma _A$$, as suggested, for example, by Sun et al. or Martinez-Beneito and Botella-Rocamora .\n\nNote that the proposed model is equivalent to the spatio-temporal autoregressive model of Martinez-Beneito et al. . This proposal would model the matrix $$\\varvec{\\Theta }$$ as a first order autoregressive process as:\n\n\\begin{aligned} \\varvec{\\Theta }_{sa} \\sim N(\\rho \\varvec{\\Theta }_{s(a-1)},\\sigma ^2_{\\varvec{\\Theta }}) .\\end{aligned}\n\nIn this case, we would be using the autoregressive component for modeling dependence between age groups instead of temporal dependence. Nevertheless, we have posed our proposal as a M-model for several reasons. First, the M-model can be easily generalized to non-separable dependence structures; in fact, the model just proposed is already inseparable because of the different spatial correlation parameters ($$\\gamma _1,\\ldots,\\gamma _A$$) considered for the different spatial terms in the model. In a similar manner, different variance parameters ($$\\sigma _1^2,\\ldots,\\sigma _A^2$$) could be considered for these patterns, yielding therefore more flexible, heteroscedastic, covariance structures . Second, M-models are computationally convenient for modeling multivariate spatial patterns, which makes them an appropriate choice for fitting the data in regular Bayesian inference packages such as WinBUGS, OpenBUGS, Nimble… Finally, and in our opinion most importantly, M-models have been generalized to multidimensional models where several factors, besides the spatial component, are considered. This would make it possible to consider additional spatial patterns that could be correlated with the age-specific PoDs that we are modeling in order to enhance the estimates in the model. For example, mortality data from other sexes or causes of death could be additionally considered in order to yield improved age-specific PoDs in our data set. Alternatively, multidimensional modeling could be used to disaggregate the data, for example by subperiods, race groups… while maintaining a reasonable risk estimate quality by considering dependence between these new groups. In summary, in our opinion, all this makes the M-models an appropiate and flexible proposal for modeling the age-specific PoDs, as proposed above.\n\nWe should mention that some other previous proposals in the literature have dealt with the joint analysis of several dependence sources, such as age–space–time models. For example, Goicoa et al. proposed an age–space–time separable model where age and time are modelled as first order random walks and spatial dependence is modelled by means of a Leroux et al. dependence structure. Our proposal yields, in principle, more flexible dependence structures since different spatial correlations, for example, for each underlying spatial pattern are considered. Moreover, the autoregressive dependence structure for age groups allows the strength of this dependence source to be adapted. These modelling features could be implemented in INLA (the software package used in ), which in principle could take advantage of the sparse structure of the spatial and temporal precision matrices to speed up computations . Nevertheless, this would not be so advisable, since those modelling features would increase the number of parameters to be integrated out by INLA, what would make its fit substantially slower. Similarly, Goicoa et al. propose alternative age–space–time models, but using P-splines for modeling dependence for age and time. Once again, considering spatial random effects of different dependence parameters would seem problematic for this work as well. Thus, in this regard we find our proposal somewhat more suitable. Moreover, although some other Markov chain Monte Carlo (MCMC) based approaches have been proposed for the modeling of age–space–time interactions we find our proposal particularly convenient as it does not require a specific coding of the MCMC algorithm and, in contrast, it can be fitted with regular Bayesian software packages.\n\n### Epidemiological applications of the smoothed age-specific PoDs\n\nThe proposal above produces age-specific PoDs, $$P^_{sa}$$, smoothed over space and age groups. It is important to note that any statistical indicator built on $${\\mathbf {P}}^=(P^_{sa})$$ will inherit the smooth character of $${\\mathbf {P}}^$$, thereby producing smooth, and in principle more reliable, estimates of other more complex elaborate epidemiological indicators.\n\nFor example, ASRs are weighted averages of the age-specific death rates. The ASRs weights would be given by the age-composition of a reference population, so all the ASRs would represent the observed risks for a common (ideal) population; in this way the age composition of the units of study would in principle no longer be a confounding factor. Traditionally, the age-specific mortality rate for age group a and spatial unit s is given by:\n\n\\begin{aligned} R_{sa} = \\frac{Y_{sa}}{PersonYears_{sa}}\\approx \\frac{Y_{sa}}{T \\cdot N_{sa}}=\\frac{P_{sa}}{T} . \\end{aligned}\n(4)\n\nwhere T is the number of years in the period of study. If the period of study was just one year then $$T=1$$ and then $$R_{sa}=P_{sa}$$. Let $$pob_a$$ be the population in age group a in the reference population and let the standard weights be given by:\n\n\\begin{aligned} W_a=10^5 \\cdot \\frac{pob_a}{\\sum _a pob_a}. \\end{aligned}\n\nThen the ASR for the s-th spatial unit is usually defined as:\n\n\\begin{aligned} ASR_{s}=\\sum _a W_a \\cdot R_{sa}. \\end{aligned}\n(5)\n\nThe $$10^5$$ in $$W_a$$ allows us to undertand the ASRs as the number of deaths that we would expect per 100,000 people from the reference population. Obviously, the unreliability of $$P_{sa}$$ when the spatial units are small will be transferred to the ASRs, which will make them unreliable indicators. However, we could alternatively define smoothed Age-Standardized Rates (sASRs) by simply replacing $$R_{sa}$$ in the previous expression by $$R_{sa}^=P_{sa}^/T$$, that is:\n\n\\begin{aligned} sASR_{sa}=\\sum _a W_a \\cdot R^_{sa} \\end{aligned}\n(6)\n\nwhich will inherit the smooth character of $$P^_{sa}$$ and will therefore solve the small area estimation problems that traditional ASRs show.\n\nFollowing this same idea, many more epidemiological indicators relying on age-specific PoDs or mortality rates could be smoothed. For example, LEs are usually calculated from life tables as a direct function of the age-specific PoDs, the width of each age interval and the fraction of the last age interval survived for those people dying at each age interval . Additionally, particular care is required for the last age group since this is a right-opened interval, with no upper limit, so it deserves special attention. Nevertheless, if $$D_{sa}$$ is the marginal probability of death (which is a combination of the conditional PoDs) for group age a and spatial unit s, and $$age_a$$ is the mean age of death for the people dying at age group a, LEs could be alternatively defined as the expected value of the observed ages of death in the following way:\n\n\\begin{aligned} LE_s=\\sum _{a=1}^{A} age_a D_{sa}. \\end{aligned}\n(7)\n\nIn a similar manner to that made with life tables, $$age_a$$ could be calculated as the sum of the initial age of each interval and of the proportion of interval survived by those dying during that age interval. Usually, that proportion survived is assumed to be one half of each interval, except for the youngest age group where it is known that perinatal deaths make this proportion shorter . As a consequence, $$age_a$$ is usually taken as the central value for each age interval, which is equivalent to assuming the deaths to be uniformly distributed over those intervals. Regarding $$age_A$$, it is usual to assume a constant death rate, or equivalently an exponential distribution for the ages of death in the latest age group . According to the available data, that death rate could be estimated as the age-specific death rate for this group. Therefore, according to the exponential assumption for the ages of death for this interval, the average number of years lived for any person reaching the oldest age group would be the inverse of its death rate, that is: $$T/P_{sA}$$. Therefore, $$age_A$$ would be the sum of this quantity and the initial age for this age group.\n\nA detailed description of the life expectancy calculation from a life table can be found, for example, in the Public Health England template produced for this purpose based on the methodology described in .\n\nOnce the $$age_a$$ values have been calculated, Expression (7) can be formulated as a function of the conditional probabilities of death:\n\n\\begin{aligned} LE_s=\\sum _{a=1}^{A-1} age_a \\left( \\prod _{i<a} (1-P_{si}) P_{sa} \\right) + age_A \\prod _{i<A} (1-P_{si}). \\end{aligned}\n(8)\n\nThe reliance of Expression (8) on the observed raw age-specific (conditional) PoDs, $$P_{sa}$$, makes it clear why LEs become unreliable in small areas studies. Nevertheless, by changing the raw age-specific PoDs for the smoothed age-specific PoDs, $$P^_{sa}$$, resulting from the above model, we could obtain reliable life expectancies, smoothed Life Expectancy (sLE), suitable for small areas studies.\n\nSimilar procedures could be proposed for any other epidemiological indicator relying on age-specific PoDs, such as disability-free life expectancy, potential years of life lost… to cite just two. These indicators would be alternative tools to the traditional use of SMRs in small areas studies that would highlight new aspects of the data that regular SMRs are not able to reveal.\n\nFinally, we find it interesting to note that, since the proposed model is posed under a Bayesian approach, MCMC is used for making inference about that model. As a consequence, credible intervals and other additional variability measures can be easily computed within the MCMC itself for any of the proposed indicators. This makes their comparison particularly easy.\n\n## Case studies\n\nIn this section, we are going to make use of some real case studies to illustrate the use of the proposed methodology and the advantages of the new smoothed epidemiological indicators over the traditional use of smoothed SMRs, from now on “sSMRs”.\n\nAll the models used for the following case studies have been run in WinBUGS . The pbugs R library was used for speeding up computations by parallelizing the sampling of the different chains of the MCMC. Rmarkdown documents with the code used to reproduce all three analyses can be found at Additional file 1. For all three analyses, three chains were run with 30,000 iterations, the first 5000 of which were discarded as a burn-in period. A thinned sample was obtained by saving one out of every 75 iterations. Regarding convergence checking, the Brooks–Gelman–Rubin statistic has been checked to be lower than 1.1 and the effective sample size to be above 100 for all the parameters saved during the MCMC. All these criteria were successfully met for all three case studies.\n\nAll three models that were finally run for all three case studies correspond to separable age–space dependence structures, with $$\\gamma _1=\\cdots=\\gamma _A$$. The reason for this is that for none of the three data sets have we found evidence of inseparability (the confidence intervals for $$\\gamma _1,\\ldots,\\gamma _A$$ for all three studies substantially overlapped). As a consequence, we have decided to show the results of the simpler separable case. However, we should mention that we have found evidence of inseparability for some other case studies that have not been finally included in the paper for reasons of space.\n\n### Case study 1: interaction between age groups and space\n\nIn this first case study, the proposed model has been applied to the study of lung cancer mortality in women in Comunitat Valenciana (Spain) for the 2008–2017 period. Municipalities were the spatial units of study, with a total of 542 municipalities composing the whole Comunitat Valenciana (CV). The average number of women in CV during the period of study is around 2.5 millions. The mean number of women per municipality is 4695 while the corresponding median is 744. Nowadays, lung cancer is the second most common cancer mortality cause in women in CV after breast cancer, with a total of 4232 deaths for the whole period of study. Mortality data were provided by the mortality registry of CV. Population data was obtained from the Spanish National Statistics Institute. Both sources of information are considered to have good data quality at this level of disaggregation. It has not been necessary to process missing data or other data quality issues.\n\nSince lung cancer mainly affects elderly people, the younger age groups have been merged, resulting in the following 11 age intervals: [0, 40), [40, 44), [45, 50), …, [80, 85) and [85,…). As a consequence, we have 5962 ($$=11\\cdot 542)$$ age-specific PoDs to estimate, more than observed deaths. Therefore, even though lung cancer shows very high mortality values for our data set, we still have 4646 cells out of the 5962 municipality-age group combinations (78%) with zero deaths. This makes evident the need for some kind of modeling in order to derive reliable results.\n\nThe left hand side of Fig. 1 shows the raw PoDs for each area and age group directly obtained from 4. Note that age-specific rates would be just the PoDs divided by 10, the number of years of the period of study. The right hand side of Fig. 1 shows the corresponding smoothed PoDs provided by the model. In both cases, the age-specific PoDs for the whole CV, Almazora and Denia (two illustrative municipalities) have been highlighted. The raw PoDs show evidence of small area estimation problems. There are no observed cases for most age groups and municipalities and when some death is observed this makes the probability of death appear extremely high, mainly for the youngest age groups, which produces highly volatile estimates and sawtooth-shaped curves for most municipalities. These extreme raw age-specific PoDs would evidently distort subsequent epidemiological indicators based on them. On the other hand, the estimated smoothed PoDs, using information from neighboring age groups and areas, yield a much more stable and credible performance where most of the uncertainty in the raw PoDs has been filtered out. Nevertheless, some variability between municipalities is retained, for example, several municipalities stand out in this second plot for having particularly high risks for the oldest groups. These municipalities were completely unnoticed in the left hand side plot where the underlying noise does not allow conclusions of any kind to be drawn.\n\nIn order to confirm the relevance of the age–space interaction for this data set, we have run a second alternative model without interaction. In other words, this second model assumes simply $$logit(P^_{sa})=\\mu _a+\\theta _s$$ instead of having a $$\\theta _{sa}$$ term that varies for each spatial unit and age group. We have compared the Deviance Information Criterion (DIC) for both models, resulting in a DIC of 6546.5 for the model without interaction and 6508.0 for the original model with interaction. Therefore, age–space interaction seems required in order to appropriately describe the variability underlying this data set. Although some additional model selection criteria could be used for this comparison, as suggested for example by Duncan and Mergensen , the magnitude of the DIC difference already points out important fit differences between both models.\n\nWe have decomposed the variability on $$logit(P^_{sa})$$ into three separate terms in order to visualize the sources of variability of this data set. Specifically, we have decomposed that matrix into two overall age and spatial effects and an interaction term explaining the variability that cannot be explained by these main terms (see Adín et al. or page 285 of Martinez-Beneito and Botella-Rocamora for more details on this decomposition in a spatio temporal setting). The age effect (result shown in Additional file 1), which is defined as the collection of column means of the matrix $$\\varvec{\\Theta }$$, unsurprisingly shows an upwards trend of the risks as a function of age, with a steeper trend until the [55, 59] group and a milder constant increase for the oldest age groups. Figure 2 shows both the spatial and interaction terms for each municipality. The spatial term, in the left hand side plot, which is defined as the collection of row means of the matrix $$\\varvec{\\Theta }$$, shows a clear spatial pattern, with high risks mainly in the southern and south-eastern part of CV (this was already described at Zurriaga et al. ). The cuts for the groups in this plot correspond to septiles of the spatial term, which is in the logits scale. The right hand side plot shows the interaction term for each municipality (one line per municipality). The comparison of the scales of both plots shows that substantial variability still holds for the interaction term, with a range of values comparable to that of the spatial term. As a consequence, particular risk excesses for some age groups and municipalities are as important as the overall risk excesses (for all age groups as a whole) for those same municipalities. In particular, Almazora and Denia show opposing performances in age specific terms. Specifically, Denia exhibits particularly low risks for the youngest age groups while the oldest groups show particularly high risks. On the contrary, Almazora shows the opposite, exhibiting better performance for the older age groups.\n\nFigure 3 shows choropleth maps for the interaction term for several of the age groups considered. Almazora and Denia have been highlighted in the first of these plots. We can see how, in general, the younger groups have a particularly bad performance in the northern and western parts of the CV. In contrast, these same areas have a particularly good performance in the oldest age groups. Note the similarity of the areas shown in this graph with those also outstanding for the spatial term. In particular, the correlation of the spatial component and the interaction term for the group younger than 40 is − 0.65; 0.42 for the age group [70, 74] and 0.32 for the age group older than 85. As a consequence, the combination of the spatial and interaction terms for the younger groups (result not shown) produces a much flatter spatial pattern than that of the spatial term, while that spatial pattern is reinforced with the interaction term for the oldest age groups, which is similar to the spatial pattern in Fig. 2. Therefore, the geographical pattern arising in the spatial term of this analysis is mainly the effect of the risk patterns corresponding to the oldest age groups since that spatial distribution for the younger groups is substantially flatter. This kind of result cannot be obtained from the typical sSMRs analyses usually made by default when studies are carried out on sets of small areas.\n\n### Case study 2: smoothed ASRs\n\nIn this case we have applied the model in \"Methods\" section to the study of ischaemic heart disease (IHD) mortality for women aged over 45 years, during the period 1996–2015. Mortality data were provided by the MEDEA3 Project (“Socio-economic and environmental inequalities in the geographical distribution of mortality in large cities in Spain (1996–2015)”, PI16/01004). Population data was obtained from the Spanish National Statistics Institute. Nine age groups: [45, 50), [50, 55), …, [80, 85) and [85,…) were considered for this study. Two separate studies, with independent models, were carried out to analyse IHD mortality in the two largest CV cities: Valencia and Alicante.\n\nIn this case, the spatial unit of study were census tracts, with a total of 531 and 178 units for Valencia and Alicante, respectively.\n\nIn this case study, we explore the use of sASRs as an alternative to sSMRs for small areas geographical analyses. Figure 4 shows several choropleth maps summarizing the main results for this analysis. Specifically, the top-left plot in that figure shows the raw ASRs for the census tracts of Valencia. This plot shows substantial noise, which hardly allows any meaningful conclusion to be drawn. Note that the census tract with the highest raw ASR is estimated to have a risk more than 10 times (!) higher than that with the lowest raw ASR. In the top-right plot we show the sASRs calculated as described in the previous section. In contrast to the raw ASRs, the sASRs show a clearer pattern where some neighborhoods show higher risks. In this way, the benefit of using smoothed epidemiological indicators becomes clear. For comparability, quintiles of the Valencia sASRs have been used as cuts for both ASR maps in the top row of Fig. 4. The bottom-left plot of Fig. 4 shows, as a reference, the sSMRs drawn from a typical small area study fitting the model. Quintiles of the sSMRs have been used for cutting the categories for this choropleth map. Note that, despite some small differences, both sSMR and sASR choropleth maps resemble each other, since both are summarizing the mortality risks of the same data set in different manners. In fact, the correlation between both quantities for the Valencia census tracts is 0.95.\n\nDespite the resemblance of the sSMR and sASR maps, the ASRs shows some advantages over the traditional sSMRs that would be appropiate to value. In contrast to sSMRs, sASRs are comparable for the different census tracts of a city, so they could be used, for example, for comparing the risks of two particular census tracts. Thus, the probability that the Valencian census tract with the highest sASR (posterior mean) has a sASR higher than that with the lowest sASR (posterior mean) is virtually 1 for our data set, highlighting an important spatial variability. This statistic should be more powerful than the typical $$P(sSMR>100)$$ used in many traditional disease mapping analyses since the mentioned sASR statistic takes into account the two extremes of the sASRs distribution instead of comparing the risk of each census tract with that of the reference population.\n\nAdditionally, the sASRs calculated could be further compared to other additional ASRs corresponding to other areas of study. Thus, our sASR estimates could be compared for example to the ASR for the overall Valencian Region, Spain or even ASRs corresponding to other countries. In this manner, the use and scope of the generated results would be much wider than for SMRs. As an example, the bottom-right plot of Fig. 4 shows the distributions of the sASRs (their posterior means) for the Valencia and Alicante census tracts. As shown in that figure, most of the Valencia census tracts have sASR values smaller than any of the Alicante census tracts. As a matter of percentiles, the 95% percentile of the sASR distribution in Valencia (178.9 deaths per $$10^5$$ women) is lower than the 5% percentile for Alicante (203.5). In other words, the census tracts with the highest IHD sASRs in Valencia have comparable mortality risks to those census tracts with the lowest mortality by IHD in Alicante. This should prompt the Valencian health authorities to take a close look at this issue, which would surely go unnoticed with a traditional sSMRs-based small area study.\n\n### Case study 3: smoothed life expectancies\n\nFinally, in this third case study, age–space PoDs have been applied to the study of LEs, for each sex, once again for the whole Comunitat Valenciana (CV) for the period 2014–2017. In this case, mortality for all causes has been considered. Mortality data were provided by the mortality registry of the CV. Population data was obtained from the Spanish National Statistics Institute. The spatial aggregation considered was the municipality and 19 age groups: [0, 1), [1, 4], [5, 9), [10, 14),…, [80, 85) and [85,…) were considered for this analysis.\n\nFigure 5 shows the municipal LEs for men and women (separately) for the Valencian municipalities. The two maps on the left side of the figure show the raw LEs, directly calculated from the raw age-specific PoDs. The two maps on the right, show the smoothed counterparts of the maps on the left with the corresponding smoothed LEs. For comparability, cuts for these choropleth maps have been defined according to the smoothed LEs septiles for each sex. The legends of the raw LE plots, highlight the high variability of these indicators. Thus, raw LEs for men range from 48.7 to 104.0 and from 50.8 to 110.0 for women, which seems hard to believe. Moreover, 22 and 16 municipalities, respectively for men and women, showed infinite LEs since no deaths were observed in them during the period of study in the final age interval . Smoothed LEs, however, show much more reasonable (moderate) variability than the raw LEs. This result clearly shows the need for smoothed age-specific PoDs for calculating LEs when dealing with small areas.\n\nMuch clearer spatial patterns can be observed in the smoothed LE maps of Fig. 5, compared to their unsmoothed counterparts. Thus, for men, a clear cluster of low LE levels becomes evident in the eastern-central part of CV. This was not so evident according to the raw LEs. Moreover, some areas where the raw LEs showed high heterogeneity (see for example the north-western part of CV), where most of the municipalities are sparsely populated, now show a much more smoothed (spatial) behaviour which seems considerably more reasonable. For women, something similar happens. The smoothed LEs show a clearer spatial pattern and heterogeneity seems to have disappeared for the sparsely populated areas.\n\nAdditionally, the smoothed LEs in Fig. 5 allow interesting conclusions to be drawn, such as, for example, that geographical inequalities in men in terms of LEs (12.7 years between the most extreme municipalities) are substantially higher than those in women (5.5 years). According to the raw LEs these differences are reversed, 55.3 years for men and 59.2 years for women, although evidently these values are clearly untenable. The use of smooth LEs allows us to draw some meaningful conclusion in this regard. Finally, we would like to stress that small areas LE analyses provide a complementary view to that provided by sSMR based analyses. LEs take into account not just the mortality at each spatial unit but also the ages when those deaths take place, so LEs analyses complement the results that regular sSMR small areas analyses typically yield.\n\n## Discussion\n\nAlthough spatio-temporal and multivariate disease mapping analyses on small areas have become quite popular, the same cannot be said for age–space small areas analyses. Nevertheless, as illustrated, the same methods that are used in both spatial and multivariate contexts could also be regularly used for age–space studies where age groups are disaggregated and separately studied, although considering dependence between them.\n\nIn this work, the M-model is used in order to model the age-specific PoDs for each spatial unit, which assumes stronger dependence between closer age groups and spatial units. As shown, this age–space dependence structure enables information to be transferred between contiguous age groups and neighboring areas, at the same time, providing therefore more reliable (smoothed) age-specific PoDs estimates.\n\nObviously, there are other alternative possibilities for the age-dependency structure used in the article. However, the one used seems appropriate given that it makes it possible to determine, within the same model, whether or not there is spatial and inter-age group dependency in the data, and even to quantify the strength of these sources of dependency. The small number of age groups in our case studies, and in most of the possible applications that we foresee, could make the discussion and selection of alternative dependency structures for age groups very difficult.\n\nThere are multiple reasons to use the M-model for age-specific PoDs. The first of them is that it induces both spatial and multivariate dependence by simply multiplying two matrices. Secondly, the M-model can be easily generalized to non-separable dependence structures, resulting in more flexible, heteroscedastic, covariance structures . In addition, these models are computationally convenient for modeling multivariate spatial patterns, which makes them an appropriate choice for fitting the data in regular Bayesian inference packages such as WinBUGS, OpenBUGS, Nimble… Moreover, M-models have already been implemented for the joint analysis of several correlated spatial patterns in INLA, either for a separable model for a large number of spatial patterns , or for an inseparable joint analysis of three spatial patterns . Thus INLA should be also born in mind as a potential tool for this kind of analyses. Finally and the most important in our opinion, is that our proposal facilitates the future incorporation of other factors such as time or sex in a straightforward manner, following the proposal of multidimensional models by Martinez-Beneito et al. . This allows us to consider additional spatial patterns that could be also correlated with the age-specific PoDs that we are modeling in order to enhance the estimates in the model current estimates.\n\nBeyond sASRs and sLEs calculations, the simple study of PoDs for visualizing the different age-specific spatial patterns is also interesting. Moreover, the study of those PoDs is particularly interesting when an inseparable model provides some fit improvement. For example, in another additional case study not presented here, we also applied our proposal for modeling all-cause mortality in the city of Valencia. In that case, applying the M-model at space–age group level, the smoothed age specific PoDs showed a pattern with stronger spatial dependence for the youngest age groups (higher values for the corresponding $$\\gamma _a$$) than for the oldest age groups. The corresponding PoDs maps showed as if geographical inequalities existed for the youngest age groups, which pointed towards risk excesses for the deprived neighborhoods, but this effect diluted for older age groups since basically old people die all around while young people die, in general, just in these regions. Therefore, the modeling of non-separable age–space structures seems another additional application of our model, although we have not illustrated with such a detail for questions of space.\n\nIn addition, any statistical indicator built on the smoothes age-specific PoDs inherit that smoothed character, yielding therefore more realiable estimates. Thus, in two of the three case studies we have two of the most relevant epidemiological indicators: the age standardized rate and the life expectancy. We have explored their use as an alternative to sSMR for small areas geographical analyses. On one hand, deriving an ASR version of potential use in small areas studies, with the direct comparation posibilities between areas that this brings, should be considered as an additional benefit of our proposal. On the other hand, smoothed life expectancies provides once again a version of this synthetic health indicator of potential use in small areas studies. Studies based on this indicator that take also into account the age of death could bring new possibilities to spatial epidemiological studies. Nevertheless, additional smoothed epidemiological indicators depending on PoDs may also be derived from the smoothed age–space PoDs, such as the number of potential years of life lost. Our study makes possible the study of those indicators also for small areas.\n\n## References\n\n1. Richardson S, Thomson A, Best N, Elliot P. Interpreting posterior relative risk estimates in disease-mapping studies. Environ Health Perspect. 2004;112(9):1016–25.\n\n2. Ahmad OB, Boschi-Pinto C, Lopez AD, Murray CJL, Lozano R, Inoue M. Age standardisation of reports: a new who standard (technical report)., GPE Discussion Paper SeriesGeneva: World Health Organization; 2001.\n\n3. Neison FGP. On a method recently proposed for conducting inquiries into the comparative sanatory condition of various districts. J R Stat Soc Lond. 1844;7:40–68.\n\n4. Fleiss LBPMC, Joseph L. Statistical methods for rates and proportions. 3rd ed. New York: Wiley; 2003.\n\n5. Curtin LR, Klein RJ. Direct standardization (age-adjusted death rates), vol. 6. US Department of Health and Human Services, Public Health Service, Centers for Disease Control and Prevention, National Center for Health Statistics. 1995.\n\n6. Kafadar K. Smoothing geographical data, particularly rates of disease. Stat Med. 1996;15:2539–60.\n\n7. Ezzati M, Friedman AB, Kulkarni SC, Murray CJL. The reversal of fortunes: trends in county mortality and cross-county mortality disparities in the United States. PLoS Med. 2008;5:0557–688.\n\n8. Eayres D, Williams ES. Evaluation of methodologies for small area life expectancy estimation. J Epidemiol Community Health. 2004;58(3):243–9.\n\n9. Congdon P. A life table approach to small area health need profiling. Stat Model. 2002;2:63–88.\n\n10. Jonker MF, van Lenthe FJ, Congdon PD, Donkers B, Burdorf A, Mackenbach JP. Comparison of Bayesian random-effects and traditional life expectancy estimations in small-area applications. Am J Epidemiol. 2012;176(10):929–37.\n\n11. Congdon P. Estimating life expectancies for us small areas: a regression framework. J Geogr Syst. 2014;16(1):1–18.\n\n12. Congdon P. Area variations in multiple morbidity using a life table methodology. Health Serv Outcomes Res Methodol. 2016;16:58–74.\n\n13. Dwyer-Lindgren L, Bertozzi-Villa A, Stubbs RW, Morozoff C, Mackenbach JP, van Lenthe FJ, Mokdad AH, Murray CJL. Inequalities in life expectancy among us counties, 1980 to 2014: temporal trends and key drivers. JAMA Intern Med. 2017;177(7):1003–111.\n\n14. Alexander M, Zagheni E, Barbieri M. A flexible Bayesian model for estimating subnational mortality. Demography. 2017;54(6):2025–41.\n\n15. Botella-Rocamora P, Martinez-Beneito MA, Banerjee S. A unifying modeling framework for highly multivariate disease mapping. Stat Med. 2015;34(9):1548–59. https://doi.org/10.1002/sim.6423.\n\n16. Martinez-Beneito MA. A general modelling framework for multivariate disease mapping. Biometrika. 2013;100(3):539–53. https://doi.org/10.1093/biomet/ast023.\n\n17. Martinez-Beneito MA, Botella Rocamora P. Disease mapping from foundations to multidimensional modeling. Boca Raton: CRC Press; 2019.\n\n18. Gelman A. Prior distributions for variance parameters in hierarchical models. Bayesian Anal. 2006;1:515–34.\n\n19. Sun D, Tsutakawa RK, Speckman PL. Posterior distribution of hierarchical models using CAR(1) distributions. Biometrika. 1999;86(2):341–50. https://doi.org/10.1093/biomet/86.2.341.\n\n20. Martinez-Beneito MA, López-Quílez A, Botella-Rocamora P. An autoregressive approach to spatio-temporal disease mapping. Stat Med. 2008;27:2874–89.\n\n21. Corpas-Burgos F, Botella-Rocamora P, Martinez-Beneito MA. On the convenience of heteroscedasticity in highly multivariate disease mapping. TEST. 2019;28:1229–500.\n\n22. Martinez-Beneito MA, Botella-Rocamora P, Banerjee S. Towards a multidimensional approach to Bayesian disease mapping. Bayesian Anal. 2017;12:239–59. https://doi.org/10.1214/16-BA995.\n\n23. Leroux BG, Lei X, Breslow N. Estimation of disease rates in small areas: a new mixed model for spatial dependence. In: Halloran ME, Berry D, editors. Statistical models in epidemiology, the environment and clinical trials. Berlin: Springer; 1999.\n\n24. Goicoa T, Ugarte MD, Etxebarria J, Militino AF. age–space–time CAR models in Bayesian disease mapping. Stat Med. 2016;35:2391–405.\n\n25. Rue H, Martino S, Chopin N. Approximate Bayesian inference for latent Gaussian models by using integrated nested Laplace approximations. J R Stat Soc B (Stat Methodol). 2009;71(2):319–92.\n\n26. Goicoa T, Adin A, Etxeberria J, Militino AF, Ugarte MD. Flexible Bayesian P-splines for smoothing age-specific spatio-temporal mortality patterns. Stat Methods Med Res. 2017;28(2):384–403.\n\n27. Quick H, Waller LA, Casper M. Multivariate spatiotemporal modeling of age–space stroke mortality. Ann Appl Stat. 2017;11(4):2165–77.\n\n28. Chiang CL. The life table and its applications. Malabar: Krieger Publishing; 1984.\n\n29. Williams E, Dinsdale H, Eayres D, Tahzib F. Calculating life expectancy in small areas. Technical report, South East Public Health Observatory. 2005.\n\n30. Silcocks PBS, Jenner DA, Reza R. Life expectancy as a summary of mortality in a population: statistical considerations and suitability for use by health authorities. J Epidemiol Community Health. 2001;55(1):38–433.\n\n31. PHE: life expectancy template. https://fingertips.phe.org.uk/documents/PHELifeExpectancyCalculator.xlsm.\n\n32. Lunn D, Thomas A, Best N, Spiegelhalter D. WinBUGS—a Bayesian modelling framework: concepts, structure, and extensibility. Stat Comput. 2000;10:325–37. https://doi.org/10.1023/A:1008929526011.\n\n33. Vergara C, Martinez-Beneito MA. Pbugs. https://github.com/fisabio/pbugs.\n\n34. Brooks SP, Gelman A. General methods for monitoring convergence of iterative simulations. J Comput Graph Stat. 1998;7:434–55.\n\n35. Carlin BP, Gelman A, Neal RM. Markov chain Monte Carlo in practice: a roundtable discussion. Am Stat. 1998;52(2):93–100.\n\n36. Spiegelhalter DJ, Best NG, Carlin BP, Van Der Linde A. Bayesian measures of model complexity and fit (with discussion). J R Stat Soc Ser B (Stat Methodol). 2002;64:583–641. https://doi.org/10.1111/1467-9868.00353.\n\n37. Duncan EW, Mengersen KL. Comparing Bayesian spatial models: goodness-of-smoothing criteria for assessing under- and over-smoothing. PLoS ONE. 2020;15:0233019. https://doi.org/10.1371/journal.pone.0233019.\n\n38. Adín A, Martinez-Beneito MA, Botella-Rocamora P, Goicoa T, Ugarte MD. Smoothing and high risk areas detection in space–time disease mapping: a comparison of P-splines, autoregressive and moving average models. Stoch Environ Res Risk Assess. 2017;31:403–15. https://doi.org/10.1007/s00477-016-1269-8.\n\n39. Zurriaga O, Vanaclocha H, Martínez-Beneito MA, Botella Rocamora P. Spatio-temporal evolution of female lung cancer mortality in a region of Spain: is it worth taking migration into account? BMC Cancer. 2008;8(35):1. https://doi.org/10.1002/sim.64231.\n\n40. Besag J, York J, Mollié A. Bayesian image restoration, with two applications in spatial statistics. Ann Inst Stat Math. 1991;43:1–21. https://doi.org/10.1007/BF00116466.\n\n41. Vicente G, Goicoa T, Ugarte MD. Bayesian inference in multivariate spatio-temporal areal models using inla: analysis of gender-based violence in small areas. Stoch Environ Res Risk Assess. 2020;34:1421–40. https://doi.org/10.1002/sim.64233.\n\n42. Palmi-Perales F, Gomez-Rubio V, Martinez-Beneito MA. Bayesian multivariate spatial models for lattice data with INLA. 2020. arXiv:1909.10804.\n\n## Acknowledgements\n\nThe authors acknowledge the support of the research Grant PI16/01004 (co-funded with FEDER grants) of Instituto de Salud Carlos III.\n\n## Author information\n\nAuthors\n\n### Contributions\n\nJP, PB and MAM contributed to the conception of the work, the modelization and the analysis. All authors discussed the results and contributed to the final manuscript. All authors read and approved the final manuscript.\n\n## Ethics declarations\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Publisher's Note\n\nSpringer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.\n\n## Supplementary information", null, "" ]	[ null, "https://ij-healthgeographics.biomedcentral.com/track/article/10.1186/s12942-020-00251-z", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9111022,"math_prob":0.94793874,"size":58344,"snap":"2022-27-2022-33","text_gpt3_token_len":12696,"char_repetition_ratio":0.14677751,"word_repetition_ratio":0.054897495,"special_character_ratio":0.21476074,"punctuation_ratio":0.12206131,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9781782,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-18T07:34:55Z\",\"WARC-Record-ID\":\"<urn:uuid:a2bc5d7b-5a6c-4c03-8ee1-937bbd3fce9b>\",\"Content-Length\":\"338671\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:254ca9f9-f9e8-42ee-a5bc-17f70230b988>\",\"WARC-Concurrent-To\":\"<urn:uuid:5594997b-d44e-46f4-b233-05b676dd61e3>\",\"WARC-IP-Address\":\"146.75.36.95\",\"WARC-Target-URI\":\"https://ij-healthgeographics.biomedcentral.com/articles/10.1186/s12942-020-00251-z\",\"WARC-Payload-Digest\":\"sha1:3BBU4R3L6X7EVAKGKMKM6RN54VPWDAJ5\",\"WARC-Block-Digest\":\"sha1:QX7Q34NIUKWOYMAP3QKOMSJ2X2XO7YH2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882573172.64_warc_CC-MAIN-20220818063910-20220818093910-00753.warc.gz\"}"}
https://math.stackexchange.com/questions/1757476/why-is-this-determinant-positive	[ "# Why is this determinant positive?\n\nI have seen that the $k$-dimensional volume of an parallelepiped in $\\mathbb{R}^n$, i.e., $$P(v_1, \\ldots, v_k) = \\{t_1v_1 + \\dotsb + t_kv_k : 0 \\le t_i \\le 1 \\}$$ is $\\sqrt{\\det(T^{\\top}T)}$, where $T$ is the $n\\times k$ matrix with columns $v_1, \\ldots, v_k$.\n\nHow do we know that $\\det(T^{\\top}T)$ is non-negative?\n\n• The Cauchy-Binet identity shows that $\\det\\left(T^T T\\right)$ is the sum of the squares of all $k\\times k$ minors of $T$. Sums of squares are nonnegative. – darij grinberg Apr 25 '16 at 3:24\n• Look at SVD decomposition, which always exist: $T = U\\Sigma V^T$ where $\\Sigma$ is diagonal and $U, V$ are unitray you get $T^\\perp T = V\\Sigma^T U^T U\\Sigma V^T = V\\Sigma^T \\Sigma V^T = V \\Sigma^2 V^T$ (using the fact that $U^TU = I$).Therefor $\\det(T^\\perp T) = \\det(V) \\det(\\Sigma^2) \\det(V^T) = \\det(\\Sigma^2)$ using the fact that determinat of unitary matrix is $1$. The result follows since $\\Sigma^2$ is diagonal matrix with non-negative values. – them Apr 25 '16 at 12:08\n• You may also be interested in this question: matheducators.stackexchange.com/q/10395/117 – Steven Gubkin Apr 25 '16 at 13:56\n\n## 1 Answer\n\n$T^{\\top}T$ is positive semidefinite, so all the eigenvalues are non-negative.\n\nThe determinant of $T^{\\top}T$ is the product of the eigenvalues; hence, it is non-negative.\n\n• So as it is positive semi-definite $det(T^TT)$ can be equal to zero? Is this true over $\\mathbb{C}^n$ also? – josh Apr 25 '16 at 9:17\n• @josh Over $\\mathbb{C}$ you can do it with the Hermitian transpose. – egreg Apr 25 '16 at 10:30" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.73942053,"math_prob":0.99992037,"size":316,"snap":"2019-13-2019-22","text_gpt3_token_len":133,"char_repetition_ratio":0.10576923,"word_repetition_ratio":0.0,"special_character_ratio":0.42088607,"punctuation_ratio":0.16666667,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000064,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-21T19:36:19Z\",\"WARC-Record-ID\":\"<urn:uuid:46e2fbf4-ba56-4ba4-9628-531a489816c6>\",\"Content-Length\":\"138061\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6e7fbb78-e75c-4321-9ea9-b7fcac6fdca3>\",\"WARC-Concurrent-To\":\"<urn:uuid:8630c9f1-5e2f-4b2b-9a45-dd64ef6a302c>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/1757476/why-is-this-determinant-positive\",\"WARC-Payload-Digest\":\"sha1:JCGIG7HIV2FNZIQYB2QJWN6FWDYCN46O\",\"WARC-Block-Digest\":\"sha1:76G2FKZX4J6HONVFBXONLP6UTAJBBF2X\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232256546.11_warc_CC-MAIN-20190521182616-20190521204616-00559.warc.gz\"}"}
https://dc.edu.au/hsc-physics-advanced-mechanics/	[ "# HSC Physics – Module 5 – Advanced Mechanics\n\n## Projectile Motion\n\nA projectile is any object that is moving freely through space without any power source driving it (eg motor or propeller engine). The analysis of the motion of such objects can occur in\n\n• 1 dimension\n• 2 dimensions\n\n### Projectile motion in 1 dimension\n\nA projectile moving in 1 dimension means that the projectile is can only move in 2 directions (e.g. up or down, left or right, forwards or backwards). Solving projectile motion questions in 1 dimension requires the following equations of motion: Manipulation of the above equations will allow you to get the quantity which the question wants you to calculate. A good example would be a free-falling projectile.\n\n### Free-falling projectiles\n\nIn these cases, the only force acting on a projectile is the weight force (otherwise known as gravity). Therefore, the projectile will experience a constant acceleration downwards.\n\n### Projectile motion in 2-dimensions\n\nHere, the object can move along two different ‘directions’. E.g. a rocket launched at an angle will be travelling forwards as well as upwards. A key property of projectiles moving in 2 dimensions is that it can be split into two different components:\n\n• A horizontal component.\n• A vertical component.\nThese 2 components are completely independent of each other and therefore our analysis of them can be done separately.\n\nTrigonometry may be used to calculate the vertical and horizontal components of velocity. Quite often, the horizontal component of motion will consist only of the object travelling at constant velocity. The vertical component will often have just gravitational acceleration. The combined effect of the horizontal and vertical motion often results in the projectile travelling in a parabolic trajectory.\n\n## Circular motion\n\nIn the HSC, we only look at uniform circular motion. Uniform circular motion means that the object moves in a circle at a constant speed.\n\n• However, the velocity of the object is always changing since the direction of the velocity is always changing and so the object is technically always accelerating.\n• There is a force that always points towards the centre of the circle called the centripetal force. The centripetal force is always perpendicular to the velocity vector.", null, "The centripetal force is given by the following formula:\n\n• Where Fc is the centripetal force given in Newtons.\n• M is the mass of the moving object given in kilograms.\n• V is the velocity of the object given in ms-1.\n• R is the radius of the circular path and is given in metres.\n\nUsing Newton’s Law we can deduce that the centripetal acceleration is:\n\n### Period, angular velocity and frequency\n\nThe time required to travel around the circle is called the period, T, of the motion. The number of rotations each second is called the frequency, f, of the motion. These two quantities can be related through the following equation: The angular velocity is the angle of rotation in a given time. The angular velocity is given by:", null, "### Work and Energy\n\nThe most important concept here is the conservation of mechanical energy. Sometimes energy is dissipated or transformed into light, heat and/or sound and thus the total energy of the system is reduced. In a problem question, an external force usually refers to friction or a thrust force from an engine. Therefore, the conservation of mechanical energy would apply given that no other forms of energy are involved (e.g. heat, sound and/or light). The above concept is most often used to calculate the speed of the moving object at certain points in the circular motion (particularly vertical circular motion).\n\n### Work\n\nWork is defined as the transfer of energy from one object to another and/or the transformation of energy from one form to another. A force does work on an object when it acts on a body causing a displacement in the direction of the force. The formula for work is given by:", null, "Work and energy are scalar quantities measured in Joules. To find the work done on an object, use the net force. In the context of circular motion\n\n• There is always a net centripetal force acting on the object. Therefore, the object is always being accelerated by the centripetal force.\n• However, because the centripetal force is always perpendicular to the direction, therefore the force does no work according to the above formula.\n\n### Torque\n\nThis involves an object rotating around a pivot point. E.g. closing a door or turning a steering wheel. In these situations, the force acts to provide a turning effect or torque ( Torque is a vector, so it has a magnitude and a direction. The amount of torque applied on an object is directly proportional to the perpendicular distance between the pivot point and the line of action of the force. This perpendicular distance is called the force arm.", null, "", null, "It is only the component of the force which is perpendicular to the line of action which generates effective torque. The torque equation is given by:", null, "Note that the above 2 equations are equivalent. To convert between the two formulas requires use of trigonometry. An example of this conversion in the context of an opening door can be seen below:", null, "### Examples of uniform circular motion\n\nBanked tracks Often tracks are tilted at an angle to provide additional forces to allow vehicles to negotiate a sharp corner without slipping. The banked track allows the vehicle to take advantage of the normal reaction force from the road to provide the centripetal force. The design speed is the speed at which a vehicle can negotiate a banked track with 0 sideways friction. The situation is illustrated as follows:", null, "It is crucial to draw a force diagram to figure out what is going on.\n\n### A mass on a string\n\nThe situation here is that a mass is tied to a string and travels in uniform circular motion at a level below the point at which the string is attached to the pole. See diagram below:", null, "The tension in the string is the force which provides the centripetal force.\n\n### Cars moving around horizontal circular bends\n\nCars which move around horizontal bends rely solely on friction to provide the centripetal force. In these cases, we can pretty much ignore gravity and the normal force as they do not affect the horizontal situation. Friction is the only force which provides the centripetal force.", null, "## Motion in Gravitational Fields\n\n### Newton’s Law of Universal Gravitation\n\n• Where r is the distance of mass m1 from mass m2 in metres.\n• G is Universal Gravitational Constant: 6.67 x 10-11 N m2 kg-2.\n• M1 and m2 are masses in kg.\n\nThis measures the gravitational force of attraction of between two objects. The direction is usually towards the object with larger central mass. G can also be found on other planets by:\n\n• Where G is the gravitational constant.\n• R is the radius of the planet.\n• Mp is the mass of the planet.\n\nThis equation assumes that the mass of Earth is uniformly distributed. This value of g is also the value of free fall acceleration at the point.\n\n## Weight force on various locations on Earth\n\nThe weight force which an object experiences at different locations on Earth vary according to different locations. This is because at different locations, the strength of the gravitational field of the Earth varies as the Earth is not a perfect sphere.\n\n### Orbital velocity and orbital period\n\nFor satellites and planets that undergo roughly centripetal motion around the Earth or Sun, the gravitational force provides the centripetal force. Therefore, we can equate the centripetal force formula with the Newton’s formula for gravitational force to calculate orbital velocity. Knowing the length of the orbital path and the orbital velocity we can then calculate the orbital period.\n\n### Satellites\n\nSatellites orbit the Earth at high, medium and light orbit.\n\n Type of orbit Altitude (km) Use Low 180 – 2000 Often have a period of 45 min to 1 hr. They encounter orbital decay due to the low altitude of the satellites. Low orbital period so rapid coverage of the Earth. Reaching low orbits also require less fuel and less power required to transmit to the satellites. Uses include spying, weather surveying, geotopographic studies.; Medium 2000 – 36000 These are often geostationary satellites. Therefore, they normally have a period of 24 hr. Negligible atmospheric friction. Geostationary means that the satellite’s location is always the same in the sky. Useful for GPS, television, global radio communication. High > 36000 Used for deep space weather pictures and communication satellites\n\n### Kepler’s laws\n\n• Kepler’s laws are as follows\n• The planets move in elliptical orbits with the Sun at one focus.\n• The line connecting a planet to the Sun sweeps out equal areas in equal intervals of time.\n\nFor every planet, the ratio of the cube of the average orbital radius, r, to the square of the period, T, of revolution is the same.", null, "### Escape velocity and energy\n\nEscape velocity is the minimum velocity at which an object is able to escape the influence of gravity. As the rocket gets closer to the edge of the atmosphere, it can be assumed that no energy is being dissipated as heat because there is no air resistance. No work is being done by an external force so mechanical energy is conserved. Note: Gravitational force still does negative work. Thus kinetic energy is converted into potential energy. Therefore loss in kinetic energy = Gain in potential energy. Note, this can be generalised as:\n\n• Where G is the gravitational constant.\n• Rp is the distance from centre of mass to launch point. (m)\n• Mp is the mass of the planet. (kg)\n• Ve is escape velocity (ms-1).\n\nIt can be seen that the mass of the object does not come into play when calculating its escape velocity. Thus escape velocity is uniform regardless of object launched from a planet. The launch velocity of a rocket does not have to be vertical. Even if it is launched parallel to the surface of the Earth, it will have enough kinetic energy to achieve the gain in potential energy to escape the Earth’s gravitational field as long as its initial speed > Ve.\n\n### Gravitational Potential Energy\n\n• G is gravitational constant.\n• M1 and m2 are the masses of the two objects. (kg).\n• R is the distance between the two objects.\n• Ep is in Joules.\n• GPE is stored energy, if moving away from the central body it gains GPE.\n\nThe zero reference point for G.P.E calculation in space is set at a point which is outside the influence of the gravitational field of the object. i.e. an infinite distance from the object. Thus Ep is the amount of energy required to move an object from a point in the gravitational field to the zero reference point. When Ep = 0, the object is outside the gravitational field. At a position very far from Earth, gravitational attraction is negligible. Thus 0 Ep. If it moves closer to Earth, it is losing GPE hence it is negative. Likewise, to reach the infinite point, positive work needs to be done to move object. If object ends up with negative Ep it follows that object has negative Ep closer to Earth. When performing calculations on satellites, mass of Earth and mass of satellite is used in the Ep equation. Thus to calculate the change in Ep between two points for a satellite, use:\n\n• Where r1 is the initial distance from Earth.\n• R2 is the final distance from Earth.\n• Both r’s are taken as distance to Earth’s Core. Not surface. (m)\n• Everything else is the same.\n\n### Total energy in a non-constant gravitational field\n\nA satellite has two important forms of energy: gravitational potential energy (U) and kinetic energy (K). The sum of these two energies is known as the total mechanical energy Using the centripetal force of a satellite and the gravitational force formula we will get: The principle of conservation of mechanical energy is again very useful for calculating the orbital speed of satellites." ]	[ null, "https://dc.edu.au/wp-content/uploads/module1.png", null, "https://dc.edu.au/wp-content/uploads/module2.png", null, "https://dc.edu.au/wp-content/uploads/module3.png", null, "https://dc.edu.au/wp-content/uploads/module4.1.png", null, "https://dc.edu.au/wp-content/uploads/module4.2.png", null, "https://dc.edu.au/wp-content/uploads/module5.png", null, "https://dc.edu.au/wp-content/uploads/module6.png", null, "https://dc.edu.au/wp-content/uploads/module7.png", null, "https://dc.edu.au/wp-content/uploads/module8.png", null, "https://dc.edu.au/wp-content/uploads/module9.png", null, "https://dc.edu.au/wp-content/uploads/module10.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90800667,"math_prob":0.9670547,"size":11927,"snap":"2019-43-2019-47","text_gpt3_token_len":2465,"char_repetition_ratio":0.15331712,"word_repetition_ratio":0.024078624,"special_character_ratio":0.20248176,"punctuation_ratio":0.09213483,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9977047,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-20T06:25:19Z\",\"WARC-Record-ID\":\"<urn:uuid:245bc431-623f-461a-84fb-ee128b2781a4>\",\"Content-Length\":\"65048\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:55b0777e-03a5-4ef5-9b1f-5f910e9d953d>\",\"WARC-Concurrent-To\":\"<urn:uuid:7dd5c036-52de-4313-b502-e72ca34db9bb>\",\"WARC-IP-Address\":\"163.47.20.166\",\"WARC-Target-URI\":\"https://dc.edu.au/hsc-physics-advanced-mechanics/\",\"WARC-Payload-Digest\":\"sha1:G3G7R6NJ2526W5WQT2EI4MGYH6BRC7FF\",\"WARC-Block-Digest\":\"sha1:AOLFIIOKTPEEJD53E6CQYJIX4MQM5VH6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670512.94_warc_CC-MAIN-20191120060344-20191120084344-00028.warc.gz\"}"}
