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 sequencelengths 1 199	image_list sequencelengths 1 199	metadata stringlengths 1.19k 3.08k
https://www.neetprep.com/question/5708-value-sf-Kp-Kp-reactionsXYZ-AB-ratio--degree-dissociation-X-equal-totalpressure-equilibrium---ratioa--b-c--d-?courseId=19	[ "NEET Questions Solved\n\nThe value sf Kp1 and Kp2 for the reactions\n\nX$⇌$Y+Z ....(1)\n\nand A$⇌$2B ......(2)\n\nare in the ratio 9:1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio:\n\n(a) 1:9 (b) 36:1\n\n(c) 1:1 (d) 3:1\n\n(b)\n\nFor X$⇌$ Y + Z for A $⇌$2B\n\n1 0 0 1 0\n\n1-$\\mathrm{\\alpha }$ $\\mathrm{\\alpha }$ $\\mathrm{\\alpha }$ 1-$\\mathrm{\\alpha }$ $\\mathrm{\\alpha }$\n\n..\n\n9 =\n\n..$\\frac{{P}_{\\mathit{1}}}{{P}_{\\mathit{2}}}$ = 36\n\nDifficulty Level:\n\n• 6%\n• 84%\n• 9%\n• 3%" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6876007,"math_prob":0.9998908,"size":316,"snap":"2019-26-2019-30","text_gpt3_token_len":134,"char_repetition_ratio":0.118589744,"word_repetition_ratio":0.0,"special_character_ratio":0.47468355,"punctuation_ratio":0.25252524,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99974304,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-20T07:56:08Z\",\"WARC-Record-ID\":\"<urn:uuid:39048179-d5c3-4a18-ab21-e378a293c80e>\",\"Content-Length\":\"96362\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8838d379-94ea-482a-891f-ebf669383f97>\",\"WARC-Concurrent-To\":\"<urn:uuid:386b93dc-f473-42b4-9a04-745f0ab4303a>\",\"WARC-IP-Address\":\"54.186.44.90\",\"WARC-Target-URI\":\"https://www.neetprep.com/question/5708-value-sf-Kp-Kp-reactionsXYZ-AB-ratio--degree-dissociation-X-equal-totalpressure-equilibrium---ratioa--b-c--d-?courseId=19\",\"WARC-Payload-Digest\":\"sha1:MZARA5C5APL5P4PITG5UWSUVDELKJDT3\",\"WARC-Block-Digest\":\"sha1:PURND4DTWVHT6HPUGMLEYGB56ISJM7EG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999163.73_warc_CC-MAIN-20190620065141-20190620091141-00350.warc.gz\"}"}
https://www.scirp.org/Journal/PaperInformation.aspx?PaperID=94656	[ "An Earthquake Model Based on Fatigue Mechanism—A Tale of Earthquake Triad\n\nEarthquakes are the result of strain build-up from without and erosion from within faults. A generic co-seismic condition includes merely three angles representing respectively fault geometry, fault strength and the ratio of fault coupling to lithostatic load. Correspondingly, gravity fluctuation, bridging effect, and granular material production/distribution form the earthquake triad. As a dynamic component of the gravity field, groundwater fluctuation is the nexus among the three intervened components and plays a pivotal role in regulating major earthquake irregularity: reducing natural (dry) inter-seismic periods and lowering magnitudes. It may act mechanical-directly (MD) through super-imposing a seismogenic lateral stress field thus aiding plate-coupling from without; or mechanical-indirectly (MI) by enhancing fault fatigue, hence weakening the fault from within. A minimum requirement for a working earthquake prediction system is stipulated and implemented into a well-vetted numerical model. This fatigue mechanism based modeling system is an important supplement to the canonical frictional theory of tectonic earthquakes. For collisional systems (e.g., peri-Tibetan Plateau regions), MD mechanism dominates, because the orographically-induced spatially highly biased precipitation is effectively channeled into deeper depth by the prevalence of through-cut faults. Droughts elsewhere also are seismogenic but likely through MI effects. For example, ENSO, as the dominant player for regional precipitation, has strong influence on the gravity field over Andes. Major earthquakes, although bearing the same 4 - 7 years occurrence frequency as ENSO, have a significant hiatus, tracing gravity fluctuations. That granular channels left behind by seamounts foster major earthquakes further aver the relevance of MI over Andes. Similarly, the stability of the Cascadia fault is found remotely affected by Californian droughts (2011-15), which created a 0.15 kPa/km stress gradient along the Pacific range, which also is the wave guide.\n\nKEYWORDS\n\n1. Earthquakes as Frictional Phenomena and Earthquake Triad\n\nTectonic earthquakes are frictional phenomena between contacting plates (Scholz, 1998; Wang & Hu, 2006; Wang & He, 1999) . It is a game of gravity- (or more generally compressive stress-) aided friction resisting against the ever-increasing plate-coupling stress ( $\\tau$ ) resulted from differential plate motions. For an ideal configuration (Figure 1(a)) of uniform fault geometry (i.e., constant slope angle $\\theta$ and unlimited width in the transverse direction) and uniform material (so that maximum frictional coefficient ${\\mu }^{\\prime }=\\mathrm{tan}\\left({\\theta }_{f}\\right)$ being a constant), the co-seismic condition is\n\n${\\varphi }_{c}=\\theta +{\\theta }_{f}$ . (1)\n\nwhere repose angle $\\varphi =\\mathrm{arctan}\\left(\\tau /G\\right)$ , with G being compressive stress load on the shear (fault) zone. Subscript c means critical value. $\\theta$ is the slope of the fault zone, determined by the geometry of the plate interface, and ${\\theta }_{f}$ is the maximum static friction angle corresponding to fault strength. Values of ${\\theta }_{f}$ depend on wall rock material properties and lithologies. During inter-seismic stage, the repose angle increases steadily and approaches the sum of fault slope angle and the maximum static friction angle. Driving stress increase and the overall resistive stress decrease (weakening of the fault, or ${\\theta }_{f}$ decrease) both are seismogenesis (Scholz, 1998) . In reality, erosion from within and burden from without likely are simultaneous. Earthquake cycle is a stress adjustment cycle. Closely related to three angles in Equation (1) are the earthquake triads: Gravity fluctuation of the overriding plate, bridging effect and fatigue. With fault strength set as constant, fault geometry and plate-coupling stress determine a natural (limiting) earthquake occurrence frequency. However, actual occurrence seldom obeys this limit and usually occurs far ahead of schedule, due to various fault-weakening factors.\n\nSince displacements caused by each earthquake (ranging between 0.5 - 3 m) are at least 3 orders of magnitude smaller than the dimension of the seismically locked fault zone, which could well be several hundred kilometers, material in the locked fault zone is well-seasoned: all having experienced numerous times of wearing and tearing. For predicting near future earthquakes, the geological backgrounds, including material properties and fault geometry, can be safely assumed to be constant. Thus, the natural limiting occurrence frequency over a fault is rather stable. The irregularity in earthquake occurrence is primarily due to fault weakening mechanism. This study focuses on more dynamic fluctuations affecting the triad of earthquake. Following discussions use subduction fault as prototype. By substituting compressive stress for lithostatic loading, same reasoning is viable for strike-slip faults. Strictly, simulating an earthquake event includes estimates of its timing (Section 2), epicenter location and magnitude (Section 2.1). Actual full 3D numerical model, with detailed parameterization outlined (Section 3) is used in the simulations discussed in Section 4. In addition to the parameterizations of material rheology (visco-elastic mantle, granular debris as produced by seismic events of all kinds, and the brittle elastic crust), Section 3 also is dedicated to the setting of the fault-following model grids, static geologic properties (e.g., plate interface geometry and physical property of each layer of medium).\n\n2. Complexity of the Earthquake Problem: Under-Determination and Interactions of the Triad\n\nDifferent segments of a realistic heterogeneous fault zone (Figure 1(b)) have different slope angles and maximum friction coefficients and also experience different loads. The driving stress distributed to each segment approaches its maximum affordable resistive stress gradually, like the saturation process of porous media. Fault “stress saturation” is thus analogously defined here as the ratio of driving stress to maximum affordable resistive stress.\n\n${S}_{i}={\\tau }_{i}/\\left[{G}_{i}\\mathrm{tan}\\left({\\theta }_{i}+{\\theta }_{f,i}\\right)\\right]=\\mathrm{tan}\\left({\\varphi }_{i}\\right)/\\mathrm{tan}\\left({\\theta }_{i}+{\\theta }_{f,i}\\right)$ (2)\n\nConsequently, an S value of 0 corresponds to full locking while an S of 1 corresponds to creeping. The overall stability of a realistic fault becomes ${\\stackrel{¯}{S}}_{i}=\\left[\\underset{i}{\\sum }{\\tau }_{i}\\right]/\\left[\\underset{i}{\\sum }{G}_{i}\\left({{\\mu }^{\\prime }}_{i}+\\mathrm{tan}{\\theta }_{i}\\right)/\\left(1-{{\\mu }^{\\prime }}_{i}\\mathrm{tan}{\\theta }_{i}\\right)\\right]\\le 1$ . S is thus a useful diagnostic index for fault stability.\n\n2.1. Multi-Solution in Seismogenesis\n\nIn reality, earthquakes involving the saturation of the entire shear zone seldom occur, because actual tectonic plates have difficulty in maintaining rigidity over extended distance, manifested as the existence of through-cut faults and many small magnitude earthquakes within close proximity to each other. Consequently, many places do not contribute their shares of resistance and depend on neighbors to stay in place. Depending on how $\\tau$ is distributed on the fault zone, there can be various “legal” combinations of saturated and partially saturated segments along the fault zone. In reality, earthquake occurrence can be compounded by any factors affecting $\\tau$ , G, and ${\\theta }_{f}$ (or fault strength ${\\mu }^{\\prime }$ ) in Equation (2). Multi-solution in seismogenesis essentially is resulted from the lack of enough information as constraints on, or an under-determination issue of lateral interactions. Realizing that earthquakes, as a result of relative motion of plates involved, inevitably have footprints on the mass (Wallace et al., 2009; Ren et al., 2015) , energy (Gao & Wang, 2014) and momentum fields ( Wang et al., 2012 and references therein), efforts for earthquake prediction are increasingly based on observational facts of these physical parameters and their fields.\n\n(a)(b)(c)(d)\n\nFigure 1. Diagram of an ideal crust front/fault (a) and a plane view of a more realistic plate interface (b). Panel (a) is a highly idealized continental plate overlying oceanic plate configuration. Black bold arrows in (a) show the opposing oceanic plate and the overlain continental plate. Stress analysis is performed for an arbitrary point (P) at the shear zone. $\\mu$ is the static friction; G is the gross gravity of the overlain plate section; $\\tau$ is the driving stress resulted from tectonic motion. Static friction and gravity work together to contest against continental driving stress. Once the driving stress overcomes gravity-aided maximum static friction, a co-seismic sliding occurs. Red-toned color shades illustrate stress saturation. Granular material, often in the form of fault gouge, inter-laces the plate-interface (b). Either because geometrical irregularity of the contacting interface, or because the existence of density anomaly (e.g., cavity) above, there are places (“pillars” or strong contact regions) that take a disproportionally larger load than the neighboring regions (Bridging effect). Plates are “riveted” together by these “sticky” spots. These strong contact regions tend to wear down faster, through generating granular debris more actively. “Asperities” and “barriers” (Wallace et al., 2004) can both be the “pillars” defined here. Panel (c) shows the vertical resistive stress σzz fields around a region with significant bridging effects (i.e., with cavity caused gravity anomaly). Note that the under-belly of the overlain plate is in weak contact with lower plate and also is a region with active granular material production (that region experiences strong tensile stress). The fault creeps primarily by wearing geometrical irregularities or heterogeneous “rises/bumps/sticky spots”. The epicenter is the ensemble (weighted by frictional heat released at these regions) mean location of the sticky points being co-seismically unlocked. Real cavity usually occurs at much shallower depths. Fault interface underneath still feels the bridging effects by producing a corresponding uneven granular material generation pattern. Panel (d) is SEGMENT simulated global distribution of inter-pate granular material thickness (m).\n\nThe co-seismic release of potential energy is transformed mainly into heat through inter-plate frictional force. A large, deep-burying area approaching saturation simultaneously, un-locking systematically and releasing the stored elastic energy coherently generally produces large magnitude earthquakes (Wang & Bilek, 2011) . The contact surface of actual tectonic plates can have sophisticated topography and may involve material heterogeneity (Figure 1(b)). Therefore, it is difficult to predict the future rupture propagation without knowing the details at the interface. Recent studies (Gao & Wang, 2014; Wang et al., 2012; Wang & Bilek, 2011) indicate that widespread and smooth contact between two relatively weaker plates harbors large earthquakes—a situation usually satisfied at the megathrusts. As an opposite case of rough contact, the subducting seamounts generally discourage the generation and propagation of large ruptures. This is because the seamounts consume a very small fractional of the plate interface, acting more effectively as a grinder for the overlain plate rather than as a blocker (Figure 1(b); Wang & Bilek, 2011 ). Downstream of these seamounts, there are stripes of granular debris composed by the material scrapped off both the sea mounts and the overlain continental plates. These existing findings learn further support to the importance of lateral coherence among different segments of the fault in determining the magnitude of the earthquakes (Equation (2)).\n\n2.2. Intertwining Earthquake Triad\n\nOne important factor causing non-uniform distribution of driving stress among segments is the existence of macroscopic crevasses of voids/cavity within the plates (Rowe et al., 2012) . In reality, bridging effects (Ren, 2014a) —load directly above can be partially carried by neighboring segments (or vise-versa)—add yet another layer of complexity. Bridging effect contributes to gross resistive stress only when the fault interface has non-uniform friction coefficients. Otherwise, if the interface has same roughness ${\\mu }^{\\prime }$ , the net contribution to resistive stress from bridging effects may be minuscule due to the canceling effects from neighboring regions that is relieved portion of their lithostatic stress.\n\nFrom Equation (2), gravity of the overriding plate always serves as a stabilizing factor. The reduction of gravity (e.g., from extended droughts) contributes to instability. Fluctuations in gravity field not only have local stability consequence, they also affect neighboring regions through bridging effect and a unique fatigue process of the fault zone: granular material (GM, Jop et al., 2006; Ren et al., 2008 ) generation and transportation, a common fault-weakening mechanism. Production of granular debris is the primary form of fatigue of the fault interface. Regions of strong contact, likely being a result of bridging effect, encourage active granular material generation. Ironically, granular material has much smaller viscosity and acts as lubricant for the plates involved. Regions carrying more loading are now footed on slippery floor (i.e., more weight is loaded on smooth contacts and less is loaded on rough contacts along the shear zone). Thus, bridging effect, through affecting granular material generation and redistribution, influences the frictional properties of the fault interface and achieves a negative spatial covariance between loading and fault strengths, being seismogenic (i.e., contributing to fault instability) according to Equation (2) (i.e. reduction of the $\\underset{i}{\\sum }{\\tau }_{i}$ ). Sources of bridging effects are plenty. For example, large scale structures in the overriding plate such as mountains and valleys signifies bridging effects at depth on the seismogenic fault zone (Bejar-Pizarro et al., 2013) . Even for regions without topographic features, existence of cavities/crevasses at depth (not necessarily all the way to the locked interface; the vertical change in loading transfers readily to horizontal stress in the fault circumstances) causes non-negligible bridging effects on faults (to be further detailed in Section 3.1.1). Bridging effects from static geological background, although may cause variations in the frictional properties of the seismogenic zone, vary slowly and are not responsible for short term variations in major earthquake occurrence. Large scale variations in groundwater, however is more dynamic and is a viable candidate for explaining earthquake occurrence irregularity on decadal to centennial time scales. Cyclic forcing is most effective in causing fatigue (Ren & Leslie, 2014) . Cyclic groundwater fluctuations, especially when acting in concord with the resonance frequency of the plates, assist granular material generation.\n\nThus, the earthquake triad is interlinked and hence reinforces one another through groundwater fluctuation as the nexus. Groundwater-aided granular material generation (G3) is the mechanism modulating earthquake cycles. Detailed discussion on fatigue in rock interface, bridging effects and their physical parameterizations in the Scalable Extensible Geofluid Modeling framework for ENvironmenT Intelligence System (SEGMENT, Ren, 2014a; Ren et al., 2008; Ren et al., 2012 ) can be found in Section 3, together with the sources of input datasets. Through solving conservation equations of mass, momentum and energy, the SEGMENT provides 3D distribution of strain, stress, and full three components of velocity in the simulation domain. How a major earthquake is identified using these prognostics is described in Section 3, irrespective of the physical domain they reside (on the main thrust interpolate or not) and the morphology (e.g., megathrust earthquakes or simultaneous rupture of several strike-slip faults as the 2012 Indian Ocean earthquake). It is noteworthy that variables in Equations (1) and (2), such as the degree of stress saturation, are merely diagnostics of SEGMENT instead of being prognostics. Through these diagnostics, one can vividly view the stress build-up during the inter-seismic stage. Section 3 also includes detailed discussions of recondite approaches in obtaining present global distribution of granular material thickness and the future evolution.\n\n3. Earthquake Triad (3.1) and Their Representation in SEGMENT Modeling System (3.2)\n\nFault geometry, although being the far-largest factor in determining the timing of major earthquakes, are looked upon as geological background here. The super-imposed gravity fluctuations are primarily responsible for earthquake irregularity. Following discussion focuses on the bridging effect and the fault weakening consequences from granular material (GM) formation. Groundwater is not singled out as a subsection because it is the nexus among the components and it is natural to discuss it as an integrated component throughout the text. The following is a description of a minimum requirement for a working earthquake prediction system for realistic 3D fault configuration—the Groundwater aided Granular material Generation (G3) framework and its implemented in the well-vetted SEGMENT system.\n\n3.1.1. Bridging Effects at the Fault Zone Make Instability “Non-Local”\n\nThat different portions of the fault/shear zone contribute different resistive stresses primarily is due to uneven loading (the variations in Gi along the shear zone). There are many causes for loading deviation from the lithostatic values. For convenience, a generic term “bridging effects” (Figure 1(c); Ren, 2014a ) is assigned to this phenomenon. It is a disturbance to the vertical resistive stress field (RZ). The “G” in Equations (1) and (2) is a special case of RZ when bridging effect is insignificant. In principle, disturbances at any depth above the fault zone can be felt at the plate interface. It must be noted that, for a point source (of horizontal dimension much less than fault dimension), the influence range of bridging effect in each horizontal direction expands linearly. So the footprint increases at a square power of depth. Consequently, the amplitude of the disturbance decreases at a rate of depth-squared. Thus, the closer the disturbance source is to the shear zone, the more effective in superposing onto RZ and contributing to motion tendency of this segment of the shear zone. According to this reasoning, localized load disturbances such as man-made infrastructure (e.g., dams, high raisers) are far less important than a cavity of similar size existing at 20 km depth in disturbing the load distribution of a shear zone at ~30 km depth. The limiting size (L) of cavity at depth H depends on rock tensile strength (fc) and density ( $\\rho$ ) according to a square root law: $L=b\\sqrt{{f}_{c}H/\\rho g}$ , with b a cavity geometry dependent constant. It is apparent that larger dimension cavities are allowed at deeper depths of the brittle crust. At deeper depth (>~30 km), the thermodynamic reduction in tensile strength of rock dominates and causes drastic reduction in L. As a result, the cavities on the locked interface are of sub-meter scale at most (Ujjie et al., 2007; Cowan et al., 2003) . Thus, bridging effects from inter-plate geometry irregularity may be only non-negligible for regions surrounding sea mounts for earthquake evolution and lifecycles. These features have clear remote-sensing signature, especially when experiencing transition in size as a result of cave-in (usually a result of continued granular material formation). Loading irregularity is thus the most prevalent form of bridging effects. Load fluctuations of shallower level, only when occurring at large enough spatial scale, affect fault stability. In this sense, regional scale groundwater fluctuations, even only resides in upper couple kilometers depth, may have significant impact on fault stability. The deeper they reach (such as around the through-cut faults in the Longmenshan Fault system (to be detailed in the discussion section) the more direct and efficient they are in affecting fault stability.\n\nThe bridging effect contributes to gross resistive stress only when the plate interface has non-uniform frictional properties. To have a net effect on resistive stress, the extra loading should be negatively coupled with the spatial variation of fault strength (i.e., their spatial co-variance being negative). Granular material (Jop et al., 2006; Ren et al., 2008) formation and transportation is such a nexus. Through bridging effects, neighbor’s weight is partially transferred over regions with strong contacts (like “pillars” of a bridge). There may not be any macroscopic voids; by “pillars” it is meant only of strong compressive stress region, or strong contact regions. These regions of strong contact encourage granular material formation (a consensus of yielding criteria of brittle rocks; Rowe et al., 2012 ). With the formation of granular material, which has viscosity (<104 Pa S) many orders of magnitude smaller than polycrystalline rocks (>1021 Pa S), the shear zone with strong contacts weakens and this is seismogenic. For example, if only 5 mm granular material is put between rock plates under the confining pressure similar to the locked fault zone, the equivalent frictional coefficient, or fault strength drops to <10−4, orders of magnitude smaller than solid-against-solid rock friction.\n\nSea mounts are a type of inter-plate “pillars/asperities” that grind upper plate and also wear out themselves. “Bridging effect” actually is the controlling factor for granular material generation. Bridging effect caused co-seismic and interseismic stress irregularities has strain/deformation consequences at all depths up to the earth surface, where the near surface displacements (directly associated with strain buildup and release) can now be accurately measured by Global Positioning System (GPS; Chlieh et al., 2011; Sun et al., 2014; Loveless & Mead, 2010 ) and Synthetic Aperture Radar interferometry (InSAR; e.g., Weston et al., 2012; Cavalie & Jonsson, 2014 and references therein) instruments. From Equation (2), it contributes to stability if more weight is loaded on rough contacts and less is loaded on smooth contacts along the shear zone. Bridging effect, through affecting granular material production rates, influences the frictional properties of the fault interface and achieves the seismogenic effect of “more weight is loaded on smooth contacts and less is loaded on rough contacts along the shear zone”. Sources of bridging effects are plenty. For example, large scale structures in the overriding plate such as mountains and valleys signify bridging effects at depth on the seismogenic fault zone (Bejar-Pizarro et al., 2013) . However, topographic features caused bridging effects and hence variations in the frictional properties of the seismogenic zone cannot explain the irregularity of earthquakes that occur on decadal to centennial scales. Large scale variations in groundwater, however is a viable candidate. Thus, fluctuations in gravity field not only have local stability consequence (Equation (2)), they also affect neighboring regions (laterally) through bridging effect and the unique fatigue process of the fault zone.\n\n3.1.2. Fatigue in Rock Interface—Generation of Granular Material\n\nTwo objects, when being pressed against each other, wear and tear at the interface, a manifestation of material fatigue. The interfaces of tectonic plates are of no exception. One proof is the existence of widespread fault gouge and pseudotachylyte melt in wall rocks (Rowe et al., 2012) . Seismic activities, irrespective magnitudes, produce granular material. The generation rate of GM however is a functional of at least three factors: 1) the degree of imbalance of the three principle stress components, in a form of yielding criteria for brittle material; 2) the existing amount of granular material or the current state of GM; and 3) forming granular material consumes elastic energy accumulated in the plates, through forming new surfaces under huge confining pressure of the fault environment. Point (1) is supported by recent findings of frictional melt along the fault surface. It happens at only a small fraction of the total rupture surface area but always begins at asperities, the effectively strongest points on the fault surface. Cyclic forcing is most effective in causing fatigue. Groundwater fluctuations, especially when acting in concord with the resonance frequency of the plate segment, enhance granular material generation. The second requirement is because, as granular material builds up at the interface, it hinders further production of granular material in exactly the same way accumulated saw dust hinders further cutting of a piece of wood. This is especially relevant for the inter-plate granular material because, compared with its parent rock it has larger porosity and easily fills up the limited inter-plate voids. Earthquakes are such an effective mechanism for removing and re-distributing existent granular material, specifically by thermal pressurization or acoustic fluidization. Granular material production and redistribution achieve a negative spatial covariance between loading and effective frictional coefficients, being seismogenic according to Equation (2).\n\nIn addition to the critical role played in granular material generation, groundwater, as an extra loading on the overriding plate, is by itself a stabilizing factor, according to Equation (2). Reduction of groundwater, usually as a result of extended wide-spread droughts, is however seismogenic. Although the weight of groundwater itself is negligible compared with the weight of the overriding plate, it is of the same order of magnitude as the residual stress (between plate-coupling compressive stress and gravity-aided friction). That is, the inter-seismic stage is actually in a very tricky balance (Wallace et al., 2017) . Fluctuation of groundwater, through superimposing a lateral/horizontal stress gradient, may play a critical role in the timing of major earthquakes. Fortuitously, fluctuations in groundwater can now be remotely sensed by gravity satellites such as the Gravity Recovery and Climate Experiment (GRACE; Famiglietti & Rodell, 2013; Ren, 2014b; Voss et al., 2013 ) and by InSAR measurements (Liu et al., 2017) .\n\nCentered on the master co-seismic criterion (Equation (1)), the triad determining earthquake irregularity is discussed. It is now clear that, because of the heterogeneity in driving stress distribution, and the overlain loading condition, unlocking is often not completed at once. Rather, different geological locations may unlock in certain orders. The ensemble center of the ruptures is the epicenter and the magnitude is directly related to the rupture size. Geological backgrounds such as fault geometry and the plate motion speeds evolve rather slowly. The irregularity in earthquake occurrence mainly is resulted from the inter-plate granular material production and transportation. Large spatial scale fluctuations in groundwater (i.e., those arising from large scale droughts and floods), through bridging effects, are a suitable mechanism in granular material generation. In a certain sense, the climate fluctuations, through affecting groundwater loading pattern, affect the earthquake cycles. Bridging effects by themselves may not cause fault weakening. But with the associated granular material generation, they weaken faults through forming a negative spatial covariance pattern between loads and friction coefficients. The generation and transportation of granular material at the plate contact reconcile above-mentioned counter-intuitive characters of major earthquakes (e.g., as enumerated in Wang (2015) ) and also are critical in orchestra of small-scale ruptures to generate major earthquakes.\n\nThis study focuses on subduction zones (e.g., Andes and Cascadia) because they are responsible for the largest earthquakes and also because they have significant footprints on the gravity (mass) and thermal (Gao & Wang, 2014) fields. The continental collisions (continental against continental, e.g., Himalayas) here are not differentiated from subductions (oceanic against continental plates). Despite the improved measurements of the Earth’s structure and rupture geometry, the timing of major earthquakes remains a scientific puzzle (Pritchard & Simons, 2006; Sawai et al., 2004) . In this research, first order schemes representing earthquake triad are implemented in a sophisticated 3D modeling system, to estimate the spatio-temporal scales of major earthquakes.