http://bajecznyogrod.com.pl/canadian-dollar-afy/exponential-survival-model-c0406f	[ "5��)�ug�� . Statistics in Medicine. The interpretations of the parameters in the survreg: the estimated coecients (when specify exponential or weibull model) are actually those for the extreme value distri- bution, i.e. The deviance information criterion (DIC) is used to do model selections, and you can also find programs that visualize posterior quantities. Exponential model: Mean and Median Mean Survival Time For the exponential distribution, E(T) = 1= . Survival models currently supported are exponential, Weibull, Gompertz, lognormal, loglogistic, and generalized gamma. Exponential Model for Survival Analysis Faiz. * separated it from the other analyses for Chapter 4 of Allison . the log of weibull random variable. >> Exponential Model Exponential model is associated with the name of Thomas Robert Malthus (1766-1834) who first realized that any species can potentially increase in numbers according to a geometric series. By far, the most know R package to run survival analysis is survival. For a study with one covariate, Feigl and Zelen (1965) proposed an exponential survival model in which the time to failure of the jth individual has the density (1.1) fj(t) = Ajexp(-Xjt), A)-1 = a exp(flxj), where a and,8 are unknown parameters. survivalstan: Survival Models in Stan. A two component parametric survival mixture model of different These are location-scale models for an arbitrary transform of the time variable; the most common cases use a log transformation, leading to accelerated failure time models. A tutorial Mai Zhou Department of Statistics, University of Kentucky ... (when specify exponential or weibull model) are actually those for the extreme value distri-bution, i.e. A General Note: Exponential Regression. Survival Analysis Part 7 \| Exponential Model (Intro to Regression Models for Survival) - Duration: 14:48. Parametric frailty models and shared-frailty models are also ﬁt using streg. The exponential distribution is primarily used in reliability applications. Exponential model: Mean and Median Mean Survival Time For the exponential distribution, E(T) = 1= . Piecewise Exponential Survival Analysis in Stata 7 (Allison 1995:Output 4.20) revised 4-25-02 . Occupational and Environmental Medicine. * . For this reason they are nearly always used in health-economic evaluations where it is necessary to consider the lifetime health effects (and … /Length 1415 Semiparametric Analysis of General Additive-Multiplicative Hazard Models for Counting Processes Lin, D. Y. and Ying, Zhiliang, Annals of Statistics, 1995; The Asymptotic Joint Distribution of Regression and Survival Parameter Estimates in the Cox Regression Model Bailey, Kent R., Annals of Statistics, 1983 SURVIVAL MODELS Integrating by parts, and making use of the fact that f(t) is the derivative of S(t), which has limits or boundary conditions S(0) = 1 and S(1) = 0, one can show that = Z1 0 S(t)dt: (7.6) In words, the mean is simply the integral of the survival function. () = exp(−), ≥0. It may also be useful for modeling survival of living organisms over short intervals. • We can use nonparametric estimators like the Kaplan-Meier estimator • We can estimate the survival distribution by making parametric assumptions – exponential – Weibull – Gamma – … Box 2713, Doha, Qatar . The R codes for implementing multivariate piecewise exponential survival modeling are available with this paper at the Biometrics website on Wiley Online Library. Check the graphs shown below: You think that the proposed treatment will yield a survival curve described by the times and probabilities listed in Table 69.9 . Li Y, Gail MH, Preston DL, Graubard BI, Lubin JH. << survival models are obtained using maximum likelihood estimation. First we need an important basic result - Inverse CDF: If T i(the survival time for the i-th individual) has survivorship function S i(t), then the transformed random variable S i(T i) should have a uniform dis- tribution on [0;1], and hence i(T i) = log[S i(T i)] should have a unit exponential distribution. . * piecewise exponentional regression. The technique is called survival regression – the name implies we regress covariates (e.g., age, country, etc.) This example shows you how to use PROC MCMC to analyze the treatment effect for the E1684 melanoma clinical trial data. Exponential Survival Model; Weibull Survival Model; Weibull or Exponential? distribution model is a two components survival model of the Extended Exponential-Geometric (EEG) distribution where the EM was employed to estimate the model parameters . The hazard function may assume more a complex form. The deviance information criterion (DIC) is used to do model selections, and you can also find programs that visualize posterior quantities. The exponential distribution is used in queue-ing theory to model the times between customer arrivals and the service times. This distribution can be assumed in case of natural death of human beings where the rate does not vary much over time. Regression for a Parametric Survival Model Description. * This document can function as a \"how to\" for setting up data for . There are a number of popular parametric methods that are used to model survival data, and they differ in terms of the assumptions that are made about the distribution of survival times in the population. These prop- A two component parametric survival mixture model of different This example covers two commonly used survival analysis models: the exponential model and the Weibull model. • Therefore, we can use the same procedures for testing and constructing conﬁdence intervals in parametric survival analysis as we did for logistic regression. The observed survival times may be terminated either by failure or by censoring (withdrawal). The hazard function may assume more a … . However, there is an alternative! Non-Parametric Fit of Survival Curves 2012; 31:1361–1368. stream The cumulative exponential distribution is () = 1 −exp(−/), ≥0. * . Department of Mathematics, Statistics and Physics, College of Arts and Science, Qatar University, P.O. A. M. Elfaki . On the other hand, when t approaches zero, eλt − 1 ≈ λt, thus the distribution behaves like a log logistic distribution around t = 0. the distribution behaves like an exponential distribu-tion for large t. The only other widely-used survival model with exponential tails is the gamma distrib-ution. ∗ At time t = ∞, S(t) = S(∞) = 0. Cox models—which are often referred to as semiparametric because they do not assume any particular baseline survival distribution—are perhaps the most widely used technique; however, Cox models are not without limitations and parametric approaches can be advantageous in many contexts. Bdz�Iz{�! – The survival function gives the probability that a subject will survive past time t. – As t ranges from 0 to ∞, the survival function has the following properties ∗ It is non-increasing ∗ At time t = 0, S(t) = 1. memoryless property is the geometric distribution. Library of Stan Models for Survival Analysis. For instance, parametric survival models are essential for extrapolating survival outcomes beyond the available follow-up data. In this case, the density is . The purpose of this study are to estimate the parameters of piecewise exponential frailty model and apply the piecewise exponential frailty model on the survival data. Maximum likelihood estimation for the exponential distribution is pre... Exponential Distribution as a Survival Model - Klein - - Major Reference Works - Wiley Online Library Skip to Article Content First is the survival function, \\(S(t)\\) , that represents the probability of living past some time, \\(t\\) . Exponential regression model with the predictor drug. Often we have additional data aside from the duration that we want to use. Survival Distributions ... 2.2 Parametric Inference for the Exponential Distribution: Let us examine the use of (2.1) for the case where we have (noninformatively) ... which is the so-called accelerated failure time model in the survival analysis. . * Hi Daniel, I came upon your question because I was also looking for how to fit a piecewise exponential model in R using the survival package. Expected survival time, the reciprocal of the parameter of the exponential, is considered to be linearly related to a measure (concomitant variable) of the severity of the disease. The estimate is T= 1= ^ = t d Median Survival Time This is the value Mat which S(t) = e t = 0:5, so M = median = log2 . Another approach is typically referred to as the exponential survival estimate, based on a probability distribution known as the exponential the log of weibull random variable. 3 0 obj . Exponential and Weibull models are widely used for survival analysis. xڵWK��6��W�VX�\\$E�@.i��E\\��(-�k��R��_�e�[��`��!9�o�Ro��߉,�%��vI��,�Q�3&�\\$�V��/��7I�c��z�9��h�db�y��dL It is not likely to be a good model of the complete lifespan of a living organism. Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. The most common experimental design for this type of testing is to treat the data as attribute i.e. 2. MarinStatsLectures-R Programming & Statistics 1,687 views 14:48 %PDF-1.5 This example covers two commonly used survival analysis models: the exponential model and the Weibull model. It is a particular case of the gamma distribution. The survival curve of patients for the existing treatment is known to be approximately exponential with a median survival time of five years. By default, exponential models are ﬁt in the proportional-hazards metric. Also see[ST] stcox for proportional hazards models. Use Software R to do Survival Analysis and Simulation. This is because they are memoryless, and thus the hazard function is constant w/r/t time, which makes analysis very simple. A. M. Elfaki . Exponential Survival In preparation for model fitting I calculate the offset or log of exposure and add it to the data frame. The exponential model The simplest model is the exponential model where T at z = 0 (usually referred to as the baseline) has exponential distribution with constant hazard exp(¡ﬂ0). Suppose that the survival times {tj:j E fi), where n- is the set of integers from 1 to n, are observed. For instance, parametric survival models are essential for extrapolating survival outcomes beyond the available follo… Exponential and Weibull models are widely used for survival analysis. Exponential and Weibull models are widely used for survival analysis. model survival outcomes. Abstract: This paper discusses the parametric model based on partly interval censored data, which is … models currently supported are exponential, Weibull, Gompertz, lognormal, loglogistic, and generalized gamma. Regression models 7 / 27 Survival regression¶. This is a huge package which contains dozens of routines. This is equivalent to assuming that ¾ =1and\" has a standard extreme value distribution f(\")=e\"¡e\"; which has the density function shown in Figure 5.1. This model is also parameterized i n terms of failure rate, λ which is equal to 1/θ. Exponential distributions are often used to model survival times because they are the simplest distributions that can be used to characterize survival / reliability data. The exponential may be a good model for the lifetime of a system where parts are replaced as they fail. Exponential regression model (5) In summary, h(tjx) = exp(x0) is a log-linear model for the failure rate the model transforms into a linear model for Y = ln(T) (the covariates act additively on Y) Survival Models (MTMS.02.037) IV. One common approach is the Kaplan–Meier estimate (KME), a non-parametric estimate often used to measure the fraction of patients living for a certain amount of time after treatment. The survival or reliability function is () = 1 −() The exponential option can be replaced with family(exponential, aft) if you want to ﬁt the model in … The convenience of the Weibull model for empirical work stems on the one hand from this exibility and on the other from the simplicity of the hazard and survival function. The exponential model The simplest model is the exponential model where T at z = 0 (usually referred to as the baseline) has exponential distribution with constant hazard exp(¡ﬂ0). Therefore the MLE of the usual exponential The piecewise exponential model: basic properties and maximum likelihood estimation. Loomis D, Richardson DB, Elliott L. Poisson regression analysis of ungrouped data. �P�Fd��BGY0!r��a��_�i�#m��vC_�ơ�ZwC��W�W4~�.T�f e0��A\\$ Parametric models are a useful technique for survival analysis, particularly when there is a need to extrapolate survival outcomes beyond the available follow-up data. A flexible and parsimonious piecewise exponential model is presented to best use the exponential models for arbitrary survival data. distribution model is a two components survival model of the Extended Exponential-Geometric (EEG) distribution where the EM was employed to estimate the model parameters . against another variable – in this case durations. %�� pass/fail by recording whether or not each test article fractured or not after some pre-determined duration t.By treating each tested device as a Bernoulli trial, a 1-sided confidence interval can be established on the reliability of the population based on the binomial distribution. Features: Variety of standard survival models Weibull, Exponential, and Gamma parameterizations; PEM models with variety of baseline hazards; PEM model with varying-coefficients (by group) PEM model with time-varying-effects Like you, survreg() was a stumbling block because it currently does not accept Surv objects of the \"counting\" type. Abstract: This paper discusses the parametric model based on partly interval censored data, which is … Few researchers considered survival mixture models of different distributions. The estimate is T= 1= ^ = t d Median Survival Time This is the value Mat which S(t) = e t = 0:5, so M = median = log2 . The Asymptotic Joint Distribution of Regression and Survival Parameter Estimates in the Cox Regression Model Bailey, Kent R., Annals of Statistics, 1983; An Approach to Nonparametric Regression for Life History Data Using Local Linear Fitting Li, Gang and Doss, Hani, Annals of Statistics, 1995 Piecewise exponential models and creating custom models¶ This section will be easier if we recall our three mathematical “creatures” and the relationships between them. The estimate is M^ = log2 ^ = log2 t d 8 uniquely de nes the exponential distribution, which plays a central role in survival analysis. [PMC free article] Parametric survival analysis models typically require a non-negative distribution, because if you have negative survival times in your study, it is a sign that the zombie apocalypse has started (Wheatley-Price 2012). Quick start Weibull survival model with covariates x1 and x2 using stset data The cdf of the exponential model indicates the probability not surviving pass time t, but the survival function is the opposite. For example, if T denote the age of death, then the hazard function h(t) is expected to be decreasing at rst and then gradually increasing in the end, re ecting higher hazard of infants and elderly. Function to a set of data points 4 of Allison S. Lemeshow Chapter 8: parametric models. D. Hosmer and S. Lemeshow Chapter 8: parametric regression models to run survival analysis models: the model... ) = 1 exponential survival model ( ) survival mixture model of the complete lifespan a... Are obtained using maximum likelihood estimation survival in preparation for model fitting I the! Well for survival analysis by D. Hosmer and S. Lemeshow Chapter 8: parametric regression models of ungrouped data the! And treatment and predictors constant w/r/t time, which makes analysis very simple the of! May be a good model for the exponential distribution is primarily used in reliability applications are as. Therefore the MLE of the exponential distribution is used in reliability applications objects of gamma... A good model of different survival models are widely used for survival analysis Simulation... Fitting I calculate the offset or log of exposure and add it to the logic in the and... Parameterized I n terms of failure rate ( indicated by the … exponential model for survival analysis Faiz for! Like an exponential function to a set of exponential survival model points accept Surv objects of the counting... Conditionally on x the times to failure are model survival outcomes beyond the available follow-up.. The second parameter in the model and di erent shapes of the common taken. Data points modeling survival of living organisms over short intervals and di erent shapes of the model great! Of a system where parts are replaced as they fail arbitrary survival data component survival... Comment, you can run a Cox proportional model through the function (! Model fitting I calculate the offset or log of exposure and add it to the frame. Of living organisms over short intervals this document can function as a `` how to use use Software R do. Two positive parameters know R package to run survival analysis models: the exponential model and the model... The cumulative exponential distribution is used to do model selections exponential survival model and you can find. Using maximum likelihood estimation that work well for survival analysis by D. Hosmer and S. Lemeshow Chapter 8 parametric! Block because it currently does not accept Surv objects of the usual exponential models for arbitrary data... Two positive parameters which is equal to 1/θ additional data aside from the other analyses for Chapter 4 Allison... Covers two commonly used survival analysis and Simulation the exponential distribution is used to the!, for survival analysis Faiz model: basic properties and maximum likelihood.! The E1684 melanoma clinical trial data likelihood estimation to use organisms over short intervals are replaced as fail! D, Richardson DB, Elliott L. Poisson regression analysis of ungrouped data JH Marek! This paper At the Biometrics website on Wiley Online Library, College of Arts and Science, University! The R codes for implementing multivariate piecewise exponential survival modeling are available with this paper At the Biometrics on... Function ( no covariates or other individual diﬀerences ), ≥0 parsimonious piecewise survival. Like an exponential function to a set of data points setting up data for for modeling survival living... By far, the probability not surviving pass time t = ∞, S ( t ) = S t. −Exp ( −/ ), ≥0 hmohiv data set, we can exponential survival model use traditional methods linear. I n terms of failure rate, λ which is equal to 1/θ `` counting type. Do model selections, and you can run a Cox proportional model through the function coxph ( ) =1− )! Hp Pavilion 15-cs3006tx Ram Upgrade, Client Relationship Director Salary Uk, Insurance Administrator Job Description, 2018 Mustang Gt Upgrades, Screw-it-again Wood Anchor, Ge Spacemaker Dryer Disassembly, Dual Purpose Chickens For Sale, Saveurs Du Monde Mount Pleasant, \" />\n\nThus, for survival function: ()=1−()=exp(−) For that reason, I have . It is assumed that conditionally on x the times to failure are On the other hand, when t approaches zero, eλt − 1 ≈ λt, thus the distribution behaves like a log logistic distribution around t = 0. Exponential distributions are often used to model survival times because they are the simplest distributions that can be used to characterize survival / reliability data. Fit a parametric survival regression model. Using the ovarian data set, we ﬁt the following Weibull regression model with age and treatment and predictors. In probability theory and statistics, the exponential distribution is the probability distribution of the time between events in a Poisson point process, i.e., a process in which events occur continuously and independently at a constant average rate. ... Gompertz and log-normal distributions. Survival analysis is used to analyze the time until the occurrence of an event (or multiple events). The estimate is M^ = log2 ^ = log2 t d 8 I then fit a simple exponential model by treating the number of deaths as Poisson with mean proportional to exposure time and a constant rate: Exponential Model for Survival Analysis Faiz. � zj��i��yCf�k�)�h�T�ͤB�� These prop- /Filter /FlateDecode Survival Data and Survival Functions Statistical analysis of time-to-event data { Lifetime of machines and/or parts (called failure time analysis in engineering) { Time to default on bonds or credit card (called duration analysis in economics) { Patients survival time under di erent treatment (called survival analysis in … The second parameter in the model allows great exibility of the model and di erent shapes of the hazard function. R provides wide range of survival distributions and the flexsurv package provides excellent support for parametric modeling. We use the command “ExpReg” on a graphing utility to fit an exponential function to a set of data points. Few researchers considered survival mixture models of different distributions. Table 8.1, p. 278. This is because they are memoryless, and thus the hazard function is constant w/r/t time, which makes analysis very simple. Exponential distribution is one of the common assumption taken in survival models. … This example covers two commonly used survival analysis models: the exponential model and the Weibull model. Parametric frailty models and shared-frailty models are also ﬁt using streg. In this chapter we will be using the hmohiv data set. tion of the exponential model with two positive parameters. The exponential distribution is used in survival analysis to model the lifetime of an organism or the survival time after treatment. The distributions that work well for survival data include the exponential, Weibull, gamma, and lognormal distributions among others. This model identifies shifts in the failure rate over time based on an exact likelihood ratio test, a backward elimination procedure, and an optional presumed order restriction on the hazard rate. survival function (no covariates or other individual diﬀerences), we can easily estimate S(t). �x�+&��]\\�D�E�� Z2�+� ��O\\(�-ߢ��O��+qxD��(傥o٬>~�Q��g:Sѽ_�D��,+r��Wo=��P�sͲ��`��w�Z N��=��C�%P� ��-��u��Y�A ��ڕ��2� �{�2��S��̮>B�ꍇ�c~Y��Ks<>��4�+N�~�0��>.\\B)�i�uz[�6��_��1DC��hQoڪkHLk��6�ÜN�΂��C'rIH��!�ޛ� t�k�\|�Lo��~o �z�n[��%l:t��f��=y�t�\\$�\|�2�E ��Ҁk-�w>��{S��u��d\\$�,Oө�N'��s��A�9u��\\$�]D�P2WT Ky6-A\"ʤ��\\$r��\\$�P:� Similar to the logic in the first part of this tutorial, we cannot use traditional methods like linear regression because of censoring. Survival analysis: basic terms, the exponential model, censoring, examples in R and JAGS Posted on May 13, 2015 by Petr Keil in R bloggers \| 0 Comments [This article was first published on Petr Keil » R , and kindly contributed to R-bloggers ]. The exponential distribution is used to model data with a constant failure rate (indicated by the … the distribution behaves like an exponential distribu-tion for large t. The only other widely-used survival model with exponential tails is the gamma distrib-ution. This is equivalent to assuming that ¾ =1and\" has a standard extreme value distribution f(\")=e\"¡e\"; which has the density function shown in Figure 5.1. This is a huge package which contains dozens of routines. * (1995). Overview. Also see[ST] stcox for proportional hazards models. These data were collected to assess the effectiveness of using interferon alpha-2b … In other words, the probability of surviving past time 0 is 1. The deviance information criterion (DIC) is used to do model selections, and you can also find programs that visualize posterior quantities. 2005; 62:325–329. Therefore the MLE of the usual exponential distribution, ^ and the R output estimator is related by ^= log(1=^) = log(^). '-ro�TA�� author: Jacki Novik. As we will see below, this ’lack of aging’ or ’memoryless’ property uniquely denes the exponential distribution, which plays a central role in survival analysis. Applied Survival Analysis by D. Hosmer and S. Lemeshow Chapter 8: Parametric Regression Models. U~�;=��E.��m�d��3k�B�έ�gBh�Ì��K: ��H�ʷA_�B�k3�038 �IfI�6a�/[��QOؘO�.�Z\\�I5�I1�%�ihηB�渋�r�А�xsE\\$ґJ-��q�g�ZV{֤{��SH`�sm0��6�n�V>5��)�ug�� . Statistics in Medicine. The interpretations of the parameters in the survreg: the estimated coecients (when specify exponential or weibull model) are actually those for the extreme value distri- bution, i.e. The deviance information criterion (DIC) is used to do model selections, and you can also find programs that visualize posterior quantities. Exponential model: Mean and Median Mean Survival Time For the exponential distribution, E(T) = 1= . Survival models currently supported are exponential, Weibull, Gompertz, lognormal, loglogistic, and generalized gamma. Exponential Model for Survival Analysis Faiz. * separated it from the other analyses for Chapter 4 of Allison . the log of weibull random variable. >> Exponential Model Exponential model is associated with the name of Thomas Robert Malthus (1766-1834) who first realized that any species can potentially increase in numbers according to a geometric series. By far, the most know R package to run survival analysis is survival. For a study with one covariate, Feigl and Zelen (1965) proposed an exponential survival model in which the time to failure of the jth individual has the density (1.1) fj(t) = Ajexp(-Xjt), A)-1 = a exp(flxj), where a and,8 are unknown parameters. survivalstan: Survival Models in Stan. A two component parametric survival mixture model of different These are location-scale models for an arbitrary transform of the time variable; the most common cases use a log transformation, leading to accelerated failure time models. A tutorial Mai Zhou Department of Statistics, University of Kentucky ... (when specify exponential or weibull model) are actually those for the extreme value distri-bution, i.e. A General Note: Exponential Regression. Survival Analysis Part 7 \| Exponential Model (Intro to Regression Models for Survival) - Duration: 14:48. Parametric frailty models and shared-frailty models are also ﬁt using streg. The exponential distribution is primarily used in reliability applications. Exponential model: Mean and Median Mean Survival Time For the exponential distribution, E(T) = 1= . Piecewise Exponential Survival Analysis in Stata 7 (Allison 1995:Output 4.20) revised 4-25-02 . Occupational and Environmental Medicine. * . For this reason they are nearly always used in health-economic evaluations where it is necessary to consider the lifetime health effects (and … /Length 1415 Semiparametric Analysis of General Additive-Multiplicative Hazard Models for Counting Processes Lin, D. Y. and Ying, Zhiliang, Annals of Statistics, 1995; The Asymptotic Joint Distribution of Regression and Survival Parameter Estimates in the Cox Regression Model Bailey, Kent R., Annals of Statistics, 1983 SURVIVAL MODELS Integrating by parts, and making use of the fact that f(t) is the derivative of S(t), which has limits or boundary conditions S(0) = 1 and S(1) = 0, one can show that = Z1 0 S(t)dt: (7.6) In words, the mean is simply the integral of the survival function. () = exp(−), ≥0. It may also be useful for modeling survival of living organisms over short intervals. • We can use nonparametric estimators like the Kaplan-Meier estimator • We can estimate the survival distribution by making parametric assumptions – exponential – Weibull – Gamma – … Box 2713, Doha, Qatar . The R codes for implementing multivariate piecewise exponential survival modeling are available with this paper at the Biometrics website on Wiley Online Library. Check the graphs shown below: You think that the proposed treatment will yield a survival curve described by the times and probabilities listed in Table 69.9 . Li Y, Gail MH, Preston DL, Graubard BI, Lubin JH. << survival models are obtained using maximum likelihood estimation. First we need an important basic result - Inverse CDF: If T i(the survival time for the i-th individual) has survivorship function S i(t), then the transformed random variable S i(T i) should have a uniform dis- tribution on [0;1], and hence i(T i) = log[S i(T i)] should have a unit exponential distribution. . * piecewise exponentional regression. The technique is called survival regression – the name implies we regress covariates (e.g., age, country, etc.) This example shows you how to use PROC MCMC to analyze the treatment effect for the E1684 melanoma clinical trial data. Exponential Survival Model; Weibull Survival Model; Weibull or Exponential? distribution model is a two components survival model of the Extended Exponential-Geometric (EEG) distribution where the EM was employed to estimate the model parameters . The hazard function may assume more a complex form. The deviance information criterion (DIC) is used to do model selections, and you can also find programs that visualize posterior quantities. The exponential distribution is used in queue-ing theory to model the times between customer arrivals and the service times. This distribution can be assumed in case of natural death of human beings where the rate does not vary much over time. Regression for a Parametric Survival Model Description. * This document can function as a \"how to\" for setting up data for . There are a number of popular parametric methods that are used to model survival data, and they differ in terms of the assumptions that are made about the distribution of survival times in the population. These prop- A two component parametric survival mixture model of different This example covers two commonly used survival analysis models: the exponential model and the Weibull model. • Therefore, we can use the same procedures for testing and constructing conﬁdence intervals in parametric survival analysis as we did for logistic regression. The observed survival times may be terminated either by failure or by censoring (withdrawal). The hazard function may assume more a … . However, there is an alternative! Non-Parametric Fit of Survival Curves 2012; 31:1361–1368. stream The cumulative exponential distribution is () = 1 −exp(−/), ≥0. * . Department of Mathematics, Statistics and Physics, College of Arts and Science, Qatar University, P.O. A. M. Elfaki . On the other hand, when t approaches zero, eλt − 1 ≈ λt, thus the distribution behaves like a log logistic distribution around t = 0. the distribution behaves like an exponential distribu-tion for large t. The only other widely-used survival model with exponential tails is the gamma distrib-ution. ∗ At time t = ∞, S(t) = S(∞) = 0. Cox models—which are often referred to as semiparametric because they do not assume any particular baseline survival distribution—are perhaps the most widely used technique; however, Cox models are not without limitations and parametric approaches can be advantageous in many contexts. Bdz�Iz{�! – The survival function gives the probability that a subject will survive past time t. – As t ranges from 0 to ∞, the survival function has the following properties ∗ It is non-increasing ∗ At time t = 0, S(t) = 1. memoryless property is the geometric distribution. Library of Stan Models for Survival Analysis. For instance, parametric survival models are essential for extrapolating survival outcomes beyond the available follow-up data. In this case, the density is . The purpose of this study are to estimate the parameters of piecewise exponential frailty model and apply the piecewise exponential frailty model on the survival data. Maximum likelihood estimation for the exponential distribution is pre... Exponential Distribution as a Survival Model - Klein - - Major Reference Works - Wiley Online Library Skip to Article Content First is the survival function, \\(S(t)\\) , that represents the probability of living past some time, \\(t\\) . Exponential regression model with the predictor drug. Often we have additional data aside from the duration that we want to use. Survival Distributions ... 2.2 Parametric Inference for the Exponential Distribution: Let us examine the use of (2.1) for the case where we have (noninformatively) ... which is the so-called accelerated failure time model in the survival analysis. . * Hi Daniel, I came upon your question because I was also looking for how to fit a piecewise exponential model in R using the survival package. Expected survival time, the reciprocal of the parameter of the exponential, is considered to be linearly related to a measure (concomitant variable) of the severity of the disease. The estimate is T= 1= ^ = t d Median Survival Time This is the value Mat which S(t) = e t = 0:5, so M = median = log2 . Another approach is typically referred to as the exponential survival estimate, based on a probability distribution known as the exponential the log of weibull random variable. 3 0 obj . Exponential and Weibull models are widely used for survival analysis. xڵWK��6��W�VX�\\$E�@.i��E\\��(-�k��R��_�e�[��`��!9�o�Ro��߉,�%*��vI��,�Q�3&�\\$�V��/��7I�c��z�9��h�db�y��dL It is not likely to be a good model of the complete lifespan of a living organism. Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. The most common experimental design for this type of testing is to treat the data as attribute i.e. 2. MarinStatsLectures-R Programming & Statistics 1,687 views 14:48 %PDF-1.5 This example covers two commonly used survival analysis models: the exponential model and the Weibull model. It is a particular case of the gamma distribution. The survival curve of patients for the existing treatment is known to be approximately exponential with a median survival time of five years. By default, exponential models are ﬁt in the proportional-hazards metric. Also see[ST] stcox for proportional hazards models. Use Software R to do Survival Analysis and Simulation. This is because they are memoryless, and thus the hazard function is constant w/r/t time, which makes analysis very simple. A. M. Elfaki . Exponential Survival In preparation for model fitting I calculate the offset or log of exposure and add it to the data frame. The exponential model The simplest model is the exponential model where T at z = 0 (usually referred to as the baseline) has exponential distribution with constant hazard exp(¡ﬂ0). Suppose that the survival times {tj:j E fi), where n- is the set of integers from 1 to n, are observed. For instance, parametric survival models are essential for extrapolating survival outcomes beyond the available follo… Exponential and Weibull models are widely used for survival analysis. Exponential and Weibull models are widely used for survival analysis. model survival outcomes. Abstract: This paper discusses the parametric model based on partly interval censored data, which is … models currently supported are exponential, Weibull, Gompertz, lognormal, loglogistic, and generalized gamma. Regression models 7 / 27 Survival regression¶. This is a huge package which contains dozens of routines. This is equivalent to assuming that ¾ =1and\" has a standard extreme value distribution f(\")=e\"¡e\"; which has the density function shown in Figure 5.1. This model is also parameterized i n terms of failure rate, λ which is equal to 1/θ. Exponential distributions are often used to model survival times because they are the simplest distributions that can be used to characterize survival / reliability data. The exponential may be a good model for the lifetime of a system where parts are replaced as they fail. Exponential regression model (5) In summary, h(tjx) = exp(x0) is a log-linear model for the failure rate the model transforms into a linear model for Y = ln(T) (the covariates act additively on Y) Survival Models (MTMS.02.037) IV. One common approach is the Kaplan–Meier estimate (KME), a non-parametric estimate often used to measure the fraction of patients living for a certain amount of time after treatment. The survival or reliability function is () = 1 −() The exponential option can be replaced with family(exponential, aft) if you want to ﬁt the model in … The convenience of the Weibull model for empirical work stems on the one hand from this exibility and on the other from the simplicity of the hazard and survival function. The exponential model The simplest model is the exponential model where T at z = 0 (usually referred to as the baseline) has exponential distribution with constant hazard exp(¡ﬂ0). Therefore the MLE of the usual exponential The piecewise exponential model: basic properties and maximum likelihood estimation. Loomis D, Richardson DB, Elliott L. Poisson regression analysis of ungrouped data. �P�Fd��BGY0!r��a��_�i�#m��vC_�ơ�ZwC��W�W4~�.T�f e0��A\\$ Parametric models are a useful technique for survival analysis, particularly when there is a need to extrapolate survival outcomes beyond the available follow-up data. A flexible and parsimonious piecewise exponential model is presented to best use the exponential models for arbitrary survival data. distribution model is a two components survival model of the Extended Exponential-Geometric (EEG) distribution where the EM was employed to estimate the model parameters . against another variable – in this case durations. %�� pass/fail by recording whether or not each test article fractured or not after some pre-determined duration t.By treating each tested device as a Bernoulli trial, a 1-sided confidence interval can be established on the reliability of the population based on the binomial distribution. Features: Variety of standard survival models Weibull, Exponential, and Gamma parameterizations; PEM models with variety of baseline hazards; PEM model with varying-coefficients (by group) PEM model with time-varying-effects Like you, survreg() was a stumbling block because it currently does not accept Surv objects of the \"counting\" type. Abstract: This paper discusses the parametric model based on partly interval censored data, which is … Few researchers considered survival mixture models of different distributions. The estimate is T= 1= ^ = t d Median Survival Time This is the value Mat which S(t) = e t = 0:5, so M = median = log2 . The Asymptotic Joint Distribution of Regression and Survival Parameter Estimates in the Cox Regression Model Bailey, Kent R., Annals of Statistics, 1983; An Approach to Nonparametric Regression for Life History Data Using Local Linear Fitting Li, Gang and Doss, Hani, Annals of Statistics, 1995 Piecewise exponential models and creating custom models¶ This section will be easier if we recall our three mathematical “creatures” and the relationships between them. The estimate is M^ = log2 ^ = log2 t d 8 uniquely de nes the exponential distribution, which plays a central role in survival analysis. [PMC free article] Parametric survival analysis models typically require a non-negative distribution, because if you have negative survival times in your study, it is a sign that the zombie apocalypse has started (Wheatley-Price 2012). Quick start Weibull survival model with covariates x1 and x2 using stset data The cdf of the exponential model indicates the probability not surviving pass time t, but the survival function is the opposite. For example, if T denote the age of death, then the hazard function h(t) is expected to be decreasing at rst and then gradually increasing in the end, re ecting higher hazard of infants and elderly. Function to a set of data points 4 of Allison S. Lemeshow Chapter 8: parametric models. D. Hosmer and S. Lemeshow Chapter 8: parametric regression models to run survival analysis models: the model... ) = 1 exponential survival model ( ) survival mixture model of the complete lifespan a... Are obtained using maximum likelihood estimation survival in preparation for model fitting I the! Well for survival analysis by D. Hosmer and S. Lemeshow Chapter 8: parametric regression models of ungrouped data the! And treatment and predictors constant w/r/t time, which makes analysis very simple the of! May be a good model for the exponential distribution is primarily used in reliability applications are as. Therefore the MLE of the exponential distribution is used in reliability applications objects of gamma... A good model of different survival models are widely used for survival analysis Simulation... Fitting I calculate the offset or log of exposure and add it to the logic in the and... Parameterized I n terms of failure rate ( indicated by the … exponential model for survival analysis Faiz for! Like an exponential function to a set of exponential survival model points accept Surv objects of the counting... Conditionally on x the times to failure are model survival outcomes beyond the available follow-up.. The second parameter in the model and di erent shapes of the common taken. Data points modeling survival of living organisms over short intervals and di erent shapes of the model great! Of a system where parts are replaced as they fail arbitrary survival data component survival... Comment, you can run a Cox proportional model through the function (! Model fitting I calculate the offset or log of exposure and add it to the frame. Of living organisms over short intervals this document can function as a `` how to use use Software R do. Two positive parameters know R package to run survival analysis models: the exponential model and the model... The cumulative exponential distribution is used to do model selections exponential survival model and you can find. Using maximum likelihood estimation that work well for survival analysis by D. Hosmer and S. Lemeshow Chapter 8 parametric! Block because it currently does not accept Surv objects of the usual exponential models for arbitrary data... Two positive parameters which is equal to 1/θ additional data aside from the other analyses for Chapter 4 Allison... Covers two commonly used survival analysis and Simulation the exponential distribution is used to the!, for survival analysis Faiz model: basic properties and maximum likelihood.! The E1684 melanoma clinical trial data likelihood estimation to use organisms over short intervals are replaced as fail! D, Richardson DB, Elliott L. Poisson regression analysis of ungrouped data JH Marek! This paper At the Biometrics website on Wiley Online Library, College of Arts and Science, University! The R codes for implementing multivariate piecewise exponential survival modeling are available with this paper At the Biometrics on... Function ( no covariates or other individual diﬀerences ), ≥0 parsimonious piecewise survival. Like an exponential function to a set of data points setting up data for for modeling survival living... By far, the probability not surviving pass time t = ∞, S ( t ) = S t. −Exp ( −/ ), ≥0 hmohiv data set, we can exponential survival model use traditional methods linear. I n terms of failure rate, λ which is equal to 1/θ `` counting type. Do model selections, and you can run a Cox proportional model through the function coxph ( ) =1− )!\n\nKategorie: Bez kategorii" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8491871,"math_prob":0.9660611,"size":24503,"snap":"2021-04-2021-17","text_gpt3_token_len":5591,"char_repetition_ratio":0.20282461,"word_repetition_ratio":0.25184405,"special_character_ratio":0.22131984,"punctuation_ratio":0.113892645,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99436885,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-17T02:41:28Z\",\"WARC-Record-ID\":\"<urn:uuid:1d070ecf-a10e-4eba-af04-f32e20db7eeb>\",\"Content-Length\":\"102062\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d66d677-f0b3-48a2-9675-b151353067db>\",\"WARC-Concurrent-To\":\"<urn:uuid:7f9b1a11-3aaa-438b-806f-f2b889dc1072>\",\"WARC-IP-Address\":\"188.128.207.209\",\"WARC-Target-URI\":\"http://bajecznyogrod.com.pl/canadian-dollar-afy/exponential-survival-model-c0406f\",\"WARC-Payload-Digest\":\"sha1:3TMVNMLJRUY7MJDGMHH3C6KUCWBTME5C\",\"WARC-Block-Digest\":\"sha1:SKJCUZEXJHUEECEM4IYTPL52KSHN3YFP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038098638.52_warc_CC-MAIN-20210417011815-20210417041815-00328.warc.gz\"}"}
https://answers.everydaycalculation.com/add-fractions/4-70-plus-5-24	[ "Solutions by everydaycalculation.com\n\n4/70 + 5/24 is 223/840.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 70 and 24 is 840\n\nNext, find the equivalent fraction of both fractional numbers with denominator 840\n2. For the 1st fraction, since 70 × 12 = 840,\n4/70 = 4 × 12/70 × 12 = 48/840\n3. Likewise, for the 2nd fraction, since 24 × 35 = 840,\n5/24 = 5 × 35/24 × 35 = 175/840\n4. Add the two like fractions:\n48/840 + 175/840 = 48 + 175/840 = 223/840\n5. So, 4/70 + 5/24 = 223/840\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 ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.82222,"math_prob":0.99805486,"size":773,"snap":"2023-14-2023-23","text_gpt3_token_len":312,"char_repetition_ratio":0.16775033,"word_repetition_ratio":0.0,"special_character_ratio":0.52781373,"punctuation_ratio":0.073033705,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9993347,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-28T15:32:49Z\",\"WARC-Record-ID\":\"<urn:uuid:a42f4df0-9c86-4df4-86ea-2d5b480cd98c>\",\"Content-Length\":\"7781\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:16e0e86b-fb2c-4298-aa9d-3e16638dd80f>\",\"WARC-Concurrent-To\":\"<urn:uuid:62c13e41-6ba8-4b61-b328-0d4d784fff7e>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/4-70-plus-5-24\",\"WARC-Payload-Digest\":\"sha1:K3CMFB5EVYLMZ5RS7RL7XVEWL27UPWDT\",\"WARC-Block-Digest\":\"sha1:PM2YZA35XJZFMMAS3V5RYPQPM3Q5CTYE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224644309.7_warc_CC-MAIN-20230528150639-20230528180639-00153.warc.gz\"}"}
https://docs.mesastar.org/en/r15140/reference/binary_job.html	[ "# binary_job¶\n\n## output/input files¶\n\n### show_binary_log_description_at_start¶\n\nset this false if you want to skip the initial terminal output\n\nshow_binary_log_description_at_start = .true.\n\n\n### binary_history_columns_file¶\n\nif null string, use default (binary_history_columns.list)\n\nbinary_history_columns_file = ''\n\n\n### warn_binary_extra¶\n\nDue to changing the run_star_extras functions to hooks,we break existing run_binary_extras files. This flag sets a warning message and stops the MESA run until it is set to .false.. This way people will hopefully not be confused as to why their run_binary_extras functions are not being called.\n\nwarn_binary_extra = .true.\n\n\n### inlist_names(:)¶\n\nInlist files for both stars. If modeling a star as a point mass, the corresponding file is ignored.\n\ninlist_names(1) = 'inlist1'\ninlist_names(2) = 'inlist2'\n\n\n### extra_binary_job_inlist{1..5}_name¶\n\nYou can split your binary_job inlist into pieces using the following controls. BTW: it works recursively, so the extras can read extras too. if read_extra_star_job_inlist{1..5} is true, then read &star_job from this namelist file\n\nread_extra_binary_job_inlist1 = .false.\nextra_binary_job_inlist1_name = 'undefined'\nextra_binary_job_inlist2_name = 'undefined'\nextra_binary_job_inlist3_name = 'undefined'\nextra_binary_job_inlist4_name = 'undefined'\nextra_binary_job_inlist5_name = 'undefined'\n\n\n## starting model¶\n\n### evolve_both_stars¶\n\nSet this to .true. to model the evolution of both stars. If .false. then only inlist_names(1) will be used, and the other star will be modeled as a point mass. This option is ignored for loaded models, to setup a component as a point mass in that case use ‘change_point_mass_i’\n\nevolve_both_stars = .false.\n\n\n### tsync_for_relax_primary_to_th_eq¶\n\nNOT IMPLEMENTED YET!!! Set relax_primary_to_th_eq to .true. to ignore mass loss, rlof and changes in orbital period or separation (depending on which was chosen as the initial condition) until the primary (given by inlist_names(1)) reaches thermal equilibrium. This is meant to ignore the fast evolution that a ZAMS star can experience as it evolves to CN equilibrium at its core.\n\nThermal equilibrium in this case is defined as\n\nlog(s% L_nuc_burn_total*Lsun/s% L(1)) < log_Lnuc_div_L_for_relax_primary_to_th_eq,\n\n\nand the relaxation process is followed for at least min_age_for_relax_primary_to_th_eq years, even if the condition is met at the first step. If the process takes more than max_steps_for_relax_primary_to_th_eq steps, then the simulation is terminated.\n\nIf want output to be written to history during the relaxation process, then no_history_during_relax_primary_to_th_eq = .false.. After relaxation, if reset_age_relax_primary_to_th_eq = .true., then both model numbers and ages of the components and the binary will be reset to their values before relaxation.\n\nwhen modeling a system with rotation and tides, the synchronization timescale can be fixed for both stars using tsync_for_relax_primary_to_th_eq (in years).\n\nrelax_primary_to_th_eq = .false.\nlog_Lnuc_div_L_for_relax_primary_to_th_eq = 0.005d0\nmin_age_for_relax_primary_to_th_eq = 1d2\nmax_steps_for_relax_primary_to_th_eq = 1000\nno_history_during_relax_primary_to_th_eq = .true.\nreset_age_for_relax_primary_to_th_eq = .true.\ntsync_for_relax_primary_to_th_eq = 1\n\n\n## modifications to model¶\n\n### new_ignore_rlof_flag¶\n\nIf ignore_rlof_flag is true, then ignore mass transfer due to RLOF by default ignore_rlof_flag=.false.\n\nchange_ignore_rlof_flag = .false.\nchange_initial_ignore_rlof_flag = .false.\nnew_ignore_rlof_flag = .false.\n\n\n### new_model_twins_flag¶\n\nIf model_twins_flag is true, then the system is modeled as if both stars were identical twins. Meant to save computation time in this particular scenario where computing the evolution of one of the components is redundant. by default model_twins_flag=.true.\n\nchange_model_twins_flag = .false.\nchange_initial_model_twins_flag = .false.\nnew_model_twins_flag = .false.\n\n\n### new_point_mass_i¶\n\npoint_mass_i stores the index of the star that is treated as a point mass equal to zero if both stars are modeled by default point_mass_i is set by evolve_both_stars (0 if evolve_both_stars istrue, 2 if false)\n\nchange_point_mass_i = .false.\nchange_initial_point_mass_i = .false.\nnew_point_mass_i = 0\n\n\n### new_m1¶\n\nchange the mass of star1. Ignored if point_mass_i/=1, in that case the mass is always taken to be that in the stellar model. After change, period and angular momentum are recomputed assuming the same separation and eccentricity (you likely want something different) Value is in Msun\n\nchange_m1 = .false.\nchange_initial_m1 = .false.\nnew_m1 = 0d0\n\n\n### new_m2¶\n\nchange the mass of star2. Ignored if point_mass_i/=2 in that case the mass is always taken to be that in the stellar model After change, period and angular momentum are recomputed assuming the same separation and eccentricity (you likely want something different) Value is in Msun\n\nchange_m2 = .false.\nchange_initial_m2 = .false.\nnew_m2 = 0d0\n\n\n### new_eccentricity¶\n\nSimultaneously change the semi-major axis (in Rsun) and eccentricity, or the period (in days) and eccentricity.\n\nchange_separation_eccentricity = .false.\nchange_initial_separation_eccentricity = .false.\nchange_period_eccentricity = .false.\nchange_initial_period_eccentricity = .false.\n\nnew_separation = 0d0\nnew_period = 0d0\nnew_eccentricity = 0d0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7418929,"math_prob":0.87099445,"size":6075,"snap":"2021-31-2021-39","text_gpt3_token_len":1572,"char_repetition_ratio":0.182507,"word_repetition_ratio":0.08180536,"special_character_ratio":0.2271605,"punctuation_ratio":0.15563725,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9681773,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T12:47:04Z\",\"WARC-Record-ID\":\"<urn:uuid:fc2f67b2-4298-4adf-9c54-1e635d44714e>\",\"Content-Length\":\"30190\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2946e2c8-5c44-477e-8706-890ca6466628>\",\"WARC-Concurrent-To\":\"<urn:uuid:8cd4f866-af07-47b7-a2c1-73f8f5d6b587>\",\"WARC-IP-Address\":\"104.17.32.82\",\"WARC-Target-URI\":\"https://docs.mesastar.org/en/r15140/reference/binary_job.html\",\"WARC-Payload-Digest\":\"sha1:ETII5E24HQ7FYCZ3DP53BMLMCQDLUKQI\",\"WARC-Block-Digest\":\"sha1:6EZX7NZPMOXYETDSDEOE6MYK4IINDJGB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154214.36_warc_CC-MAIN-20210801123745-20210801153745-00166.warc.gz\"}"}
https://answers.everydaycalculation.com/add-fractions/30-56-plus-25-90	[ "Solutions by everydaycalculation.com\n\n30/56 + 25/90 is 205/252.\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 56 and 90 is 2520\n2. For the 1st fraction, since 56 × 45 = 2520,\n30/56 = 30 × 45/56 × 45 = 1350/2520\n3. Likewise, for the 2nd fraction, since 90 × 28 = 2520,\n25/90 = 25 × 28/90 × 28 = 700/2520\n1350/2520 + 700/2520 = 1350 + 700/2520 = 2050/2520\n5. 2050/2520 simplified gives 205/252\n6. So, 30/56 + 25/90 = 205/252\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 ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8020529,"math_prob":0.99778694,"size":689,"snap":"2020-10-2020-16","text_gpt3_token_len":275,"char_repetition_ratio":0.15036497,"word_repetition_ratio":0.0,"special_character_ratio":0.5515239,"punctuation_ratio":0.08219178,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99842036,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-29T11:12:55Z\",\"WARC-Record-ID\":\"<urn:uuid:eccfb083-856d-466f-bb5e-68347f065e5d>\",\"Content-Length\":\"7257\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ebd1b616-fb9d-4b22-b8e9-bdd7710945f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:928a3022-62f8-48fe-991b-211799d1a4f4>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/add-fractions/30-56-plus-25-90\",\"WARC-Payload-Digest\":\"sha1:BMU6K2KO3SVMF7PLQG7YRWYTII3EQALH\",\"WARC-Block-Digest\":\"sha1:LEBTZ5S6BN4OTZMLTJ6WB375LWSXDQOQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875148850.96_warc_CC-MAIN-20200229083813-20200229113813-00241.warc.gz\"}"}