\n\n3.2. Numerical Model Setup, Parameterization of the Earthquake Triads, and Data Acquisition\n\nAbove discussed theoretical concepts, or the Groundwater aided Granular material Generation—G3 hypothesis is implemented into a global spherical crown version of the SEGMENT, with the stress decomposition into driving stress and resistive stress follows the convention of Van der Veen and Whillans (1989) . The SEGMENT solves the equations of mass, momentum and energy conservation in a 3D spherical geometry. The model prognostics include three components of velocity, temperature, and six components of the deviotoric stresses. Their evolution is depicted by the time marching of the governing equations. Variables in Equations (1) and (2), such as the degree of stress saturation, are just diagnostics of SEGMENT instead of being prognostics. Through these diagnostics, one can vividly view the stress build-up during the inter-seismic stage. A major earthquake in model is identified as more than 104 km2 adjacent model grids in the horizontal dimension possess macroscopic motion velocities. Once an event is identified, the released energy is estimated within a time window fully covers the time period with macroscopic motion. An epicenter is diagnosed as the geological locations of the saturated grids, weighted by the respective energy release. Degree of stress saturation S is not directly used to identify earthquakes also because the grid elements are not isolated; there are interactions among adjacent grids. If the unlocking/rupture at a grid point releases a large amount of energy, a neighboring grid point, even still quite away from saturation, can be unlocked and set into motion. In fact, some very large earthquakes are formed because there are some sections of the fault (not necessarily adjacent to each other) get simultaneously ruptured and the released energy triggered neighboring sections, resulting in an upward spiral that causes a domino effect in propagating away of the rupture region.\n\n3.2.1. Grid Stencil of SEGMENT-Fault Zone Oriented Coordinate System\n\nFor this study, the original 3D spherical model of the upper ~4 km medium is now applied into a 500 km thick lithospheric material (including crust and upper mantle). The horizontal resolution is generally 30 km at the earth surface and refined to 2 km at subduction zones (Figure 2). Vertical resolution varies as 101 stretched vertical layers are used to represent the 500 km depth. The stretching scheme is used so that the shear zone gets better represented. In setting up grids in the model simulation domain, special attention is paid so that the upper surfaces of the oceanic plates are locally parallel to the model grids spanned surface (there always is a vertical level that has surface parallel to the oceanic plate’s upper surface).\n\n3.2.2. Static Geological Parameters and Viscosity Parameterizations\n\nLike other 3D tectonic models (e.g. subduction zone models), an accurate representation of the fault geometry (e.g., interface geometry such as subduction thrust geometry). This requires high resolution geodic and seismic data. The CRUST1.0 (Laske et al., 2013) is used to set up the density values over the grids within the crust layer (oceanic and continental crusts). Underneath bottom of the crust layer (shown in CRUST 1.0) is assumed to be upper mantle material (silicates) and given a uniform viscosity of ${\\eta }_{0}=\\nu ={10}^{19}\\text{\\hspace{0.17em}}\\text{Pa}\\cdot \\text{S}$ and a density of 3.3 × 103 kg/m3 as base values and a pressure- and temperature-dependent modifier (to be detailed very soon). In setting up the grids above present sea level, the Advanced Space-borne Thermal Emission and Reflection Radiometer (ASTER) global digital elevation map (SATER DEM, asterweb.jpl.nasa.gov/) and ETOPO1 (downloadable from https://www.ngdc.noaa.gov/mgg/global/global.html) bathymetry are referenced to. For faults with intensive gravity measurements, for example the Longmenshan Fault Zone (LFZ) of the eastern margin of the Tibetan Plateau, there was a detailed gravity survey after the Mw7.9 Wenchuan earthquake on 12 May 2008. Finer resolution data on elastic plate thickness, density and rigidity profiles are available and used in replacement of the CRUST1.0-derived parameters. Similarly, the Andes survey (Andes Geological Survey, 2006) and data pointed out by references therein provided information also is merged into the coarse resolution CRUST 1.0. For the Hikurangi subduction zone (New Zealand), the revised interface geometry (Williams et al., 2013) are used instead of the CRUST 1.0. For the Puysegur trench, we adapted the geometry data from Liu and Bird (2002) . For the Cascadia, Mexico, and Japan Trench, the higher resolution interface geometry are obtained from Gao and Wang (2017) . Personal communication with Z. Liu also provided higher resolution geometry data along the west frontal range of N. America. Other regions we have to tolerate the lower resolution interface geometry provided by SLAB 1.0 (Hayes et al., 2012) .\n\nBased on the fact that the co-seismic ruptures consume only a portion of the plate interface (Lay, 2015) to the center and the down-dip section creeps, the upper and lower portions of the plate should be treated respectively as elastic\n\nFigure 2. Horizontally and vertically stretched (so that the fault region and especially the plate interface zone is better represented) grids in SEGMENT, as delineated by black vertical lines and horizontal curves. The “locked” region gets the finest spatial resolution. In the corresponding calculational space, two shape factors (i.e., Jacobian operators) are involved in the governing equations. Background map showing subduction fault geometry is adapted from Wang (2015) .\n\nand visco-elastic material. Grids in the upper crust are considered elastic and elastic moduli are specified using canonical values. Grids in the lower crust and upper mantle away from the shear zone are considered to be visco-elastic (Wang et al., 2012) but getting more rigid upwards intending a smooth transition in the vertical domain. This is achieved by considering the thermal structure (Gao & Wang, 2014) in the parameterization of material viscosity\n\n$\\eta \\left(T,P\\right)={\\eta }_{0}D\\mathrm{exp}\\left[\\frac{Q+PV}{RT}\\right]$ . Here P is confining pressure, T is absolute\n\ntemperature, and D depends on the rock composition, grain size and fluid content. Here Arrhenius’ activation energy Q, specific volume V and universal gas constant R were assumed constants in the model. In response to different stress conditions, wall rocks can be compressed or stretched within a narrow range before onset of brittle fatigue. Arrays are allocated to record the amount of granular material formed at each grid cube. The shear zone is a granular zone of variable thickness (Figure 1(d)). In most cases, it is only a small friction of the shear zone layer that is granular material (also recorded in arrays). The parameterization of granular material viscosity follows Jop et al. (2006) , as implemented in SEGMENT (Ren et al., 2008) . The shear-thinning viscosity structure (Equation 8 in Ren et al., 2008 ) implies a shear-localizing effect from granular material accumulation. The present amount of granular material (Figure 1(d)) is deduced from surface velocity using the method in Ren et al. (2012) . Future variation of granular material is diagnosed from Equation (3) (described in next subsection on granular material generation; immediately follows). The granular ensemble size is deduced from its parent rock type and the associated grain sizes. In addition to the brittle interface of oceanic and continental crusts, GM also exist in the underbelly of the overlying continental crust just updip of the mantle wedge corner, a byproduct of (forearc) mantle wedge serpentinization (e.g., Hyndman et al., 2015; Audet & Burgmann, 2014 ). The GM material amount in this zone can be estimated from the clustering of episodic tremor and accompanying slow slip events and their magnitudes. Re-mapping of the grid stencils are automatically performed only if granular material accumulated exceeds 20 m or earthquake caused displacement over 20 m.\n\nSetup of the lower boundary stress conditions is critical for the simulation of stress buildup inside the simulation domain. The GPS measurements of plate motion speeds $V$ are used to fine tune the parameters of upper mantle layers so that lateral stress (basal drag) exerted on the lowest model layer at the active zones (mantle driven plate motion) and passive zones (plate driven mantle motion) nearly balance. From the vertical profile of the horizontal velocity components, it is clear that for most regions, the mantle motion is passively forced by the overlain plates and only in the active thermal hot spots (for Cascadia, the Yellowstone area), the boundary layer flow structure indicates that it is the mantle flow that exerts a strong drag to the overlain plate. Reflected in the flow structure, mantle flow speeds increase away from the lithosphere-asthenosphere boundary (LAB) rather than being a maximum at the LAB. Therefore, the driving stress comes primarily from lateral stress from neighboring grids within the lithosphere, rather than from the local basal drag from mantle (ultimately it is from the convective mantle but local to subduction zones it is not). The upper boundary condition is assumed to be stress-free (i.e., $\\tau \\cdot n=0$ at upper surface). The model spin-up from ad hoc initial conditions is aided by advanced data assimilation schemes using the remotely sensed plate motion, gravity (weather signal pre-filtered) signal and thermal flux measurements available. Accurate flow field is critical for an accurate estimation of the stress build-up between plates. Figure 3 shows present plate motion velocities at two vertical levels: 0 - 50 km and 100 - 250 km depth averaged. Compared with GPS measurements, globally, errors in meridional velocity is <0.02 mm/yr and in longitudinal direction is less than 0.03 mm/yr. For the interested Andes region, the error is slightly larger than global average (0.035 and 0.03 mm/yr). Over the Cascadia region, however, the errors are smaller than global average (0.027 and 0.015 mm/yr respectively). Based on the optimized solution, further experiments are performed on the sensitivity of groundwater perturbation and the dynamic evolution of the fault strength and the associated earthquake cycles. Although the governing thermos-dynamic equations of SEGMENT are very similar to those of the Advanced Solver for Problems in Earth’s ConvecTion (ASPECT, Kronbichler et al. (2012) ), it should be noted that the ASPECT model, due to its intended research purpose, has coarse resolution and cannot sufficiently represent surface topography. In fact, surface topography-induced creeping is a significant component for GPS sites in mountainous regions and on islands. This results in discrepancies in model top-level velocity and GPS measurements. Through skilled numerical setting, SEGMENT sidestepped this issue and significantly improved the velocity simulation of the near surface (<100 km depth) layers, where the stress build-up is critical for earthquakes (Figure 3(a)).\n\nFigure 3. Present day near surface (a) and deep (b) level plate motion velocities (blue arrows, in m/yr). GPS measurements (black and red (for regions of interest only) arrows) are from ftp://data-out.unavco.org/pub/products/velocity/pbo.final_nam08.vel (for North American Plate) and http://sideshow.jpl.nasa.gov/post/series.html (for global coverage). In Panel (b), resonance frequencies of global tectonic plates at the interfaces are labeled (years in red font). Hollow red arrows indicate the rupture tendency for the upcoming major earthquakes that are temporally in sequence. This also is the relocating direction of their epicenters.\n\n3.2.3. Groundwater Time Series\n\nFrom Equations (1) and (2), groundwater, through affecting the gravity field, contribute to fault instability. Its spatial fluctuation, through enhancing granular material generation, also weakens the fault from within (Section 3.1.2). To obtain the groundwater fluctuations over the past 15 years (2000-2015), a continuous time marching of the land surface scheme is performed with 30-min time step. Atmospheric parameters such as near surface air temperature, humidity, pressure, winds, and radiative components are from NCEP/NCAR reanalyses, and precipitation information from the Global Historical Climatology Network (GHCN, 1900-2015), GPCP, and The Tropical Rainfall Measuring Mission (TRMM, 1998-2015) are taken as input during the observational periods to drive the land surface scheme in SEGMENT to diagnose how much precipitation is percolated into ground water reservoir. For the recent decade (2003 onward), GRACE measurements provide a reliable means for the groundwater recharge/discharge. Model simulations are satisfactory when verified against GRACE measurements. GRACE is of very coarse resolution and model can provide spatial distribution of groundwater at much finer resolution. Obtaining groundwater distribution in a warmer climate is challenging. A small ensemble of climate models (27 climate models listed in Table 1) and an innovative approach in estimating extreme precipitation (Ren, 2014a) is used for the prediction of future occurrence of earthquakes. The partition into surface runoff, evapotranspiration and ground water percolation also is performed within SEGMENT.\n\nTable 1. Twenty seven GCM models used in this study to provide atmospheric forcing parameters to SEGMENT.\n\n3.2.4. Granular Material Generation and Transportation\n\nIn analogous to the state equation in Scholz (1998) , the following force-restore relationship is proposed here for granular material generation:\n\n$\\frac{\\partial a}{\\partial t}=-{c}_{1}a+{c}_{2}\\delta \\Delta E+ADV$ (3)\n\nwhere a is the thickness of the granularized rock layer, starting from 0, t is time, c1 is the e-fold time scale, and c2 is the Paris coefficient (Timoshenko & Gere, 1963) . Apparently, c1 signifies the negative feedback as granular material accumulates. The unit surficial energy density of rocks, J (<300 mJ/m2, Gilman (1960) ), is inverse proportional to c2. To form new surfaces at high confining pressure (~0.1 GPa), the energy need far surpasses the surficial energy. The c2 thus also is inversely proportional to confining pressure. Thermal effects also are included in c2. The coefficient $\\delta$ is 0 when the stress intensity change falls within the elastic range of the parent rock and is 1 when exceeding the elasticity range. $\\Delta E$ is the energy input rate when stress perturbation caused imbalance in principal stress exceeds the elastic range. External (e.g. groundwater fluctuation) tides with resonance frequencies are most effective in transferring distortion energy into the plates to cause fatigue and influence the granular material production rates.\n\n$\\Delta E={b}_{1}{\\text{e}}^{{b}_{0}\|\\nu -{\\nu }_{p}\|}{\\left(\\Delta K\\right)}^{4}$ (4)\n\nwhere the Paris coefficient ${b}_{1}$ depends on material property such as unit surficial energy and porosity of fault rocks. The coefficient ${b}_{0}$ is a negative number indicating the exponential damping of the groundwater tidal wearing when it is not synchronized with the natural frequency of the crust plate, $\\nu$ is the groundwater variation frequency, ${\\nu }_{p}$ is the resonance frequency of the crust plate, and $\\Delta K$ is the range of stress intensity change (proportional to the groundwater variations’ amplitude squared). From 2003 onward, GRACE measurements are available for improved SEGMENT estimates of the groundwater fluctuation amplitude. Plates at their interface assume complex configurations and determining the resonance frequency ${\\nu }_{p}$ is achieved by solving the deflecting equation, with realistic plate configurations and distributed density and temperature from local geology maps and recent observational studies (e.g., the Andes Survey 2006). The estimated inherit resonant frequencies are labeled in Figure 4. The framework of Equations (3) and (4) also is applicable to the silica granules “floating” above the mantle wedge corner. In that circumstances, the $\\Delta E$ is parameterized as thermal control: a thermal switch is designed as a function of the temperature gradients around the mantle wedge and the plate dipped in. In that manner, GM generation is shut-off for cold-slab subduction zones such as the Japan Trench and Hikurangi; whereas being active around warm-slab subduction zones such as the Mexico and Nankai subduction zones.\n\nFigure 4. SEGMENT diagnosed time scales of plate resonance frequency to groundwater fluctuation forcing (numbers in red). Background map was obtained from sideshow.jpl.nasa.gov/mbh/series.html and has been cleared for public release.\n\nThe inter-plate granular material chute system, as left behind by seamounts, removes granular deposits through advection process (ADV term in Equation (3)) exactly the same way as the surface runoff moves debris (Dietrich & Perron 2006) . Transportation of granular material is left to the Navier-Stokes solver of SEGMENT. Although granular materials are ubiquitous between contacting plates, under the huge confining pressure, the mean free path, in analogous to Prandtl mixing length in fluid dynamics, of the granular particles are very limited for most places lacking the parallel granular stripes left behind by seamounts or strong-contact “asperities” (Figure 1(b)). In the absence of macroscopic tunnels between plates, granular material redistribution resembles the diffusion of heat: moving along the potential gradient (included in coefficient c1). Earthquakes, primarily through p-wave shaking, increase the mean free path of the granular particle and make it more like fluids, and facilitate its downstream dispatch. Advection dominates the co-seismic down-dip transport and diffusion processes works steadily during the inter-seismic period. Earthquakes also are the mechanism shedding off the “saw dust” and facilitating generation of new granular material. Granular material’s formation is a fatigue/damage process (Bercovici & Ricard, 2014) . As cyclic variations in forcing favors fatigue (first two terms on the RHS of Equation (3)), especially when the forcing frequency is in resonance with the segment of the plate. In this sense, groundwater fluctuation, which possesses frequency on annual to multi-decadal scales, likely in concord with the fatigue of the inter-plate strong contacts. Thus, large spatial scale droughts and floods contribute to the irregularity of earthquake cycle.\n\nAlthough in principle healing effects can be included into c1, we did not involve the healing mechanism in Equation (3) because it is only applied in the brittle crust zone near subduction zones, where temperature are lower due to the double insulation from oceanic and continental plates. The best way to heal a fatigue, the phase changes (e.g., melt) are not relevant here. The released elastic energy during an earthquake is primarily dissipated as heat at the interface. The remainder propagates away as P and S-wave energy. Earthquake involves long distance displacements may cause melt and partially heal the fatigues interface (Rowe et al., 2012) . For the inter-seismic periods, Equation (3) is safely applicable. Clarifying these points about granular material also helps as numerical modeling of faults are now feasible with super-computers and more detailed observations (e.g., INSAR and GPS measurements of displacements; GRACE measurements of mass changes and numerous ways of thermal heat fluxes).\n\nTransportation of granular material is left to the Navier-Stokes solver of SEGMENT. The motion of the granular material inside the chute system at the plate interface is primarily low Reynolds number (Re) laminar flow. The gravity potential associated with the self-weight of granular material is negligible compared with its dominance at the earth surface flow systems. Conversely, confining pressure, the granular viscosity, and the interaction of granular with the channel boundary become important. These forces cause a passive pumping of granular material in the inter-plate chute systems. The thermal process also is simplified by the laminar flow because convective mixing in regular free fluids are non-existent and only diffusion kinetics are dominant. In many aspects, it is similar to micro-fluids in the capillary pipes. Confining pressure becomes the driver in guiding the flow direction.\n\nAs in Figure 1(c), the strong contact regions have the most active granular material generation. Its existence in the locking zone assists the unlocking because the shear resistance granular material can provide at the strain rate of the locking stage is equivalent to a ${\\mu }^{\\prime }$ value of 10−3, four orders of magnitudes smaller than solid-solid contact, representing a lubricant between the two elastic plates. The bridging effects by themselves may not cause fault weakening. But with the associated granular material generation, they definitely weaken faults through forming a negative spatial covariance between loading and friction coefficients. During the past two decades, it is being gradually realized that the maximum apparent frictional coefficient decreases with time (Scholz, 1998) . Granular material formation is one feasible explanation for slide weakening. The generation and transport of granular material at the plate contact can reconcile many counter-intuitive characteristics of major earthquakes ( Wang, 2015 and references therein) and also is critical in organizing small-scale ruptures to generate major earthquakes. Proper treatment of present granular material (Figure 1(d)) between plates also may provide a time frame for the next major earthquake on the same fault. From Figure 1(d), it seems, globally, the granular material thickness distribution is positively correlated with major earthquakes’ occurrence frequency. Following sections are discussion of earthquakes over two selected regions, as actual applications of the numerical models.\n\n4. Results\n\nThere are three interlaced mechanisms working together to make the earthquake prediction tricky for existing rubrics. The mechanisms we identified after sedulous essays using the singular SEGMENT model and remote sensing data are whetted against 73 historical cases, all within the reanalyses period with quality precipitation data. The reproduced earthquakes are viable, especially for cases after 2003, the starting point of the GRACE measurements, and before 2015, the aging of the current GRACE satellites. Basic model background verifications are available in Section 3, including simulated surface velocities against GPS measurements. Following sub-sections summarize model verification and simulations over three representative regions where groundwater fluctuations play significant role in affecting occurrence of major earthquakes. Although this study attempts a projection of future major earthquakes, the results cannot be regarded too literally. This is because major earthquakes are rare events and the existing information suitable for training the dynamical model still is scarce. Consequently, many ad-hoc assumptions are still involved in the parameterizations and interpretation of the input data, making the predictions of spatial and temporal of the future occurrences only of reference values.\n\n4.1. Model Verification\n\nBy simulating the time scale of the evolution of strain and stress in the fault zone, using SEGMENT, with appropriate parameter setting, assisted by remote sensing information, the timing and location of major earthquakes (Mw > 5 globally) can be estimated, with uncertainties in timing reduced down to a year and location error within 120 km (root mean squared error for historical records during 1979-2015). The primary limitation on the prediction accuracy resides with the resolution of CRUST-1.0 model, depending on which SEGMENT resolves fault geometry and composing material properties. As with all prediction system, the uncertainty grows exponentially with forecasting time. In the following discussion, the uncertainty level will be presented as necessary.\n\nAlthough global results are obtained, the following discussion focuses on three regions: The Tibetan Plateau and surrounding regions (TP), the Cascadia, and the Andes subduction faults. All three regions are adequately covered by GPS measurements and have large enough land areas that can be sensed by GRACE for the mass fluctuations around the faults. Primarily from different fault geometries, plate opposing motion speeds, and the distribution patterns of inter-plate granular material, earthquakes over the selected three regions are of various morphologies. Major earthquakes over the Andes and the TP refer respectively to those with magnitude > Mw7.5 and >Mw5.5. Without explicit statements, >Mw5 over other world regions is regarded as major earthquakes.\n\n4.2. Earthquake Activity in the TP Region: Correlations with Groundwater Fluctuations\n\nLarge scale (e.g., thousand kilometers) groundwater fluctuations exert extra stress on fault interface. Almost all major (>Mw5.5) earthquakes during 2003-2016 over the Tibetan Plateau occurred during low groundwater phases (figure not shown). Although the fluctuation of groundwater is secondary to plate motion in causing stress build up, it is an “extra”, and usually “the last straw that breaks the camel”. Analyzing the stress field clearly indicates that once two plates are coupled together, the underlain plate drags the lighter overriding plate downward and they go in tandem deeper into the Earth, pushing away the higher density upper mantle material. This causes mass loss around the fault region. This process, although only causing mass loss at a rate two orders of magnitude smaller than groundwater discharge, is however aided when the region is experiencing mass loss due to groundwater reduction (e.g., when evaporation steadily exceeds precipitation during extended droughts). The recent Nepal earthquakes of April and early May, 2015, and a third earthquake in Burma a year later all resides inside the mass loss region. The temporal sequence of the three major earthquakes and the ruptures’ spatial pattern (starting from northwest and extending to southeast) can be explained by the granular material production and transportation condition as proposed here. They represent key stages of strain energy release in the lateral direction, along the transverse direction of the fault by unlocking the granular sticky spots scattered over the plate interface. The order of unlocking strictly followed the granular material accumulation on the shear zone.\n\nThe three epicenter-consisting patches (consisting the three respective epicenters) were all about to approach saturation by April 2015, after ~72-year’s stress accumulation (Bollinger et al., 2014; Men & Zhao, 2016) . The reason the west-most one ruptured first is because the aiding of groundwater reduction. Based on SEGMENT simulation, the groundwater recharge during the past 80 years has a decreasing trend over the Bay of Bengal region (74-104˚E; 15-35˚N as the region of interest), agree with the monsoonal precipitation trends as identified in Wang and Ding (2006) . On average, the groundwater reduction rate is ~31 Gt/yr. The super-imposed stress field produces ~85 km2 saturated area (as defined in Equation (2)) annually. The matured area is not spatially contiguous on the fault interface. Rather they scatter sporadically over the inter-seismically locked interface. Connection of these sporadically scattered saturated patches is critical for generating earthquakes. For example, the occurrence of the first earthquake (April 25, 2015; M7.8; epicenter at 28.147˚N, 84.708˚E) redistributed the granular material due underneath and re-assumes a stronger rock-rock contact. This lessens the degree of saturation of the second major earthquake (occurred on May 12, 2015 of M7.3 magnitude and epicenter located 200 km to its southeast at 27.973˚N, 85.963˚E) region and it took another month for its occurrence. The earthquakes over Burma (Mw6.9 on April 12, 2016 at 23.13˚N, 94.9˚E and Mw6.8 on August 25, 2016 at 20.9˚N, 94.6˚E), further to the southeast, were not directly related to these two events. They were a result of the lightening of the Bay of Bengal region, to be further addressed later. Unlike some recent studies (e.g., Men & Zhao, 2016 ) that predicted the future earthquake (after the Nepal earthquake 2015) would be to the north-west sector, SEGMENT simulation indicates that the sector to the south-east also is becoming unstable. Actually the maturation pattern is a jumping between the eastern sector and the western sector, with cluster centers (30˚N, 84.4˚E) and (27.5˚N; 86.2˚E). Consequently, major earthquakes also occur alternatively (between these two regions in a bi-polar mode).\n\nThe Longmenshan fault zone (LFZ, 102-106˚E; 30-33˚N) is a vivid incarnation of the “non-unique solution” configuration as presented in Section 2. The LFZ defines the eastern margin of the Tibetan Plateau. Very recent synthetic approach (gravity survey, seismic reflection profile and earthquake focal mechanism data, and sediment analysis; personal communication with Z-X. Li, 2016) indicated that the LFZ is roughly composed of two segments of variant formations. The central-northern section, formed ~40 Myr ago (Jiang & Li, 2014; Richardson et al., 2008) , is a lithospheric through-cut fault zone that possesses low elastic strength but with strong dextral strike-slip motions. In contrast, the southern sector is a crustal thrust zone dominated by shallow-angle thrust motion of the TP over the moderately stiff South China Plate. This Mobius-twisted fault plane and contrasting behaviors along the LFZ, in addition to providing kinematically favorable conditions for instability, have also profound influence on groundwater reservoir. The Mw7.8 Wenchuan earthquake further lent evidence of the importance of gravity fields in fostering major earthquakes.\n\nExamining the mass change fields averaged between 2003 and 2008, it is clear that Wenchuan resides at the center of a saddle region (Figure 3 in Ren et al. (2015) ). For earthquakes fostered along the same fault (but temporally consecutive), the tectonic plate motion caused compressing usually invokes very similar saturation patterns. The primary reason for the randomness of epicenters is due to more transient perturbations such as groundwater fluctuation. The super imposed field cause a specific point to reach saturation first and propagate away to cause a large rupture area. The rupture can grow because, to seize the motion of the unstable plates, neighboring unsaturated regions may be activated. If the propagation encountered too many unsaturated regions, the momentum is quickly lost and this strongly limits the magnitude of the earthquake. On the other hand, if the propagation starts from a region surrounded by a ripen neighborhood, a positive feedback forms and resulted into a large magnitude earthquake. By initiating the rupture near Wenchuan, a more efficient energy release is achieved. From SEGMENT simulation with future precipitation scenarios under A1B scenario provided by a small-volume climate model ensemble (as listed in Table 1), the next major earthquake will be within Ganshu province (epicenter located along 104.8˚E longitudinal arc between 33 and 35˚N) to the north east, ~120 ± 45 years later of magnitude ~Mw7. For the far-larger TP region, the India-Eurasin collisional system, while thickening the elastic lithosphere, created numerous vertically extended crevasses. These crevasses are very effective in channeling groundwater into depth. Combined with TP’s orographic effects, precipitation variation transfers readily into extra lateral stress on the faults. Therefore, groundwater fluctuation becomes a dynamic factor affecting the epicenter location and the magnitude of major earthquakes. Because the persistent gravity reduction in the Bay of Bengal area and Southwest China, coupled with insignificant changes over the plateau and the northern Tarim and very arid Qaidam basins, a saddle-shape super-imposed stress field is formed and maintained over the past half century. The natural, geological background-deduced recurrence frequency (“dry” frequency) is greatly reduced over the region, especially over the Southeast Asia peninsula. The fault zone over Burma likely becomes unstable in about two years. Very interestingly, the epicenters of the upcoming earthquakes, affected by the weakening of monsoonal precipitation, are located along 96 ± 1˚E parallel between 17-28˚N (most frequently 17-23˚N sector), roughly parallel the Ayeyarwady river basin.\n\nFactors controlling earthquake irregularity over Cascadia and Andean share similarity with the TP region but have their respective specialties. Here provides a brief summary before detailed discussion. The granular material filled tunnels left behind by seamounts located on top of the Nazca and Juan Fernandez ridges foster many major Andean earthquakes. Because of the proximity of ENSO, the groundwater fluctuation over the Peru and Chilean coasts bear apparent 4 - 7 year fluctuations. However, unlike the collisional system of TP, there lack through-cut faults to channel precipitation into deeper depth (closer to the locked fault zone). From Section 3.1.1, the deeper the groundwater reservoir resides, the more efficient its mechanical effects. As a result, groundwater fluctuations have to exert effects mechanical-indirectly through aiding granular material generation and distribution. Consequently, earthquakes occur after a significant hiatus. Andes earthquakes are co-controlled by groundwater fluctuation and existing granular material distribution patterns.\n\nCascadia is similar to Andean subduction faults such as Nazca and Juan Fernandez in that they are weak inter-plate coupling and the natural occurrence frequency of earthquakes are rather low. Also, there lack secondary through-cut faults on the continental plates. Cascadia fault is unique in that local precipitation, although being of highest rate in North America, lacks annual to decadal variabilities. That explains the ~600 year natural major seismic cycle. To its south, along the pacific coast cordillera (ranges), lies the Mediterranean climate California, which precipitation is affected by teleconnection patterns such as ENSO and PDO and turned out to be sensitive to climate warming (Ren et al., 2011) . The Cascadia fault is thus more sensitive to Californian precipitation than local precipitation.\n\n4.3. Andes Earthquakes: Co-Controlled by Granular Material Pattern and Groundwater Fluctuation\n\nFigure 5 illustrates the present plates interface geometry at coastal Andes, as CRUST1.0 provided to SEGMENT. Panel (b) shows the ongoing base state creeping structure. The black arrows are horizontal flow speeds. The convergence in this direction is apparent but the total convergence is not apparent because the horizontal velocities rotate counterclockwise in the horizontal plane and the magnitudes are not changing so drastically. The coupling is weak as to the compressive stress between continental and oceanic plates. Especially, the ocean-continental subduction here does not lead to thickening of the continental plate. Baby et al. (1997) suggested that the subduction of oceanic lithosphere coupled with under-plating and a brief episode of gravity spreading contributed to crustal thickening in the back arc of the Central Andes. There still lacks a uniform theory that explains how very thick crusts develop (e.g., anomalies in the map of Christensen and Mooney (1995) ).\n\n4.3.1. Footprints of Seamounts-Role of Granular Material in Andes Earthquakes Spatial Pattern\n\nGranular accumulation along the Peru coast (South of Lima) extends deep inland (Figure 1(d)). The pattern reflects the way the Nazca Ridge passed through the continental plate. Because the thrust is at an acute angle to the continental plate motion, the channels on the overriding plate left by the seamounts is oriented at an angle to the Nazca Ridge’s moving direction (as shown on Figure 6).\n\nFigure 5. Present plates interface geometry at coastal Andes (represented in the SEGMENT model using CRUST 1.0 data). Panel (a) is crust thickness distribution. Panel (b) is a vertical cross-section along the red line in panel (a). The black arrows are horizontal flow speeds. The convergence in this direction is apparent but the total convergence is not apparent because the horizontal velocities rotate counterclockwise and the magnitudes are not changing so drastically. Panel (c) is the density structure in the same vertical cross-section (as in (b)). Over this fault, the motion is driven by mantle flow from underneath. The coupling is weak as to the compressive stress between continental and oceanic plates. Especially, we would like to emphasize that ocean-continental subduction here does not lead to thickening of the continental plate. Geometry data obtained from CRUST1.0 (Laske et al., 2013) .\n\nFigure 6. Granular material (shades are thickness in meter) around the subducting Nazca Ridge under the advancing South American plate. Currently locked regions (A, B to the left) with thick granular material accumulation are connected through macroscopic tunnels with seamounts (A, B and C on the right, over the Nazca Ridge). These seamounts grind the overriding continental plate and created the slanted tunnels. These tunnels collapsed and had earth surface footprints. Bold blue arrow is the moving direction of the flat oceanic slab with sea mounts A-C sitting on. The Hollow red arrow indicates the direction of the “tunnels” on the overriding continental plate plowed by seamounts. Note that the Mw8.1 Chilean 2014 event occurred exactly on the granular belt left by the “handle” of the Juan Fernandez Ridge.\n\nThis is just like chiseling a rotating disk. The granular material leftover also is distributed in stripes oriented the same way. In fact, all stripes originated from the sea mounts extends to the left flanks, due to the relative motion of the underlying and overriding plates. The granular material to the west (A’) ages as old as 14 Ma, whereas granular material at A (on the current flat lab) is newly generated. As the channels get older, they collapse and get wider but shallower, following the mechanisms as described in subsections 1.1 and 2.4 of the SM. These granular stripes become the soft belly of the fault and are the locales of many major earthquakes. Juan Fernandez to the south is very similar to Nazca ridge at present but are very different 3 Ma ago (Yanez et al., 2001; Yuan et al., 2002) . This chain of seamounts plowed beneath Puna at ~10-6 Ma ago like a hockey stick. At present, the “head” of the stick had melt into the upper mantle and only the “handle” left. However, the granular material grinded off by the “head and shaft” of the stick still acts as locales of earthquakes, especially at the seismic section of the subducting fault at the joint of Peru and Chile (at 20S; 70W). Also, the channels and the granular stripes left behind by the seamounts on the Juan Fernandez Ridge are more south-north oriented than those left behind by the Nazca Ridge. The epicenters of major earthquakes after 1800 correspond very well with the granular stripes. The granular stripes left by Juan Fernandez Ridge foster larger (but less frequent) earthquakes. For both seamount ridges, their left (northern) flanks foster larger earthquakes, because the granular accumulation patterns.\n\n4.3.2. Groundwater’s Indirect Control (With Apparent Hiatus) of Andes Earthquakes’ Spatio-Temporal Patterns\n\nFor the Peru and Chilean coast, El Nino and the opposite phase La Nina ocean-atmospheric teleconnection patterns (ENSO) play a critical role in determining the phase of groundwater. Earthquakes, especially those greater than Mw 5.5 but less than Mw7, do not necessarily in phase with the ENSO. This means that, unlike for TP region, the direct mechanical effect, such as indicated by Equation (1), may not be the dominant factor. However, the occurrence frequency of the major earthquakes is at ~4 - 7 years, same as ENSO. Model simulation confirmed that it is the fluctuation, rather than the actual weight of groundwater that is critical for the granular material generation and ensuing earthquakes. In other words, for this region, groundwater caused loading fluctuation affects earthquakes through granular material generation and evolution, rather than through the direct mechanical mechanism (Equation (2)).\n\nEven there are apparent temporal hiatus for most earthquakes between MW5.5 and MW7, all major earthquakes greater than Mw7.5 occurred at dry stages of the groundwater fluctuation cycles. For example, the 1998, 2003 and 2010 earthquakes all occurred at the nadir of the regional gravity field. In a certain sense, ENSO lessened the severity of the earthquakes over the Andes. Without ENSO, all the scattered < Mw7.5 earthquakes will occur as less frequent but more severe 1960-like super earthquakes. Through affecting precipitation partition into evaporation, runoff and infiltration into groundwater reservoir, topographic features have strong correlation of seismogenic zone segmentation (Bejar-Pizarro et al., 2013) . However, it simply is an illusion that surface topographic feature corresponds to ruptures at depth. Instead, the precipitation and the ensuing groundwater distribution contribute to earthquake irregularity. This is primarily because the input of strain energy from the periodic recharge and discharge of groundwater in a large regional area increases the prospect of locking formation, through granular fatigue. Monitoring the fluctuation tides of groundwater thus is a practical way for understanding earthquake cycle. Gravity mapping satellites such as GRACE provide essential assistance.\n\n4.4. Cascadia Subduction Zone’s Stability Affected by Californian Droughts\n\nDetailed discussion of Cascadia subduction fault, such as effective strength ( ${\\mu }^{\\prime }$ ), geometrical characteristics and unique geological conditions can be found in Wang and He (1999) . Here we would like to discuss the extra stress field super-imposed by precipitation fluctuations. Californian droughts (2011-2015), even though hundreds kilometers apart, significantly enhanced the instability of the Cascadia subduction zone. Due to the lack of vertically extended crevasses such as resulted from the India-Eurasia collisional system, groundwater fluctuations take effect through aiding granular material formation. Therefore there is a hiatus between droughts and the seismogenic effects. Rather than a direct temporal correspondence, there is a multiple year hiatus before the seismogenic effects from droughts to set on. For example, the extended Californian drought may contribute to fault instability 2 (southern side Oregon-Washington coast) to 10 years (northern BC coast) later. In addition, the Cascadia fault’s locking is special in that the oceanic plate still is creeping, which is clear from examining the velocity profile. It is locked in a sense that the neighboring regions have large gradients in motion speeds, signifying stress build-up. For example, the Olympic National Park coast of the Cascadia subduction zone is now inter-seismically locked (Figure 5). At the contact, the speed of the continental plate is close to zero. The oceanic plate, however, still is in motion. The continental plate is accumulating elastic potential energy at an annual rate of 7.6 × 1013 J/yr. The estimated apparent frictional coefficient is 0.12. If there is no perturbation on the loading field, the natural frequency of seismic cycle is ~600 year. By releasing elastic energy accumulated, an earthquake of Mw7.8 will be produced (of energy of ~45.7 PJ). However, from SEGMENT simulations, it is apparent that even the Californian drought exerted a cyclic stress on the interface and enhances the fatigue of the interface. If the same hydrological cycle over the past 50 years are representative for the past 800 years, the recurrence frequency is reduced to ~446 ± 70 years. As a result, a >Mw7.5 (~34 PJ) earthquake is expected within the upcoming decade. From the granular material distribution pattern, the rupture likely starts from the southern sector, or between the Oregon coast and the BC, Canada coast.\n\nThe logic of relating California droughts to future Cascadia earthquakes seems counter-intuitive since Cascadia has a lot of its own hydrological processes (it enjoys some of the highest precipitation rates in North America). The reason the fault is more sensitive to Californian precipitation is because of the fluctuation in precipitation, not the absolute amount. The high latitude precipitation over Cascadia region of Canada does not have large inter-annual to inter-decadal variation as in the Californian region, which is dually influenced by ENSO and PDO. More importantly, the Mediterranean climate is sensitive to climate warming (Ren et al., 2011) . Local Canadian precipitation fluctuation would be more direct but the magnitude of fluctuations truly is smaller than that from Californian precipitation. The North American plate cannot keep full rigidity over such a distance. The waveguide for propagating the bridging effect zig-zags ~5.5 wave lengths roughly along the pacific coast cordillera (ranges).\n\n5. Discussion\n\nThe quest to understand Earthquakes is an ancient and universal question. The new earthquake theory and its precepts presented here are a convergence of several factors: recent advances in remote sensing technology; recent, tectonically active stage motivated many researchers to re-evaluate previous assumptions (Wang, 2015) , sometimes to a radical degree, and most critically, a well-vetted landslide model as prototype of an advanced numerical modeling system to handle the mechanics of inter-plate earthquakes and to verify the original hypotheses proposed here.\n\nEarthquake occurs when the plate-coupling stress determined repose angle reaches the sum of two angles representing respectively fault geometry and fault strength. Accurate representation of the evolution of fault strength and repose angle thus is critical for earthquake simulation and prediction. The earthquake triad is a generic framework that applies unanimously. Groundwater fluctuations, although being small, play a pivotal role in connecting the three components together in regulating earthquake irregularity. Groundwater may influence fault instability mechanical-directly by super-imposing a seismogenic lateral stress field that works in synergy with the plate coupling stress; and mechanical-indirectly, or weakens the fault from within, achieved by enhancing fault fatigue through granular material generation. Mechanical-directly alone, groundwater fluctuation superimposed stress field is on the same order as the residual of gravity-aided frictional stress and plate coupling stress thus is non-negligible in affecting the timing of major earthquakes.\n\nClosely centered on the earthquake triad, a minimum requirement for a working earthquake prediction system for realistic 3D fault configuration, the Groundwater aided Granular material Generation framework is proposed here and implemented in the well-vetted SEGMENT numerical modeling system. By exploring existing records of earthquakes using SEGMENT, it is found that for collisional systems, mechanical-direct effects from groundwater control the irregularity of major earthquakes. This is because the orographically spatially highly biased precipitation is effectively channeled into deeper depth by the prevalence of through-cut faults. Droughts elsewhere also are seismogenic but likely take effects mechanical-indirectly. For example, ENSO, as the dominant player for regional precipitation, has strong influence on the gravity field over the Andes region. The occurrence of major earthquakes, although bearing the same 4 - 7 years occurrence frequency of ENSO, has a significant hiatus when compared with the fluctuation in the gravity field. Similarly, the stability of the Cascadia fault is remotely affected by Californian droughts. The input of strain energy from the periodic recharge and discharge of groundwater in a large regional area increases the prospect of un-locking of seismically coupled plates. In a certain sense, climate fluctuations, through extended droughts determining groundwater loading pattern, affect the earthquake cycles.\n\nThis study is a complement to the prevailing Coulomb fracture criterion and Anderson’s theory of faulting (Jaeger, 1963; Fossen, 2010) . The earthquake triad concept reconciles and rectifies some existing conceptual errors in the field. All previous earthquake models considered one or at most two aspects of the triad hence lacked predictive capability. Solution non-uniqueness essentially roots in information paucity. Remotely sensed data found their use in deducing interpolate granular material distribution as well as verifying model simulated precipitation drainage into groundwater reservoir, greatly improved model performance. Although we still cannot make punctilious predictions in timing and location of epicenter, the performance of the model is trenchant enough and deserves further improvements, especially as we are entering a big data era.\n\nAcknowledgements\n\nWe are thankful to Drs Z-X Li and K. Wang for discussion on earthquake mechanisms and various modern approaches for monitoring stress build up. Dr Z-X Li also is appreciated for providing the Andes Geological Survey details. Drs Lance M Leslie, M. Lynch and J Bornman proof-read the first draft of this manuscript. The data used are all publicly available.\n\nThe author declares no competing financial interests.\n\nAppendix 1: Velocity Transformation between SEGMENT’s Sphere Crown Model and Cartesian System of ASPECT\n\nFor generic curvilinear system, we have the following identities.\n\n$\\text{d}\\Re ={H}_{i}{e}_{i}\\text{d}{q}_{i}$ (A1)\n\n$V=\\text{d}\\Re /\\text{d}t={H}_{i}{e}_{i}\\text{d}{q}_{i}/\\text{d}t={V}_{qi}{e}_{i}$ (A2)\n\n$V=\\text{d}\\Re /\\text{d}t=u{i}_{0}+v{j}_{0}+w{k}_{0}$ (A3)\n\nUsing (A2),\n\n$\\left\\{\\begin{array}{l}u=\\text{d}x/\\text{d}t=\\frac{\\partial x}{\\partial {q}_{1}}\\text{d}{q}_{1}/\\text{d}t+\\frac{\\partial x}{\\partial {q}_{2}}\\text{d}{q}_{2}/\\text{d}t+\\frac{\\partial x}{\\partial {q}_{3}}\\text{d}{q}_{3}/\\text{d}t=\\frac{\\partial x}{\\partial {q}_{i}}\\frac{{V}_{qi}}{{H}_{i}}\\\\ v=\\text{d}y/\\text{d}t=\\frac{\\partial y}{\\partial {q}_{1}}\\text{d}{q}_{1}/\\text{d}t+\\frac{\\partial y}{\\partial {q}_{2}}\\text{d}{q}_{2}/\\text{d}t+\\frac{\\partial y}{\\partial {q}_{3}}\\text{d}{q}_{3}/\\text{d}t=\\frac{\\partial y}{\\partial {q}_{i}}\\frac{{V}_{qi}}{{H}_{i}}\\\\ w=\\text{d}z/\\text{d}t=\\frac{\\partial z}{\\partial {q}_{1}}\\text{d}{q}_{1}/\\text{d}t+\\frac{\\partial z}{\\partial {q}_{2}}\\text{d}{q}_{2}/\\text{d}t+\\frac{\\partial z}{\\partial {q}_{3}}\\text{d}{q}_{3}/\\text{d}t=\\frac{\\partial z}{\\partial {q}_{i}}\\frac{{V}_{qi}}{{H}_{i}}\\end{array}$ (A4)\n\nApplying to a spherical coordinate system (Figure A1) with the forward transformation equation of\n\nFigure A1. Geometry representation of the spherical coordinate system as in SEGMENT. Panels (a)-(c) show respectively the projections of unit directional vector ${i}_{0},{j}_{0}$ and ${k}_{0}$ into the spherical coordinates ${\\theta }_{0},{\\lambda }_{\\text{0}}$ and ${r}_{0}$ . The geometrical relationships in (d) is repeatedly used in deriving the cosine relationships of angles composed by the intersection line and two vectors residing respectively in the two mutually perpendicular planes. The projection relationship between the unit cosine directions are summarized in (a)-(c).\n\n$\\left\\{\\begin{array}{l}x=r\\mathrm{cos}\\lambda \\mathrm{cos}\\theta \\\\ y=r\\mathrm{cos}\\lambda \\mathrm{sin}\\theta \\\\ z=r\\mathrm{sin}\\lambda \\end{array}$ (A5)\n\nwhere in geoscience convention, $\\theta$ is longitude, $\\lambda$ is latitude and r is the distance from the earth center (Figure A1). Still in right-handed coordinate system, lets define ${q}_{1}=\\theta ,{q}_{2}=\\lambda ,{q}_{3}=r$ . The corresponding Lame operators:\n\n$\\left\\{\\begin{array}{l}{H}_{\\theta }=\\sqrt{{\\left({{x}^{\\prime }}_{\\theta }\\right)}^{2}+{\\left({{y}^{\\prime }}_{\\theta }\\right)}^{2}+{\\left({{z}^{\\prime }}_{\\theta }\\right)}^{2}}=r\\mathrm{cos}\\lambda \\\\ {H}_{\\lambda }=\\sqrt{{\\left({{x}^{\\prime }}_{\\lambda }\\right)}^{2}+{\\left({{y}^{\\prime }}_{\\lambda }\\right)}^{2}+{\\left({{z}^{\\prime }}_{\\lambda }\\right)}^{2}}=r\\\\ {H}_{r}=\\sqrt{{\\left({{x}^{\\prime }}_{r}\\right)}^{2}+{\\left({{y}^{\\prime }}_{r}\\right)}^{2}+{\\left({{z}^{\\prime }}_{r}\\right)}^{2}}=1\\end{array}$ (A6)\n\nFrom Equation (A2),\n\n$\\left\\{\\begin{array}{l}{V}_{\\theta }=r\\mathrm{cos}\\lambda \\frac{\\text{d}\\theta }{\\text{d}t}\\\\ {V}_{\\lambda }=r\\frac{\\text{d}\\lambda }{\\text{d}t}\\\\ {V}_{r}=\\frac{\\text{d}r}{\\text{d}t}\\end{array}$ (A7)\n\nFrom Equation (A4), we get\n\n$\\left\\{\\begin{array}{l}u=\\frac{\\partial x}{\\partial {q}_{i}}\\frac{{V}_{qi}}{{H}_{i}}=\\frac{\\partial x}{\\partial \\theta }\\frac{{V}_{\\theta }}{{H}_{\\theta }}+\\frac{\\partial x}{\\partial \\lambda }\\frac{{V}_{\\lambda }}{{H}_{\\lambda }}+\\frac{\\partial x}{\\partial r}\\frac{{V}_{r}}{{H}_{r}}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{ }=\\left(-r\\mathrm{cos}\\lambda \\mathrm{sin}\\theta \\right)\\frac{{V}_{\\theta }}{r\\mathrm{cos}\\lambda }+\\left(-r\\mathrm{sin}\\lambda \\mathrm{cos}\\theta \\right)\\frac{{V}_{\\lambda }}{r}+\\left(\\mathrm{cos}\\lambda \\mathrm{cos}\\theta \\right)\\frac{{V}_{r}}{1}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{ }=-\\mathrm{sin}\\theta {V}_{\\theta }-\\mathrm{sin}\\lambda \\mathrm{cos}\\theta {V}_{\\lambda }+\\mathrm{cos}\\lambda \\mathrm{cos}\\theta {V}_{r}\\\\ v=\\frac{\\partial y}{\\partial {q}_{i}}\\frac{{V}_{qi}}{{H}_{i}}=\\frac{\\partial y}{\\partial \\theta }\\frac{{V}_{\\theta }}{{H}_{\\theta }}+\\frac{\\partial y}{\\partial \\lambda }\\frac{{V}_{\\lambda }}{{H}_{\\lambda }}+\\frac{\\partial y}{\\partial r}\\frac{{V}_{r}}{{H}_{r}}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{ }=\\left(r\\mathrm{cos}\\lambda \\mathrm{cos}\\theta \\right)\\frac{{V}_{\\theta }}{r\\mathrm{cos}\\lambda }+\\left(-r\\mathrm{sin}\\lambda \\mathrm{sin}\\theta \\right)\\frac{{V}_{\\lambda }}{r}+\\left(\\mathrm{cos}\\lambda \\mathrm{sin}\\theta \\right)\\frac{{V}_{r}}{1}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{ }=\\mathrm{cos}\\theta {V}_{\\theta }-\\mathrm{sin}\\lambda \\mathrm{sin}\\theta {V}_{\\lambda }+\\mathrm{cos}\\lambda \\mathrm{sin}\\theta {V}_{r}\\\\ w=\\frac{\\partial z}{\\partial {q}_{i}}\\frac{{V}_{qi}}{{H}_{i}}=\\frac{\\partial z}{\\partial \\theta }\\frac{{V}_{\\theta }}{{H}_{\\theta }}+\\frac{\\partial z}{\\partial \\lambda }\\frac{{V}_{\\lambda }}{{H}_{\\lambda }}+\\frac{\\partial z}{\\partial r}\\frac{{V}_{r}}{{H}_{r}}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{ }=0\\frac{{V}_{\\theta }}{r\\mathrm{cos}\\lambda }+\\left(r\\mathrm{cos}\\lambda \\right)\\frac{{V}_{\\lambda }}{r}+\\left(\\mathrm{sin}\\lambda \\right)\\frac{{V}_{r}}{1}\\\\ \\text{\\hspace{0.17em}}\\text{\\hspace{0.17em}}\\text{ }=\\mathrm{cos}\\lambda {V}_{\\lambda }+\\mathrm{sin}\\lambda {V}_{r}\\end{array}$ (A8)\n\nEquation (A8) is forward transformation, or using velocity in spherical coordinates (SEGMENT) to express velocity components in Cartesian system.\n\nDefine matrix $\\begin{array}{c}{I}_{1}=\\left(\\begin{array}{c}{i}_{0}\\\\ {j}_{0}\\\\ {k}_{0}\\end{array}\\right)\\left({\\theta }_{0},{\\lambda }_{\\text{0}},{r}_{0}\\right)=\\left(\\begin{array}{ccc}\\mathrm{cos}\\left({i}_{0},{\\theta }_{0}\\right)& \\mathrm{cos}\\left({i}_{0},{\\lambda }_{\\text{0}}\\right)& \\mathrm{cos}\\left({i}_{0},{r}_{0}\\right)\\\\ \\mathrm{cos}\\left({j}_{0},{\\theta }_{0}\\right)& \\mathrm{cos}\\left({j}_{0},{\\lambda }_{\\text{0}}\\right)& \\mathrm{cos}\\left({j}_{0},{r}_{0}\\right)\\\\ \\mathrm{cos}\\left({k}_{0},{\\theta }_{0}\\right)& \\mathrm{cos}\\left({k}_{0},{\\lambda }_{\\text{0}}\\right)& \\mathrm{cos}\\left({k}_{0},{r}_{0}\\right)\\end{array}\\right)\\\\ =\\left(\\begin{array}{ccc}-\\mathrm{sin}\\theta & -\\mathrm{cos}\\theta \\mathrm{sin}\\lambda & \\mathrm{cos}\\theta \\mathrm{cos}\\lambda \\\\ \\mathrm{cos}\\theta & -\\mathrm{sin}\\theta \\mathrm{sin}\\lambda & \\mathrm{sin}\\theta \\mathrm{cos}\\lambda \\\\ 0& \\mathrm{cos}\\lambda & \\mathrm{sin}\\lambda \\end{array}\\right),\\end{array}$ then Equation (A8) can be expressed as\n\n$\\left(\\begin{array}{c}u\\\\ v\\\\ w\\end{array}\\right)={I}_{1}\\left(\\begin{array}{c}{V}_{\\theta }\\\\ {V}_{\\lambda }\\\\ {V}_{r}\\end{array}\\right)$ (A9).\n\nInverse matrix of I1 is\n\n$\\begin{array}{c}{\\left({I}_{1}\\right)}^{-1}=\\left(\\begin{array}{c}{\\theta }_{0}\\\\ {\\lambda }_{\\text{0}}\\\\ {r}_{0}\\end{array}\\right)\\left({i}_{0},{j}_{0},{k}_{0}\\right)=\\left(\\begin{array}{ccc}\\mathrm{cos}\\left({\\theta }_{0},{i}_{0}\\right)& \\mathrm{cos}\\left({\\theta }_{0},{j}_{0}\\right)& \\mathrm{cos}\\left({\\theta }_{0},{k}_{0}\\right)\\\\ \\mathrm{cos}\\left({\\lambda }_{\\text{0}},{i}_{0}\\right)& \\mathrm{cos}\\left({\\lambda }_{\\text{0}},{j}_{0}\\right)& \\mathrm{cos}\\left({\\lambda }_{\\text{0}},{k}_{0}\\right)\\\\ \\mathrm{cos}\\left({r}_{0},{i}_{0}\\right)& \\mathrm{cos}\\left({r}_{0},{j}_{0}\\right)& \\mathrm{cos}\\left({r}_{0},{k}_{0}\\right)\\end{array}\\right)\\\\ =\\left(\\begin{array}{ccc}-\\mathrm{sin}\\theta & \\mathrm{cos}\\theta & 0\\\\ -\\mathrm{cos}\\theta \\mathrm{sin}\\lambda & -\\mathrm{sin}\\theta \\mathrm{sin}\\lambda & \\mathrm{cos}\\lambda \\\\ \\mathrm{cos}\\theta \\mathrm{cos}\\lambda & \\mathrm{sin}\\theta \\mathrm{cos}\\lambda & \\mathrm{sin}\\lambda \\end{array}\\right)\\end{array}$ (A10).\n\nNotice also that transpose of I1 also is inverse of I1 (I1 is thus an orthogonal matrix). In addition, I1 also is unitary and normal matrix (det(I1) = I).\n\nFrom Equation (A9), $\\left(\\begin{array}{c}{V}_{\\theta }\\\\ {V}_{\\lambda }\\\\ {V}_{r}\\end{array}\\right)={\\left({I}_{1}\\right)}^{-1}\\left(\\begin{array}{c}u\\\\ v\\\\ w\\end{array}\\right)$ , that is\n\n$\\left\\{\\begin{array}{l}{V}_{\\theta }=\\left(-\\mathrm{sin}\\theta \\right)u+\\left(\\mathrm{cos}\\theta \\right)v\\\\ {V}_{\\lambda }=\\left(-\\mathrm{cos}\\theta \\mathrm{sin}\\lambda \\right)u-\\left(\\mathrm{sin}\\theta \\mathrm{sin}\\lambda \\right)v+\\left(\\mathrm{cos}\\lambda \\right)w\\\\ {V}_{r}=\\left(\\mathrm{cos}\\theta \\mathrm{cos}\\lambda \\right)u+\\left(\\mathrm{sin}\\theta \\mathrm{cos}\\lambda \\right)v+\\left(\\mathrm{sin}\\lambda \\right)w\\end{array}$ (A11)\n\nNote also that I1 and its inverse matrix also denote the projection of the unit vectors between the curvilinear and Cartesian coordinate system. For example, simply look at the geometry in Figure (A1), it is apparent that the projection of ${\\lambda }_{\\text{0}}$ in the $\\left({i}_{0},{j}_{0},{k}_{0}\\right)$ directions is exactly the second row in the inverse matrix (Equation (A10)). Similarly, projection of ${j}_{0}$ in the curvilinear coordinate system is the second row in I1, or, ${j}_{0}=\\mathrm{cos}\\theta {\\theta }_{0}-\\mathrm{sin}\\lambda \\mathrm{sin}\\theta {\\lambda }_{\\text{0}}+\\mathrm{cos}\\lambda \\mathrm{sin}\\theta {r}_{0}$ . Notice the similarity with Equation (A8). Actually, plugging these unit vectors’ projection expressions into Equations (A2) and (A3), the velocity transformation equations can be similarly obtained. For example,\n\n$\\begin{array}{c}v=V\\cdot {j}_{0}=\\left({V}_{\\theta }{\\theta }_{0}+{V}_{\\lambda }{\\lambda }_{\\text{0}}+{V}_{r}{r}_{0}\\right)\\left(\\mathrm{cos}\\theta {\\theta }_{0}-\\mathrm{sin}\\lambda \\mathrm{sin}\\theta {\\lambda }_{\\text{0}}+\\mathrm{cos}\\lambda \\mathrm{sin}\\theta {r}_{0}\\right)\\\\ ={V}_{\\theta }\\mathrm{cos}\\theta -\\mathrm{sin}\\lambda \\mathrm{sin}\\theta {V}_{\\lambda }+\\mathrm{cos}\\lambda \\mathrm{sin}\\theta {V}_{r}.\\end{array}$ This is\n\nexactly the second equation of Equation (A8). Similarly, using the direction-cosine relationships, for example, using the projections of ${\\theta }_{0}$ in the original Cartesian coordinate system:\n\n$\\mathrm{cos}\\left({\\theta }_{0},{i}_{0}\\right)=-\\mathrm{sin}\\theta ;\\mathrm{cos}\\left({\\theta }_{0},{j}_{0}\\right)=\\mathrm{cos}\\theta ;\\mathrm{cos}\\left({\\theta }_{0},{k}_{0}\\right)=0$ . It is straightforward to obtain that\n\n${V}_{\\theta }=V\\cdot {\\theta }_{0}=\\left(u{i}_{0}+v{j}_{0}+w{k}_{0}\\right)\\left(-\\mathrm{sin}\\theta {i}_{0}+\\mathrm{cos}\\theta {j}_{0}+0{k}_{0}\\right)=-\\mathrm{sin}\\theta u+\\mathrm{cos}\\theta v$ , exactly the first equation of Equation (A11).\n\nAppendix 2: Viscoelastic Relaxation after Major Earthquakes\n\nRheology of mantle material is generally assumed to be viscoelastic. A combination of Maxwell fluid and Kelvin solid in the form of Burgers rheology is commonly accepted in the geophysics community.\n\nIn Figure A2, the Burgers rheology is represented by a serial connection of a Maxwell fluid of viscosity ${\\eta }_{M}$ and rigidity ${\\mu }_{M}$ and a Kelvin solid of viscosity ${\\eta }_{K}$ and rigidity ${\\mu }_{K}$ . ${\\tau }_{M}$ and ${\\tau }_{K}$ are Maxwell and Kelvin relaxation times.