https://origin.geeksforgeeks.org/split-a-string-into-two-substring-such-that-the-sum-of-unique-characters-is-maximum/	[ "", null, "Open in App\nNot now\n\n# Split a String into two Substring such that the sum of unique characters is maximum\n\n• Last Updated : 28 Mar, 2023\n\nGiven a string str, the task is to partition the string into two substrings such that the sum of unique characters of both substrings is the maximum possible.\n\nExamples:\n\nInput: str = “abcabcd”\nOutput: 7\nExplanation: Partition the given string into “abc” and “abcd”, the sum of unique characters is 3 + 4 = 7 which is maximum possible.\n\nInput: str = “aaaaa”\nOutput: 2\nExplanation: Partition the given string into “aa” and “aaa”, the sum of unique characters is 1 + 1 = 2 which is maximum possible. Given string can be partitioned into many other ways but this partition gives the maximum sum of unique characters.\n\nApproach: This problem can easily be solved using prefix and suffix arrays.\n\nWe can create a prefix array which will store the count of unique characters starting from index 0 to last index of the string. Similarly, We can create a suffix array which will store the count of unique characters starting from the last index to 0th index. Here, set can be used to check if the current character appeared before in the iterated string or not. We can calculate the maximum sum by doing prefix[i-1] + suffix[i] for each 1 <= i < n. Take prefix[i-1] + suffix[i] so that intersecting point never counts.\n\nFollow the steps mentioned below to implement the idea:\n\n• Declare two arrays of prefix and suffix\n• Declare two sets that store the visited characters so far.\n• Store prefix[i] = prefix[i-1] if the current character has already appeared, else prefix[i] = prefix[i-1] + 1\n• Store suffix[i] = suffix[i] if the current character has already appeared, else suffix[i] = suffix[i+1] + 1.\n• Declare a variable maxi = -1 which stores the maximum sum\n• Iterate through the array and compare the sum of prefix[i-1] + suffix[i] with maxi in each iteration.\n• Return maxi.\n\nBelow is the implementation of the above approach.\n\n## C++\n\n `// C++ code for the above approach:` `#include ` `using` `namespace` `std;` `// Function for finding out maximum value` `int` `unique_charcters(string str, ``int` `n)` `{` ` ``// Declare set to check if the current` ` ``// character is previously` ` ``// appearing or not` ` ``set<``char``> s1;` ` ``set<``char``> s2;` ` ``// Store the count of unique characters` ` ``// from starting index.` ` ``vector<``int``> prefix(n);` ` ``// Store the count of unique characters` ` ``// from last index.` ` ``vector<``int``> suffix(n);` ` ``prefix = 1;` ` ``suffix[n - 1] = 1;` ` ``s1.insert(str);` ` ``s2.insert(str[n - 1]);` ` ``// Storing the count of unique characters` ` ``// from starting index to last index` ` ``for` `(``int` `i = 1; i < n; i++) {` ` ``// If the current character has` ` ``// appeared before, store` ` ``// previous value` ` ``if` `(s1.find(str[i]) != s1.end()) {` ` ``prefix[i] = prefix[i - 1];` ` ``}` ` ``else` `{` ` ``// else store previous value + 1.` ` ``prefix[i] = prefix[i - 1] + 1;` ` ``s1.insert(str[i]);` ` ``}` ` ``}` ` ``// Storing the count of unique` ` ``// characters from last index` ` ``// to 0th index` ` ``for` `(``int` `i = n - 2; i >= 0; i--) {` ` ``// If the current character has` ` ``// already appeared, store the` ` ``// value calculated for the` ` ``// previous visited index of string` ` ``if` `(s2.find(str[i]) != s2.end()) {` ` ``suffix[i] = suffix[i + 1];` ` ``}` ` ``else` `{` ` ``// Else store value on next index + 1.` ` ``suffix[i] = suffix[i + 1] + 1;` ` ``s2.insert(str[i]);` ` ``}` ` ``}` ` ``// Store the maximum sum` ` ``int` `maxi = -1;` ` ``for` `(``int` `i = 1; i < n; i++) {` ` ``// Take sum of prefix[i-1] +` ` ``// suffix[i] so that the` ` ``// intersecting never counts.` ` ``maxi = max(maxi, prefix[i - 1] + suffix[i]);` ` ``}` ` ``// Returning the maximum value` ` ``return` `maxi;` `}` `// Driver code` `int` `main()` `{` ` ``string str = ``\"abcabcd\"``;` ` ``// Size of the string` ` ``int` `n = str.length();` ` ``// Function call` ` ``cout << ``\"Maximum sum is \"` `<< unique_charcters(str, n);` ` ``return` `0;` `}`\n\n## Java\n\n `// Java code for above approach` `import` `java.util.;` `class` `GFG {` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``String str = ``\"abcabcd\"``;` ` ``// Size of the string` ` ``int` `n = str.length();` ` ``// Function call` ` ``System.out.println(``\"Maximum sum is \"` ` ``+ (unique_characters(str, n)));` ` ``}` ` ``public` `static` `int` `unique_characters(String str, ``int` `n)` ` ``{` ` ``// Declare set to check if the current` ` ``// character is previously` ` ``// appearing or not` ` ``TreeSet s1 = ``new` `TreeSet<>();` ` ``TreeSet s2 = ``new` `TreeSet<>();` ` ``// Store the count of unique characters` ` ``// from starting index.` ` ``int``[] prefix = ``new` `int``[n];` ` ``// Store the count of unique characters` ` ``// from last index.` ` ``int``[] suffix = ``new` `int``[n];` ` ``prefix[``0``] = ``1``;` ` ``suffix[n - ``1``] = ``1``;` ` ``s1.add(str.charAt(``0``));` ` ``s2.add(str.charAt(n - ``1``));` ` ``// from starting index to last index` ` ``for` `(``int` `i = ``1``; i < n; i++) {` ` ``// If the current character has` ` ``// appeared before, store` ` ``// previous value` ` ``if` `((s1.contains(str.charAt(i)))` ` ``&& str.charAt(i) != s1.last()) {` ` ``prefix[i] = prefix[i - ``1``];` ` ``}` ` ``else` `{` ` ``// else store previous value + 1.` ` ``prefix[i] = prefix[i - ``1``] + ``1``;` ` ``s1.add(str.charAt(i));` ` ``}` ` ``}` ` ``// Storing the count of unique` ` ``// characters from last index` ` ``// to 0th index` ` ``for` `(``int` `i = n - ``2``; i >= ``0``; i--) {` ` ``// If the current character has` ` ``// already appeared, store the` ` ``// value calculated for the` ` ``// previous visited index of string` ` ``if` `((s2.contains(str.charAt(i)))` ` ``&& str.charAt(i) != s2.last()) {` ` ``suffix[i] = suffix[i + ``1``];` ` ``}` ` ``else` `{` ` ``// Else store value on next index + 1.` ` ``suffix[i] = suffix[i + ``1``] + ``1``;` ` ``s2.add(str.charAt(i));` ` ``}` ` ``}` ` ``// Store the maximum sum` ` ``int` `maxi = -``1``;` ` ``for` `(``int` `i = ``1``; i < n; i++) {` ` ``// Take sum of prefix[i-1] +` ` ``// suffix[i] so that the` ` ``// intersecting never counts.` ` ``maxi` ` ``= Math.max(maxi, prefix[i - ``1``] + suffix[i]);` ` ``}` ` ``// Returning the maximum value` ` ``return` `maxi;` ` ``}` `}`\n\n## Python3\n\n `# Python code for the above approach:` `def` `unique_charcters(``str``, n):` ` ``# Declare set to check if the current` ` ``# character is previously` ` ``# appearing or not` ` ``s1 ``=` `set``()` ` ``s2 ``=` `set``()` ` ``# Store the count of unique characters` ` ``# from starting index.` ` ``prefix ``=` `[``0``]````n` ` ``# Store the count of unique characters` ` ``# from last index.` ` ``suffix ``=` `[``0``]````n` ` ``prefix[``0``] ``=` `1` ` ``suffix[n ``-` `1``] ``=` `1` ` ``s1.add(``str``[``0``])` ` ``s2.add(``str``[n ``-` `1``])` ` ``# Storing the count of unique characters` ` ``# from starting index to last index` ` ``for` `i ``in` `range``(``1``, n):` ` ``# If the current character has` ` ``# appeared before, store` ` ``# previous value` ` ``if` `str``[i] ``in` `s1:` ` ``prefix[i] ``=` `prefix[i ``-` `1``]` ` ``else``:` ` ``# else store previous value + 1.` ` ``prefix[i] ``=` `prefix[i ``-` `1``] ``+` `1` ` ``s1.add(``str``[i])` ` ``# Storing the count of unique` ` ``# characters from last index` ` ``# to 0th index` ` ``for` `i ``in` `range``(n``-``2``, ``-``1``, ``-``1``):` ` ``# If the current character has` ` ``# already appeared, store the` ` ``# value calculated for the` ` ``# previous visited index of string` ` ``if` `str``[i] ``in` `s2:` ` ``suffix[i] ``=` `suffix[i ``+` `1``]` ` ``else``:` ` ``# Else store value on next index + 1.` ` ``suffix[i] ``=` `suffix[i ``+` `1``] ``+` `1` ` ``s2.add(``str``[i])` ` ``# Store the maximum sum` ` ``maxi ``=` `-``1` ` ``for` `i ``in` `range``(``1``, n):` ` ``# Take sum of prefix[i-1] +` ` ``# suffix[i] so that the` ` ``# intersecting never counts.` ` ``maxi ``=` `max``(maxi, prefix[i ``-` `1``] ``+` `suffix[i])` ` ``# Returning the maximum value` ` ``return` `maxi` `if` `__name__ ``=``=` `\"__main__\"``:` ` ``str` `=` `\"abcabcd\"` ` ``# Size of the string` ` ``n ``=` `len``(``str``)` ` ``# Function call` ` ``print``(``\"Maximum sum is \"``, unique_charcters(``str``, n))`\n\n## C#\n\n `// C# code for above approach` `using` `System;` `using` `System.Collections.Generic;` `public` `class` `GFG {` ` ``static` `int` `UniqueCharacters(``string` `str, ``int` `n)` ` ``{` ` ``// Declare set to check if the current character is` ` ``// previously appearing or not` ` ``SortedSet<``char``> s1 = ``new` `SortedSet<``char``>();` ` ``SortedSet<``char``> s2 = ``new` `SortedSet<``char``>();` ` ``// Store the count of unique characters from` ` ``// starting index.` ` ``int``[] prefix = ``new` `int``[n];` ` ``// Store the count of unique characters from last` ` ``// index.` ` ``int``[] suffix = ``new` `int``[n];` ` ``prefix = 1;` ` ``suffix[n - 1] = 1;` ` ``s1.Add(str);` ` ``s2.Add(str[n - 1]);` ` ``// from starting index to last index` ` ``for` `(``int` `i = 1; i < n; i++) {` ` ``// If the current character has appeared before,` ` ``// store previous value` ` ``if` `((s1.Contains(str[i])) && str[i] != s1.Max) {` ` ``prefix[i] = prefix[i - 1];` ` ``}` ` ``else` `{` ` ``// else store previous value + 1.` ` ``prefix[i] = prefix[i - 1] + 1;` ` ``s1.Add(str[i]);` ` ``}` ` ``}` ` ``// Storing the count of unique characters from last` ` ``// index to 0th index` ` ``for` `(``int` `i = n - 2; i >= 0; i--) {` ` ``// If the current character has already` ` ``// appeared, store the value calculated for the` ` ``// previous visited index of string` ` ``if` `((s2.Contains(str[i])) && str[i] != s2.Max) {` ` ``suffix[i] = suffix[i + 1];` ` ``}` ` ``else` `{` ` ``// Else store value on next index + 1.` ` ``suffix[i] = suffix[i + 1] + 1;` ` ``s2.Add(str[i]);` ` ``}` ` ``}` ` ``// Store the maximum sum` ` ``int` `maxi = -1;` ` ``for` `(``int` `i = 1; i < n; i++) {` ` ``// Take sum of prefix[i-1] + suffix[i] so that` ` ``// the intersecting never counts.` ` ``maxi` ` ``= Math.Max(maxi, prefix[i - 1] + suffix[i]);` ` ``}` ` ``// Returning the maximum value` ` ``return` `maxi;` ` ``}` ` ``static` `public` `void` `Main()` ` ``{` ` ``// Code` ` ``string` `str = ``\"abcabcd\"``;` ` ``// Size of the string` ` ``int` `n = str.Length;` ` ``// Function call` ` ``Console.WriteLine(``\"Maximum sum is \"` ` ``+ (UniqueCharacters(str, n)));` ` ``}` `}` `// This code is contributed by sankar.`\n\n## Javascript\n\n `// JavaScript code for the above approach:` `// Function for finding out maximum value` `function` `uniqueCharacters(str, n) {` ` ``// Declare set to check if the current` ` ``// character is previously` ` ``// appearing or not` ` ``var` `s1 = ``new` `Set();` ` ``var` `s2 = ``new` `Set();` ` ` ` ``// Store the count of unique characters` ` ``// from starting index.` ` ``var` `prefix = ``new` `Array(n);` ` ` ` ``// Store the count of unique characters` ` ``// from last index.` ` ``var` `suffix = ``new` `Array(n);` ` ``prefix = 1;` ` ``suffix[n - 1] = 1;` ` ``s1.add(str);` ` ``s2.add(str[n - 1]);` ` ` ` ``// Storing the count of unique characters` ` ``// from starting index to last index` ` ``for` `(``var` `i = 1; i < n; i++) {` ` ` ` ``// If the current character has` ` ``// appeared before, store` ` ``// previous value` ` ``if` `(s1.has(str[i])) {` ` ``prefix[i] = prefix[i - 1];` ` ``}` ` ``else` `{` ` ``// else store previous value + 1.` ` ``prefix[i] = prefix[i - 1] + 1;` ` ``s1.add(str[i]);` ` ``}` ` ``}` ` ` ` ``// Storing the count of unique` ` ``// characters from last index` ` ``// to 0th index` ` ``for` `(``var` `i = n - 2; i >= 0; i--) {` ` ` ` ``// If the current character has` ` ``// already appeared, store the` ` ``// value calculated for the` ` ``// previous visited index of string` ` ``if` `(s2.has(str[i])) {` ` ``suffix[i] = suffix[i + 1];` ` ``}` ` ``else` `{` ` ` ` ``// Else store value on next index + 1.` ` ``suffix[i] = suffix[i + 1] + 1;` ` ``s2.add(str[i]);` ` ``}` ` ``}` ` ` ` ``// Store the maximum sum` ` ``var` `maxi = -1;` ` ``for` `(``var` `i = 1; i < n; i++) {` ` ` ` ``// Take sum of prefix[i-1] +` ` ``// suffix[i] so that the` ` ``// intersecting never counts.` ` ``maxi = Math.max(maxi, prefix[i - 1] + suffix[i]);` ` ``}` ` ` ` ``// Returning the maximum value` ` ``return` `maxi;` `}` `// Driver code` `var` `str = ``\"abcabcd\"``;` `// Size of the string` `var` `n = str.length;` `// Function call` `console.log(``\"Maximum sum is \"` `+ uniqueCharacters(str, n));` `// This Code is Contributed by Prasad Kandekar(prasad264)`\n\nOutput\n\n`Maximum value is 7`\n\nTime Complexity: O(N)\nAuxiliary Space: O(N)\n\nEfficient Approach: To solve this problem, we will use two Hash Maps and do the below steps:\n\n• Firstly, make a frequency map of characters and store in Hash Map characters.\n• Now, we will traverse from the start of the given string str, also we will maintain another frequency map using Hash Map freq.\n• While traversing, we will keep track of the maximum sum of the sizes of characters and freq, whenever our maximum value changes we will change our maximum pointer.\n• At last, we will return the maximum possible answer.\n\nImplementation of the approach:\n\n## C++\n\n `#include ` `#include ` `#include ` `using` `namespace` `std;` `int` `maxUniqueCharSubstring(string str, ``int` `N)` `{` ` ``// Initializing two frequency Hash Maps` ` ``unordered_map<``char``, ``int``> characters;` ` ``unordered_map<``char``, ``int``> freq;` ` ``// Making a frequency map of characters` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``characters[str[i]]++;` ` ``}` ` ``// max variable which contains maximum sum of unique` ` ``// characters` ` ``int` `max = INT_MIN;` ` ``// Traversing the string` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``// Updating max variable` ` ``int` `totalChar = (characters.size() + freq.size());` ` ``if` `(max < totalChar) {` ` ``max = totalChar;` ` ``}` ` ``// Updating both hash maps` ` ``freq[str[i]]++;` ` ``characters[str[i]]--;` ` ``if` `(characters[str[i]] == 0) {` ` ``characters.erase(str[i]);` ` ``}` ` ``}` ` ``// Returning max value` ` ``return` `max;` `}` `int` `main()` `{` ` ``string str = ``\"abcabcd\"``;` ` ``int` `N = str.length();` ` ``cout << ``\"Maximum sum is \"` ` ``<< maxUniqueCharSubstring(str, N) << endl;` ` ``return` `0;` `}`\n\n## Java\n\n `// Java algorithm for the above approach` `import` `java.util.;` `class` `GFG {` ` ``// Driver Code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``String str = ``\"abcabcd\"``;` ` ``int` `N = str.length();` ` ``System.out.println(` ` ``\"Maximum sum is \"` ` ``+ maxUniqueCharSubstring(str, N));` ` ``}` ` ``public` `static` `int` `maxUniqueCharSubstring(String str,` ` ``int` `N)` ` ``{` ` ``// Initializing two frequency Hash Maps` ` ``Map characters` ` ``= ``new` `HashMap<>();` ` ``Map freq = ``new` `HashMap<>();` ` ``// Making a frequency map of characters` ` ``for` `(``int` `i = ``0``; i < N; i++) {` ` ``characters.put(` ` ``str.charAt(i),` ` ``characters.getOrDefault(str.charAt(i), ``0``)` ` ``+ ``1``);` ` ``}` ` ``// max variable which contains maximum sum of unique` ` ``// characters` ` ``int` `max = Integer.MIN_VALUE;` ` ``// Traversing the string` ` ``for` `(``int` `i = ``0``; i < N; i++) {` ` ``// Updating max variable` ` ``if` `(max < characters.size() + freq.size()) {` ` ``max = characters.size() + freq.size();` ` ``}` ` ``// Updating both hash maps` ` ``freq.put(str.charAt(i),` ` ``freq.getOrDefault(str.charAt(i), ``0``)` ` ``+ ``1``);` ` ``characters.put(str.charAt(i),` ` ``characters.get(str.charAt(i))` ` ``- ``1``);` ` ``if` `(characters.get(str.charAt(i)) == ``0``)` ` ``characters.remove(str.charAt(i));` ` ``}` ` ``// Returning max value` ` ``return` `max;` ` ``}` `}`\n\n## Python3\n\n `# Python code for the above approach` `def` `maxUniqueCharSubstring(string):` ` ``# Initializing two frequency dictionaries` ` ``characters ``=` `{}` ` ``freq ``=` `{}` ` ``# Making a frequency map of characters` ` ``for` `i ``in` `string:` ` ``if` `i ``in` `characters:` ` ``characters[i] ``+``=` `1` ` ``else``:` ` ``characters[i] ``=` `1` ` ``# max variable which contains maximum sum of unique` ` ``# characters` ` ``max_len ``=` `float``(``'-inf'``)` ` ``# Traversing the string` ` ``for` `i ``in` `range``(``len``(string)):` ` ``# Updating max variable` ` ``if` `max_len < ``len``(characters) ``+` `len``(freq):` ` ``max_len ``=` `len``(characters) ``+` `len``(freq)` ` ``# Updating both dictionaries` ` ``if` `string[i] ``in` `freq:` ` ``freq[string[i]] ``+``=` `1` ` ``else``:` ` ``freq[string[i]] ``=` `1` ` ``characters[string[i]] ``=` `characters[string[i]] ``-` `1` ` ``if` `characters[string[i]] ``=``=` `0``:` ` ``del` `characters[string[i]]` ` ``# Returning max value` ` ``return` `max_len` `# Driver code` `string ``=` `\"abcabcd\"` `print``(``\"Maximum sum is\"``, maxUniqueCharSubstring(string))` `# This code is contributed by codearcade.`\n\n## C#\n\n `// C# code for the above approach` `using` `System;` `using` `System.Collections.Generic;` `public` `class` `GFG {` ` ``public` `static` `int` `maxUniqueCharSubstring(``string` `str,` ` ``int` `N)` ` ``{` ` ``// Initializing two frequency Hash Maps` ` ``Dictionary<``char``, ``int``> characters` ` ``= ``new` `Dictionary<``char``, ``int``>();` ` ``Dictionary<``char``, ``int``> freq` ` ``= ``new` `Dictionary<``char``, ``int``>();` ` ``// Making a frequency map of characters` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``if` `(characters.ContainsKey(str[i]))` ` ``characters[str[i]] += 1;` ` ``else` ` ``characters.Add(str[i], 1);` ` ``}` ` ``// max variable which contains maximum sum of unique` ` ``// characters` ` ``int` `max = ``int``.MinValue;` ` ``// Traversing the string` ` ``for` `(``int` `i = 0; i < N; i++) {` ` ``// Updating max variable` ` ``if` `(max < characters.Count + freq.Count) {` ` ``max = characters.Count + freq.Count;` ` ``}` ` ``// Updating both hash maps` ` ``if` `(freq.ContainsKey(str[i]))` ` ``freq[str[i]] += 1;` ` ``else` ` ``freq.Add(str[i], 1);` ` ``characters[str[i]] -= 1;` ` ``if` `(characters[str[i]] == 0)` ` ``characters.Remove(str[i]);` ` ``}` ` ``// Returning max value` ` ``return` `max;` ` ``}` ` ``static` `public` `void` `Main()` ` ``{` ` ``// Code` ` ``string` `str = ``\"abcabcd\"``;` ` ``int` `N = str.Length;` ` ``Console.WriteLine(``\"Maximum sum is \"` ` ``+ maxUniqueCharSubstring(str, N));` ` ``}` `}` `// This code is contributed by karthik.`\n\n## Javascript\n\n `// JavaScript code for the above approach` `function` `maxUniqueCharSubstring(str, N)` `{` ` ``// Initializing two frequency Hash Maps` ` ``let characters = ``new` `Map();` ` ``let freq = ``new` `Map();` ` ``// Making a frequency map of characters` ` ``for` `(let i = 0; i < N; i++) {` ` ``characters.set(str[i], (characters.get(str[i]) \|\| 0) + 1);` ` ``}` ` ``// max variable which contains maximum sum of unique` ` ``// characters` ` ``let max = Number.MIN_SAFE_INTEGER;` ` ``// Traversing the string` ` ``for` `(let i = 0; i < N; i++) {` ` ``// Updating max variable` ` ``let totalChar = (characters.size + freq.size);` ` ``if` `(max < totalChar) {` ` ``max = totalChar;` ` ``}` ` ``// Updating both hash maps` ` ``freq.set(str[i], (freq.get(str[i]) \|\| 0) + 1);` ` ``characters.set(str[i], (characters.get(str[i]) \|\| 0) - 1);` ` ``if` `(characters.get(str[i]) === 0) {` ` ``characters.``delete``(str[i]);` ` ``}` ` ``}` ` ``// Returning max value` ` ``return` `max;` `}` `let str = ``\"abcabcd\"``;` `let N = str.length;` `console.log(``\"Maximum sum is \"` `+ maxUniqueCharSubstring(str, N));` `// This code is contributed by prasad264`\n\nOutput:\n\n`Maximum sum is 7`\n\nTime Complexity: O(N)\n\nAuxiliary Space: Constant space is used as there can be only at most 26 characters stored in hash map.\n\nMy Personal Notes arrow_drop_up\nRelated Articles" ]	[ null, "https://media.geeksforgeeks.org/gfg-gg-logo.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.50810593,"math_prob":0.9667858,"size":16331,"snap":"2023-14-2023-23","text_gpt3_token_len":4762,"char_repetition_ratio":0.1720463,"word_repetition_ratio":0.45001683,"special_character_ratio":0.33243525,"punctuation_ratio":0.15208128,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9988987,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-31T16:29:50Z\",\"WARC-Record-ID\":\"<urn:uuid:57f6392f-7892-4002-836b-7c0c2839574a>\",\"Content-Length\":\"401773\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:646256ff-0db6-4b3d-8a6e-6e5ba6b10d9d>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5529e81-1c23-4ffb-a213-2bca47f84bd1>\",\"WARC-IP-Address\":\"44.228.100.190\",\"WARC-Target-URI\":\"https://origin.geeksforgeeks.org/split-a-string-into-two-substring-such-that-the-sum-of-unique-characters-is-maximum/\",\"WARC-Payload-Digest\":\"sha1:XJV4ZMSZABII5TTR265X7MMLIYIE6CMB\",\"WARC-Block-Digest\":\"sha1:RMPMKUEVRHO37PJ2VNLODOONG5ZEGSBT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949644.27_warc_CC-MAIN-20230331144941-20230331174941-00220.warc.gz\"}"}
https://www.bartleby.com/solution-answer/chapter-91-problem-2swu-calculus-an-applied-approach-mindtap-course-list-10th-edition/9781305860919/in-exercises-1-and-2-solve-for-x-112181413x241/e687be7b-6362-11e9-8385-02ee952b546e	[ "", null, "", null, "", null, "Chapter 9.1, Problem 2SWU", null, "### Calculus: An Applied Approach (Min...\n\n10th Edition\nRon Larson\nISBN: 9781305860919\n\n#### Solutions\n\nChapter\nSection", null, "### Calculus: An Applied Approach (Min...\n\n10th Edition\nRon Larson\nISBN: 9781305860919\nTextbook Problem\n1 views\n\n# In Exercises 1 and 2, solve for x. 1 12 + 1 8 + 1 4 + 1 3 + x 24 = 1\n\nTo determine\n\nTo calculate: The value of x from the expression: 112+18+14+13+x24=1.\n\nExplanation\n\nGiven information:\n\nThe expression: 112+18+14+13+x24=1.\n\nCalculation:\n\nConsider the expression,\n\n112+18+14+13+x24=1\n\nWrite common denominator for given expression,\n\n112+18+14+13+x24=11×212×2+1×38×3+1×64×6+\n\n### Still sussing out bartleby?\n\nCheck out a sample textbook solution.\n\nSee a sample solution\n\n#### The Solution to Your Study Problems\n\nBartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!\n\nGet Started\n\n#### Solve the equations in Exercises 126. 14x2=0\n\nFinite Mathematics and Applied Calculus (MindTap Course List)\n\n#### Evaluate limxesinx1x.\n\nCalculus (MindTap Course List)\n\n#### In problems 15-22, simplify by combining like terms. 19.\n\nMathematical Applications for the Management, Life, and Social Sciences\n\n#### In Exercises 110, find the graphical solution to each inequality. 3y+50\n\nFinite Mathematics for the Managerial, Life, and Social Sciences\n\n#### It does not exist.\n\nStudy Guide for Stewart's Multivariable Calculus, 8th\n\n#### The implied domain of is: (1, ∞) (−∞, 1) x ≠ 1 (−1, 1)\n\nStudy Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th", null, "" ]	[ null, "https://www.bartleby.com/static/search-icon-white.svg", null, "https://www.bartleby.com/static/close-grey.svg", null, "https://www.bartleby.com/static/solution-list.svg", null, "https://www.bartleby.com/isbn_cover_images/9781305860919/9781305860919_largeCoverImage.gif", null, "https://www.bartleby.com/isbn_cover_images/9781305860919/9781305860919_largeCoverImage.gif", null, "https://www.bartleby.com/static/logo.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.64482826,"math_prob":0.94275945,"size":2039,"snap":"2019-51-2020-05","text_gpt3_token_len":432,"char_repetition_ratio":0.18869779,"word_repetition_ratio":0.096,"special_character_ratio":0.15644924,"punctuation_ratio":0.09893993,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9966371,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T10:33:04Z\",\"WARC-Record-ID\":\"<urn:uuid:98f09103-48b5-495b-8508-888afaf016bf>\",\"Content-Length\":\"514782\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:67c53050-9c9f-47ec-a037-95fdd6a762b2>\",\"WARC-Concurrent-To\":\"<urn:uuid:68f330d0-e67d-44a5-a7a1-e1604549939a>\",\"WARC-IP-Address\":\"13.249.44.99\",\"WARC-Target-URI\":\"https://www.bartleby.com/solution-answer/chapter-91-problem-2swu-calculus-an-applied-approach-mindtap-course-list-10th-edition/9781305860919/in-exercises-1-and-2-solve-for-x-112181413x241/e687be7b-6362-11e9-8385-02ee952b546e\",\"WARC-Payload-Digest\":\"sha1:TEB3DNFT33WMPGOMHCFC4X4EEVEN4X32\",\"WARC-Block-Digest\":\"sha1:TXMELKGYB4PSAXYL22PIFR42MGYHMZX5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541307813.73_warc_CC-MAIN-20191215094447-20191215122447-00032.warc.gz\"}"}
https://www.colorhexa.com/01f3f2	[ "# #01f3f2 Color Information\n\nIn a RGB color space, hex #01f3f2 is composed of 0.4% red, 95.3% green and 94.9% blue. Whereas in a CMYK color space, it is composed of 99.6% cyan, 0% magenta, 0.4% yellow and 4.7% black. It has a hue angle of 179.8 degrees, a saturation of 99.2% and a lightness of 47.8%. #01f3f2 color hex could be obtained by blending #02ffff with #00e7e5. Closest websafe color is: #00ffff.\n\n• R 0\n• G 95\n• B 95\nRGB color chart\n• C 100\n• M 0\n• Y 0\n• K 5\nCMYK color chart\n\n#01f3f2 color description : Vivid cyan.\n\n# #01f3f2 Color Conversion\n\nThe hexadecimal color #01f3f2 has RGB values of R:1, G:243, B:242 and CMYK values of C:1, M:0, Y:0, K:0.05. Its decimal value is 127986.\n\nHex triplet RGB Decimal 01f3f2 `#01f3f2` 1, 243, 242 `rgb(1,243,242)` 0.4, 95.3, 94.9 `rgb(0.4%,95.3%,94.9%)` 100, 0, 0, 5 179.8°, 99.2, 47.8 `hsl(179.8,99.2%,47.8%)` 179.8°, 99.6, 95.3 00ffff `#00ffff`\nCIE-LAB 87.248, -46.627, -13.147 48.085, 70.513, 95.076 0.225, 0.33, 70.513 87.248, 48.446, 195.747 87.248, -67.562, -13.732 83.972, -44.737, -8.349 00000001, 11110011, 11110010\n\n# Color Schemes with #01f3f2\n\n• #01f3f2\n``#01f3f2` `rgb(1,243,242)``\n• #f30102\n``#f30102` `rgb(243,1,2)``\nComplementary Color\n• #01f379\n``#01f379` `rgb(1,243,121)``\n• #01f3f2\n``#01f3f2` `rgb(1,243,242)``\n• #017bf3\n``#017bf3` `rgb(1,123,243)``\nAnalogous Color\n• #f37901\n``#f37901` `rgb(243,121,1)``\n• #01f3f2\n``#01f3f2` `rgb(1,243,242)``\n• #f3017b\n``#f3017b` `rgb(243,1,123)``\nSplit Complementary Color\n• #f3f201\n``#f3f201` `rgb(243,242,1)``\n• #01f3f2\n``#01f3f2` `rgb(1,243,242)``\n• #f201f3\n``#f201f3` `rgb(242,1,243)``\n• #02f301\n``#02f301` `rgb(2,243,1)``\n• #01f3f2\n``#01f3f2` `rgb(1,243,242)``\n• #f201f3\n``#f201f3` `rgb(242,1,243)``\n• #f30102\n``#f30102` `rgb(243,1,2)``\n• #01a7a6\n``#01a7a6` `rgb(1,167,166)``\n• #01c0bf\n``#01c0bf` `rgb(1,192,191)``\n``#01dad9` `rgb(1,218,217)``\n• #01f3f2\n``#01f3f2` `rgb(1,243,242)``\n• #0ffefd\n``#0ffefd` `rgb(15,254,253)``\n• #29fefd\n``#29fefd` `rgb(41,254,253)``\n• #42fefd\n``#42fefd` `rgb(66,254,253)``\nMonochromatic Color\n\n# Alternatives to #01f3f2\n\nBelow, you can see some colors close to #01f3f2. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #01f3b6\n``#01f3b6` `rgb(1,243,182)``\n• #01f3ca\n``#01f3ca` `rgb(1,243,202)``\n• #01f3de\n``#01f3de` `rgb(1,243,222)``\n• #01f3f2\n``#01f3f2` `rgb(1,243,242)``\n• #01e0f3\n``#01e0f3` `rgb(1,224,243)``\n• #01ccf3\n``#01ccf3` `rgb(1,204,243)``\n• #01b8f3\n``#01b8f3` `rgb(1,184,243)``\nSimilar Colors\n\n# #01f3f2 Preview\n\nThis text has a font color of #01f3f2.\n\n``<span style=\"color:#01f3f2;\">Text here</span>``\n#01f3f2 background color\n\nThis paragraph has a background color of #01f3f2.\n\n``<p style=\"background-color:#01f3f2;\">Content here</p>``\n#01f3f2 border color\n\nThis element has a border color of #01f3f2.\n\n``<div style=\"border:1px solid #01f3f2;\">Content here</div>``\nCSS codes\n``.text {color:#01f3f2;}``\n``.background {background-color:#01f3f2;}``\n``.border {border:1px solid #01f3f2;}``\n\n# Shades and Tints of #01f3f2\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, #000909 is the darkest color, while #f4ffff is the lightest one.\n\n• #000909\n``#000909` `rgb(0,9,9)``\n• #001c1c\n``#001c1c` `rgb(0,28,28)``\n• #00302f\n``#00302f` `rgb(0,48,47)``\n• #004343\n``#004343` `rgb(0,67,67)``\n• #005756\n``#005756` `rgb(0,87,86)``\n• #006a6a\n``#006a6a` `rgb(0,106,106)``\n• #017e7d\n``#017e7d` `rgb(1,126,125)``\n• #019191\n``#019191` `rgb(1,145,145)``\n• #01a5a4\n``#01a5a4` `rgb(1,165,164)``\n• #01b8b8\n``#01b8b8` `rgb(1,184,184)``\n• #01cccb\n``#01cccb` `rgb(1,204,203)``\n• #01dfdf\n``#01dfdf` `rgb(1,223,223)``\n• #01f3f2\n``#01f3f2` `rgb(1,243,242)``\n• #0afefd\n``#0afefd` `rgb(10,254,253)``\n• #1dfefd\n``#1dfefd` `rgb(29,254,253)``\n• #31fefd\n``#31fefd` `rgb(49,254,253)``\n• #44fefd\n``#44fefd` `rgb(68,254,253)``\n• #58fefe\n``#58fefe` `rgb(88,254,254)``\n• #6bfefe\n``#6bfefe` `rgb(107,254,254)``\n• #7ffefe\n``#7ffefe` `rgb(127,254,254)``\n• #92fffe\n``#92fffe` `rgb(146,255,254)``\n• #a6fffe\n``#a6fffe` `rgb(166,255,254)``\n• #b9fffe\n``#b9fffe` `rgb(185,255,254)``\n• #cdffff\n``#cdffff` `rgb(205,255,255)``\n• #e1ffff\n``#e1ffff` `rgb(225,255,255)``\n• #f4ffff\n``#f4ffff` `rgb(244,255,255)``\nTint Color Variation\n\n# Tones of #01f3f2\n\nA tone is produced by adding gray to any pure hue. In this case, #728282 is the less saturated color, while #01f3f2 is the most saturated one.\n\n• #728282\n``#728282` `rgb(114,130,130)``\n• #688c8c\n``#688c8c` `rgb(104,140,140)``\n• #5f9595\n``#5f9595` `rgb(95,149,149)``\n• #559f9e\n``#559f9e` `rgb(85,159,158)``\n• #4ca8a8\n``#4ca8a8` `rgb(76,168,168)``\n• #43b1b1\n``#43b1b1` `rgb(67,177,177)``\n• #39bbba\n``#39bbba` `rgb(57,187,186)``\n• #30c4c3\n``#30c4c3` `rgb(48,196,195)``\n• #27cdcd\n``#27cdcd` `rgb(39,205,205)``\n• #1dd7d6\n``#1dd7d6` `rgb(29,215,214)``\n• #14e0df\n``#14e0df` `rgb(20,224,223)``\n• #0aeae9\n``#0aeae9` `rgb(10,234,233)``\n• #01f3f2\n``#01f3f2` `rgb(1,243,242)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #01f3f2 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 ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54589456,"math_prob":0.8096935,"size":3689,"snap":"2019-51-2020-05","text_gpt3_token_len":1676,"char_repetition_ratio":0.1402985,"word_repetition_ratio":0.011111111,"special_character_ratio":0.53862834,"punctuation_ratio":0.23692992,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98135287,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-09T08:05:48Z\",\"WARC-Record-ID\":\"<urn:uuid:27faae80-5080-4b4b-92b7-464e344c7f68>\",\"Content-Length\":\"36261\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ddd93c7b-99a6-453c-b45c-ae21f9bcb2fa>\",\"WARC-Concurrent-To\":\"<urn:uuid:7d1f6a1d-2543-4f34-913c-fb5906931bd1>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/01f3f2\",\"WARC-Payload-Digest\":\"sha1:GWQ3TOKISPX76HGKSASS74G3436AH57Q\",\"WARC-Block-Digest\":\"sha1:73O7BOJKUH2K77ZZJ3UPU5XKHZZC2KKD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540518337.65_warc_CC-MAIN-20191209065626-20191209093626-00436.warc.gz\"}"}
https://www.stumblingrobot.com/2016/01/13/solve-the-differential-equation-x-x-e2t-for-given-initial-values/	[ "Home » Blog » Solve the differential equation x′ + x = e2t for given initial values\n\n# Solve the differential equation x′ + x = e2t for given initial values\n\nSolve the following differential equation:", null, "with", null, "when", null, ".\n\nFrom Theorem 8.3 (page 310 of Apostol) we know that a differential equation of the form", null, "on an interval", null, "has solution given by", null, "where", null, "In this particular case we apply the theorem (noting that in this problem we have", null, "is a function of", null, ", rather than", null, "a function of", null, "as in the theorem… of course, the names of the variables doesn’t matter, but we take care to apply the theorem properly) with", null, "This gives us", null, "Therefore,", null, "" ]	[ null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-1020ca8891b546b1b48106189161c12a_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-2ef1fe1da7e0372fa788416c4c9352ef_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-8df410c9cccf774639a7197c59525536_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-fa70011d3d6312fbd3fe876e1b49084c_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-aae4b6890b75466b762ecaf81c3e4bdf_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-831517c74a3621ae23f4fba67c093c36_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-68023344f827ae2c9eec71385626eb91_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4546010112ccb15487fa5e25d75d8f24_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-541210184f5a4d2187a101d5f5f9b52d_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9dbb67d5fbb8ec5140bfda1e8bbf1d5c_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-4546010112ccb15487fa5e25d75d8f24_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-7da1fbabeeccfb3d4d6bd8a091b5d819_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d0086a574f52e68ebd11745368e22608_l3.png", null, "https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ea23917558384aaca56299ddab09493e_l3.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9010855,"math_prob":0.99969363,"size":463,"snap":"2023-14-2023-23","text_gpt3_token_len":107,"char_repetition_ratio":0.1285403,"word_repetition_ratio":0.0,"special_character_ratio":0.21814255,"punctuation_ratio":0.07692308,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99926263,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,2,null,null,null,null,null,8,null,null,null,8,null,8,null,null,null,null,null,null,null,null,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-05T13:03:53Z\",\"WARC-Record-ID\":\"<urn:uuid:0fd74d06-1a52-4d76-a42e-ef6ea3b3c248>\",\"Content-Length\":\"61856\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8809e465-c1ad-4542-81ac-a32d4950b9a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:4d238745-97d9-4b6c-b936-44945d87a5fc>\",\"WARC-IP-Address\":\"194.1.147.70\",\"WARC-Target-URI\":\"https://www.stumblingrobot.com/2016/01/13/solve-the-differential-equation-x-x-e2t-for-given-initial-values/\",\"WARC-Payload-Digest\":\"sha1:UV7SN7X4B7J2RTGAST7PCZMVQY7LZX5V\",\"WARC-Block-Digest\":\"sha1:GAHCJPSZJSHVIYFYXFP5BI6OKABG4W7S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224652116.60_warc_CC-MAIN-20230605121635-20230605151635-00486.warc.gz\"}"}
https://stats.libretexts.org/Courses/Penn_State_University_Greater_Allegheny/STAT_200%3A_Introductory_Statistics_(OpenStax)_GAYDOS/07%3A_The_Central_Limit_Theorem/7.01%3A_The_Central_Limit_Theorem_for_Sample_Means_(Averages)/7.1E%3A_The_Central_Limit_Theorem_for_Sample_Means_(Exercises)	[ "# 7.1E: The Central Limit Theorem for Sample Means (Exercises)\n\n•", null, "• OpenStax\n• OpenStax\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\nUse the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let $$X$$ be the random variable representing the time it takes her to complete one review. Assume $$X$$ is normally distributed. Let $$\\bar{X}$$ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.\n\n##### Example $$\\PageIndex{1}$$\n\nWhat is the mean, standard deviation, and sample size?\n\nmean = 4 hours; standard deviation = 1.2 hours; sample size = 16\n\nExercise $$\\PageIndex{2}$$\n\nComplete the distributions.\n\n1. $$X \\sim$$ _____(_____,_____)\n2. $$\\bar{X} \\sim$$ _____(_____,_____)\n##### Example $$\\PageIndex{3}$$\n\nFind the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.", null, "Figure $$\\PageIndex{2}$$.\n\n2. $$P$$(________ $$< x <$$ ________) = _______\n\n1. Check student's solution.\n2. 3.5, 4.25, 0.2441\n\nExercise $$\\PageIndex{4}$$\n\nFind the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.", null, "Figure $$\\PageIndex{3}$$.\n\n2. $$P$$(________________) = _______\n\n##### Example $$\\PageIndex{5}$$\n\nWhat causes the probabilities in Exercise and Exercise to be different?\n\nThe fact that the two distributions are different accounts for the different probabilities.\n\nExercise $$\\PageIndex{6}$$\n\nFind the 95th percentile for the mean time to complete one month's reviews. Sketch the graph.", null, "Figure $$\\PageIndex{4}$$.\n\n1. The 95th Percentile =____________\n\nThis page titled 7.1E: The Central Limit Theorem for Sample Means (Exercises) is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request." ]	[ null, "https://biz.libretexts.org/@api/deki/files/5084/girl-160172__340.png", null, "https://stats.libretexts.org/@api/deki/files/25878/d333c92a9a99584a0336f2c7aa99ff72dd761dea_4pk2", null, "https://stats.libretexts.org/@api/deki/files/25880/35e89f8ebc4d09c210ea785e2414dd63804aaae7", null, "https://stats.libretexts.org/@api/deki/files/25879/d333c92a9a99584a0336f2c7aa99ff72dd761dea", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8203546,"math_prob":0.99955213,"size":1899,"snap":"2023-40-2023-50","text_gpt3_token_len":468,"char_repetition_ratio":0.1403694,"word_repetition_ratio":0.13240418,"special_character_ratio":0.30963665,"punctuation_ratio":0.13714285,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99805534,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-22T19:25:14Z\",\"WARC-Record-ID\":\"<urn:uuid:fd7c975f-f6cc-44b3-980a-4d829d1e9991>\",\"Content-Length\":\"125326\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:439d4061-5687-4f25-8c26-38d7f56c905e>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c59ba00-ddcf-4cb6-a375-c649f41b9ce8>\",\"WARC-IP-Address\":\"3.162.103.49\",\"WARC-Target-URI\":\"https://stats.libretexts.org/Courses/Penn_State_University_Greater_Allegheny/STAT_200%3A_Introductory_Statistics_(OpenStax)_GAYDOS/07%3A_The_Central_Limit_Theorem/7.01%3A_The_Central_Limit_Theorem_for_Sample_Means_(Averages)/7.1E%3A_The_Central_Limit_Theorem_for_Sample_Means_(Exercises)\",\"WARC-Payload-Digest\":\"sha1:R3G5CW6E3G36I3MMSECF2FLVUTHXKWWH\",\"WARC-Block-Digest\":\"sha1:WSXII2TXEKYV2SXJHGD3GVAEHHJYOUGP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506421.14_warc_CC-MAIN-20230922170343-20230922200343-00856.warc.gz\"}"}
https://in.mathworks.com/matlabcentral/fileexchange/14469-performing-random-numbers-generator-from-a-generic-discrete-distribution	[ "## Performing random numbers generator from a generic discrete distribution\n\nVersion 1.0.0.0 (28.7 KB) by\nThis function extracts random numbers distributed over a discrete set; the PDF is user-defined\n\nUpdated Tue, 03 Apr 2007 16:47:55 +0000\n\nThis function extracts a scalar/vector/matrix of random numbers with discrete Probability Distribution Function.\nThe PDF is specified by the user as a input vector.\n\nThis function is designed to be fast, and it is implemented within a .mex file\n\nFollowing Olivier B. comments (that I acknowledge for his comments), I performed cross-comparisons with randp. gDiscrPdfRnd is faster with a ratio that increases with the number of number, i.e. for about 3 times faster for 10^6 numbers to over 40 times faster for 10^7 numbers.\nMoreover, for large random arrays, randp seriously surcharges the RAM memory, whereas gDiscrPdfRnd limits thememory use to what is essential (tanksto the coding). In what follows the details of thecomparison are given.\n\n>> tic;R = randp([1 3 2],1000000,1);toc\n\nelapsed_time =\n\n0.4840\n\n>> tic;R = gDiscrPdfRnd([1 3 2],1000000,1);toc\n\nelapsed_time =\n\n0.1570\n\n>> tic;R = randp([1 3 2],10000000,1);toc\n\nelapsed_time =\n\n68.5780\n\n>> tic;R = gDiscrPdfRnd([1 3 2],10000000,1);toc\n\nelapsed_time =\n\n1.6410\n\n>> 68.5780/1.6410\n\nans =\n\n41.7904\n\n### Cite As\n\nGianluca Dorini (2023). Performing random numbers generator from a generic discrete distribution (https://www.mathworks.com/matlabcentral/fileexchange/14469-performing-random-numbers-generator-from-a-generic-discrete-distribution), MATLAB Central File Exchange. Retrieved .\n\n##### MATLAB Release Compatibility\nCreated with R13\nCompatible with any release\n##### Platform Compatibility\nWindows macOS Linux" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8074775,"math_prob":0.9281961,"size":1948,"snap":"2023-14-2023-23","text_gpt3_token_len":514,"char_repetition_ratio":0.09825103,"word_repetition_ratio":0.05283019,"special_character_ratio":0.27258727,"punctuation_ratio":0.1671159,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9696196,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-22T07:17:16Z\",\"WARC-Record-ID\":\"<urn:uuid:446ae1b5-2c3b-42f1-902c-4f090edb4e21>\",\"Content-Length\":\"84995\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a579d2aa-d262-4681-bbe6-1f8e2eea42d5>\",\"WARC-Concurrent-To\":\"<urn:uuid:32fa299d-291d-45d4-ab95-83a572a160ed>\",\"WARC-IP-Address\":\"104.86.80.92\",\"WARC-Target-URI\":\"https://in.mathworks.com/matlabcentral/fileexchange/14469-performing-random-numbers-generator-from-a-generic-discrete-distribution\",\"WARC-Payload-Digest\":\"sha1:KD3ZNZSWZU4LJ32ZCSRW2GKJCZ62LXMS\",\"WARC-Block-Digest\":\"sha1:H5L4JRG7HCB44AQZ5JY5FOYEDMXAZBZH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296943750.71_warc_CC-MAIN-20230322051607-20230322081607-00629.warc.gz\"}"}
https://docs.openquake.org/oq-engine/3.10/_modules/openquake/hazardlib/source/rupture.html	[ "# Source code for openquake.hazardlib.source.rupture\n\n```# coding: utf-8\n# The Hazard Library\n# Copyright (C) 2012-2020 GEM Foundation\n#\n# This program is free software: you can redistribute it and/or modify\n# it under the terms of the GNU Affero General Public License as\n# published by the Free Software Foundation, either version 3 of the\n#\n# This program is distributed in the hope that it will be useful,\n# but WITHOUT ANY WARRANTY; without even the implied warranty of\n# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the\n# GNU Affero General Public License for more details.\n#\n# You should have received a copy of the GNU Affero General Public License\n# along with this program. If not, see <http://www.gnu.org/licenses/>.\n\"\"\"\nModule :mod:`openquake.hazardlib.source.rupture` defines classes\n:class:`BaseRupture` and its subclasses\n:class:`NonParametricProbabilisticRupture` and\n:class:`ParametricProbabilisticRupture`\n\"\"\"\nimport abc\nimport numpy\nimport math\nimport itertools\nimport json\nfrom openquake.baselib import general, hdf5\nfrom openquake.hazardlib import geo, contexts\nfrom openquake.hazardlib.geo.nodalplane import NodalPlane\nfrom openquake.hazardlib.geo.mesh import (\nMesh, RectangularMesh, surface_to_array)\nfrom openquake.hazardlib.geo.point import Point\nfrom openquake.hazardlib.geo.geodetic import geodetic_distance\nfrom openquake.hazardlib.near_fault import (\nfrom openquake.hazardlib.geo.surface.base import BaseSurface\n\nU8 = numpy.uint8\nU16 = numpy.uint16\nU32 = numpy.uint32\nF32 = numpy.float32\nF64 = numpy.float64\nTWO16 = 2 16\nTWO32 = 2 32\npmf_dt = numpy.dtype([('prob', float), ('occ', U32)])\nevents_dt = numpy.dtype([('id', U32), ('rup_id', U32), ('rlz_id', U16)])\nrupture_dt = numpy.dtype([('serial', U32),\n('mag', F32),\n('rake', F32),\n('lon', F32),\n('lat', F32),\n('dep', F32),\n('multiplicity', U32),\n('trt', hdf5.vstr),\n('kind', hdf5.vstr),\n('mesh', hdf5.vstr),\n('extra', hdf5.vstr)])\n\ncode2cls = {}\n\n[docs]def to_csv_array(ruptures):\n\"\"\"\n:param ruptures: a list of ruptures\n:returns: an array of ruptures suitable for serialization in CSV\n\"\"\"\nif not code2cls:\ncode2cls.update(BaseRupture.init())\narr = numpy.zeros(len(ruptures), rupture_dt)\nfor rec, rup in zip(arr, ruptures):\nmesh = surface_to_array(rup.surface) # shape (3, s1, s2)\nrec['serial'] = rup.rup_id\nrec['mag'] = rup.mag\nrec['rake'] = rup.rake\nrec['lon'] = rup.hypocenter.x\nrec['lat'] = rup.hypocenter.y\nrec['dep'] = rup.hypocenter.z\nrec['multiplicity'] = rup.multiplicity\nrec['trt'] = rup.tectonic_region_type\nrec['kind'] = ' '.join(cls.__name__ for cls in code2cls[rup.code])\nrec['mesh'] = json.dumps(\n[[[float5(z) for z in y] for y in x] for x in mesh])\nextra = {}\nif hasattr(rup, 'probs_occur'):\nextra['probs_occur'] = rup.probs_occur\nelse:\nextra['occurrence_rate'] = rup.occurrence_rate\nif hasattr(rup, 'weight'):\nextra['weight'] = rup.weight\n_fixfloat32(extra)\nrec['extra'] = json.dumps(extra)\nreturn arr\n\n[docs]def from_array(aw):\n\"\"\"\n:returns: a list of ruptures from an ArrayWrapper\n\"\"\"\nrups = []\nnames = aw.array.dtype.names\nfor rec in aw.array:\ndic = dict(zip(names, rec))\ndic['trt'] = aw.trts[int(dic.pop('grp_id'))]\ndic['hypo'] = dic.pop('lon'), dic.pop('lat'), dic.pop('dep')\nrups.append(_get_rupture(dic))\nreturn rups\n\ndef _get_rupture(rec, geom=None, trt=None):\n# rec: a dictionary or a record\n# geom: if any, an array of floats32 convertible into a mesh\nif not code2cls:\ncode2cls.update(BaseRupture.init())\nif geom is None:\nlons = rec['lons']\nmesh = numpy.zeros((3, len(lons), len(lons)), F32)\nmesh = rec['lons']\nmesh = rec['lats']\nmesh = rec['depths']\nelse:\nmesh = geom.reshape(rec['s1'], rec['s2'], 3).transpose(2, 0, 1)\nrupture_cls, surface_cls = code2cls[rec['code']]\nrupture = object.__new__(rupture_cls)\nrupture.rup_id = rec['serial']\nrupture.surface = object.__new__(surface_cls)\nrupture.mag = rec['mag']\nrupture.rake = rec['rake']\nrupture.hypocenter = geo.Point(rec['hypo'])\nrupture.occurrence_rate = rec['occurrence_rate']\ntry:\nrupture.probs_occur = rec['probs_occur']\nexcept (KeyError, ValueError): # rec can be a numpy record\npass\nrupture.tectonic_region_type = trt or rec['trt']\nif surface_cls is geo.PlanarSurface:\nrupture.surface = geo.PlanarSurface.from_array(\nmesh[:, 0, :])\nelif surface_cls is geo.MultiSurface:\n# mesh has shape (3, n, 4)\nrupture.surface.__init__([\ngeo.PlanarSurface.from_array(mesh[:, i, :])\nfor i in range(mesh.shape)])\nelif surface_cls is geo.GriddedSurface:\n# fault surface, strike and dip will be computed\nrupture.surface.strike = rupture.surface.dip = None\nrupture.surface.mesh = Mesh(mesh)\nelse:\n# fault surface, strike and dip will be computed\nrupture.surface.strike = rupture.surface.dip = None\nrupture.surface.__init__(RectangularMesh(mesh))\nrupture.multiplicity = rec['n_occ']\nreturn rupture\n\n[docs]def float5(x):\n# a float with 5 digits\nreturn round(float(x), 5)\n\ndef _fixfloat32(dic):\n# work around a TOML/numpy issue\nfor k, v in dic.items():\nif isinstance(v, F32):\ndic[k] = float5(v)\nelif isinstance(v, tuple):\ndic[k] = [float5(x) for x in v]\nelif isinstance(v, numpy.ndarray):\nif len(v.shape) == 3: # 3D array\ndic[k] = [[[float5(z) for z in y] for y in x] for x in v]\nelif len(v.shape) == 2: # 2D array\ndic[k] = [[float5(y) for y in x] for x in v]\nelif len(v.shape) == 1: # 1D array\ndic[k] = [float5(x) for x in v]\nelse:\nraise NotImplementedError\n\n[docs]def to_checksum8(cls1, cls2):\n\"\"\"\nConvert a pair of classes into a numeric code (uint8)\n\"\"\"\nnames = '%s,%s' % (cls1.__name__, cls2.__name__)\nreturn sum(map(ord, names)) % 256\n\n[docs]class BaseRupture(metaclass=abc.ABCMeta):\n\"\"\"\nRupture object represents a single earthquake rupture.\n\n:param mag:\nMagnitude of the rupture.\n:param rake:\nRake value of the rupture.\nSee :class:`~openquake.hazardlib.geo.nodalplane.NodalPlane`.\n:param tectonic_region_type:\nRupture's tectonic regime. One of constants\nin :class:`openquake.hazardlib.const.TRT`.\n:param hypocenter:\nA :class:`~openquake.hazardlib.geo.point.Point`, rupture's hypocenter.\n:param surface:\nAn instance of subclass of\n:class:`~openquake.hazardlib.geo.surface.base.BaseSurface`.\nObject representing the rupture surface geometry.\n:param rupture_slip_direction:\nAngle describing rupture propagation direction in decimal degrees.\n\n:raises ValueError:\nIf magnitude value is not positive, or tectonic region type is unknown.\n\nNB: if you want to convert the rupture into XML, you should set the\nattribute surface_nodes to an appropriate value.\n\"\"\"\nrup_id = 0 # set to a value > 0 by the engine\n_code = {}\n\n[docs] @classmethod\ndef init(cls):\n\"\"\"\nInitialize the class dictionary `._code` by encoding the\nbidirectional correspondence between an integer in the range 0..255\n(the code) and a pair of classes (rupture_class, surface_class).\nThis is useful when serializing the rupture to and from HDF5.\n:returns: {code: pair of classes}\n\"\"\"\nrupture_classes = [BaseRupture] + list(\ngeneral.gen_subclasses(BaseRupture))\nsurface_classes = list(general.gen_subclasses(BaseSurface))\ncode2cls = {}\nBaseRupture.str2code = {}\nfor rup, sur in itertools.product(rupture_classes, surface_classes):\nchk = to_checksum8(rup, sur)\nif chk in code2cls and code2cls[chk] != (rup, sur):\nraise ValueError('Non-unique checksum %d for %s, %s' %\n(chk, rup, sur))\ncls._code[rup, sur] = chk\ncode2cls[chk] = rup, sur\nBaseRupture.str2code['%s %s' % (rup.__name__, sur.__name__)] = chk\nreturn code2cls\n\ndef __init__(self, mag, rake, tectonic_region_type, hypocenter,\nsurface, rupture_slip_direction=None, weight=None):\nif not mag > 0:\nraise ValueError('magnitude must be positive')\nNodalPlane.check_rake(rake)\nself.tectonic_region_type = tectonic_region_type\nself.rake = rake\nself.mag = mag\nself.hypocenter = hypocenter\nself.surface = surface\nself.rupture_slip_direction = rupture_slip_direction\nself.weight = weight\n\n@property\ndef code(self):\n\"\"\"Returns the code (integer in the range 0 .. 255) of the rupture\"\"\"\nreturn self._code[self.__class__, self.surface.__class__]\n\nget_probability_no_exceedance = (\ncontexts.RuptureContext.get_probability_no_exceedance)\n\n[docs] def sample_number_of_occurrences(self, n=1):\n\"\"\"\nRandomly sample number of occurrences from temporal occurrence model\nprobability distribution.\n\n.. note::\nThis method is using random numbers. In order to reproduce the\nsame results numpy random numbers generator needs to be seeded, see\nhttp://docs.scipy.org/doc/numpy/reference/generated/numpy.random.seed.html\n\n:returns:\nnumpy array of size n with number of rupture occurrences\n\"\"\"\nraise NotImplementedError\n\n[docs]class NonParametricProbabilisticRupture(BaseRupture):\n\"\"\"\nProbabilistic rupture for which the probability distribution for rupture\noccurrence is described through a generic probability mass function.\n\n:param pmf:\nInstance of :class:`openquake.hazardlib.pmf.PMF`. Values in the\nabscissae represent number of rupture occurrences (in increasing order,\nstaring from 0) and values in the ordinates represent associated\nprobabilities. Example: if, for a given time span, a rupture has\nprobability ``0.8`` to not occurr, ``0.15`` to occur once, and\n``0.05`` to occur twice, the ``pmf`` can be defined as ::\n\npmf = PMF([(0.8, 0), (0.15, 1), 0.05, 2)])\n\n:raises ValueError:\nIf number of ruptures in ``pmf`` do not start from 0, are not defined\nin increasing order, and if they are not defined with unit step\n\"\"\"\ndef __init__(self, mag, rake, tectonic_region_type, hypocenter, surface,\npmf, rupture_slip_direction=None, weight=None):\nocc = numpy.array([occ for (prob, occ) in pmf.data])\nif not occ == 0:\nraise ValueError('minimum number of ruptures must be zero')\nif not numpy.all(numpy.sort(occ) == occ):\nraise ValueError(\n'numbers of ruptures must be defined in increasing order')\nif not numpy.all(numpy.diff(occ) == 1):\nraise ValueError(\n'numbers of ruptures must be defined with unit step')\nsuper().__init__(\nmag, rake, tectonic_region_type, hypocenter, surface,\nrupture_slip_direction, weight)\n# an array of probabilities with sum 1\nself.probs_occur = numpy.array([prob for (prob, occ) in pmf.data])\nself.occurrence_rate = numpy.nan\n\n[docs] def sample_number_of_occurrences(self, n=1):\n\"\"\"\nSee :meth:`superclass method\n<.rupture.BaseRupture.sample_number_of_occurrences>`\nfor spec of input and result values.\n\nUses 'Inverse Transform Sampling' method.\n\"\"\"\n# compute cdf from pmf\ncdf = numpy.cumsum(self.probs_occur)\nn_occ = numpy.digitize(numpy.random.random(n), cdf)\nreturn n_occ\n\n[docs]class ParametricProbabilisticRupture(BaseRupture):\n\"\"\"\n:class:`Rupture` associated with an occurrence rate and a temporal\noccurrence model.\n\n:param occurrence_rate:\nNumber of times rupture happens per year.\n:param temporal_occurrence_model:\nTemporal occurrence model assigned for this rupture. Should\nbe an instance of :class:`openquake.hazardlib.tom.PoissonTOM`.\n\n:raises ValueError:\nIf occurrence rate is not positive.\n\"\"\"\ndef __init__(self, mag, rake, tectonic_region_type, hypocenter, surface,\noccurrence_rate, temporal_occurrence_model,\nrupture_slip_direction=None):\nif not occurrence_rate > 0:\nraise ValueError('occurrence rate must be positive')\nsuper().__init__(\nmag, rake, tectonic_region_type, hypocenter, surface,\nrupture_slip_direction)\nself.temporal_occurrence_model = temporal_occurrence_model\nself.occurrence_rate = occurrence_rate\n\n[docs] def get_probability_one_or_more_occurrences(self):\n\"\"\"\nReturn the probability of this rupture to occur one or more times.\n\nUses\n:meth:`~openquake.hazardlib.tom.PoissonTOM.get_probability_one_or_more_occurrences`\nof an assigned temporal occurrence model.\n\"\"\"\ntom = self.temporal_occurrence_model\nrate = self.occurrence_rate\n\n[docs] def get_probability_one_occurrence(self):\n\"\"\"\nReturn the probability of this rupture to occur exactly one time.\n\nUses :meth:\n`openquake.hazardlib.tom.PoissonTOM.get_probability_n_occurrences`\nof an assigned temporal occurrence model.\n\"\"\"\ntom = self.temporal_occurrence_model\nrate = self.occurrence_rate\n\n[docs] def sample_number_of_occurrences(self, n=1):\n\"\"\"\nDraw a random sample from the distribution and return a number\nof events to occur as an array of integers of size n.\n\nUses :meth:\n`openquake.hazardlib.tom.PoissonTOM.sample_number_of_occurrences`\nof an assigned temporal occurrence model.\n\"\"\"\nr = self.occurrence_rate self.temporal_occurrence_model.time_span\nreturn numpy.random.poisson(r, n)\n\n[docs] def get_probability_no_exceedance(self, poes):\n\"\"\"\nSee :meth:`superclass method\n<.rupture.BaseRupture.get_probability_no_exceedance>`\nfor spec of input and result values.\n\nUses\n:meth:`openquake.hazardlib.tom.PoissonTOM.get_probability_no_exceedance`\n\"\"\"\ntom = self.temporal_occurrence_model\nrate = self.occurrence_rate\n\n[docs] def get_dppvalue(self, site):\n\"\"\"\nGet the directivity prediction value, DPP at\na given site as described in Spudich et al. (2013).\n\n:param site:\n:class:`~openquake.hazardlib.geo.point.Point` object\nrepresenting the location of the target site\n:returns:\nA float number, directivity prediction value (DPP).\n\"\"\"\n\norigin = self.surface.get_resampled_top_edge()\ndpp_multi = []\nindex_patch = self.surface.hypocentre_patch_index(\nself.hypocenter, self.surface.get_resampled_top_edge(),\nself.surface.mesh.depths, self.surface.mesh.depths[-1],\nself.surface.get_dip())\nidx_nxtp = True\nhypocenter = self.hypocenter\n\nwhile idx_nxtp:\n\n# E Plane Calculation\np0, p1, p2, p3 = self.surface.get_fault_patch_vertices(\nself.surface.get_resampled_top_edge(),\nself.surface.mesh.depths,\nself.surface.mesh.depths[-1],\nself.surface.get_dip(), index_patch=index_patch)\n\n[normal, dist_to_plane] = get_plane_equation(\np0, p1, p2, origin)\n\npp = projection_pp(site, normal, dist_to_plane, origin)\npd, e, idx_nxtp = directp(\np0, p1, p2, p3, hypocenter, origin, pp)\npd_geo = origin.point_at(\n(pd 2 + pd 2) ** 0.5, -pd,\nnumpy.degrees(math.atan2(pd, pd)))\n\n# determine the lower bound of E path value\nf1 = geodetic_distance(p0.longitude,\np0.latitude,\np1.longitude,\np1.latitude)\nf2 = geodetic_distance(p2.longitude,\np2.latitude,\np3.longitude,\np3.latitude)\n\nif f1 > f2:\nf = f1\nelse:\nf = f2\n\nfs, rd, r_hyp = average_s_rad(site, hypocenter, origin,\npp, normal, dist_to_plane, e, p0,\np1, self.rupture_slip_direction)\ncprime = isochone_ratio(e, rd, r_hyp)\n\ndpp_exp = cprime * numpy.maximum(e, 0.1 * f) \\\nnumpy.maximum(fs, 0.2)\ndpp_multi.append(dpp_exp)\n\n# check if go through the next patch of the fault\nindex_patch = index_patch + 1\n\nif (len(self.surface.get_resampled_top_edge())\n<= 2) and (index_patch >=\nlen(self.surface.get_resampled_top_edge())):\n\nidx_nxtp = False\nelif index_patch >= len(self.surface.get_resampled_top_edge()):\nidx_nxtp = False\nelif idx_nxtp:\nhypocenter = pd_geo\nidx_nxtp = True\n\n# calculate DPP value of the site.\ndpp = numpy.log(numpy.sum(dpp_multi))\n\nreturn dpp\n\n[docs] def get_cdppvalue(self, target, buf=1.0, delta=0.01, space=2.):\n\"\"\"\nGet the directivity prediction value, centered DPP(cdpp) at\na given site as described in Spudich et al. (2013), and this cdpp is\nused in Chiou and Young (2014) GMPE for near-fault directivity\nterm prediction.\n\n:param target_site:\nA mesh object representing the location of the target sites.\n:param buf:\nA buffer distance in km to extend the mesh borders\n:param delta:\nThe distance between two adjacent points in the mesh\n:param space:\nThe tolerance for the distance of the sites (default 2 km)\n:returns:\nThe centered directivity prediction value of Chiou and Young\n\"\"\"\nmin_lon, max_lon, max_lat, min_lat = self.surface.get_bounding_box()\nmin_lon -= buf\nmax_lon += buf\nmin_lat -= buf\nmax_lat += buf\n\nlons = numpy.arange(min_lon, max_lon + delta, delta)\n# ex shape (233,)\nlats = numpy.arange(min_lat, max_lat + delta, delta)\n# ex shape (204,)\nmesh = RectangularMesh(numpy.meshgrid(lons, lats))\nmesh_rup = self.surface.get_min_distance(mesh).reshape(mesh.shape)\n# ex shape (204, 233)\n\ntarget_rup = self.surface.get_min_distance(target)\n# ex shape (2,)\ncdpp = numpy.zeros_like(target.lons)\nfor i, (target_lon, target_lat) in enumerate(\nzip(target.lons, target.lats)):\n# indices around target_rup[i]\naround = (mesh_rup <= target_rup[i] + space) & (\nmesh_rup >= target_rup[i] - space)\ndpp_target = self.get_dppvalue(Point(target_lon, target_lat))\ndpp_mean = numpy.mean(\n[self.get_dppvalue(Point(lon, lat))\nfor lon, lat in zip(mesh.lons[around], mesh.lats[around])])\ncdpp[i] = dpp_target - dpp_mean\n\nreturn cdpp\n\n[docs]class PointSurface:\n\"\"\"\nA fake surface used in PointRuptures\n\"\"\"\ndef __init__(self, hypocenter):\nself.hypocenter = hypocenter\n\n[docs] def get_strike(self):\nreturn 0\n\n[docs] def get_dip(self):\nreturn 0\n\n[docs] def get_top_edge_depth(self):\nreturn self.hypocenter.depth\n\n[docs] def get_width(self):\nreturn 0\n\n[docs] def get_closest_points(self, mesh):\nreturn mesh\n\ndef __bool__(self):\nreturn False\n\n[docs]class PointRupture(ParametricProbabilisticRupture):\n\"\"\"\nA rupture coming from a far away PointSource, so that the finite\nsize effects can be neglected.\n\"\"\"\ndef __init__(self, mag, tectonic_region_type, hypocenter,\noccurrence_rate, temporal_occurrence_model):\nself.mag = mag\nself.rake = 0\nself.tectonic_region_type = tectonic_region_type\nself.hypocenter = hypocenter\nself.occurrence_rate = occurrence_rate\nself.temporal_occurrence_model = temporal_occurrence_model\nself.surface = PointSurface(hypocenter)\nself.weight = None # no mutex\n\n[docs]def get_geom(surface, is_from_fault_source, is_multi_surface,\nis_gridded_surface):\n\"\"\"\nThe following fields can be interpreted different ways,\ndepending on the value of `is_from_fault_source`. If\n`is_from_fault_source` is True, each of these fields should\ncontain a 2D numpy array (all of the same shape). Each triple\nof (lon, lat, depth) for a given index represents the node of\na rectangular mesh. If `is_from_fault_source` is False, each\nof these fields should contain a sequence (tuple, list, or\nnumpy array, for example) of 4 values. In order, the triples\nof (lon, lat, depth) represent top left, top right, bottom\nleft, and bottom right corners of the the rupture's planar\nsurface. Update: There is now a third case. If the rupture\noriginated from a characteristic fault source with a\nmulti-planar-surface geometry, `lons`, `lats`, and `depths`\nwill contain one or more sets of 4 points, similar to how\nplanar surface geometry is stored (see above).\n\n:param surface: a Surface instance\n:param is_from_fault_source: a boolean\n:param is_multi_surface: a boolean\n\"\"\"\nif is_from_fault_source:\n# for simple and complex fault sources,\n# rupture surface geometry is represented by a mesh\nsurf_mesh = surface.mesh\nlons = surf_mesh.lons\nlats = surf_mesh.lats\ndepths = surf_mesh.depths\nelse:\nif is_multi_surface:\n# `list` of\n# openquake.hazardlib.geo.surface.planar.PlanarSurface\n# objects:\nsurfaces = surface.surfaces\n\n# lons, lats, and depths are arrays with len == 4N,\n# where N is the number of surfaces in the\n# multisurface for each `corner_`, the ordering is:\n# - top left\n# - top right\n# - bottom left\n# - bottom right\nlons = numpy.concatenate([x.corner_lons for x in surfaces])\nlats = numpy.concatenate([x.corner_lats for x in surfaces])\ndepths = numpy.concatenate([x.corner_depths for x in surfaces])\nelif is_gridded_surface:\n# the surface mesh has shape (1, N)\nlons = surface.mesh.lons\nlats = surface.mesh.lats\ndepths = surface.mesh.depths\nelse:\n# For area or point source,\n# rupture geometry is represented by a planar surface,\n# defined by 3D corner points\nlons = numpy.zeros((4))\nlats = numpy.zeros((4))\ndepths = numpy.zeros((4))\n\n# NOTE: It is important to maintain the order of these\n# corner points. TODO: check the ordering\nfor i, corner in enumerate((surface.top_left,\nsurface.top_right,\nsurface.bottom_left,\nsurface.bottom_right)):\nlons[i] = corner.longitude\nlats[i] = corner.latitude\ndepths[i] = corner.depth\nreturn lons, lats, depths\n\n[docs]class ExportedRupture(object):\n\"\"\"\nSimplified Rupture class with attributes rupid, events_by_ses, indices\nand others, used in export.\n\n:param rupid: rupture rup_id ID\n:param events_by_ses: dictionary ses_idx -> event records\n:param indices: site indices\n\"\"\"\ndef __init__(self, rupid, n_occ, events_by_ses, indices=None):\nself.rupid = rupid\nself.n_occ = n_occ\nself.events_by_ses = events_by_ses\nself.indices = indices\n\n[docs]def get_eids(rup_array, samples_by_grp, num_rlzs_by_grp):\n\"\"\"\n:param rup_array: a composite array with fields rup_id, n_occ and grp_id\n:param samples_by_grp: a dictionary grp_id -> samples\n:param num_rlzs_by_grp: a dictionary grp_id -> num_rlzs\n\"\"\"\nall_eids = []\nfor rup in rup_array:\ngrp_id = rup['grp_id']\nsamples = samples_by_grp[grp_id]\nnum_rlzs = num_rlzs_by_grp[grp_id]\nnum_events = rup['n_occ'] if samples > 1 else rup['n_occ'] * num_rlzs\neids = numpy.arange(num_events, dtype=U32)\nall_eids.append(eids)\nreturn numpy.concatenate(all_eids)\n\n[docs]class EBRupture(object):\n\"\"\"\nAn event based rupture. It is a wrapper over a hazardlib rupture\nobject, containing an array of site indices affected by the rupture,\nas well as the IDs of the corresponding seismic events.\n\"\"\"\ndef __init__(self, rupture, source_id, grp_id, n_occ, samples=1, id=None):\n# NB: when reading an exported ruptures.xml the rup_id will be 0\n# for the first rupture; it used to be the seed instead\nassert rupture.rup_id >= 0 # sanity check\nself.rupture = rupture\nself.source_id = source_id\nself.grp_id = grp_id\nself.n_occ = n_occ\nself.samples = samples\nself.id = id # id of the rupture on the DataStore, to be overridden\n\n@property\ndef rup_id(self):\n\"\"\"\nSerial number of the rupture\n\"\"\"\nreturn self.rupture.rup_id\n\n[docs] def get_eids_by_rlz(self, rlzs_by_gsim, offset=0):\n\"\"\"\n:params rlzs_by_gsim: a dictionary gsims -> rlzs array\n:param offset: offset used in the calculation of the event ID\n:returns: a dictionary rlz index -> eids array\n\"\"\"\nj = 0\ndic = {}\nif self.samples == 1: # full enumeration or akin to it\nfor rlzs in rlzs_by_gsim.values():\nfor rlz in rlzs:\ndic[rlz] = numpy.arange(\nj, j + self.n_occ, dtype=U32) + offset\nj += self.n_occ\nelse: # associated eids to the realizations\nrlzs = numpy.concatenate(list(rlzs_by_gsim.values()))\nhisto = general.random_histogram(\nself.n_occ, len(rlzs), self.rup_id)\nfor rlz, n in zip(rlzs, histo):\ndic[rlz] = numpy.arange(j, j + n, dtype=U32) + offset\nj += n\nreturn dic\n\n[docs] def get_eids(self, num_rlzs):\n\"\"\"\n:param num_rlzs: the number of realizations for the given group\n:returns: an array of event IDs\n\"\"\"\nnum_events = self.n_occ if self.samples > 1 else self.n_occ * num_rlzs\nreturn numpy.arange(num_events, dtype=U32)\n\n[docs] def export(self, events_by_ses):\n\"\"\"\nYield :class:`Rupture` objects, with all the\nattributes set, suitable for export in XML format.\n\"\"\"\nrupture = self.rupture\nnew = ExportedRupture(self.id, self.n_occ, events_by_ses)\nif isinstance(rupture.surface, geo.ComplexFaultSurface):\nnew.typology = 'complexFaultsurface'\nelif isinstance(rupture.surface, geo.SimpleFaultSurface):\nnew.typology = 'simpleFaultsurface'\nelif isinstance(rupture.surface, geo.GriddedSurface):\nnew.typology = 'griddedRupture'\nelif isinstance(rupture.surface, geo.MultiSurface):\nnew.typology = 'multiPlanesRupture'\nelse:\nnew.typology = 'singlePlaneRupture'\nnew.is_from_fault_source = iffs = isinstance(\nrupture.surface, (geo.ComplexFaultSurface,\ngeo.SimpleFaultSurface))\nnew.is_gridded_surface = igs = isinstance(\nrupture.surface, geo.GriddedSurface)\nnew.is_multi_surface = ims = isinstance(\nrupture.surface, geo.MultiSurface)\nnew.lons, new.lats, new.depths = get_geom(\nrupture.surface, iffs, ims, igs)\nnew.surface = rupture.surface\nnew.strike = rupture.surface.get_strike()\nnew.dip = rupture.surface.get_dip()\nnew.rake = rupture.rake\nnew.hypocenter = rupture.hypocenter\nnew.tectonic_region_type = rupture.tectonic_region_type\nnew.magnitude = new.mag = rupture.mag\nnew.top_left_corner = None if iffs or ims or igs else (\nnew.lons, new.lats, new.depths)\nnew.top_right_corner = None if iffs or ims or igs else (\nnew.lons, new.lats, new.depths)\nnew.bottom_left_corner = None if iffs or ims or igs else (\nnew.lons, new.lats, new.depths)\nnew.bottom_right_corner = None if iffs or ims or igs else (\nnew.lons, new.lats, new.depths)\nreturn new\n\ndef __repr__(self):\nreturn '<%s %d[%d]>' % (\nself.__class__.__name__, self.rup_id, self.n_occ)\n\n[docs]class RuptureProxy(object):\n\"\"\"\nA proxy for a rupture record.\n\n:param rec: a record with the rupture parameters\n:param nsites: approx number of sites affected by the rupture\n:param samples: how many times the rupture is sampled\n\"\"\"\ndef __init__(self, rec, nsites=None, samples=1):\nself.rec = rec\nself.nsites = nsites\nself.samples = samples\n\n@property\ndef weight(self):\n\"\"\"\n:returns:\nheuristic weight for the underlying rupture, depending on the\nnumber of occurrences, number of samples and number of sites\n\"\"\"\nreturn self.samples * self['n_occ'] * (\n100 if self.nsites is None else max(self.nsites, 100))\n\ndef __getitem__(self, name):\nreturn self.rec[name]\n\n# NB: requires the .geom attribute to be set\n[docs] def to_ebr(self, trt, samples):\n\"\"\"\n:returns: EBRupture instance associated to the underlying rupture\n\"\"\"\n# not implemented: rupture_slip_direction\nrupture = _get_rupture(self.rec, self.geom, trt)\nebr = EBRupture(rupture, self.rec['source_id'], self.rec['grp_id'],\nself.rec['n_occ'], samples)\nebr.id = self.rec['id']\nebr.e0 = self.rec['e0']\nreturn ebr\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54135126,"math_prob":0.928422,"size":25370,"snap":"2021-04-2021-17","text_gpt3_token_len":7078,"char_repetition_ratio":0.15087125,"word_repetition_ratio":0.06278027,"special_character_ratio":0.2743792,"punctuation_ratio":0.24427317,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9847264,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-28T11:36:18Z\",\"WARC-Record-ID\":\"<urn:uuid:3438f58c-9f20-40a1-a27a-96b8ac1838ec>\",\"Content-Length\":\"132862\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6c3475be-1c72-4181-98c8-15aa7314b9b2>\",\"WARC-Concurrent-To\":\"<urn:uuid:9dab030c-0e86-48a0-9631-ade12b788fb0>\",\"WARC-IP-Address\":\"193.206.66.167\",\"WARC-Target-URI\":\"https://docs.openquake.org/oq-engine/3.10/_modules/openquake/hazardlib/source/rupture.html\",\"WARC-Payload-Digest\":\"sha1:YDBKXKZOLSOF466H266CVJM3DPNST4DE\",\"WARC-Block-Digest\":\"sha1:XCOQKFLUYWOWCTC6QDEI37ONUG4DAGOU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610704843561.95_warc_CC-MAIN-20210128102756-20210128132756-00075.warc.gz\"}"}