\n\n1) A maxwell fluid is a viscoelastic material having the properties of both elasticity and viscosity. It can be represented by a purely viscous damper and a purely elastic spring connected in series. In this configuration, under applied axial stress, the total stress, ${\\sigma }_{Total}$ and the total strain ${\\epsilon }_{Total}$ can be defined as follows. ${\\sigma }_{Total}={\\sigma }_{D\\left(amper\\right)}={\\sigma }_{S\\left(pring\\right)}$ ; ${\\epsilon }_{Total}={\\epsilon }_{D\\left(amper\\right)}+{\\epsilon }_{S\\left(pring\\right)}$ . Where the subscript D indicates the strain/stress in the damper and the substract S indicates the stress/strain in the spring. For Newtonian fluids, we have\n\n$\\frac{\\text{d}{\\epsilon }_{Dl}}{\\text{d}t}={\\sigma }_{D}{\\eta }_{M}={\\sigma }_{T}{\\eta }_{M}$ ; and from Hook’s law for elastic modulus form $\\frac{\\text{d}{\\epsilon }_{S}}{\\text{d}t}=\\frac{1}{E}\\frac{\\text{d}{\\sigma }_{S}}{\\text{d}t}=\\frac{1}{E}\\frac{\\text{d}{\\sigma }_{T}}{\\text{d}t}$ . The total strain rate thus follows\n\n$\\frac{\\text{d}{\\epsilon }_{T}}{\\text{d}t}=\\frac{\\text{d}{\\epsilon }_{D}}{\\text{d}t}+\\frac{\\text{d}{\\epsilon }_{S}}{\\text{d}t}=\\frac{{\\sigma }_{T}}{\\eta }+\\frac{1}{E}\\frac{\\text{d}{\\sigma }_{T}}{\\text{d}t}.$ (A12)\n\nIn a Maxwell material, stress $\\sigma$ , strain $\\epsilon$ , and their rates of change w.r.t. time are governed by equations of the form $\\stackrel{˙}{\\epsilon }=\\frac{\\sigma }{\\eta }+\\frac{\\stackrel{˙}{\\sigma }}{E}.$\n\nEquation (A12) can be applied either to the shear stress or to the uniform tension in a material. In the former case, the viscous component sorresponds to that for a Newtonian fluid. In the latter case, it has a slightly different meaning in relating stress and rate of strain. The Maxwell model usually is applied to the case of small deformations. For large deformations one should include some geometrical non-linearity. The following we discuss two special scenarios: a) effect of a sudden deformation, or the stress evolution after a sudden deformation; and b) Effect of a sudden stress, or how the strain evolves upon a sudden applied stress in the axial direction.\n\na) Effect of a sudden deformation\n\nIf an M material is suddenly deformed and held to a strain of ${\\epsilon }_{0}$ , then the stress decays with a characteristic time of $\\frac{\\eta }{E}$ .\n\nA brief proof of this starts from the governing equation (Equation A12) with initial condition of $\\epsilon \\left(t=0\\right)={\\epsilon }_{0}$ and strain rate term $\\stackrel{˙}{\\epsilon }=0$ . That is\n\n$\\left\\{\\begin{array}{l}0=\\frac{\\sigma }{\\eta }+\\frac{\\stackrel{˙}{\\sigma }}{E}\\\\ {\\epsilon \|}_{t=0}={\\epsilon }_{0},\\text{}\\sigma =\\sigma \\left({\\epsilon }_{\\text{0}}\\right)={\\sigma }_{0};\\text{\\hspace{0.17em}}{\\epsilon }_{0}={\\sigma }_{0}/E\\end{array}$ (A13).\n\nUsing the generic solution for the first order ordinary differential equation (Appendix 3), it is straightforward to obtain the solution for Equation (A13).\n\n$\\sigma \\left(t\\right)={\\sigma }_{0}{\\text{e}}^{-\\frac{E}{\\eta }t}$ (A14)\n\nFigure A2. Burgers fluid as a serial connection of a Maxwell fluid (subscripts M) and a Kelvin solid (subscripts K). The Maxwell fluid and Kelvin solid are respectively serial and parallel connection of pure spring and pure viscous damper filled with Newtonian fluids.\n\nIf we then free the material at time t1, then the elastic element will sprong back by the value of\n\n${\\epsilon }_{back}=-\\frac{\\sigma \\left({t}_{1}\\right)}{E}={\\epsilon }_{0}{\\text{e}}^{-\\frac{E}{\\eta }{t}_{1}}$ (A15)\n\nThis is because at time t1, from Equation (A14), keeping the existing deformation need a confining pressure of $\\sigma \\left({t}_{1}\\right)={\\sigma }_{0}{\\text{e}}^{-\\frac{E}{\\eta }{t}_{1}}={\\sigma }_{0}\\mathrm{exp}\\left(-\\frac{E}{\\eta }{t}_{1}\\right)$ . If remove the confining pressure at this moment, the elastic recovery would be $\\frac{\\sigma \\left({t}_{1}\\right)}{E}=\\frac{1}{E}{\\sigma }_{0}\\mathrm{exp}\\left(-\\frac{E}{\\eta }{t}_{1}\\right)\\stackrel{\\text{usingEq}\\text{.}\\left(\\text{A13}\\right)’\\text{sinitialconditionexpression}}{=}{\\epsilon }_{0}\\mathrm{exp}\\left(-\\frac{E}{\\eta }{t}_{1}\\right)$ . As the viscous element would not return to its original length, the irreversible component of deformation can be simplified as\n\n${\\epsilon }_{irreversible}={\\epsilon }_{0}-{\\epsilon }_{back}={\\epsilon }_{0}\\left(1-\\mathrm{exp}\\left(-\\frac{E}{\\eta }{t}_{1}\\right)\\right)$ (A16)\n\nb) Effect of a sudden stress\n\nIf an M material is suddenly subject to a stress ${\\sigma }_{0}$ , then the elastic element would suddenly deform and the viscous element would deform with a constant rate\n\n$\\epsilon \\left(t\\right)=\\underset{reversible}{\\frac{{\\sigma }_{0}}{E}}+\\underset{irreversible}{\\frac{{\\sigma }_{0}}{\\eta }t}$ . (A17)\n\n(Recall the definition of Newtonian fluid as $\\stackrel{˙}{\\epsilon }=\\frac{\\sigma }{\\eta }$ , part of Equation (A12)).\n\nIf at some time t1 we would release the material, then the deformation of the elastic element would be the spring back deformation and the deformation to the viscous part would not change back. That is at any arbitrary time t1,\n\n${\\epsilon }_{reversible}=\\frac{{\\sigma }_{0}}{E};{\\epsilon }_{irreversible}=\\frac{{\\sigma }_{0}}{\\eta }{t}_{1}$ .\n\nConflicts of Interest\n\nThe authors declare no conflicts of interest.\n\nCite this paper\n\nRen, D. (2019) An Earthquake Model Based on Fatigue Mechanism—A Tale of Earthquake Triad. Journal of Geoscience and Environment Protection, 7, 290-326. doi: 10.4236/gep.2019.78021.\n\n Andes Geological Survey (2006). Audet, P., & Burgmann, R. (2014). Possible Control of Subduction Zone Slow Earthquake Periodicity by Silica Enrichment. Nature, 510, 389-392.https://doi.org/10.1038/nature13391 Baby, P., Rochat, P., Mascle, G., & Hérail, G. (1997). Neogene Shortening Contribution to Crustal Thickening in the Back Arc of the Central Andes. Geology, 25, 883-886. https://doi.org/10.1130/0091-7613(1997)025<0883:NSCTCT>2.3.CO;2 Bejar-Pizarro, M., Socquet, A., Armijo, R., Carrizo, D., & Genrich, J. (2013). Andean Structural Control on Interseismic Coupling in the North Chile Subduction Zone. Nature Geoscience, 6, 462-467. https://doi.org/10.1038/ngeo1802 Bercovici, D., & Ricard, Y. (2014). Plate Tectonics, Damage and Inheritance. Nature, 508, 513-516. https://doi.org/10.1038/nature13072 Bollinger, L., Sapkota, S., Tapponnier, P., Klinger, Y., Rizza, M., Van der Woerd, J., Tiwari, D., Pandey, R., Bitri, A., & Bes de Berc, S. (2014). Estimating the Return Times of Great Himalayan Earthquakes in Eastern Nepal: Evidence from the Patu and Bardibas Strands of the Main Frontal Thrust. Journal of Geophysical Research: Solid Earth, 119, 7123-7163. https://doi.org/10.1002/2014JB010970 Cavalie, O., & Jónsson, S. (2014). Block-Like Plate Movements in Eastern Anatolia Observed by InSAR. Geophysical Research Letters, 41, 26-31.https://doi.org/10.1002/2013GL058170 Chlieh, M., Perfettini, H., Tavera, H., Avouac, J. P., Remy, D., Nocquet, J. M., Rolandone, F., Bondoux, F., Gabalda, G., & Bonvalot, S. (2011). Interseismic Coupling and Seismic Potential along the Central Andes Subduction Zone. Journal of Geophysical Research: Solid Earth, 116, B12405. https://doi.org/10.1029/2010JB008166 Christensen, N., & Mooney, W. (1995). Seismic Velocity Structure and Composition of the Continental Crust: A Global View. Journal of Geophysical Research: Solid Earth, 100, 9761-9788. https://doi.org/10.1029/95JB00259 Cowan, D., Cladouhos, T., & Morgan, J. (2003). Structural Geology and Kinematic History of Rocks Formed along Low-Angle Faults, Death Valley, California. Geological Society of America Bulletin, 115, 1230-1248. https://doi.org/10.1130/B25245.1 Dietrich, W., & Perron, J. (2006). The Search for a Topographic Signature of Life. Nature, 439, 411-418. https://doi.org/10.1038/nature04452 Famiglietti, J., & Rodell, M. (2013). Water in the Balance. Science, 340, 1300-1301.https://doi.org/10.1126/science.1236460 Fossen, H. (2010). Structural Geology (463 p). Cambridge: Cambridge University Press. https://doi.org/10.1017/CBO9780511777806 Gao, X., & Wang, K. (2014). Strength of Stick-Slip and Creeping Subduction Megathrusts from Heat Flow Observations. Science, 345, 1038-1041. https://doi.org/10.1126/science.1255487 Gao, X., & Wang, K. (2017). Rheological Separation of the Megathrust Seismogenic Zone and Episodic Tremor and Slip. Nature, 543, 416-419.https://doi.org/10.1038/nature21389 Gilman, J. (1960). Direct Measurements of the Surface Energy of Crystals. Journal of Applied Physics, 31, 2208-2218. https://doi.org/10.1063/1.1735524 Hayes, G., Wald, D., & Johnson, R. (2012). Slab 1.0: A Three-Dimensional Model of Global Subduction Zone Geometries. Journal of Geophysical Research: Solid Earth, 117, B01302. https://doi.org/10.1029/2011JB008524 Hyndman, R., McCrory, P., Wech, A., Kao, H., & Ague, J. (2015). Cascadia Subducting Plate Fluids Channeled to Fore-Arc Mantle Corner: ETS and Silica Deposition. Journal of Geophysical Research: Solid Earth, 120, 4344-4358. https://doi.org/10.1002/2015JB011920 Jaeger, J. (1963). Extension Failures in Rocks Subject to Fluid Pressure. Journal of Geophysical Research, 68, 6066-6067. https://doi.org/10.1029/JZ068i021p06066 Jiang, X., & Li, Z. (2014). Seismic Reflection Data Support Episodic and Simultaneous Growth of the Tibetan Plateau Since 25 Myr. Nature Communications, 5, 5453.https://doi.org/10.1038/ncomms6453 Jop, P., Forterre, Y., & Pouliquen, O. (2006). A Constitutive Law for Dense Granular Flows. Nature, 441, 727-730. https://doi.org/10.1038/nature04801 Kronbichler, M., Heister, T., & Bangerth, W. (2012). High Accuracy Mantle Convection Simulation through Modern Numerical Methods. Geophysics Journal International, 191, 12-29. https://doi.org/10.1111/j.1365-246X.2012.05609.x Laske, G., Masters, G., Ma, Z., & Pasyanos, M. (2013). Update on CRUST1.0—A 1-Degree Global Model of Earth’s Crust. Lay, T. (2015). The Surge of Great Earthquakes from 2004 to 2014. Earth and Planetary Science Letters, 409, 133-146. https://doi.org/10.1016/j.epsl.2014.10.047 Liu, X., Wang, Y., & Yan, S. (2017). Ground Deformation Associated with Exploitation of Deep Groundwater in Cangzhou City Measured by Multi-Sensor Synthetic Aperture Radar Images. Environmental Earth Sciences, 76, 6. https://doi.org/10.1007/s12665-016-6311-0 Liu, Z., & Bird, P. (2002). Finite Element Modeling of Neotectonics in New Zealand. Journal of Geophysical Research: Solid Earth, 107, ETG 1-1-ETG 1-18. https://doi.org/10.1029/2001JB001075 Loveless, J., & Meade, B. (2010). Geodetic Imaging of Plate Motions, Slip Rates, and Partitioning of Deformation in Japan. Journal of Geophysical Research: Solid Earth, 115, B09301. https://doi.org/10.1029/2008JB006248 Men, K., & Zhao, K. (2016). The 2015 Nepal M8.1 Earthquake and the Prediction for M ≥ 8 Earthquakes in West China. Natural Hazards, 82, 1767-1777. https://doi.org/10.1007/s11069-016-2268-2 Pritchard, M., & Simons, M. (2006). An Aseismic Slip Pulse in Northern Chile and along-Strike Variations in Seismogenic Behavior. Journal of Geophysical Research: Solid Earth, 111, B08405. https://doi.org/10.1029/2006JB004258 Ren, D. (2014a). Storm-Triggered Landslides in Warmer Climates. New York: Springer. Ren, D. (2014b). The Devastating Zhouqu Storm-Triggered Debris Flow of August 2010: Likely Causes and Possible Trends in a Future Warming Climate. Journal of Geophysical Research: Atmospheres, 119, 3643-3662. https://doi.org/10.1002/2013JD020881 Ren, D., & Leslie, L. (2014). Effects of Waves on Tabular Ice-Shelf Calving. Earth Interactions, 18, 1-28. https://doi.org/10.1175/EI-D-14-0005.1 Ren, D., Fu, R., Leslie, L. M., & Dickinson, R. (2011). Predicting Storm-Triggered Landslides. Bulletin of the American Meteorological Society, 92, 129-139.https://doi.org/10.1175/2010BAMS3017.1 Ren, D., Leslie, L. M., & Karoly, D. (2008). Landslide Risk Analysis Using a New Constitutive Relationship for Granular Flow. Earth Interactions, 12, 1-16.https://doi.org/10.1175/2007EI237.1 Ren, D., Leslie, L. M., & Lynch, M. (2012). Verification of Model Simulated Mass Balance, Flow Fields and Tabular Calving Events of the Antarctic Ice Sheet against Remotely Sensed Observations. Climate Dynamics, 40, 2617-2636. https://doi.org/10.1007/s00382-012-1464-3 Ren, D., Leslie, L., Shen, X., Hong, Y., Duan, Q., Mahmood, R., Li, Y., Huang, G., Guo, W., & Lynch, M. (2015). The Gravity Environment of Zhouqu Debris Flow of August 2010 and Its Implication for Future Recurrence. International Journal of Geosciences, 6, 317-325. https://doi.org/10.4236/ijg.2015.64025 Richardson, N., Densmore, A., Seward, D., Fowler, A., Wipf, M., Ellis, M., Yong, L., & Zhang, Y. (2008). Extraordinary Denudation in the Sichuan Basin: Insights from Low-Temperature Thermochronology Adjacent to the Eastern Margin of the Tibetan Plateau. Journal of Geophysical Research: Solid Earth, 113, B04409.https://doi.org/10.1029/2006JB004739 Rowe, C., Kirkpatrick, J., & Brodsky, E. (2012). Fault Rock Injections Record Paleo-Earthquakes. Earth and Planetary Science Letters, 335-336, 154-166. Sawai, Y., Satake, K., Kamataki, T., Nasu, H., Shishikura, M., Atwater, B., Horton, B., Kelsey, H., Nagumo, T., & Yamaguchi, M. (2004). Transient Uplift after a 17th-Century Earthquake along the Kuril Subduction Zone. Science, 306, 1918-1920. https://doi.org/10.1126/science.1104895 Scholz, C. (1998). Earthquakes and Friction Laws. Nature, 391, 37-42.https://doi.org/10.1038/34097 Sun, T., Wang, K., Iinuma, T., Hino, R., He, J., Fujimoto, H., Kido, M., Osada, Y., Miura, S., Ohta, Y., & Hu, Y. (2014). Prevalence of Viscoelastic Relaxation after the 2011 Tohoku-Oki Earthquake. Nature, 514, 84-87. https://doi.org/10.1038/nature13778 Timoshenko, S., & Gere, J. (1963). Theory of Elastic Stability. New York: McGraw-Hill. Ujjie, K., Yamaguchi, A., Kimura, G., & Toh, S. (2007). Fluidization of Granular Material in a Subduction Thrust at Seismogenic Depth. Earth and Planetary Science Letters, 259, 307-318. https://doi.org/10.1016/j.epsl.2007.04.049 Van der Veen, C., & Whillans, I. (1989). Force Budget: I. Theory and Numerical Methods. Journal of Glaciology, 35, 53-60. https://doi.org/10.3189/002214389793701581 Voss, K., Famiglietti, J., Lo, M., de Linage, C., Rodell, M., & Swenson, S. (2013). Groundwater Depletion in the Middle East from GRACE with Implications for Transboundary Water Management in the Tigris-Euphrates-Western Iran Region. Water Resources Research, 49, 904-914. https://doi.org/10.1002/wrcr.20078 Wallace, L., Beavan, J., & McCaffrey, R. (2004). Subduction Zone Coupling and Tectonic Block Rotation in the North Island, New Zealand. Journal of Geophysical Research: Solid Earth, 109. https://doi.org/10.1029/2004JB003241 Wallace, L., Ellis, S., & Mann, P. (2009). Collisional Model for Rapid Fore-Arc Block Rotations, Arc Curvature, and Episodic Back-Arc Rifting in Subduction Settings. Geochemistry, Geophysics, Geosystems, 10, Q05001. https://doi.org/10.1029/2008GC002220 Wallace, L., Kaneko, Y., Hreinsdottir, S., Hamling, I., Peng, Z., Bartlow, N., D’Anastasio, E., & Fry, B. (2017). Large-Scale Dynamic Triggering of Shallow Slow Slip Enhanced by Overlying Sedimentary Wedge. Nature Geoscience, 10, 765-770. https://doi.org/10.1038/ngeo3021 Wang, B., & Ding, Q. (2006). Changes in Global Monsoon Precipitation over the Past 56 Years. Geophysical Research Letters, 33, L06711. https://doi.org/10.1029/2005GL025347 Wang, K. (2015). Subduction Faults as We See Them in the 21st Century. AGU Birch Lecture. Wang, K., & Bilek, S. (2011). Do Subducting Seamounts Generate or Stop Large Earthquakes. Geology, 39, 819-822. https://doi.org/10.1130/G31856.1 Wang, K., & He, J. (1999). Mechanics of Low-Stress Forearcs: Nankai and Cascadia. Journal of Geophysical Research: Solid Earth, 104, 15191-15205. https://doi.org/10.1029/1999JB900103 Wang, K., & Hu, Y. (2006). Accretionary Prisms in Subduction Earthquake Cycles: The Theory of Dynamic Coulomb Wedge. Journal of Geophysical Research: Solid Earth, 111, B06410. https://doi.org/10.1029/2005JB004094 Wang, K., Hu, Y., & He, J. (2012). Deformation Cycles of Subduction Earthquakes in a Viscoelastic Earth. Nature, 484, 327-332. https://doi.org/10.1038/nature11032 Weston, J., Ferreira, A., & Funning, G. (2012). Systematic Comparisons of Earthquake Source Models Determined Using In SAR and Seismic Data. Tectonophysics, 532-535, 61-81. https://doi.org/10.1016/j.tecto.2012.02.001 Williams, C., Eberhart-Phillips, D., Bannister, S., Barker, D., Henrys, S., Reyners, M., & Sutherland, R. (2013). Revised Interface Geometry for the Hikurangi Subduction Zone, New Zealand. Seismological Research Letters, 84, 1066-1073. https://doi.org/10.1785/0220130035 Yanez, G., Ranero, C., Huene, R., & Diaz, J. (2001). Magnetic Anomaly Interpretation across the Southern Central Andes (32-34 S): The Role of the Juan Fernandez Ridge in the Late Tertiary Evolution of the Margin. Journal of Geophysical Research, 106, 6325-6345. https://doi.org/10.1029/2000JB900337 Yuan, X., Sobolev, S., & Kind, R. (2002). Moho Topography in the Central Andes and Its Geodynamic Implications. Earth and Planetary Science Letters, 199, 389-402. https://doi.org/10.1016/S0012-821X(02)00589-7", null, "" ]	[ null, "https://www.scirp.org/Images/ccby.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90519917,"math_prob":0.96556383,"size":79492,"snap":"2019-43-2019-47","text_gpt3_token_len":16325,"char_repetition_ratio":0.15992351,"word_repetition_ratio":0.04191919,"special_character_ratio":0.19520204,"punctuation_ratio":0.107611164,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9751732,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-24T06:25:11Z\",\"WARC-Record-ID\":\"<urn:uuid:5bbf2ed1-8b89-4fca-9771-9e6ae08a16b5>\",\"Content-Length\":\"362153\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:436f1a0f-57da-42a0-a767-89570cdc315c>\",\"WARC-Concurrent-To\":\"<urn:uuid:2387651b-0960-4631-9efd-a1707b0943fb>\",\"WARC-IP-Address\":\"173.208.146.83\",\"WARC-Target-URI\":\"https://www.scirp.org/Journal/PaperInformation.aspx?PaperID=94656\",\"WARC-Payload-Digest\":\"sha1:YMOITQH2TDHNBC7OWVHZGDKWSJMMIJZD\",\"WARC-Block-Digest\":\"sha1:QDM22W7ZNZYDS2ZJNOCVLVGYZ7UNFAD3\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570987841291.79_warc_CC-MAIN-20191024040131-20191024063631-00126.warc.gz\"}"}
https://www.ozgrid.com/VBA/sort-alphanumeric.htm	[ "# Excel VBA Video Training", null, "/ EXCEL DASHBOARD REPORTS", null, "# Sort Alphanumeric Text\n\n## Sort Cells In Excel With Text & Numbers\n\nGot any Excel Questions? Excel Help .\n\nSee Normal Excel Sort and Extract Numbers From Text\n\nSorting Alphanumeric\n\nExcel has a problem trying to sort alphanumeric cells in cells by the number portion only. The reason is simply because Excels Sort feature evaluates each cell value by reading left to right. However, we can over-come this in a few ways with the aid of Excel Macros .\n\nFixed Length Text\n\nIf the alphanumeric cells are all a fixed length we can simply use another column to extract out the numeric portion of the alphanumeric text and then sort by the new column. For example, say we can alphanumeric text in Column A like ABC196, FRH564 etc. We can simply add the formula below to Column B.\n\n=--RIGHT(A1,3)\n\nOR\n\n=--Left(A1,3) for fixed length alphanumeric text like 196GFT\n\nOR\n\n=--MID(A1,5,4) for alphanumeric text like a-bg1290rqty where you know the number Start s at the 5th character and has 4 numbers\n\nThen Fill Down as far as needed. Then we can select Column B, copy and Edit>Paste Special - Values. Next we sort Columns A & B by Column B and then delete Column B.\n\nNOTE: the double negative (--) ensures the number returned is seen as a true number.\n\nSort Alphanumeric\n\nAny Length Alphanumeric Text\n\nA problem comes about when the numeric portion and/or the text portion can be any length. In these cases a macro is best. The code below should be copied to any standard Module (Insert>Module). Then simply run the SortAlphaNumerics Procedure.\n\nIt should be noted that the ExtractNumber Function has 2 optional arguments (Take_decimal and Take_negative). These are both False if omitted. See the table below to see how alphanumeric text is treated.\n\n Alphanumeric Text Formula Result a-bg-12909- =ExtractNumber(A1,,TRUE) -12909 a-bg-12909- =ExtractNumber(A2) 12909 a.a1.2... =ExtractNumber(A3,TRUE) 1.2 a.a1.2... =ExtractNumber(A4) 12 a.a-1.2.... =ExtractNumber(A5,TRUE,TRUE) -1.2 abg1290.11 =ExtractNumber(A6,TRUE) 1290.11 abg129013Agt =ExtractNumber(A7) 129013 abg129012 =ExtractNumber(A8) 129013\n\nAlphanumeric Sorting Code\n\n```'MUST be at top of same public module housing _\n\nSub SortAlphaNumerics and Function ExtractNumber\n\nDim bDec As Boolean, bNeg As Boolean\n\nSub SortAlphaNumerics()\n\nDim wSheetTemp As Worksheet\n\nDim wsStart As Worksheet\n\nDim lLastRow As Long, lReply As Long\n\nDim rSort As Range\n\n''''''''''''''''''''''''''''''''''''''''''\n\n'www.ozgrid.com\n\n'Sorts Alphanumerics Of Column \"A\" of active Sheet.\n\n'ExtractNumber Function REQUIRED\n\n'http://www.ozgrid.com/VBA/ExtractNum.htm\n\n''''''''''''''''''''''''''''''''''''''''''\n\nSet wsStart = ActiveSheet\n\nOn Error Resume Next\n\nSet rSort = Application.InputBox _\n\n(\"Select range to sort. Any heading should be bolded and included\", \"ALPHANUMERIC SORT\", _\nActiveCell.CurrentRegion.Columns(1).Address, , , , , 8)\n\nIf rSort Is Nothing Then Exit Sub\n\nIf rSort.Columns.Count > 1 Then\n\nMsgBox \"Single Column Only\"\n\nSortAlphaNumerics\n\nEnd If\n\n'Application.ScreenUpdating = False\n\nSet rSort = Range(Cells(rSort.Cells(1, 1).Row, rSort.Column), _\nCells(Rows.Count, rSort.Column).End(xlUp))\n\nlReply = MsgBox(\"Include Decimals within numbers\", vbYesNoCancel, \"OZGRID ALPHANUMERIC SORT\")\n\nIf lReply = vbCancel Then Exit Sub\n\nlReply = MsgBox(\"Include negative signs within numbers\", vbYesNoCancel, \"OZGRID ALPHANUMERIC SORT\")\n\nIf lReply = vbCancel Then Exit Sub\n\nlLastRow = rSort.Cells(rSort.Rows.Count).Row\n\nrSort.Copy wSheetTemp.Range(\"A1\")\n\nWith wSheetTemp.Range(\"B1:B\" & lLastRow)\n\n.FormulaR1C1 = \"=ExtractNumber(RC[-1],\" & bDec & \",\" & bNeg & \")\"\n\n.Copy\n\n.PasteSpecial xlPasteValues\n\nApplication.CutCopyMode = False\n\nEnd With\n\nbNeg = False\n\nbDec = False\n\nWith wSheetTemp.UsedRange\n\nOrderCustom:=1, MatchCase:=False, Orientation:=xlTopToBottom, _\nDataOption1:=xlSortNormal\n\nEnd With\n\nwSheetTemp.Delete\n\nApplication.ScreenUpdating = True\n\nEnd Sub\n\nFunction ExtractNumber(rCell As Range, _\nOptional Take_decimal As Boolean, Optional Take_negative As Boolean) As Double\n\nDim iCount As Integer, i As Integer, iLoop As Integer\n\nDim sText As String, strNeg As String, strDec As String\n\nDim lNum As String\n\nDim vVal, vVal2\n\n''''''''''''''''''''''''''''''''''''''''''\n\n'www.ozgrid.com\n\n'Extracts a number from a cell containing text and numbers.\n\n''''''''''''''''''''''''''''''''''''''''''\n\nsText = rCell\n\nIf Take_decimal = True And Take_negative = True Then\n\nstrNeg = \"-\" 'Negative Sign MUST be before 1st number.\n\nstrDec = \".\"\n\nElseIf Take_decimal = True And Take_negative = False Then\n\nstrNeg = vbNullString\n\nstrDec = \".\"\n\nElseIf Take_decimal = False And Take_negative = True Then\n\nstrNeg = \"-\"\n\nstrDec = vbNullString\n\nEnd If\n\niLoop = Len(sText)\n\nFor iCount = iLoop To 1 Step -1\n\nvVal = Mid(sText, iCount, 1)\n\nIf IsNumeric(vVal) Or vVal = strNeg Or vVal = strDec Then\n\ni = i + 1\n\nlNum = Mid(sText, iCount, 1) & lNum\n\nIf IsNumeric(lNum) Then\n\nIf CDbl(lNum) < 0 Then Exit For\n\nElse\n\nlNum = Replace(lNum, Left(lNum, 1), \"\", , 1)\n\nEnd If\n\nEnd If\n\nIf i = 1 And lNum <> vbNullString Then lNum = CDbl(Mid(lNum, 1, 1))\n\nNext iCount\n\nExtractNumber = CDbl(lNum)\n\nEnd Function```\n\nExcel Dashboard Reports & Excel Dashboard Charts 50% Off Become an ExcelUser Affiliate & Earn Money\n\nSpecial! Free Choice of Complete Excel Training Course OR Excel Add-ins Collection on all purchases totaling over \\$64.00. ALL purchases totaling over \\$150.00 gets you BOTH! Purchases MUST be made via this site. Send payment proof to [email protected] 31 days after purchase date." ]	[ null, "http://ad.linksynergy.com/fs-bin/show", null, "http://www.ozgrid.com/images/logo/ozgrid.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5923442,"math_prob":0.9289412,"size":6349,"snap":"2023-40-2023-50","text_gpt3_token_len":1691,"char_repetition_ratio":0.13806146,"word_repetition_ratio":0.04366812,"special_character_ratio":0.27295637,"punctuation_ratio":0.15575221,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99163324,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-28T15:23:21Z\",\"WARC-Record-ID\":\"<urn:uuid:26689448-6636-4312-8ef9-50e0edb9e69b>\",\"Content-Length\":\"15029\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3524d152-ee9e-4f4d-9188-3e1c3f97933d>\",\"WARC-Concurrent-To\":\"<urn:uuid:2eaa2f7a-5a19-4583-ab18-e917dfaceb11>\",\"WARC-IP-Address\":\"104.26.6.95\",\"WARC-Target-URI\":\"https://www.ozgrid.com/VBA/sort-alphanumeric.htm\",\"WARC-Payload-Digest\":\"sha1:IJPADM5X3VH6PIWSRQCCSOJB4C35H5W7\",\"WARC-Block-Digest\":\"sha1:XT235U3EY2CYBEYO735CZ3UH25U3FQVD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510412.43_warc_CC-MAIN-20230928130936-20230928160936-00799.warc.gz\"}"}
http://fricas.github.io/api/Poset.html	[ "Poset S¶\n\nholds a complete set together with a structure to codify the partial order. for more documentation see: http://www.euclideanspace.com/prog/scratchpad/mycode/discrete/logic/index.htm Date Created: Aug 2015 Basic Operations: Related packages: UserDefinedPartialOrdering in setorder.spad Related categories: PartialOrder in catdef.spad Related Domains: DirectedGraph in graph.spad Also See: AMS Classifications:\n\n+: (%, %) -> %\n\nfrom FiniteGraph S\n\n=: (%, %) -> Boolean\n\nfrom BasicType\n\n~=: (%, %) -> Boolean\n\nfrom BasicType\n\naddArrow!: (%, NonNegativeInteger, NonNegativeInteger) -> %\n\naddArrow!(s, nm, n1, n2) adds an arrow to the graph s, where: n1 is the index of the start object n2 is the index of the end object This is done in a non-mutable way, that is, the original poset is not changed instead a new one is constructed.\n\naddArrow!: (%, Record(name: String, arrType: NonNegativeInteger, fromOb: NonNegativeInteger, toOb: NonNegativeInteger, xOffset: Integer, yOffset: Integer, map: List NonNegativeInteger)) -> %\n\nfrom FiniteGraph S\n\naddArrow!: (%, String, NonNegativeInteger, NonNegativeInteger) -> %\n\nfrom FiniteGraph S\n\naddArrow!: (%, String, NonNegativeInteger, NonNegativeInteger, List NonNegativeInteger) -> %\n\nfrom FiniteGraph S\n\naddArrow!: (%, String, S, S) -> %\n\nfrom FiniteGraph S\n\naddObject!: (%, Record(value: S, posX: NonNegativeInteger, posY: NonNegativeInteger)) -> %\n\nfrom FiniteGraph S\n\naddObject!(s, n) adds object with coordinates n to the graph s. This is done in a non-mutable way, that is, the original poset is not changed instead a new one is constructed.\n\nfrom FiniteGraph S\n\narrowName: (%, NonNegativeInteger, NonNegativeInteger) -> String\n\nfrom FiniteGraph S\n\narrowsFromArrow: (%, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\narrowsFromNode: (%, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\narrowsToArrow: (%, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\narrowsToNode: (%, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\ncoerce: % -> OutputForm\ncompleteReflexivity: % -> %\n\nReflexivity requires forall(x): x<=x This function enforces this by making sure that every element has arrow to itself. That is, the leading diagonal is true.\n\ncompleteTransitivity: % -> %\n\nTransitivity requires forall(x, y, z): x<=y and y<=z implies x<=z This function enforces this by making sure that the composition of any two arrows is also an arrow.\n\ncoverMatrix: % -> IncidenceAlgebra(Integer, S)\n\nthe covering matrix of a list of elements from a comparison function the list is assumed to be topologically sorted, i.e. w.r. to a linear extension of the comparison function f This function is based on code by Franz Lehner. Notes by Martin Baker on the webpage here: url{http://www.euclideanspace.com/prog/scratchpad/mycode/discrete/logic/moebius/}\n\ncreateWidth: NonNegativeInteger -> NonNegativeInteger\n\nfrom FiniteGraph S\n\ncreateX: (NonNegativeInteger, NonNegativeInteger) -> NonNegativeInteger\n\nfrom FiniteGraph S\n\ncreateY: (NonNegativeInteger, NonNegativeInteger) -> NonNegativeInteger\n\nfrom FiniteGraph S\n\ncycleClosed: (List S, String) -> %\n\nfrom FiniteGraph S\n\ncycleOpen: (List S, String) -> %\n\nfrom FiniteGraph S\n\ndeepDiagramSvg: (String, %, Boolean) -> Void\n\nfrom FiniteGraph S\n\ndiagramHeight: % -> NonNegativeInteger\n\nfrom FiniteGraph S\n\ndiagramsSvg: (String, List %, Boolean) -> Void\n\nfrom FiniteGraph S\n\ndiagramSvg: (String, %, Boolean) -> Void\n\nfrom FiniteGraph S\n\ndiagramWidth: % -> NonNegativeInteger\n\nfrom FiniteGraph S\n\ndistance: (%, NonNegativeInteger, NonNegativeInteger) -> Integer\n\nfrom FiniteGraph S\n\ndistanceMatrix: % -> Matrix Integer\n\nfrom FiniteGraph S\n\nfinitePoset: (List S, (S, S) -> Boolean) -> %\n\nconstructor where the set and structure is supplied. The structure is supplied as a predicate function.\n\nfinitePoset: (List S, List List Boolean) -> %\n\nconstructor where the set and structure is supplied.\n\nflatten: DirectedGraph % -> %\n\nfrom FiniteGraph S\n\ngetArr: % -> List List Boolean\n\ngetArr(s) returns a list of all the arrows (or edges) Note: different from getArrows(s) which is inherited from FiniteGraph(S)\n\ngetArrowIndex: (%, NonNegativeInteger, NonNegativeInteger) -> NonNegativeInteger\n\nfrom FiniteGraph S\n\ngetArrows: % -> List Record(name: String, arrType: NonNegativeInteger, fromOb: NonNegativeInteger, toOb: NonNegativeInteger, xOffset: Integer, yOffset: Integer, map: List NonNegativeInteger)\n\nfrom FiniteGraph S\n\ngetVert: % -> List S\n\ngetVert(s) returns a list of all the vertices (or objects) of the graph s. Note: different from getVertices(s) which is inherited from FiniteGraph(S)\n\ngetVertexIndex: (%, S) -> NonNegativeInteger\n\nfrom FiniteGraph S\n\ngetVertices: % -> List Record(value: S, posX: NonNegativeInteger, posY: NonNegativeInteger)\n\nfrom FiniteGraph S\n\nglb: (%, List NonNegativeInteger) -> Union(NonNegativeInteger, failed)\n\n‘greatest lower bound’ or ‘infimum’ In this version of glb nodes are represented as index values. Not every subset of a poset will have a glb in which case “failed” will be returned as an error indication.\n\nhash: % -> SingleInteger\n\nfrom SetCategory\n\nhashUpdate!: (HashState, %) -> HashState\n\nfrom SetCategory\n\nimplies: (%, NonNegativeInteger, NonNegativeInteger) -> Boolean\n\nincidenceMatrix: % -> Matrix Integer\n\nfrom FiniteGraph S\n\ninDegree: (%, NonNegativeInteger) -> NonNegativeInteger\n\nfrom FiniteGraph S\n\nindexToObject: (%, NonNegativeInteger) -> S\n\nreturns the object at a given index\n\ninitial: () -> %\n\nfrom FiniteGraph S\n\nisAcyclic?: % -> Boolean\n\nfrom FiniteGraph S\n\nisAntiChain?: % -> Boolean\n\nis a subset of a partially ordered set such that any two elements in the subset are incomparable\n\nisAntisymmetric?: % -> Boolean\n\nAntisymmetric requires forall(x, y): x<=y and y<=x iff x=y Returns true if this is the case for every element.\n\nisChain?: % -> Boolean\n\nis a subset of a partially ordered set such that any two elements in the subset are comparable\n\nisDirected?: () -> Boolean\n\nfrom FiniteGraph S\n\nisDirectSuccessor?: (%, NonNegativeInteger, NonNegativeInteger) -> Boolean\n\nfrom FiniteGraph S\n\nisFixPoint?: (%, NonNegativeInteger) -> Boolean\n\nfrom FiniteGraph S\n\nisFunctional?: % -> Boolean\n\nfrom FiniteGraph S\n\nisGreaterThan?: (%, NonNegativeInteger, NonNegativeInteger) -> Boolean\n\nfrom FiniteGraph S\n\njoinIfCan: (%, List NonNegativeInteger) -> Union(NonNegativeInteger, failed)\n\nreturns the join of a subset of lattice given by list of elements\n\njoinIfCan: (%, NonNegativeInteger, NonNegativeInteger) -> Union(NonNegativeInteger, failed)\n\nreturns the join of ‘a’ and 'b' In this version of join nodes are represented as index values. In the general case, not every poset will have a join in which case “failed” will be returned as an error indication.\n\nkgraph: (List S, String) -> %\n\nfrom FiniteGraph S\n\nlaplacianMatrix: % -> Matrix Integer\n\nfrom FiniteGraph S\n\nlatex: % -> String\n\nfrom SetCategory\n\nle: (%, NonNegativeInteger, NonNegativeInteger) -> Boolean\n\nfrom Preorder S\n\nloopsArrows: % -> List Loop\n\nfrom FiniteGraph S\n\nloopsAtNode: (%, NonNegativeInteger) -> List Loop\n\nfrom FiniteGraph S\n\nloopsNodes: % -> List Loop\n\nfrom FiniteGraph S\n\nlooseEquals: (%, %) -> Boolean\n\nfrom FiniteGraph S\n\nlowerSet: % -> %\n\na subset U with the property that, if x is in U and x >= y, then y is in U\n\nlub: (%, List NonNegativeInteger) -> Union(NonNegativeInteger, failed)\n\n‘least upper bound’ or ‘supremum’ In this version of lub nodes are represented as index values. Not every subset of a poset will have a lub in which case “failed” will be returned as an error indication.\n\nmap: (%, List NonNegativeInteger, List S, Integer, Integer) -> %\n\nfrom FiniteGraph S\n\nmapContra: (%, List NonNegativeInteger, List S, Integer, Integer) -> %\n\nfrom FiniteGraph S\n\nmax: % -> NonNegativeInteger\n\nfrom FiniteGraph S\n\nmax: (%, List NonNegativeInteger) -> NonNegativeInteger\n\nfrom FiniteGraph S\n\nmeetIfCan: (%, List NonNegativeInteger) -> Union(NonNegativeInteger, failed)\n\nreturns the meet of a subset of lattice given by list of elements\n\nmeetIfCan: (%, NonNegativeInteger, NonNegativeInteger) -> Union(NonNegativeInteger, failed)\n\nreturns the meet of ‘a’ and 'b' In this version of meet nodes are represented as index values. In the general case, not every poset will have a meet in which case “failed” will be returned as an error indication.\n\nmerge: (%, %) -> %\n\nfrom FiniteGraph S\n\nmin: % -> NonNegativeInteger\n\nfrom FiniteGraph S\n\nmin: (%, List NonNegativeInteger) -> NonNegativeInteger\n\nfrom FiniteGraph S\n\nmoebius: % -> IncidenceAlgebra(Integer, S)\n\nmoebius incidence matrix for this poset This function is based on code by Franz Lehner. Notes by Martin Baker on the webpage here: url{http://www.euclideanspace.com/prog/scratchpad/mycode/discrete/logic/moebius/}\n\nnodeFromArrow: (%, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\nnodeFromNode: (%, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\nnodeToArrow: (%, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\nnodeToNode: (%, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\nobjectToIndex: (%, S) -> NonNegativeInteger\n\nreturns the index of a given object\n\nopposite: % -> %\n\nconstructs the opposite in the category theory sense of reversing all the arrows\n\noutDegree: (%, NonNegativeInteger) -> NonNegativeInteger\n\nfrom FiniteGraph S\n\npowerSetStructure: S -> %\n\npowerSetStructure(set) is a constructor for a Poset where each element is a set (implemented as a list) and with a subset structure. requires S to be a list.\n\nrouteArrows: (%, NonNegativeInteger, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\nrouteNodes: (%, NonNegativeInteger, NonNegativeInteger) -> List NonNegativeInteger\n\nfrom FiniteGraph S\n\nsetArr: (%, List List Boolean) -> Void\n\nsets the list of all arrows (or edges)\n\nsetVert: (%, List S) -> Void\n\nsets the list of all vertices (or objects)\n\nspanningForestArrow: % -> List Tree Integer\n\nfrom FiniteGraph S\n\nspanningForestNode: % -> List Tree Integer\n\nfrom FiniteGraph S\n\nspanningTreeArrow: (%, NonNegativeInteger) -> Tree Integer\n\nfrom FiniteGraph S\n\nspanningTreeNode: (%, NonNegativeInteger) -> Tree Integer\n\nfrom FiniteGraph S\n\nsubdiagramSvg: (Scene SCartesian 2, %, Boolean, Boolean) -> Void\n\nfrom FiniteGraph S\n\nterminal: S -> %\n\nfrom FiniteGraph S\n\nunit: (List S, String) -> %\n\nfrom FiniteGraph S\n\nupperSet: % -> %\n\na subset U with the property that, if x is in U and x <= y, then y is in U\n\nzetaMatrix: % -> IncidenceAlgebra(Integer, S)\n\nzetaMatrix(P) returns the matrix of the zeta function This function is based on code by Franz Lehner. Notes by Martin Baker on the webpage here: url{http://www.euclideanspace.com/prog/scratchpad/mycode/discrete/logic/moebius/}\n\nBasicType\n\nSetCategory" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.669631,"math_prob":0.54890895,"size":10298,"snap":"2022-05-2022-21","text_gpt3_token_len":2734,"char_repetition_ratio":0.33417526,"word_repetition_ratio":0.37483355,"special_character_ratio":0.24266848,"punctuation_ratio":0.2033582,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9903874,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-28T16:50:42Z\",\"WARC-Record-ID\":\"<urn:uuid:9f31da3b-55e3-4f31-b8fe-94bd6f0f6141>\",\"Content-Length\":\"77289\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e8fca2f2-d8df-4917-be12-f3bffbb7634b>\",\"WARC-Concurrent-To\":\"<urn:uuid:443712ac-d9e2-4cd7-b006-3102e1eb02b5>\",\"WARC-IP-Address\":\"185.199.109.153\",\"WARC-Target-URI\":\"http://fricas.github.io/api/Poset.html\",\"WARC-Payload-Digest\":\"sha1:AMQ6UWVTITQZLXSEQDBVANOUAT6OEQ7M\",\"WARC-Block-Digest\":\"sha1:PZ6F2Z5YW66AOEZKDIUB2AM2CM6OKM6L\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320306301.52_warc_CC-MAIN-20220128152530-20220128182530-00126.warc.gz\"}"}
https://tipcalc.net/how-much-is-a-20-percent-tip-on-70.31	[ "# Tip Calculator\n\nHow much is a 20 percent tip on \\$70.31?\n\nTIP:\n\\$ 0\nTOTAL:\n\\$ 0\n\nTIP PER PERSON:\n\\$ 0\nTOTAL PER PERSON:\n\\$ 0\n\n## How much is a 20 percent tip on \\$70.31? How to calculate this tip?\n\nAre you looking for the answer to this question: How much is a 20 percent tip on \\$70.31? Here is the answer.\n\nLet's see how to calculate a 20 percent tip when the amount to be paid is 70.31. Tip is a percentage, and a percentage is a number or ratio expressed as a fraction of 100. This means that a 20 percent tip can also be expressed as follows: 20/100 = 0.2 . To get the tip value for a \\$70.31 bill, the amount of the bill must be multiplied by 0.2, so the calculation is as follows:\n\n1. TIP = 70.3120% = 70.310.2 = 14.062\n\n2. TOTAL = 70.31+14.062 = 84.372\n\n3. Rounded to the nearest whole number: 84\n\nIf you want to know how to calculate the tip in your head in a few seconds, visit the Tip Calculator Home.\n\n## So what is a 20 percent tip on a \\$70.31? The answer is 14.06!\n\nOf course, it may happen that you do not pay the bill or the tip alone. A typical case is when you order a pizza with your friends and you want to split the amount of the order. For example, if you are three, you simply need to split the tip and the amount into three. In this example it means:\n\n1. Total amount rounded to the nearest whole number: 84\n\n2. Split into 3: 28\n\nSo in the example above, if the pizza order is to be split into 3, you’ll have to pay \\$28 . Of course, you can do these settings in Tip Calculator. You can split the tip and the total amount payable among the members of the company as you wish. So the TipCalc.net page basically serves as a Pizza Tip Calculator, as well.\n\n## Tip Calculator Examples (BILL: \\$70.31)\n\nHow much is a 5% tip on \\$70.31?\nHow much is a 10% tip on \\$70.31?\nHow much is a 15% tip on \\$70.31?\nHow much is a 20% tip on \\$70.31?\nHow much is a 25% tip on \\$70.31?\nHow much is a 30% tip on \\$70.31?" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.959624,"math_prob":0.99352044,"size":2593,"snap":"2021-43-2021-49","text_gpt3_token_len":856,"char_repetition_ratio":0.31517962,"word_repetition_ratio":0.23529412,"special_character_ratio":0.40185115,"punctuation_ratio":0.16235632,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99777305,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-11-29T10:14:56Z\",\"WARC-Record-ID\":\"<urn:uuid:324ee658-8154-496e-aabb-2bb62e05df90>\",\"Content-Length\":\"10951\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:0ecbcf01-8ede-488b-8647-c2d8d8429862>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad331300-afd0-4bdc-847c-556364a6503d>\",\"WARC-IP-Address\":\"161.35.97.186\",\"WARC-Target-URI\":\"https://tipcalc.net/how-much-is-a-20-percent-tip-on-70.31\",\"WARC-Payload-Digest\":\"sha1:575RXO5WDKR2LKV2V5KEDKTDWVJIE77B\",\"WARC-Block-Digest\":\"sha1:STTB6VWZP527E63QR7GF7EEPKGPVSCUV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964358702.43_warc_CC-MAIN-20211129074202-20211129104202-00194.warc.gz\"}"}
https://stattrek.com/statistics/dictionary.aspx?definition=probability%20density%20function	[ "# Statistics Dictionary\n\nTo see a definition, select a term from the dropdown text box below. The statistics dictionary will display the definition, plus links to related web pages.\n\nSelect term:\n\n### Probability Density Function\n\nMost often, the equation used to describe a continuous probability distribution is called a probability density function . Sometimes, it is referred to as a density function, a PDF, or a pdf. For a continuous probability distribution, the density function has the following properties:\n\n• Since the continuous random variable is defined over a continuous range of values (called the domain of the variable), the graph of the density function will also be continuous over that range.\n• The area bounded by the curve of the density function and the x-axis is equal to 1, when computed over the domain of the variable.\n• The probability that a random variable assumes a value between a and b is equal to the area under the density function bounded by a and b.\n\nFor example, consider the probability density function shown in the graph below. Suppose we wanted to know the probability that the random variable X was less than or equal to a. The probability that X is less than or equal to a is equal to the area under the curve bounded by a and minus infinity - as indicated by the shaded area.", null, "Note: The shaded area in the graph represents the probability that the random variable X is less than or equal to a. This is a cumulative probability . However, the probability that X is exactly equal to a would be zero. A continuous random variable can take on an infinite number of values. The probability that it will equal a specific value (such as a) is always zero.\n\n See also: Tutorial: Discrete and Continuous Random Variables \| Probability Distributions" ]	[ null, "https://stattrek.com/img/normalprob1.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9034902,"math_prob":0.99692285,"size":1681,"snap":"2019-13-2019-22","text_gpt3_token_len":353,"char_repetition_ratio":0.17173524,"word_repetition_ratio":0.072164945,"special_character_ratio":0.19809636,"punctuation_ratio":0.09090909,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998994,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-26T10:56:24Z\",\"WARC-Record-ID\":\"<urn:uuid:c5a1f6ec-f49a-47ee-9748-65a00801a96b>\",\"Content-Length\":\"91948\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9dd58bea-89d9-42dc-b101-7aa8e97ce952>\",\"WARC-Concurrent-To\":\"<urn:uuid:87b862c9-7884-40bc-a2cf-83ca32874633>\",\"WARC-IP-Address\":\"34.237.101.58\",\"WARC-Target-URI\":\"https://stattrek.com/statistics/dictionary.aspx?definition=probability%20density%20function\",\"WARC-Payload-Digest\":\"sha1:NF5XAQTMIMHJ7ZSCXM7YOJW45RIGUB4Q\",\"WARC-Block-Digest\":\"sha1:SJSX3QMRP6O44NJF5MH4KO63UQIHO2CZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232259126.83_warc_CC-MAIN-20190526105248-20190526131248-00246.warc.gz\"}"}
https://documen.tv/question/a-carbon-resistor-is-to-be-used-as-a-thermometer-on-a-winter-day-when-the-temperature-is-4-00-c-15012693-38/	[ "## A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00 ∘C, the resistance of the carbon resistor is\n\nQuestion\n\nA carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00 ∘C, the resistance of the carbon resistor is 217.0 Ω . What is the temperature on a spring day when the resistance is 215.1 Ω ? Take the temperature coefficient of resistivity for carbon to be α = −5.00×10−4 C−1 .\n\nin progress 0\n1 year 2021-09-02T14:56:01+00:00 1 Answers 57 views 0\n\n21.5 °C.\n\nExplanation:\n\nGiven:\n\nα = −5.00 × 10−4 °C−1\n\nTo = 4°C\n\nRo = 217 Ω\n\nRt = 215.1 Ω\n\nRt/Ro = 1 + α(T – To)\n\n215.1/217 = 1 + (-5 × 10^−4) × (T – 4)\n\n-0.00876 = -5 × 10^−4 × (T – 4)\n\n17.5 = (T – 4)\n\nT = 21.5 °C." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7331478,"math_prob":0.9986235,"size":789,"snap":"2022-40-2023-06","text_gpt3_token_len":307,"char_repetition_ratio":0.1388535,"word_repetition_ratio":0.29113925,"special_character_ratio":0.46007603,"punctuation_ratio":0.14925373,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996096,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-01-29T21:45:43Z\",\"WARC-Record-ID\":\"<urn:uuid:6c09afe1-33e4-4152-91bc-9b7d036aa64c>\",\"Content-Length\":\"92290\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:78a8b366-2ceb-4e43-a31e-b24ba77c075b>\",\"WARC-Concurrent-To\":\"<urn:uuid:c5e79c80-ca30-47a1-80bf-fd39d4f789bf>\",\"WARC-IP-Address\":\"5.78.45.21\",\"WARC-Target-URI\":\"https://documen.tv/question/a-carbon-resistor-is-to-be-used-as-a-thermometer-on-a-winter-day-when-the-temperature-is-4-00-c-15012693-38/\",\"WARC-Payload-Digest\":\"sha1:VOI6RHMA3FWBGLH7YX6ZWAYAEUW7RW5L\",\"WARC-Block-Digest\":\"sha1:7HAPEEBSBQIQJKW7STCLFTKH77E5BV44\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499768.15_warc_CC-MAIN-20230129211612-20230130001612-00600.warc.gz\"}"}
https://www.lumoslearning.com/llwp/resources/educational-videos-k-12-elementary-middle-school.html?id=1849479	[ "The Remainder Theorem - Example 1 - Free Educational videos for Students in K-12 \| Lumos Learning\n\n## The Remainder Theorem - Example 1 - Free Educational videos for Students in k-12\n\nTranscript\n00:0\nSummarizer\n\n#### DESCRIPTION:\n\nYoutube Presents The Remainder Theorem - Example 1 an educational video resources on english language arts.\n\n#### OVERVIEW:\n\nThe Remainder Theorem - Example 1 is a free educational video by PatrickJMT.It helps students in grades 9,10,11,12 practice the following standards HSA.APR.B.2.\n\nThis page not only allows students and teachers view The Remainder Theorem - Example 1 but also find engaging Sample Questions, Apps, Pins, Worksheets, Books related to the following topics.\n\n1. HSA.APR.B.2 : Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x - a is p(a), so p(a) = 0 if and only if (x - a) is a factor of p(x).\n\n9\n10\n11\n12\n\nHSA.APR.B.2\n\n### RELATED VIDEOS\n\nRate this Video?\n0\n\n0 Ratings & 0 Reviews\n\n5\n0\n0\n4\n0\n0\n3\n0\n0\n2\n0\n0\n1\n0\n0", null, "EdSearch WebSearch", null, "" ]	[ null, "https://statc.lumoslearning.com/llwp/wp-content/uploads/2018/01/edSearch_Logo_LowRes-e1517334931747.png", null, "https://googleads.g.doubleclick.net/pagead/viewthroughconversion/1066999506/", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8040768,"math_prob":0.71094394,"size":807,"snap":"2021-31-2021-39","text_gpt3_token_len":212,"char_repetition_ratio":0.14819427,"word_repetition_ratio":0.060606062,"special_character_ratio":0.25402725,"punctuation_ratio":0.17159763,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9649594,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-25T13:30:47Z\",\"WARC-Record-ID\":\"<urn:uuid:773ef018-bb34-4fc0-85c2-b4870bdd7334>\",\"Content-Length\":\"255637\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d004fa22-5be8-4836-b4bd-cab4798755ef>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2046805-1dc5-4a36-9608-50677882d75d>\",\"WARC-IP-Address\":\"3.212.85.120\",\"WARC-Target-URI\":\"https://www.lumoslearning.com/llwp/resources/educational-videos-k-12-elementary-middle-school.html?id=1849479\",\"WARC-Payload-Digest\":\"sha1:3P5L6G7NW3NDMY63NXP2PXVDCAIAFKOX\",\"WARC-Block-Digest\":\"sha1:HRWRQLJQ6LHFTGDDU2VFCKILDF6WHZLJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057622.15_warc_CC-MAIN-20210925112158-20210925142158-00267.warc.gz\"}"}
https://riverkc.com/6016/worksheet/beginner-algebra/	[ "# Beginner Algebra", null, "Division Elementary Algebra Worksheet Printable Elementary Worksheets Basic Math Worksheets Elementary Algebra", null, "", null, "For Dummies Pre Algebra Basic Math Algebra", null, "Algebra For Beginners Worksheet Education Com Algebra Worksheets Basic Algebra Pre Algebra", null, "Pre Algebra Beginner Level Taught By Nurten C Afterschoollearning Teachers Bayarea Dublin Sanfrancisco Oakland Hayward Basic Math Pre Algebra Sat Math", null, "An Explanation Of Basic Algebra Terms And Terminology Operations Terms Variables Con Algebraic Expressions Vocabulary Basic Algebra Mathematical Expression", null, "Worksheets For Kids Free Printables Combining Like Terms Like Terms Algebra Worksheets", null, "Important Algebraic Formula Studying Math Basic Math Learning Mathematics", null, "23 Basic Algebra Worksheets Pdf Beginning Algebra Worksheets Openlayers Basic Algebra Basic Algebra Worksheets Algebra Worksheets", null, "Algebra Cheat Sheet Logarithm 80k Views Algebra Cheat Sheet Algebra Cheat Sheets", null, "Simple Algebra Worksheet Free Printable Educational Worksheet Algebra Worksheets Algebra Equations Worksheets Basic Algebra Worksheets", null, "Newbie Electronics Hobbyist Reference Poster Bundle Print Download Basic Algebra Math Formulas Math Methods", null, "Worksheetfun Free Printable Worksheets Algebra Worksheets Algebra Printable Worksheets", null, "Worksheet Basic Algebra 1 Basic Algebra Algebra Algebra 1", null, "Algebra For Beginners Worksheet Education Com Algebra Worksheets Basic Algebra Pre Algebra", null, "Basic Algebra I Havewho Has Game By Thatmathlady Via Slideshare Basic Algebra Algebra I Algebra Games", null, "Algebra 1 Weekly1 Warm Up Starter Algebra Algebra 1 Basic Math Skills", null, "8th Grade Math Worksheets Algebra Worksheets 8th Grade Math Worksheets Mathematics Worksheets", null, "Introduction To Algebra Vocabulary Worksheet Vocabulary Worksheets Algebra Vocabulary Algebra Worksheets" ]	[ null, "https://i.pinimg.com/originals/9f/5d/5a/9f5d5a01a7b41bae5a8bf63c8d29d48d.png", null, "https://i.pinimg.com/originals/20/3e/15/203e15f04395d527e7687e631876f0ea.png", null, "https://i.pinimg.com/originals/9a/0a/5a/9a0a5a54039c6c26101819fb55c3c96d.jpg", null, "https://riverkc.com/wp-content/uploads/2021/02/15e9da1a58947052f1b6cb96d954927e-1.gif", null, "https://i.pinimg.com/originals/20/3e/15/203e15f04395d527e7687e631876f0ea.png", null, "https://i.pinimg.com/originals/d7/b4/21/d7b4215939ea354a904ffa99550789fd.png", null, "https://i.pinimg.com/originals/84/20/40/842040eb4dbf78f3e93bdf0a30b2a8e6.gif", null, "https://i.pinimg.com/originals/e1/79/16/e17916d690d46d52880364f6d6c4e735.jpg", null, "https://i.pinimg.com/736x/e5/25/e7/e525e7e053416c46603e3fdabaf4222e.jpg", null, "https://i.pinimg.com/170x/ab/fd/a4/abfda4006f5c5ccf951d3f6d9886eb01.jpg", null, "https://riverkc.com/wp-content/uploads/2021/02/ec9f366a89615afc8121dccb9adb9d5b-7.png", null, "https://i.pinimg.com/originals/37/15/8f/37158f84c29703a1c3770ecf53e34722.png", null, "https://i.pinimg.com/originals/ca/7d/f9/ca7df956a6d0375e0a437be9e57ddea7.png", null, "https://i.pinimg.com/originals/16/26/d2/1626d21315d6e19d65461e29a3b3fd94.jpg", null, "https://i.pinimg.com/originals/46/e3/22/46e322b1e00ed537a0231c35ff09b0aa.gif", null, "https://i.pinimg.com/originals/d1/00/fc/d100fc97e815495bd8e6bf5250adb3b4.jpg", null, "https://i.pinimg.com/600x315/43/f8/9c/43f89cd978f0b5637c594546931f85b0.jpg", null, "https://i.pinimg.com/564x/5b/b0/9b/5bb09bad7c6aee2e63ef01bf0c2605d6.jpg", null, "https://i.pinimg.com/474x/b6/08/cc/b608cc35d698ab1d17440f6ac5ccae8d.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54631686,"math_prob":0.44304508,"size":1773,"snap":"2021-43-2021-49","text_gpt3_token_len":295,"char_repetition_ratio":0.36065573,"word_repetition_ratio":0.07079646,"special_character_ratio":0.13536379,"punctuation_ratio":0.0,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.995517,"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],"im_url_duplicate_count":[null,null,null,2,null,2,null,2,null,2,null,5,null,null,null,1,null,9,null,1,null,1,null,2,null,null,null,1,null,null,null,1,null,1,null,null,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-16T08:38:02Z\",\"WARC-Record-ID\":\"<urn:uuid:7711ee52-695f-4e50-8f59-5855f06985bb>\",\"Content-Length\":\"43758\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ae8d1276-4def-4c80-845b-15a97a2f42d1>\",\"WARC-Concurrent-To\":\"<urn:uuid:cf962cb0-bd82-4b0b-8afd-e78625df0466>\",\"WARC-IP-Address\":\"104.21.96.47\",\"WARC-Target-URI\":\"https://riverkc.com/6016/worksheet/beginner-algebra/\",\"WARC-Payload-Digest\":\"sha1:ESKYFK3KMIZU325IY72V2P22SPJ5QMFW\",\"WARC-Block-Digest\":\"sha1:V3PMOG6DZHB456YXJSHJIKPI24FY43ZD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323584554.98_warc_CC-MAIN-20211016074500-20211016104500-00121.warc.gz\"}"}
https://hakanyurdakul.com/category/math/	[ "## Curve Point Distribution by Simpson Rule\n\nC++. This project is still going on; it needs for some improvements. Its purpose is to equally distribute desired number of points on a curve. To do that, I applied Simpson rule by sweeping small […]\n\n## Latitude Point Distribution\n\nC++. This is for symmetrical distribution of points in each latitude on a sphere. As an input, different number of points for each latitude can be assigned. Top and bottom coordinates, of course, are the […]\n\n## Deepest Pit\n\nC++. This is a question from Codility. Question: A non-empty zero-indexed array A consisting of N integers is given. A pit in this array is any triplet of integers (P, Q, R) such that: 0 ≤ P < […]\n\n## Top Cut of Convex Hull\n\nC++. I developed an algorithm based on “three penny algorithm” to find the top cut of a convex hull for dense circle shaped data. It takes the most left point as an entry point like Jarvis’s […]\n\n## Totient Function\n\nC++. This is the Problem 69 from Project Euler. Question: Euler’s Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n. […]\n\n## Pell Equation\n\nC++. This is the Problem 66 from Project Euler. Question: Consider quadratic Diophantine equations of the form: x2 – Dy2 = 1 For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1. It can be assumed that there are […]\n\n## Pentagonal Numbers\n\nC++. This is the Problem 44 from Project Euler. Question: Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten pentagonal numbers are: 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, … It […]\n\n## Huge Multiplication\n\nC++, Matlab. This was one of the assignments in the algorithm course I took before. Inputs are two big integers like 60 digits each. I implemented traditional multiplication method into my algorithm. Multiplication stage takes […]\n\n## Digit Fifth Powers\n\nC++, Matlab. This is the Problem 30 from Project Euler. Question: Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits: 1634 = 1^4 + 6^4 + 3^4 + […]\n\n## Numbers on Spiral Diagonals\n\nC++, Matlab. This is the Problem 28 from Project Euler website. Question: Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is formed as follows: 21 22 […]\n\n1 2" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91184485,"math_prob":0.96955824,"size":2332,"snap":"2023-14-2023-23","text_gpt3_token_len":577,"char_repetition_ratio":0.12972508,"word_repetition_ratio":0.19138756,"special_character_ratio":0.27744424,"punctuation_ratio":0.14285715,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9976782,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-04T19:58:58Z\",\"WARC-Record-ID\":\"<urn:uuid:97f591be-5608-4fd4-a888-46a1c5cd35a9>\",\"Content-Length\":\"49530\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b98136bd-8b58-464c-bc64-2c82891af6cc>\",\"WARC-Concurrent-To\":\"<urn:uuid:1397e6fd-3ef9-4933-990b-fc4be485f130>\",\"WARC-IP-Address\":\"92.249.44.254\",\"WARC-Target-URI\":\"https://hakanyurdakul.com/category/math/\",\"WARC-Payload-Digest\":\"sha1:GXNOXOLQN4EQJSJR3JIAIECNL6PIMSOE\",\"WARC-Block-Digest\":\"sha1:SZN7FNAMZEMFPEQJXQY6QDFTYDIKHKCX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224650264.9_warc_CC-MAIN-20230604193207-20230604223207-00532.warc.gz\"}"}
https://www.aimsciences.org/journal/1078-0947/2007/17/3	[ "", null, "", null, "", null, "", null, "", null, "ISSN:\n1078-0947\n\neISSN:\n1553-5231\n\n## Journal Home\n\n• Open Access Articles", null, "All Issues\n\n## Discrete & Continuous Dynamical Systems - A\n\nAugust 2007 , Volume 17 , Issue 3\n\nSelect all articles\n\nExport/Reference:\n\n2007, 17(3): 449-480 doi: 10.3934/dcds.2007.17.449 +[Abstract](1572) +[PDF](336.5KB)\nAbstract:\nIn this paper we prove the existence of a solution in Lloc$(\\Omega)$ to the Euler-Lagrange equation for the variational problem\n\ninf$\\bar u + W^{1,\\infty}_0(\\Omega)$$\\int_{\\Omega} (\\I_D(\\nabla u) + g(u)) dx,\\ \\ \\$ (0.1)\n\nwith $D$ convex closed subset of $\\R^n$ with non empty interior. We next show that if D* is strictly convex, then the Euler-Lagrange equation can be reduced to an ODE along characteristics, and we deduce that the solution to Euler-Lagrange is different from $0$ a.e. and satisfies a uniqueness property. Using these results, we prove a conjecture on the existence of variations on vector fields .\n\n2007, 17(3): 481-500 doi: 10.3934/dcds.2007.17.481 +[Abstract](2088) +[PDF](281.9KB)\nAbstract:\nWe study the relations between the global dynamics of the 3D Leray-$\\alpha$ model and the 3D Navier-Stokes system. We prove that time shifts of bounded sets of solutions of the Leray-$\\alpha$ model converges to the trajectory attractor of the 3D Navier-Stokes system as time tends to infinity and $\\alpha$ approaches zero. In particular, we show that the trajectory attractor of the Leray-$\\alpha$ model converges to the trajectory attractor of the 3D Navier-Stokes system when $\\alpha \\rightarrow 0\\+.$\n2007, 17(3): 501-507 doi: 10.3934/dcds.2007.17.501 +[Abstract](1711) +[PDF](125.7KB)\nAbstract:\nWe give an example where for an open set of Lagrangians on the n-torus there is at least one cohomology class c with at least n different ergodic c-minimizing measures. One of the problems posed by Ricardo Mañé in his paper 'Generic properties and problems of minimizing measures of Lagrangian systems' (Nonlinearity, 1996) was the following:\nIs it true that for generic Lagrangians every minimizing measure is uniquely ergodic?\nA weaker statement is that for generic Lagrangians every cohomology class has exactly one minimizing measure, which of course will be ergodic. Our example shows that this can't be true and as a consequence one can hope to prove at most that for a generic Lagrangian, for every cohomology class there are at most n corresponding ergodic minimizing measures, where n is the dimension of the first cohomology group.\n2007, 17(3): 509-528 doi: 10.3934/dcds.2007.17.509 +[Abstract](1781) +[PDF](256.9KB)\nAbstract:\nIn this paper, we study various dissipative mechanics associated with the Boussinesq systems which model two-dimensional small amplitude long wavelength water waves. We will show that the decay rate for the damped one-directional model equations, such as the KdV and BBM equations, holds for some of the damped Boussinesq systems.\n2007, 17(3): 529-540 doi: 10.3934/dcds.2007.17.529 +[Abstract](2096) +[PDF](167.9KB)\nAbstract:\nWe consider the Lorenz system $\\dot x = \\s (y-x)$, $\\dot y =rx -y-xz$ and $\\dot z = -bz + xy$; and the Rössler system $\\dot x = -(y+z)$, $\\dot y = x +ay$ and $\\dot z = b-cz + xz$. Here, we study the Hopf bifurcation which takes place at $q_{\\pm}=(\\pm\\sqrt{br-b},\\pm\\sqrt{br-b},r-1),$ in the Lorenz case, and at $s_{\\pm}=(\\frac{c+\\sqrt{c^2-4ab}}{2},-\\frac{c+\\sqrt{c^2-4ab}}{2a}, \\frac{c\\pm\\sqrt{c^2-4ab}}{2a})$ in the Rössler case. As usual this Hopf bifurcation is in the sense that an one -parameter family in ε of limit cycles bifurcates from the singular point when ε=0. Moreover, we can determine the kind of stability of these limit cycles. In fact, for both systems we can prove that all the bifurcated limit cycles in a neighborhood of the singular point are either a local attractor, or a local repeller, or they have two invariant manifolds, one stable and the other unstable, which locally are formed by two $2$-dimensional cylinders. These results are proved using averaging theory. The method of studying the Hopf bifurcation using the averaging theory is relatively general and can be applied to other $3$- or $n$-dimensional differential systems.\n2007, 17(3): 541-552 doi: 10.3934/dcds.2007.17.541 +[Abstract](1446) +[PDF](192.1KB)\nAbstract:\nIn this paper we consider some systems of ordinary differential equations which are related to coagulation-fragmentation processes. In particular, we obtain explicit solutions $\\{c_k(t)\\}$ of such systems which involve certain coefficients obtained by solving a suitable algebraic recurrence relation. The coefficients are derived in two relevant cases: the high-functionality limit and the Flory-Stockmayer model. The solutions thus obtained are polydisperse (that is, $c_k(0)$ is different from zero for all $k \\ge 1$) and may exhibit monotonically increasing or decreasing total mass. We also solve a monodisperse case (where $c_1(0)$ is different from zero but $c_k(0)$ is equal to zero for all $k \\ge 2$) in the high-functionality limit. In contrast to the previous result, the corresponding solution is now shown to display a sol-gel transition when the total initial mass is larger than one, but not when such mass is less than or equal to one.