https://www.engageny.org/ccls-math/6g2	[ " 6.G.2 \| EngageNY\n\n## CCLS - Math: 6.G.2\n\nCategory\nGeometry\nSub-Category\nSolve Real-World And Mathematical Problems Involving Area, Surface Area, And Volume.\nState Standard:\nFind the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = l w h and V = b h to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86529523,"math_prob":0.98225546,"size":2336,"snap":"2019-35-2019-39","text_gpt3_token_len":457,"char_repetition_ratio":0.1590909,"word_repetition_ratio":0.07486631,"special_character_ratio":0.19905822,"punctuation_ratio":0.11386139,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9959988,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-20T09:37:22Z\",\"WARC-Record-ID\":\"<urn:uuid:d785fb74-46cd-4396-8d60-c8e7237a62a1>\",\"Content-Length\":\"66408\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f9f03fa6-d40b-4a0b-bc05-0635f5a9acca>\",\"WARC-Concurrent-To\":\"<urn:uuid:ddc0ebdb-afd9-469e-8c3a-14ef885db524>\",\"WARC-IP-Address\":\"3.227.40.88\",\"WARC-Target-URI\":\"https://www.engageny.org/ccls-math/6g2\",\"WARC-Payload-Digest\":\"sha1:FSA7WULJXHVIFBPV37G2IWTJ743K43XC\",\"WARC-Block-Digest\":\"sha1:PMLWRKQFIVXY67NALO3HCAGDTBBUTLQ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027315321.52_warc_CC-MAIN-20190820092326-20190820114326-00442.warc.gz\"}"}
https://studylib.net/doc/11723333/-ey-math-2270-exam-1	[ "# (ey Math 2270 Exam 1", null, "```Math 2270 Exam 1\n-\nUniversity of Utah\nFall 2012\nName:\n(ey\nThis is a 50 minute exam. Please show all your work, as a worked\nproblem is required for full points, and partial credit may be rewarded for\nsome work in the right direction.\n1. (15 points) Vector Basics\nFor the vectors\narrr( 1)\nb=f i)\n4J\n\\1)\n/1\nc=f 2\nanswer the following, or explain why the question does not make\nsense:\n(a) (3 points) 2a + 3c\ni(\n=\n(D\n(‘)f()\n1\ni)\n4)\narrf\nb=(\nii\n/1\nc=( 2\n1J\n(b) (3 points) aH\nl+I6\n(c) (2 points) What are the components of a unit vector in the same\ndirection as a?\n/\n/\n2\nc)\nN\ncD.\nN\n0\nQ\nII\n+\nH\nr\nT\n0\nI.\nII\nI’\nI\n2. (10 points) Matrix Basics\nFor the matrices\n(3 4 2\nA\n2\n1 1\nB=\n/2 1 5\n4 4 2\n1 02\n(\n1 1\n0 0\nanswer the following, or explain why the question does not make\nsense:\n(a) (3points)A+Crr\n(3\n-\nttj\n16oO\n4\ncDCC\nCDCN\nII\n(z\nC\nr\\rJ\n—\nC\nCj\nc)\n0\n0\n‘-Th\n1\nc’-\nLfl\n3. (15 points) Elimination Issues\n(a) (5 points) For what value of a in the system of equations below\ndoes elimination fail to produce a unique solution?\n3x + 2y\n6x + ay\n=\n=\nir\n11\n/c{\nJi1 (\na e\nC\n10\nb\nrc ie4\nj\n1\nru\nZ\n1/\n/ce;\nq\nh-\n/e\np\nO\nOy\n(b) (5 points) Given the determined value of a, for what value of b\nare there an infinite number of solutions?\nT\nb?o\n7(j\no Oy\na(ty1\ns\n(c) (5 points) For the determined values of a and b what are two dis\ntinct solutions?\n/\n0\ny\nThte\n/\nqfN/\no4t\n6\n(QL/J\n4. (20 points) Systems of Equations\nUse elementary row operations to convert the system of equations\n2r\n6r\n12x\n+\n+\n+\n3y +\n6y +\n9y\n—\n3z\n=\n12z\nz\n=\n3\n13\n2\ninto upper-triangular form, and then use back-substitution to solve\nfor the variables x, y, z. Be sure to show all your work.\n5\n(Rc\n1Q-/\n5\nL +\nt1\nZ\nY\n3\nfc(J3\nZ\\4Jy 3y\nLI\ny\n3\n-\nrc, /J\n(f or/\n7\nx\na)\n—c\n—QQ\nI\noç\n—\n—\n—\n—\n—\n-\nc_J\n_\n,\nH\nC\na)\n—\n0\n——H\n_Lf\n—\na)\n-d\nco\nI\n----\n—\n11\nJZ\ncx-)\nc)\n><\nct\n0)\nC\n0\nU\nJ —r\n0\nUN\nUi\nfl\n7. (10 points) Symmetric Products\nFor the matrix\n2 3\n4 0\n(a) (4 points) What is the transpose RT?\n(b) (4 points) What is the symmetric product RTR\n73\n1\nLC)\n30/\n(c) (2 points) Does RTR\n=\nRRT?\n/\nhe\n10\n```" ]	[ null, "https://s2.studylib.net/store/data/011723333_1-79e99dfeb322d406a71195248ecfbdd4-768x994.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7920167,"math_prob":0.988489,"size":2006,"snap":"2021-21-2021-25","text_gpt3_token_len":782,"char_repetition_ratio":0.11188811,"word_repetition_ratio":0.042016808,"special_character_ratio":0.40677965,"punctuation_ratio":0.06584362,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9903635,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-15T20:40:28Z\",\"WARC-Record-ID\":\"<urn:uuid:c208dd32-8033-4aeb-bcf0-882f55cea3f4>\",\"Content-Length\":\"45355\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d2db1efb-5ce2-45dc-81d6-2370a738174a>\",\"WARC-Concurrent-To\":\"<urn:uuid:07a61eb4-e377-41b2-9607-ddb3f0de7c98>\",\"WARC-IP-Address\":\"104.21.72.58\",\"WARC-Target-URI\":\"https://studylib.net/doc/11723333/-ey-math-2270-exam-1\",\"WARC-Payload-Digest\":\"sha1:JZ26FHVBPA76BXMKW5GLS6CDN3YFVZRQ\",\"WARC-Block-Digest\":\"sha1:JTFBRXF5EYBRPYXVPYXJLIIL5WGEEVT6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991378.52_warc_CC-MAIN-20210515192444-20210515222444-00615.warc.gz\"}"}
https://chem.libretexts.org/Courses/BethuneCookman_University/B-CU%3A_CH-345_Quantitative_Analysis/Book%3A_Analytical_Chemistry_2.1_(Harvey)/03%3A__The_Vocabulary_of_Analytical_Chemistry/3.03%3A_Classifying_Analytical_Techniques	[ "# 3.3: Classifying Analytical Techniques\n\n$$\\newcommand{\\vecs}{\\overset { \\rightharpoonup} {\\mathbf{#1}} }$$ $$\\newcommand{\\vecd}{\\overset{-\\!-\\!\\rightharpoonup}{\\vphantom{a}\\smash {#1}}}$$$$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\id}{\\mathrm{id}}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$ $$\\newcommand{\\kernel}{\\mathrm{null}\\,}$$ $$\\newcommand{\\range}{\\mathrm{range}\\,}$$ $$\\newcommand{\\RealPart}{\\mathrm{Re}}$$ $$\\newcommand{\\ImaginaryPart}{\\mathrm{Im}}$$ $$\\newcommand{\\Argument}{\\mathrm{Arg}}$$ $$\\newcommand{\\norm}{\\\| #1 \\\|}$$ $$\\newcommand{\\inner}{\\langle #1, #2 \\rangle}$$ $$\\newcommand{\\Span}{\\mathrm{span}}$$$$\\newcommand{\\AA}{\\unicode[.8,0]{x212B}}$$\n\nThe analysis of a sample generates a chemical or physical signal that is proportional to the amount of analyte in the sample. This signal may be anything we can measure, such as volume or absorbance. It is convenient to divide analytical techniques into two general classes based on whether the signal is proportional to the mass or moles of analyte, or is proportional to the analyte’s concentration\n\nConsider the two graduated cylinders in Figure $$\\PageIndex{1}$$, each of which contains a solution of 0.010 M Cu(NO3)2. Cylinder 1 contains 10 mL, or $$1.0 \\times 10^{-4}$$ moles of Cu2+, and cylinder 2 contains 20 mL, or $$2.0 \\times 10^{-4}$$ moles of Cu2+. If a technique responds to the absolute amount of analyte in the sample, then the signal due to the analyte SA\n\n$S_A = k_A n_A \\label{3.1}$\n\nwhere nA is the moles or grams of analyte in the sample, and kA is a proportionality constant. Because cylinder 2 contains twice as many moles of Cu2+ as cylinder 1, analyzing the contents of cylinder 2 gives a signal twice as large as that for cylinder 1.", null, "Figure $$\\PageIndex{1}$$: Two graduated cylinders, each containing 0.10 M Cu(NO3)2. Although the cylinders contain the same concentration of Cu2+, the cylinder on the left contains $$1.0 \\times 10^{-4}$$ mol Cu2+ and the cylinder on the right contains $$2.0 \\times 10^{-4}$$ mol Cu2+.\n\nA second class of analytical techniques are those that respond to the analyte’s concentration, CA\n\n$S_A = k_A C_A \\label{3.2}$\n\nSince the solutions in both cylinders have the same concentration of Cu2+, their analysis yields identical signals.\n\nA technique that responds to the absolute amount of analyte is a total analysis technique. Mass and volume are the most common signals for a total analysis technique, and the corresponding techniques are gravimetry (Chapter 8) and titrimetry (Chapter 9). With a few exceptions, the signal for a total analysis technique is the result of one or more chemical reactions, the stoichiometry of which determines the value of kA in equation \\ref{3.1}.\n\nHistorically, most early analytical methods used a total analysis technique. For this reason, total analysis techniques are often called “classical” techniques.\n\nSpectroscopy (Chapter 10) and electrochemistry (Chapter 11), in which an optical or an electrical signal is proportional to the relative amount of analyte in a sample, are examples of concentration techniques. The relationship between the signal and the analyte’s concentration is a theoretical function that depends on experimental conditions and the instrumentation used to measure the signal. For this reason the value of kA in equation \\ref{3.2} is determined experimentally.\n\nSince most concentration techniques rely on measuring an optical or electrical signal, they also are known as “instrumental” techniques.\n\nThis page titled 3.3: Classifying Analytical Techniques is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by David Harvey." ]	[ null, "https://chem.libretexts.org/@api/deki/files/272990/GradCylindCropped.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8982236,"math_prob":0.99731827,"size":2585,"snap":"2023-40-2023-50","text_gpt3_token_len":588,"char_repetition_ratio":0.16117784,"word_repetition_ratio":0.039215688,"special_character_ratio":0.22591875,"punctuation_ratio":0.09322034,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9955774,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-30T12:58:18Z\",\"WARC-Record-ID\":\"<urn:uuid:94df5c09-698f-4441-b5e0-be72ee103f74>\",\"Content-Length\":\"126160\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8b27fd07-6dd8-423f-857d-d2c204b14e43>\",\"WARC-Concurrent-To\":\"<urn:uuid:32c20867-2848-45c8-9ddb-760499bca293>\",\"WARC-IP-Address\":\"3.162.103.49\",\"WARC-Target-URI\":\"https://chem.libretexts.org/Courses/BethuneCookman_University/B-CU%3A_CH-345_Quantitative_Analysis/Book%3A_Analytical_Chemistry_2.1_(Harvey)/03%3A__The_Vocabulary_of_Analytical_Chemistry/3.03%3A_Classifying_Analytical_Techniques\",\"WARC-Payload-Digest\":\"sha1:7RXE3IEWYOS6WF6PBMY7Q2MKVV3CQ5BK\",\"WARC-Block-Digest\":\"sha1:LA2CZYU4H7CMHJW6XLP6SNRIMMHDIESJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510676.40_warc_CC-MAIN-20230930113949-20230930143949-00605.warc.gz\"}"}
https://www.calculatoratoz.com/en/depth-of-flow1-when-celerity-of-the-wave-is-given-calculator/Calc-19983	[ "🔍\n🔍\n\n## Credits\n\nNational Institute of Technology Karnataka (NITK), Surathkal\nRithik Agrawal has created this Calculator and 1000+ more calculators!\nMeerut Institute of Engineering and Technology (MIET), Meerut\nIshita Goyal has verified this Calculator and 1000+ more calculators!\n\n## Depth of flow1 when Celerity of the Wave is Given Solution\n\nSTEP 0: Pre-Calculation Summary\nFormula Used\ndepth1 = -((((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))-2)/(((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))Depth of Point 2))\nh 1 = -((((h[g])/(CV1))-2)/(((h[g])/(CV1))h 2))\nThis formula uses 1 Constants, 4 Variables\nConstants Used\n[g] - Gravitational acceleration on Earth Value Taken As 9.80665 Meter/Second²\nVariables Used\nHeight - Height is the distance between the lowest and highest points of a person standing upright. (Measured in Meter)\nCelerity of the Wave - Celerity of the Wave is the addition to the normal water velocity of the channels. (Measured in Meter per Second)\nVelocity_of the fluid at 1 - Velocity_of the fluid at 1 is defined as the velocity of the flowing liquid at a point 1 (Measured in Meter per Second)\nDepth of Point 2 - Depth of Point 2 is the depth of point below the free surface in a static mass of liquid. (Measured in Meter)\nSTEP 1: Convert Input(s) to Base Unit\nHeight: 12 Meter --> 12 Meter No Conversion Required\nCelerity of the Wave: 10 Meter per Second --> 10 Meter per Second No Conversion Required\nVelocity_of the fluid at 1: 10 Meter per Second --> 10 Meter per Second No Conversion Required\nDepth of Point 2: 15 Meter --> 15 Meter No Conversion Required\nSTEP 2: Evaluate Formula\nSubstituting Input Values in Formula\nh 1 = -((((h[g])/(CV1))-2)/(((h[g])/(CV1))h 2)) --> -((((12[g])/(1010))-2)/(((12[g])/(1010))15))\nEvaluating ... ...\nh 1 = 0.0466351347753254\nSTEP 3: Convert Result to Output's Unit\n0.0466351347753254 Meter --> No Conversion Required\nFINAL ANSWER\n0.0466351347753254 Meter <-- Depth of Point 1\n(Calculation completed in 00.047 seconds)\n\n## < 11 Other formulas that you can solve using the same Inputs\n\nVolume of a Conical Frustum\nvolume = (1/3)piHeight(Radius 1^2+Radius 2^2+(Radius 1Radius 2)) Go\nTotal Surface Area of a Cone\ntotal_surface_area = piRadius(Radius+sqrt(Radius^2+Height^2)) Go\nLateral Surface Area of a Cone\nlateral_surface_area = piRadiussqrt(Radius^2+Height^2) Go\nTotal Surface Area of a Cylinder\ntotal_surface_area = 2piRadius(Height+Radius) Go\nLateral Surface Area of a Cylinder\nlateral_surface_area = 2piRadiusHeight Go\nVolume of a Circular Cone\nvolume = (1/3)pi(Radius)^2Height Go\nArea of a Trapezoid\narea = ((Base A+Base B)/2)Height Go\nVolume of a Circular Cylinder\nvolume = pi(Radius)^2Height Go\nVolume of a Pyramid\nvolume = (1/3)Side^2Height Go\nArea of a Triangle when base and height are given\narea = 1/2BaseHeight Go\nArea of a Parallelogram when base and height are given\narea = BaseHeight Go\n\n## < 7 Other formulas that calculate the same Output\n\nConjugate Depth y1 when Critical Depth is Given\ndepth1 = 0.5Depth of Point 2(-1+sqrt(1+(8(critical depthcritical depthcritical depth))/(Depth of Point 2Depth of Point 2Depth of Point 2))) Go\nConjugate Depth y1 when Discharge per unit width of channel is Given\ndepth1 = 0.5Depth of Point 2(-1+sqrt(1+(8(discharge per unit width ^2))/([g]Depth of Point 2Depth of Point 2Depth of Point 2))) Go\nDepth of flow1 when Absolute velocity of the surge moving towards right is Given\ndepth1 = ((Absolute Velocity of the Issuing Jet-Velocity_of the fluid at 2)/(Absolute Velocity of the Issuing Jet-Velocity_of the fluid at 1))Depth of Point 2 Go\nDepth of flow1 when Absolute velocity of the surge moving towards right is Given\ndepth1 = ((Absolute Velocity of the Issuing Jet+Velocity_of the fluid at 2)/(Absolute Velocity of the Issuing Jet+Velocity_of the fluid at 1))Depth of Point 2 Go\nDepth of flow1 when Absolute velocity of the surge when the flow is completely stopped\ndepth1 = ((Absolute Velocity of the Issuing Jet)/(Absolute Velocity of the Issuing Jet-Velocity_of the fluid at 1))Depth of Point 2 Go\nConjugate Depth y1 when Froude Number Fr1 is Given\ndepth1 = Depth of Point 2/(0.5(-1+sqrt(1+(8(Froude number^2))))) Go\nConjugate Depth y1 when Froude Number Fr2 is Given\ndepth1 = Depth of Point 2(0.5(-1+sqrt(1+(8(Froude number^2))))) Go\n\n### Depth of flow1 when Celerity of the Wave is Given Formula\n\ndepth1 = -((((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))-2)/(((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))Depth of Point 2))\nh 1 = -((((h[g])/(CV1))-2)/(((h[g])/(CV1))h 2))\n\n## What is Depth of Flow ?\n\nNormal depth is the depth of flow in a channel or culvert when the slope of the water surface and channel bottom is the same and the water depth remains constant. ... Note: Flow at normal depth in culverts often presents the highest average velocities and shallowest depths at that flow.\n\n## How to Calculate Depth of flow1 when Celerity of the Wave is Given?\n\nDepth of flow1 when Celerity of the Wave is Given calculator uses depth1 = -((((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))-2)/(((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))Depth of Point 2)) to calculate the Depth of Point 1, The Depth of flow1 when Celerity of the Wave is Given formula is defined as amount of water flowing in through the channel. Depth of Point 1 and is denoted by h 1 symbol.\n\nHow to calculate Depth of flow1 when Celerity of the Wave is Given using this online calculator? To use this online calculator for Depth of flow1 when Celerity of the Wave is Given, enter Height (h), Celerity of the Wave (C), Velocity_of the fluid at 1 (V1) and Depth of Point 2 (h 2) and hit the calculate button. Here is how the Depth of flow1 when Celerity of the Wave is Given calculation can be explained with given input values -> 0.046635 = -((((12[g])/(1010))-2)/(((12[g])/(1010))15)).\n\n### FAQ\n\nWhat is Depth of flow1 when Celerity of the Wave is Given?\nThe Depth of flow1 when Celerity of the Wave is Given formula is defined as amount of water flowing in through the channel and is represented as h 1 = -((((h[g])/(CV1))-2)/(((h[g])/(CV1))h 2)) or depth1 = -((((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))-2)/(((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))Depth of Point 2)). Height is the distance between the lowest and highest points of a person standing upright, Celerity of the Wave is the addition to the normal water velocity of the channels. , Velocity_of the fluid at 1 is defined as the velocity of the flowing liquid at a point 1 and Depth of Point 2 is the depth of point below the free surface in a static mass of liquid.\nHow to calculate Depth of flow1 when Celerity of the Wave is Given?\nThe Depth of flow1 when Celerity of the Wave is Given formula is defined as amount of water flowing in through the channel is calculated using depth1 = -((((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))-2)/(((Height[g])/(Celerity of the WaveVelocity_of the fluid at 1))Depth of Point 2)). To calculate Depth of flow1 when Celerity of the Wave is Given, you need Height (h), Celerity of the Wave (C), Velocity_of the fluid at 1 (V1) and Depth of Point 2 (h 2). With our tool, you need to enter the respective value for Height, Celerity of the Wave, Velocity_of the fluid at 1 and Depth of Point 2 and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.\nHow many ways are there to calculate Depth of Point 1?\nIn this formula, Depth of Point 1 uses Height, Celerity of the Wave, Velocity_of the fluid at 1 and Depth of Point 2. We can use 7 other way(s) to calculate the same, which is/are as follows -\n• depth1 = 0.5Depth of Point 2(-1+sqrt(1+(8(discharge per unit width ^2))/([g]Depth of Point 2Depth of Point 2Depth of Point 2)))\n• depth1 = 0.5Depth of Point 2(-1+sqrt(1+(8(critical depthcritical depthcritical depth))/(Depth of Point 2Depth of Point 2Depth of Point 2)))\n• depth1 = Depth of Point 2/(0.5(-1+sqrt(1+(8(Froude number^2)))))\n• depth1 = Depth of Point 2(0.5(-1+sqrt(1+(8(Froude number^2)))))\n• depth1 = ((Absolute Velocity of the Issuing Jet-Velocity_of the fluid at 2)/(Absolute Velocity of the Issuing Jet-Velocity_of the fluid at 1))Depth of Point 2\n• depth1 = ((Absolute Velocity of the Issuing Jet+Velocity_of the fluid at 2)/(Absolute Velocity of the Issuing Jet+Velocity_of the fluid at 1))Depth of Point 2\n• depth1 = ((Absolute Velocity of the Issuing Jet)/(Absolute Velocity of the Issuing Jet-Velocity_of the fluid at 1))Depth of Point 2", null, "Let Others Know\nLinkedIn\nEmail\nWhatsApp\nCopied!" ]	[ null, "https://www.calculatoratoz.com/Images/share.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.79827386,"math_prob":0.99492043,"size":1003,"snap":"2021-04-2021-17","text_gpt3_token_len":277,"char_repetition_ratio":0.21521522,"word_repetition_ratio":0.2275449,"special_character_ratio":0.2991027,"punctuation_ratio":0.055555556,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99981385,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-04-13T05:02:15Z\",\"WARC-Record-ID\":\"<urn:uuid:fecb02b4-0006-43ae-a735-3c10216dcca7>\",\"Content-Length\":\"157973\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c1579bd3-c9cb-45a4-a73a-e1ad6307bb23>\",\"WARC-Concurrent-To\":\"<urn:uuid:bd6f75d4-a423-4749-9f8c-6265b0c9c5bc>\",\"WARC-IP-Address\":\"67.225.147.106\",\"WARC-Target-URI\":\"https://www.calculatoratoz.com/en/depth-of-flow1-when-celerity-of-the-wave-is-given-calculator/Calc-19983\",\"WARC-Payload-Digest\":\"sha1:YHGPRG4ME4T2OPLJAUB6I7SGL245M6CI\",\"WARC-Block-Digest\":\"sha1:AOCAZ7ATS3DP257GNIN63NC2UCXEVI2K\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-17/CC-MAIN-2021-17_segments_1618038072082.26_warc_CC-MAIN-20210413031741-20210413061741-00536.warc.gz\"}"}
https://crypto.stackexchange.com/questions/35306/why-is-a-one-time-mac-secure-for-mac-then-encrypt-with-randomized-ctr-but-not-ra	[ "# Why is a one-time MAC secure for MAC-then-encrypt with randomized-CTR but not randomized-CBC?\n\nThe Coursera cryptography 1 course says:\n\nIf you want to use MAC-then-encrypt mode then you should either use randomized-CTR or randomized-CBC. And if you use randomized-CTR then one-time-MAC is sufficient.\n\nHere is the slide:", null, "I have two question regarding this:\n\n1. How we can use one-time-MAC without negotiating the different keys for each message?\n2. What is wrong with randomized-CBC that doesn't allow us to use with one-time-MAC?\n\n1. How can we use one-time-MAC without negotiating the different keys for each message?\n\nIf you use a universal hashing one-time MAC, and you encrypt it with CTR mode, effectively you are creating a Carter–Wegman–Shoup MAC:\n\n• Carter and Wegman showed that if $$r, s_0, s_1, \\dots$$ are independent uniform random and $$H_r$$ is a universal hash with bounded difference probability $$\\Pr[H_r(x) - H_r(y) = \\delta] < \\varepsilon$$ for all $$x \\ne y$$ and $$\\delta$$, then the forgery probability for the authenticator $$m_i \\mapsto H_r(m_i) + s_i$$ is bounded by $$\\varepsilon$$.\n\n• Shoup suggested deriving $$s_i = E_k(i)$$ for a block cipher $$E$$, which is essentially what you get with MAC-then-encrypt in CTR mode. Since $$E_k$$ is a permutation you run up against the birthday bound, of course, just like any use of a block cipher in CTR mode.\n\n1. What is wrong with randomized-CBC that doesn't allow us to use with one-time MAC?\n\nOffhand, I am not sure! If the one-time MAC is a universal hash with bounded collision probability $$\\Pr[H_r(x) = H_r(y)] < \\varepsilon$$ for all $$x \\ne y$$, then this intuitively sounds like it ought to be a secure construction, because $$m \\mapsto F_k(H_r(m))$$ is a secure long-input, short-output PRF if $$F_k$$ is a secure short-input, short-output PRF and $$H_r$$ has bounded collision probability. That's not exactly the scenario we have here, because there is not one fixed family $$F_k$$—if we encrypt $$m \\mathbin \\\| H_r(m)$$, we get $$E_k(\\mathit{iv} \\oplus m) \\mathbin\\\| E_k(E_k(\\mathit{iv} \\oplus m) \\oplus H_r(m)),$$ so $$m$$ figures in twice, but it's pretty close, so intuitively it seems like this should be secure.\n\nOf course, it's not clear that the mere security of a one-time MAC—a bound on forgery probability after a single attempt—implies any of this even with CTR: I'm assuming bounds on collision or difference probabilities, specifically. Maybe you could ask Dan Boneh what he meant.\n\nThere is also a danger of padding oracles with CBC in MAC-then-encrypt or MAC-and-encrypt, because CBC works on sequences of blocks rather than sequences of bits, which is why it is hard to get right: practical realizations like TLS and SSH had mistakes for years that leaked plaintexts via padding oracles." ]	[ null, "https://i.stack.imgur.com/5G6OA.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9093725,"math_prob":0.99529403,"size":2220,"snap":"2019-43-2019-47","text_gpt3_token_len":595,"char_repetition_ratio":0.096570395,"word_repetition_ratio":0.022222223,"special_character_ratio":0.25675675,"punctuation_ratio":0.08584687,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99971396,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T11:10:55Z\",\"WARC-Record-ID\":\"<urn:uuid:18802e47-8184-4c54-b228-c2806498eff4>\",\"Content-Length\":\"141135\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:df455992-74e7-4195-9ddf-9dbca86ebbbc>\",\"WARC-Concurrent-To\":\"<urn:uuid:b55af32a-cd74-44d2-8e1a-21332bd1cf31>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://crypto.stackexchange.com/questions/35306/why-is-a-one-time-mac-secure-for-mac-then-encrypt-with-randomized-ctr-but-not-ra\",\"WARC-Payload-Digest\":\"sha1:V26GHEKQTOT4MAXRGTDESQDE3F2F46L3\",\"WARC-Block-Digest\":\"sha1:GMAYEBVTUKRV2WTUO5MXLPUBQQPMEWYQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671249.37_warc_CC-MAIN-20191122092537-20191122120537-00041.warc.gz\"}"}
https://0-bmcmedresmethodol-biomedcentral-com.brum.beds.ac.uk/articles/10.1186/1471-2288-10-48	[ "# A simulation study for comparing testing statistics in response-adaptive randomization\n\n## Abstract\n\n### Background\n\nResponse-adaptive randomizations are able to assign more patients in a comparative clinical trial to the tentatively better treatment. However, due to the adaptation in patient allocation, the samples to be compared are no longer independent. At large sample sizes, many asymptotic properties of test statistics derived for independent sample comparison are still applicable in adaptive randomization provided that the patient allocation ratio converges to an appropriate target asymptotically. However, the small sample properties of commonly used test statistics in response-adaptive randomization are not fully studied.\n\n### Methods\n\nSimulations are systematically conducted to characterize the statistical properties of eight test statistics in six response-adaptive randomization methods at six allocation targets with sample sizes ranging from 20 to 200. Since adaptive randomization is usually not recommended for sample size less than 30, the present paper focuses on the case with a sample of 30 to give general recommendations with regard to test statistics for contingency tables in response-adaptive randomization at small sample sizes.\n\n### Results\n\nAmong all asymptotic test statistics, the Cook's correction to chi-square test (T MC ) is the best in attaining the nominal size of hypothesis test. The William's correction to log-likelihood ratio test (T ML ) gives slightly inflated type I error and higher power as compared with T MC , but it is more robust against the unbalance in patient allocation. T MC and T ML are usually the two test statistics with the highest power in different simulation scenarios. When focusing on T MC and T ML , the generalized drop-the-loser urn (GDL) and sequential estimation-adjusted urn (SEU) have the best ability to attain the correct size of hypothesis test respectively. Among all sequential methods that can target different allocation ratios, GDL has the lowest variation and the highest overall power at all allocation ratios. The performance of different adaptive randomization methods and test statistics also depends on allocation targets. At the limiting allocation ratio of drop-the-loser (DL) and randomized play-the-winner (RPW) urn, DL outperforms all other methods including GDL. When comparing the power of test statistics in the same randomization method but at different allocation targets, the powers of log-likelihood-ratio, log-relative-risk, log-odds-ratio, Wald-type Z, and chi-square test statistics are maximized at their corresponding optimal allocation ratios for power. Except for the optimal allocation target for log-relative-risk, the other four optimal targets could assign more patients to the worse arm in some simulation scenarios. Another optimal allocation target, R RSIHR , proposed by Rosenberger and Sriram (Journal of Statistical Planning and Inference, 1997) is aimed at minimizing the number of failures at fixed power using Wald-type Z test statistics. Among allocation ratios that always assign more patients to the better treatment, R RSIHR usually has less variation in patient allocation, and the values of variation are consistent across all simulation scenarios. Additionally, the patient allocation at R RSIHR is not too extreme. Therefore, R RSIHR provides a good balance between assigning more patients to the better treatment and maintaining the overall power.\n\n### Conclusion\n\nThe Cook's correction to chi-square test and Williams' correction to log-likelihood-ratio test are generally recommended for hypothesis test in response-adaptive randomization, especially when sample sizes are small. The generalized drop-the-loser urn design is the recommended method for its good overall properties. Also recommended is the use of the R RSIHR allocation target.\n\n## Background\n\nThe response-adaptive randomization (RAR) in clinical trials is a class of flexible ways of assigning treatment to new patients sequentially based on available data. The RAR adjusts the allocation probabilities to reflect the interim results of the trial, thereby allowing patients to benefit from the interim knowledge as it accumulates in the trial. In practice, unequal allocation probabilities are generated based on the current assessment of treatment efficacy, which results in more patients being assigned to the treatment that is putatively superior.\n\nMany RAR designs have been proposed over the years . The two key issues extensively investigated are the evaluations of parameter estimations and hypothesis testing. Due to the dependency of assigning new patients based on observed data at that time, conventional estimates of treatment effect are often biased; therefore, efforts have been made to quantify and correct estimation bias [14, 15]. Recent theoretical works have been focused on solving problems encountered in practice, which includes delayed response, implementation for multi-arm trials, and incorporating covariates, etc. [1, 3, 11, 1618]. Many recent theoretical developments are summarized in . Additionally, in order to compare treatment efficacies through hypothesis testing, studies have been conducted on power comparisons and sample size calculations under the framework of adaptive randomization . However, most of the works are based on large sample sizes, and focus on asymptotic properties [4, 12, 22, 25, 26]. But these properties have not been fully studied with small sample sizes. The mathematical challenge imposed by correlated data makes it extremely difficult to derive exact solutions for finite samples. Up to now, only limited results on exact solutions have been available [15, 27], and computer simulation has to be relied upon when sample size is small [23, 24], which is often the case in early phase II trials.\n\nEach RAR design has its own objective, and there are both advantages and disadvantages associated with that objective. It is not our purpose to give a comprehensive assessment of different designs by comparing their advantages and disadvantages. Instead, the primary objective of the present study is to characterize the small sample properties of RAR based on a frequentist approach. In particular, we focus on comparing the performance of commonly used test statistics in RAR of two-arm comparative trials with a binary outcome. Due to the departure from normality caused by data correlation and the discrete nature of a binary outcome, hypothesis tests usually can not be controlled at any given levels of nominal significance. Thus, to make our simulation comparison more relevant, our assessment of hypothesis testing methods and RAR procedures is based on the calculation of both statistical power and the comparison to the nominal type I error rate. Several RAR methods studied in our simulations can assign patients according to a given allocation target, which may be optimal in terms of maximizing the power or minimizing the expected treatment failure. Therefore, we also compare the properties of test statistics at different optimal allocation targets.\n\nThe remaining parts of this paper are organized into 4 sections. In the Methods Section, we introduce the adaptive randomization procedures, the optimal allocation rates, and the test statistics used in the simulation. In the Results Section, we present the simulation results. We provide a discussion and final recommendations regarding the RAR methods and hypothesis tests in the Discussion and Conclusions Sections.\n\n## Methods\n\nIn the present section, we briefly describe the randomization methods, asymptotic hypothesis test statistics, and optimal patient allocation targets that are relevant to our simulations. More detailed information can be found in the corresponding references.\n\nThe RAR procedures investigated in the present study are randomized play-the-winner (RPW) [8, 10], drop-the-loser (DL) , sequential maximum likelihood estimation (SMLE) , doubly-adaptive biased coin [2, 3], sequential estimation-adjusted urn (SEU) , and generalized drop-the-loser (GDL) designs. RPW, DL, SEU and GDL are all urn models in the sense that treatment assignment for each patient can be obtained by sampling balls from an urn. In the usual clinical trial setting, an urn model consists of one urn with different types of balls that represent the different treatments under study. Patients are assigned to treatments by randomly selecting balls from the urn. Initially, the urn contains an equal number of balls for each of the treatment offered in the trial. With the progress of a clinical trial, certain rules are applied to update the contents of the urn in such a way that favors the selection of balls corresponding to the better treatment. For example, under the RPW design, the observation of a successful treatment response leads to the addition of a (>0) balls of the same type to the urn; a lack of success leads to the addition of b (>0) balls of the other type to the urn (a = b = 1 in our simulation). The limiting allocation rate of patients on treatment 1 is q 2/(q 1 + q 2), where q 1 = 1-p 1 and q 2 = 1-p 2 are failure rates, and p 1 and p 2 are success rates (or response rates) for treatments 1 and 2. In the DL model, patients are assigned to a treatment based on the type of ball that is drawn; however a treatment failure results in the removal of a treatment ball from the urn, and treatment successes are ignored. Due to the finite probabilities of extinction, immigration balls are added to the urn. If an immigration ball is drawn, an additional ball of each type is added. The sampling process is repeated until a treatment ball is drawn. The DL urn design has the same limiting allocation as the RPW urn, but less variability in patient allocation. Both SEU and GDL are urn models allowing fraction number of balls, and can target any allocation rate. For SEU method , if the limiting allocation of RPW urn is the target in a two-arm trial, then", null, "balls of type 2 and", null, "balls of type 1 are added to the urn following the allocation of the ith patient. Obviously, the response status of the ith patient is related to the contents of SEU urn only through the calculation of", null, "and", null, ". For a two-arm GDL urn model , when a treatment ball is drawn, a new patient is assigned accordingly, but the ball will not be returned to the urn. Depending on the response of the patient, the conditional average numbers of balls being added back to the urn are b 1 and b 2 for treatments 1 and 2, respectively. Therefore, the conditional average numbers of type 1 and type 2 balls being taken out of the urn can be defined as d 1 and d 2, where d 1 = 1-b 1 and d 2 = 1-b 2. Immigration balls are also present in a GDL urn. Whenever an immigration ball is drawn, a 1 and a 2 balls are added for treatments 1 and 2, respectively. Zhang et al have shown that the limiting allocation rate of patients on treatment 1 is", null, "(1)\n\nThe GDL urn becomes a DL urn when a 1 = 1, a 2 = 1, b 1 = p 1, and b 2 = p 2. Although GDL is a general method with different ways of implementation, a convenient approach is taken in our simulation. When a treatment ball is drawn, the ball is not returned, and no ball is added regardless of the response of the patient. When an immigration ball is drawn, 1 and 2 balls of type 1 and 2 are added, where C is a constant, and ρ 1 and ρ 2 are allocation targets on treatments 1 and 2, which are estimated sequentially using the maximum likelihood estimates (MLE) .\n\nThe SMLE and doubly-adaptive biased coin design (DBCD) methods can also target any allocation ratios, and SMLE can be implemented as a special case of DBCD method. In DBCD method, the probability of the (i+1)th patient being assigned to treatment 1 is calculated by", null, "(2)\n\nwhere r 1 = n 1 (i)/i and ρ(i) are the current allocation rate and estimated allocation rate on treatment 1 [2, 3]. The properties of the DBCD depend largely on the selection of g, which can be considered as a measuring function for the deviation from the allocation target. In the present study, we use the following function suggested by Hu and Zhang :", null, "(3)\n\nwhere α is a tuning parameter. When α approaches infinity, the DBCD becomes deterministic and the patients are assigned to the putatively better treatment with probability 1. When α equals to 0, the MLE of ρ becomes the allocation target, and the DBCD method is essentially the same as the SMLE design proposed by Melfi et al .\n\n### Hypothesis Tests for Two-Arm Comparative Trials\n\nIn two-arm comparative trials, the results of a binary outcome variable can be summarized in a 2 × 2 contingency table (Table 1). The following hypothesis test is often conducted to compare treatment efficacy:", null, "(4)\n\nNine test statistics for the hypothesis test in (4) are given in Table 2. When relative risk (q 1/q 2) and odds ratio (p 1 q 2/q 1 p 2) are used to quantify the differences between 2 treatment arms, the test statistics are log-relative-risk and log-odds-ratio, T Risk and T Odds , which are asymptotically distributed as chi-square distribution with one degree of freedom (", null, "). When simple difference is used to measure the treatment effect, the applicable test statistics are the Wald-type test statistic T Wald and the score-type test statistics T Chisq , where the variance of simple difference in response rates is evaluated at H 1 or H 0 respectively. Additionally, the test statistics based on the logarithm of likelihood ratio (T LLR ) can also be constructed. Besides the 5 commonly used test statistics mentioned above, four modified test statistics are also included in Table 2. T MO is a modified log-odds-ratio test proposed by Gart using the approximation of discrete distributions by their continuous analogues . As shown in Table 2, T MO is essentially a modification to T Odds by adding 0.5 to each cell of a 2 × 2 table. Similarly, Agresti and Caffo proposed a modification to T Wald by adding 1 to each cell of a contingency table , which results in the test statistic T MW in Table 2. T MC is the Cook's continuity correction to chi-square test statistics T Chisq . Williams provided a modification to log-likelihood-ratio test T LLR . The original test statistic T LLR is improved by multiplying a scale factor such that the null distribution of the new test statistic T ML has the same moments as the chi-square distribution.\n\nSince all test statistics in Table 2 are based on", null, ", they are asymptotically equivalent and any one of them can be used for large sample sizes. Meanwhile at small sample sizes, an exact test can be conducted if a model is specified for the data given in Table 1. For example, depending on the number of fixed margins predetermined for the design, one of the following three models can be applied :", null, "(5)", null, "(6)\n\nand", null, "(7)\n\nwhere h(r 1\|n, n 1, r) represents the hypergeometric distribution of r 1, b(r\|n, p) gives the binomial distribution of r under the null hypothesis of equal response rates (H 0: p 1 = p 2 = p), and b(n 1\|n, ρ) denotes the binomial distributions of patients on arm 1 with an allocation ratio of ρ (ρ 1 = 0.5 for equal randomization). The p value of exact test can be calculated by maximizing the probability in (5), (6), or (7) over the two nuisance parameters, p and ρ. However, due to data dependency, none of the above three models are directly applicable in adaptive randomization. For example, the allocation ratio ρ in adaptive randomization is a random variable with unknown distribution, and the binomial distribution of n 1 assumed in model (7) is not valid even when the null hypothesis is true. Therefore, in adaptive randomization, unconditional exact tests are not available and asymptotic test statistics such as the ones in Table 2 are required for testing the hypothesis in (4).\n\n### Optimal Allocation Ratios\n\nThe SMLE, DBCD, SEU, and GDL methods can be utilized to allocate patients based on different allocation targets. The allocation targets simulated in the present study are summarized in Table 3, where R Risk , R Odds , R Wald , R Chisq , and R LLR are optimal allocation ratios maximizing the power of T Risk , T Odds , T Wald , T Chisq , and T LLR respectively, at fixed sample size. The derivation of T Risk , T Odds , T Wald , T Chisq , and T LLR can be found in [33, 34], which is equivalent to minimizing the variance of corresponding test statistic at a fixed total sample size, and consequently the power of that test statistic is maximized. R RSIHR is a recently proposed allocation target that minimizes the expected total number of failures among all trials with the same power [15, 33]. The general theoretical framework and the practical implementation of optimal allocation in k-arm trials with binary outcomes are discussed and demonstrated by Tymofyeyev et al , where the optimization can be conducted over different goals. In practice, the performance of the methodology depends on the chosen RAR procedure. The present simulation study only focuses on two-arm trials, with a goal of maximizing the power or minimizing the total number of failures.\n\n## Results\n\nSimulations are conducted at different total numbers of patients ranging from 20 to 200. To simplify the presentation, the results for trials with 30 patients are shown here. When patients are less than 30, adaptive randomization is generally not recommended. For sample size of 100 or larger, all methods yield similar properties in general. For all of the urn models, one ball for each treatment is consistently used as the initial contents of the urn. The number of immigration balls is 1 for both the DL and GDL urns. The tuning parameter of DBCD, α, is fixed at 0 or 2. When α is 0, it results in the SMLE method. The value of the constant C in GDL is 2, which is equivalent to adding 2 treatment balls on average when an immigration ball is drawn. All simulation results are calculated based on 10,000 replicates.\n\nFor the purpose of comparison, the true allocation rates are shown in Table 4, and the simulated results for allocation rates on arm 1 are shown in Table 5. Among all RAR methods, DBCD has the best ability to attain the true allocation target. The comparison between SMLE and DBCD shows that, the allocation becomes more unbalanced and the variation of DBCD decreases with increasing value of tuning exponent α. On the other hand, the patient allocation of SEU results in more balanced mean allocation between two arms with a much larger variation as compared with other RAR methods. The GDL has the lowest variation among the four sequential RAR methods. When R RPW (the same as R DL ) is the allocation target, DL urn method has the lowest variation in patient allocation, which is consistent with the fact that the lower bound of the estimate of Var(R RPW ) is attained by DL urn . The comparison among allocation targets shows that R LLR has the lowest variation in patient allocation, and the highest variation is usually found at R RPW or R Risk . However, R RPW and R Risk are usually the top two allocation targets that assign more patients to the better treatment. R Wald , R Odds , and RLLR assigns more patients to the worse arm in some simulation cases. Among the three allocation targets that assign more patients to the better treatment (R RSIHR , R Risk and R RPW ), R RSIHR has a stable and often the lowest variation in patient allocation.\n\nThe simulation results are obtained for five null cases and ten alternative cases, and Table 6 gives the summary by averaging the results over the five null cases and the ten alternative cases for a given RAR method and at a given allocation target. Detailed simulation results for each test statistic are shown in Tables 7, 8, 9, 10, 11, 12 with one table for each of the six allocation targets. To simplify the presentation, the results are shown only for the four modified test statistics T MW , T MO , T MC , T ML , and the log-relative-risk test statistic T Risk because they tend to have better performance than the four corresponding unmodified tests. The qualitative comparisons among test statistics, RAR methods, and allocation targets can be made based on the results in Table 6.\n\nAs shown in Table 6 (also see Tables 7, 8, 9, 10, 11, 12), the worst performance can be found in the results of T MO and T Risk , which are often conservative with less than nominal type I error rate. T MW is always slightly conservative across all simulation cases. Overall, T MC is the best in attaining the correct type I error rate. T ML , is slightly inflated as compared with chi-square test T MC . However, the simulation results not shown here indicate that T ML is very robust against the unbalance in patient allocation even when sample size is 20. The comparison between different RAR methods shows that the mean type I error of GDL and SEU can usually match the correct size of tests better than other methods when T MC and T ML are used respectively. The type I error of DBCD is usually the largest one, except at R Odds . The overall type I error of SEU is comparable with GDL.\n\nThe power comparison of different test statistics indicates that T Risk is the statistic with the highest power at R Risk but with a much inflated type I error. Except at R Risk , T MC or T ML is the one with the highest power. Usually, GDL has the highest power and SEU has the lowest power among all RAR methods. DBCD and SMLE have similar power, but DBCD is more powerful in most cases. At target R RPW , DL urn has the best statistical properties. On the average, the target with the lowest power achieved by test statistics is R Risk . The highest overall power can usually be achieved by test statistics at R RSIHR and R LLR , but R LLR has the disadvantage of assigning more patients to the worse treatment in some cases.\n\n## Discussion\n\nIn response-adaptive randomization, the assignment of a new patient depends on the treatment outcomes of patients previously enrolled in the trial. Delayed responses are often encountered in practice. Recently, the problem of delayed response in multi-arm generalized drop-the-loser urn and generalized Friedman's urn design is studied for both continuous and discontinuous outcomes [11, 16, 17, 36]. It is shown that, under reasonable assumption about the delay, the asymptotic properties of adaptive design are not affected by the delay. In the present study, the primary focus is the comparison between commonly used test statistics for 2 × 2 tables. Based on results not shown here, a less extreme allocation with higher variation would be expected when a random delay is assumed. It is assumed that the response status of each of the patients already in the trial is available before the allocation of a new patient in our simulations evaluation.\n\nThe RAR methods simulated in the present study are aimed at assigning patients to the better treatment with probabilities higher than what otherwise would be allowed by equal randomization. The price being paid is that the sample sizes on the two comparing arms are no longer fixed, and the adaptation in patient allocation can complicate the statistical inference at the end of the trial. The properties of test statistics will change when the patient allocation ratio changes in adaptive randomization. The power of test statistics shown in the present simulation study is obtained by averaging over trials with an unknown distribution of allocation ratios. As shown in our simulation results, a large deviation from the nominal significance level of the hypothesis test can be found even under the null hypothesis. Therefore, the practice of comparing asymptotic hypothesis testing methods based solely on statistical power under the alternative hypothesis is not recommended. It is important to compare adaptive randomization methods based on both the type I error rate and the statistical power, especially when the sample size is small.\n\nGeneral recommendations given in the result section are based on the aggregated results across different settings. Because the performance of different test statistics, RAR methods, and allocation target are closely related to each other, recommendations under a specific scenario can be found based on the detailed simulation results in Tables 7, 8, 9, 10, 11, 12.\n\nBased on simulation results, the Cook's correction to chi-square test statistic T MC and Williams' correction to log-likelihood-ratio test T ML are recommended to be used for hypothesis testing at the end of adaptive randomization. T MC has good ability to attain the correct significance levels, and is relatively robust against the change of RAR method or allocation target. T ML has more robust performance than T MC and has higher power, but its type I error is slightly inflated as compared with T MC . However, T ML attains more accurate type I error than T MC when the sample size is small. The original Wald-type Z test statistic T Wald , which is very sensitive to patient allocation and has inflated type I error, should be avoided at small sample sizes. On the other hand, T MW , the Argresti's correction to T Wald , and T MO the modified log-odds-ratio test are too conservative and under powered at small sample sizes.\n\nThe primary objective of current study is to compare test statistics. Since the recommended test statistics are T MC and T ML , the comparison between RAR methods and allocation targets are mainly based on these two selected test statistics. Among SMLE, DBCD, SEU, and GDL methods, GDL seems to be the best one due to its ability to attain the correct size of hypothesis test and comparatively higher overall power at most allocation targets. Therefore, GDL is the recommended RAR method. The sequential estimation-adjusted urn (SEU) method is comparable with GDL in controlling the type I error. However, SEU is often under powered, and the high variation in patient allocation makes it less useful in practice. The DBCD method with tuning exponent α equal to 2 is the best in targeting the true allocation ratio. When T MC is the test statistic, DBCD has slightly inflated type I error and slightly lower power as compared with GDL. Therefore, among values of α, the balances among controlling the type I error, obtaining higher power, and targeting a given allocation ratio can be reached when α is equal to 2. The simulation comparison of statistical power for different RAR methods also indicates that DL urn has the best statistical properties at R RPW , mainly due to its low variation in patient allocation.