\n2007, 17(3): 553-560 doi: 10.3934/dcds.2007.17.553 +[Abstract](1503) +[PDF](145.6KB)\nAbstract:\nThe sectional-hyperbolic sets constitute a class of partially hyperbolic sets introduced in to describe robustly transitive singular dynamics on $n$-manifolds (e.g. the multidimensional Lorenz attractor ). Here we prove that a transitive sectional-hyperbolic set with singularities contains no local strong stable manifold through any of its points. Hence a transitive, isolated, sectional-hyperbolic set containing a local strong stable manifold is a hyperbolic saddle-type repeller. In addition, a proper transitive sectional-hyperbolic set on a compact $n$-manifold has empty interior and topological dimension $\\leq n-1$. It follows that a singular-hyperbolic attractor with singularities on a compact $3$-manifold has topological dimension $2$. Hence such an attractor is expanding, i.e., its topological dimension coincides with the dimension of its central subbundle. These results apply to the robustly transitive sets considered in , and also to the Lorenz attractor in the Lorenz equation .\n2007, 17(3): 561-587 doi: 10.3934/dcds.2007.17.561 +[Abstract](1572) +[PDF](316.7KB)\nAbstract:\nWe study chain transitivity and Morse decompositions of discrete and continuous-time semiflows on fiber bundles with emphasis on (generalized) flag bundles. In this case an algebraic description of the chain transitive sets is given. Our approach consists in embedding the semiflow in a semigroup of continuous maps to take advantage of the good properties of the semigroup actions on the flag manifolds.\n2007, 17(3): 589-627 doi: 10.3934/dcds.2007.17.589 +[Abstract](1238) +[PDF](507.3KB)\nAbstract:\nWe present a model illustrating heterodimensional cycles (i.e., cycles associated to saddles having different indices) as a mechanism leading to the collision of hyperbolic homoclinic classes (of points of different indices) and thereafter to the persistence of (heterodimensional) cycles. The collisions are associated to secondary (saddle-node) bifurcations appearing in the unfolding of the initial cycle.\n2007, 17(3): 629-652 doi: 10.3934/dcds.2007.17.629 +[Abstract](1620) +[PDF](629.6KB)\nAbstract:\nWe present a topological method of obtaining the existence of infinite number of symmetric periodic orbits for systems with reversing symmetry. The method is based on covering relations. We apply the method to a four-dimensional reversible map.\n2007, 17(3): 653-670 doi: 10.3934/dcds.2007.17.653 +[Abstract](1430) +[PDF](242.9KB)\nAbstract:\nProperties of the sequence ind$(\\infty,\\id-f^{m}),$ $m=1,2,\\ldots$ where $f^{m}$ is the $m$-th iterate of the mapping $f:R^{d}\\to R^{d}$, and ind denotes the Kronecker index are investigated. The case when $f$ has the asymptotic derivative $A$ at infinity and some eigenvalues of $A$ are roots of unity is of primary interest.\n2007, 17(3): 671-689 doi: 10.3934/dcds.2007.17.671 +[Abstract](1578) +[PDF](567.6KB)\nAbstract:\nPerron-Frobenius operators and their eigendecompositions are increasingly being used as tools of global analysis for higher dimensional systems. The numerical computation of large, isolated eigenvalues and their corresponding eigenfunctions can reveal important persistent structures such as almost-invariant sets, however, often little can be said rigorously about such calculations. We attempt to explain some of the numerically observed behaviour by constructing a hyperbolic map with a Perron-Frobenius operator whose eigendecomposition is representative of numerical calculations for hyperbolic systems. We explicitly construct an eigenfunction associated with an isolated eigenvalue and prove that a special form of Ulam's method well approximates the isolated spectrum and eigenfunctions of this map.\n2007, 17(3): 690-690 doi: 10.3934/dcds.2007.17.690 +[Abstract](1225) +[PDF](45.9KB)\nAbstract:\nN/A\n\n2018 Impact Factor: 1.143" ]	[ null, "https://www.aimsciences.org:443/style/web/images/white_google.png", null, "https://www.aimsciences.org:443/style/web/images/white_facebook.png", null, "https://www.aimsciences.org:443/style/web/images/white_twitter.png", null, "https://www.aimsciences.org:443/style/web/images/white_linkedin.png", null, "https://www.aimsciences.org/fileAIMS/journal/img/cover/book_cover.png", null, "https://www.aimsciences.org:443/style/web/images/OA.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89119476,"math_prob":0.9812513,"size":7991,"snap":"2019-51-2020-05","text_gpt3_token_len":1920,"char_repetition_ratio":0.10229123,"word_repetition_ratio":0.02242525,"special_character_ratio":0.21761982,"punctuation_ratio":0.08067941,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9942216,"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\":\"2020-01-19T19:50:12Z\",\"WARC-Record-ID\":\"<urn:uuid:c4ab3dfd-7a36-4697-b9e3-44981a11d7f7>\",\"Content-Length\":\"123655\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a5e8000e-a3a5-447a-a514-2a0b8e3d14c3>\",\"WARC-Concurrent-To\":\"<urn:uuid:bf834e7e-9057-4541-bb55-907bb3f3409d>\",\"WARC-IP-Address\":\"216.227.221.143\",\"WARC-Target-URI\":\"https://www.aimsciences.org/journal/1078-0947/2007/17/3\",\"WARC-Payload-Digest\":\"sha1:3BE5IHRAJHD3QMEQFMU7H4D3IKSQV5QF\",\"WARC-Block-Digest\":\"sha1:MVP2L7REGGQ6VLZ4TXV75RDDWQNTAW7N\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250594705.17_warc_CC-MAIN-20200119180644-20200119204644-00218.warc.gz\"}"}
https://escholarship.org/uc/item/1cw491jk	[ "", null, "Open Access Publications from the University of California\n\n## Last Passage Percolation and the Slow Bond Problem\n\n• Author(s): Sarkar, Sourav\n• et al.\nAbstract\n\nLast passage percolation models are fundamental examples in statistical mechanics where the energy of a path is maximized over all directed paths with given endpoints in a random environment, and the maximizing paths are called {\\em geodesics}. Here we consider the Poissonian last passage percolation (LPP) and the exponential directed last passage percolation\n\n(DLPP), the latter having a standard coupling with another classical interacting particle system, the totally asymmetric simple exclusion process or TASEP. These belong to the so-called KPZ universality class, for which exact algebraic formulae have led to precise results for fluctuations and scaling limits. However, such formulae are not very robust and studying the geometry of the geodesics can often provide new insights into these models. Here we consider three problems in each of these three models; exponential DLPP, TASEP and Poissonian LPP, and see how geometric and probabilistic techniques solve such problems.\n\nIn the first problem, we study finer properties of the coalescence structure of finite and semi-infinite geodesics for exactly solvable models of last passage percolation. We consider directed last passage percolation on $\\Z^2$ with i.i.d. exponential weights on the vertices. Fix two points $v_1=(0,0)$ and $v_2=(0, \\lfloor k^{2/3} \\rfloor)$ for some $k>0$, and consider the maximal paths $\\Gamma_1$ and $\\Gamma_2$ starting at $v_1$ and $v_2$ respectively to the point $(n,n)$ for $n\\gg k$. Our object of study is the point of coalescence, i.e., the point $v\\in \\Gamma_1\\cap \\Gamma_2$ with smallest $\|v\|_1$. We establish that the distance to coalescence $\|v\|_1$ scales as $k$, by showing the upper tail bound $\\P(\|v\|_1> Rk) \\leq R^{-c}$ for some $c>0$. We also consider the problem of coalescence for semi-infinite geodesics. For the almost surely unique semi-infinite geodesics in the direction $(1,1)$ starting from $v_3=(-\\lfloor k^{2/3} \\rfloor , \\lfloor k^{2/3}\\rfloor)$ and $v_4=(\\lfloor k^{2/3} \\rfloor ,- \\lfloor k^{2/3}\\rfloor)$, we establish the optimal tail estimate $\\P(\|v\|_1> Rk) \\asymp R^{-2/3}$, for the point of coalescence $v$. This answers a question left open by Pimentel(2016) who proved the corresponding lower bound.\n\nNext, we study the slow bond\" model, where the totally asymmetric simple exclusion process (TASEP) on $\\Z$ is modified by adding a slow bond at the origin. The slow bond increases the particle density immediately to its left and decreases the particle density immediately to its right. Whether or not this effect is detectable in the macroscopic current started from the step initial condition has attracted much interest over the years and this question was settled recently in Basu-Sidoravicius-Sly (2014) where it was shown that the current is reduced even for arbitrarily small strength of the defect. Following non-rigorous physics arguments in Janowsky-Lebowitz(1992,1994) and some unpublished works by Bramson, a conjectural description of properties of invariant measures of TASEP with a slow bond at the origin was provided by Liggett in his book (1999). We establish Liggett's conjectures and in particular show that, starting from step initial condition, TASEP with a slow bond at the origin, as a Markov process, converges in law to an invariant measure that is asymptotically close to product measures with different densities far away from the origin towards left and right. Our proof exploits the correspondence between TASEP and the last passage percolation on $\\Z^2$ with exponential weights and uses the understanding of geometry of maximal paths in those models.\n\nFinally, we study the modulus of continuity of polymer fluctuations and weight profiles in Poissonian LPP. The geodesics and their energy in Poissonian LPP can be scaled so that transformed geodesics cross unit distance and have fluctuations and scaled energy of unit order, and we refer to scaled geodesics\n\nas {\\em polymers} and\n\ntheir scaled energies as {\\em weights}. Polymers may be viewed as random functions of the vertical coordinate and, when they are, we show that they have modulus of continuity whose order is at most $t^{2/3}\\big(\\log t^{-1}\\big)^{1/3}$.\n\nThe power of one-third in the logarithm may be expected to be sharp and in a related problem we show that it is: among polymers in the unit box whose endpoints have vertical separation $t$ (and a horizontal separation of the same order),\n\nthe maximum transversal fluctuation has order $t^{2/3}\\big(\\log t^{-1}\\big)^{1/3}$.\n\nRegarding the orthogonal direction, in which growth occurs, we show that, when one endpoint of the polymer is fixed at $(0,0)$ and the other is varied vertically over $(0,z)$, $z\\in [1,2]$, the resulting random weight profile has sharp modulus of continuity of order $t^{1/3}\\big(\\log t^{-1}\\big)^{2/3}$.\n\nIn this way, we identify exponent pairs of $(2/3,1/3)$\n\nand $(1/3,2/3)$ in power law and polylogarithmic correction, respectively for polymer fluctuation, and polymer weight under vertical endpoint perturbation.\n\nThe two exponent pairs describe [Hammond(2012, 2012, 2011)] the fluctuation of the boundary separating two phases in subcritical planar random cluster models." ]	[ null, "https://escholarship.org/images/logo_eschol-mobile.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87680197,"math_prob":0.9929334,"size":5393,"snap":"2019-43-2019-47","text_gpt3_token_len":1335,"char_repetition_ratio":0.0957506,"word_repetition_ratio":0.035409037,"special_character_ratio":0.24105322,"punctuation_ratio":0.08895706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.997257,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-19T07:14:12Z\",\"WARC-Record-ID\":\"<urn:uuid:7ee5db3b-1274-4bf9-99ba-14ed17bba82d>\",\"Content-Length\":\"66222\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:dea4c09a-7053-43a3-bd57-4a506e1bc37d>\",\"WARC-Concurrent-To\":\"<urn:uuid:5ec6a7c8-74e4-444f-a42e-e79da233473f>\",\"WARC-IP-Address\":\"13.225.62.103\",\"WARC-Target-URI\":\"https://escholarship.org/uc/item/1cw491jk\",\"WARC-Payload-Digest\":\"sha1:7PYYGMWW2GWIKR3JGA4W5A6KHOL7CV4B\",\"WARC-Block-Digest\":\"sha1:HIFEJPPRDKOLSB7M3ESYJYFVU3LNAFIK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670036.23_warc_CC-MAIN-20191119070311-20191119094311-00111.warc.gz\"}"}
https://www.autoitscript.com/forum/topic/141207-sorting-array-using-main-and-secondary-index/	[ "## Recommended Posts\n\nHi!\n\nThis is my small contribution.\n\nThis code will first generate a 2 column array, and random some values. It will then sort the array using the 2nd column, and then sort some more using the 1st column. I have searched a lot and didnt found anything so i build mine!\n\nTO DO: Use UBOUND(ARRAY) to find the array size\n\n```#include <array.au3>\n;create Sample Array\nLocal \\$Array\n\\$Array = \"index\"\nFor \\$i = 1 To 99\n\\$Array[\\$i] = Int(Random() * 5) * 10\n\\$Array[\\$i] = Int(Random() * 5) * 10\nNext\n\\$Array = 33\n\\$Array = 33\n\\$Array = 22\n\\$Array = 22\n_ArrayDisplay(\\$Array,\"Before\")\n\\$Main_Index = 2\n\\$Secondary_Index = 1\n_ArraySort(\\$Array, 0, 0, 0, \\$Main_Index)\n\\$start = 1\nDo\n\\$i = \\$start\nIf \\$Array[\\$i][\\$Main_Index] = \\$Array[\\$i + 1][\\$Main_Index] Then\nDo\n\\$i += 1\nUntil \\$i = 99 Or \\$Array[\\$i][\\$Main_Index] <> \\$Array[\\$i + 1][\\$Main_Index]\n_ArraySort(\\$Array, 1, \\$start, \\$i, \\$Secondary_Index);small sort secondary index\nEndIf\n\\$start = \\$i + 1\nUntil \\$start >= 98 ;Array size -1\n_ArrayDisplay(\\$Array,\"After\")```\n\nHope it helps\n\n## Create an account\n\nRegister a new account\n\n• ### Recently Browsing 0 members\n\n×\n\n• Wiki\n\n• Back\n\n• #### Beta\n\n• Git\n• FAQ\n• Our Picks\n×\n• Create New..." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5689483,"math_prob":0.97832525,"size":1120,"snap":"2020-45-2020-50","text_gpt3_token_len":377,"char_repetition_ratio":0.16129032,"word_repetition_ratio":0.022222223,"special_character_ratio":0.38035715,"punctuation_ratio":0.10599078,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9737011,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-30T23:35:39Z\",\"WARC-Record-ID\":\"<urn:uuid:27772740-1472-44ab-abb2-3bd8be15c8df>\",\"Content-Length\":\"74032\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aa31b191-15d6-4282-8b84-5ca7d1e60269>\",\"WARC-Concurrent-To\":\"<urn:uuid:bcee909a-ded3-42c2-a528-16af22ac84ed>\",\"WARC-IP-Address\":\"212.227.91.231\",\"WARC-Target-URI\":\"https://www.autoitscript.com/forum/topic/141207-sorting-array-using-main-and-secondary-index/\",\"WARC-Payload-Digest\":\"sha1:VBDOGGID7EEQW4K6BQ3WPQVZXWCWUJZF\",\"WARC-Block-Digest\":\"sha1:GJCC6KGLI37OJF3GDMOPLVJLWFSWX7KX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141515751.74_warc_CC-MAIN-20201130222609-20201201012609-00646.warc.gz\"}"}
https://www.ginac.de/ginac.git/?p=ginac.git;a=blob;f=check/check_matrices.cpp;h=91f4b5da4b64ace42691890fbf614f9268d0a467;hb=91f4b5da4b64ace42691890fbf614f9268d0a467	[ "]> www.ginac.de Git - ginac.git/blob - check/check_matrices.cpp\n91f4b5da4b64ace42691890fbf614f9268d0a467\n1 /** @file check_matrices.cpp\n2 \n3 Here we test manipulations on GiNaC's symbolic matrices. They are a\n4 * well-tried resource for cross-checking the underlying symbolic\n5 * manipulations. /\n7 /\n8 * GiNaC Copyright (C) 1999-2003 Johannes Gutenberg University Mainz, Germany\n9 \n10 This program is free software; you can redistribute it and/or modify\n11 * it under the terms of the GNU General Public License as published by\n12 * the Free Software Foundation; either version 2 of the License, or\n13 * (at your option) any later version.\n14 \n15 This program is distributed in the hope that it will be useful,\n16 * but WITHOUT ANY WARRANTY; without even the implied warranty of\n17 * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the\n18 * GNU General Public License for more details.\n19 \n20 You should have received a copy of the GNU General Public License\n21 * along with this program; if not, write to the Free Software\n22 * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA\n23 /\n25 #include \"checks.h\"\n27 / determinants of some sparse symbolic matrices with coefficients in\n28 * an integral domain. /\n29 static unsigned integdom_matrix_determinants()\n30 {\n31 unsigned result = 0;\n32 symbol a(\"a\");\n34 for (unsigned size=3; size<22; ++size) {\n35 matrix A(size,size);\n36 // populate one element in each row:\n37 for (unsigned r=0; r<size-1; ++r)\n38 A.set(r,unsigned(rand()%size),dense_univariate_poly(a,5));\n39 // set the last row to a linear combination of two other lines\n40 // to guarantee that the determinant is zero:\n41 for (unsigned c=0; c<size; ++c)\n42 A.set(size-1,c,A(0,c)-A(size-2,c));\n43 if (!A.determinant().is_zero()) {\n44 clog << \"Determinant of \" << size << \"x\" << size << \" matrix \"\n45 << endl << A << endl\n46 << \"was not found to vanish!\" << endl;\n47 ++result;\n48 }\n49 }\n51 return result;\n52 }\n54 / determinants of some symbolic matrices with multivariate rational function\n55 * coefficients. /\n56 static unsigned rational_matrix_determinants()\n57 {\n58 unsigned result = 0;\n59 symbol a(\"a\"), b(\"b\"), c(\"c\");\n61 for (unsigned size=3; size<9; ++size) {\n62 matrix A(size,size);\n63 for (unsigned r=0; r<size-1; ++r) {\n64 // populate one or two elements in each row:\n65 for (unsigned ec=0; ec<2; ++ec) {\n66 ex numer = sparse_tree(a, b, c, 1+rand()%4, false, false, false);\n67 ex denom;\n68 do {\n69 denom = sparse_tree(a, b, c, rand()%2, false, false, false);\n70 } while (denom.is_zero());\n71 A.set(r,unsigned(rand()%size),numer/denom);\n72 }\n73 }\n74 // set the last row to a linear combination of two other lines\n75 // to guarantee that the determinant is zero:\n76 for (unsigned co=0; co<size; ++co)\n77 A.set(size-1,co,A(0,co)-A(size-2,co));\n78 if (!A.determinant().is_zero()) {\n79 clog << \"Determinant of \" << size << \"x\" << size << \" matrix \"\n80 << endl << A << endl\n81 << \"was not found to vanish!\" << endl;\n82 ++result;\n83 }\n84 }\n86 return result;\n87 }\n89 / Some quite funny determinants with functions and stuff like that inside. /\n90 static unsigned funny_matrix_determinants()\n91 {\n92 unsigned result = 0;\n93 symbol a(\"a\"), b(\"b\"), c(\"c\");\n95 for (unsigned size=3; size<8; ++size) {\n96 matrix A(size,size);\n97 for (unsigned co=0; co<size-1; ++co) {\n98 // populate one or two elements in each row:\n99 for (unsigned ec=0; ec<2; ++ec) {\n100 ex numer = sparse_tree(a, b, c, 1+rand()%3, true, true, false);\n101 ex denom;\n102 do {\n103 denom = sparse_tree(a, b, c, rand()%2, false, true, false);\n104 } while (denom.is_zero());\n105 A.set(unsigned(rand()%size),co,numer/denom);\n106 }\n107 }\n108 // set the last column to a linear combination of two other columns\n109 // to guarantee that the determinant is zero:\n110 for (unsigned ro=0; ro<size; ++ro)\n111 A.set(ro,size-1,A(ro,0)-A(ro,size-2));\n112 if (!A.determinant().is_zero()) {\n113 clog << \"Determinant of \" << size << \"x\" << size << \" matrix \"\n114 << endl << A << endl\n115 << \"was not found to vanish!\" << endl;\n116 ++result;\n117 }\n118 }\n120 return result;\n121 }\n123 / compare results from different determinant algorithms.*/\n124 static unsigned compare_matrix_determinants()\n125 {\n126 unsigned result = 0;\n127 symbol a(\"a\");\n129 for (unsigned size=2; size<8; ++size) {\n130 matrix A(size,size);\n131 for (unsigned co=0; co<size; ++co) {\n132 for (unsigned ro=0; ro<size; ++ro) {\n133 // populate some elements\n134 ex elem = 0;\n135 if (rand()%(size/2) == 0)\n136 elem = sparse_tree(a, a, a, rand()%3, false, true, false);\n137 A.set(ro,co,elem);\n138 }\n139 }\n140 ex det_gauss = A.determinant(determinant_algo::gauss);\n141 ex det_laplace = A.determinant(determinant_algo::laplace);\n142 ex det_divfree = A.determinant(determinant_algo::divfree);\n143 ex det_bareiss = A.determinant(determinant_algo::bareiss);\n144 if ((det_gauss-det_laplace).normal() != 0 \|\|\n145 (det_bareiss-det_laplace).normal() != 0 \|\|\n146 (det_divfree-det_laplace).normal() != 0) {\n147 clog << \"Determinant of \" << size << \"x\" << size << \" matrix \"\n148 << endl << A << endl\n149 << \"is inconsistent between different algorithms:\" << endl\n150 << \"Gauss elimination: \" << det_gauss << endl\n151 << \"Minor elimination: \" << det_laplace << endl\n152 << \"Division-free elim.: \" << det_divfree << endl\n153 << \"Fraction-free elim.: \" << det_bareiss << endl;\n154 ++result;\n155 }\n156 }\n158 return result;\n159 }\n161 static unsigned symbolic_matrix_inverse()\n162 {\n163 unsigned result = 0;\n164 symbol a(\"a\"), b(\"b\"), c(\"c\");\n166 for (unsigned size=2; size<6; ++size) {\n167 matrix A(size,size);\n168 do {\n169 for (unsigned co=0; co<size; ++co) {\n170 for (unsigned ro=0; ro<size; ++ro) {\n171 // populate some elements\n172 ex elem = 0;\n173 if (rand()%(size/2) == 0)\n174 elem = sparse_tree(a, b, c, rand()%2, false, true, false);\n175 A.set(ro,co,elem);\n176 }\n177 }\n178 } while (A.determinant() == 0);\n179 matrix B = A.inverse();\n180 matrix C = A.mul(B);\n181 bool ok = true;\n182 for (unsigned ro=0; ro<size; ++ro)\n183 for (unsigned co=0; co<size; ++co)\n184 if (C(ro,co).normal() != (ro==co?1:0))\n185 ok = false;\n186 if (!ok) {\n187 clog << \"Inverse of \" << size << \"x\" << size << \" matrix \"\n188 << endl << A << endl\n189 << \"erroneously returned: \"\n190 << endl << B << endl;\n191 ++result;\n192 }\n193 }\n195 return result;\n196 }\n198 unsigned check_matrices()\n199 {\n200 unsigned result = 0;\n202 cout << \"checking symbolic matrix manipulations\" << flush;\n203 clog << \"---------symbolic matrix manipulations:\" << endl;\n205 result += integdom_matrix_determinants(); cout << '.' << flush;\n206 result += rational_matrix_determinants(); cout << '.' << flush;\n207 result += funny_matrix_determinants(); cout << '.' << flush;\n208 result += compare_matrix_determinants(); cout << '.' << flush;\n209 result += symbolic_matrix_inverse(); cout << '.' << flush;\n211 if (!result) {\n212 cout << \" passed \" << endl;\n213 clog << \"(no output)\" << endl;\n214 } else {\n215 cout << \" failed \" << endl;\n216 }\n218 return result;\n219 }" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5729338,"math_prob":0.9453031,"size":5738,"snap":"2023-14-2023-23","text_gpt3_token_len":1751,"char_repetition_ratio":0.1573073,"word_repetition_ratio":0.16759777,"special_character_ratio":0.38741723,"punctuation_ratio":0.20319432,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9964519,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-02T04:08:05Z\",\"WARC-Record-ID\":\"<urn:uuid:d8bfa2c4-831b-4e9f-8789-bbe37e814832>\",\"Content-Length\":\"81510\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a41dea58-643c-4b46-8064-7b169ee8c841>\",\"WARC-Concurrent-To\":\"<urn:uuid:8df790ac-f340-45a6-8855-2a0cc6c44dcc>\",\"WARC-IP-Address\":\"188.68.41.94\",\"WARC-Target-URI\":\"https://www.ginac.de/ginac.git/?p=ginac.git;a=blob;f=check/check_matrices.cpp;h=91f4b5da4b64ace42691890fbf614f9268d0a467;hb=91f4b5da4b64ace42691890fbf614f9268d0a467\",\"WARC-Payload-Digest\":\"sha1:IZPQQPCHUOJYFBF4WE6A7QJ5RPIODH53\",\"WARC-Block-Digest\":\"sha1:N2ZPL7YOQFQAHJFFK33CF6ODJFMN7BLD\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224648322.84_warc_CC-MAIN-20230602040003-20230602070003-00295.warc.gz\"}"}
http://www.koreascience.or.kr/search.page?keywords=Linear+programming+%28LP%29&pageSize=10&pageNo=1	[ "• Title, Summary, Keyword: Linear programming (LP)\n\n### Problem Solution of Linear Programming based Neural Network\n\n• Son, Jun-Hyug;Seo, Bo-Hyeok\n• Proceedings of the KIEE Conference\n• /\n• /\n• pp.98-101\n• /\n• 2004\n• Linear Programming(LP) is the term used for defining a wide range of optimization problems in which the objective function to be minimized or maximized is linear in the unknown variables and the constraints are a combination of linear equalities and inequalities. LP problems occur in many real-life economic situations where profits are to be maximized or costs minimized with constraint limits on resources. While the simplex method introduced in a later reference can be used for hand solution of LP problems, computer use becomes necessary even for a small number of variables. Problems involving diet decisions, transportation, production and manufacturing, product mix, engineering limit analysis in design, airline scheduling, and so on are solved using computers. This technique is called Sequential Linear Programming (SLP). This paper describes LP's problems and solves a LP's problems using the neural networks.\n\n### Forest Management Planning by Linear Programming - Timber Harvest Scheduling of a Korean Pine stand - (Linear Programming에 의한 삼림경영계획(森林經營計劃) - 잣나무임분(林分)의 삼림수확계획(森林收穫計劃)을 중심으로 -)\n\n• Woo, Jong Choon\n• Journal of Korean Society of Forest Science\n• /\n• v.80 no.4\n• /\n• pp.427-435\n• /\n• 1991\n• Linear programming(LP) is a well-known method in optimizing timber harvest schedules. This paper describes a linear programming formulation of korean pine stands for timber harvest scheduling problems. Simulation technique and LP were applied to optimize the time and space distribution of the sustained yield for the 10-year forest management planning horizon. Growthfunction of korean pine stands in study area was derived with the yield table. This growthfunction was contained to the simulation model in estimating of changing stand volume conditions for the planning horizon. These estimated values were served as the basic data of LP model, and LP model was formulated with the maximum of periodical harvest volume calculated by the classical yield regulation method (Paulsen-Hundeshagen formula) and the maximum of periodical harvest area calculated for the normal age distribution. The timber harvest schedule was established periodically for each subcompartment of korean pine stands in experiment forest of College of Forestry in Kangweon National University with the here developed LP model.\n\n### Deformable Surface 3D Reconstruction from a Single Image by Linear Programming\n\n• Ma, Wenjuan;Sun, Shusen\n• KSII Transactions on Internet and Information Systems (TIIS)\n• /\n• v.11 no.6\n• /\n• pp.3121-3142\n• /\n• 2017\n• We present a method for 3D shape reconstruction of inextensible deformable surfaces from a single image. The key of our approach is to represent the surface as a 3D triangulated mesh and formulate the reconstruction problem as a sequence of Linear Programming (LP) problems. The LP problem consists of data constraints which are 3D-to-2D keypoint correspondences and shape constraints which are designed to retain original lengths of mesh edges. We use a closed-form method to generate an initial structure, then refine this structure by solving the LP problem iteratively. Compared with previous methods, ours neither involves smoothness constraints nor temporal consistency, which enables us to recover shapes of surfaces with various deformations from a single image. The robustness and accuracy of our approach are evaluated quantitatively on synthetic data and qualitatively on real data.\n\n### Analysis of the Methodology for Linear Programming Optimality Analysis using Metamodelling Techniques\n\n• Lee, Young-Hae;Jeong, Chan-Seok\n• Journal of the military operations research society of Korea\n• /\n• v.25 no.2\n• /\n• pp.1-14\n• /\n• 1999\n• Metamodels using response surface methodology (RSM) are used for the optimality analysis of linear programming (LP). They have the form of a simple polynomial, and predict the optimal objective function value of an LP for various levels of the constraints. The metamodelling techniques for optimality analysis of LP can be applied to large-scale LP models. What is needed is some large-scale application of the techniques to verify how accurate they are. In this paper, we plan to use the large scale LP model, strategic transport optimal routing model (STORM). The developed metamodels of the large scale LP can provide some useful information.\n\n### Development of an LP integrated environment software under MS-DOS (MS-DOS용 선형계획법 통합환경 소프트웨어의 개발)\n\n• 설동렬;박찬규;서용원;박순달\n• Korean Management Science Review\n• /\n• v.12 no.1\n• /\n• pp.125-138\n• /\n• 1995\n• This paper is to develop an integrated environment software on MS-DOS for linear programming. For the purpose, First, the linear programming integrated environment software satisfying both the educational purpose and the professional purpose was designed and constructed on MS-DOS. Second, the text editor with big capacity was developed. The arithmetic form analyser was also developed and connected to the test editor so that users can input data in the arithmetic form. As a result, users can learn and perform linear programming in the linear programming integrated environment software.\n\n### Linear Programming Applications to Managerial Accounting Decision Makings (선형계획법을 이용한 관리회계적 의사결정)\n\n• Song, Han-Sik;Choi, Min-Cheol\n• /\n• v.9 no.4\n• /\n• pp.99-117\n• /\n• 2018\n• This study has investigated Linear Programming (LP) applications to special decision making problems in managerial accounting with the help of spreadsheet Solver tools. It uses scenario approaches to case examples having three products and three resources in make-and-supply business operations, which is applicable to cases having more variables and constraints. Integer Programmings (IP) are applied in order to model situations when products are better valued in integer values or logical constraints are required. Three cases in one-time-only special order decisions include Goal Programming approach, Knapsack problems with 0/1 selections, and fixed-charge 0/1 integer modelling techniques for set-up operation costs. For the decisions in outsourcing problems, opportunity-costs of resources expressed by shadow-prices are considered to determine their precise contributions. It has also shown that the improvement in work-shop operation for an unprofitable product must overcome its 'reduced cost' by the sum of direct manufacturing cost savings and its shadow-price contributions. This paper has demonstrated how various real situations of special decision problem in managerial accounting can be approached without mistakes by using LP's and IP's, and how students both in accounting and management science can acquire LP skills in their education.\n\n### A Mathematical Structure and Formulation of Bottom-up Model based on Linear Programming (온실가스감축정책 평가를 위한 LP기반 상향식 모형의 수리 구조 및 정식화에 대한 연구)\n\n• Kim, Hu Gon\n• Journal of Energy Engineering\n• /\n• v.23 no.4\n• /\n• pp.150-159\n• /\n• 2014\n• Since the release of mid-term domestic GHG goals until 2020, in 2009, some various GHG reduction policies have been proposed. There are two types of modeling approaches for identifying options required to meet greenhouse gas (GHG) abatement targets and assessing their economic impacts: top-down and bottom-up models. Examples of the bottom-up optimization models include MARKAL, MESSAGE, LEAP, and AIM, all of which are developed based on linear programming (LP) with a few differences in user interface and database utilization. In this paper, we suggest a simplified LP formulation and how can build it through step-by-step procedures.\n\n### A Case Study on the Application of Decomposition Principle to a Linear Programming Problem (분할기법(分割技法)을 이용한 선형계획법(線型計劃法)의 응용(應用)에 관한 사례 연구(事例硏究))\n\n• Yun, In-Jung;Kim, Seong-In\n• IE interfaces\n• /\n• v.1 no.1\n• /\n• pp.1-7\n• /\n• 1988\n• This paper discusses the applicability of the decomposition principle to an LP (Linear Programming) problem. Through a case study on product mix problems in a chemical process of Korean Steel Chemical Co., Ltd., the decomposition algorithm, LP Simplex method and a modified method are compared and evaluated in computation time and number of iterations.\n\n### On Solving the Fuzzy Goal Programming and Its Extension (불분명한 북표계확볍과 그 확장)\n\n• 정충영\n• Journal of the Korean Operations Research and Management Science Society\n• /\n• v.11 no.2\n• /\n• pp.79-87\n• /\n• 1986\n• This paper illustrates a new method to solve the fuzzy goal programming (FGP) problem. It is proved that the FGP proposed by Narasimhan can be solved on the basis of linear programming(LP) model. Narasimhan formulated the FGP problem as a set of $S^{K}$LP problems, each containing 3K constraints, where K is the number of fuzzy goals/constraints. Whereas Hanna formulated the FGP problem as a single LP problem with only 2K constraints and 2K + 1 additional variables. This paper presents that the FGP problem can be transformed with easy into a single LP model with 2K constraints and only one additional variables. And we propose extended FGP :(1) FGP with weights associated with individual goals, (2) FGP with preemptive prioities. The extended FGP has a framework that is identical to that of conventional goal programming (GP), such that the extended FGP can be applied with fuzzy concept to the all areas where GP can be applied.d.\n\n### A New Approach for Forest Management Planning : Fuzzy Multiobjective Linear Programming (삼림경영계획(森林經營計劃)을 위한 새로운 접근법(接近法) : 퍼지 다목표선형계획법(多目標線型計劃法))\n\n• Woo, Jong Choon\n• Journal of Korean Society of Forest Science\n• /\n• v.83 no.3\n• /\n• pp.271-279\n• /\n• 1994\n• This paper descbibes a fuzzy multiobjective linear programming, which is a relatively new approach in forestry in solving forest management problems. At first, the fuzzy set theory is explained briefly and the fuzzy linear programming(FLP) and the fuzzy multiobjective linear programming(FMLP) are introduced conceptionally. With the information obtained from the study area in Thailand, a standard linear programming problem is formulated, and optimal solutions (present net worth) are calculated for four groups of timber price by this LP model, respectively. This LP model is reformulated to a fuzzy multiobjective linear programming model to accommodate uncertain timber values and with this FMLP model a compromise solution is attained. Optimal solutions of four objective functions for four timber price groups and the compromise solution are compared and discussed." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9272438,"math_prob":0.78794587,"size":922,"snap":"2020-34-2020-40","text_gpt3_token_len":161,"char_repetition_ratio":0.11220044,"word_repetition_ratio":0.0,"special_character_ratio":0.17028199,"punctuation_ratio":0.08496732,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9560654,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-09-28T03:19:54Z\",\"WARC-Record-ID\":\"<urn:uuid:568679ae-fca8-4bec-a1a1-8ed279df27b9>\",\"Content-Length\":\"268580\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f8df0070-2977-47b6-80fa-c0c83f7778f3>\",\"WARC-Concurrent-To\":\"<urn:uuid:26bf7dc5-6612-426a-ac87-958ced1ff75d>\",\"WARC-IP-Address\":\"203.250.218.22\",\"WARC-Target-URI\":\"http://www.koreascience.or.kr/search.page?keywords=Linear+programming+%28LP%29&pageSize=10&pageNo=1\",\"WARC-Payload-Digest\":\"sha1:LTMZS524KHYG2FLEFF7HLLJ5FFYNOSKU\",\"WARC-Block-Digest\":\"sha1:543WAEYFFRVGLYG4EMSGLCLSQ5NHBDVG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-40/CC-MAIN-2020-40_segments_1600401583556.73_warc_CC-MAIN-20200928010415-20200928040415-00367.warc.gz\"}"}