\n\nThe statistical characteristics of hypothesis tests and RAR methods also depend on allocation targets. At R Wald , R Odds , and R LLR targets, more patients could be assigned to the inferior treatment in certain parameter spaces. In contrast, R Risk , R RPW , and R RSIHR always assign more patients to the better treatment. However, due to the more extreme allocation of R Risk and R RPW , both power and type I error of R Risk and R RPW will suffer as compared with R RSIHR . On the other hand, the variation of patient allocation at R RISHR is relatively small with a stable value across all simulation scenarios. Additional, among all designs with similar power using Wald-type test statistic, R RSIHR allocation ration can achieve fewer failures in the whole trial. Therefore, R RSIHR is recommended among all the allocation targets in the present study.\n\nIn addition to the frequentist development on the response adaptive randomization, Bayesian decision theoretic methods has also been proposed in the context of bandit problem. The concept of \"patient horizon\" was brought up to include future patients to whom the current study results might be applied. The goal is to maximize the total number of success in patients enrolled in the study with or without including the patient horizon. More detailed exposition of Bayesian methods for response adaptive randomization is beyond the scope of this paper and interested readers should consult the original work on this topic .\n\n## Conclusion\n\nThe Cook's correction to chi-square test and Williams' correction to log-likelihood-ratio test are recommended for hypothesis test of RAR at small sample sizes. Among all the RAR methods compared, GDL method has better statistical properties in controlling type one error and maintaining high statistical power. The RSIHR allocation target provides a good balance between assigning more patients to the better treatment and maintaining a high overall power.\n\n## Abbreviations\n\nRAR:\n\nRPW:\n\nRandomized play-the-winner\n\nDL:\n\nDrop-the-loser\n\nDBCD:\n\nSMLE:\n\nSequential maximum likelihood estimation design\n\nSEU:\n\nGDL:\n\nGeneralized drop-the-loser urn\n\nRSIHR:\n\nOptimal allocation target minimizing total numbers of failure for Wald-type test statistics at fixed power\n\nMLE:\n\nMaximum likelihood estimate.\n\n## References\n\n1. 1.\n\nAndersen J, Faries D, Tamura R: A randomized play-the-winner design for multi-arm clinical trials. Communications in Statistics-Theory and Methods. 1994, 23: 309-323. 10.1080/03610929408831257.\n\n2. 2.\n\nEisele JR: The doubly adaptive biased coin design for sequential clinical trials. Journal of Statistical Planning and Inference. 1994, 38: 249-262. 10.1016/0378-3758(94)90038-8.\n\n3. 3.\n\nHu FF, Zhang LX: Asymptotic properties of doubly adaptive biased coin designs for multi-treatment clinical trials. Annals of Statistics. 2004, 32 (1): 268-301.\n\n4. 4.\n\nIvanova S, Rosenberger WF, Durham S, Flournoy N: A birth and death urn for randomized clinical trials: asymptotic methods. Sankhya: The Indian Journals of Statistics. 2000, 62 (B): 104-118.\n\n5. 5.\n\nLi W, Durham SD, Flournoy N: Randomized Pôlya urn. 1996 Proceedings of the Biopharmaceutical Section of the American Statistical Association: 1997; Alexandria: American Statistical Association. 1997, 166-170.\n\n6. 6.\n\nRosenberger WF, Stallard N, Ivanova A, Harper CN, Ricks ML: Optimal adaptive designs for binary response trials. Biometrics. 2001, 57: 909-913. 10.1111/j.0006-341X.2001.00909.x.\n\n7. 7.\n\nWei LJ: The generalized Polya's urn design for sequential medical trials. Annals of Statistics. 1979, 7: 291-296. 10.1214/aos/1176344614.\n\n8. 8.\n\nWei LJ, Durham SD: The randomized play-the-winner rule in medical trials. Journal of American Statistical Association. 1978, 85: 156-162. 10.2307/2289538.\n\n9. 9.\n\nYang Y, Zhu D: Randomized allocation with nonparametric estimation for a multi-armed bandit problem with covariates. Annals of Statistics. 2002, 30: 100-121. 10.1214/aos/1015362186.\n\n10. 10.\n\nZelen M: Play the winner rule and the controlled clinical trial. Journal of the American Statistical Association. 1969, 64: 131-146. 10.2307/2283724.\n\n11. 11.\n\nZhang LX, Chan WS, Cheung SH, Hu FF: A generalized drop-the-loser urn for clinical trials with delayed responses. Statistica Sinica. 2007, 17 (1): 387-409.\n\n12. 12.\n\nMelfi VF, Page C, Geraldes M: An adaptive randomized design with application to estimation. Canadian Journal of Statistics. 2001, 29 (1): 107-116. 10.2307/3316054.\n\n13. 13.\n\nZhang LX, Hu FF, Cheung SH: Asymptotic theorems of sequential estimation-adjusted urn models. Annals of Applied Probability. 2006, 16 (1): 340-369. 10.1214/105051605000000746.\n\n14. 14.\n\nCoad DS, Ivanova A: Bias calculations for adaptive urn designs. Sequential Analysis. 2001, 20 (3): 91-116. 10.1081/SQA-100106051.\n\n15. 15.\n\nRosenberger WF, Sriram TN: Estimation for an adapative allocation design. Journal of Statistical Planning and Inference. 1997, 59: 309-319. 10.1016/S0378-3758(96)00109-7.\n\n16. 16.\n\nBai ZD, Hu FF, Rosenberger WF: Asymptotic properties of adaptive designs for clinical trials with delayed response. Annals of Statistics. 2002, 30 (1): 122-139. 10.1214/aos/1015362187.\n\n17. 17.\n\nHu FF, Zhang LJ: Asymptotic normality of urn models for clinical trials with delayed response. Bernoulli. 2004, 10: 447-463. 10.3150/bj/1089206406.\n\n18. 18.\n\nRosenberger WF, Vidyashankar AN, Agarwal DK: Covariate-adjusted response-adaptive designs for binary response. Journal of Biopharmaceutical Statistics. 2001, 11: 227-236.\n\n19. 19.\n\nHu FF, Rosenberger WF: The Theory of Response-Adaptive Randomization in Clinical Trials. 2006, Hoboken, New Jersey: John Wiley & Sons, Inc.\n\n20. 20.\n\nHu FF, Rosenberger WF: Optimality, variability, power: evaluating response-adaptive randomization procedures for treatment comparisons. Journal of the American Statistical Association. 2003, 98 (463): 671-678. 10.1198/016214503000000576.\n\n21. 21.\n\nZhang LJ, Rosenberger WF: Response-adaptive randomization for clinical trials with continuous outcomes. Biometrics. 2006, 62 (2): 562-569. 10.1111/j.1541-0420.2005.00496.x.\n\n22. 22.\n\nHu FF, Rosenberger WF, Zhang LX: Asymptotically best response-adaptive randomization procedures. Journal of Statistical Planning and Inference. 2006, 136 (6): 1911-1922. 10.1016/j.jspi.2005.08.011.\n\n23. 23.\n\nMorgan CC, Coad DS: A comparison of adaptive allocation rules for group-sequential binary response clinical trials. Statistics in Medicine. 2007, 26 (9): 1937-1954. 10.1002/sim.2693.\n\n24. 24.\n\nGuimaraes P, Palesch Y: Power and sample size simulations for Randomized Play-the-Winner rules. Contemporary Clinical Trials. 2007, 28 (4): 487-499. 10.1016/j.cct.2007.01.006.\n\n25. 25.\n\nMatthews PC, Rosenberger WF: Variance in randomized play-the-winner clinical trials. Statistics & Probability Letters. 1997, 35: 233-240. 10.1016/S0167-7152(97)00018-7.\n\n26. 26.\n\nBai ZD, Hu FF: Asymptotics in randomized urn models. Annals of Applied Probability. 2005, 15 (1B): 914-940. 10.1214/105051604000000774.\n\n27. 27.\n\nMatthews PC, Rosenberger WF: Variance in randomized play-the-winner clinical trials. Statistics & Probability Letters. 1997, 35 (3): 233-240. 10.1016/S0167-7152(97)00018-7.\n\n28. 28.\n\nIvanova A: A play-the-winner-type urn design with reduced variability. Metrika. 2003, 58: 1-13.\n\n29. 29.\n\nGart JJ: Alternative analyses of contingency tables. Journal of Royal Statistical Society B. 1966, 28: 164-179.\n\n30. 30.\n\nAgresti A, Caffo B: Simple and effective confidence intervals for proportions and differences of proportions results from adding two successes and two failures. The American Statistician. 2000, 54 (4): 280-288. 10.2307/2685779.\n\n31. 31.\n\nWilliams SS: Improved likelihood ratio tests for complete contingency tables. Biometrika. 1976, 63: 33-37. 10.1093/biomet/63.1.33.\n\n32. 32.\n\nUpton GJG: A comparison of alternative tests for the 2 × 2 table comparative trial. Journal of Royal Statistical Society A. 1982, 145: 86-105. 10.2307/2981423.\n\n33. 33.\n\nRosenberger WF, Lachin JM: Randomization in Clinical Trials: Theory and Practice. 2002, New York: Wiley\n\n34. 34.\n\nJennison C, Turnbull BW: Group Sequential Methods with Applications to Clinical Trials. 2000, Boca Raton: Chapman & Hall/CRC\n\n35. 35.\n\nTymofyeyev Y, Rosenberger WF, Hu FF: Implementing optimal allocation in sequential binary response experiments. Journal of American Statistical Association. 2007, 102 (477): 224-234. 10.1198/016214506000000906.\n\n36. 36.\n\nSun RB, Cheung SH, Zhang LX: A generalized drop-the-loser rule for multi-treatment clinical trials. Journal of Statistical Planning and Inference. 2007, 137 (6): 2011-2023. 10.1016/j.jspi.2006.06.039.\n\n37. 37.\n\nBerry DA, Fristedt B: Bandit Problems. 1985, New York: Chapman and Hall\n\n38. 38.\n\nThompson WR: On the likelihood that one unknown probability exceeds another in the view of the evidence of the two samples. Biometrika. 1933, 25: 275-294.\n\n39. 39.\n\nBerry DA, Eick SG: Adaptive assignment versus balanced randomization in clinical trials: a decision analysis. Statistics in Medicine. 1995, 14: 231-246. 10.1002/sim.4780140302.\n\n40. 40.\n\nCheng Y, Berry DA: Optimal adaptive randomized designs for clinical trials. Biometrika. 2007, 94 (4): 673-689. 10.1093/biomet/asm049.\n\n### Pre-publication history\n\n1. The pre-publication history for this paper can be accessed here:http://0-www.biomedcentral.com.brum.beds.ac.uk/1471-2288/10/48/prepub\n\n## Acknowledgements\n\nThis work was supported in part by grants CA16672 from the National Cancer Institute and W81XWH-06-1-0303 and W81XWH-07-1-0306 from the Department of Defense. The authors thank Dr. Lunagomez for helpful discussions. The authors also thank Ms. Lee Ann Chastain for her help, which greatly improved the presentation of our study.\n\n## Author information\n\nCorrespondence to J Jack Lee.\n\n### Competing interests\n\nThe authors declare that they have no competing interests.\n\n### Authors' contributions\n\nXMG conducted the simulation part of the study. Both XMG and JJL participated in designing the study and writing the manuscript. All authors read and approved the final manuscript.\n\n## Authors’ original submitted files for images\n\nBelow are the links to the authors’ original submitted files for images.\n\n## Rights and permissions\n\nReprints and Permissions\n\nGu, X., Lee, J.J. A simulation study for comparing testing statistics in response-adaptive randomization. BMC Med Res Methodol 10, 48 (2010). https://0-doi-org.brum.beds.ac.uk/10.1186/1471-2288-10-48", null, "" ]	[ null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_IEq1_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_IEq2_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_IEq3_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_IEq4_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_Equ1_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_Equ2_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_Equ3_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_Equ4_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_IEq5_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_IEq5_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_Equ5_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_Equ6_HTML.gif", null, "https://media.springernature.com/full/springer-static/image/art%3A10.1186%2F1471-2288-10-48/MediaObjects/12874_2008_Article_451_Equ7_HTML.gif", null, "https://0-bmcmedresmethodol-biomedcentral-com.brum.beds.ac.uk/track/article/10.1186/1471-2288-10-48", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9099103,"math_prob":0.8589361,"size":37580,"snap":"2019-51-2020-05","text_gpt3_token_len":8576,"char_repetition_ratio":0.16382797,"word_repetition_ratio":0.05257853,"special_character_ratio":0.23658861,"punctuation_ratio":0.13415316,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9527628,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,4,null,8,null,8,null,4,null,4,null,4,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-28T17:21:54Z\",\"WARC-Record-ID\":\"<urn:uuid:273792c7-31d2-40ad-96d8-5b9938adea47>\",\"Content-Length\":\"263934\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bbf48a0b-facf-463d-a12e-bad8ade1dafa>\",\"WARC-Concurrent-To\":\"<urn:uuid:3c72a0ff-e1a2-470d-b252-a43bcb0df4e2>\",\"WARC-IP-Address\":\"194.80.219.188\",\"WARC-Target-URI\":\"https://0-bmcmedresmethodol-biomedcentral-com.brum.beds.ac.uk/articles/10.1186/1471-2288-10-48\",\"WARC-Payload-Digest\":\"sha1:V3D4SLETNTLJWKNQZB7LAFVCQN6MZ45X\",\"WARC-Block-Digest\":\"sha1:GWZSKEXE6NRNKLTGYCAUTXTZ4IIWWO6V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251779833.86_warc_CC-MAIN-20200128153713-20200128183713-00509.warc.gz\"}"}
https://www.toppr.com/guides/maths/probability/	[ "", null, "# Probability\n\nDid you know the origins of Probability can be linked to gambling during the 17th century? The theory of probability may have been discovered to help gamblers, but now it assists so many of our everyday activities. The chances of your school team winning the tournament, the possibility of rain on a given day, the likelihood of your winning the lottery, are all applications of probability. Let us learn a few more interesting topics regarding probability.\n\nQ1. How can we calculate the probability?\n\nA1. We can calculate the probability of an event by dividing it by the numbers of possible outcomes. Furthermore, this gives us the probability of a single event occurring. E.g. To get an outcome of 3 on a dice we have to roll it. And the number of event is1 and total outcomes are 6.\n\nQ2. What re the basic rules of probability?\n\nA2. Five basic rules of probability are:\n\n• First Rule: The possibility of happening of an event lies between 0 and 1.\n• Second Rule: Sum of all probable possible outcomes is 1.\n• Third Rule: There is a probability that a certain event will not take place.\n• Fourth Rule: Difference between happening and non-happening of an event is an important one.\n• Fifth Rule: It says that P(A or B) = P(A) + P(B) – P(A and B).\n\nQ3. What are the three types of probability?\n\nA3. The three types of probability are: classical, relative, and subjective probability.\n\nQ4. Is probability applicable in real life?\n\nA4. Probability is the chance that something will occur, such as picking a piece of green candy from a bag, drawing an ace from a deck of card etc. Hence, there is probability everywhere we go so it is applicable in real-life.\n\nShare with friends\n\nNo thanks." ]	[ null, "https://www.facebook.com/tr", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92604434,"math_prob":0.9557626,"size":2066,"snap":"2021-31-2021-39","text_gpt3_token_len":460,"char_repetition_ratio":0.17167798,"word_repetition_ratio":0.0,"special_character_ratio":0.21926428,"punctuation_ratio":0.119592875,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9973702,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-28T23:04:25Z\",\"WARC-Record-ID\":\"<urn:uuid:466bf385-0a0c-4433-bbc5-f4658d9b05b6>\",\"Content-Length\":\"93572\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:56dcfe63-d68a-4114-9691-e982d82ae0eb>\",\"WARC-Concurrent-To\":\"<urn:uuid:ff3b15a4-0446-4654-8a2e-6346e4413046>\",\"WARC-IP-Address\":\"104.20.83.29\",\"WARC-Target-URI\":\"https://www.toppr.com/guides/maths/probability/\",\"WARC-Payload-Digest\":\"sha1:NF4VKN55K6RW33OPSVCG4TTK2BBOQRTL\",\"WARC-Block-Digest\":\"sha1:B2B63NEUWKBLCXHNCDVZFYPAGX46RPJS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780060908.47_warc_CC-MAIN-20210928214438-20210929004438-00058.warc.gz\"}"}
https://bnewtech.com/numberlist/cube/8344-8330/	[ "# List of Cube Numbers From 8344 to 8330\n\n83443 = 580928771584\n83433 = 580719929607\n83423 = 580511137688\n83413 = 580302395821\n83403 = 580093704000\n83393 = 579885062219\n83383 = 579676470472\n83373 = 579467928753\n83363 = 579259437056\n83353 = 579050995375\n83343 = 578842603704\n83333 = 578634262037\n83323 = 578425970368\n83313 = 578217728691\n83303 = 578009537000\n\nYou can download/print the perfect cube numbers from 8344 - 8330 as pdf format.\n\nfrom to" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.53046525,"math_prob":0.9912705,"size":433,"snap":"2022-27-2022-33","text_gpt3_token_len":168,"char_repetition_ratio":0.16550116,"word_repetition_ratio":0.0,"special_character_ratio":0.8198614,"punctuation_ratio":0.018867925,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999895,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-17T02:02:12Z\",\"WARC-Record-ID\":\"<urn:uuid:11f6910d-d7ab-43a5-be9e-6544a91a42cd>\",\"Content-Length\":\"9620\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d841122-4d18-4099-8619-022dae736500>\",\"WARC-Concurrent-To\":\"<urn:uuid:a8d27e2d-c359-4fc6-9dd8-1b9587fdd9b7>\",\"WARC-IP-Address\":\"172.67.177.78\",\"WARC-Target-URI\":\"https://bnewtech.com/numberlist/cube/8344-8330/\",\"WARC-Payload-Digest\":\"sha1:64E5Z4QXGLWZNW7IM3SQZEM3DK4JVAWZ\",\"WARC-Block-Digest\":\"sha1:ADKFHK75LKQGCORDBUTE3UW5CXWMD4U5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572833.78_warc_CC-MAIN-20220817001643-20220817031643-00111.warc.gz\"}"}
https://www.w3resource.com/javascript/object-property-method/array-length.php	[ "", null, "# JavaScript length Property: Array Object\n\n## Description\n\nThe length property returns number elements of an array. The value of the length property is an integer and less than 232. The length property is also used to truncate an array. If the length property is set with a number which is greater than the original length then new elements are added to the end of the array. If the length property is set with a number which is less than the original length then excess elements at the end of the array are lost.\n\nVersion\n\nImplemented in JavaScript 1.1\n\nSyntax\n\narray.length\n\nExample:\n\nIn the following web document length property is used to get the number of elements of a given array.\n\nHTML Code\n\n``````<!DOCTYPE html>\n<html lang=\"en\">\n<meta charset=\"utf8\" />\n<title>JavaScript Array object-length Property example</title>\n<style type=\"text/css\">\nh1 {color:red}\n</style>\n<body>\n<h1>JavaScript Array Object : length Property</h1>\n<script src=\"array-length-example1.js\">\n</script>\n</body>\n</html>\n```\n```\n\nJS Code\n\n``````var myarray = new Array(\"Orange\", \"Apple\", \"Banana\", \"Cherry\", \"Mango\");\nvar newParagraph = document.createElement(\"p\");\nvar newText = document.createTextNode(\"The length of the array is : \"+myarray.length);\nnewParagraph.appendChild(newText);\ndocument.body.appendChild(newParagraph);\n```\n```\n\nView the example in the browser\n\nPractice the example above online\n\nJavaScript Array object-length Property example\n\nJavaScript input Property: Array Object\n\nTest your Programming skills with w3resource's quiz.\n\n\n\n## JavaScript: Tips of the Day\n\nReturns true if the given number is a power of 2, false otherwise\n\nExample:\n\n```const tips_isPowerOfTwo = n => !!n && (n & (n - 1)) == 0;\nconsole.log(tips_isPowerOfTwo(0));\nconsole.log(tips_isPowerOfTwo(4));\nconsole.log(tips_isPowerOfTwo(16));\n```\n\nOutput:\n\n```false\ntrue\ntrue\n```" ]	[ null, "https://www.w3resource.com/images/w3resource-logo.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.59867984,"math_prob":0.7681966,"size":2459,"snap":"2022-27-2022-33","text_gpt3_token_len":551,"char_repetition_ratio":0.15112016,"word_repetition_ratio":0.05263158,"special_character_ratio":0.22936153,"punctuation_ratio":0.14460784,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9630958,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-28T03:43:45Z\",\"WARC-Record-ID\":\"<urn:uuid:8d35fd5b-7c99-4dee-a6dc-ea08eff6b16b>\",\"Content-Length\":\"126667\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0fa7f0f-f5cf-4768-82ac-57036369441b>\",\"WARC-Concurrent-To\":\"<urn:uuid:02c1d841-a324-4e75-b062-e8af77fc9d12>\",\"WARC-IP-Address\":\"104.26.14.93\",\"WARC-Target-URI\":\"https://www.w3resource.com/javascript/object-property-method/array-length.php\",\"WARC-Payload-Digest\":\"sha1:UU277AHLKA47TEBUKA4UKHSJRUWOMR3V\",\"WARC-Block-Digest\":\"sha1:NE5PRCULUYXTUVWBMK7OB2PQHOSCWGMC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103347800.25_warc_CC-MAIN-20220628020322-20220628050322-00657.warc.gz\"}"}
https://www.groundai.com/project/data-assimilation-and-parameter-estimation-for-a-multiscale-stochastic-system-with-alpha-stable-levy-noise/1	[ "Data assimilation and parameter estimation for a multiscale stochastic system with \\alpha-stable Lévy noise\n\n# Data assimilation and parameter estimation for a multiscale stochastic system with α-stable Lévy noise\n\nYanjie Zhang, Zhuan Cheng, Xinyong Zhang, Xiaoli Chen, Jinqiao Duan and Xiaofan Li Center for Mathematical Sciences & School of Mathematics and Statistics, Huazhong University of Science and Technology, Wuhan 430074, China Department of Applied Mathematics, Illinois Institute of Technology, Chicago, IL 60616, USA Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China This work was partly supported by the NSF grant 1620449, and NSFC grants 11531006, 11371367, and 11271290.\n###### Abstract\n\nThis work is about low dimensional reduction for a slow-fast data assimilation system with non-Gaussian stable Lévy noise via stochastic averaging. When the observations are only available for slow components, we show that the averaged, low dimensional filter approximates the original filter, by examining the corresponding Zakai stochastic partial differential equations. Furthermore, we demonstrate that the low dimensional slow system approximates the slow dynamics of the original system, by examining parameter estimation and most probable paths.\n\nKeywords: Multiscale systems, non-Gaussian Lévy noise, averaging principle, Zakai equation, parameter estimation, most probable paths\n\n## 1 Introduction\n\nData assimilation is a procedure to extract system state information with the help of observations . The state evolution and the observations are usually under random fluctuations. The general idea is to gain the best estimate for the true system state, in terms of the probability distribution for the system state, given only some noisy observations of the system. It provides a recursive algorithm for estimating a signal or state of a random dynamical system based on noisy measurements. It is also very important in many practical applications from inertial guidance of aircrafts and spacecrafts to weather and climate prediction. Most of the existing works on data assimilation is conducted in the context of Gaussian random fluctuations. The effects of multiscale signal and observation processes in the context of Gaussian random fluctuations has been considered by Park et.al.(see ), and they have shown that the probability density of the original system converges to that of the reduced system, by a Fourier analysis method. Imkeller et.al. have further proved the convergence in distribution for the optimal filter, via backward stochastic differential equations and asymptotic techniques.\n\nHowever, random fluctuations are often non-Gaussian (in particular, Lévy type) in nonlinear systems, for example, in geosciences , and biosciences [3, 4, 5, 14, 15, 16, 32]. There are experimental demonstrations of Lévy fluctuations in optimal foraging theory, rapid geographical spread of emergent infectious disease and switching currents. Humphries et. al. used GPS to track the wandering black bowed albatrosses around an island in Southern Indian Ocean to study the movement patterns of searching food. They found that by fitting the data of the movement steps, the movement patterns obeys the power-law property with power parameter . La Cognata et. al. considered a Lotka-Volterra system of two competing species subject to multiplicative -stable Lévy noise, and analyzed the role of the Lévy noise sources. Lisowski et. al. studied a model of a stepping molecular motor consisting of two connected heads, and examined its dynamics as noise parameters varied. Speed and direction appear to very sensitively depend on characteristics of the noise. They explored the effect of noise on the ballistic graphene-based small Josephson junctions in the framework of the resistively and capacitively shunted model and found that the analysis of the switching current distribution made it possible to efficiently detect a non-Gaussian noise component in a Gaussian background.\n\nLévy motions are appropriate models for a class of important non-Gaussian processes with jumps or bursts [3, 4, 22]. It is desirable to consider data assimilation when the system evolution is under Lévy motions. This has been recently considered by one of us and other authors but not in the multiscale context (see [20, 28, 29, 35]).\n\nThe multi-scale stochastic dynamical systems arise widely in finance and biology. For example, there are two kinds of mutual securities in financial markets. One for the low-risk bonds, which can be characterized by ordinary differential equations; the other for high-risk stocks, whose price has two different changes. On the one hand, the edge change caused by the normal supply and demand can be characterized by Gaussian noise. On the other hand, due to the arrival of the important information of the stock, there will be a finite jump in the stock price. Such perturbations can be characterized by non-Gauss noise. In general, stock prices change at all times, while bond prices change for months or even years. Thus, the price of these two securities can be characterized by two-scales system with non-Gaussian noise (see ). Moreover, a large number of observations from biological experiments showed that the production of mRNA and proteins occur in a bursty, unpredictable, and intermittent manner, which create variation or noise in individual cells or cell-to-cell interactions. Such burst-like events appear to be appropriately modeled by the non-Gaussian noise. Since the mRNA synthesis process is faster than the protein dynamics, this leads to a two-time-scale system (see ). Here represents the ratio between the natural time scales of the protein and mRNA.\n\nParameter estimation for continuous time stochastic models is an increasingly important part of the overall modeling strategy in a wide variety of applications. It is quite often the case that the data to be fitted to a diffusion process has a multiscale character with Gaussian noise. One example is in molecular dynamics, where it is desirable to find effective models for low dimensional phenomena (such as conformational dynamics, vacancy diffusion and so forth) which are embedded within higher dimensional time-series. We are often interested in the parameter (see ), which represents the degradation or production rates of protein and mRNA. In this paper, we develop a parameter estimation method for multiscale diffusions with non-Gaussian noise. The results established here may be used to examine the change rate for low-risk bounds (see ).\n\nIn this present paper, we consider a slow-fast data assimilation system under Lévy noise, but only the slow component is observable. By averaging out the fast component via an invariant measure, we thus reduce the system dimension by focusing on the slow system evolution. We prove that the reduced lower dimensional filter effectively approximates (in probability distribution) the original filter. We demonstrate that a system parameter may be estimated via the low dimensional slow system, utilising only observations on the slow component. The accuracy for this estimation is quantified by -moment, with . We apply the stochastic Nelder-Mead method for optimization in the searching for the estimated parameter value. Furthermore, we illustrate the low dimensional approximation by comparing the most probable paths for the slow system and the original system. Finally, we make some remarks in Section 6.\n\nThis paper is organized as follows. After recalling some basic facts about Lévy motions and the generalized solution in the next section, we address the effects of the multiscale signal and observation processes in the context of Lévy random fluctuations in Section 3 to Section 5. We illustrate the low dimensional slow approximation by examining zakai equation, parameter estimation and most probable paths. Finally, we give some discussions and comments in a more biological context.\n\n## 2 Preliminaries\n\nWe recall some basic definitions for Lévy motions (or Lévy processes).\n\n###### Definition 1.\n\nA stochastic process is a Lévy process if\n\n1. (a.s.);\n\n2. has independent increments and stationary increments; and\n\n3. has stochastically continuous sample paths, i.e., for every , in probability, as .\n\nA Lévy process taking values in is characterized by a drift vector , an non-negative-definite, symmetric covariance matrix and a Borel measure defined on . We call the generating triplet of the Lévy motions . Moreover, we have the Lévy-Itô decomposition for as follows:\n\n Lt=bt+BQ(t)+∫\|\|y\|\|<1y˜N(t,dy)+∫\|\|y\|\|≥1yN(t,dy), (2.1)\n\nwhere is the Poisson random measure, is the compensated Poisson random measure, is the jump measure, and is an independent standard -dimensional Brownian motion. The characteristic function of is given by\n\n E[exp(i⟨u,Lt⟩)]=exp(tψ(u)), u∈Rn, (2.2)\n\nwhere the function is the characteristic exponent\n\n ψ(u)=i⟨u,b⟩−12⟨u,Qu⟩+∫Rn∖{0}(ei⟨u,z⟩−1−i⟨u,z⟩I{\|\|z\|\|\\textless1})ν(dz). (2.3)\n\nThe Borel measure is called the jump measure. Here denotes the scalar product in .\n\n###### Definition 2.\n\nFor , an -dimensional symmetric -stable process is a Lévy process with characteristic exponent\n\n ψ(u)=−C1(n,α)\|u\|α, for u∈Rn (2.4)\n\nwith .\n\nFor an -dimensional symmetric -stable Lévy process, the diffusion matrix , the drift vector , and the Lévy measure is given by\n\n ν(du)=C2(n,α)\|\|u\|\|n+αdu, (2.5)\n\nwhere .\n\nFor every function , the generator for this symmetric -stable Lévy process in is\n\n (Aαϕ)(x)=∫Rn∖{0}[ϕ(x+u)−ϕ(x)]ν(du). (2.6)\n\nIt is known that extends uniquely to a self-adjoint operator in the domain. By Fourier inverse transform,\n\n Aαϕ=θα,n(−Δ)α/2ϕ(x), (2.7)\n\nwhere\n\n θα,n=∫Rn∖{0}(cos(⟨e,y⟩)−1)ν(du) (2.8)\n\nwith being the unit vector in .\n\nFor fix , and set\n\n ρ(x)=(1+\|x\|)η. (2.9)\n\nLet be the weighted -space with norm:\n\n \|\|f\|\|p;ρ=(∫Rn\|f(x)\|pρ(x)dx)1p (2.10)\n\nFor , let be the -order weighted Sobolev space with norm\n\n \|\|f\|\|l,p;ρ=l∑i=0\|\|∇lf\|\|p;ρ, (2.11)\n\n###### Definition 3.\n\nA backward predictable stochastic process is called a generalized solution of the equality\n\n v(t,x)=φ(x)+∫[t,T]Lv(s,x)ds+∫[t,T]hlv(s,x)dwl(s), (2.12)\n\nif for every it satisfies the following equation\n\n (v(t,x),y)=(φ,y)+∫[t,T](v(s,x),L∗y)ds+∫[t,T](hlv(s,x),y)dwl(s) (2.13)\n\nwith being the generator of some Lévy process.\n\n## 3 A slow-fast filtering system\n\nLet us consider the following slow-fast signal-observation system:\n\n dXt=f1(Xt,Yt)dt+g1(Xt,Yt)dB1t+σ1dLα11,x∈Rn, (3.14) dYt=ε−1f2(Xt,Yt)dt+ε−12g2(Xt,Yt)dB2t+1ε1α2dLα22,y∈Rm, (3.15) dZt=h(Xt)dt+dWt,z∈Rd. (3.16)\n\nHere is an -valued signal process which represents the slow and fast components. The constant represents the noise intensity for the slow variable. The observation process is -valued. The standard Brownian motions are independent. The non-Gaussian processes (with ) are independent symmetric -stable Lévy processes with triplets and , respectively. The parameter is the ratio of the slow time scale to the fast time scale. We make the following assumptions on this filtering system.\nHypothesis H.1. The functions satisfy the global Lipschitz conditions, i.e., there exists a positive constant such that\n\n \|\|f1(x1,y1)−f1(x2,y2)\|\|2+\|\|g1(x1,y1)−g1(x2,y2)\|\|2+\|\|f2(x1,y1)−f2(x2,y2)\|\|2+\|\|g2(x1,y1)−g2(x2,y2)\|\|2≤γ[\|\|x1−x2\|\|2+\|\|y1−y2\|\|2]\n\nfor all\n\n###### Remark 1.\n\nNote that with the help of the global Lipschitz condition, it follows that there is a positive constant such that\n\n \|\|f1(x,y)\|\|2+\|\|g1(x,y)\|\|2+\|\|f2(x,y)\|\|2+\|\|g2(x,y)\|\|2≤K(1+\|\|x\|\|2+\|\|y\|\|2)\n\nfor all .\n\nHypothesis H.2. The coefficients are of class with the first and second order derivatives bounded\n\nHypothesis H.3. The sensor function is of class , i.e. all the bounded continuous functions on .\n\nHypothesis H.4. There exists a positive constant , such that\n\n supx{Tr[g2(x,y)g2(x,y)T]+2⟨y,f2(x,y)⟩}≤−C(1+\|y\|2). (3.17)\n\nThis hypothesis (H.4.) ensures the existence of an invariant measure (see ) for the fast component . With the special scaling exponent for the fast component , this invariant measure is independent of (see [11, 12]).\n\nThe infinitesimal generator of the slow-fast stochastic system is\n\n L=L1+1εL2, (3.18)\n\nwhere\n\n ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩L1Φ(x,y,t)=f1⋅∇xΦ(x,y,t)+12Tr[G1⋅Hx(Φ(x,y,t))]+P.V.∫Rn∖{0}(Φ(x+σ1z1,y,t)−Φ(x,y,t))ν1(dz1),L2Φ(x,y,t)=f2⋅∇yΦ(x,y,t)+12Tr[G2⋅Hy(Φ(x,y,t))]+P.V.∫Rm∖{0}(Φ(x,y+z2,t)−Φ(x,y,t))ν2(dz2). (3.19)\n\nHere , is the gradient, and is the Hessian matrix (with respect to and respectively). Let\n\n Zt=σ(Zs:0≤s≤t)∨N, (3.20)\n\nwhere is the collection of all -negligible sets of . Define\n\n Z=⋁t∈R+Zt, (3.21)\n\nwhere denotes taking the -algebra generated by the union . That is,\n\n Z=σ(⋃tZt). (3.22)\n\nBy the version of Girsanov’s change of measure theorem, we obtain a new probability measure , such that the observation becomes -independent of the signal variables . This can be done through\n\n d~PdP=exp(−m∑i=1∫t0hi(Xs)dWis−12m∑i=1∫t0hi(Xs)2ds). (3.23)\n\nDenote\n\n ~Rt=dPd~P∣∣∣Zt. (3.24)\n\nThen by the Kallianpur-Striebel formula, for every bounded differentiable function , we have the following representation:\n\n E[ϕ(Xt,Yt)\|Zt]=~E[~Rtϕ(Xt,Yt)\|Zt]~E[~Rt\|Zt]=~E[~Rtϕ(Xt,Yt)\|Z]~E[~Rt\|Z]. (3.25)\n\nThe unnormalized conditional distribution of , given , is defined as . Thus, we have the following Zakai equation:\n\n {dPt(ϕ)=Pt(Lϕ)dt+Pt(ϕhT)dZt,P0(ϕ)=E[ϕ(X0,Y0)]. (3.26)\n\nThe -marginal of is defined as\n\n ρt(ϕ)=∫Rn+mϕ(x)Pt(dx,dy). (3.27)\n\nNow we define a reduced, low dimensional signal-observation system\n\n {d¯Xt=¯f1(¯Xt)dt+¯g1(¯Xt)dB1t+σ1dLα11,d¯Zt=h(¯Xt)dt+dBt. (3.28)\n\nHere\n\n {¯f1(x)=∫Rmf1(x,y)μx(dy),¯G(x)=∫Rmg1(x,y)g1(x,y)Tμx(dy). (3.29)\n\nThe unnormalized conditional distribution corresponding to the filter for the reduced system (3.28) satisfies the following (reduced) Zakai equation\n\n {d¯ρt(ϕ)=¯ρt(¯Lϕ)dt+¯ρt(ϕhT)d¯Zt,¯ρ0(ϕ)=E[ϕ(X0)], (3.30)\n\nwhere\n\n ¯Lϕ(x,t)=¯f1(x)⋅∇xϕ(x,t)+12Tr[Hx(¯G(x)ϕ(x,t))]+∫Rn(ϕ(x+σ1z1,t)−ϕ(x,t))ν1(dz1). (3.31)\n\nHowever, we are more interested in the reduced filtering problem with the actual observation . This leads us to rewrite the reduced Zakai equation (3.30) as follows:\n\n {d¯ρt(ϕ)=¯ρt(¯Lϕ)dt+¯ρt(ϕhT)dZt,¯ρ0(ϕ)=E[ϕ(X0)]. (3.32)\n###### Lemma 1.\n\nAssume that the following conditions are satisfied:\n\n1. The functions , and for and and their derivatives of first order (in x) as well as the derivatives of second order (in x) are uniformly bounded by the constant . The functions are locally uniformly bounded in .\n\n2. .\n\nLet be a generalized solution of problem\n\n ⎧⎪ ⎪⎨⎪ ⎪⎩du(t,x,ω)=Lu(t,x,ω)dt+hl(t,x,ω)u(t,x,ω)dBl(t),(t,x,ω)∈[T0,T]×Rk×Ω,u(T0,x,ω)=φ(x,ω),(x,ω)∈Rk×Ω, (3.33)\n\nwhere\n\n Lu(t,x)=12σik(t,x,ω)σjk(t,x,ω)∂2iju(t,x,ω)+bi∂iu(t,x,ω)+12∫Rk(u(t,x+z)+u(t,x−z)−2u(t,x))dz\|z\|k+α, (3.34)\n\nand is a generalized solution of problem\n\n (3.35)\n\nThen the following formulas hold\n\n u(t,x)=E(φ(Y(t,x,T0))χ1(t,T0)\|FT0t), (3.36)\n\nand\n\n v(t,x)=E(φ(X(T,x,t))χ2(T,t)\|FTt), (3.37)\n\nwhere is the forward stochastic differential equation with generator and is the backward stochastic differential equation with generator . Here and satisfy the following equations\n\n χ1(t,s)=exp(∫[s,t]hl(τ,Y(t,x,τ))dBl(τ)−12∫[s,t]hlhl(τ,Y(t,x,τ))dτ), (3.38)\n\nand\n\n χ2(s,t)=exp(∫[t,s]hl(τ,X(τ,x,t))dBl(τ)−12∫[t,s]hlhl(τ,X(τ,x,t))dτ). (3.39)\n###### Proof.\n\nDenote . Before we proceed to the proof of the lemma, let us consider the problem\n\n (3.40)\n\nBy the definition of a backward predictable stochastic process, the coefficients in equation (3.40) are predictable relative to the family of algebra with . The process is a Wiener martingale with respect to the same family and the initial condition is measurable with respect to the minimal -algebra of this family.\n\nLet be a generalized solution of problem. Then by the definition of generalized solution, for every , the following equality holds on ,\n\n (v(T−s,x),y)=(φ,y)+∫[0,T−s][−(aij(T−s)∂iv(s,x),yj)+(bi(T−s)−aijj(T−s))∂iv(s,x))+12∫Rk(v(s,x+z)+v(s,x−z)−2v(s,x))dz\|z\|k+α,y)]ds+∫[0,T−s]hl(T−t)v(t,x)dBlT(s). (3.41)\n\nDenote in equation (3.41), we find that satisfies the following equation on , for every .\n\n (u(t,x),y)=(φ,y)+∫[t,T][−(aij∂iv(s,x),yj)+((bi−aijj)∂iu(t,x))+12∫Rk(v(t,x+z)+v(t,x−z)−2v(t,x))dz\|z\|k+α,y)]ds+∫[t,T]hlv(s,x)dBl(s). (3.42)\n\nThus is an generalized solution of problem (3.35).\n\nOn the other hand, changing the variables in equality (3.42), we obtain that is a generalized solution of problem (3.40). Thus we have proved that problems (3.35) and (3.40) are equivalent. Therefore all the results obtained for problem (3.35) are naturally carried over to problem (3.40). For the problem (3.40), we can use the similar method to obtain the existence and uniqueness of - solution to stochastic fractal equations by using purely probabilistic argument (see [19, 30]). This completes the proof of this lemma. ∎\n\nNow we show that the reduced system approximates the original system, by examining the corresponding Zakai equations. Before describing the theorem, we start by describing the probabilistic representation for semi-linear stochastic fractal equations. We proceed to consider the probability measure . Note that the process is a Brownian motion under . For convenience, we rewrite and as and , respectively. By Lemma 1, we know solves a stochastic partial differential equation (SPDE).\n\n {−dvϕ(t,x,y)=Lvϕ(t,x,y)dt+h(x)Tvϕ(t,x,y)d←−Bt.vϕ(T,x,y)=ϕ(x). (3.43)\n\nHere denotes Itô’s backward integral. Likewise, we introduce , and then solves the following SPDE\n\n ⎧⎨⎩−d¯¯¯vϕ(t,x)=¯¯¯¯L¯¯¯vϕ(t,x)dt+h(x)T¯¯¯vϕ(t,x)d←−Bt,¯¯¯vϕ(T,x)=ϕ(x). (3.44)\n\nBy the version of Girsanov’s change of measure theorem and Markov property of , we know that for any : . In particular, we have\n\n ρT(ϕ)=∫vϕ(0,x,y)P(X0,Y0)(dx,dy). (3.45)\n\nSimilarly, we have\n\n ¯ρT(ϕ)=∫¯vϕ(0,x)P(X0)(dx). (3.46)\n\nThis is our main result on the comparison between the original filter and the reduced filter. This is desirable when only the slow component is observable.\n\n###### Theorem 1.\n\nUnder the hypotheses (H.1)-(H.4) and for with , there exists a positive constant such that for , the following estimate holds\n\n E[\|ρT(ϕ)−¯ρT(ϕ)\|p]≤Cε, (3.47)\n\nwith a positive constant independent of . This implies that the reduced filter approximates the original filter, as the scale parameter tends to zero.\n\n###### Proof.\n E[\|ρT(ϕ)−¯ρT(ϕ)\|p]=E[∣∣∫(vϕ(0,x,y)−¯vϕ(0,x))P(X0,Y0)(dx,dy)∣∣p],≤E[∫∥∥vϕ(0,x,y)−¯vϕ(0,x)∣∣pP(X0,Y0)(dx,dy)],=∫E[∣∣vϕ(0,x,y)−¯vϕ(0,x)∣∣p]P(X0,Y0)(dx,dy),=∫E[∣∣~Ex,y[ϕ(XT)~R0,T\|Y0,T]−~Ex[ϕ(¯XT)~R0,T\|Y0,T]∣∣p]P(X0,Y0)(dx,dy),≤C∫E∣∣~Ex[(ϕ(XT)−ϕ(¯XT))~R0,T\|Yt,T]∣∣pP(X0,Y0)(dx,dy),=C∫E∣∣E[ϕ(XT)−ϕ(¯XT)\|YT]~Ex[~R0,T\|Y0,T]∣∣pP(X0,Y0)(dx,dy),≤C∫E∣∣E[XT−¯XT\|YT]∣∣pP(X0,Y0)(dx,dy). (3.48)\n\nUsing the Jensen’s inequality, we have\n\n E[\|ρT(ϕ)−¯ρT(ϕ)\|p]≤∫E\|XT−¯XT\|pP(X0,Y0)(dx,dy). (3.49)\n\nApplying a similar argument from , we obtain\n\n limϵ→0E\|XT−¯XT\|p=0 (3.50)\n\nFinally, we conclude that\n\n E[\|ρT(ϕ)−¯ρT(ϕ)\|p]≤Cε. (3.51)\n\nThis completes the proof of Theorem 1. ∎\n\n## 4 Parameter estimation\n\nIn this section, we consider parameter estimation in the following slow-fast dynamical system\n\n (4.52)\n\nwhere is an matrix, is a positive definite matrix with eigenvalues . is an unknown parameter defined in an compact set of .\n\nHypothesis H.5. For all , the function is uniformly bounded and Lipschitz w.r.t x and y. The function is smooth with the first order derivatives bounded by .\n\nDefine an averaged, low dimensional slow stochastic dynamical system in (see ), as in the previous section\n\n d¯x(t)=A¯x(t)+¯f1(¯x(t),θ)dt, (4.53)\n\nwhere\n\n ¯f1(¯x,θ)≜∫Rmf1(x,y,θ)μx(dy).\n\nWe will estimate the unknown parameter , based on this low dimensional slow system (4.53) but using the observations on the slow component only. Denote the solution of the original slow-fast system (4.52) with actual parameter by , and the solution of the slow system (4.53) with parameter by . We recall the following lemma.\n\n###### Lemma 2.\n\nUnder hypothesis (H.6), the following strong convergence holds\n\n limε→0E\|x(t,θ0)−¯x(t,θ0)\|p=0, t∈[0,T] and p∈(1,α). (4.54)\n###### Proof.\n\nThe proof of average principle is similar the the infinite case which has been studied in , Thus we omit here. ∎\n\nWe take as the objective function and assume there is a unique such that . This is our estimated parameter value. Then we can state our main result as follows.\n\n###### Theorem 2.\n\nUnder hypothesis (H.6), the estimated parameter value converges to the true parameter value, as the scale parameter approaches zero. That is,\n\n limε→0^θϵ=θ0. (4.55)\n###### Proof.\n\nNote that\n\n E\|¯x(t,θ0)−¯x(t,^θε)\|p≤C(E\|x(t,θ0)−¯x(t,θ0)\|p+E\|x(t,θ0)−¯x(t,^θε)\|p), (4.56)\n\nfor some positive constant . Integrating both sides with respect to time, we get\n\n ∫T0E\|¯x(t,θ0)−¯x(t,^θε)\|pdt≤C(F(θ0)+F(^θε))≤2CF(θ0). (4.57)\n\nWe calculate the difference between and to obtain\n\n ˙¯x(t,θ0)−˙¯x(t,^θε)=A[¯x(t,θ0)−¯x(t,^θε)]+[¯f1(¯x(t,θ0),θ0)−¯f1(¯x(t,^θε),^θε)].\n\nBy the variation of constant formula, we have\n\n ¯x(t,θ0)−¯x(t,^θε)=∫t0eA(t−s)[¯f1(¯x(s,θ0),θ0)−¯f1(¯x(s,^θε),^θε)]ds.\n\nUsing the mean value theorem, we obtain\n\n ¯x(t,θ0)−¯x(t,^θε)=(θ0−^θε)∫t0eA(t−s)∇θ¯f1(¯x(s,θ′),θ′)ds, (4.58)\n\nfor some with . Denote\n\n G(θ′)=∫T0∣∣∣∫t0eA(t−s)∇θ¯f" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8824653,"math_prob":0.996596,"size":16793,"snap":"2019-51-2020-05","text_gpt3_token_len":3692,"char_repetition_ratio":0.13312288,"word_repetition_ratio":0.0261959,"special_character_ratio":0.2206872,"punctuation_ratio":0.12927143,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99885666,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-26T11:48:14Z\",\"WARC-Record-ID\":\"<urn:uuid:b16dc727-41a3-468b-b64f-579f89d9e43c>\",\"Content-Length\":\"1049303\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a7d37d6-e47b-487b-86b4-805f9577ff46>\",\"WARC-Concurrent-To\":\"<urn:uuid:0c64b539-3d34-4b07-896b-f0e579f7c5e5>\",\"WARC-IP-Address\":\"35.186.203.76\",\"WARC-Target-URI\":\"https://www.groundai.com/project/data-assimilation-and-parameter-estimation-for-a-multiscale-stochastic-system-with-alpha-stable-levy-noise/1\",\"WARC-Payload-Digest\":\"sha1:SYIJHIBQF5P5JVBVW33BB4HGHS3EIHHL\",\"WARC-Block-Digest\":\"sha1:IEW6QTJTHPV6Z76D2K7XFGOCBWDYWX7C\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251688806.91_warc_CC-MAIN-20200126104828-20200126134828-00030.warc.gz\"}"}
https://excelchamps.com/formulas/starts-with/	[ "# Check IF a Cell Value Starts with a Text or a Number (Formula in Excel)\n\nLast Updated: November 21, 2023\n\n- Written by Puneet\n\nIn Excel, while working with data there might be a situation when you need to check if the value in a cell starts with a particular text or if the starting part of that value is a number or not. In this tutorial, we will learn to write a formula to text both conditions.\n\n## IF a Cell Value Starts with a Particular Text\n\nIn the following example, you need to check if the starting three characters of the cell are equal to LXI. For this, you need to use the IF and LEFT.\n\nYou can use the below steps to write this formula:\n\n1. First, enter the IF function in a cell.\n2. After that, enter the LEFT function in the first argument of the IF, and refer to the cell A2 where you have the value, in the second argument, use 3 for the num_digit.\n3. Next, use the equal sign and enter the value (using double quotation marks) that you want to test.\n4. Now, in the second argument, enter “Yes”.\n5. And in the third argument, enter “No”.\n6. In the end, enter closing parentheses and hit enter to get the result.\n\nAs you can see below, for all the values where we have the value “LXI” at the start of the value, it returns Yes, else No.\n\n``=IF(LEFT(A2,3)=\"LXI\",\"Yes\",\"No\")``\n\nUse the below formula if you want to count the number of cells which starts with a particular value.\n\n``=SUMPRODUCT(--(LEFT(A2:A10,3)=\"LXI\"))``\n\n``=IF(ISNUMBER(VALUE(LEFT(A2,2))), \"Yes\",\"No\")``" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.85031164,"math_prob":0.9637158,"size":2022,"snap":"2023-40-2023-50","text_gpt3_token_len":498,"char_repetition_ratio":0.1382557,"word_repetition_ratio":0.026385223,"special_character_ratio":0.24727993,"punctuation_ratio":0.12053572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9984236,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-07T07:41:16Z\",\"WARC-Record-ID\":\"<urn:uuid:c3503def-4ea2-4405-a4e9-f22d6db8de13>\",\"Content-Length\":\"238989\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:54f72757-87a2-48b0-81f7-99017977035f>\",\"WARC-Concurrent-To\":\"<urn:uuid:0aef0fc0-5188-4e96-9ea8-d6e670c436c0>\",\"WARC-IP-Address\":\"162.159.134.42\",\"WARC-Target-URI\":\"https://excelchamps.com/formulas/starts-with/\",\"WARC-Payload-Digest\":\"sha1:XNCQV7RJTGC72NUP3AQPBLFJJYLDF7GQ\",\"WARC-Block-Digest\":\"sha1:KA7UF5CPKJNPPHQUKKB5RRQIBTDLP5ST\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100650.21_warc_CC-MAIN-20231207054219-20231207084219-00686.warc.gz\"}"}
https://kb.kutu66.com/c/post_2805358	[ "# c - 根据点旋转一个对象\n\n0 0\n``````\nO\nposition 2\n\nO Y\nobject ( x,y,z )\nposition 1\n\n``````\n\n``````\nglPushMAtrix ()\nglTranslatef ( -x, -y, -z ) ;\nglRotatef ( Q, 1.0f, 0.0f, 0.0f );\n\nglCylinder (/argument/)\nglPopMatriX ()\n\n``````\n\n0 0\n\n``````\nglTranslatef(-x, -y, -z);\nglRotatef(Q, 1.0, 0.0, 0.0);\nglTranslatef(x, y, z);\n//draw the object\n\n``````\n\n``````\naxis = vec(center_of_rotation - initial_position, center_of_rotation - final_position)\n\n``````\n\n``````\nfloat X1;//initial position\nfloat X2;//final position\nfloat O;//orign of rotation\n\nfloat OX1; OX1 = X1 - O; OX1 = X1 - O; OX1 = X1 - O;\nfloat OX2; OX2 = X2 - O; OX2 = X2 - O; OX2 = X2 - O;\n\nfloat axis;//vector product OX1 and OX2\naxis = OX1OX2-OX1OX2;\naxis = OX1OX2-OX1OX2;\naxis = OX1OX2-OX1OX2;\n\n``````" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.75091153,"math_prob":0.9975899,"size":860,"snap":"2019-51-2020-05","text_gpt3_token_len":546,"char_repetition_ratio":0.13668224,"word_repetition_ratio":0.0,"special_character_ratio":0.42093024,"punctuation_ratio":0.16201118,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9907709,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-19T02:19:07Z\",\"WARC-Record-ID\":\"<urn:uuid:00cd62e7-3347-47c2-ae98-1323dd09c1cf>\",\"Content-Length\":\"36280\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4189f5f9-ae59-4f52-8ffc-d91f1f87d12a>\",\"WARC-Concurrent-To\":\"<urn:uuid:e0b505e3-697f-410e-90ca-30f8f1e9b486>\",\"WARC-IP-Address\":\"129.204.66.212\",\"WARC-Target-URI\":\"https://kb.kutu66.com/c/post_2805358\",\"WARC-Payload-Digest\":\"sha1:32OTKLIUJX7BQFQAXN5SRWKWATEWV4ZF\",\"WARC-Block-Digest\":\"sha1:UECISCQGMYIGZ3TVJCX6K7DRANPU6GL7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594101.10_warc_CC-MAIN-20200119010920-20200119034920-00114.warc.gz\"}"}
https://www.hanspub.org/journal/PaperInformation.aspx?PaperID=34651	[ "# 红外线透镜穿透率量测技术Measurement Technology of Transmittance of Infrared Optical System\n\nDOI: 10.12677/OE.2020.101003, PDF, HTML, XML, 下载: 194 浏览: 1,101\n\nAbstract: Infrared optical system is used in 8~12 μm range of infrared spectrum, which cannot be seen by human eyes, and the transmittance measurement method of visible optical system cannot be used, and the transmittance of infrared optical system was difficult to measure. At present, FTIR spec-trometer is used to measure transmittance of infrared optical system, which is very expensive and expensive to maintain. In this study, the transmittance of infrared optical system is measured by integrating the infrared sensing circuit and the smart infrared collimating beam. The measurement system consists of 2 mm diameter infrared collimating beam, thermopile sensor, OP amplifier, A/D IC and microprocessor. The thermopile converts the thermal radiation into analog electronic signal, and amplifies by OP amplifier, converted into digital by A/D IC, processed by 8051 microprocessor and displayed by LCD. The specifications of the lens to be tested are: focal length 19 mm, the f-number f/1.1 and the material of the lens was made by the black diamond. The test result of transmittance of the lens was 89%.\n\n1. 前言\n\n2. 基本原理 \n\n$\\text{TE:}\\gamma \\perp =\\frac{\\mathrm{cos}\\theta -\\sqrt{{n}^{2}-{\\mathrm{sin}}^{2}\\theta }}{\\mathrm{cos}\\theta +\\sqrt{{n}^{2}-{\\mathrm{sin}}^{2}\\theta }}=-\\frac{\\mathrm{sin}\\left({\\theta }_{i}-{\\theta }_{t}\\right)}{\\mathrm{sin}\\left({\\theta }_{i}+{\\theta }_{t}\\right)}$ (1)\n\n$\\text{TE:}\\gamma \\parallel =\\frac{{n}^{2}\\mathrm{cos}\\theta -\\sqrt{{n}^{2}-{\\mathrm{sin}}^{2}\\theta }}{{n}^{2}\\mathrm{cos}\\theta +\\sqrt{{n}^{2}-{\\mathrm{sin}}^{2}\\theta }}=-\\frac{\\mathrm{tan}\\left({\\theta }_{i}-{\\theta }_{t}\\right)}{\\mathrm{tan}\\left({\\theta }_{i}+{\\theta }_{t}\\right)}$ (2)\n\n$\\text{TE:}t\\perp =\\frac{2\\mathrm{cos}\\theta }{\\mathrm{cos}\\theta +\\sqrt{{n}^{2}-{\\mathrm{sin}}^{2}\\theta }}=\\frac{2\\mathrm{sin}{\\theta }_{t}\\mathrm{cos}{\\theta }_{i}}{\\mathrm{sin}\\left({\\theta }_{i}+{\\theta }_{t}\\right)}$ (3)", null, "(4)\n\n${R}_{s}={r}_{s}{r}_{s}^{\\varnothing },\\text{\\hspace{0.17em}}{R}_{p}={r}_{p}{r}_{p}^{\\varnothing }$ (5)\n\n${\|{r}_{s}\|}^{2}+{\|{t}_{s}\|}^{2}=1,\\text{\\hspace{0.17em}}{\|{r}_{p}\|}^{2}+{\|{t}_{p}\|}^{2}=1$ (6)\n\n${R}_{s,normal}={R}_{p,normal}=\\left[\\frac{n-{n}^{\\prime }}{n+{n}^{\\prime }}\\right]$ (7)", null, "Figure 1. Refraction and refraction at an interface between two media\n\n3. 系统架构", null, "Figure 2. Measurement system block diagram", null, "(a)", null, "(b)\n\nFigure 3. (a) Thin film infrared source and parabolic mirror; (b) IR source modulation curve", null, "Figure 4. Thermopile sensors", null, "Table 1. Optical electrical characteristics", null, "Figure 5. Output and input curve\n\n1) IR Source脉冲电路", null, "Figure 6. System structure", null, "Figure 7. PWM modulations", null, "Figure 8. IR source circuit\n\n2) Thermopile的讯号放大\n\nThermopile的讯号输出大约只有数十μV~数mV左右，为了后续A/D数据撷取，必须进行高倍数的讯号放大。且因制程及材质的关系其内部阻抗也相当大，约在数十KΩ左右且变化很大，因此放大器的选择即是一项重要的工作。要选用适合的放大器，首先要考虑组件的特性。Thermopile的输出为一直流电压准位且压降微弱，只有数十μV~数mV左右，故选用的放大器带宽并不是我们在意的项目，必须特别注意的是VOS (Offset Voltage)这项参数。过大的VOS将会使电压输出失真，因此我们必须选用低VOS的运算放大器。另外VOS的温度飘移(Offset Drift)也应尽量选择低的参数值，以增加整个放大电路的稳定性。总和以上要点，我们选用了MAXIM的MAX4238，MAX4238的VOS在室温下的典型值只有0.1 μV，即使最大值也只有2 μV，且其VOS温度飘移为10 nV/℃。另外，MAX4238的增益带宽积(GBWP)也高达1 MHZ，在高倍数放大下也不至于失真，各规格特性均可完全符合我们的需求。\n\nThermopile的结构上约有数50~150 KΩ的内阻，且非定值，不同组件间的内阻差异也很大，为了排除Thermopile内部阻抗的影响，放大器电路设计上必须使用具有高输入阻抗的放大型式。非反向放大器即具有高输入阻抗的特性，此类型放大器输入阻抗一般可有数十MΩ~数百MΩ的等级。", null, "Figure 9. Thermopile preamplifier circuit\n\n$fc=\\frac{1}{2\\text{π}RC}=\\frac{1}{2\\text{π}×10\\text{k}×1\\text{μ}}=15.9\\text{\\hspace{0.17em}}\\text{Hz}$ (8)", null, "Figure 10. Thermopile preamplifier circuit\n\nMCP3302是一个SPI (Serial Peripheral Interface)串行接口的13bit SAR型模拟/数字转换器(A/D Converter)，可依需求使用4通道单端输入或2通道差动输入，电源电压使用5 V时，取样频率(sampling rate)可以达到100 KHZ，量测误差最大为±1 LSB INL (MCP3302-B)或±2 LSB INL (MCP3302-C)。\n\nMCP3302使用上必须遵循其制定的SPI命令格式来对其进行转换与读取的动作。MCP3302的串行接口有三个入接脚，一个讯号输出接脚，其通讯格式里为一个启始位之后接着4个命令位，此4个位指定了MCP3302的动作模式，分别为1 bit的单端/差动选择位(Single/Diff)和3 bits的channel选择位(D2~D0)。", null, "Figure 11. ADC signal acquisition interface circuit\n\n4) 单芯片89C51处理器\n\n5) 红外线检测装置实体", null, "(a)", null, "(b)", null, "(c)\n\nFigure 12. (a) System hardware diagram; (b) IR Source pulse circuit; (c) Thermopile measurement circuit\n\n4. 实验方法与结果\n\n1) 测试物品：红外镜组焦距19 mm，其光圈f/1.1，光学材料为黑钻石硫属化物玻璃(Black Diamond™ chalcogenide glass)，其光学特性如表2所列。黑钻石硫属化物玻璃如AMTIR-1，IG6，BD2，GASIR-1及GE等其穿透率光谱范围0.7~20.0 μm如图13所示。图14所示为红外线镜组外观图。\n\n2) 精密镜片夹具：要求垂直度与平行度。\n\n3) 红外光学系统穿透率量测系统。\n\n4) 实验方法\n\na. 确认红外光学系统穿透率量测系统在最佳运作状况。\n\nb. Alignment红外准直光束与Thermopile在同一水平面及同一光轴，使Thermopile输出信号最大。\n\nc. 将红外镜组置于红外光学系统穿透率量测系统装置中。\n\nd. 调整红外镜组垂直于远红外准直光束，平移Thermopile，使Thermopile接受到穿透信号最大。确认Thermopile接受到最大信号之后，进行穿透率量测。\n\ne. 改变红外光源强弱，量取不同穿透信号。\n\nf. 进行红外镜组穿透率分析。", null, "Figure 13. Transmittance spectrum of the chalcogenide glass", null, "Table 2. Optical characteristics of infrared lens", null, "Figure 14. Infrared lens\n\n5) 红外镜组穿透率(T)的量测结果", null, "Table 3. Test data of transmittance", null, "Figure 15. Statistical curve of test data\n\n5. 结论", null, "Table 4. Comparison of this method with FTIR spectrometer detection\n\n Hecht, E. (2016) Optics. 5th Edition, Addison-Wesley Longman, Inc., Reading. Smith, W.J. (1992) Modern Lens Design. McGraw Hill, New York. Index of /wp-content/uploads/Linear/IRSources. https://www.boselec.com/wp-content/uploads/Linear/IRSources PerkinElmer Optoelectronics, Inc. (2003) “TPS 334—Thermopile Detector” Datasheet. Wolfe, W.L., Kruse, P.W. and Bass, I.M. (1995) “Thermal detectors” in Handbook of Optics. McGraw-Hill, New York. https://www.hamamatsu.com/resources/pdf/ssd/e07_handbook_Thermal_detectors, Chp. 7, pp. 3-7. Nam, M., Washer, J. and Oh, J. (2015) Breaking the Mold: Overcoming Manufacturing Challenges of Chal-cogenide Glass Optics. https://www.photonics.com/Articles/Breaking_the_Mold_Overcoming_Manufacturing/a57309" ]	[ null, "https://html.hanspub.org/file/3-2140218x13_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x17_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x18_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x20_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x19_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x21_hanspub.png", null, "https://www.hanspub.org/journal/Images/Table_Tmp.jpg", null, "https://html.hanspub.org/file/3-2140218x22_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x23_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x24_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x25_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x26_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x28_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x29_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x31_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x30_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x32_hanspub.png", null, "https://html.hanspub.org/file/3-2140218x33_hanspub.png", null, "https://www.hanspub.org/journal/Images/Table_Tmp.jpg", null, "https://html.hanspub.org/file/3-2140218x34_hanspub.png", null, "https://www.hanspub.org/journal/Images/Table_Tmp.jpg", null, "https://html.hanspub.org/file/3-2140218x35_hanspub.png", null, "https://www.hanspub.org/journal/Images/Table_Tmp.jpg", null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.86317796,"math_prob":0.98484874,"size":8041,"snap":"2020-34-2020-40","text_gpt3_token_len":7211,"char_repetition_ratio":0.081622496,"word_repetition_ratio":0.0,"special_character_ratio":0.19450317,"punctuation_ratio":0.068206824,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.98686016,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,null,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,null,null,3,null,null,null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-26T09:30:09Z\",\"WARC-Record-ID\":\"<urn:uuid:225cc1cf-93b2-4722-8ae7-58fda8975335>\",\"Content-Length\":\"98488\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9155f2aa-cf45-4da3-b5a0-4093ca0f8109>\",\"WARC-Concurrent-To\":\"<urn:uuid:5d385479-caca-4084-8d76-33b14cf24528>\",\"WARC-IP-Address\":\"128.14.148.27\",\"WARC-Target-URI\":\"https://www.hanspub.org/journal/PaperInformation.aspx?PaperID=34651\",\"WARC-Payload-Digest\":\"sha1:DKV7YBXEEQ6BGML6PJTFF2SLHU4BWQDK\",\"WARC-Block-Digest\":\"sha1:7R4P7GWHCYA5GNUTMDXQKKWUMXYXIMJP\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600400238038.76_warc_CC-MAIN-20200926071311-20200926101311-00715.warc.gz\"}"}
http://www.abuildersengineer.com/2014/07/	[ "## Wednesday, July 16, 2014\n\n### SHEAR STRESS IN STEEL BEAM\n\nThis beam, supporting a column point load of 96 k over a door, is a composite beam consisting of a wide-flange base beam with 8x½ in plates welded to top and bottom flanges. The beam is analyzed with and without plates. As shown before, for steel beams shear stress is assumed to be resisted by the web only, computed as fv = V/Av. The base beam is a W10x49 [10 in (254 mm) nominal depth, 49 lbs/ft (6.77 kg/m) DL] with a moment of inertia Ixx= 272 in^4 (11322 cm^4). Shear in the welds connecting the plates to the beam is found using the shear flow formula q = VQ/(I).\n\n1 Beam of L= 6 ft (1.83 m) span with P = 96 k point load\n2 Composite wide-flange beam W10x49 with 8x½ inch stiffener plates\n\nShear force V = P/2 = 96/2 V = 48 k\nBending moment M = 48(3) M = 144 k’\n\nWide-flange beam\n\nSince the beam would fail in bending, a composite beam is used.\n\nComposite beam\n\nSince the shear force remains unchanged, the web shear stress is still ok.\n\nShear flow q in welded plate connection\n\nSince there are two welds, each resists half the total shear flow\n\nNote: in this steel beam, bending is stress is more critical than shear stress; this is typical for steel beams, except very short ones.\n\n## Monday, July 7, 2014\n\n### SHEAR STRESS IN WOOD I-BEAM\n\nSince this is not a rectangular beam, shear stress must be computed by the general shear formula. The maximum shear stress at the neutral axis as well as shear stress at the intersection between flange and web (shear plane As) will be computed. The latter gives the shear stress in the glued connection. To compare shear- and bending stress the latter is also computed.\n\nBeam of L= 10 ft length, with uniform load w= 280plf (W = 2800 lbs)\nCross-section of wood I-beam\n\nFor the formula v= VQ/(Ib) we must find the moment of inertia of the entire cross-section. We could use the parallel axis theorem of Appendix A. However, due to symmetry, a simplified formula is possible, finding the moment of inertia for the overall dimensions as rectangular beam minus that for two rectangles on both sides of the web.\n\nNote c= 10/2 = 5 (half the beam depth due to symmetry)\n\nStatic moment Q of flange about the neutral axis:\n\nShear stress at flange/web intersection:\n\nStatic moment Q of flange plus upper half of web about the neutral axis\n\nMaximum shear stress at neutral axis:\n\nNote: Maximum shear stress reaches almost the allowable stress limit, but bending stress is well below allowable bending stress because the beam is very short. We can try at what span the beam approaches allowable stress, assuming L= 30 ft, using the same total load W = 2800 lbs to keep shear stress constant:\n\nAt 30 ft span bending stress is just over the allowable stress of 1450 psi. This shows that in short beams shear governs, but in long beams bending or deflection governs.\n\n## Wednesday, July 2, 2014\n\n### SHEAR STRESS IN WOOD AND STEEL BEAMS\n\nBased on the forgoing general derivation of shear stress, the formulas for shear stress in rectangular wood beams and flanged steel beams is derived here. The maximum stress in those beams is customarily defined as fv instead of v in the general shear formula.\n\nShear at neutral axis of rectangular beam (maximum stress),\nNote: this is the same formula derived for maximum shear stress before\n\nShear stress at the bottom of rectangular beam. Note that y= 0 since the centroid of the area above the shear plane (bottom) coincides with the neutral axis of the entire section. Thus Q= Ay = (bd/2) 0 = 0, hence\n\nv = V 0/(I b) = 0 = fv, thus\nfv = 0\n\nNote: this confirms an intuitive interpretation that suggests zero stress since no fibers below the beam could resist shear\n\n3 Shear stress at top of rectangular beam. Note A = 0b = 0 since the depth of the shear area above the top of the beam is zero. Thus\n\nQ = Ay = 0 d/2 = 0, hence v = V 0/(I b) = 0 = fv, thus\nfv = 0\n\nNote: this, too, confirms an intuitive interpretation that suggests zero stress since no fibers above the beam top could resist shear.\n\n4\nShear stress distribution over a rectangular section is parabolic as implied by the formula Q=b(d^2)/8 derived above.\n\n5 Shear stress in a steel beam is minimal in the flanges and parabolic over the web.\n\nThe formula v = VQ/(I b) results in a small stress in the flanges since the width b of flanges is much greater than the web thickness. However, for convenience, shear stress in steel beams is computed as “average” by the simplified formula:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92495054,"math_prob":0.9795067,"size":1544,"snap":"2021-31-2021-39","text_gpt3_token_len":380,"char_repetition_ratio":0.15194805,"word_repetition_ratio":0.13937283,"special_character_ratio":0.24481866,"punctuation_ratio":0.077922076,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9955832,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-18T00:31:15Z\",\"WARC-Record-ID\":\"<urn:uuid:7f46f6df-fa75-45e5-a4b7-420ecbd2a9a9>\",\"Content-Length\":\"90582\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6b548d3a-36e0-4888-871f-ac210b09d9ee>\",\"WARC-Concurrent-To\":\"<urn:uuid:db4332de-488c-4dea-8589-f1c8ac520354>\",\"WARC-IP-Address\":\"142.251.33.211\",\"WARC-Target-URI\":\"http://www.abuildersengineer.com/2014/07/\",\"WARC-Payload-Digest\":\"sha1:ZXWMLA2B5HIXEW5JU7MJONBLFIJOKGTA\",\"WARC-Block-Digest\":\"sha1:E4CZMEPTNXIMY3OROVRQ3CCGKSN4HQGD\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780056120.36_warc_CC-MAIN-20210918002951-20210918032951-00086.warc.gz\"}"}
https://mathematica.stackexchange.com/questions/105079/plotting-eigenvalues-with-devoted-level-coloring	[ "# plotting eigenvalues with devoted level coloring\n\nI am about to plot eigenvalues of a matrix versus varyingj's.\n\nHma[j_Integer] = {{0.1j^2, 0, 0, 0.2j, 0}, {0, 0.1j, 0, 0.2j^3,\n0}, {0, 0.1j, 0, 0, 0.2j}, {0.1j^2, 0, 0, 0, 0.2j}, {0, 0,\n0.1j, 0.2j, 0}};\nListPlot[data]", null, "I am trying to have a plot with specified color as I manually showed above. (higher eigenvalues (first ones) with a same color (for example Red above), seconds with same color (and different from the first ones for example Yellow))\n\nFor the data in the question, the following also works.\n\nListPlot[Transpose[Split[data, First[#1] === First[#2] &]],\nPlotStyle -> {Blue, Green, Purple, Orange, Red}]", null, "This, of course, assumes that data already is ordered by size. If not, use\n\nListPlot[SortBy[Transpose[Split[data, First[#1] === First[#2] &]], Last],\nPlotStyle -> {Blue, Green, Purple, Orange, Red}]\n\n\nAnd, if larger markers (or other such presentation changes) are desired, use\n\nListPlot[SortBy[Transpose[Split[data, First[#1] === First[#2] &]], Last],\nPlotStyle -> ({#, PointSize[Large]} & /@ {Blue, Green, Purple, Orange, Red})]", null, "Note that the Green points are scarcely visible, because they approximately coincide with the Purple points.\n\n n=5;", null, "" ]	[ null, "https://i.stack.imgur.com/e0KzR.png", null, "https://i.stack.imgur.com/zyRYF.png", null, "https://i.stack.imgur.com/xZYf0.png", null, "https://i.stack.imgur.com/XpwZf.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6372406,"math_prob":0.9678874,"size":528,"snap":"2023-14-2023-23","text_gpt3_token_len":212,"char_repetition_ratio":0.15839694,"word_repetition_ratio":0.0,"special_character_ratio":0.42045453,"punctuation_ratio":0.27044025,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97639245,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-05-28T12:27:05Z\",\"WARC-Record-ID\":\"<urn:uuid:9fc9177f-283e-4a6a-9ef4-9244c7581222>\",\"Content-Length\":\"160060\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d001ffba-2c57-4719-baf4-be4a10119b42>\",\"WARC-Concurrent-To\":\"<urn:uuid:e6001260-2b00-4d80-8499-50d984790bf9>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/105079/plotting-eigenvalues-with-devoted-level-coloring\",\"WARC-Payload-Digest\":\"sha1:HGUCASAF47XFZTG5EZFTPNBKVOAHNEHM\",\"WARC-Block-Digest\":\"sha1:LIYDTZ23S3VSUPB4D3VWGEZDPKBVSI2A\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224643784.62_warc_CC-MAIN-20230528114832-20230528144832-00332.warc.gz\"}"}
https://m.wxwm.cc/nansheng/shanggan/	[ "• 琴弦低声唱离别。\n\n• 再美的曾经、我只能回忆\n\n• 硪卟數蒵丗庎\n\n• _⺌詮劇終.﹎＂\n\n• 风゛化了我们的回忆\n\n• 只剩下背影\n\n• 柠檬太酸i\n\n• 她求我滚i\n\n• 无伴人\n\n• ξ 回不去的苍白\n\n• 一个人高傲的独活\n\n• 无言以对 ζ\n\n• 若终要失去、不如放开\n\n• 惹尘埃是非\n\n• 花醉花思泪\n\n• 胭脂　※\n\n• 凄惨的回忆＂\n\n• 呼吸时，心都会痛。\n\n• 原来爱情这么难\n\n• 深夜在被窝里偷哭\n\n• 你凭什么不打扰我\n\n• 任这空虚沸腾\n\n• 锁心，锁\n\n• 筋疲力尽的追逐，只是徒劳\n\n• 爱人别走\n\n• 输不起的感情\n\n• 空寂\n\n• oo-┈→凉心℡~\n\n• 都叫兽别走i\n\n• 再见、亲爱的\n\n• 承载着曾经。\n\n• ┗從未拥冇、談何矢佉┓\n\n• ﹎ゞ没有信任的爱情\u0007/╱°\n\n• 夜盲症\n\n• 壹朵薔薇花\n\n• 苦撑\n\n• 划伤了心_ジ\n\n• 选择放手\n\n• 听海的哭泣\n\n• ╰天使的伤\n\n• 冷态度i\n\n• 孤独是世界绝症！\n\n• 傷、待續\n\n• 习惯了∥与影子的对话﹌\n\n• 笑着、说再见\n\n• 弦断心凉繁华逝\n\n• 独活\n\n• 费劲心思爱过\n\n• 没有未来。\n\n• ︶ㄣ x1n痛の兹味\n\n• ~隐隐的伤感 /\n\n• 情到深处竟是离别i\n\n• 不狸不气的。爱你\n\n• ﹌你给的诺言、都记得″\n\n• 婲開、終須落\n\n• 有一种想念叫避而不见@\n\n• 是你让我学会熬、\n\n• 心碎见蓝天\n\n• 那悲傷、凋落\n\n• 〆雨滴落丶浅似留殇\n\n• 天堂の劇終℡\n\n• 繁华里的沧桑ぢ\n\n• ζ眼泪在失落\n\n• 带泪怎笑.\n\n• 笑着流泪、我学不会じ\n\n• 最终的最终、终究是戏\n\n• 我没变、只是你厌倦了\n\n• 欲言又止的难受\n\n• 残阳照、墨痕\n\n• 看见你心都碎了\n\n• 在給我Ⅰ丶勇气\n\n• 你萌化了我的心i\n\n• 一切只是我希望i\n\n• 想念、是会【致命】的痛\n\n• [ 不如初 ]\n\n• 强势控局\n\n• 读书凶铃！\n\n• 別ㄋ、那纏綿\n\n• ▲黑色瞳孔丶满带伤痛\n\n• 〆青春献给了课桌\n\n• 一次性抹掉春冬\n\n• 。- ‘梦毁千百次’.\n\n• 回忆像章鱼扑上来\n\n• 眼泪替我卸妆\n\n• 等你归\n\n• 转圈圈\n\n• 笑出的泪 =\n\n• 心酸不已。\n\n• 纠结、回忆的痛\n\n• 一个人的承诺\n\n• 宁做不懂是是非非的孩子\n\n• 深秋、知涼\n\n• 我深知我不配\n\n• 一秒的安慰、\n\n• 他說的愛過是錯過…\n\n• 々被你所伤\n\n• 心里住了个不可能的人i\n\n• 花露水。防蚊の\n\n• 红丶╮A\n\n• 容我爱你可好\n\n• 说㈠半旳爱\n\n• 执手曲已凉i\n\n• 无法诠释的永恒\n\n• 回来好么\n\n• 相爱太早，不能终老。\n\n• 伱给的爱太牵强\n\n• 伴君悲歌\n\n• 孤独解酒\n\n• ╯遗失的繁华\n\n• シ独守空城づ\n\n• 血溅了白纱。\n\n• 一曲断肠i\n\n• The city of the sky\n\n• 你的爱、太虚伪。\n\n• 莪的悲伤仍在继续\n\n• 旧心酸。\n\n• ︶ㄣ鈊痛\|戀落泪\n\n• 没有你\n\n• 一个人的浪漫/\n\n• 疯狂的丶丿石头\n\n• 和寂寞有染 ∥\n\n• 他故事中的我、\n\n• 何时秋风悲画扇\n\n• 糖果控、\n\n• ︶心痛的玩笑\n\n• 本是古典何须时尚\n\n• 你名毁她命\n\n• 回不到从前@\n\n• 时间能让我忘了你吗\n\n• 美人性情\n\n• 悲痛\n\n• 微笑要带着眼泪才耐看づ\n\n• 多余的⌒温柔\n\n• 爲何離開↘莪\n\n• 想哭无泪\n\n• 认命扮矮人\n\n• 沧桑的岁月浮华了谁\n\n• 逼着放弃最痛苦\n\n• 那时年少，\n\n• 柠檬落泪都是酸;\n\n• 爱情的旋律\n\n• 刺心谎言,\n\n• 真相背后的残忍\n\n• 爱上夜的黑、迷失了我的爱\n\n• 心痛像条线\n\n• ＊﹏無心不心痛〆\n\n• 凄寒半世留殇铭\n\n• 荒废过的心存有你\n\n• 我忘不掉你！\n\n• 有些歌，听得沉醉了\n\n• ㄨ_花叹、\n\n• 情话再多、只是一种敷衍\n\n• 作业变成蝴蝶飞走了@\n\n• 爱情已剧终\n\n• 我与影子孤独终老\n\n• 你不过是个过客而已@\n\n• ち潮货↙总裁\n\n• ——摇摆世界\n\n• 安徒生的谎言、\n\n• 离你我心不碎\n\n• 矫情不是我的范#\n\n• 不爱，滚i\n\n• 放肆演绎青春i\n\n• 吃货总比痴货好!\n\n• 骑着蜗牛，拽天下ㄟ\n\n• 爱的分界线。\n\n• 回忆太细\n\n• 初衷的味道\n\n• 拥抱漏风\n\n• 思念埋在发间\n\n• 半分笑——〆\n\n• 一万种快乐\n\n• 梦好沉\n\n• 几度温柔,\n\n• 笑脸是悲伤的面纱.\n\n• 厌情症°\n\n• 醋，是一种爱i\n\n• 墨香染城\n\n• 玉砌似画染\n\n• 旧夏浅入梦°\n\n• 想赢就别喊停。\n\n• 对伱微笑╮纯属伱很可笑\n\n• 旧街凉风°\n\n• 撑起一片天\n\n• 倾听心跳de旋律\n\n• 若爱、请深爱。\n\n• 时间把苦熬成甜\n\n• 不合脚的鞋再美也不要\n\n• 阳光总在风雨后\n\n• 我若不努力，谁给我未来\n\n• 安于得失、淡于成败\n\n• 曾经的你早已不复存在\n\n• 深情在睫,孤意在眉\n\n• 浅嗄ゞ写卟完的温柔\n\n• 关于爱情只字不提\n\n• 江南烟雨〆相思醉つ\n\n• 花静幽然\n\n• 许你三千笔墨画我绝世倾城\n\n• 我像谁°\n\n• 多情人\n\n• 海屿荒\n\n• 后来呢\n\n• 顾与我i\n\n• 攻你心i\n\n• 陪你闹i\n\n• 久伴我i\n\n• 我会暖i\n\n• 一抹高傲旳笑。\n\n• 走开、暧昧\n\n• 钻石般耀眼\n\n• 完美洒脱\n\n• 膜拜的对象≈\n\n• 快乐是选择\n\n• 高调。转身\n\n• 不期待.再重来\n\n• 无所谓的魅惑\n\n• 释放シ生命的光华\n\n• 兄弟是拿心来对待的\n\n• 脉搏跳动着高傲ヅ\n\n• 非买品ю\n\n• 演绎つ只属于我们的未来\n\n• 尘缘而已。\n\n• 分了、我来合。\n\n• 止步爱情°\n\n• ——活出自己的精彩.℡\n\n• 美男子\n\n• 微笑面对未来\n\n• 光彩夺目的高傲\n\n• 耀眼的光芒\n\n• 晋豫达人\n\n• 拒绝翻版.\n\n• 丶限量版帅仔\n\n• 爱就一辈子\n\n• 漫天游ゝ\n\n• 亲吻ωǒ资本\n\n• 继续丿辉煌、\n\n• 〞将军っ\n\n• 淡淡的优雅\n\n• ∝耀出彩\n\n• ╰ 笑看人生\n\n• 你好优秀!\n\n• 炫丽德青春、\n\n• 活出高姿态、\n\n• 炫彩死神\n\n• 、信者得爱\n\n• 玉衡逍龙.+\n\n• 青春爱无赦 \n\n• ワ逆转结局\n\n• 梦的怒放╮\n\n• 永不退缩\n\n• 活着便精采°\n\n• 受大气，成大器i\n\n• 男心凉\n\n• ╰︶男爵\n\n• 那个位置，叫做刻骨铭心\n\n• 时间把苦熬成甜\n\n• 别吵赢道理却输了感情\n\n• 给时间时间，让过去过去\n\n• 想赢就别喊停\n\n• 梦想不只是梦与想\n\n• 笑脸赢人脉\n\n• 一边得到一边失去\n\n• 我若不努力，谁给我未来\n\n• 安于得失、淡于成败\n\n• 把每天当末日来过\n\n• 柠檬不该羡慕西瓜的甜\n\n• 你若盛开、蝴蝶自来\n\n• 不要活给别人看\n\n• 若无过客丶何来人生\n\n• 每一天都是限量版\n\n• 挫折是化了妆的祝福\n\n• 成熟不在于年龄而在于经历\n\n• 得之坦然、失之淡然\n\n• 心是晴朗的人生就没有雨天\n\n• 不努力，拿什么说明天\n\n• 密码里隐藏着秘密\n\n• 珍惜〆眼前的幸福╮\n\n• 至少我还有梦\n\n• 欲戴王冠必承其重\n\n• 不忘初心方得终始\n\n• 孤独是成长的必修课\n\n• 今天不走明天要跑\n\n• 因为知足，所以幸福\n\n• 念你情深\n\n• 人生如梦、梦如烟\n\n• 相爱就爱，别怕阻碍\n\n• 北岛旅客\n\n• 一夕意相左\n\n• 旧光影里的格子少年\n\n• 陪笑\n\n• 跪下喊爷!\n\n• 吻妳是不想失去妳\n\n• 一步踏尽一树白\n\n• 你知道我长短\n\n• 执念于你\n\n• 赠你深爱与久伴\n\n• 醉挽清风丶已成空\n\n• 念念相忘\n\n• 心不大只有你\n\n• 纤纤公子\n\n• 独手毁天下\n\n• 年轻不拽世界何以精彩\n\n• 风雨都扛过\n\n• 劳资最烦起名了\n\n• 酷似你爹\n\n• 命里不缺狗，你要走就走\n\n• 星战@\n\n• 将神@\n\n• 龙将@\n\n• 邪恶%\n\n• 新仙剑\n\n• 大将军\n\n• 夜店之王\n\n• 斗破乾坤\n\n• 坦克世界\n\n• 修真世界\n\n• 傲视天地\n\n• 光辉终结\n\n• 现代战争\n\n• 卡通农场\n\n• 公子泺尘\n\n• 神将三国\n\n• 王者天下\n\n• 橘萝卜蹲@\n\n• 苍空的蓝耀\n\n• 旋风小陀螺\n\n• 秋风追猎者\n\n• 迷你大城市\n\n• 噬血魅影@\n\n• 暴君与猫@\n\n• 战神将军@\n\n• 柔媚——妖瞳\n\n• 夏眠i\n\n• 乱斗堂\n\n• 弑鬼神\n\n• 战无极\n\n• 天下白\n\n• 苏伯陵\n\n• 安如山\n\n• 刺情°\n\n• 微生墨\n\n• 霸剑皇侯\n\n• 似梦似幻\n\n• 很傻、很固执\n\n• 卖萌的小行家\n\n• 阳光男孩\n\n• yí銗、镎命嫒ぐ\n\n• 劳资独宠一方\n\n• 。赫颜\n\n• 酷似你爹\n\n• 渣爷!\n\n• 小屁孩先森^\n\n• ￡霸王↘爷战神三\n\n• 爷ワ゛给钮淡定\n\n• 行星美男\n\n• 謌践踏爱綪\n\n• 欠你的幸福\n\n• 三好不良少年°\n\n• 竹·小马‖儒\n\n• 无处话凄凉\n\n• 滁ㄋ故亊涐一兂所冇\n\n• 烈酒配孤独\n\n• 渺远的断云□\n\n• ’楠軕蒗児ヽ\n\n• 又被自己帅到了\n\n• 荒世独拥\n\n• 任性男神\n\n• 苍苍迟暮\n\n• 静守己心，看淡浮华\n\n• 胖次超人\n\n• 跑到坟场去吓鬼\n\n• 久挽不留\n\n• 劳资独宠一方\n\n• 荒笹独拥\n\n• 怨风离愁人\n\n• 孤骨\n\n• 濄徃谩谩说\n\n• 承诺早已泛黄\n\n• 廻忆里菂馀温\n\n• 风月瘦如刀，催人老\n\n• 无处寄笺\n\n• 情话至深\n\n• 旅人泪\n\n• 一句爱的晚安语\n\n• ←狱邪魂断ぐ\n\n• 竖起中指操起生活\n\n• 阳光宅男\n\n• ——鲭风痴鵟/／\n\n• 无能\n\n• 时间煮雨是他的倾诉\n\n• 清尊素影\n\n• 长愿相随\n\n• 再闹硪憱垱众亲祢\n\n• 拿命珍惜她\n\n• 为你我受冷风吹\n\n• 孤独是我妻\n\n• 谁会发光我就灭谁的光\n\n• 我曾经掏空了心\n\n• 人前给面子，人后扇耳光\n\n• 猎鹿人\n\n• 风度翩翩°\n\n• 龙爷无敌°\n\n• 那一季、那点伤\n\n• Mr°骚年ぃ\n\n• 陷阱里的王子\n\n• い第一先生が\n\n• 跳梁小丑\n\n• 蜀国犀利哥\n\n• 老师、她亲硪\n\n• 怪咖男渣!\n\n• 岁月枯荣\n\n• ╰嘞蚗儅哖\n\n• 未来家庭主男。\n\n• 挑眉少年°\n\n• 英雄♀奶油♂\n\n• ＼親愛dé咾婆ゞ \|\n\n• 坏男孩\n\n• 苏三大爷\n\n• 喜旧之人——\n\n• 爵士少年.﹌‖\n\n• 上学穿拖鞋是汉子@\n\n• 餦街舊魜灬\n\n• 丿灬哥的范儿丶\n\n• 小小少爷i\n\n• 少年未老心已凉√\n\n• 亦枫\n\n• 孤男寡旅\n\n• 辣条先森i\n\n• 老纳曾是一朵花i\n\n• 哥贱到你烦\n\n• 零帝\n\n• 胡渣硬男。\n\n• 指尖星光在流浪\n\n• 仮蒎の尐夥姅\n\n• 萌漢紙\n\n• 仯年丕變杺ヽ\n\n• 瑷俄dê請呼χí←\n\n• 做你的男人。\n\n• 小萌男！\n\n• 学弟°\n\n• 我是Bb哥\n\n• 傲视群雄\n\n• 暮然回首脖子疼\n\n• ヤrú菓méi姷ài簻\n\n• ←栢褩缃厭oО\n\n• 沬粜遈沬箌粜\n\n• 璐蒤鎐遠、匢朢初吢\n\n• tū嘫鋒悧の徊yì…\n\n• 风扑进怀\n\n• 旧事甚歉\n\n• 无尽空虚\n\n• 贱是一种态度\n\n• 單裑等鱫\n\n• ⺷壹筆縴連ず\n\n• zhí戳xīn扉/／\n\n• 爱人想离我怎挽\n\n• 剑起苍穹云归处\n\n• ╰嘞蚗儅哖\n\n• ◇喰種，蠶飠吣靈\n\n• 殺手\n\n• ﹎.茵为她值得⺌\n\n• 酒醉夜未阑\n\n• 痴心错付\n\n• ﹏風流∨成癮Ο\n\n• ╰→嗳棏呔遲ㄞ\n\n• 练习拥抱\n\n• 泡面君ˇ\n\n• 两包辣条约吗\n\n• 独酌陈酿\n\n• 各走各路〃\n\n• 也许，我只是一个过客\n\n• 你是我在青春里辗转的歌\n\n• 三千青丝散\n\n• 只為ㄋ遇見你\n\n• 活着别让心遭罪\n\n• 丢了薇笑ヽ嗱什麼伪装\n\n• 幸福加载中…\n\n• 总有一个人先走\n\n• 说爱太烫嘴ァ\n\n• 想你的时候*落泪忧伤\n\n• 心动心痛丶\n\n• 冗长梦毁\n\n• 你可知道我的梦i\n\n• 好了伤疤忘了痛\n\n• ″ 葬你身边\n\n• 比纸薄的情ゝ\n\n• 情缘隔世\n\n• 私の心は撩乱\n\n• 还在等\n\n• 一句分手，都是泡沫\n\n• 人心真脏@\n\n• 忧伤倒数≈\n\n• - 對ai情禁言。、\n\n• 哭到失控\n\n• 爱是泡沫。\n\n• 死撑着装坚强\n\n• 男神离我好远@\n\n• 柠檬太酸i\n\n• 海伤°\n\n• 执著心却凉ι\n\n• 离北°\n\n• 那一抹忧伤。\n\n• 哪有胖子不忧伤丶\n\n• 该笑吗.\n\n• 远离碍i\n\n• 够狠心i\n\n• 知心碍人i\n\n• 放不下你是我活该\n\n• 太揪心\n\n• 他心无我！\n\n• 一千零一句谎言\n\n• 怪我冷i\n\n• 我该拿什么换回你，\n\n• 爱情谷底\n\n• 他不会回来了。\n\n• 我已开始怀念\n\n• 今年夏天我们就散了。\n\n• 我算什么@\n\n• 童话破灭。\n\n• 一个人失忆\n\n• 我本冷情怎暖你心i\n\n• 救赎@\n\n• 抱歉我无梦赠你\n\n• 灰白色充斥记忆i\n\n• 想说的话压在心底——.\n\n• 〆﹏落日般的忧伤\n\n• 追爱成忆i\n\n• 太熟悉了会陌生°\n\n• 郁闷中……\n\n• ℡ 晴天有点孤单、\n\n• 寂寞的心。\n\n• 半城繁華半城傷づ\n\n• 一个人的孤单情歌。\n\n• 离心愁i\n\n• 感情已褪色\n\n• 殇城似觞\n\n• 回忆那么疼！\n\n• 时间把回忆埋了\n\n• 回忆再美、也带着伤\n\n• 梦醒人走茶凉↘\n\n• 痛而不言i\n\n• 会好的i\n\n• 人心不可猜透@\n\n• 眼中的忧伤如同仓皇的落日\n\n• 微笑、掩盖悲伤\n\n• 眼里的寂寞。\n\n• 梦散了i\n\n• 用微笑掩盖痛\n\n• 三千弱水化作眼泪！\n\n• 抓不紧 i\n\n• 我会怕i\n\n• 左眼会陪右眼哭i\n\n• 殘留的温柔@\n\n• 分手理由ヾ\n\n• 心瞎,\n\n• 孤伴i\n\n• 盐不及泪咸 i\n\n• 不曾拥有你i\n\n• 低凋、进行曲\n\n• 笑、淡化了那伤\n\n• 我年轻丶心不定\n\n• の),茈娚孓\n\n• ㄗs銱児啷当\n\n• 风蓅Иé尙\n\n• /.赱私娚孒o.\n\n• メ霸气爷ャ\n\n• 尐样贼≯坏\n\n• →伤你无鈊^-\n\n• 對伱啲依賴\n\n• 为幸福向前冲\n\n• 丶柔情小男人\n\n• 温习你的温柔．\n\n• 連椄 ф簖╮\n\n• 情在⒈夜之间﹏\n\n• 霸道D温柔\n\n• 心照不宣\n\n• 贩卖幸福。\n\n• 丢不掉的思念\n\n• 不稀罕你的爱\n\n• 念念不忘︶——\n\n• 无力的，微笑\n\n• 哥、寂寞成伤\n\n• 無法回頭、\n\n• 午夜，醉情歌\n\n• 斩碎星空\n\n• 單身リ天涯\n\n• 一切从头。\n\n• ╭冷眼旁观\n\n• 坦白\n\n• 怡然自得。\n\n• 哼着情歌、享受寂寞\n\n• 无声无息。安之若素╮\n\n• 站在高岗\n\n• 暴雨的洗礼\n\n• 你只是道听途说つ\n\n• 做人要有风范\n\n• 〆岁月成『风』\n\n• ￠无可替代的友谊﹌\n\n• 我们性格不合\n\n• 用眼角去看中国人" ]	[ null ]	{"ft_lang_label":"__label__zh","ft_lang_prob":0.6767719,"math_prob":0.570266,"size":4355,"snap":"2022-27-2022-33","text_gpt3_token_len":6188,"char_repetition_ratio":0.01861641,"word_repetition_ratio":0.0,"special_character_ratio":0.21905856,"punctuation_ratio":0.035952065,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9884763,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-16T09:38:40Z\",\"WARC-Record-ID\":\"<urn:uuid:03c9c497-257e-431c-8976-56f84b7157ab>\",\"Content-Length\":\"25350\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:44a52d21-9838-404a-96e2-24604a9ed867>\",\"WARC-Concurrent-To\":\"<urn:uuid:b11790a9-1ce7-4004-888b-807b04b1e6be>\",\"WARC-IP-Address\":\"172.67.175.8\",\"WARC-Target-URI\":\"https://m.wxwm.cc/nansheng/shanggan/\",\"WARC-Payload-Digest\":\"sha1:6EIAM6MANT2QMX4STMUCYEM4AXIPXAX2\",\"WARC-Block-Digest\":\"sha1:FTZIWTZXQWPQ7OE36PIZTCOSPZVSQX4W\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572286.44_warc_CC-MAIN-20220816090541-20220816120541-00040.warc.gz\"}"}
https://www.managementtutors.com/Recent_Question/37331/RESEARCH-SCHOOL-OF-FINANCE-ACTUARIAL-STUDIESAND-STATISTICSINTRODUCTORY	[ "Recent Question/Assignment\n\nRESEARCH SCHOOL OF FINANCE, ACTUARIAL STUDIES\nAND STATISTICS\nINTRODUCTORY MATHEMATICAL STATISTICS PRINCIPLES OF MATHEMATICAL STATISTICS\n(STAT2001/6039)\nAssignment 1 (2018 Semester 1)\nYour solutions to this assignment should be placed in the appropriate box in the RSFAS School foyer by the due time and date (as provided in the Course Outline and on Wattle). Attach a cover sheet (as provided on Wattle) which has your ANU ID number. The assignment is out of 100 and is worth 10% of your overall course mark.\nProblem 1 (Total 20 marks)\nSuppose that P AB( ) = 0.1, P(B ? =C) 1 and P B( - =A) P A( ? =B) 2 (P A- B). Find P AB P AC P ABC P A P( )+ ()- ()- ( )+ ( )B -PC P ABC( )- ( ), or determine the range of possible values for this quantity if it cannot be calculated as a single number.\nProblem 2 (Total 30 marks)\nHomer and Marge play a game by taking turns rolling a standard six-sided die, starting with Homer, until there occurs a sequence of k or more 4s immediately followed by a 5 or 6 (e.g. 45, 245 or 5446, if k = 1). The last person to roll wins the game. (E.g., if k = 2, then each of the sequences 446, 64445 and 465443445 results in Homer winning.) Find the probability that Homer wins, for each k = 0, 1, 2, 3 and 99.\nProblem 3 (Total 50 marks)\nA stack of ten cards has four hearts (which are red), three diamonds (which are red), two spades (which are black) and one club (which is black). Four cards are sampled from the stack, randomly and without replacement, and placed in a box. Then one card is randomly selected from the box. If that card is a heart then all of the other three cards in the box are burnt. Otherwise, two cards are sampled, randomly and without replacement, from the other three cards in the box, and then these two cards are burnt. Find the probability that exactly k red cards are burnt, for each possible value of k.\nPage 1 of 1" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91432256,"math_prob":0.9694733,"size":1897,"snap":"2019-43-2019-47","text_gpt3_token_len":527,"char_repetition_ratio":0.09508716,"word_repetition_ratio":0.011111111,"special_character_ratio":0.2888772,"punctuation_ratio":0.11278196,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9934575,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-22T06:24:55Z\",\"WARC-Record-ID\":\"<urn:uuid:32eb334e-9fb3-44e5-a0af-f4803ccc5439>\",\"Content-Length\":\"28773\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ef281c3a-dd58-42ac-97a8-cb27ece08a08>\",\"WARC-Concurrent-To\":\"<urn:uuid:90f654a9-02c8-442a-a2a2-2c81290cd582>\",\"WARC-IP-Address\":\"35.200.226.70\",\"WARC-Target-URI\":\"https://www.managementtutors.com/Recent_Question/37331/RESEARCH-SCHOOL-OF-FINANCE-ACTUARIAL-STUDIESAND-STATISTICSINTRODUCTORY\",\"WARC-Payload-Digest\":\"sha1:L4HEW7NDEYOGOVUUK6GYMIPXAASZ72VA\",\"WARC-Block-Digest\":\"sha1:JUEATZ5RYNTQSW4RPA4ATL26HUBXVO3Z\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987803441.95_warc_CC-MAIN-20191022053647-20191022081147-00549.warc.gz\"}"}
https://jutge.org/problems/P70756_en	[ "# Partitions P70756\n\nStatement", null, "html\n\nWrite a program that, given n different words s1, …, sn and a number p, prints all the ways to share the words between p subsets.\n\nInput\n\nInput consists of a natural number n > 0, followed by s1, …, sn, followed by a natural number p > 0.\n\nOutput\n\nPrint all the ways to share the words between p subsets. The elements of each set must appear in the same order than in the input. Print an empty line after each partition.\n\nObservation\n\nStrictly speaking, a partition cannot have empty subsets, but we forget about that restriction in this exercise.\n\nYou can print the solutions to this exercise in any order.\n\nPublic test cases\n• Input\n\n```2\nhello bye\n2\n```\n\nOutput\n\n```subset 1: {hello,bye}\nsubset 2: {}\n\nsubset 1: {hello}\nsubset 2: {bye}\n\nsubset 1: {bye}\nsubset 2: {hello}\n\nsubset 1: {}\nsubset 2: {hello,bye}\n\n```\n• Information\nAuthor" ]	[ null, "https://jutge.org/problems/P70756_en/png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8311587,"math_prob":0.62061,"size":765,"snap":"2021-31-2021-39","text_gpt3_token_len":204,"char_repetition_ratio":0.15243101,"word_repetition_ratio":0.08695652,"special_character_ratio":0.2640523,"punctuation_ratio":0.16363636,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97866356,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T11:09:12Z\",\"WARC-Record-ID\":\"<urn:uuid:f937288e-d996-4925-9fed-8dadc2451942>\",\"Content-Length\":\"23268\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:71a88fc2-65e2-48d1-9c8b-8434849270c4>\",\"WARC-Concurrent-To\":\"<urn:uuid:7e51f39a-a769-4106-8b4e-42c37bf06596>\",\"WARC-IP-Address\":\"147.83.29.168\",\"WARC-Target-URI\":\"https://jutge.org/problems/P70756_en\",\"WARC-Payload-Digest\":\"sha1:UNLNU6HGUOUS6AEK3KGJSOAXRVUI6XJS\",\"WARC-Block-Digest\":\"sha1:SULUBOWY4G2ARXZYXSVAWWR4H6YMSDVK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057421.82_warc_CC-MAIN-20210923104706-20210923134706-00352.warc.gz\"}"}
https://www.1onepsilon.com/single-post/2018/04/27/Can-the-truth-be-tabulated	[ "© 2016 by One on Epsilon PTY LTD\n\n# Can the truth be tabulated?\n\nApril 26, 2018\n\nHow often during the day do you think about what is true or what is false? What does it mean for a statement to be true? Are some statements more true than others? Think about the last time someone asked you a question. When you responded, was there a clear distinction between a true answer and a false answer?\n\nIn regular day to day communication, often the issue of true vs. false can become quite a grey area. What about truth in mathematics? In mathematical logic, statements are taken as either true or false. In this post, we consider some examples of circumstances where this thinking can lead to a contradiction. Enter the paradox.\n\nA paradox can mean many things in mathematics. Sometimes it refers to a result that is counter-intuitive or unexpected. Other times it refers to a situation that results in an incorrect conclusion because of a subtle error or incorrect assumption made somewhere along the way.\n\nHere, we would like to explore a well-known paradox that occurs in the context of mathematical logic known as Curry’s Paradox. Before we delve into Curry’s Paradox, let's consider a simpler logical paradox.\n\nConsider the following statement:\n\nThis sentence is false.\n\nThink about this for a moment. If the sentence is actually false, as it claims to be, then the sentence must be true. Alternatively, if the sentence is true, then the sentence must be false! Hang on, we’re going in circles here! This statement is an example of a logical paradox, that is, a statement that is inconsistent with logic. The paradox arises due to the statement referring to the value of its own truth.\n\nIn using the word logic here, we mean that every statement is either true or false, but it cannot be both true and false. Furthermore, if a statement is not true, then it is false, and vice versa.\n\nIn order to carefully examine and understand Curry’s Paradox, let’s first explore compound statements. A compound statement is one that is made up of two or more statements. Here is an example of a compound statement:\n\nIt is raining or the kids are playing outside.\n\nWe can use brackets to separate out the distinct statements.\n\n(It is raining) or (the kids are playing outside).\n\nLet’s set A to represent the statement “it is raining” and B to represent the statement “the kids are playing outside.” Our sentence is then:\n\nA or B.\n\nOne way to analyze a compound statement is by using a truth table. This is a table that keeps a record of all possibilities of the truth-value of each statement. These values are used to determine the value of the compound statement.\n\nThere are two possible values for statement A, true or false. There are two possible values of statement B, true or false. The compound statement, \"A or B\" uses the Boolean operator OR. We can think of it as, \"When A and/or B is true, this statement is true.\" Here is the truth table for \"A or B\":\n\nNote that the compound statement \"A or B\" is only false if A and B are both false.\n\nTwo statements are logically equivalent if their truth tables are the same. It might seem strange at first, but another way of conveying the meaning of that original sentence is the following logical conditional statement:\n\nIf (not A) then B\n\nLet’s look at this a bit more closely. If “not A” (the hypothesis) is true, then the entire statement can only be true if B (the conclusion) is also true.\n\nWe must be very clear to take the statement at face value and not interpret anything further than what is being said. The overall statement makes no claim in the case the premise is false. Look at the actual statement again: “If it is not raining then the kids are playing outside.” No claim is being made about where the kids play if it is raining. So, when it is raining, or \"not A\" is false, the compound statement is not false, and is therefore true.\n\nThe situation represented by these logically equivalent statements can be summarized in the following diagram:\n\nNow we can have fun with Curry’s Paradox!\n\nWe find ourselves in quite a logical mess if we make a statement about that very statement's truth or falsehood in a compound logical statement, especially one with a nonsensical conclusion. Consider:\n\nThis sentence is false or zero is equal to one.\n\nSet C to represent the statement “this sentence is false” and D to represent the statement “zero is equal to one.” The form of the sentence simplifies to:\n\nC or D.\n\nLooking through our mathematical lens, we produce this truth table as before:\n\nHere is where the logical paradox arises. In rows 1 and 2, C is true. However, C is a statement that refers to its own value. When C is true, this corresponds to saying that the statement “this sentence is false” is true, which means the sentence is false. That is inconsistent with the value in the right-most column.\n\nSomething similar occurs in row 4, where C is false. This corresponds to saying that the statement “this sentence is false” is false, which means that the sentence is true. Again, we see in the right-most column, in row 4, that the full sentence is false in this case. So far, it has all been inconsistent.\n\nWhat about row 3? Well, the entries for \"C\" and \"C or D\" seem to be logically consistent. Great! What about the entry in the column for \"D\"? That corresponds to D being true. Unfortunately, that would mean that the statement \"zero is equal to one\" is true. Eeek! This goes against our broader experience and knowledge, and indeed the foundations of mathematics! So everything is a mess: Indeed, it's a paradox!\n\nBy the way, another common presentation of this paradox is the conditional statement:\n\nIf this sentence is true then zero is equal to one.\n\nIf you are interested in looking at other types of mathematical paradoxes, check out this video on the Banach-Tarski paradox by VSauce:\n\nYou can also check out the “proof” that pi = 4 by Vihart:\n\nAlthough neither of these are a logical paradox like Curry's Paradox, they demonstrate that mathematics is more than just a set of rules that we follow to make intuitive deductions. There is so much more richness to mathematics!\n\nDo you have a favourite mathematical paradox? Is it just impossible to decide? Well, the word “impossible” is not in our vocabulary!\n\nSearch Tags:\n\n###### Featured Posts\n\nOur blog posts moved to Epsilon Stream\n\nJanuary 3, 2019\n\n1/1" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9444869,"math_prob":0.9079895,"size":6160,"snap":"2019-51-2020-05","text_gpt3_token_len":1319,"char_repetition_ratio":0.16845354,"word_repetition_ratio":0.022038568,"special_character_ratio":0.21217532,"punctuation_ratio":0.11298077,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9797496,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-27T06:02:38Z\",\"WARC-Record-ID\":\"<urn:uuid:07764474-fa56-4112-a8ef-4dfb0e3a108e>\",\"Content-Length\":\"1051608\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9d1bb6c9-4ef2-4614-84e8-7fc318fcdc09>\",\"WARC-Concurrent-To\":\"<urn:uuid:005c657a-e60f-46ad-9bce-c112a547473a>\",\"WARC-IP-Address\":\"185.230.60.195\",\"WARC-Target-URI\":\"https://www.1onepsilon.com/single-post/2018/04/27/Can-the-truth-be-tabulated\",\"WARC-Payload-Digest\":\"sha1:QLFAO4KQXEVNBT6FIJ25WURQI5WWKMK7\",\"WARC-Block-Digest\":\"sha1:FH3I27KYRP6XBLVHWRGMOHFNWSNBFHXF\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579251694908.82_warc_CC-MAIN-20200127051112-20200127081112-00403.warc.gz\"}"}
https://mathoverflow.net/questions/236353/compressing-a-hypersurface-on-the-sphere	[ "# Compressing a hypersurface on the sphere\n\nLet $M^n$ be a compact, connected, orientable hypersurface of the unit sphere $S^{n+1} \\subset \\mathbb{R}^{n+2}$. Suppose $M$ is contained in the northern hemisphere $S_+^{n+1}$ and has nonzero principal curvatures everywhere, i.e., has nonzero gaussian curvature everywhere. It seems to me that if we compress $M$ somehow in the direction of the north pole, its principal curvatures will, in absolute value, get arbitrarily large.\n\nIs it true? If so, is there a way to show it without explicitly exhibiting a map $S_+^{n+1} \\to S_+^{n+1}$ that does the job? Instead of compressing $M$ towards the north pole, let us think more generally; is it true that there exists a diffeomorphic copy of $M$ in $S_+^{n+1}$ having prinicipal curvatures say, in absolute value bigger than 1?\n\nThanks for your help and thoughts!\n\n• You can write out the Ptolemaic coordinates on the sphere with an explicit expression for the Riemannian metric, so that dilation in those coordinates gives a very explicit compression toward the north pole. You can see explicitly an orthonormal frame and how it transforms in those coordinates, I think, so you should be able to see how shape operators transform. – Ben McKay Apr 15 '16 at 20:48\n• @BenMcKay What are Ptolemaic coordinates? – Eduardo Longa Apr 15 '16 at 23:25\n• I am not sure which one Ben McKay means specifically, but you can do his construction in either stereographic projection or orthographic projection. – Willie Wong Apr 29 '16 at 19:06" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90389234,"math_prob":0.96533936,"size":811,"snap":"2020-10-2020-16","text_gpt3_token_len":222,"char_repetition_ratio":0.11524164,"word_repetition_ratio":0.0,"special_character_ratio":0.26017264,"punctuation_ratio":0.12195122,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9818594,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-04T22:37:07Z\",\"WARC-Record-ID\":\"<urn:uuid:73f90098-a829-4a0c-806c-81c9ea8769e7>\",\"Content-Length\":\"116617\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:28399d38-f35f-474b-8821-7e529d3924bd>\",\"WARC-Concurrent-To\":\"<urn:uuid:9daa10d8-2f5c-41bf-822f-0e5f75b3d6b0>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/236353/compressing-a-hypersurface-on-the-sphere\",\"WARC-Payload-Digest\":\"sha1:3IJLHUN5X4ATSGY27J6IFISIGJBMSRSP\",\"WARC-Block-Digest\":\"sha1:ZJ3F7CK7NW4ILH2N7A37WETSAMJPZKMF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585370525223.55_warc_CC-MAIN-20200404200523-20200404230523-00319.warc.gz\"}"}
https://www.splashlearn.com/s/math-worksheets/add-multiples-of-10-and-100-vertical-addition	[ "Home > Math > Add Multiples of 10 and 100: Vertical Addition Worksheet\n\n## Reinforce math concepts by practicing to add multiples of 10 and 100.", null, "Adding a number to 0 results in the number itself. Students love to apply this concept when they add numbers ending in zeros. This worksheet strengthens their accuracy with a set of problems on add multiples of 10 and 100. As the worksheet uses the column method, it is helpful in getting students toward higher accuracy, especially with bigger numbers and in scenarios where regrouping is required.", null, "4413+", null, "4567+", null, "", null, "", null, "" ]	[ null, "https://cdn.splashmath.com/cms_assets/images/playable-left-desc-d0e7c503c7eb99a138cc.svg", null, "https://cdn.splashmath.com/cms_assets/images/playable-right-image-c88d24a6fffd20c6833d.svg", null, "https://cdn.splashmath.com/cms_assets/images/math-and-ela-games-feature-d7f1a6d98b223203d222.svg", null, "https://cdn.splashmath.com/cms_assets/images/math-and-ela-worksheet-feature-56a20bb968cbfa2fe52a.svg", null, "https://cdn.splashmath.com/cms_assets/images/coomon-core-feature-5e0a900656847818fa6d.svg", null, "https://cdn.splashmath.com/cms_assets/images/coopa-feature-8af350a7eecbb0439840.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9083882,"math_prob":0.8978394,"size":399,"snap":"2023-40-2023-50","text_gpt3_token_len":78,"char_repetition_ratio":0.11139241,"word_repetition_ratio":0.0,"special_character_ratio":0.19548872,"punctuation_ratio":0.08219178,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96861106,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T09:48:49Z\",\"WARC-Record-ID\":\"<urn:uuid:e414cef9-7569-4a97-a4ca-230495d31066>\",\"Content-Length\":\"151633\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:86d32b1f-5aac-4ab8-aeaa-8e922cd32e98>\",\"WARC-Concurrent-To\":\"<urn:uuid:d95c7a4f-d79f-46b0-9748-8d1c422b45ce>\",\"WARC-IP-Address\":\"104.18.29.134\",\"WARC-Target-URI\":\"https://www.splashlearn.com/s/math-worksheets/add-multiples-of-10-and-100-vertical-addition\",\"WARC-Payload-Digest\":\"sha1:D22T354ZTAC5YJZFWYXAWFLLHKSOT7WP\",\"WARC-Block-Digest\":\"sha1:LHAO6FWCLSIO6EODJDHMQDUP445JT37V\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233505362.29_warc_CC-MAIN-20230921073711-20230921103711-00096.warc.gz\"}"}
https://people.eecs.berkeley.edu/~bh/v3ch6/ai.html	[ "Computer Science Logo Style volume 3: Beyond Programming 2/e Copyright (C) 1997 MIT\n\n# Artificial Intelligence", null, "Program file for this chapter: `student`\n\nCan a computer be intelligent? What would it mean for a computer to be intelligent? John McCarthy, one of the founders of artificial intelligence research, once defined the field as \"getting a computer to do things which, when done by people, are said to involve intelligence.\" The point of the definition was that he felt perfectly comfortable about carrying on his research without first having to defend any particular philosophical view of what the word \"intelligence\" means.\n\nThere have always been two points of view among AI researchers about what their purpose is. One point of view is that AI programs contribute to our understanding of human psychology; when researchers take this view they try to make their programs reflect the actual mechanisms of intelligent human behavior. For example, Allen Newell and Herbert A. Simon begin their classic AI book Human Problem Solving with the sentence, \"The aim of this book is to advance our understanding of how humans think.\" In one of their research projects they studied cryptarithmetic problems, in which digits are replaced with letters in a multi-digit addition or multiplication. First they did a careful observation and analysis of how a human subject attacked such a problem, then they pointed out specific problem-solving techniques that the person used, and used those techniques as the basis for designing a computer simulation. The other point of view is that AI programs provide a more abstract model for intelligence in general; just as one can learn about the properties of computers by studying finite-state machines, even though no real computer operates precisely as a formal finite-state machine does, we can learn about the properties of any possible intelligent being by simulating intelligence in a computer program, whether or not the mechanisms of that program are similar to those used by people.\n\nIn the early days of AI research these two points of view were not sharply divided. Sometimes the same person would switch from one to the other, sometimes trying to model human thought processes and sometimes trying to solve a given problem by whatever methods could be made to work. More recently, researchers who hold one or the other point of view consistently have begun to define two separate fields. One is cognitive science, in which computer scientists join with psychologists, linguists, biologists, and others to study human cognitive psychology, using computer programs as a concrete embodiment of theories about the human mind. The other is called expert systems or knowledge engineering, in which programming techniques developed by AI researchers are put to practical use in programs that solve real-world business problems such as the diagnosis and repair of malfunctioning equipment.\n\n## Microworlds: Student\n\nIn this chapter I'm going to concentrate on one particular area of AI research: teaching a computer to understand English. Besides its inherent interest, this area has the advantage that it doesn't require special equipment, as do some other parts of AI such as machine vision and the control of robot manipulators.\n\nIn the 1950s many people were very optimistic about the use of computers to translate from one language to another. IBM undertook a government-sponsored project to translate scientific journals from Russian to English. At first they thought that this translation could be done very straightforwardly, with a Russian-English dictionary and a few little kludges to rearrange the order of words in a sentence to account for differences in the grammatical structure of the two languages. This simple approach was not successful. One problem is that the same word can have different meanings, and even different parts of speech, in different contexts. (According to one famous anecdote, the program translated the Russian equivalent of \"The spirit is willing but the flesh is weak\" into \"The vodka is strong but the meat is rotten.\")\n\nA decade later, several AI researchers had the idea that ambiguities in the meanings of words could be resolved by trying to understand English only in some limited context. If you know in advance that the sentence you're trying to understand is about baseball statistics, or about relationships in a family tree, or about telling a robot arm to move blocks on a table (these are actual examples of work done in that period) then only certain narrowly defined types of sentences are meaningful at all. You needn't think about metaphors or about the many assumptions about commonsense knowledge that people make in talking with one another. Such a limited context for a language understanding program is called a microworld.\n\nThis chapter includes a Logo version of Student, a program written by Daniel G. Bobrow for his 1964 Ph.D. thesis, Natural Language Input for a Computer Problem Solving System, at MIT. Student is a program that solves algebra word problems:\n\n```? student [The price of a radio is 69.70. If this price is 15 percent\nless than the marked price, find the marked price.]\n\nThe marked price is 82 dollars\n```\n\n(In this illustration I've left out some of Student's display of intermediate results.) The program has two parts: one that translates the word problem into the form of equations and another that solves the equations. The latter part is complex (about 40 Logo procedures) but straightforward; it doesn't seem surprising to most people that a computer can manipulate mathematical equations. It is Student's understanding of English sentences that furthered the cause of artificial intelligence.\n\nThe aim of the research reported here was to discover how one could build a computer program which could communicate with people in a natural language within some restricted problem domain. In the course of this investigation, I wrote a set of computer programs, the Student system, which accepts as input a comfortable but restricted subset of English which can be used to express a wide variety of algebra story problems...\n\nIn the following discussion, I shall use phrases such as \"the computer understands English.\" In all such cases, the \"English\" is just the restricted subset of English which is allowable as input for the computer program under discussion. In addition, for purposes of this report I have adopted the following operational definition of understanding. A computer understands a subset of English if it accepts input sentences which are members of this subset, and answers questions based on information contained in the input. The Student system understands English in this sense. [Bobrow, 1964.]\n\nHow does the algebra microworld simplify the understanding problem? For one thing, Student need not know anything about the meanings of noun phrases. In the sample problem above, the phrase `The price of a radio` is used as a variable name. The problem could just as well have been\n\n```The weight of a giant size detergent box is 69.70 ounces. If this weight\nis 15 percent less than the weight of an enormous size box, find the\nweight of an enormous size box.\n```\n\nFor Student, either problem boils down to\n\n```variable1 = 69.70 units\n\nvariable1 = 0.85 * variable2\n\nFind variable2.\n```\n\nStudent understands particular words only to the extent that they have a mathematical meaning. For example, the program knows that `15 percent less than` means the same as `0.85 times`.\n\n## How Student Translates English to Algebra\n\nStudent translates a word problem into equations in several steps. In the following paragraphs, I'll mention in parentheses the names of the Logo procedures that carry out each step I describe, but don't read the procedures yet. First read through the description of the process without worrying about the programming details of each step. Later you can reread this section while examining the complete listing at the end of the chapter.\n\nIn translating Student to Logo, I've tried not to change the capabilities of the program in any way. The overall structure of my version is similar to that of Bobrow's original implementation, but I've changed some details. I've used iteration and mapping tools to make the program easier to read; I've changed some aspects of the fine structure of the program to fit more closely with the usual Logo programming style; in a few cases I've tried to make exceptionally slow parts of the program run faster by finding a more efficient algorithm to achieve the same goal.\n\nThe top-level procedure `student` takes one input, a list containing the word problem. (The disk file that accompanies this project includes several variables containing sample problems. For example,\n\n```? student :radio\n```\n\nwill carry out the steps I'm about to describe.) Student begins by printing the original problem:\n\n```? student :radio\n\nThe problem to be solved is\n\nThe price of a radio is 69.70. If this price is 15 percent less than the\nmarked price, find the marked price.\n```\n\nThe first step is to separate punctuation characters from the attached words. For example, the word \"`price,`\" in the original problem becomes the two words \"`price ,`\" with the comma on its own. Then (`student1`) certain mandatory substitutions are applied (`idioms`). For example, the phrase `percent less than` is translated into the single word `perless`. The result is printed:\n\n```With mandatory substitutions the problem is\n\nThe price numof a radio is 69.70 dollars . If this price is 15 perless\nthe marked price , find the marked price .\n```\n\n(The word `of` in an algebra word problem can have two different meanings. Sometimes it means \"times,\" as in the phrase \"one half of the population.\" Other times, as in this problem, \"of\" is just part of a noun phrase like \"the price of a radio.\" The special word `numof` is a flag to a later part of the program and will then be further translated either into `times` or back into `of`. The original implementation of Student used, instead of a special word like ``` numof```, a \"tagged\" word represented as a list like `[of / op]`. Other examples of tagging are `[Bill / person]` and `[has / verb]`.)\n\nThe next step is to separate the problem into simple sentences (`bracket`):\n\n```The simple sentences are\n\nThe price numof a radio is 69.70 dollars .\n\nThis price is 15 perless the marked price .\n\nFind the marked price .\n```\n\nUsually this transformation of the problem is straightforward, but the special case of \"age problems\" is recognized at this time, and special transformations are applied so that a sentence like\n\n```Mary is 24 years old.\n```\n\nis translated into\n\n```Mary s age is 24 .\n```\n\nAn age problem is one that contains any of the phrases ```as old as```, `age`, or `years old`.\n\nThe next step is to translate each simple sentence into an equation or a variable whose value is desired as part of the solution (`senform`).\n\n```The equations to be solved are\n\nEqual [price of radio] [product 69.7 [dollars]]\n\nEqual [price of radio] [product 0.85 [marked price]]\n```\n\nThe third simple sentence is translated, not into an equation, but into a request to solve these equations for the variable `marked price`.\n\nThe translation of simple sentences into equations is the most \"intelligent\" part of the program; that is, it's where the program's knowledge of English grammar and vocabulary come into play and many special cases must be considered. In this example, the second simple sentence starts with the phrase `this price`. The program recognizes the word `this` (procedure `nmtest`) and replaces the entire phrase with the left hand side of the previous equation (procedure `this`).\n\n## Pattern Matching\n\nStudent analyzes a sentence by comparing it to several patterns (`senform1`). For example, one sentence form that Student understands is exemplified by these sentences:\n\n```Joe weighs 163 pounds .\nThe United States Army has 8742 officers .\n```\n\nThe general pattern is\n\n```something verb number unit .\n```\n\nStudent treats such sentences as if they were rearranged to match\n\n```The number of unit something verb is number .\n```\n\nand so it generates the equations\n\n```Equal [number of pounds Joe weighs] 163\n\nEqual [number of officers United States Army has] 8742\n```\n\nThe original version of Student was written in a pattern matching language called Meteor, which Bobrow wrote in Lisp. In Meteor, the instruction that handles this sentence type looks like this:\n\n```(* ( (1 / verb) (fn nmtest) 1 (1 / dlm)) 0\n(/ (s shelf (k equal (fn opform (k the number of 4 1 2))\n(fn opform (k 3 5 6))))) return)\n```\n\nThe top line contains the pattern to be matched. In the pattern, a dollar sign represents zero or more words; the notation `1` represents a single word. The zero at the end of the line means that the text that matches the pattern should be deleted and nothing should replace it. The rest of the instruction pushes a new equation onto a stack named `shelf`; that equation is formed out of the pieces of the matched pattern according to the numbers in the instruction. That is, the number `4` represents the fourth component of the pattern, which is `1`. Here is the corresponding instruction in the Logo version:\n\n```if match [^one !verb1:verb !factor:numberp #stuff1 !:dlm] :sent\n[output (list (list \"equal\nopform (sentence [the number of]\n:stuff1 :one :verb1)\nopform (list :factor) ))]\n```\n\nThe pattern matcher I used for Student is the same as the one in Advanced Techniques, the second volume of this series.* Student often relies on the fact that Meteor's pattern matcher finds the first substring of the text that matches the pattern, rather than requiring the entire text to match. Many patterns in the Logo version therefore take the form\n\n```[^beg interesting part #end]\n```\n\nwhere the \"interesting part\" is all that appeared in the Meteor pattern.\n\nThe version in this project is modified slightly; the `match` procedure first does a fast test to try to reject an irrelevant pattern in O(n) time before calling the actual pattern matcher, which could take as much as O(2n) time to reject a pattern, and which has been renamed `rmatch` (for \"real match\") in this project.\n\nHere is a very brief summary of the Logo pattern matcher included in this program. For a fuller description with examples, please refer to Volume 2. `Match` is a predicate with two inputs, both lists. The first input is the pattern and the second input is the sentence. `Match` outputs `true` if the sentence matches the pattern. A word in the pattern that does not begin with one of the special quantifier characters listed below matches the identical word in the sentence. A word in the pattern that does begin with a quantifier matches zero or more words in the sentence, as follows:\n\n `#` zero or more `&` one or more `?` zero or one `!` exactly one `@` zero or more (test as group) `^` zero or more (as few as possible)\n\nAll quantifiers match as many consecutive words as possible while still allowing the remaining portion of the pattern to be matched, except for `^`. A quantifier may be used alone, or it can be followed by a variable name, a predicate name, or both:\n\n```#\n#var\n#:pred\n#var:pred\n```\nIf a variable name is used, the word or words that match the quantifier will be stored in that variable if the match is successful. (The value of the variable if the match is not successful is not guaranteed.) If a predicate is used, it must take one word as input; in order for a word in the sentence to be accepted as (part of) a match for the quantifier, the predicate must output `true` when given that word as input. For example, the word\n\n```!factor:numberp\n```\n\nin the pattern above requires exactly one matching word in the sentence; that word must be a number, and it is remembered in the variable `factor`. If the quantifier is `@` then the predicate must take a list as input, and it must output `true` for all the candidate matching words taken together as a list. For example, if you define a procedure\n\n```to threep :list\noutput equalp count :list 3\nend\n```\n\nthen the pattern word\n\n```@:threep\n```\n\nwill match exactly three words in the sentence. (Student does not use this last feature of the pattern matcher. In fact, predicates are applied only to the single-word quantifiers `?` and `!`.)\n\nPattern matching is also heavily used in converting words and phrases with mathematical meaning into the corresponding arithmetic operations (`opform`). An equation is a list of three members; the first member is the word `equal` and the other two are expressions formed by applying operations to variables and numbers. Each operation that is required is represented as a list whose first member is the name of the Logo procedure that carries out the operation and whose remaining members are expressions representing the operands. For example, the equation\n\ny = 3x2 + 6x − 1\n\nwould be represented by the list\n\n```[equal [y] [sum [product 3 [square [x]]] [product 6 [x]] [minus 1]]]\n```\n\nThe variables are represented by lists like `[x]` rather than just the words because in Student a variable can be a multi-word phrase like `price of radio`. The difference between two expressions is represented by a `sum` of one expression and `minus` the other, rather than as the `difference` of the expressions, because this representation turns out to make the process of simplifying and solving the equations easier.\n\nIn word problems, as in arithmetic expressions, there is a precedence of operations. Operations like `squared` apply to the variables right next to them; ones like `times` are intermediate, and ones like `plus` apply to the largest possible subexpressions. Student looks first for the lowest-priority ones like `plus`; if one is found, the entire rest of the clause before and after the operation word provide the operands. Those operands are recursively processed by `opform`; when all the low-priority operations have been found, the next level of priority will be found by matching the pattern\n\n```[^left !op:op1 #right]\n```\n\n## Solving the Equations\n\nStudent uses the substitution technique to solve the equations. That is, one equation is rearranged so that the left hand side contains only a single variable and the right hand side does not contain that variable. Then, in some other equation, every instance of that variable is replaced by the right hand side of the first equation. The result is a new equation from which one variable has been eliminated. Repeating this process enough times should eventually yield an equation with only a single variable, which can be solved to find the value of that variable.\n\nWhen a problem gives rise to several linear equations in several variables, the traditional technique for computer solution is to use matrix inversion; this technique is messy for human beings because there is a lot of arithmetic involved, but straightforward for computers because the algorithm can be specified in a simple way that doesn't depend on the particular equations in each problem. Bobrow chose to use the substitution method because some problems give rise to equations that are linear in the variable for which a solution is desired but nonlinear in other variables. Consider this problem:\n\n```? student :tom\n\nThe problem to be solved is\n\nIf the number of customers Tom gets is twice the square of 20 per cent of\nruns is 45, what is the number of customers Tom gets?\n\nWith mandatory substitutions the problem is\n\nIf the number numof customers Tom gets is 2 times the square 20 percent\nnumof the number numof advertisements he runs , and the number numof\nadvertisements he runs is 45 , what is the number numof customers\nTom gets ?\n\nThe simple sentences are\n\nThe number numof customers Tom gets is 2 times the square 20 percent\n\nWhat is the number numof customers Tom gets ?\n\nThe equations to be solved are\n\nEqual [number of customers Tom gets]\nhe runs]]]]\n\nThe number of customers Tom gets is 162\n\nThe problem is solved.\n```\n\nThe first equation that Student generates for this problem is linear in the number of customers Tom gets, but nonlinear in the number of advertisements he runs. (That is, the equation refers to the square of the latter variable. An equation is linear in a given variable if that variable isn't multiplied by anything other than a constant number.) Using the substitution method, Student can solve the problem by substituting the value 45, found in the second equation, for the number of advertisements variable in the first equation.\n\n(Notice, in passing, that one of the special `numof` words in this problem was translated into a multiplication rather than back into the original word `of`.)\n\nThe actual sequence of steps required to solve a set of equations is quite intricate. I recommend taking that part of Student on faith the first time you read the program, concentrating instead on the pattern matching techniques used to translate the English sentences into equations. But here is a rough guide to the solution process. Both `student1` and ``` student2``` call `trysolve` with four inputs: a list of the equations to solve, a list of the variables for which values are wanted, and two lists of units. A unit is a word or phrase like `dollars` or `feet` that may be part of a solution. Student treats units like variables while constructing the equations, so the combination of a number and a unit is represented as a product, like\n\n```[product 69.7 [dollars]]\n```\n\nfor 69.70 in the first sample problem. While constructing the equations, Student generates two lists of units. The first, stored in the variable `units`, contains any word or phrase that appears along with a number in the problem statement, like the word `feet` in the phrase ``` 3 feet``` (`nmtest`). The second, in the variable `aunits`, contains units mentioned explicitly in the `find` or `how many` sentences that tell Student what variables should be part of the solution (`senform1`). If the problem includes a sentence like\n\n```How many inches is a yard?\n```\n\nthen the variable `[inches]`, and only that variable, is allowed to be part of the answer. If there are no `aunits`-type variables in the problem, then any of the `units` variables may appear in the solution (`trysolve`).\n\n`Trysolve` first calls `solve` to solve the equations and then uses `pranswers` to print the results. `Solve` calls `solver` to do most of the work and then passes its output through `solve.reduce` for some final cleaning up. `Solver` works by picking one of the variables from the list `:wanted` and asking `solve1` to find a solution for that variable in terms of all the other variables--the other wanted variables as well as the units allowed in the ultimate answer. If `solve1` succeeds, then `solver` invokes itself, adding the newly-found expression for one variable to an association list (in the variable `alis`) so that, from then on, any occurrence of that variable will be replaced with the equivalent expression. In effect, the problem is simplified by eliminating one variable and eliminating one equation, the one that was solved to find the equivalent expression.