https://stackoverflow.com/questions/2702564/how-can-i-quickly-sum-all-numbers-in-a-file/13073486	[ "How can I quickly sum all numbers in a file?\n\nI have a file which contains several thousand numbers, each on it's own line:\n\n``````34\n42\n11\n6\n2\n99\n...\n``````\n\nI'm looking to write a script which will print the sum of all numbers in the file. I've got a solution, but it's not very efficient. (It takes several minutes to run.) I'm looking for a more efficient solution. Any suggestions?\n\n• What was your slow solution? Maybe we can help you figure out what was slow about it. :) – brian d foy Apr 23 '10 at 23:59\n• @brian d foy, I'm too embarrassed to post it. I know why it's slow. It's because I call \"cat filename \| head -n 1\" to get the top number, add it to a running total, and call \"cat filename \| tail...\" to remove the top line for the next iteration... I have a lot to learn about programming!!! – Mark Roberts Apr 24 '10 at 1:22\n• That's...very systematic. Very clear and straight forward, and I love it for all that it is a horrible abomination. Built, I assume, out of the tools that you knew when you started, right? – dmckee Apr 24 '10 at 2:43\n• full duplicate: stackoverflow.com/questions/450799/… – codeholic Apr 26 '10 at 11:39\n• @MarkRoberts It must have taken you a long while to work that out. It's a very cleaver problem solving technique, and oh so wrong. It looks like a classic case of over think. Several of Glen Jackman's solutions shell scripting solutions (and two are pure shell that don't use things like `awk` and `bc`). These all finished adding a million numbers up in less than 10 seconds. Take a look at those and see how it can be done in pure shell. – David W. Aug 22 '13 at 14:24\n\nFor a Perl one-liner, it's basically the same thing as the `awk` solution in Ayman Hourieh's answer:\n\n`````` % perl -nle '\\$sum += \\$_ } END { print \\$sum'\n``````\n\nIf you're curious what Perl one-liners do, you can deparse them:\n\n`````` % perl -MO=Deparse -nle '\\$sum += \\$_ } END { print \\$sum'\n``````\n\nThe result is a more verbose version of the program, in a form that no one would ever write on their own:\n\n``````BEGIN { \\$/ = \"\\n\"; \\$\\ = \"\\n\"; }\nLINE: while (defined(\\$_ = <ARGV>)) {\nchomp \\$_;\n\\$sum += \\$_;\n}\nsub END {\nprint \\$sum;\n}\n-e syntax OK\n``````\n\nJust for giggles, I tried this with a file containing 1,000,000 numbers (in the range 0 - 9,999). On my Mac Pro, it returns virtually instantaneously. That's too bad, because I was hoping using `mmap` would be really fast, but it's just the same time:\n\n``````use 5.010;\nuse File::Map qw(map_file);\n\nmap_file my \\$map, \\$ARGV;\n\n\\$sum += \\$1 while \\$map =~ m/(\\d+)/g;\n\nsay \\$sum;\n``````\n• Wow, that shows a deep understanding on what code -nle actually wraps around the string you give it. My initial thought was that you shouldn't post while intoxicated but then I noticed who you were and remembered some of your other Perl answers :-) – paxdiablo Apr 23 '10 at 23:52\n• -n and -p just put characters around the argument to -e, so you can use those characters for whatever you want. We have a lot of one-liners that do interesting things with that in Effective Perl Programming (which is about to hit the shelves). – brian d foy Apr 23 '10 at 23:56\n• Nice, what are these non-matching curly braces about? – Frank Apr 24 '10 at 6:00\n• -n adds the `while { }` loop around your program. If you put `} ... {` inside, then you have `while { } ... { }`. Evil? Slightly. – jrockway Apr 24 '10 at 8:35\n• Big bonus for highlighting the `-MO=Deparse` option! Even though on a separate topic. – conny Nov 4 '11 at 12:47\n\nYou can use awk:\n\n``````awk '{ sum += \\$1 } END { print sum }' file\n``````\n• program exceeded: maximum number of field sizes: 32767 – leef Nov 2 '12 at 3:50\n• Simple, and doesn't require perl. This is the best answer. – Eddified Jul 8 '14 at 21:12\n• With the `-F '\\t'` option if your fields contain spaces and are separated by tabs. – Ethan Furman Jul 17 '14 at 22:17\n• Please mark this as the best answer. It also works if you want to sum the first value in each row, inside a TSV (tab-separated value) file. – Andrea Oct 27 '16 at 10:42\n\nNone of the solution thus far use `paste`. Here's one:\n\n``````paste -sd+ filename \| bc\n``````\n\nAs an example, calculate Σn where 1<=n<=100000:\n\n``````\\$ seq 100000 \| paste -sd+ \| bc -l\n5000050000\n``````\n\n(For the curious, `seq n` would print a sequence of numbers from `1` to `n` given a positive number `n`.)\n\n• Very nice! And easy to remember – Brendan Maguire Jul 30 '14 at 11:45\n• very unix style solution =) – Peter K May 27 '15 at 11:15\n• `seq 100000 \| paste -sd+ - \| bc -l` on Mac OS X Bash shell. And this is by far the sweetest and the unixest solution! – Simo A. Mar 2 at 15:54\n\nJust for fun, let's benchmark it:\n\n``````\\$ for ((i=0; i<1000000; i++)) ; do echo \\$RANDOM; done > random_numbers\n\n\\$ time perl -nle '\\$sum += \\$_ } END { print \\$sum' random_numbers\n16379866392\n\nreal 0m0.226s\nuser 0m0.219s\nsys 0m0.002s\n\n\\$ time awk '{ sum += \\$1 } END { print sum }' random_numbers\n16379866392\n\nreal 0m0.311s\nuser 0m0.304s\nsys 0m0.005s\n\n\\$ time { { tr \"\\n\" + < random_numbers ; echo 0; } \| bc; }\n16379866392\n\nreal 0m0.445s\nuser 0m0.438s\nsys 0m0.024s\n\n\\$ time { s=0;while read l; do s=\\$((s+\\$l));done<random_numbers;echo \\$s; }\n16379866392\n\nreal 0m9.309s\nuser 0m8.404s\nsys 0m0.887s\n\n\\$ time { s=0;while read l; do ((s+=l));done<random_numbers;echo \\$s; }\n16379866392\n\nreal 0m7.191s\nuser 0m6.402s\nsys 0m0.776s\n\n\\$ time { sed ':a;N;s/\\n/+/;ta' random_numbers\|bc; }\n^C\n\nreal 4m53.413s\nuser 4m52.584s\nsys 0m0.052s\n``````\n\nI aborted the sed run after 5 minutes\n\nI've been diving to , and it is speedy:\n\n``````\\$ time lua -e 'sum=0; for line in io.lines() do sum=sum+line end; print(sum)' < random_numbers\n16388542582.0\n\nreal 0m0.362s\nuser 0m0.313s\nsys 0m0.063s\n``````\n\nand while I'm updating this, ruby:\n\n``````\\$ time ruby -e 'sum = 0; File.foreach(ARGV.shift) {\|line\| sum+=line.to_i}; puts sum' random_numbers\n16388542582\n\nreal 0m0.378s\nuser 0m0.297s\nsys 0m0.078s\n``````\n\nHeed Ed Morton's advice: using `\\$1`\n\n``````\\$ time awk '{ sum += \\$1 } END { print sum }' random_numbers\n16388542582\n\nreal 0m0.421s\nuser 0m0.359s\nsys 0m0.063s\n``````\n\nvs using `\\$0`\n\n``````\\$ time awk '{ sum += \\$0 } END { print sum }' random_numbers\n16388542582\n\nreal 0m0.302s\nuser 0m0.234s\nsys 0m0.063s\n``````\n• +1: For coming up with a bunch of solutions, and benchmarking them. – David W. Aug 22 '13 at 14:24\n• time cat random_numbers\|paste -sd+\|bc -l real 0m0.317s user 0m0.310s sys 0m0.013s – rafi wiener Jul 3 '17 at 10:21\n• that should be just about identical to the `tr` solution. – glenn jackman Jul 7 '17 at 12:47\n• Your awk script should execute a bit faster if you use `\\$0` instead of `\\$1` since awk does field splitting (which obviously takes time) if any field is specifically mentioned in the script but doesn't otherwise. – Ed Morton Mar 18 at 13:56\n\nThis works:\n\n``````{ tr '\\n' +; echo 0; } < file.txt \| bc\n``````\n• what's the reason for adding `echo 0` after `tr`? – Dhiraj Jul 5 '18 at 17:17\n• Out of `tr` you end up with a trailing `+`: `1+2+3+4+` That would be a syntax error to `bc`. So echo `0` to fix up the syntax: `1+2+3+4+0` – Mark L. Smith Jul 6 '18 at 14:02\n\nAnother option is to use `jq`:\n\n``````\\$ seq 10\|jq -s add\n55\n``````\n\n`-s` (`--slurp`) reads the input lines into an array.\n\nThis is straight Bash:\n\n``````sum=0\ndo\n(( sum += line ))\ndone < file\necho \\$sum\n``````\n• this willl work as long as there are no decimals – ghostdog74 Apr 24 '10 at 5:09\n• And it's probably one of the slowest solutions and therefore not so suitable for large amounts of numbers. – David Apr 23 at 16:06\n\nHere's another one-liner\n\n``````( echo 0 ; sed 's/\\$/ +/' foo ; echo p ) \| dc\n``````\n\nThis assumes the numbers are integers. If you need decimals, try\n\n``````( echo 0 2k ; sed 's/\\$/ +/' foo ; echo p ) \| dc\n``````\n\nAdjust 2 to the number of decimals needed.\n\nI prefer to use GNU datamash for such tasks because it's more succinct and legible than perl or awk. For example\n\n``````datamash sum 1 < myfile\n``````\n\nwhere 1 denotes the first column of data.\n\n• This does not appear to be a standard component as I do not see it in my Ubuntu installation. Would like to see it benchmarked, though. – Steven the Easily Amused Jan 29 '18 at 17:59\n``````cat nums \| perl -ne '\\$sum += \\$_ } { print \\$sum'\n``````\n\n(same as brian d foy's answer, without 'END')\n\nJust for fun, lets do it with PDL, Perl's array math engine!\n\n``````perl -MPDL -E 'say rcols(shift)->sum' datafile\n``````\n\n`rcols` reads columns into a matrix (1D in this case) and `sum` (surprise) sums all the element of the matrix.\n\n• How fix Can't locate PDL.pm in @INC (you may need to install the PDL module) (@INC contains: /etc/perl /usr/local/lib/x86_64-linux-gnu/perl/5.22.1 ?)) for fun of course=) – Fortran Apr 29 '17 at 15:01\n• You have to install PDL first, it isn't a Perl native module. – Joel Berger May 1 '17 at 20:06\n\nHere is a solution using python with a generator expression. Tested with a million numbers on my old cruddy laptop.\n\n``````time python -c \"import sys; print sum((float(l) for l in sys.stdin))\" < file\n\nreal 0m0.619s\nuser 0m0.512s\nsys 0m0.028s\n``````\n• A simple list comprehension with a named function is a nice use-case for `map()`: `map(float, sys.stdin)` – sevko May 1 '15 at 19:50\n\nI prefer to use R for this:\n\n``````\\$ R -e 'sum(scan(\"filename\"))'\n``````\n``````sed ':a;N;s/\\n/+/;ta' file\|bc\n``````\n``````\\$ perl -MList::Util=sum -le 'print sum <>' nums.txt\n``````\n\nMore succinct:\n\n``````# Ruby\nruby -e 'puts open(\"random_numbers\").map(&:to_i).reduce(:+)'\n\n# Python\npython -c 'print(sum(int(l) for l in open(\"random_numbers\")))'\n``````\n\nPerl 6\n\n``````say sum lines\n``````\n``````~\\$ perl6 -e '.say for 0..1000000' > test.in\n\n~\\$ perl6 -e 'say sum lines' < test.in\n500000500000\n``````\n\nI have not tested this but it should work:\n\n``````cat f \| tr \"\\n\" \"+\" \| sed 's/+\\$/\\n/' \| bc\n``````\n\nYou might have to add \"\\n\" to the string before bc (like via echo) if bc doesn't treat EOF and EOL...\n\n• It doesn't work. `bc` issues a syntax error because of the trailing \"+\" and lack of newline at the end. This will work and it eliminates a useless use of `cat`: `{ tr \"\\n\" \"+\" \| sed 's/+\\$/\\n/'\| bc; } < numbers2.txt` or `<numbers2.txt tr \"\\n\" \"+\" \| sed 's/+\\$/\\n/'\| bc` – Dennis Williamson Apr 24 '10 at 2:18\n• `tr \"\\n\" \"+\" <file \| sed 's/+\\$/\\n/' \| bc` – ghostdog74 Apr 24 '10 at 5:23\n\nAnother for fun\n\n``````sum=0;for i in \\$(cat file);do sum=\\$((sum+\\$i));done;echo \\$sum\n``````\n\nor another bash only\n\n``````s=0;while read l; do s=\\$((s+\\$l));done<file;echo \\$s\n``````\n\nBut awk solution is probably best as it's most compact.\n\nWith Ruby:\n\n``````ruby -e \"File.read('file.txt').split.inject(0){\|mem, obj\| mem += obj.to_f}\"\n``````\n• Another option (when input is from STDIN) is `ruby -e'p readlines.map(&:to_f).reduce(:+)'`. – nisetama Feb 24 at 22:30\n\nI don't know if you can get a lot better than this, considering you need to read through the whole file.\n\n``````\\$sum = 0;\nwhile(<>){\n\\$sum += \\$_;\n}\nprint \\$sum;\n``````\n• what's the \\$_ mean? – Mark Roberts Apr 23 '10 at 23:39\n• Very readable. For perl. But yeah, it's going to have to be something like that... – dmckee Apr 23 '10 at 23:40\n• `\\$_` is the default variable. The line input operator, `<>`, puts it's result in there by default when you use `<>` in `while`. – brian d foy Apr 23 '10 at 23:53\n• @Mark, `\\$_` is the topic variable--it works like the 'it'. In this case `<>` assigns each line to it. It gets used in a number of places to reduce code clutter and help with writing one-liners. The script says \"Set the sum to 0, read each line and add it to the sum, then print the sum.\" – daotoad Apr 23 '10 at 23:59\n• @Stefan, with warnings and strictures off, you can skip declaring and initializing `\\$sum`. Since this is so simple, you can even use a statement modifier `while`: `\\$sum += \\$_ while <>; print \\$sum;` – daotoad Apr 24 '10 at 0:00\n\nHere's another:\n\n``````open(FIL, \"a.txt\");\n\nmy \\$sum = 0;\nforeach( <FIL> ) {chomp; \\$sum += \\$_;}\n\nclose(FIL);\n\nprint \"Sum = \\$sum\\n\";\n``````\n\nC always wins for speed:\n\n``````#include <stdio.h>\n#include <stdlib.h>\n\nint main(int argc, char *argv) {\nchar line = NULL;\nsize_t len = 0;\ndouble sum = 0.0;\n\nwhile (read = getline(&line, &len, stdin) != -1) {\nsum += atof(line);\n}\n\nprintf(\"%f\", sum);\nreturn 0;\n}\n``````\n\nTiming for 1M numbers (same machine/input as my python answer):\n\n``````\\$ gcc sum.c -o sum && time ./sum < numbers\n5003371677.000000\nreal 0m0.188s\nuser 0m0.180s\nsys 0m0.000s\n``````\n\nYou can do it with Alacon - command-line utility for Alasql database.\n\nIt works with Node.js, so you need to install Node.js and then Alasql package:\n\nTo calculate sum from TXT file you can use the following command:\n\n``````> node alacon \"SELECT VALUE SUM() FROM TXT('mydata.txt')\"\n``````\n\nIt is not easier to replace all new lines by `+`, add a `0` and send it to the `Ruby` interpreter?\n\n``````(sed -e \"s/\\$/+/\" file; echo 0)\|irb\n``````\n\nIf you do not have `irb`, you can send it to `bc`, but you have to remove all newlines except the last one (of `echo`). It is better to use `tr` for this, unless you have a PhD in `sed` .\n\n``````(sed -e \"s/\\$/+/\" file\|tr -d \"\\n\"; echo 0)\|bc\n``````\n\nJust to be ridiculous:\n\n``````cat f \| tr \"\\n\" \"+\" \| perl -pne chop \| R --vanilla --slave\n``````\n• This one eventually died with \"Error: evaluation nested too deeply: infinite recursion / options(expressions=)?\" for my tests. I would have thought R could do this all by itself. – brian d foy Apr 24 '10 at 0:54\n• Haha nice solution. – Frank Apr 24 '10 at 5:58\n\ncat f \| tr \"\\n\" \"+\" \| perl -pne chop \| R --vanilla --slave\n\nprotected by InianFeb 14 at 10:58\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count)." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8753501,"math_prob":0.7925725,"size":1041,"snap":"2019-26-2019-30","text_gpt3_token_len":316,"char_repetition_ratio":0.08100289,"word_repetition_ratio":0.049751244,"special_character_ratio":0.3506244,"punctuation_ratio":0.15517241,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9791008,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-19T20:10:46Z\",\"WARC-Record-ID\":\"<urn:uuid:74c85a87-d2ef-4d0c-b60d-298d3d001d05>\",\"Content-Length\":\"295345\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ab457767-3a9b-4ea0-9b40-a1e0208c3c07>\",\"WARC-Concurrent-To\":\"<urn:uuid:988e2c1b-5fb9-42b1-900c-390719f4fb97>\",\"WARC-IP-Address\":\"151.101.65.69\",\"WARC-Target-URI\":\"https://stackoverflow.com/questions/2702564/how-can-i-quickly-sum-all-numbers-in-a-file/13073486\",\"WARC-Payload-Digest\":\"sha1:DWTMZ73JPTH5DVRQYQJPKAA6IF6ZXK5N\",\"WARC-Block-Digest\":\"sha1:EU4UPZY5JYBD7L3R7BGFJHFVZDUXYIMS\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627999040.56_warc_CC-MAIN-20190619184037-20190619210037-00016.warc.gz\"}"}
https://www.electrondepot.com/fpga/questions-about-counter-in-vhdl-38087-.htm	[ "# Questions about counter in VHDL\n\n• posted\n\nI have implemented a 8-bit synchronous counter by VHDL. The result is that the LED display show continuously running the count from 0 to F(in Hex). Now, I need to change the result which the LED display can count from 0 to 9 only. How can I change in the VHDL code? Can anyone answer me? Thanks a lot!! The following VHDL code are about 8-bit synchronous counter: Library IEEE; USE IEEE.std_logic_1164.all; USE IEEE.std_logic_arith.all; USE IEEE.std_logic_unsigned.all;\n\nENTITY counter_eg IS PORT( PB1, clk_25MHz : IN std_logic; led0, led1, led2, led3, led4, led5, led6 : OUT std_logic); END counter_eg;\n\nARCHITECTURE c OF counter_eg IS\n\nCOMPONENT counter PORT( Clock, Reset : IN std_logic; count : OUT std_logic_vector(3 DOWNTO 0));\n\nEND COMPONENT;\n\nCOMPONENT dec_7seg PORT( hex_digit : IN std_logic_vector(3 DOWNTO 0); segment_a, segment_b, segment_c, segment_d, segment_e, segment_f, segment_g : OUT std_logic); END COMPONENT;\n\nCOMPONENT clk_div PORT( clock_25MHz : IN std_logic; clock_1MHz : OUT std_logic; clock_100KHz : OUT std_logic; clock_10KHz : OUT std_logic; clock_1KHz : OUT std_logic; clock_100Hz : OUT std_logic; clock_10Hz : OUT std_logic; clock_1Hz : OUT std_logic); END COMPONENT;\n\nSIGNAL count : std_logic_vector(3 DOWNTO 0); SIGNAL clk_1KHz : std_logic; BEGIN c0: clk_div port map (clock_25MHz=> clk_25MHz, clock_1KHz=>clk_1KHz); c1: counter port map (clock=>clk_1KHz, reset=>PB1, count=> count); c2: dec_7seg port map (count, led0, led1, led2, led3, led4, led5, led6); END c;\n\n• posted\n\nI'm not going to do your homework for you but here's a suggestion, how about decoding when the count equals 9 and using this to reset your counter. For your percentage I'm sure you can figure out how to code that in your vhdl.\n\nAlan\n\n• posted\n\nI'll give you a hint, the counter is not in this file, only the intantiation of the counter. but you can decode count=10 and trigger a reset. (in this file) as Alan just said.\n\nAurash\n\nThe result is\n\n• posted\n\nHow about gating the clock to something really fast when the count is 10 -\n\n15 inclusive. The A-F will whizz by so fast, no-one will notice. If you get that to work, you deserve extra marks. HTH, Syms.\n• posted\n\nAlan, I want to ask how to decode in vhdl? When the count is at 9, press reset button is go back to 0 again.\n\nAlan Myler =E5=AF=AB=E9=81=93=EF=BC=9A\n\nHz);\n\nt);\n\n• posted\n\ngating the clcok?? i don't understand!! Can you explain? Thanks!!\n\n• posted\n\nSymon, gating the clcok?? i don't understand!! Can you explain? Thanks!!\n\n• posted\n\n• posted\n\nThat would only appear to count 0..9, a smart marker might say it still shows A..F just that your eye cannot resolve it... I think the answer is to re-write the FONT ROM code, so it only displays 0..9 - This would even pass a cursory test. ;) - and it appears to fully meet the wording of the spec...\n\n-jg\n\n• posted\n\nhi, Aurelian Lazarut I still don't understand what you means. How to decode in VHDL? The following are my counter VHDL code: Library IEEE; USE IEEE.std_logic_1164.all; USE IEEE.std_logic_arith.all; USE IEEE.std_logic_unsigned.all;\n\nEntity Counter IS Port ( Clock, Reset :IN std_logic; count : OUT std_logic_vector(3 DOWNTO 0)); END Counter;\n\nARCHITECTURE a OF Counter IS SIGNAL internal_count: std_logic_vector(3 DOWNTO 0);\n\nBEGIN count intantiation of the counter.\n\nHz);\n\nt);\n\n• posted\n\nJim, Of course! That's an excellent solution. Yet another example of why it's unnecessary to gate clocks. :-) cheers mate, Syms.\n\n• posted\n\nJim, I don't understand what you means. what is Font Rom code? Laura\n\nSymon =E5=AF=AB=E9=81=93=EF=BC=9A\n\nys\n\n• posted\n\nIn your first post, you mentioned a LED display - I assumed 7 Segment as you said 0..F. This needs a lookup table, to convert/map the 4 bit binary, to required LEDs - that table is what I call the Font-Rom. It is missing from the examples you have posted thus far.\n\nHave a look at the ISE webpack, WATCHxx.zip examples : these are a 3 digit stopwatch, and they have all you will need.\n\n-jg\n\n• posted" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5802835,"math_prob":0.5401259,"size":1488,"snap":"2023-40-2023-50","text_gpt3_token_len":449,"char_repetition_ratio":0.21698113,"word_repetition_ratio":0.028301887,"special_character_ratio":0.3030914,"punctuation_ratio":0.2857143,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9824454,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-25T03:49:37Z\",\"WARC-Record-ID\":\"<urn:uuid:cedbb73c-35b7-40af-9ecb-b05b9908330d>\",\"Content-Length\":\"97056\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:704b000e-a54b-46bb-87a5-955c058de11d>\",\"WARC-Concurrent-To\":\"<urn:uuid:55c33bd5-8155-491f-ba09-8cfb7f2c73c7>\",\"WARC-IP-Address\":\"104.131.180.133\",\"WARC-Target-URI\":\"https://www.electrondepot.com/fpga/questions-about-counter-in-vhdl-38087-.htm\",\"WARC-Payload-Digest\":\"sha1:PYXOQNN5TT7FVIFUGAMJKXLPK3AA4UYX\",\"WARC-Block-Digest\":\"sha1:NV66TJLD6ZVMQKRDAEB7N7ROLMHDGQR7\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506676.95_warc_CC-MAIN-20230925015430-20230925045430-00752.warc.gz\"}"}
https://apboardsolutions.guru/ap-board-6th-class-maths-notes-chapter-6/	[ "Students can go through AP Board 6th Class Maths Notes Chapter 6 Basic Arithmetic to understand and remember the concepts easily.\n\n## AP State Board Syllabus 6th Class Maths Notes Chapter 6 Basic Arithmetic\n\n→ If a comparison is made by finding the difference between two quantities, it is called comparison by difference.\nEg: Age of Harshita is 11 years and age of Srija is 8 years. Harshita is (11 – 8 = 3) 3 years older than Srija or Srija is 3 years younger than Harshita.\n\n→ If a comparison is made by division it makes more sense than the comparison made by taking the difference.\nEg: If cost a key pad cell phone is Rs. 3000 and another smart phone is Rs. 15000, then the cost of the second phone is five times the cost of the first phone.", null, "→ Ratio: Comparison of two quantities of the same type by virtue of division is called ratio. Eg: The weight of Ramu is 24 kg and the weight of the Gopi is 36 kg., then the ratio of weights is 24/36. It can also be written as 24:36 and read as 24 is to 36.\nThe ratio of two numbers ‘a’ and ‘b’ (b ≠ 0) is a ÷ b or a/b or $$\\frac{a}{b}$$ and is denoted as a : b and is read as a is to b.\nIn the ratio a : b the quantities a and b are called the terms of the ratio.\nIn the ratio a : b the quantity a is called the first term or antecedent and b is called the second term or the consequent of the ratio.\nThe value of a fraction remains the same if the numerator and the denominators are multiplied or divided by the same non-zero number so is the ratio.\nThat is if the first term and the second term of a ratio are multiplied or divided by . the same non-zero number.\n3 : 4 = 3 × 5 : 4 × 5 = 15 : 20\nAlso 36 : 24 = 36 – 4 : 24 – 4 = 9 : 6.\n\n→ Ratio in the simplest form or in the lowest terms:\nA ratio a : b is said to be in its simplest form if its terms have no factors in common other than 1. A ratio in the simplest form is also called the ratio in its lowest terms. Generally ratios are expressed in their lowest terms.\nTo express a given ratio in its simplest term, we cancel H.C.F. from both its terms. To find the ratio of two terms, we express the both terms in the same units.\nEg: Ratio of 3 hours and 120 minutes is 3 : 2 (as 120 minutes = 2 hours) or 180 : 120 (as 3 hours = 180 minutes)\nA ratio has no units or it is independent of units used in the quantities compared. The order of terms in a ratio a : b is important a : b ≠ b : a.\n\n→ Equivalent ratio:\nA ratio obtained by multiplying or dividing the antecedent and consequent of a given ratio by the same number is called its equivalent ratio.\nEg: 3 : 4 = 3 × 5 : 4 × 5 = 15 : 20. Here 3 : 4 & 15 : 20 are called equivalent ratios.\nAlso 36 : 24 = 36 ÷ 4 : 24 ÷ 4 = 9 : 6. Here 36 : 24 & 9 : 6 are called equivalent ratios.", null, "→ Comparison of ratios: To compare two ratios\na) First express them as fractions\nb) Now convert them to like fractions\nc) Compare the like fractions\n\n→ Proportion:\nIf two ratios are equal, then the four terms of these ratios are said to be in proportion. If a : b = c : d, then a, b, c and d are said to be in proportion.\nThis is represented as a : b :: c : d and read as a is b is as c is d.\nThe equality of ratios is called proportion.\nConversely in the proportion a : b :: c : d , the terms a and d are called extremes and b and c are called means.\nIf four quantities are in proportion, then\nProduct of extremes = Product of means .\nIf a : b :: c : d, then a × d = b × c\nFrom this we have", null, "→ Unitary method:\nThe method in which first we find the value of one unit and then the value of required number of units is known as unitary method.\nEg: If the cost of 8 books Rs.96, then find the cost of 15 books.\nCost of one book = 96/8 = 12 Cost of 15 books = 12 × 15 = 180\nDistance travelled in a given time = speed × time From this we have", null, "", null, "→ Percentage:\nThe word per cent means for every hundred or out of hundred. The word percentage is derived from the Latin language. The % symbol is uses to represent percent.\nEg: 5% is read as five percent\n5% = $$\\frac{5}{100}$$ = 0.05\n38% = $$\\frac{38}{100}$$ = 0.38\n\n→ To convert a percentage into a fraction:\na) Drop the % symbol\nb) Divide the number by 100\nEg: 25% = $$\\frac{25}{100}$$ = 0.25 = $$\\frac{1}{4}$$\n\n→ To convert a fraction into percentage:\na) Assign the percentage symbol %\nb) Multiply the given fraction with 100\nEg: $$\\frac{3}{4}$$ = $$\\frac{3}{4}$$ × 100% = 75% = 0.75" ]	[ null, "https://apboardsolutions.guru/wp-content/uploads/2020/12/AP-Board-Solutions.png", null, "https://apboardsolutions.guru/wp-content/uploads/2020/12/AP-Board-Solutions.png", null, "https://apboardsolutions.guru/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Notes-Chapter-6-Basic-Arithmetic-1.png", null, "https://apboardsolutions.guru/wp-content/uploads/2021/04/AP-Board-6th-Class-Maths-Notes-Chapter-6-Basic-Arithmetic-2.png", null, "https://apboardsolutions.guru/wp-content/uploads/2020/12/AP-Board-Solutions.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.94214386,"math_prob":0.9988071,"size":4347,"snap":"2022-05-2022-21","text_gpt3_token_len":1248,"char_repetition_ratio":0.12502879,"word_repetition_ratio":0.06903991,"special_character_ratio":0.31930068,"punctuation_ratio":0.12372188,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99994934,"pos_list":[0,1,2,3,4,5,6,7,8,9,10],"im_url_duplicate_count":[null,null,null,null,null,2,null,2,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-19T11:31:43Z\",\"WARC-Record-ID\":\"<urn:uuid:e959c883-9a93-4b35-afa6-9137473730ee>\",\"Content-Length\":\"36758\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:054b8075-f411-4b87-a071-41014d055f7d>\",\"WARC-Concurrent-To\":\"<urn:uuid:563eb782-899d-4500-859b-d047e8005a66>\",\"WARC-IP-Address\":\"143.244.140.115\",\"WARC-Target-URI\":\"https://apboardsolutions.guru/ap-board-6th-class-maths-notes-chapter-6/\",\"WARC-Payload-Digest\":\"sha1:IHNTJNLUCYOMYXBTSW3646ZEGQHATDFG\",\"WARC-Block-Digest\":\"sha1:BZCZCD3DJEADQBCSF5P5D2WL7SVESQ4K\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662527626.15_warc_CC-MAIN-20220519105247-20220519135247-00193.warc.gz\"}"}