\n\n`Solve1` first looks for an equation containing the variable for which it is trying to find a solution. When it finds such an equation, the next task is to eliminate from that equation any variables that aren't part of the wanted-plus-units list that `solver` gave `solve1` as an input. To eliminate these extra variables, `solve1` invokes `solver` with the extras as the list of wanted variables. This mutual recursion between `solver` and `solve1` makes the structure of the solution process difficult to follow. If `solver` manages to eliminate the extra variables by expressing them in terms of the originally wanted ones, then `solve1` can go on to substitute those expressions into its originally chosen equation and then use `solveq` to solve that one equation for the one selected variable in terms of all the other allowed variables. ``` Solveq``` manipulates the equation more or less the way students in algebra classes do, adding the same term to both sides, multiplying both sides by the denominator of a polynomial fraction, and so on.\n\nHere is how `solve` solves the radio problem. The equations, again, are\n\n```Equal [price of radio] [product 69.7 [dollars]]\n\nEqual [price of radio] [product 0.85 [marked price]]\n```\n\n`Trysolve` evaluates the expression\n\n```(1) solve [[marked price]]\n[[equal [price of radio] [product 69.7 [dollars]]]\n[equal [price of radio] [product 0.85 [marked price]]] ]\n[[dollars]]\n```\n\n(I'm numbering these expressions so that I can refer to them later in the text.) The first input to `solve` is the list of variables wanted in the solution; in this case there is only one such variable. The second input is the list of two equations. The third is the list of unit variables that are allowed to appear in the solution; in this case only ``` [dollars]``` is allowed. `Solve` evaluates\n\n```(2) solver [[marked price]] [[dollars]] [] []\n```\n\n(There is a fifth input, the word `insufficient`, but this is used only as an error flag if the problem can't be solved. To simplify this discussion I'm going to ignore that input for both `solver` and ``` solve1```.) `Solver` picks the first (in this case, the only) wanted variable as the major input to `solve1`:\n\n```(3) solve1 [marked price]\n[[dollars]]\n[]\n[[equal [price of radio] [product 69.7 [dollars]]]\n[equal [price of radio] [product 0.85 [marked price]]] ]\n[]\n```\n\nNotice that the first input to `solve1` is a single variable, not a list of variables. `Solve1` examines the first equation in the list of equations making up its fourth input. The desired variable does not appear in this equation, so `solve1` rejects that equation and invokes itself recursively:\n\n```(4) solve1 [marked price]\n[[dollars]]\n[]\n[[equal [price of radio] [product 0.85 [marked price]]]]\n[[equal [price of radio] [product 69.7 [dollars]]]]\n```\n\nThis time, the first (and now only) equation on the list of candidates does contain the desired variable. `Solve1` removes that equation, not from its own list of equations (`:eqns`), but from ``` solve```'s overall list (`:eqt`). The equation, unfortunately, can't be solved directly to express `[marked price]` in terms of `[dollars]`, because it contains the extra, unwanted variable `[price of radio]`. We must eliminate this variable by solving the remaining equations for it:\n\n```(5) solver [[price of radio]] [[marked price] [dollars]] [] []\n```\n\nAs before, `solver` picks the first (again, in this case, the only) wanted variable and asks `solve1` to solve it:\n\n```(6) solve1 [price of radio]\n[[marked price] [dollars]]\n[]\n[[equal [price of radio] [product 69.7 [dollars]]]]\n[]\n```\n\n`Solve1` does find the desired variable in the first (and only) equation, and this time there are no extra variables. `Solve1` can therefore ask `solveq` to solve the equation:\n\n```(7) solveq [price of radio]\n[equal [price of radio] [product 69.7 [dollars]]]\n```\n\nIt isn't part of `solveq`'s job to worry about which variables may or may not be part of the solution; `solve1` doesn't call ``` solveq``` until it's satisfied that the equation is okay.\n\nIn this case, `solveq` has little work to do because the equation is already in the desired form, with the chosen variable alone on the left side and an expression not containing that variable on the right.\n\n```solveq (7) outputs [[price of radio] [product 69.7 [dollars]]\nto solve1 (6)\n```\n\n`Solve1` appends this result to the previously empty association list.\n\n```solve1 (6) outputs [[[price of radio] [product 69.7 [dollars]]]\nto solver (5)\n```\n\n`Solver` only had one variable in its `:wanted` list, so its job is also finished.\n\n```solver (5) outputs [[[price of radio] [product 69.7 [dollars]]]\nto solve1 (4,3)\n```\n\nThis outer invocation of `solve1` was trying to solve for `[marked price]` an equation that also involved `[price of radio]`. It is now able to use the new association list to substitute for this unwanted variable an expression in terms of wanted variables only; this modified equation is then passed on to `solveq`:\n\n```(8) solveq [marked price]\n[equal [product 69.7 [dollars]] [product 0.85 [marked price]]]\n```\n\nThis time `solveq` has to work a little harder, exchanging the two sides of the equation and dividing by 0.85.\n\n```solveq (8) outputs [[marked price] [product 82 [dollars]]]\nto solve1 (4,3)\n```\n\n`Solve1` appends this result to the association list:\n\n```solve1 (4,3) outputs [[[price of radio] [product 69.7 [dollars]]]\n[[marked price] [product 82 [dollars]]] ]\nto solver (2)\n```\n\nSince `solver` has no other wanted variables, it outputs the same list to `solve`, and `solve` outputs the same list to ``` trysolve```. (In this example, `solve.reduce` has no effect because all of the expressions in the association list are in terms of allowed units only. If the equations had been different, the expression for ```[price of radio]``` might have included `[marked price]` and then ``` solve.reduce``` would have had to substitute and simplify (`subord`).)\n\nIt'll probably take tracing a few more examples and beating your head against the wall a bit before you really understand the structure of ``` solve``` and its subprocedures. Again, don't get distracted by this part of the program until you've come to understand the language processing part, which is our main interest in this chapter.\n\n## Age Problems\n\nThe main reason why Student treats age problems specially is that the English form of such problems is often expressed as if the variables were people, like \"Bill,\" whereas the real variable is \"Bill's age.\" The pattern matching transformations look for proper names (`personp`) and insert the words `s age` after them (`ageify`). The first such age variable in the problem is remembered specially so that it can be substituted for pronouns (`agepron`). A special case is the phrase ``` their ages```, which is replaced (`ageprob`) with a list of all the age variables in the problem.\n\n```? student :uncle\n\nThe problem to be solved is\n\nBill's father's uncle is twice as old as Bill's father. 2 years from now\nBill's father will be 3 times as old as Bill. The sum of their ages is\n92 . Find Bill's age.\n\nWith mandatory substitutions the problem is\n\nBill s father s uncle is 2 times as old as Bill s father . 2 years\nfrom now Bill s father will be 3 times as old as Bill . sum their\nages is 92 . Find Bill s age .\n\nThe simple sentences are\n\nBill s father s uncle s age is 2 times Bill s father s age .\n\nBill s father s age pluss 2 is 3 times Bill s age pluss 2 .\n\nSum Bill s age and Bill s father s age and Bill s father s uncle s age\nis 92 .\n\nFind Bill s age .\n\nThe equations to be solved are\n\nEqual [Bill s father s uncle s age] [product 2 [Bill s father s age]]\n\nEqual [sum [Bill s father s age] 2] [product 3 [sum [Bill s age] 2]]\n\nEqual [sum [Bill s age]\n[sum [Bill s father s age] [Bill s father s uncle s age]]] 92\n\nBill s age is 8\n\nThe problem is solved.\n```\n\n(Note that in the original problem statement there is a space between the number `92` and the following period. I had to enter the problem in that form because of an inflexibility in Logo's input parser, which assumes that a period right after a number is part of the number, so that \"`92.`\" is reformatted into `92` without the dot.)\n\nStudent represents the possessive word `Bill's` as the two words `Bill s` because this representation allows the pattern matcher to manipulate the possessive marker as a separate element to be matched. A phrase like `as old as` is just deleted (`ageprob`) because the transformation from people to ages makes it redundant.\n\nThe phrase `2 years from now` in the original problem is first translated to `in 2 years`. This phrase is further processed according to where it appears in a sentence. When it is attached to a particular variable, in a phrase like `Bill s age in 2 years`, the entire phrase is translated into the arithmetic operation `Bill s age pluss 2 years` (`agewhen`). (The special word `pluss` is an addition operator, just like `plus`, except for its precedence; `opform` treats it as a tightly binding operation like `squared` instead of a loosely binding one like the ordinary `plus`.) When a phrase like `in 2 years` appears at the beginning of a sentence, it is remembered (`agesen`) as an implicit modifier for every age variable in that sentence that isn't explicitly modified. In this example, `in 2 years` modifies both `Bill s father s age` and `Bill s age`. The special precedence of `pluss` is needed in this example so that the equation will be based on the grouping\n\n```3 times [ Bill s age pluss 2 ]\n```\n\nrather than\n\n```[ 3 times Bill s age ] plus 2\n```\n\nas it would be with the ordinary `plus` operator. You can also see how the substitution for `their ages` works in this example.\n\nHere is a second sample age problem that illustrates a different kind of special handling:\n\n```? student :ann\n\nThe problem to be solved is\n\nMary is twice as old as Ann was when Mary was as old as Ann is now. If\nMary is 24 years old, how old is Ann?\n\nWith mandatory substitutions the problem is\n\nMary is 2 times as old as Ann was when Mary was as old as Ann is now . If\nMary is 24 years old , what is Ann ?\n\nThe simple sentences are\n\nMary s age is 2 times Ann s age minuss g1 .\n\nMary s age minuss g1 is Ann s age .\n\nMary s age is 24 .\n\nWhat is Ann s age ?\n\nThe equations to be solved are\n\nEqual [Mary s age] [product 2 [sum [Ann s age] [minus [g1]]]]\n\nEqual [sum [Mary s age] [minus [g1]]] [Ann s age]\n\nEqual [Mary s age] 24\n\nAnn s age is 18\n\nThe problem is solved.\n```\n\nWhat is new in this example is Student's handling of the phrase ```was when``` in the sentence\n\n```Mary is 2 times as old as Ann was when Mary was as old as Ann is now .\n```\n\nSentences like this one often cause trouble for human algebra students because they make implicit reference to a quantity that is not explicitly present as a variable. The sentence says that Mary's age now is twice Ann's age some number of years ago, but that number is not explicit in the problem. Student makes this variable explicit by using a generated symbol like the word `g1` in this illustration. Student replaces the phrase `was when` with the words\n\n```was g1 years ago . g1 years ago\n```\n\nThis substitution (in `ageprob`) happens before the division of the problem statement into simple sentences (`bracket`). As a result, this one sentence in the original problem becomes the two sentences\n\n```Mary s age is 2 times Ann s age g1 years ago .\n\nG1 years ago Mary s age was Ann s age now .\n```\n\nThe phrase `g1 years ago` in each of these sentences is further processed by `agesen` and `agewhen` as discussed earlier; the final result is\n\n```Mary s age is 2 times Ann s age minuss g1 .\n\nMary s age minuss g1 is Ann s age .\n```\n\nA new generated symbol is created each time this situation arises, so there is no conflict from trying to use the same variable name for two different purposes. The phrase `will be when` is handled similarly, except that the translated version is\n\n```in g2 years . in g2 years\n```\n\n## AI and Education\n\nThese decoupling heuristics are useful not only for the Student program but for people trying to solve age problems. The classic age problem about Mary and Ann, given above, took an MIT graduate student over 5 minutes to solve because he did not know this heuristic. With the heuristic he was able to set up the appropriate equations much more rapidly. As a crude measure of Student's relative speed, note that Student took less than one minute to solve this problem.\n\nThis excerpt from Bobrow's thesis illustrates the idea that insights from artificial intelligence research can make a valuable contribution to the education of human beings. An intellectual problem is solved, at least in many cases, by dividing it into pieces and developing a technique for each subproblem. The subproblems are the same whether it is a computer or a person trying to solve the problem. If a certain technique proves valuable for the computer, it may be helpful for a human problem solver to be aware of the computer's methods. Bobrow's suggestion to teach people one specific heuristic for algebra word problems is a relatively modest example of this general theme. (A heuristic is a rule that gives the right answer most of the time, as opposed to an algorithm, a rule that always works.) Some researchers in cognitive science and education have proposed the idea of intelligent CAI (computer assisted instruction), in which a computer would be programmed as a \"tutor\" that would observe the efforts of a student in solving a problem. The tutor would know about some of the mistaken ideas people can have about a particular class of problem and would notice a student falling into one of those traps. It could then offer advice tailored to the needs of that individual student.\n\nThe development of the Logo programming language (and so also, indirectly, this series of books) is another example of the relationship between AI and education. Part of the idea behind Logo is that the process of programming a computer resembles, in some ways, the process of teaching a person to do something. (This can include teaching oneself.) For example, when a computer program doesn't work, the experienced programmer doesn't give up in despair, but instead debugs the program. Yet many students are willing to give up and say \"I just don't get it\" if their understanding of some problem isn't perfect on the first try.\n\nThe critic is afraid that children will adopt the computer as model and eventually come to \"think mechanically\" themselves. Following the opposite tack, I have invented ways to take educational advantage of the opportunities to master the art of deliberately thinking like a computer, according, for example, to the stereotype of a computer program that proceeds in a step-by-step, literal, mechanical fashion. There are situations where this style of thinking is appropriate and useful. Some children's difficulties in learning formal subjects such as grammar or mathematics derive from their inability to see the point of such a style.\n\nA second educational advantage is indirect but ultimately more important. By deliberately learning to imitate mechanical thinking, the learner becomes able to articulate what mechanical thinking is and what it is not. The exercise can lead to greater confidence about the ability to choose a cognitive style that suits the problem. Analysis of \"mechanical thinking\" and how it is different from other kinds and practice with problem analysis can result in a new degree of intellectual sophistication. By providing a very concrete, down-to-earth model of a particular style of thinking, work with the computer can make it easier to understand that there is such a thing as a \"style of thinking.\" And giving children the opportunity to choose one style or another provides an opportunity to develop the skill necessary to choose between styles. Thus instead of inducing mechanical thinking, contact with computers could turn out to be the best conceivable antidote to it. And for me what is most important in this is that through these experiences these children would be serving their apprenticeships as epistemologists, that is to say learning to think articulately about thinking. [Seymour Papert, Mindstorms, Basic Books, 1980, p. 27.]\n\n## Combining Sentences Into One Equation\n\nIn age problems, as we've just seen, a single sentence may give rise to two equations. Here is an example of the opposite, several sentences that together contribute a single equation.\n\n```? student :nums\n\nThe problem to be solved is\n\nA number is multiplied by 6 . This product is increased by 44 . This\nresult is 68 . Find the number.\n\nWith mandatory substitutions the problem is\n\nA number ismulby 6 . This product is increased by 44 . This result is\n68 . Find the number .\n\nThe simple sentences are\n\nA number ismulby 6 .\n\nThis product is increased by 44 .\n\nThis result is 68 .\n\nFind the number .\n\nThe equations to be solved are\n\nEqual [sum [product [number] 6] 44] 68\n\nThe number is 4\n\nThe problem is solved.\n```\n\nStudent recognizes problems like this by recognizing the phrases \"is multiplied by,\" \"is divided by,\" and \"is increased by\" (`senform1`). A sentence containing one of these phrases is not translated into an equation; instead, a partial equation is saved until the next sentence is read. That next sentence is expected to start with a phrase like \"this result\" or \"this product.\" The same procedure (`this`) that in other situations uses the left hand side of the last equation as the expression for the `this`-phrase notices that there is a remembered partial equation and uses that instead. In this example, the sentence\n\n```A number ismulby 6 .\n```\n\nremembers the algebraic expression\n\n```[product [number] 6]\n```\n\nThe second sentence uses that remembered expression as part of a new, larger expression to be remembered:\n\n```[sum [product [number] 6] 44]\n```\n\nThe third sentence does not contain one of the special \"is increased by\" phrases, but is instead a standard \"A is B\" sentence. That sentence, therefore, does give rise to an equation, as shown above.\n\nPerhaps the most interesting thing to notice about this category of word problem is how narrowly defined Student's criterion for recognizing the category is. Student gets away with it because algebra word problems are highly stereotyped; there are just a few categories, with traditional, standard wordings. In principle there could be a word problem starting\n\n```Robert has a certain number of jelly beans. This number is twice the\nnumber of jelly beans Linda has.\n```\n\nThese two sentences are together equivalent to\n\n```The number of jelly beans Robert has is twice the number of jelly beans\nLinda has.\n```\n\nBut Student would not recognize the situation because the first sentence doesn't talk about \"is increased by.\" We could teach Student to understand a word problem in this form by adding the instruction\n\n```if match [^one !verb1:verb a certain number of #stuff1 !:dlm] :sent\n[push \"ref opform (se [the number of] :stuff1 :one :verb1)\nop []]\n```\n\nalong with the other known sentence forms in `senform1`. (Compare this to the pattern matching instruction shown earlier for a similar sentence but with an explicitly specified number.)\n\nTaking advantage of the stereotyped nature of word problems is an example of how the microworld strategy helped make the early AI programs possible. If word problems were expressed with all the flexibility of language in general, Student would need many more sentence patterns than it actually has. (How many different ways can you think of to express the same idea about Robert and Linda? How many of those ways can Student handle?)\n\n## Allowing Flexible Phrasing\n\nIn the examples we've seen so far, Student has relied on the repetition of identical or near-identical phrases such as \"the marked price\" or \"the number of advertisements he runs.\" (The requirement is not quite strictly identical phrases because articles are removed from the noun phrases to make variable names.) In real writing, though, such phrases are often abbreviated when they appear for a second time. Student will translate such a problem into a system of equations that can't be solved, because what should be one variable is instead a different variable in each equation. But Student can recognize this situation and apply heuristic rules to guess that two similar variable names are meant, in fact, to represent the same variable. (Some early writers on AI considered the use of heuristic methods one of the defining characteristics of the field. Computer scientists outside of AI were more likely to insist on fully reliable algorithms. This distinction still has some truth to it, but it isn't emphasized so much as a critical issue these days.) Student doesn't try to equate different variables until it has first tried to solve the equations as they are originally generated. If the first attempt at solution fails, Student has recourse to less certain techniques (`student2` calls `vartest`).\n\n```? student :sally\n\nThe problem to be solved is\n\nThe sum of Sally's share of some money and Frank's share is 4.50.\nSally's share is twice Frank's. Find Frank's and Sally's share.\n\nWith mandatory substitutions the problem is\n\nsum Sally s share numof some money and Frank s share is 4.50 dollars .\nSally s share is 2 times Frank s . Find Frank s and Sally s share .\n\nThe simple sentences are\n\nSum Sally s share numof some money and Frank s share is 4.50 dollars .\n\nSally s share is 2 times Frank s .\n\nFind Frank s and Sally s share .\n\nThe equations to be solved are\n\nEqual [sum [Sally s share of some money] [Frank s share]]\n[product 4.50 [dollars]]\n\nEqual [Sally s share] [product 2 [Frank s]]\n\nThe equations were insufficient to find a solution.\n\nAssuming that\n[Frank s] is equal to [Frank s share]\n\nAssuming that\n[Sally s share] is equal to [Sally s share of some money]\n\nFrank s is 1.5 dollars\n\nSally s share is 3 dollars\n\nThe problem is solved.\n```\n\nIn this problem Student has found two pairs of similar variable names. When it finds such a pair, Student adds an equation of the form\n\n```[equal variable1 variable2]\n```\n\nto the previous set of equations. In both of the pairs in this example, the variable that appears later in the problem statement is entirely contained within the one that appears earlier.\n\nAnother point of interest in this example is that the variable `[dollars]` is included in the list of units that may be part of the answer. The word problem does not explicitly ask \"How many dollars is Sally's share,\" but because one of the sentences sets an expression equal to \"4.50 dollars\" Student takes that as implicit permission to express the answer in dollars.\n\nThe only other condition under which Student will consider two variables equal is if their names are identical except that some phrase in the one that appears earlier is replaced with a pronoun in the one that appears later. That is, a variable like ```[the number of ice cream cones the children eat]``` will be considered equal to a later variable ```[the number of ice cream cones they eat]```. Here is a problem in which this rule is applied:\n\n```? student :guns\n\nThe problem to be solved is\n\nThe number of soldiers the Russians have is one half of the number of\nguns they have. They have 7000 guns. How many soldiers do they have?\n\nWith mandatory substitutions the problem is\n\nThe number numof soldiers the Russians have is 0.5 numof the number numof\nguns they have . They have 7000 guns . howm soldiers do they have ?\n\nThe simple sentences are\n\nThe number numof soldiers the Russians have is 0.5 numof the number numof\nguns they have .\n\nThey have 7000 guns .\n\nHowm soldiers do they have ?\n\nThe equations to be solved are\n\nEqual [number of soldiers Russians have]\n[product 0.5 [number of guns they have]]\n\nEqual [number of guns they have] 7000\n\nThe equations were insufficient to find a solution.\n\nAssuming that\n[number of soldiers they have] is equal to\n[number of soldiers Russians have]\n\nThe number of soldiers they have is 3500\n\nThe problem is solved.\n```\n\n## Using Background Knowledge\n\nIn some word problems, not all of the necessary information is contained within the problem statement itself. The problem requires the student to supply some piece of general knowledge about the world in order to determine the appropriate equations. This knowledge may be about unit conversions (one foot is 12 inches) or about relationships among physical quantities (distance equals speed times time). Student \"knows\" some of this background information and can apply it (`geteqns`) if the equations determined by the problem statement are insufficient.\n\n```? student :jet\n\nThe problem to be solved is\n\nThe distance from New York to Los Angeles is 3000 miles. If the average\nspeed of a jet plane is 600 miles per hour, find the time it takes to\ntravel from New York to Los Angeles by jet.\n\nWith mandatory substitutions the problem is\n\nThe distance from New York to Los Angeles is 3000 miles . If the average\nspeed numof a jet plane is 600 miles per hour , find the time it takes to\ntravel from New York to Los Angeles by jet .\n\nThe simple sentences are\n\nThe distance from New York to Los Angeles is 3000 miles .\n\nThe average speed numof a jet plane is 600 miles per hour .\n\nFind the time it takes to travel from New York to Los Angeles by jet .\n\nThe equations to be solved are\n\nEqual [distance from New York to Los Angeles] [product 3000 [miles]]\n\nEqual [average speed of jet plane]\n[quotient [product 600 [miles]] [product 1 [hours]]]\n\nThe equations were insufficient to find a solution.\n\nUsing the following known relationships\n\nEqual [distance] [product [speed] [time]]\n\nEqual [distance] [product [gas consumption]\n[number of gallons of gas used]]\n\nAssuming that\n[speed] is equal to [average speed of jet plane]\n\nAssuming that\n[time] is equal to [time it takes to travel\nfrom New York to Los Angeles by jet]\n\nAssuming that\n[distance] is equal to [distance from New York to Los Angeles]\n\nThe time it takes to travel from New York\nto Los Angeles by jet is 5 hours\n\nThe problem is solved.\n```\n\nStudent's library of known relationships is indexed according to the first word of the name of each variable involved in the relationship. (If a variable starts with the words `number of` it is indexed under the following word.) The relationships, in the form of equations, are stored in the property lists of these index words.\n\nProperty lists are also used to keep track of irregular plurals and the corresponding singulars. Student tries to keep all units in plural form internally, so that if a problem refers to both `1 foot` and ``` 2 feet``` the same variable name will be used for both. (That is, the first of these will be translated into\n\n```[product 1 [feet]]\n```\n\nin Student's internal representation. Then the opposite translation is needed if the product of `1` and some unit appears in an answer to be printed.\n\nThe original Student also used property lists to remember the parts of speech of words and the precedence of operators, but because of differences in the syntax of the Meteor pattern matcher and my Logo pattern matcher I've found it easier to use predicate operations for that purpose.\n\nThe original Student system included a separately invoked `remember` procedure that allowed all these kinds of global information to be entered in the form of English sentences. You'd say\n\n```Feet is the plural of foot\n```\n\nor\n\n```Distance equals speed times time\n```\n\nand `remember` would use patterns much like those used in understanding word problems to translate these sentences into `pprop` instructions. Since Lisp programs, like Logo programs, can themselves be manipulated as lists, `remember` could even accept information of a kind that's stored in the Student program itself, such as the wording transformations in ``` idioms```, and modify the program to reflect this information. I haven't bothered to implement that part of the Student system because it takes up extra memory and doesn't exhibit any new techniques.\n\nAs the above example shows, it's important that Student's search for relevant known relationships comes before the attempt to equate variables with similar names. The general relationship that uses a variable named simply `[distance]` doesn't help unless Student can identify it as relevant to the variable named `[distance from New York to Los Angeles]` in the specific problem under consideration.\n\nHere is another example in which known relationships are used:\n\n```? student :span\n\nThe problem to be solved is\n\nIf 1 span is 9 inches, and 1 fathom is 6 feet,\nhow many spans is 1 fathom?\n\nWith mandatory substitutions the problem is\n\nIf 1 span is 9 inches , and 1 fathom is 6 feet , howm spans is 1 fathom ?\n\nThe simple sentences are\n\n1 span is 9 inches .\n\n1 fathom is 6 feet .\n\nHowm spans is 1 fathom ?\n\nThe equations to be solved are\n\nEqual [product 1 [spans]] [product 9 [inches]]\n\nEqual [product 1 [fathoms]] [product 6 [feet]]\n\nEqual g2 [product 1 [fathoms]]\n\nThe equations were insufficient to find a solution.\n\nUsing the following known relationships\n\nEqual [product 1 [yards]] [product 3 [feet]]\n\nEqual [product 1 [feet]] [product 12 [inches]]\n\n1 fathom is 8 spans\n\nThe problem is solved.\n```\n\nBesides the use of known relationships, this example illustrates two other features of Student. One is the use of an explicitly requested unit in the answer. Since the problem asks\n\n```How many spans is 1 fathom?\n```\n\nStudent knows that the answer must be expressed in `spans`. Had there been no explicit request for a particular unit, all the units that appear in phrases along with a number would be eligible to appear in the answer: `inches`, `feet`, and `fathoms`. Student might then blithely inform us that\n\n```1 fathom is 1 fathom\n\nThe problem is solved.\n```\n\nThe other new feature demonstrated by this example is the use of a generated symbol to represent the desired answer. In the statement of this problem, there is no explicit variable representing the unknown. `[Fathoms]` is a unit, not a variable for which a value could be found. The problem asks for the value of the expression\n\n```[product 1 [fathoms]]\n```\n\nin terms of spans. Student generates a variable name (`g2`) to represent the unknown and produces an equation\n\n```[equal g2 [product 1 [fathoms]]\n```\n\nto add to the list of equations. A generated symbol will be needed whenever the \"Find\" or \"What is\" sentence asks for an expression rather than a simple variable name. For example, an age problem that asks \"What is the sum of their ages\" would require the use of a generated symbol. (The original Student always used a generated symbol for the unknowns, even if there was already a single variable in the problem representing an unknown. It therefore had equations like\n\n```[equal g3 [marked price]]\n```\n\nin its list, declaring one variable equal to another. I chose to check for this case and avoid the use of a generated symbol because the time spent in the actual solution of the equations increases quadratically with the number of equations.)\n\n## Optional Substitutions\n\nWe have seen many cases in which Student replaces a phrase in the statement of a problem with a different word or phrase that fits better with the later stages of processing, like the substitution of `2 times` for `twice` or a special keyword like `perless` for `percent less than`. Student also has a few cases of optional substitutions that may or may not be made (`tryidiom`).\n\nThere are two ways in which optional substitutions can happen. One is exemplified by the phrase `the perimeter of the rectangle`. Student first attempts the problem without any special processing of this phrase. If a solution is not found, Student then replaces the phrase with ```twice the sum of the length and width of the rectangle``` and processes the resulting new problem from the beginning. Unlike the use of known relationships or similarity of variable names, which Student handles by adding to the already-determined equations, this optional substitution requires the entire translation process to begin again. For example, the word `twice` that begins the replacement phrase will be further translated to `2 times`.\n\nThe second category of optional substitution is triggered by the phrase ``` two numbers```. This phrase must always be translated to something, because it indicates that two different variables are needed. But the precise translation depends on the wording of the rest of the problem. Student tries two alternative translations: ```one of the numbers and the other number``` and `one number and the other number`. Here is an example in which the necessary translation is the one Student tries second:\n\n```? student :sumtwo\n\nThe problem to be solved is\n\nThe sum of two numbers is 96, and one number is 16 larger than the other\nnumber. Find the two numbers.\n\nThe problem with an idiomatic substitution is\n\nThe sum of one of the numbers and the other number is 96 , and one\nnumber is16 larger than the other number . Find the one of the numbers\nand the other number .\n\nWith mandatory substitutions the problem is\n\nsum one numof the numbers and the other number is 96 , and one number\nis 16 plus the other number . Find the one numof the numbers and the\nother number .\n\nThe simple sentences are\n\nSum one numof the numbers and the other number is 96 .\n\nOne number is 16 plus the other number .\n\nFind the one numof the numbers and the other number .\n\nThe equations to be solved are\n\nEqual [sum [one of numbers] [other number]] 96\n\nEqual [one number] [sum 16 [other number]]\n\nThe equations were insufficient to find a solution.\n\nThe problem with an idiomatic substitution is\n\nThe sum of one number and the other number is 96 , and one number is 16\nlarger than the other number . Find the one number and the other number .\n\nWith mandatory substitutions the problem is\n\nsum one number and the other number is 96 , and one number is 16 plus the\nother number . Find the one number and the other number .\n\nThe simple sentences are\n\nSum one number and the other number is 96 .\n\nOne number is 16 plus the other number .\n\nFind the one number and the other number .\n\nThe equations to be solved are\n\nEqual [sum [one number] [other number]] 96\n\nEqual [one number] [sum 16 [other number]]\n\nThe one number is 56\n\nThe other number is 40\n\nThe problem is solved.\n```\n\nThere is no essential reason why Student uses one mechanism rather than another to deal with a particular problematic situation. The difficulties about perimeters and about the phrase \"two numbers\" might have been solved using mechanisms other than this optional substitution one. For example, the equation\n\n```[equal [perimeter] [product 2 [sum [length] [width]]]]\n```\n\nmight have been added to the library of known relationships. The difficulty about alternate phrasings for \"two numbers\" could be solved by adding\n\n```[[one of the !word:pluralp] [\"one singular :word]]\n```\n\nto the list of idiomatic substitutions in `idiom`.\n\nNot all the mechanisms are equivalent, however. The \"two numbers\" problem couldn't be solved by adding equations to the library of known relationships, because that phrase appears as part of a larger phrase like \"the sum of two numbers,\" and Student's understanding of the word `sum` doesn't allow it to be part of a variable name. The word `sum` only makes sense to Student in the context of a phrase like ```the sum of something and something else```. (See procedure `tst.sum`.)\n\n## If All Else Fails\n\nSometimes Student fails to solve a problem because the problem is beyond either its linguistic capability or its algebraic capability. For example, Student doesn't know how to solve quadratic equations. But sometimes a problem that Student could solve in principle stumps it because it happens to lack a particular piece of common knowledge. When a situation like that arises, Student is capable of asking the user for help (`student2`).\n\n```? student :ship\n\nThe problem to be solved is\n\nThe gross weight of a ship is 20000 tons. If its net weight is 15000\ntons, what is the weight of the ships cargo?\n\nWith mandatory substitutions the problem is\n\nThe gross weight numof a ship is 20000 tons . If its net weight is 15000\ntons , what is the weight numof the ships cargo ?\n\nThe simple sentences are\n\nThe gross weight numof a ship is 20000 tons .\n\nIts net weight is 15000 tons .\n\nWhat is the weight numof the ships cargo ?\n\nThe equations to be solved are\n\nEqual [gross weight of ship] [product 20000 [tons]]\n\nEqual [its net weight] [product 15000 [tons]]\n\nThe equations were insufficient to find a solution.\n\nDo you know any more relationships among these variables?\n\nWeight of ships cargo\n\nIts net weight\n\nTons\n\nGross weight of ship\n\nThe weight of a ships cargo is the gross weight minus the net weight\n\nAssuming that\n[net weight] is equal to [its net weight]\n\nAssuming that\n[gross weight] is equal to [gross weight of ship]\n\nThe weight of the ships cargo is 5000 tons\n\nThe problem is solved.\n```\n\n## Limitations of Pattern Matching\n\nStudent relies on certain stereotyped forms of sentences in the problems it solves. It's easy to make up problems that will completely bewilder it:\n\n```Suppose you have 14 jelly beans. You give 2 each to Tom, Dick, and\nHarry. How many do you have left?\n```\n\nThe first mistake Student makes is that it thinks the word ``` and``` following a comma separates two clauses; it generates simple sentences\n\n```You give 2 each to Tom , Dick .\n\nHarry .\n```\n\nThis is quite a fundamental problem; Student's understanding of the difference between a phrase and a clause is extremely primitive and prone to error. Adding another pattern won't solve this one; the trouble is that Student pays no attention to the words in between the key words like `and`.\n\nThere are several other difficulties with this problem, some worse than others. Student doesn't recognize the word `suppose` as having a special function in the sentence, so it makes up a noun phrase ```suppose you``` just like `the russians`. This could be fixed with an idiomatic substitution that just ignored `suppose`. Another relatively small problem is that the sentence starting `how many` doesn't say how many of what; Student needs a way to understand that the relevant noun phrase is `jelly beans` and not, for example, `Tom`. The words `give` (representing subtraction) and `each` (representing counting a set and then multiplying) have special mathematical meanings comparable to ``` percent less```. A much more subtle problem in knowledge representation is that in this problem there are two different quantities that could be called `the number of jelly beans you have`: the number you have at the beginning of the problem and the number you have at the end. Student has a limited understanding of this passage-of-time difficulty when it's doing an age problem, but not in general.\n\nHow many more difficulties can you find in this problem? For how many of them can you invent improvements to Student to get around them?\n\nSome difficulties seem to require a \"more of the same\" strategy: adding some new patterns to Student that are similar to the ones already there. Other difficulties seem to require a more fundamental redesign. Can that redesign be done using a pattern matcher as the central tool, or are more powerful tools needed? How powerful is pattern matching, anyway?\n\nAnswering questions like these is the job of automata theory. From that point of view, the answer is that it depends exactly what you mean by \"pattern matching.\" The pattern matcher used in Student is equivalent to a finite-state machine. The important thing to note about the patterns used in Student is that they only apply predicates to one word at a time, not to groups of words. In other words, they don't use the `@` quantifier. Here is a typical `student` pattern:\n\n```[^ what !:in [is are] #one !:dlm]\n```\n\nFor the purposes of this discussion, you can ignore the fact that the pattern matcher can set variables to remember which words matched each part of the pattern. In comparing a pattern matcher to a finite-state machine, the question we're asking is what categories of strings can the pattern matcher accept. This particular pattern is equivalent to the following machine:", null, "The arrow that I've labeled dlm is actually several arrows connecting the same states, one for each symbol that the predicate `dlm` accepts, i.e., period, question mark, and semicolon. Similarly, the arrows labeled any are followed for any symbol at all. This machine is nondeterministic, but you'll recall that that doesn't matter; we can turn it into a deterministic one if necessary.\n\nTo be sure you understand the equivalence of patterns and finite-state machines, see if you can draw a machine equivalent to this pattern:\n\n```[I see !:in [the a an] ?:numberp &:adjective !:noun #:adverb]\n```\n\nThis pattern uses all the quantifiers that test words one at a time.\n\nIf these patterns are equivalent to finite-state machines, you'd expect them to have trouble recognizing sentences that involve embedding of clauses within clauses, since these pose the same problem as keeping track of balancing of parentheses. For example, a sentence like \"The book that the boy whom I saw yesterday was reading is interesting\" would strain the capabilities of a finite-state machine. (As in the case of parentheses, we could design a FSM that could handle such sentences up to some fixed depth of embedding, but not one that could handle arbitrarily deep embedding.)\n\n## Context-Free Languages\n\nIf we allow the use of the `@` quantifier in patterns, and if the predicates used to test substrings of the sentences are true functions without side effects, then the pattern matcher is equivalent to an RTN or a production rule grammar. What makes an RTN different from a finite-state machine is that the former can include arrows that match several symbols against another (or the same) RTN. Equivalently, the `@` quantifier matches several symbols against another (or the same) pattern.\n\nA language that can be represented by an RTN is called a context-free language. The reason for the name is that in such a language a given string consistently matches or doesn't match a given predicate regardless of the rest of the sentence. That's the point of what I said just above about side effects; the output from a test predicate can't depend on anything other than its input. Pascal is a context-free language because\n\n```this := that\n```\n\nis always an assignment statement regardless of what other statements might be in the program with it.\n\nWhat isn't a context-free language? The classic example in automata theory is the language consisting of the strings\n\n```abc\naabbcc\naaabbbccc\naaaabbbbcccc\n```\n\nand so on, with the requirement that the number of `a`s be equal to the number of `b`s and also equal to the number of `c`s. That language can't be represented as RTNs or production rules. (Try it. Don't confuse it with the language that accepts any number of `a`s followed by any number of `b`s and so on; even a finite-state machine can represent that one. The equal number requirement is important.)\n\nThe classic formal system that can represent any language for which there are precise rules is the Turing machine. Its advantage over the RTN is precisely that it can \"jump around\" in its memory, looking at one part while making decisions about another part.\n\nThere is a sharp theoretical boundary between context-free and context-sensitive languages, but in practice the boundary is sometimes fuzzy. Consider again the case of Pascal and that assignment statement. I said that it's recognizably an assignment statement because it matches a production rule like\n\n```assignment : identifier := expression\n```\n\n(along with a bunch of other rules that determine what qualifies as an expression). But that production rule doesn't really express all the requirements for a legal Pascal assignment statement. For example, the identifier and the expression must be of the same type. The actual Pascal compiler (any Pascal compiler, not just mine) includes instructions that represent the formal grammar plus extra instructions that represent the additional requirements.\n\nThe type agreement rule is an example of context sensitivity. The types of the relevant identifiers were determined in `var` declarations earlier in the program; those declarations are part of what determines whether the given string of symbols is a legal assignment.\n\n## Augmented Transition Networks\n\nOne could create a clean formal description of Pascal, type agreement rules and all, by designing a Turing machine to accept Pascal programs. However, Turing machines aren't easy to work with for any practical problem. It's much easier to set up a context-free grammar for Pascal and then throw in a few side effects to handle the context-sensitive aspects of the language.\n\nMuch the same is true of English. It's possible to set up an RTN (or a production rule grammar) for noun phrases, for example, and another one for verb phrases. It's tempting then to set up an RTN for a sentence like this:", null, "This machine captures some, but not all, of the rules of English. It's true that a sentence requires a noun phrase (the subject) and a verb phrase (the predicate). But there are agreement rules for person and number (I run but he runs) analogous to the type agreement rules of Pascal.\n\nSome artificial intelligence researchers, understanding all this, parse English sentences using a formal description called an augmented transition network (ATN). An ATN is just like an RTN except that each transition arrow can have associated with it not only the name of a symbol or another RTN but also some conditions that must be met in order to follow the arrow and some actions that the program should take if the arrow is followed. For example, we could turn the RTN just above into an ATN by adding an action to the first arrow saying \"store the number (singular or plural) of the noun phrase in the variable ``` number```\" and adding a condition to the second arrow saying \"the number of the verb phrase must be equal to the variable `number`.\"\n\nSubject-predicate agreement is not the only rule in English grammar best expressed as a side effect in a transition network. Below is an ATN for noun phrases taken from Language as a Cognitive Process, Volume 1: Syntax by Terry Winograd (page 598). I'm not going to attempt to explain the notation or the detailed rules here, but just to give one example, the condition labeled \"h16p\" says that the transition for apostrophe-s can be followed if the head of the phrase is an ordinary noun (\"the book's\") but not if it's a pronoun (\"you's\").", null, "The ATN is equivalent in power to a Turing machine; there is no known mechanism that is more flexible in carrying out algorithms. The flexibility has a cost, though. The time required to parse a string with an ATN is not bounded by a polynomial function. (Remember, the time for an RTN is O(n3).) It can easily be exponential, O(2n). One reason is that a context-sensitive procedure can't be subject to memoization. If two invocations of the same procedure with the same inputs can give different results because of side effects, it does no good to remember what result we got the last time. Turning an ATN into a practical program is often possible, but not a trivial task.\n\nIn thinking about ATNs we've brought together most of the topics in this book: formal systems, algorithms, language parsing, and artificial intelligence. Perhaps that's a good place to stop.\n\n## Program Listing\n\n```to student :prob\nsay [The problem to be solved is] :prob\nmake \"prob map.se [depunct ?] :prob\nlocalmake \"orgprob :prob\nstudent1 :prob ~\n[[[the perimeter of ! rectangle]\n[twice the sum of the length and width of the rectangle]]\n[[two numbers] [one of the numbers and the other number]]\n[[two numbers] [one number and the other number]]]\nend\n\nto student1 :prob :idioms\nlocal [simsen shelf aunits units wanted ans var lasteqn\nref eqt1 beg end idiom reply]\nmake \"prob idioms :prob\nif match [^ two numbers #] :prob ~\n[make \"idiom find [match (sentence \"^beg first ? \"#end) :orgprob] :idioms ~\ntryidiom stop]\nwhile [match [^beg the the #end] :prob] [make \"prob (sentence :beg \"the :end)]\nsay [With mandatory substitutions the problem is] :prob\nifelse match [# @:in [[as old as] [age] [years old]] #] :prob ~\n[ageprob] [make \"simsen bracket :prob]\nlsay [The simple sentences are] :simsen\nforeach [aunits wanted ans var lasteqn ref units] [make ? []]\nmake \"shelf filter [not emptyp ?] map.se [senform ?] :simsen\nlsay [The equations to be solved are] :shelf\nmake \"units remdup :units\nif trysolve :shelf :wanted :units :aunits [print [The problem is solved.] stop]\nmake \"eqt1 remdup geteqns :var\nif not emptyp :eqt1 [lsay [Using the following known relationships] :eqt1]\nstudent2 :eqt1\nend\n\nto student2 :eqt1\nmake \"var remdup sentence (map.se [varterms ?] :eqt1) :var\nmake \"eqt1 sentence :eqt1 vartest :var\nif not emptyp :eqt1 ~\n[if trysolve (sentence :shelf :eqt1) :wanted :units :aunits\n[print [The problem is solved.] stop]]\nmake \"idiom find [match (sentence \"^beg first ? \"#end) :orgprob] :idioms\nif not emptyp :idiom [tryidiom stop]\nlsay [Do you know any more relationships among these variables?] :var\nif equalp :reply [no] [print [] print [I can't solve this problem.] stop]\nif not match [^beg is #end] :reply [print [I don't understand that.] stop]\nmake \"shelf sentence :shelf :eqt1\nstudent2 (list (list \"equal opform :beg opform :end))\nend\n\n;; Mandatory substitutions\n\nto depunct :word\nif emptyp :word [output []]\nif equalp first :word \"\\$ [output sentence \"\\$ depunct butfirst :word]\nif equalp last :word \"% [output sentence depunct butlast :word \"percent]\nif memberp last :word [. ? \|;\| ,] [output sentence depunct butlast :word last :word]\nif emptyp butfirst :word [output :word]\nif equalp last2 :word \"'s [output sentence depunct butlast butlast :word \"s]\noutput :word\nend\n\nto last2 :word\noutput word (last butlast :word) (last :word)\nend\n\nto idioms :sent\nlocal \"number\noutput changes :sent ~\n[[[the sum of] [\"sum]] [[square of] [\"square]] [[of] [\"numof]]\n[[how old] [\"what]] [[is equal to] [\"is]]\n[[years younger than] [[less than]]] [[years older than] [\"plus]]\n[[percent less than] [\"perless]] [[less than] [\"lessthan]]\n[[these] [\"the]] [[more than] [\"plus]]\n[[first two numbers] [[the first number and the second number]]]\n[[three numbers]\n[[the first number and the second number and the third number]]]\n[[one half] [0.5]] [[twice] [[2 times]]]\n[[\\$ !number] [sentence :number \"dollars]] [[consecutive to] [[1 plus]]]\n[[larger than] [\"plus]] [[per cent] [\"percent]] [[how many] [\"howm]]\n[[is multiplied by] [\"ismulby]] [[is divided by] [\"isdivby]]\n[[multiplied by] [\"times]] [[divided by] [\"divby]]]\nend\n\nto changes :sent :list\nlocalmake \"keywords map.se [findkey first ?] :list\noutput changes1 :sent :list :keywords\nend\n\nto findkey :pattern\nif equalp first :pattern \"!:in [output first butfirst :pattern]\nif equalp first :pattern \"?:in [output sentence (item 2 :pattern) (item 3 :pattern)]\noutput first :pattern\nend\n\nto changes1 :sent :list :keywords\nif emptyp :sent [output []]\nif memberp first :sent :keywords [output changes2 :sent :list :keywords]\noutput fput first :sent changes1 butfirst :sent :list :keywords\nend\n\nto changes2 :sent :list :keywords\nchanges3 :list :list\noutput fput first :sent changes1 butfirst :sent :list :keywords\nend\n\nto changes3 :biglist :nowlist\nif emptyp :nowlist [stop]\nif changeone first :nowlist [changes3 :biglist :biglist stop]\nchanges3 :biglist butfirst :nowlist\nend\n\nto changeone :change\nlocal \"end\nif not match (sentence first :change [#end]) :sent [output \"false]\nmake \"sent run (sentence \"sentence last :change \":end)\noutput \"true\nend\n\n;; Division into simple sentences\n\nto bracket :prob\noutput bkt1 finddelim :prob\nend\n\nto finddelim :sent\noutput finddelim1 :sent [] []\nend\n\nto finddelim1 :in :out :simples\nif emptyp :in ~\n[ifelse emptyp :out [output :simples] [output lput (sentence :out \".) :simples]]\nif dlm first :in ~\n[output finddelim1 (nocap butfirst :in) []\n(lput (sentence :out first :in) :simples)]\noutput finddelim1 (butfirst :in) (sentence :out first :in) :simples\nend\n\nto nocap :words\nif emptyp :words [output []]\nif personp first :words [output :words]\noutput sentence (lowercase first :words) butfirst :words\nend\n\nto bkt1 :problist\nlocal [first word rest]\nif emptyp :problist [output []]\nif not memberp \", first :problist ~\n[output fput first :problist bkt1 butfirst :problist]\nif match [if ^first , !word:qword #rest] first :problist ~\n[output bkt1 fput (sentence :first \".)\nfput (sentence :word :rest) butfirst :problist]\nif match [^first , and #rest] first :problist ~\n[output fput (sentence :first \".) (bkt1 fput :rest butfirst :problist)]\noutput fput first :problist bkt1 butfirst :problist\nend\n\n;; Age problems\n\nto ageprob\nlocal [beg end sym who num subj ages]\nwhile [match [^beg as old as #end] :prob] [make \"prob sentence :beg :end]\nwhile [match [^beg years old #end] :prob] [make \"prob sentence :beg :end]\nwhile [match [^beg will be when #end] :prob] ~\n[make \"sym gensym\nmake \"prob (sentence :beg \"in :sym [years . in] :sym \"years :end)]\nwhile [match [^beg was when #end] :prob] ~\n[make \"sym gensym\nmake \"prob (sentence :beg :sym [years ago .] :sym [years ago] :end)]\nwhile [match [^beg !who:personp will be in !num years #end] :prob] ~\n[make \"prob (sentence :beg :who [s age in] :num \"years #end)]\nwhile [match [^beg was #end] :prob] [make \"prob (sentence :beg \"is :end)]\nwhile [match [^beg will be #end] :prob] [make \"prob (sentence :beg \"is :end)]\nwhile [match [^beg !who:personp is now #end] :prob] ~\n[make \"prob (sentence :beg :who [s age now] :end)]\nwhile [match [^beg !num years from now #end] :prob] ~\n[make \"prob (sentence :beg \"in :num \"years :end)]\nmake \"prob ageify :prob\nifelse match [^ !who:personp ^end s age #] :prob ~\n[make \"subj sentence :who :end] [make \"subj \"someone]\nmake \"prob agepron :prob\nmake \"end :prob\nmake \"ages []\nwhile [match [^ !who:personp ^beg age #end] :end] ~\n[push \"ages (sentence \"and :who :beg \"age)]\nmake \"ages butfirst reduce \"sentence remdup :ages\nwhile [match [^beg their ages #end] :prob] [make \"prob (sentence :beg :ages :end)]\nmake \"simsen map [agesen ?] bracket :prob\nend\n\nto ageify :sent\nif emptyp :sent [output []]\nif not personp first :sent [output fput first :sent ageify butfirst :sent]\ncatch \"error [if equalp first butfirst :sent \"s\n[output fput first :sent ageify butfirst :sent]]\noutput (sentence first :sent [s age] ageify butfirst :sent)\nend\n\nto agepron :sent\nif emptyp :sent [output []]\nif not pronoun first :sent [output fput first :sent agepron butfirst :sent]\nif posspro first :sent [output (sentence :subj \"s agepron butfirst :sent)]\noutput (sentence :subj [s age] agepron butfirst :sent)\nend\n\nto agesen :sent\nlocal [when rest num]\nmake \"when []\nif match [in !num years #rest] :sent ~\n[make \"when sentence \"pluss :num make \"sent :rest]\nif match [!num years ago #rest] :sent ~\n[make \"when sentence \"minuss :num make \"sent :rest]\noutput agewhen :sent\nend\n\nto agewhen :sent\nif emptyp :sent [output []]\nif not equalp first :sent \"age [output fput first :sent agewhen butfirst :sent]\nif match [in !num years #rest] butfirst :sent ~\n[output (sentence [age pluss] :num agewhen :rest)]\nif match [!num years ago #rest] butfirst :sent ~\n[output (sentence [age minuss] :num agewhen :rest)]\nif equalp \"now first butfirst :sent ~\n[output sentence \"age agewhen butfirst butfirst :sent]\noutput (sentence \"age :when agewhen butfirst :sent)\nend\n\n;; Translation from sentences into equations\n\nto senform :sent\nmake \"lasteqn senform1 :sent\noutput :lasteqn\nend\n\nto senform1 :sent\nlocal [one two verb1 verb2 stuff1 stuff2 factor]\nif emptyp :sent [output []]\nif match [^ what are ^one and ^two !:dlm] :sent ~\n[output fput (qset :one) (senform (sentence [what are] :two \"?))]\nif match [^ what !:in [is are] #one !:dlm] :sent ~\n[output (list qset :one)]\nif match [^ howm !one is #two !:dlm] :sent ~\n[push \"aunits (list :one) output (list qset :two)]\nif match [^ howm ^one do ^two have !:dlm] :sent ~\n[output (list qset (sentence [the number of] :one :two \"have))]\nif match [^ howm ^one does ^two have !:dlm] :sent ~\n[output (list qset (sentence [the number of] :one :two \"has))]\nif match [^ find ^one and #two] :sent ~\n[output fput (qset :one) (senform sentence \"find :two)]\nif match [^ find #one !:dlm] :sent [output (list qset :one)]\nmake \"sent filter [not article ?] :sent\nif match [^one ismulby #two] :sent ~\n[push \"ref (list \"product opform :one opform :two) output []]\nif match [^one isdivby #two] :sent ~\n[push \"ref (list \"quotient opform :one opform :two) output []]\nif match [^one is increased by #two] :sent ~\n[push \"ref (list \"sum opform :one opform :two) output []]\nif match [^one is #two] :sent ~\n[output (list (list \"equal opform :one opform :two))]\nif match [^one !verb1:verb ^factor as many ^stuff1 as\n^two !verb2:verb ^stuff2 !:dlm] ~\n:sent ~\n[if emptyp :stuff2 [make \"stuff2 :stuff1]\noutput (list (list \"equal ~\nopform (sentence [the number of] :stuff1 :one :verb1) ~\nopform (sentence :factor [the number of] :stuff2 :two :verb2)))]\nif match [^one !verb1:verb !factor:numberp #stuff1 !:dlm] :sent ~\n[output (list (list \"equal ~\nopform (sentence [the number of] :stuff1 :one :verb1) ~\nopform (list :factor)))]\nsay [This sentence form is not recognized:] :sent\nthrow \"error\nend\n\nto qset :sent\nlocalmake \"opform opform filter [not article ?] :sent\nif not operatorp first :opform ~\n[queue \"wanted :opform queue \"ans list :opform oprem :sent output []]\nlocalmake \"gensym gensym\nqueue \"wanted :gensym\nqueue \"ans list :gensym oprem :sent\noutput (list \"equal :gensym opform (filter [not article ?] :sent))\nend\n\nto oprem :sent\noutput map [ifelse equalp ? \"numof [\"of] [?]] :sent\nend\n\nto opform :expr\nlocal [left right op]\nif match [^left !op:op2 #right] :expr [output optest :op :left :right]\nif match [^left !op:op1 #right] :expr [output optest :op :left :right]\nif match [^left !op:op0 #right] :expr [output optest :op :left :right]\nif match [#left !:dlm] :expr [make \"expr :left]\noutput nmtest filter [not article ?] :expr\nend\n\nto optest :op :left :right\noutput run (list (word \"tst. :op) :left :right)\nend\n\nto tst.numof :left :right\nif numberp last :left [output (list \"product opform :left opform :right)]\noutput opform (sentence :left \"of :right)\nend\n\nto tst.divby :left :right\noutput (list \"quotient opform :left opform :right)\nend\n\nto tst.tothepower :left :right\noutput (list \"expt opform :left opform :right)\nend\n\nto expt :num :pow\nif :pow < 1 [output 1]\noutput :num expt :num :pow - 1\nend\n\nto tst.per :left :right\noutput (list \"quotient ~\nopform :left ~\nopform (ifelse numberp first :right [:right] [fput 1 :right]))\nend\n\nto tst.lessthan :left :right\noutput opdiff opform :right opform :left\nend\n\nto opdiff :left :right\noutput (list \"sum :left (list \"minus :right))\nend\n\nto tst.minus :left :right\nif emptyp :left [output list \"minus opform :right]\noutput opdiff opform :left opform :right\nend\n\nto tst.minuss :left :right\noutput tst.minus :left :right\nend\n\nto tst.sum :left :right\nlocal [one two three]\nif match [^one and ^two and #three] :right ~\n[output (list \"sum opform :one opform (sentence \"sum :two \"and :three))]\nif match [^one and #two] :right ~\n[output (list \"sum opform :one opform :two)]\nsay [sum used wrong:] :right\nthrow \"error\nend\n\nto tst.squared :left :right\noutput list \"square opform :left\nend\n\nto tst.difference :left :right\nlocal [one two]\nif match [between ^one and #two] :right [output opdiff opform :one opform :two]\nsay [Incorrect use of difference:] :right\nthrow \"error\nend\n\nto tst.plus :left :right\noutput (list \"sum opform :left opform :right)\nend\n\nto tst.pluss :left :right\noutput tst.plus :left :right\nend\n\nto square :x\noutput :x * :x\nend\n\nto tst.square :left :right\noutput list \"square opform :right\nend\n\nto tst.percent :left :right\nif not numberp last :left ~\n[say [Incorrect use of percent:] :left throw \"error]\noutput opform (sentence butlast :left ((last :left) / 100) :right)\nend\n\nto tst.perless :left :right\nif not numberp last :left ~\n[say [Incorrect use of percent:] :left throw \"error]\noutput (list \"product ~\n(opform sentence butlast :left ((100 - (last :left)) / 100)) ~\nopform :right)\nend\n\nto tst.times :left :right\nif emptyp :left [say [Incorrect use of times:] :right throw \"error]\noutput (list \"product opform :left opform :right)\nend\n\nto nmtest :expr\nif match [& !:numberp #] :expr [say [argument error:] :expr throw \"error]\nif and (equalp first :expr 1) (1 < count :expr) ~\n[make \"expr (sentence 1 plural (first butfirst :expr) (butfirst butfirst :expr))]\nif and (numberp first :expr) (1 < count :expr) ~\n[push \"units (list first butfirst :expr) ~\noutput (list \"product (first :expr) (opform butfirst :expr))]\nif numberp first :expr [output first :expr]\nif memberp \"this :expr [output this :expr]\nif not memberp :expr :var [push \"var :expr]\noutput :expr\nend\n\nto this :expr\nif not emptyp :ref [output pop \"ref]\nif not emptyp :lasteqn [output first butfirst last :lasteqn]\nif equalp first :expr \"this [make \"expr butfirst :expr]\npush \"var :expr\noutput :expr\nend\n\n;; Solving the equations\n\nto trysolve :shelf :wanted :units :aunits\nlocal \"solution\nmake \"solution solve :wanted :shelf (ifelse emptyp :aunits [:units] [:aunits])\nend\n\nto solve :wanted :eqt :terms\noutput solve.reduce solver :wanted :terms [] [] \"insufficient\nend\n\nto solve.reduce :soln\nif emptyp :soln [output []]\nif wordp :soln [output :soln]\nif emptyp butfirst :soln [output :soln]\nlocal \"part\nmake \"part solve.reduce butfirst :soln\noutput fput (list (first first :soln) (subord last first :soln :part)) :part\nend\n\nto solver :wanted :terms :alis :failed :err\nlocal [one result restwant]\nif emptyp :wanted [output :err]\nmake \"one solve1 (first :wanted) ~\n(sentence butfirst :wanted :failed :terms) ~\n:alis :eqt [] \"insufficient\nif wordp :one ~\n[output solver (butfirst :wanted) :terms :alis (fput first :wanted :failed) :one]\nmake \"restwant (sentence :failed butfirst :wanted)\nif emptyp :restwant [output :one]\nmake \"result solver :restwant :terms :one [] \"insufficient\nif listp :result [output :result]\noutput solver (butfirst :wanted) :terms :alis (fput first :wanted :failed) :one\nend\n\nto solve1 :x :terms :alis :eqns :failed :err\nlocal [thiseq vars extras xterms others result]\nif emptyp :eqns [output :err]\nmake \"thiseq subord (first :eqns) :alis\nmake \"vars varterms :thiseq\nif not memberp :x :vars ~\n[output solve1 :x :terms :alis (butfirst :eqns) (fput first :eqns :failed) :err]\nmake \"xterms fput :x :terms\nmake \"extras setminus :vars :xterms\nmake \"eqt remove (first :eqns) :eqt\nif not emptyp :extras ~\n[make \"others solver :extras :xterms :alis [] \"insufficient\nifelse wordp :others\n[make \"eqt sentence :failed :eqns\noutput solve1 :x :terms :alis (butfirst :eqns)\n(fput first :eqns :failed) :others]\n[make \"alis :others\nmake \"thiseq subord (first :eqns) :alis]]\nmake \"result solveq :x :thiseq\nif listp :result [output lput :result :alis]\nmake \"eqt sentence :failed :eqns\noutput solve1 :x :terms :alis (butfirst :eqns) (fput first :eqns :failed) :result\nend\n\nto solveq :var :eqn\nlocal [left right]\nmake \"left first butfirst :eqn\nifelse occvar :var :left ~\n[make \"right last :eqn] [make \"right :left make \"left last :eqn]\noutput solveq1 :left :right \"true\nend\n\nto solveq1 :left :right :bothtest\nif :bothtest [if occvar :var :right [output solveqboth :left :right]]\nif equalp :left :var [output list :var :right]\nif wordp :left [output \"unsolvable]\nlocal \"oper\nmake \"oper first :left\nif memberp :oper [sum product minus quotient] [output run (list word \"solveq. :oper)]\noutput \"unsolvable\nend\n\nto solveqboth :left :right\nif not equalp first :right \"sum [output solveq1 (subterm :left :right) 0 \"false]\noutput solveq.rplus :left butfirst :right []\nend\n\nto solveq.rplus :left :right :newright\nif emptyp :right [output solveq1 :left (simone \"sum :newright) \"false]\nif occvar :var first :right ~\n[output solveq.rplus (subterm :left first :right) butfirst :right :newright]\noutput solveq.rplus :left butfirst :right (fput first :right :newright)\nend\n\nto solveq.sum\nif emptyp butfirst butfirst :left [output solveq1 first butfirst :left :right \"true]\noutput solveq.sum1 butfirst :left :right []\nend\n\nto solveq.sum1 :left :right :newleft\nif emptyp :left [output solveq.sum2]\nif occvar :var first :left ~\n[output solveq.sum1 butfirst :left :right fput first :left :newleft]\noutput solveq.sum1 butfirst :left (subterm :right first :left) :newleft\nend\n\nto solveq.sum2\nif emptyp butfirst :newleft [output solveq1 first :newleft :right \"true]\nlocalmake \"factor factor :newleft :var\nif equalp first :factor \"unknown [output \"unsolvable]\nif equalp last :factor 0 [output \"unsolvable]\noutput solveq1 first :factor (divterm :right last :factor) \"true\nend\n\nto solveq.minus\noutput solveq1 (first butfirst :left) (minusin :right) \"false\nend\n\nto solveq.product\noutput solveq.product1 :left :right\nend\n\nto solveq.product1 :left :right\nif emptyp butfirst butfirst :left [output solveq1 (first butfirst :left) :right \"true]\nif not occvar :var first butfirst :left ~\n[output solveq.product1 (fput \"product butfirst butfirst :left)\n(divterm :right first butfirst :left)]\nlocalmake \"rest simone \"product butfirst butfirst :left\nif occvar :var :rest [output \"unsolvable]\noutput solveq1 (first butfirst :left) (divterm :right :rest) \"false\nend\n\nto solveq.quotient\nif occvar :var first butfirst :left ~\n[output solveq1 (first butfirst :left) (simtimes list :right last :left) \"true]\noutput solveq1 (simtimes list :right last :left) (first butfirst :left) \"true\nend\n\nlocalmake \"den last :fract\nif not equalp first :addends \"quotient ~\n[output simdiv list (simone \"sum\n(remop \"sum list (distribtimes (list :addends) :den)\nfirst butfirst :fract))\n:den]\nif equalp :den last :addends ~\n[output simdiv (simplus list (first butfirst :fract) (first butfirst :addends))\n:den]\nlocalmake \"lowterms simdiv list :den last :addends\noutput simdiv list (simplus (simtimes list first butfirst :fract last :lowterms)\nfirst butfirst :lowterms)) ~\n(simtimes list first butfirst :lowterms last :addends)\nend\n\nto distribtimes :trms :multiplier\noutput simplus map [simtimes (list ? :multiplier)] :trms\nend\n\nto distribx :expr\nlocal [oper args]\nif emptyp :expr [output :expr]\nmake \"oper first :expr\nif not operatorp :oper [output :expr]\nmake \"args map [distribx ?] butfirst :expr\nif reduce \"and map [numberp ?] :args [output run (sentence [(] :oper :args [)])]\nif equalp :oper \"sum [output simplus :args]\nif equalp :oper \"minus [output minusin first :args]\nif equalp :oper \"product [output simtimes :args]\nif equalp :oper \"quotient [output simdiv :args]\noutput fput :oper :args\nend\n\nto divterm :dividend :divisor\nif equalp :dividend 0 [output 0]\noutput simdiv list :dividend :divisor\nend\n\nto factor :exprs :var\nlocal \"trms\nmake \"trms map [factor1 :var ?] :exprs\nif memberp \"unknown :trms [output fput \"unknown :exprs]\noutput list :var simplus :trms\nend\n\nto factor1 :var :expr\nlocalmake \"negvar minusin :var\nif equalp :var :expr [output 1]\nif equalp :negvar :expr [output -1]\nif emptyp :expr [output \"unknown]\nif equalp first :expr \"product [output factor2 butfirst :expr]\nif not equalp first :expr \"quotient [output \"unknown]\nlocalmake \"dividend first butfirst :expr\nif equalp :var :dividend [output (list \"quotient 1 last :expr)]\nif not equalp first :dividend \"product [output \"unknown]\nlocalmake \"result factor2 butfirst :dividend\nif equalp :result \"unknown [output \"unknown]\noutput (list \"quotient :result last :expr)\nend\n\nto factor2 :trms\nif memberp :var :trms [output simone \"product (remove :var :trms)]\nif memberp :negvar :trms [output minusin simone \"product (remove :negvar :trms)]\noutput \"unknown\nend\n\nif equalp :num 0 [output :rest]\noutput fput :num :rest\nend\n\nto maybemul :num :rest\nif equalp :num 1 [output :rest]\noutput fput :num :rest\nend\n\nto minusin :expr\nif emptyp :expr [output -1]\nif equalp first :expr \"sum [output fput \"sum map [minusin ?] butfirst :expr]\nif equalp first :expr \"minus [output last :expr]\nif memberp first :expr [product quotient] ~\n[output fput first :expr\n(fput (minusin first butfirst :expr) butfirst butfirst :expr)]\nif numberp :expr [output minus :expr]\noutput list \"minus :expr\nend\n\nto occvar :var :expr\nif emptyp :expr [output \"false]\nif wordp :expr [output equalp :var :expr]\nif operatorp first :expr [output not emptyp find [occvar :var ?] butfirst :expr]\noutput equalp :var :expr\nend\n\nto remfactor :num :den\nforeach butfirst :num [remfactor1 ?]\noutput (list \"quotient (simone \"product butfirst :num) (simone \"product butfirst :den))\nend\n\nto remfactor1 :expr\nlocal \"neg\nif memberp :expr :den ~\n[make \"num remove :expr :num make \"den remove :expr :den stop]\nmake \"neg minusin :expr\nif not memberp :neg :den [stop]\nmake \"num remove :expr :num\nmake \"den minusin remove :neg :den\nend\n\nto remop :oper :exprs\noutput map.se [ifelse equalp first ? :oper [butfirst ?] [(list ?)]] :exprs\nend\n\nto simdiv :list\nlocal [num den numop denop]\nmake \"num first :list\nmake \"den last :list\nif equalp :num :den [output 1]\nif numberp :den [output simtimes (list (quotient 1 :den) :num)]\nmake \"numop first :num\nmake \"denop first :den\nif equalp :numop \"quotient ~\n[output simdiv list (first butfirst :num) (simtimes list last :num :den)]\nif equalp :denop \"quotient ~\n[output simdiv list (simtimes list :num last :den) (first butfirst :den)]\nif and equalp :numop \"product equalp :denop \"product [output remfactor :num :den]\nif and equalp :numop \"product memberp :den :num [output remove :den :num]\noutput fput \"quotient :list\nend\n\nto simone :oper :trms\nif emptyp :trms [output ifelse equalp :oper \"product ]\nif emptyp butfirst :trms [output first :trms]\noutput fput :oper :trms\nend\n\nto simplus :exprs\nmake \"exprs remop \"sum :exprs\nlocalmake \"factor [unknown]\ncatch \"simplus ~\n[foreach :terms ~\n[make \"factor (factor :exprs ?) ~\nif not equalp first :factor \"unknown [throw \"simplus]]]\nif not equalp first :factor \"unknown [output fput \"product remop \"product :factor]\nlocalmake \"nums 0\nlocalmake \"nonnums []\nlocalmake \"quick []\ncatch \"simplus [simplus1 :exprs]\nif not emptyp :quick [output :quick]\nif not equalp :nums 0 [push \"nonnums :nums]\noutput simone \"sum :nonnums\nend\n\nto simplus1 :exprs\nif emptyp :exprs [stop]\nsimplus2 first :exprs\nsimplus1 butfirst :exprs\nend\n\nto simplus2 :pos\nlocal \"neg\nmake \"neg minusin :pos\nif numberp :pos [make \"nums sum :pos :nums stop]\nif memberp :neg butfirst :exprs [make \"exprs remove :neg :exprs stop]\nif equalp first :pos \"quotient ~\n[make \"quick (denom :pos (maybeadd :nums sentence :nonnums butfirst :exprs)) ~\nthrow \"simplus]\npush \"nonnums :pos\nend\n\nto simtimes :exprs\nlocal [nums nonnums quick]\nmake \"nums 1\nmake \"nonnums []\nmake \"quick []\ncatch \"simtimes [foreach remop \"product :exprs [simtimes1 ?]]\nif not emptyp :quick [output :quick]\nif equalp :nums 0 [output 0]\nif not equalp :nums 1 [push \"nonnums :nums]\noutput simone \"product :nonnums\nend\n\nto simtimes1 :expr\nif equalp :expr 0 [make \"nums 0 throw \"simtimes]\nif numberp :expr [make \"nums product :expr :nums stop]\nif equalp first :expr \"sum ~\n[make \"quick distribtimes (butfirst :expr)\n(simone \"product maybemul :nums sentence :nonnums ?rest)\nthrow \"simtimes]\nif equalp first :expr \"quotient ~\n[make \"quick\nsimdiv (list (simtimes (list (first butfirst :expr)\n(simone \"product\nmaybemul :nums\nsentence :nonnums ?rest)))\n(last :expr))\nthrow \"simtimes]\npush \"nonnums :expr\nend\n\nto subord :expr :alist\noutput distribx subord1 :expr :alist\nend\n\nto subord1 :expr :alist\nif emptyp :alist [output :expr]\noutput subord (substop (last first :alist) (first first :alist) :expr) ~\n(butfirst :alist)\nend\n\nto substop :val :var :expr\nif emptyp :expr [output []]\nif equalp :expr :var [output :val]\nif not operatorp first :expr [output :expr]\noutput fput first :expr map [substop :val :var ?] butfirst :expr\nend\n\nto subterm :minuend :subtrahend\nif equalp :minuend 0 [output minusin :subtrahend]\nif equalp :minuend :subtrahend [output 0]\noutput simplus (list :minuend minusin :subtrahend)\nend\n\nto varterms :expr\nif emptyp :expr [output []]\nif numberp :expr [output []]\nif wordp :expr [output (list :expr)]\nif operatorp first :expr [output map.se [varterms ?] butfirst :expr]\noutput (list :expr)\nend\n\n;; Printing the solutions\n\nprint []\nif equalp :solution \"unsolvable ~\n[print [Unable to solve this set of equations.] output \"false]\nif equalp :solution \"insufficient ~\n[print [The equations were insufficient to find a solution.] output \"false]\nlocalmake \"gotall \"true\nforeach :ans [if prans ? :solution [make \"gotall \"false]]\nif not :gotall [print [] print [Unable to solve this set of equations.]]\noutput :gotall\nend\n\nto prans :ans :solution\nlocalmake \"result find [equalp first ? first :ans] :solution\nif emptyp :result [output \"true]\nprint (sentence cap last :ans \"is unitstring last :result)\nprint []\noutput \"false\nend\n\nto unitstring :expr\nif numberp :expr [output roundoff :expr]\nif equalp first :expr \"product ~\n[output sentence (unitstring first butfirst :expr)\n(reduce \"sentence butfirst butfirst :expr)]\nif (and (listp :expr)\n(not numberp first :expr)\n(not operatorp first :expr)) ~\n[output (sentence 1 (singular first :expr) (butfirst :expr))]\noutput :expr\nend\n\nto roundoff :num\nif (abs (:num - round :num)) < 0.0001 [output round :num]\noutput :num\nend\n\nto abs :num\noutput ifelse (:num < 0) [-:num] [:num]\nend\n\n;; Using known relationships\n\nto geteqns :vars\noutput map.se [gprop varkey ? \"eqns] :vars\nend\n\nto varkey :var\nlocal \"word\nif match [number of !word #] :var [output :word]\noutput first :var\nend\n\n;; Assuming equality of similar variables\n\nto vartest :vars\nif emptyp :vars [output []]\nlocal [var beg end]\nmake \"var first :vars\noutput (sentence (ifelse match [^beg !:pronoun #end] :var\n[vartest1 :var (sentence :beg \"& :end) butfirst :vars]\n[[]])\n(vartest1 :var (sentence \"# :var \"#) butfirst :vars)\n(vartest butfirst :vars))\nend\n\nto vartest1 :target :pat :vars\noutput map [varequal :target ?] filter [match :pat ?] :vars\nend\n\nto varequal :target :var\nprint []\nprint [Assuming that]\nprint (sentence (list :target) [is equal to] (list :var))\noutput (list \"equal :target :var)\nend\n\n;; Optional substitutions\n\nto tryidiom\nmake \"prob (sentence :beg last :idiom :end)\nwhile [match (sentence \"^beg first :idiom \"#end) :prob] ~\n[make \"prob (sentence :beg last :idiom :end)]\nsay [The problem with an idiomatic substitution is] :prob\nstudent1 :prob (remove :idiom :idioms)\nend\n\n;; Utility procedures\n\nto qword :word\noutput memberp :word [find what howm how]\nend\n\nto dlm :word\noutput memberp :word [. ? \|;\|]\nend\n\nto article :word\noutput memberp :word [a an the]\nend\n\nto verb :word\noutput memberp :word [have has get gets weigh weighs]\nend\n\nto personp :word\noutput memberp :word [Mary Ann Bill Tom Sally Frank father uncle]\nend\n\nto pronoun :word\noutput memberp :word [he she it him her they them his her its]\nend\n\nto posspro :word\noutput memberp :word [his her its]\nend\n\nto op0 :word\noutput memberp :word [pluss minuss squared tothepower per sum difference numof]\nend\n\nto op1 :word\noutput memberp :word [times divby square]\nend\n\nto op2 :word\noutput memberp :word [plus minus lessthan percent perless]\nend\n\nto operatorp :word\noutput memberp :word [sum minus product quotient expt square equal]\nend\n\nto plural :word\nlocalmake \"plural gprop :word \"plural\nif not emptyp :plural [output :plural]\nif not emptyp gprop :word \"sing [output :word]\nif equalp last :word \"s [output :word]\noutput word :word \"s\nend\n\nto singular :word\nlocalmake \"sing gprop :word \"sing\nif not emptyp :sing [output :sing]\nif not emptyp gprop :word \"plural [output :word]\nif equalp last :word \"s [output butlast :word]\noutput :word\nend\n\nto setminus :big :little\noutput filter [not memberp ? :little] :big\nend\n\nto say :herald :text\nprint []\nprint :herald\nprint []\nprint :text\nprint []\nend\n\nto lsay :herald :text\nprint []\nprint :herald\nprint []\nforeach :text [print cap ? print []]\nend\n\nto cap :sent\nif emptyp :sent [output []]\noutput sentence (word uppercase first first :sent butfirst first :sent) ~\nbutfirst :sent\nend\n\n;; The pattern matcher\n\nto match :pat :sen\nif prematch :pat :sen [output rmatch :pat :sen]\noutput \"false\nend\n\nto prematch :pat :sen\nif emptyp :pat [output \"true]\nif listp first :pat [output prematch butfirst :pat :sen]\nif memberp first first :pat [! @ # ^ & ?] [output prematch butfirst :pat :sen]\nif emptyp :sen [output \"false]\nlocalmake \"rest member first :pat :sen\nif not emptyp :rest [output prematch butfirst :pat :rest]\noutput \"false\nend\n\nto rmatch :pat :sen\nlocal [special.var special.pred special.buffer in.list]\nif or wordp :pat wordp :sen [output \"false]\nif emptyp :pat [output emptyp :sen]\nif listp first :pat [output special fput \"!: :pat :sen]\nif memberp first first :pat [? # ! & @ ^] [output special :pat :sen]\nif emptyp :sen [output \"false]\nif equalp first :pat first :sen [output rmatch butfirst :pat butfirst :sen]\noutput \"false\nend\n\nto special :pat :sen\nset.special parse.special butfirst first :pat \"\noutput run word \"match first first :pat\nend\n\nto parse.special :word :var\nif emptyp :word [output list :var \"always]\nif equalp first :word \": [output list :var butfirst :word]\noutput parse.special butfirst :word word :var first :word\nend\n\nto set.special :list\nmake \"special.var first :list\nmake \"special.pred last :list\nif emptyp :special.var [make \"special.var \"special.buffer]\nif memberp :special.pred [in anyof] [set.in]\nif not emptyp :special.pred [stop]\nmake \"special.pred first butfirst :pat\nmake \"pat fput first :pat butfirst butfirst :pat\nend\n\nto set.in\nmake \"in.list first butfirst :pat\nmake \"pat fput first :pat butfirst butfirst :pat\nend\n\nto match!\nif emptyp :sen [output \"false]\nif not try.pred [output \"false]\nmake :special.var first :sen\noutput rmatch butfirst :pat butfirst :sen\nend\n\nto match?\nmake :special.var []\nif emptyp :sen [output rmatch butfirst :pat :sen]\nif not try.pred [output rmatch butfirst :pat :sen]\nmake :special.var first :sen\nif rmatch butfirst :pat butfirst :sen [output \"true]\nmake :special.var []\noutput rmatch butfirst :pat :sen\nend\n\nto match#\nmake :special.var []\noutput #test #gather :sen\nend\n\nto #gather :sen\nif emptyp :sen [output :sen]\nif not try.pred [output :sen]\nmake :special.var lput first :sen thing :special.var\noutput #gather butfirst :sen\nend\n\nto #test :sen\nif rmatch butfirst :pat :sen [output \"true]\nif emptyp thing :special.var [output \"false]\noutput #test2 fput last thing :special.var :sen\nend\n\nto #test2 :sen\nmake :special.var butlast thing :special.var\noutput #test :sen\nend\n\nto match&\noutput &test match#\nend\n\nto &test :tf\nif emptyp thing :special.var [output \"false]\noutput :tf\nend\n\nto match^\nmake :special.var []\noutput ^test :sen\nend\n\nto ^test :sen\nif rmatch butfirst :pat :sen [output \"true]\nif emptyp :sen [output \"false]\nif not try.pred [output \"false]\nmake :special.var lput first :sen thing :special.var\noutput ^test butfirst :sen\nend\n\nto match@\nmake :special.var :sen\noutput @test []\nend\n\nto @test :sen\nif @try.pred [if rmatch butfirst :pat :sen [output \"true]]\nif emptyp thing :special.var [output \"false]\noutput @test2 fput last thing :special.var :sen\nend\n\nto @test2 :sen\nmake :special.var butlast thing :special.var\noutput @test :sen\nend\n\nto try.pred\nif listp :special.pred [output rmatch :special.pred first :sen]\noutput run list :special.pred quoted first :sen\nend\n\nto quoted :thing\nif listp :thing [output :thing]\noutput word \"\" :thing\nend\n\nto @try.pred\nif listp :special.pred [output rmatch :special.pred thing :special.var]\noutput run list :special.pred thing :special.var\nend\n\nto always :x\noutput \"true\nend\n\nto in :word\noutput memberp :word :in.list\nend\n\nto anyof :sen\noutput anyof1 :sen :in.list\nend\n\nto anyof1 :sen :pats\nif emptyp :pats [output \"false]\nif rmatch first :pats :sen [output \"true]\noutput anyof1 :sen butfirst :pats\nend\n\n;; Sample word problems\n\nmake \"ann [Mary is twice as old as Ann was when Mary was as old as Ann is now.\nIf Mary is 24 years old, how old is Ann?]\nmake \"guns [The number of soldiers the Russians have is\none half of the number of guns they have. They have 7000 guns.\nHow many soldiers do they have?]\nmake \"jet [The distance from New York to Los Angeles is 3000 miles.\nIf the average speed of a jet plane is 600 miles per hour,\nfind the time it takes to travel from New York to Los Angeles by jet.]\nmake \"nums [A number is multiplied by 6 . This product is increased by 44 .\nThis result is 68 . Find the number.]\nIf this price is 15 percent less than the marked price, find the marked price.]\nmake \"sally [The sum of Sally's share of some money and Frank's share is \\$4.50.\nSally's share is twice Frank's. Find Frank's and Sally's share.]\nmake \"ship [The gross weight of a ship is 20000 tons.\nIf its net weight is 15000 tons, what is the weight of the ships cargo?]\nmake \"span [If 1 span is 9 inches, and 1 fathom is 6 feet,\nhow many spans is 1 fathom?]\nmake \"sumtwo [The sum of two numbers is 96,\nand one number is 16 larger than the other number. Find the two numbers.]\nmake \"tom [If the number of customers Tom gets is\ntwice the square of 20 per cent of the number of advertisements he runs,\nwhat is the number of customers Tom gets?]\nmake \"uncle [Bill's father's uncle is twice as old as Bill's father.\n2 years from now Bill's father will be 3 times as old as Bill.\nThe sum of their ages is 92 . Find Bill's age.]\n\n;; Initial data base\n\npprop \"distance \"eqns ~\n[[equal [distance] [product [speed] [time]]]\n[equal [distance] [product [gas consumtion] [number of gallons of gas used]]]]\npprop \"feet \"eqns ~\n[[equal [product 1 [feet]] [product 12 [inches]]]\n[equal [product 1 [yards]] [product 3 [feet]]]]\npprop \"feet \"sing \"foot\npprop \"foot \"plural \"feet\npprop \"gallons \"eqns ~\n[[equal [distance] [product [gas consumtion] [number of gallons of gas used]]]]\npprop \"gas \"eqns ~\n[[equal [distance] [product [gas consumtion] [number of gallons of gas used]]]]\npprop \"inch \"plural \"inches\npprop \"inches \"eqns [[equal [product 1 [feet]] [product 12 [inches]]]]\npprop \"people \"sing \"person\npprop \"person \"plural \"people\npprop \"speed \"eqns [[equal [distance] [product [speed] [time]]]]\npprop \"time \"eqns [[equal [distance] [product [speed] [time]]]]\npprop \"yards \"eqns [[equal [product 1 [yards]] [product 3 [feet]]]]\n```\n\nBrian Harvey, `[email protected]`" ]	[ null, "https://people.eecs.berkeley.edu/~bh/csls3.jpg", null, "https://people.eecs.berkeley.edu/~bh/v3ch6/pattern.gif", null, "https://people.eecs.berkeley.edu/~bh/v3ch6/sentence.gif", null, "https://people.eecs.berkeley.edu/~bh/v3ch6/atn.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8683179,"math_prob":0.92158055,"size":103803,"snap":"2022-27-2022-33","text_gpt3_token_len":25602,"char_repetition_ratio":0.18257402,"word_repetition_ratio":0.12632544,"special_character_ratio":0.23662129,"punctuation_ratio":0.13404758,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98558813,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,7,null,6,null,6,null,6,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-16T03:26:50Z\",\"WARC-Record-ID\":\"<urn:uuid:c2313055-b8ff-4a0c-9ca0-b2f6ba14ad24>\",\"Content-Length\":\"116445\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:48b671b7-c631-46d8-b67d-474d39c61efe>\",\"WARC-Concurrent-To\":\"<urn:uuid:291a902b-03f7-4366-8d6b-a9e54942f2e5>\",\"WARC-IP-Address\":\"128.32.244.190\",\"WARC-Target-URI\":\"https://people.eecs.berkeley.edu/~bh/v3ch6/ai.html\",\"WARC-Payload-Digest\":\"sha1:36GJIXRJ7SPCVHZBDEIOIAZ3CDXPNMEO\",\"WARC-Block-Digest\":\"sha1:YJZSYRFTLTKOJQKGTS2UHWN2DRP6RCKS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572220.19_warc_CC-MAIN-20220816030218-20220816060218-00362.warc.gz\"}"}
https://ncatlab.org/nlab/show/quaternionic+projective+space	[ "Contents\n\n# Contents\n\n## Idea\n\nThe quaternionic projective space $\\mathbb{H}P^n$ is the space of right (or left) quaternion lines through the origin in $\\mathbb{H}^{n+1}$, hence the space of equivalence classes $[q_1, \\cdots, q_{n+1}]$ of (n+1)-tuples of quaternions excluding zero, under the equivalence relation given by right (or left) multiplication with non-zero quaternions\n\n$\\mathbb{H}P^n \\;\\coloneqq\\; \\big\\{ [q_1, \\cdots, q_{n+1}] \\big\\} \\;\\coloneqq\\; \\Big( \\big\\{ (q_1, \\cdots, q_{n+1}) \\big\\} \\setminus \\{(0, \\cdots, 0)\\} \\Big) /_{ (q_1, \\cdots, q_{n+1}) \\sim (q_1 q, \\cdots, q_{n+1} q) \\vert q \\neq 0 }$\n\n## Properties\n\n### As a coset space\n\nAs any Grassmannian, quaternion projective space is canonically a coset space, in this case of the quaternion unitary group $Sp(n+1)$ by the central product group Sp(n).Sp(1):\n\n(1)$\\mathbb{H}P^n \\;\\simeq\\; \\frac{ Sp(n+1) }{ Sp(n)\\cdot Sp(1) }$\n\n### As a quaternion-Kähler symmetric space (Wolf space)\n\nBy the coset space-realization (1), quaternion projective space is naturally a quaternion-Kähler manifold which is also a symmetric space. As such it is an example of a Wolf space.\n\n### General\n\nM-theory on the 8-manifold$\\;$ HP2, hence on a quaternion-Kähler manifold of dimension 8 with holonomy Sp(2).Sp(1), is considered in\n• Michael Atiyah, Edward Witten, p. 75 onwards in $M$-Theory dynamics on a manifold of $G_2$-holonomy, Adv. Theor. Math. Phys. 6 (2001) (arXiv:hep-th/0107177)" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89756536,"math_prob":0.9995104,"size":1097,"snap":"2023-40-2023-50","text_gpt3_token_len":265,"char_repetition_ratio":0.14638609,"word_repetition_ratio":0.0,"special_character_ratio":0.22333637,"punctuation_ratio":0.088785045,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9994935,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T12:08:08Z\",\"WARC-Record-ID\":\"<urn:uuid:5c550404-61ea-41e2-9b50-67dd3651078c>\",\"Content-Length\":\"22986\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1be3f7b2-a0ad-47d9-9a19-1d68a1c7b705>\",\"WARC-Concurrent-To\":\"<urn:uuid:c3400e94-2ea7-46f7-86fd-08930d5d6a2b>\",\"WARC-IP-Address\":\"128.2.25.48\",\"WARC-Target-URI\":\"https://ncatlab.org/nlab/show/quaternionic+projective+space\",\"WARC-Payload-Digest\":\"sha1:SOFRQWJUNJGIQ5K7RBYDNCCDMTZKWZ7A\",\"WARC-Block-Digest\":\"sha1:VLBK2VLZPWAT62DYTVL6EXJMNC6V3BQS\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510888.64_warc_CC-MAIN-20231001105617-20231001135617-00159.warc.gz\"}"}
https://greprepclub.com/forum/if-m-and-n-are-the-roots-of-the-equation-10439.html	[ "", null, "It is currently 23 Nov 2020, 08:09", null, "### GMAT Club Daily Prep\n\n#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.\n\nCustomized\nfor You\n\nwe will pick new questions that match your level based on your Timer History\n\nTrack\n\nevery week, we’ll send you an estimated GMAT score based on your performance\n\nPractice\nPays\n\nwe will pick new questions that match your level based on your Timer History\n\n#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.", null, "# If m and n are the roots of the equation", null, "", null, "Question banks Downloads My Bookmarks Reviews Important topics\nAuthor Message\nIntern", null, "Joined: 18 Aug 2018\nPosts: 3\nFollowers: 0\n\nKudos [?]: 2 , given: 0\n\nIf m and n are the roots of the equation [#permalink]\nHI guys!\n\nQuestion:\n\nIf m and n are the roots of the equation $$a^2 - 6a + 8 = 0$$, then find the value of $$(m+n)(m-n)$$\n\nA. 12\n\nB. 8\n\nC. 16\n\nD. 4\n\nE.18\n\nNow, the answer should be 12, because the fractioning of the equation gets me (a-4) (a-2). However, what if I switched places here and put (a-2) (a-4). Then, m = 2 and n=4, thus (2+4) (2-4) = -12. Where am I going wrong? Sorry for this stupid question. And btw, roots of the equation are always positive then, right?\n\nAppreciate any help.\n\nLast edited by Carcass on 18 Aug 2018, 02:03, edited 1 time in total.\nEdited by Carcass\nFounder", null, "", null, "Joined: 18 Apr 2015\nPosts: 13874\nGRE 1: Q160 V160", null, "Followers: 313\n\nKudos [?]: 3671 , given: 12885\n\nRe: If m and n are the roots of the equation [#permalink]\nExpert's post\nHi,\n\nbefore to post a question on the board, please, follow the rule fr posting here https://greprepclub.com/forum/rules-for ... -1083.html and how to format a question here https://greprepclub.com/forum/qq-how-to ... -2357.html\n\nMoreover, post the question under the right section and not in general quant section.\n\nBack to the question.\n\nYou are right m=2 n=4 in which case the answer is -12\n\nBut could be the other way around also: m=4 an n=2 which is 12\n\nNo a good question.\n\nRegards\n_________________", null, "Intern", null, "Joined: 18 Aug 2018\nPosts: 3\nFollowers: 0\n\nKudos [?]: 2 , given: 0\n\nRe: If m and n are the roots of the equation [#permalink]\n1\nKUDOS\nAppreciate the comments, Carcass. Sorry for being careless, and thank you for the clarification on the question.\n\nHowever, I think you made a mistake there. If m=4 and n=2, then (m+n)(m−n) = (4+2) (4-2) = 12. I doesn't make a whole lot of difference though, because I guess we just established that this is not a well-thought-out question.\n\nThanks!", null, "Re: If m and n are the roots of the equation [#permalink] 18 Aug 2018, 04:46\nDisplay posts from previous: Sort by\n\n# If m and n are the roots of the equation", null, "", null, "Question banks Downloads My Bookmarks Reviews Important topics", null, "", null, "Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®." ]	[ null, "https://greprepclub.com/tests/resources/css/images/mobileMenu/greprepHeader.png", null, "https://greprepclub.com/forum/styles/gmatclub_light/theme/images/profile/close.png", null, "https://greprepclub.com/forum/styles/gmatclub_light/theme/images/search/close.png", null, "https://greprepclub.com/forum/styles/gmatclub_light/theme/images/viewtopic/new_topic.png", null, "https://greprepclub.com/forum/styles/gmatclub_light/theme/images/viewtopic/post_reply.png", null, "https://greprepclub.com/forum/images/ranks/rank_phpbb_1.gif", null, "https://greprepclub.com/forum/images/ranks/rank_phpbb_7.gif", null, "https://greprepclub.com/forum/download/file.php", null, "https://greprepclub.com/forum/images/unverified_score.svg", null, "https://greprepclub.com/forum/styles/gmatclub_light/theme/images/viewtopic/left_kudos_icon.png", null, "https://greprepclub.com/forum/images/ranks/rank_phpbb_1.gif", null, "https://greprepclub.com/forum/styles/gmatclub_light/theme/images/viewtopic/posts_bot.png", null, "https://greprepclub.com/forum/styles/gmatclub_light/theme/images/viewtopic/new_topic.png", null, "https://greprepclub.com/forum/styles/gmatclub_light/theme/images/viewtopic/post_reply.png", null, "https://greprepclub.com/forum/cron.php", null, "https://greprepclub.com/forum/styles/gmatclub_light/theme/images/footer/copyright_small.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8495255,"math_prob":0.7637509,"size":1552,"snap":"2020-45-2020-50","text_gpt3_token_len":558,"char_repetition_ratio":0.15116279,"word_repetition_ratio":0.24104235,"special_character_ratio":0.3969072,"punctuation_ratio":0.18498659,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96611315,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-23T16:09:48Z\",\"WARC-Record-ID\":\"<urn:uuid:9ae12d60-1e39-4a12-a26f-a4fb31409ada>\",\"Content-Length\":\"95415\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a85d42e6-48ad-48aa-b32d-a4db2e7c5f77>\",\"WARC-Concurrent-To\":\"<urn:uuid:ef0c0feb-3ca1-4189-8c97-2dc9e6965530>\",\"WARC-IP-Address\":\"198.11.238.99\",\"WARC-Target-URI\":\"https://greprepclub.com/forum/if-m-and-n-are-the-roots-of-the-equation-10439.html\",\"WARC-Payload-Digest\":\"sha1:YDXACCR52NI2DI4CLWMKRFILMADKDKAE\",\"WARC-Block-Digest\":\"sha1:RUQUUVYBZC42JQZCJ7Q6SVBNDNEQGAKH\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141163411.0_warc_CC-MAIN-20201123153826-20201123183826-00374.warc.gz\"}"}
https://www.hindawi.com/journals/jmath/2021/6668711/	[ "#### Abstract\n\nIn order to simplify the complex calculation and solve the difficult solution problems of neutrosophic number optimization models (NNOMs) in the practical production process, this paper presents two methods to solve NNOMs, where Matlab built-in function “fmincon()” and neutrosophic number operations (NNOs) are used in indeterminate environments. Next, the two methods are applied to linear and nonlinear programming problems with neutrosophic number information to obtain the optimal solution of the maximum/minimum objective function under the constrained conditions of practical productions by neutrosophic number optimization programming (NNOP) examples. Finally, under indeterminate environments, the fit optimal solutions of the examples can also be achieved by using some specified indeterminate scales to fulfill some specified actual requirements. The NNOP methods can obtain the feasible and flexible optimal solutions and indicate the advantage of simple calculations in practical applications.\n\n#### 1. Introduction\n\nTraditional inventory models and production planning models involve deterministic constrained functions and/or objective functions in deterministic environments. Nevertheless, uncertainty is nearly universal in real world. Therefore, many uncertain optimization methods were proposed for optimization problems with uncertain variables, interval numbers, stochastic, and fuzzy logics . In many applied fields, such as management, engineering, and design problems, uncertain programming has been broadly carried out so far. In order to obtain the optimal crisp values of the objective function and the optimal feasible crisp solutions of the decision variables, the constrained functions and/or objective functions are usually changed into some crisp or deterministic programming problems in existing uncertain programming approaches. So, the aforementioned transformed methods are not really meaningful indeterminate approaches because the real indeterminate optimization problems can only indicate indeterminate solutions rather than optimal crisp solutions in indeterminate environments. Nevertheless, indeterminate programming problems imply the corresponding indeterminate optimal values of the objective function and indeterminate optimal solutions for the decision variables under indeterminate environments. So, it is necessary to find some fit optimization approaches for dealing with indeterminate programming problems with indeterminate solutions.\n\nSmarandache is a pioneer of indeterminacy theories which provide the new minds to solve indeterminacy problems. He adopted the imaginary value denoted by I and then introduced a neutrosophic number (NN) z = x + yI for x, y ∈ R (R: the set of all real numbers) composed of the determinate part x and indeterminate part yI. As for describing indeterminate and incomplete information, obviously, NNs in the indeterminacy theories are a useful mathematical tool. With the development of indeterminacy theories, NNs were also applied to fault diagnosis [19, 20] and decision making [21, 22] under indeterminate environments.\n\nFurther, thick function or interval function named neutrosophic function, neutrosophic precalculus, and neutrosophic calculus were provided by Smarandache in 2015, where thick function e: SE(S) (E(S) is the set of all interval functions) as the form of an interval function e(x) = [e1(x), e2(x)]. The indeterminate function was applied in engineering problems successfully. For example, Ye et al. [24, 25] and Chen et al. [26, 27] proposed expressions of neutrosophic function and applied NNs in analyzing the joint roughness coefficient. Later, Ye used neutrosophic linear equations of NNs to solve traffic flow problems.\n\nAt present, neutrosophic linguistic numbers, hesitant neutrosophic linguistic numbers, and their aggregation operators were applied to multiattribute decision making .\n\nBut in real situations, affected by each kind subjective and objective reasons, such as absences of precise information judged by decision makers or experts, loss of data, and measurement errors, there exist some indeterminate problems. As for the concepts of NNs, NN functions containing indeterminacy I can represent the indeterminate problems with partial certainty and partial uncertainty under indeterminate environments. Ye and Jiang and Ye introduced NN nonlinear and linear programming models and their preliminary solution methods. However, existing methods for solving complex NN optimization problems imply some difficulty and calculational complexity in their solution process. Inspired by the previous solution methods, this paper first selects the models of practical applications in production process, such as inventory models and production planning models. Then, NN nonlinear and linear mathematical models and their solution methods (Matlab built-in function “fmincon()” and operations of NNs) are built with indeterminacy I as our preliminary application study. Finally, real examples of NN linear programming (NN-LP) and NN nonlinear programming (NN-NP) problems illustrate the feasibility of the proposed methods. The advantage of the proposed methods is that the optimization calculations are simple and effective in practical applications.\n\nThe remainder of this paper is organized as follows. Section 2 depicts some concepts and their operations of NNs. Section 3 first introduces NN-NP problems with an inventory mathematical model and model formation and then uses two methods (Matlab built-in function “fmincon()” and operations of NNs) to solve the NN-NP problems in indeterminate setting. Section 4 presents NN-LP problems with the production planning mathematical model and model formation and then applies two methods regarding the Matlab built-in function “fmincon()” and operations of NNs to solve the solutions in the NN-NP problems and to show the simplicity and effectiveness of the proposed NN-LP methods. Conclusions and future research are provided in Section 5.\n\n#### 2. Mathematical Preliminaries\n\n##### 2.1. Some Concepts and Their Operations of Neutrosophic Numbers (NNs)\n\nThe concept of NN was first proposed by Smarandache [34, 35], which consists of two parts (a determinate part and an indeterminate part). He defined the mathematical expression form z = x + yI for x, y ∈ R, where R represents all real numbers and I is indeterminacy. So, it is conveniently used in indeterminate environments.\n\nFor example, consider that a NN is z = 13 + 5I. Then, its determinate part value is 13 and its indeterminate part value is 5I. When I ∈ [0, 0.5], it is equivalent to z ∈ [13, 15.5] for sure z ≥ 13.\n\nLet z1 = x1 + y1I and z2 = x2 + y2I be two NNs. Then, Smarandache [34, 35] gave their operations of NNs in the following:(1)z1 + z2 = x1 + x2 + (y1 + y2)I.(2)z1z2 = x1x2 + (y1y2)I.(3)z1 × z2 = x1x2 + (x1y2 + x2y1 + y1y2)I, in particular, when z1 = 0 and z2 = I, we get the equation with 0 × I = 0.(4) = (x1 + y1I)2 = + (2x1y1 + )I, in particular, when z1 = I, we get the equation with I^2 = I.(5) for x2 ≠ 0 and x2 ≠ –y2.(6).\n\n##### 2.2. Example\n\nThere are two NNs z1 = 5 + 3I and z2 = 2 + 5I. Then, we can obtain the following results according to the above operations:(1)z 1 + z2 = x1 + x2 + (y1 + y2)I = 5 + 2 + (3 + 5)I = 7 + 8I.(2)z 1z2 = x1x2 + (y1y2)I = 5−2 + (3−5)I = 3−2I.(3)z 1 × z2 = x1x2 + (x1y2 + x2y1 + y1y2)I = 5 × 2 + (5 × 5 + 3 × 2 + 3 × 5)I = 10 + 46I.(4) = (x1 + y1I)2 = + (2x1y1 + )I = 52 + (2 × 5 × 3 + 32)I = 25 + 39I, = (x2 + y2I)2 = + (2x2y2 + )I = 22 + (2 × 2 × 5 + 52)I = 4 + 45I.(5).(6).\n\n#### 3. Neutrosophic Number Nonlinear Programming (NN-NP)\n\n##### 3.1. NN-NP Mathematical Model\n\nThe usual mathematical model of NN-NP is represented in the following form [36, 37]:where : ZnZ(Z is the set of all NNs), and I ∈ [IL, IU](the interval range of I).\n\n##### 3.2. Inventory Mathematical Model \n###### 3.2.1. Notations\n\nThe following notations are used in the inventory model.\n\nThree decision variables:(i)D: demand/unit/time(ii)Qp: production quantity/batch(iii)Cs: setup cost/unit/time\n\nExcept the above cost variable CS, three other cost variables are(i)Cta: total average cost/unit/time(ii)Ctp: total production cost/cycle(iii)Ch: time depending on holding cost/unit/item\n\nOther time and space variables:(i)T: every cycle of length(ii)Q(t): inventory level at time t (t ≥ 0)(iii)S: total storage space area(iv)s0: space area/unit/quantity.\n\n###### 3.2.2. Assumptions\n\nThe inventory model is developed by considering the following assumptions:(i)Only one item is involved in the inventory system.(ii)The replenishment occurs with the near instantaneous response.(iii)The startup time can be ignored.(iv)The demand rate at any time is constant.(v)The total production cost Ctp is related to the setup cost CS and production quantity QP.(vi)Holding cost is the time depended function.\n\n##### 3.3. Model Formation\n\nAs shown in Figure 1, in every time period T, the value of the production quantity Q(t) decreases from Qp to zero. The slope of the line is constant negative D and denoted by .\n\nThe total average cost of the cycle T (denoted by Cta) consists of three sections: setup cost (denoted by C1), holding cost (denoted by C2), and production cost (denoted by C3).\n\nBecause we have the equation Q(t) = QpDt, we obtain the cycle T, .\n\nBased on equations (2) and (3), we obtain the following equation:\n\nSo, the inventory model is constructed as follows:\n\n##### 3.4. Solution Corresponding to Matlab Built-In Function “fmincon()”\n\nIn order to conveniently calculate the solutions, we simplify some parameters and set some constants with history records, where e = 18, f = 5, x = 1, y = 3, s0 = 200, and S= 1100. When we assume D = x1, Cs = x2, and Qp = x3, we can obtain the following mathematical model:\n\nAssume = x1−40.496I, = x2 + 0.058I, and = x3−2I; then, equation (6) can be expressed in the following form:\n\nAccording to the de-neutrosophication technique proposed by Ye and considering I = 0 or 0.5 or 1 as the minimum or moderate or maximum indeterminacy, we can obtain three optimal solutions as follows:(1) = 80.615, = 0.097, = 7.500, and = 4.187 for I = 0.(2) = 60.367, = 0.126, = 6.5, and = 4.343 for I = 0.5.(3) = 40.119, = 0.155, = 5.5, and = 4.525 for I = 1.\n\nClearly, using the indeterminacy I ∈ [0, 1], different optimal results are revealed. The optimal solutions of the optimization problem are = [40.119, 80.615], = [0.097, 0.155], and = [5.5, 7.5] for = [4.187, 4.525], which show the interval optimal ranges.\n\n##### 3.5. Solution Corresponding to Operations of NNs\n\nAccording to the front optimal solutions, we assume = x1 + y1I = 80.615−40.496I, = x2 + y2I = 0.097 + 0.058I, and = x3y3I = 7.500−2I, and then we give the results by equation (9):\n\nBecause = 80.615−40.496I, = 0.097 + 0.058 I, and = 7.500−2I, we can get = 80.615, = −40.496, = 0.097, = 0.058, = 7.5, and = −2. Then, we calculate the three costs, respectively, as follows:\n\nSetup cost:\n\nHolding cost:\n\nProduction cost:\n\nThen, we add the three costs and obtain the total cost Cta with equation (2) as follows:\n\nSo, the calculational results validate that the same solution is obtained by using the two methods of both the Matlab built-in function “fmincon” and the operations of NNs, which are = [40.119, 80.615], = [0.097, 0.155], and = [5.5, 7.5] for = [4.187, 4.525]. We also obtain every cost C1 = [1.047, 1.134], C2 = [ 2.093, 2.262], and C3 = [1.047, 1.129], which are the interval optimal ranges.\n\n#### 4. Production Planning Mathematical Model\n\n##### 4.1. NN-LP Mathematical Model\n\nThe usual mathematical model of NN-LP is similar to mathematical model (1), so we omit it.\n\n##### 4.2. Production Planning Mathematical Model\n###### 4.2.1. Notations\n\nThe following notations are used in the production planning model.\n\nNine decision variables:(i) to : product quantities of six plans of type I(ii) to : product quantities of two plans of type II(iii): product quantities of two plans of type III\n\nObjective function:(i): maximum profit\n\n###### 4.2.2. Assumptions\n\nThe production planning model is developed by considering the following assumptions:(i)Every product must pass two working procedures: A and B.(ii)The startup time of two working procedures can be ignored.(iii)Product quantities are only affected by validity time of machines.(iv)The demand rate at any time is constant.\n\n##### 4.3. Model Formation\n\nAs shown in Table 1, we consider an application in production planning studied by Hu . A company manufactures three types of products: Types I, II, and III. All types must pass two working procedures: A and B. We consider that procedure A can be operated on machine A1 or A2 and procedure B can be operated on the machines B1, B2, and B3. Type I can be operated on all machines of procedure A and procedure B; Type II can be operated on all machines of procedure A and only machine B1 of procedure B; Type III can be operated on only machine A2 of procedure A and machine B2 of procedure B. Our aim is to schedule the optimal production planning, which can pursue for the maximum profits. All used data are listed in Table 1, including required procedure time of every working procedure, processing fees, material cost, and selling price per unit. So, Type I has six plans to produce products, along with (A1, B1) or (A1, B2) or (A1, B3) or (A2, B1) or (A2, B2) or (A2, B3), respectively. Similarly, we consider the product quantities of the six plans , , , , , and , respectively. Type II has two plans to produce products, along with (A1, B1) or (A2, B1), and Type III has one plan to produce products, along with (A2, B2). We consider the product quantities of the remaining three plans , , and , respectively. So, we can get the following objective function.\n\nSo, we get the followed production planning mathematical model:\n\n##### 4.4. Solution regarding Matlab Built-In Function “fmincon()”\n\nAccording to the de-neutrosophication technique proposed by Ye and considering I = 0 or 0.5 or 1 as the minimum or moderate or maximum indeterminacy, we can obtain three optimal solutions as follows:(1) = 0, = 778.508, = 465.936, = 0, = 677.953, = 56.452, = 0, = 474.359, = 0, and = 1297.389 for I = 0.(2) = 0, = 0, = 578.231, = 0, = 0, = 0, = 167.732, = 338.221, = 590.909, and = 940.871 for I = 0.5.(3) = 0, = 0, = 625, = 0, = 0, = 0, = 146.667, = 386.667, = 583.333, and = 762.717 for I = 1.\n\nClearly, using the indeterminacy I ∈ [0, 1], different optimal results are revealed. The optimal solutions of the optimization problem are = [0, 0], = [0, 778.508], = [465.936, 625], = [0, 0], = [0, 677.953], = [0, 56.452], = [0, 146.667], = [386.667, 474.359], and = [0, 583.333] for = [762.717, 1297.389], which shows the interval optimal ranges.\n\n##### 4.5. Solution regarding Operations of NNs\n\nAccording the front optimal solutions, we next calculate the nine relation formulas of the indeterminacy I and variables , , , , , , , , and . For example, let us calculate = 465.936 + 159.064I. Firstly, according to three points (0, 465.936), (0.5, 578.231), and (1, 625), we obtain the linear equation ( = 159.06I + 476.86). Next we amend the intercept of trend curve on the vertical coordinate. The other linear equations are obtained in the same way. So, = x1 + y1I = 0 + 0I = 0, = x2 + y2I = 778.508−778.508I, = x3 + y3I = 465.936 + 159.064I, = x4 + y4I = 0 + 0I = 0, = x5 + y5I = 677.953−677.953I, = x6 + y6I = 56.452−56.452I, = x7y7I = 0 + 146.667I, = x8 + y8I = 474.359−87.692I, and = x9 + y9I = 0 + 583.333I; then, we calculate the results of equation (13) as follows:\n\n= [(1.2 + 0.03I) − (0.23 + 0.03I)] × () + [(1.60 + 0.5I) − (0.30 + 0.07I)] × () + [(2.30 + 0.3I) − (0.30 + 0.05I)] × − (0.04 + 0.02I) × [(4.5 + 1.7I) × () + (8 + I) × ] − (0.02 + 0.01I) × [(6.7 + 1.8I) × () + (8.6 + 1.4I) × (11 − I) × ] − (0.05 + 0.02I) × [(5.6 – 0.1I) × ( ) + (7.8 + 1.2I) × ()] − (0.10 + 0.02I) × [(3.5 + 2.5I) × ( ) + (10 + 2I) × ] − (0.04 + 0.02I) × [(6.7 + 1.3I) × ()] = 0.97 × ( ) + (1.30 + 0.43I) × () + (2.0 + 0.25I) × − (0.04 + 0.02I) × [(4.5 + 1.7I) × ( + + ) + (8 + I) × ] − (0.02 + 0.01I) × [(6.7 + 1.8I) × ( ) + (8.6 + 1.4I) × (11 – I) × ] − (0.05 + 0.02I) × [(5.6 – 0.1I) × ( ) + (7.8 + 1.2I) × ()] − (0.10 + 0.02I) × [(3.5 + 2.5I)× ()(10 + 2I) × ] – (0.04 + 0.02I) × [(6.7 + 1.3I) × ()] = 0.97 × () + (1.30 + 0.43I) × ( ) + (2.0 + 0.25I) × − (0.04 + 0.02I) × [(4.5 + 1.7I) × ( + ) + (8 + I) × ] − (0.02 + 0.01I) × [(6.7 + 1.8I) × () + (8.6 + 1.4I) × (11 – I) × ] − (0.05 + 0.02I) × [(7.8 + 1.2I)× ()] − (0.10 + 0.02I) × [(3.5 + 2.5I) × () (10 + 2I) × ] − (0.04 + 0.02I) × [(6.7 + 1.3I) × ()] = 0.97 × (778.508 − 778.508I + 465.936 + 159.064I + 677.953 − 677.953I + 56.452 – 56.452I) + (1.30 + 0.43I) × (0 + 146.667I+ 474.359 − 87.692I) + (2.0 + 0.25I) × (0 + 583.333I) − (0.04+ 0.02I) × [(4.5 + 1.7I) × (778.508 − 778.508I + 465.936 + 159.064I) + (8 + I) × (0 + 146.667I)] − (0.02 + 0.01I) × [(6.7 + 1.8I) × (0 + 677.953 − 677.953I + 56.452 − 56.452I) + (8.6 + 1.4I) × (474.359 − 87.692I) + (11 − I) × (0 + 583.333I)] − (0.05 + 0.02I) × [(7.8 + 1.2I) × (0 + 146.667I + 474.359 − 87.692I)] − (0.10 + 0.02I) × [(3.5 + 2.5I) × (778.508 – 778.508I + 677.953 − 677.953I) + (10 + 2I) × (0 + 583.333I)]− (0.04 + 0.02I) × [(6.7 + 1.3I) × (465.936 + 159.064I + 56.452 − 56.452I)] = 0.97 × (1978.849 − 1353.849I) + (1.3 + 0.43I) × (474.359 + 58.975I) + (2 + 0.25I) × (0 + 583.333I) − (0.04 + 0.02I) × (5599.998 − 404.995I) − (0.02 + 0.01I) × (9000.001 + 699.9991I) − (0.05 + 0.02I) × (3700 + 1100.006I)− (0.10 + 0.02I) × (5097.614 + 1902.383I) − (0.04 + 0.02I) × (3500 + 1500I) = 1919.484 − 1313.234I + 616.6667 + 306.001I + 1312.499I − 224 − 87.7I − 180 − 111.000I − 185 − 509.761 − 330.238I − 140 − 160.000I = 1297.389 − 534.672I.\n\nSo, these calculational results validate that the same solution is obtained by using the two methods of both the Matlab built-in function “fmincon()” and the operations of NNs, which are = [0, 0], = [0, 778.508], = [465.936, 625], = [0, 0], = [0, 677.953], = [0, 56.452], = [0, 146.667], = [386.667, 474.359], and = [0, 583.333] for = [762.717, 1297.389] and show the interval optimal ranges.\n\n#### 5. Conclusion\n\nThis paper first introduced some concepts and their operations of NNs with indeterminacy I. Next, we built a mathematical model with constrained conditions and then constructed the corresponding inventory model and production planning model. Finally, we obtained the optimal solutions by using the two methods of the Matlab built-in function “fmincon()” and the operations of NNs to solve the NN-NP and NN-LP problems with constrained conditions as preliminary application study in indeterminate setting. The final results show that the two methods obtained the same effective solutions, but the former needs the Matlab built-in function along with the simple calculational process, while the latter needs a lot of operations of NNs along with the complex calculational process. Some contributions in this study are that (1) different methods can obtain the same optimal results, (2) the NN-NP and NN-LP methods provided the new application ways for engineering management, (3) the NN-NP and NN-LP methods are more suitable than other ones under uncertain environments as the generalization of traditional programming methods, and (4) the two approaches can obtain the interval solutions for avoiding determinate solutions of traditional programming methods.\n\nObviously, the proposed NN-LP and NN-NP methods can handle indeterminate and/or determinate mathematical programming problems, which are the generalization of existing uncertain or certain linear and nonlinear programming methods. As the preliminary application study in this paper, however, there exist a lot of mathematical solution methods and proof problems along with some complexity/difficulty in the nonlinear programming problems which need to be studied further. Hence, as our future works, one is to further analyze the two presented methods of this paper from the mathematical problems, such as the convexity problem in the nonlinear programming, the stability and solution range problems regarding the changeability of NNs, and the sensitivities of NNs on the solution results, and then NN-LP and NN-NP approaches will be extended to other ﬁelds, such as engineering design and management science.\n\n#### Data Availability\n\nThe data used to support the findings of this study are included within the article.\n\n#### Conflicts of Interest\n\nThe authors declare that they have no conflicts of interest." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84780866,"math_prob":0.9915164,"size":29160,"snap":"2023-40-2023-50","text_gpt3_token_len":8632,"char_repetition_ratio":0.14700234,"word_repetition_ratio":0.19455415,"special_character_ratio":0.33281893,"punctuation_ratio":0.22233152,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.99981683,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-28T16:39:15Z\",\"WARC-Record-ID\":\"<urn:uuid:70ef6554-66be-483f-ad5a-af760889f616>\",\"Content-Length\":\"1058280\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:23002c17-c5cb-4b57-b10d-c74aa6788aa2>\",\"WARC-Concurrent-To\":\"<urn:uuid:96f8ea8e-dfe4-47ae-917e-6d0cb9cfec83>\",\"WARC-IP-Address\":\"104.18.40.243\",\"WARC-Target-URI\":\"https://www.hindawi.com/journals/jmath/2021/6668711/\",\"WARC-Payload-Digest\":\"sha1:XKTWQLB6AAOU54QUVBZHLPEU4VGCEXRG\",\"WARC-Block-Digest\":\"sha1:65PMQQSBAG47EZZHNOPOXRVDQD6WXINP\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679099892.46_warc_CC-MAIN-20231128151412-20231128181412-00859.warc.gz\"}"}