https://publications.iitm.ac.in/publication/student-course-allocation-with-constraints	[ "", null, "X\nStudent Course Allocation with Constraints\nPublished in Springer\n2019\nVolume: 11544 LNCS\n\nPages: 51 - 68\nAbstract\nReal-world matching scenarios, like the matching of students to courses in a university setting, involve complex downward-feasible constraints like credit limits, time-slot constraints for courses, basket constraints (say, at most one humanities elective for a student), in addition to the preferences of students over courses and vice versa, and class capacities. We model this problem as a many-to-many bipartite matching problem where both students and courses specify preferences over each other and students have a set of downward-feasible constraints. We propose an Iterative Algorithm Framework that uses a many-to-one matching algorithm and outputs a many-to-many matching that satisfies all the constraints. We prove that the output of such an algorithm is Pareto-optimal from the student-side if the many-to-one algorithm used is Pareto-optimal from the student side. For a given matching, we propose a new metric called the Mean Effective Average Rank (MEAR), which quantifies the goodness of allotment from the side of the students or the courses. We empirically evaluate two many-to-one matching algorithms with synthetic data modeled on real-world instances and present the evaluation of these two algorithms on different metrics including MEAR scores, matching size and number of unstable pairs. © 2019, Springer Nature Switzerland AG.\nJournal Data powered by TypesetLecture Notes in Computer Science (including subseries Lecture Notes in Artificial Intelligence and Lecture Notes in Bioinformatics) Data powered by TypesetSpringer 03029743 No\nAuthors (1)\n•", null, "Concepts (11)\n•", null, "Iterative methods\n•", null, "Pareto principle\n•", null, "Bipartite matching problems\n•", null, "COURSE ALLOCATION\n•", null, "Iterative algorithm\n•", null, "MANY TO MANY\n•", null, "MATCHING ALGORITHM\n•", null, "Pareto-optimal\n•", null, "Synthetic data\n•", null, "UNIVERSITY SETTINGS\n•", null, "Students" ]	[ null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/ico-hamburger.svg", null, "https://typeset-partner-institution.s3-us-west-2.amazonaws.com/iit-madras/authors/Meghana-Nasre.png", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null, "https://d5a9y5rnan99s.cloudfront.net/tds-static/images/Concept-default.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9018555,"math_prob":0.6960551,"size":1350,"snap":"2022-27-2022-33","text_gpt3_token_len":254,"char_repetition_ratio":0.14710252,"word_repetition_ratio":0.0,"special_character_ratio":0.17925926,"punctuation_ratio":0.07725322,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.967807,"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],"im_url_duplicate_count":[null,null,null,8,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\":\"2022-08-09T11:11:03Z\",\"WARC-Record-ID\":\"<urn:uuid:45cd2fda-d4fe-445d-9e30-37fab9b58801>\",\"Content-Length\":\"130075\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:02e62f00-2207-4ab7-980f-69c3d50c6e5e>\",\"WARC-Concurrent-To\":\"<urn:uuid:e21df782-e3ad-49f8-9424-27276701e030>\",\"WARC-IP-Address\":\"54.218.38.100\",\"WARC-Target-URI\":\"https://publications.iitm.ac.in/publication/student-course-allocation-with-constraints\",\"WARC-Payload-Digest\":\"sha1:2QE5PYUQRQA2PM2LXQIFHOEIT7MD2PUZ\",\"WARC-Block-Digest\":\"sha1:LQQOARZTY25W2A3WH2IX44XS5VV7YLEW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570921.9_warc_CC-MAIN-20220809094531-20220809124531-00411.warc.gz\"}"}
https://www.vackergroup.ae/our-products/heaters/electric-heater/	[ "# Electric heater\n\nThe Electric heaters use electric coils to produce heat. They have thermostats to control the required temperature. These electric heaters are for buildings, warehouses, tents, construction sites, accommodation camps etc.\n\n## Our models of Electric Heaters\n\nOur major models of Electric heaters are listed below:\n\n1. ### Specifications of Electric Heater Model VAC-TEH 20 T\n\n1. This electric heater has a heating capacity of 2.5 kW in single-stage control.\n2. The maximum capacity stage is 2,150 kCal.", null, "3. The capacity of this electric heater at the maximum stage is 2.5 kW.\n4. The maximum amount of air in the maximum stage is 300 m³/h.\n5. Air pressure in the maximum stage is 40 Pa.\n6. The maximum temperature increase for this electric heater is 40 °C.\n7. It has an axial fan with one stage.\n8. It operates in an Electrical Mains connection of 230 V/50 Hz.\n9. The nominal current consumption is 10.9 Amps.\n10. Recommended fusing is 16 Amps.\n11. The power input of this electric heater is 2.5 kW.\n12. Electric connection is through a connection plug type CEE 7/7.\n13. The cable length provided along with the heater is 2.6 meters.\n14. It has a sound of 63 dB(A) at a distance of 1 meter.\n15. It has a Hose connector & Hose connection\n16. The Diameter of the hose of this electric heater is 155 mm.\n17. The maximum length of the hose is 7 meters.\n18. The dimensions of this electric heater are 280 mm x 260 mm x 300mm.\n19. The weight of this electric heater is 10.6 kg without packing.\n20. It has an adjustable thermostat to control the heat.\n2. ### Specifications of Electric Heater Model VAC-TEH 30 T\n\n1. This electric heater has a heating capacity of 3.3 kW in single-stage control.\n2. The maximum capacity stage is 2,837 kCal.", null, "3. The capacity of this electric heater at the maximum stage is 3.3 kW.\n4. The maximum amount of air in the maximum stage is 300 m³/h.\n5. Air pressure in the maximum stage is 40 Pa.\n6. The maximum temperature increase for this electric heater is 54 °C.\n7. It has an axial fan with one stage.\n8. It operates in an Electrical Mains connection of 230 V/50 Hz.\n9. The nominal current consumption is 14.4 Amps.\n10. Recommended fusing is 16 Amps.\n11. The power input of this electric heater is 3.3 kW.\n12. Electric connection is through a connection plug type CEE 7/7.\n13. The cable length provided along with the heater is 2.6 meters.\n14. It has a sound of 63 dB(A) at a distance of 1 meter.\n15. It has a Hose connector & Hose connection\n16. The Diameter of the hose of this electric heater is 155 mm.\n17. The maximum length of the hose is 7 meters.\n18. The dimensions of this electric heater are 280 mm x 260 mm x 300mm.\n19. The weight of this electric heater is 10.6 kg without packing.\n20. It has an adjustable thermostat to control the heat.\n3. ### Specifications of Electric Heater Model VAC-TEH 70 T\n\n1. This electric heater has a Heating capacity of 6,9 & 12 in three-stage control.\n2. The maximum capacity stage is 10,300 kCal.", null, "3. The capacity of this electric heater at the maximum stage is 12 kW.\n4. The maximum amount of air in the maximum stage is 1,258m³/h.\n5. Air pressure in the maximum stage is 110 Pa.\n6. The maximum temperature increase for this electric heater is 35 °C.\n7. It has an axial fan with one stage.\n8. It operates in an Electrical Mains connection of 440 V/50 Hz.\n9. The nominal current consumption is 18 Amps.\n10. Recommended fusing is 32 Amps.\n11. The power input of this electric heater is 12 kW.\n12. Electric connection is through a connection plug type CEE 32A – 5 Pin.\n13. The cable length provided along with the heater is 2.6 meters.\n14. It has a sound of 66 dB(A) at a distance of 1 meter.\n15. It has a Hose connector & Hose connection\n16. The Diameter of the hose of this electric heater is 300 mm.\n17. The maximum length of the hose is 15 meters.\n18. The dimensions of this electric heater are 400 mm x 400 mm x 480mm.\n19. The weight of this electric heater is 29.5 kg without packing.\n20. It has an adjustable thermostat to control the heat.\n4. ### Specifications of Electric Heater Model VAC-TEH 100\n\n1. This electric heater has a Heating capacity of 9,13.5 & 18 in three-stage control.\n2. The maximum capacity stage is 15,400 kCal.", null, "3. The capacity of this electric heater at the maximum stage is 18 kW.\n4. The maximum amount of air in the maximum stage is 1,785 m³/h.\n5. Air pressure in the maximum stage is 130 Pa.\n6. The maximum temperature increase for this electric heater is 38 °C.\n7. It has an axial fan with one stage.\n8. It operates in an Electrical Mains connection of 440 V/50 Hz.\n9. The nominal current consumption is 27.2 Amps.\n10. Recommended fusing is 32 Amps.\n11. The power input of this electric heater is 18 kW.\n12. Electric connection is through a connection plug type CEE 32A – 5 Pin.\n13. The cable length provided along with the heater is 2.6 meters.\n14. It has a sound of 69 dB(A) at a distance of 1 meter.\n15. It has a Hose connector & Hose connection\n16. The Diameter of the hose of this electric heater is 300 mm.\n17. The maximum length of the hose is 15 meters.\n18. The dimensions of this electric heater are 400 mm x 400 mm x 480mm.\n19. The weight of this electric heater is 29.5 kg without packing.\n20. It has an adjustable thermostat to control the heat.", null, "#### Comparisons of different models of Electric Heaters\n\n Model Nos: VAC-TEH 20 T VAC-TEH 30 T VAC-TEH 70 VAC-TEH 100 Heating capacity Stage 1 [kW] 2.5 3.3 0 0 Stage 2 [kW] 6 9 Stage 3 [kW] 9 13.5 Stage 4 [kW] 12 18 Stage Max. [kcal] 2,150 2,837 10,300 15,400 Stage Max. [kW] 2.5 12 18 Amount of air Stage Max. [m³/h] 300 300 1,258 1,785 Air pressure Stage Max. [Pa] 40 40 110 130 Temperature increase in [°C] 40 54 35 38 Fans Axial Yes Yes Yes Yes Radial Number of fan stages 1 1 1 1 Electrical values Mains connection 230 V/50 Hz 230 V/50 Hz 400 V/50 Hz 400 V/50 Hz Nominal current consumption [A] 10.9 14.4 18 27.2 Recommended fusing [A] 16 16 32 32 Power input [kW] 2.5 3.3 12 18 Electric connection Connection plug CEE 7/7 CEE 7/7 CEE 32 A, 5-pin CEE 32 A, 5-pin Provided by the customer Cable length [m] 2.6 2.6 Sound values Distance 1 m [dB(A)] 63 63 66 69 Hose connector Hose connection Diameter [mm] 155 155 300 300 Max. hose length [m] 7 7 15 15 Dimensions Length (packaging excluded) [mm] 280 280 400 400 Width (packaging excluded) [mm] 260 260 400 400 Height (packaging excluded) [mm] 300 300 480 480 Weight (packaging excluded) [kg] 10.6 10.6 29.5 29.5 EQUIPMENT, FEATURES AND FUNCTIONS Operation Thermostat Image", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "## Electric Heaters by Master, Italy (Dantherm Group)\n\nMaster Climate Solutions (Italy) is part of Dantherm Group and is a group company of Dantherm. Their major models are listed below:\n\n1. ### Specification of Electric heater model VAC-B2PTC for poultry farm, warehouses\n\n1. It has a heating power of 1/2 kilowatts.\n2. (3,400-6,800 Btu per hour and 860-1,720 kcal per hour.)\n3. The air displacement is 97 m³ per hour.\n4. It works with 230 Volts and 50 Hertz power supply.\n5. The male plug from the heater side is 16A/3P.\n6. The rated current is 8.7 A.\n7. Switch Pos. 1 is 1 kilowatt and the switch pos. 2 is 2 kilowatt.\n8. It has a thermostat control.\n9. The temperature range is 0-40 °C.\n10. The electronic box protection is IP21.\n11. Box size (l x w x h) is 200 x 200 x 200 mm.\n12. With the net/gross weight of 1.9/2.1 kilograms.\n13. One pallet comes with 160 pcs.\n2. ### Specification of Electric heater model VAC-B 3PTC for warehouses, greenhouses, indoor farming.\n\n1. It has a heating power of 1.5/3 kilowatts.\n2. (3,400-10,200 Btu per hour and 860-2,580 kcal per hour.)\n3. The air displacement is 97 m³ per hour.\n4. It works with 230 Volts and 50 Hertz power supply.\n5. The male plug from the heater side is 16A/3P.\n6. The rated current is 13 A.\n7. Switch Pos. 1 is off and the switch pos. 2 is 1.5 kilowatt.\n8. Switch Pos. 3/4 is 3 kilowatts.\n9. It has a thermostat control.\n10. The temperature range is 0-40 °C.\n11. The electronic box protection is IP21.\n12. Box size (l x w x h) is 244 x 240 x 250 mm.\n13. With the net/gross weight of 3.4/3.7 kilograms.\n14. One pallet comes with 160 pcs.\n3. ### Specification of Electric heater model VAC-B2 for poultry farm, livestock, dairy farm.\n\n1. It has a heating power of 1/2 kilowatts.\n2. (3,400-6,800 Btu per hour and 860-1,720 kcal per hour.)\n3. The air displacement is 184 m³ per hour.\n4. It works with 230 Volts and 50-60 Hertz power supply.\n5. The male plug from the heater side is 16A/3P.\n6. The rated current is 8.7 A.\n7. Switch Pos. 1 is off and the switch pos. 2 is a fan.\n8. Switch Pos. 3 is 1.0/2.0 kilowatts.\n9. It has a thermostat control.\n10. The temperature range is 5-35 °C.\n11. The electronic box protection is IP24.\n12. Product size (l x w x h) is 200 x 200 x 330 mm.\n13. Box size (l x w x h) is 235 x 210 x 340 mm.\n14. With the net/gross weight of 3.7/4.2 kilograms.\n15. One pallet comes with 75 pcs.\n4. ### Specification of Electric heater model VAC-B3 for construction sites\n\n1. It has a heating power of 1.65/3.3 kilowatts.\n2. (5,630-11,260 Btu per hour and 1,430-2,860 kcal per hour.)\n3. The air displacement is 510 m³ per hour.\n4. It works with 230 Volts and 50-60 Hertz power supply.\n5. The male plug from the heater side is 16A/3P.\n6. The rated current is 14.5 A.\n7. Switch Pos. 1 is off and the switch pos. 2 is a fan.\n8. Switch Pos. 3 is 1.65/3.3 kilowatts.\n9. It has a thermostat control.\n10. The temperature range is 5-35 °C.\n11. The electronic box protection is IP24.\n12. Product size (l x w x h) is 260 x 260 x 410 mm.\n13. Box size (l x w x h) is 280 x 270 x 440 mm.\n14. With the net/gross weight of 5.1/5.7 kilograms.\n15. One pallet comes with 48 pcs.\n5. ### Specification of Electric heater model VAC-B5 for construction sites\n\n1. It has a heating power of 2.5/5 kilowatts.\n2. (8,530-17,000 Btu per hour and 2,150-4,300 kcal per hour.)\n3. The air displacement is 510 m³ per hour.\n4. It works with 3 phase 400 Volts and 50 Hertz power supply.\n5. The male plug from the heater side is 16A/5P.\n6. The rated current is 7.2 A.\n7. Switch Pos. 1 is off and the switch pos. 2 is a fan.\n8. Switch Pos. 3 is 2.5/5.0 kilowatts.\n9. It has a thermostat control.\n10. The temperature range is 5-35 °C.\n11. The electronic box protection is IP24.\n12. Product size (l x w x h) is 310 x 360 x 380 mm.\n13. Box size (l x w x h) is 380 x 330 x 440 mm.\n14. With a net/gross weight of 6.4/6.8 kilograms.\n15. One pallet comes with 24 pcs.\n6. ### Specification of Electric heater model VAC-B9 for warehouses, poultry farms, dairy farms, livestock\n\n1. It has a heating power of 4.5/9 kilowatts.\n2. (15,350-30,700 Btu per hour and 3,870-7,740 kcal per hour.)\n3. The air displacement is 800 m³ per hour.\n4. It works with 3 phase 400 Volts and 50 Hertz power supply.\n5. The male plug from the heater side is 16A/5P.\n6. The rated current is 13 A.\n7. Switch Pos. 1 is off and the switch pos. 2 is a fan.\n8. Switch Pos. 3 is 4.5/9.0 kilowatts.\n9. It has a thermostat control.\n10. The temperature range is 5-35 °C.\n11. The electronic box protection is IP24.\n12. Product size (l x w x h) is 340 x 420 x 440 mm.\n13. Box size (l x w x h) is 355 x 440 x 490 mm.\n14. With the net/gross weight of 9.3/10.8 kilograms.\n15. One pallet comes with 20 pcs.\n7. ### Specification of Electric heater model VAC-B15\n\n1. It has a heating power of 7.5/15 kilowatts.\n2. (25,600-51,200 Btu per hour and 6,450-12,900 kcal per hour.)\n3. The air displacement is 1700 m³ per hour.\n4. It works with 3 phase 400 Volts and 50 Hertz power supply.\n5. The male plug from the heater side is 32A/5P.\n6. The rated current is 22 A.\n7. Switch Pos. 1 is off and the switch pos. 2 is a fan.\n8. Switch Pos. 3 is 7.5/15 kilowatts.\n9. It has a thermostat control.\n10. The temperature range is 5-35 °C.\n11. The electronic box protection is IP24.\n12. Product size (l x w x h) is 350 x 470 x 490 mm.\n13. Box size (l x w x h) is 370 x 480 x 530 mm.\n14. With a net/gross weight of 15/15.9 kilograms.\n15. One pallet comes with 12 pcs.", null, "#### Product Description of Electrical Heater\n\n1. Brief Title of the device: Electric heaters for buildings, warehouses, tents, construction sites etc.\n2. Brief Description of the device: Electric heaters for increasing ambient temperature using Electric coils for Industrial and Commercial applications. Useful for buildings, warehouses, tents, construction sites etc.\n3. Model number: VAC-TEH70\n4. Brand: VackerGlobal\n5. Seller: VackerGlobal\n6. SKU Number: 1003000075\n7. Price (AED): 4550.00\n8. Price Validity: 30 June 2023" ]	[ null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/uploads/2018/10/Electric-heater-UAE-Saudi-Qatar-Oman-Kuwait-Africa-VACTEH20T-e1538930049883-150x150.jpg", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/uploads/2018/10/Electric-heater-UAE-Saudi-Qatar-Oman-Kuwait-Africa-VACTEH30T-150x150.jpg", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/uploads/2018/10/Electric-heater-UAE-Saudi-Qatar-Oman-Kuwait-Africa-VACTEH70-e1538929856552-150x150.jpg", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/uploads/2018/10/Electric-heater-UAE-Saudi-Qatar-Oman-Kuwait-Africa-VACTEH100-e1538930003740-150x150.jpg", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null, "https://www.vackergroup.ae/wp-content/plugins/a3-lazy-load/assets/images/lazy_placeholder.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7734102,"math_prob":0.98456925,"size":11459,"snap":"2023-40-2023-50","text_gpt3_token_len":3281,"char_repetition_ratio":0.18315147,"word_repetition_ratio":0.6577273,"special_character_ratio":0.318876,"punctuation_ratio":0.14110187,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95569795,"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],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,5,null,null,null,5,null,null,null,5,null,null,null,5,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T00:28:25Z\",\"WARC-Record-ID\":\"<urn:uuid:17f4cccd-c00a-4426-a954-3d574b8478d6>\",\"Content-Length\":\"119928\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4a221f6e-366a-4952-8318-250e2f7fe0d0>\",\"WARC-Concurrent-To\":\"<urn:uuid:ebff6dc9-f203-42fa-bbb7-70dc4d868884>\",\"WARC-IP-Address\":\"139.59.85.15\",\"WARC-Target-URI\":\"https://www.vackergroup.ae/our-products/heaters/electric-heater/\",\"WARC-Payload-Digest\":\"sha1:MALNOFG4GIB2XULPNILYWVRRN76T2VHZ\",\"WARC-Block-Digest\":\"sha1:KBUNNJLJQAZGXAICQPYD3G3TISKKACTE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511023.76_warc_CC-MAIN-20231002232712-20231003022712-00440.warc.gz\"}"}
http://colgatephys111.blogspot.com/2016/11/this-week-when-i-was-laying-around-at.html	[ "Friday, November 25, 2016\n\nPhysics in New Guinness World Record\n\nPhysics in New Guinness World Record\n\nThis week when I was laying around at home I was scrolling my Facebook feed and came across a video of a new Guinness World Record that was recently set. A man from the United Kingdom completed the highest bungee dunk. After watching the video, the first thing that came to mind was the amount of physics that was involved in order to precisely land this jump.\n\nThe physics principles that are involved are conservation of mechanical energy and springs.\nIt says that the total jump was 73.41 m, the man weighs approximately 95 kg, and the spring coefficient of the bungee cord is 60 N/m and the unstretched length is 20 m.\n\nWe can figure out the equilibrium length if the man is just hanging on it:\n\nmg/k = (95 kg)(9.8 m/s)/60 N/m = 15.5 m\n\nWe can also figure out what his fastest speed will be:\n\nmgho = 1/2 mv^2 + mghm + 1/2 kx^2\n\n(95 kg)(9.8m /s^2 )(73.41m) = 1/2 (95 kg) v^2 +(95kg)(9.8m /s^2 )(73.41m − 20m − 15.5m )+ 1/2(60N/m)(15.5)^2\n\nv= 23.3 m/s or 52.2 mph\n\nWe can also figure out how far from the ground he will be:\n\nmgho = 1/2 kx^2 + mghm\n\n(95kg)(9.8m/s^2 )(73.41m) = 1/2(60 N/m)(x)^2 +(95kg)(9.8m/s^2 )(73.41m − 20m − x)\n0=17.5x^2-931x-18620\nx=44.9 m + 20 =64.9 m --> 8.51 m off the ground\n\nHis Max acceleration:\nFmax = mamax = −kx − mg = −(60N ×(−44.9m)) −(95kg× 9.8m/s^2 ) = 1763 N\nF = ma → (95kg)a = 1763N → a = 18.6 m/s^2\nor 3.8 g's" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91645974,"math_prob":0.9928617,"size":1499,"snap":"2019-43-2019-47","text_gpt3_token_len":552,"char_repetition_ratio":0.098996654,"word_repetition_ratio":0.014440433,"special_character_ratio":0.39626417,"punctuation_ratio":0.113513514,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99252695,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-15T02:24:51Z\",\"WARC-Record-ID\":\"<urn:uuid:2641b762-3442-45c3-82a0-88d46eca06a6>\",\"Content-Length\":\"63830\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:cecd30d1-96bb-476e-a7f8-1c29506bba6e>\",\"WARC-Concurrent-To\":\"<urn:uuid:b47c2531-652f-4edb-a377-91ec046a5a82>\",\"WARC-IP-Address\":\"172.217.7.225\",\"WARC-Target-URI\":\"http://colgatephys111.blogspot.com/2016/11/this-week-when-i-was-laying-around-at.html\",\"WARC-Payload-Digest\":\"sha1:64RWAWMEJMCIWBHCHMOMDXEY4RWV2CEQ\",\"WARC-Block-Digest\":\"sha1:SQ4V3OIWOLXXFRH2XT63CV4K2VP2GXWT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986655735.13_warc_CC-MAIN-20191015005905-20191015033405-00390.warc.gz\"}"}
https://barpsi.com/9.6-bar-to-psi	[ "", null, "# 9.6 Bar to Psi\n\n## Convert 9.6 Bar to Psi. 9.6 Bar to Psi conversion.\n\n9.6\n\n=\n\n139.236192\n\nLooking to find what is 9.6 Bar in Psi? Want to convert 9.6 Bar units to Psi units?\n\nUsing a simple formula, 9.6 Bar units are equal to 139.236192 Psi units.\n\nWant to convert 9.6 Bar into other Bar units?\n\nBar, Psi, Bar to Psi, Bar in Psi, 9.6 Bar to Psi, 9.6 Bar in Psi" ]	[ null, "https://barpsi.com/assets/google-play-bar-to-psi.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7706198,"math_prob":0.6628042,"size":284,"snap":"2019-43-2019-47","text_gpt3_token_len":98,"char_repetition_ratio":0.23214285,"word_repetition_ratio":0.036363635,"special_character_ratio":0.38732395,"punctuation_ratio":0.22093023,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9745503,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-15T17:37:28Z\",\"WARC-Record-ID\":\"<urn:uuid:9ce9c9b9-a60b-4ece-b932-189096e5f75b>\",\"Content-Length\":\"24202\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2924dff2-4836-4a34-af52-ccc5a2840184>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea4f3a06-41aa-4a25-813b-27c22aacd82b>\",\"WARC-IP-Address\":\"142.93.178.22\",\"WARC-Target-URI\":\"https://barpsi.com/9.6-bar-to-psi\",\"WARC-Payload-Digest\":\"sha1:J6TVGGI4J6JBVGOHZDQPV5NIPZBHLLC3\",\"WARC-Block-Digest\":\"sha1:DX42HGJESFDEJVVNL7325T6ZRPEFLMVE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668699.77_warc_CC-MAIN-20191115171915-20191115195915-00483.warc.gz\"}"}
https://www.tweag.io/blog/2020-01-16-data-vs-control/	[ "Haskell’s Data and Control module hierarchies have always bugged me. They feel arbitrary. There’s Data.Functor and Control.Monad—why? Monads are, after all, functors. They should belong to the same hierarchy!\n\nI’m not that person anymore. Now, I understand that the intuition behind the Data/Control separation is rooted in a deep technical justification. But—you rightly insist—monads are still functors! So what’s happening here? Well, the truth is that there are two different kinds of functors. But you could never tell them apart because they coincide in regular Haskell.\n\nBut they are different—so let’s split them into two kinds: data functors and control functors. We can use linear-types to show why they are different. Let’s get started.\n\n## Data functors\n\nIf you haven’t read about linear types, you may want to check out Tweag’s other posts on the topic. Notwithstanding, here’s a quick summary: linear types introduce a new type a ⊸ b of linear functions. A linear function is a function that, roughly, uses its argument exactly once.\n\nWith that in mind, let’s consider a prototypical functor: lists.\n\ninstance Functor [] where\nfmap :: (a -> b) -> [a] -> [b]\nfmap f [] = []\nfmap f (a:l) = (f a) : (fmap f l)\n\nHow could we give it a linear type?\n\n• Surely, it’s ok to take a linear function as an argument (if fmap works on any function, it will work on functions which happen to be linear).\n• The f function is, on the other hand, not used linearly: it’s used once per element of a list (of which there can be many!). So the second arrow must be a regular arrow.\n• However, we are calling f on each element of the list exactly once. So it makes sense to make the rightmost arrow linear—exactly once.\n\nSo we get the following alternative type for list’s fmap:\n\nfmap :: (a ⊸ b) -> [a] ⊸ [b]\n\nList is a functor because it is a container of data. It is a data functor.\n\nclass Data.Functor f where\nfmap :: (a ⊸ b) -> f a ⊸ f b\n\nSome data functors can be extended to applicatives:\n\nclass Data.Applicative f where\npure :: a -> f a\n(<>) :: f (a ⊸ b) ⊸ f a ⊸ f b\n\nThat means that containers of type f a can be zipped together. It also constrains the type of pure: I typically need more than one occurrence of my element to make a container that can be zipped with something else. Therefore pure can’t be linear.\n\nAs an example, vectors of size 2 are data applicatives:\n\ndata V2 a = V2 a a\n\ninstance Data.Functor f where\nfmap f (V2 x y) = V2 (f x) (f y)\n\ninstance Data.Applicative f where\npure x = V2 x x\n(V2 f g) <> (V2 x y) = V2 (f x) (g y)\n\nLists would almost work, too, but there is no linear way to zip together two lists of different sizes. Note: such an instance would correspond to the Applicative instance of ZipList in base, the Applicative instance for [], in base is definitely not linear (left as an exercise to the reader).\n\n## Control functors\n\nThe story takes an interesting turn when considering monads. There is only one reasonable type for a linear monadic bind:\n\n(>>=) :: m a ⊸ (a ⊸ m b) ⊸ m b\n\nAny other choice of linearization and you will either get no linear values at all (if the continuation is given type a -> m b), or you can’t use linear values anywhere (if the two other arrows are non-linear). In short: if you want the do-notation to work, you need monads to have this precise type.\n\nNow, you may remember that, from (>>=) alone, it is possible to derive fmap:\n\nfmap :: (a ⊸ b) ⊸ m a ⊸ m b\nfmap f x = x >>= (\\a -> return (f a))\n\nBut wait! Something happened here: all the arrows are linear! We’ve just discovered a new kind of functor! Rather than containing data, we see them as wrapping a result value with an effect. They are control functors.\n\nclass Control.Functor m where\nfmap :: (a ⊸ b) ⊸ m a ⊸ m b\n\nclass Control.Applicative m where\npure :: a ⊸ m a -- notice how pure is linear, but Data.pure wasn't\n(<*>) :: m (a ⊸ b) ⊸ m a ⊸ m b\n\n(>>=) :: (>>=) :: m a ⊸ (a ⊸ m b) ⊸ m b\n\nLists are not one of these. Why? Because you cannot map over a list with a single use of the function! (Neither is Maybe because you may drop the function altogether, which is not permitted either.)\n\nThe prototypical example of a control functor is linear State\n\nnewtype State s a = State (s ⊸ (s, a))\n\ninstance Control.Functor (State s) where\nfmap f (State act) = \\s -> on2 f (act s)\nwhere\non2 :: (a ⊸ b) ⊸ (s, a) ⊸ (s, b)\non2 g (s, a) = (s, g b)\n\n## Conclusion\n\nThere you have it. There indeed are two kinds of functors: data and control.\n\n• Data functors are containers: they contain many values; some are data applicatives that let you zip containers together.\n• Control functors contain a single value and are all about effects; some are monads that the do-notation can chain.\n\nThat is all you need to know. Really.\n\nBut if you want to delve deeper, follow me to the next section because there is, actually, a solid mathematical foundation behind it all. It involves a branch of category theory called enriched category theory.\n\nEither way, I hope you enjoyed the post and learned lots. Thanks for reading!\n\n## Appendix: The maths behind it all\n\nBriefly, in a category, you have a collection of objects and sets of morphisms between them. Then, the game of category theory is to replace sets in some part of mathematics, by objects in some category. For example, one can substitute “set” in the definition of group by topological space (topological group) or by smooth manifold (Lie group).\n\nEnriched category theory is about playing this game on the definition of categories itself: a category enriched in $\\mathcal{C}$ has a collection of objects and objects-of-$\\mathcal{C}$ of morphisms between them.\n\nFor instance, we can consider categories enriched in abelian groups: between each pair of objects there is an abelian group of morphisms. In particular, there is at least one morphism, 0, between each pair of objects. The category of vector spaces over a given field (and, more generally, of modules over a given ring) is enriched in abelian groups. Categories enriched in abelian groups are relevant, for instance, to homology theory.\n\nThere is a theorem that all symmetric monoidal closed categories (of which the category of abelian groups is an example) are enriched in themselves. Therefore, the category of abelian groups itself is another example of a category enriched in abelian groups. Crucially for us, the category of types and linear functions is also symmetric monoidal closed. Hence is enriched in itself!\n\nFunctors can either respect this enrichment (in which case we say that they are enriched functors) or not. In the category Hask (seen as a proxy for the category of sets), this theorem is just saying that all functors are enriched because “Set-enriched functor” means the same as “regular functor”. That’s why Haskell without linear types doesn’t need a separate enriched functor type class.\n\nIn the category of abelian groups, the functor which maps $A$ to $A\\otimes A$ is an example of a functor which is not enriched: the map from $A → B$ to $A\\otimes A → B\\otimes B$, which maps $f$ to $f\\otimes f$ is not a group morphism. But the functor from $A$ to $A\\oplus A$ is.\n\nControl functors are the enriched functors of the category of linear functions, while data functors are the regular functors.\n\nHere’s the last bit of insight: why isn’t there a Data.Monad? The mathematical notion of a monad does apply perfectly well to data functors—it just wouldn’t be especially useful in Haskell. We need the monad to be strong for things like the do-notation to work correctly. But, as it happens, a strong functor is the same as an enriched functor, so data monads aren’t strong. Except in Hask, of course, where data monads and control monads, being the same, are, in particular, strong." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8972539,"math_prob":0.9499492,"size":7599,"snap":"2020-45-2020-50","text_gpt3_token_len":1942,"char_repetition_ratio":0.13285056,"word_repetition_ratio":0.043010753,"special_character_ratio":0.24121594,"punctuation_ratio":0.12987843,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.996313,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-21T13:29:05Z\",\"WARC-Record-ID\":\"<urn:uuid:605aa472-0f8b-413f-b8c4-3e30319442d8>\",\"Content-Length\":\"283397\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:908daec3-4a9b-4d86-8631-153c77161cd8>\",\"WARC-Concurrent-To\":\"<urn:uuid:ef2f778a-3869-406b-8234-4bf0cc37b163>\",\"WARC-IP-Address\":\"104.31.69.163\",\"WARC-Target-URI\":\"https://www.tweag.io/blog/2020-01-16-data-vs-control/\",\"WARC-Payload-Digest\":\"sha1:Q37KFFT2SQXQMCWKMJQES43OZVMKMACE\",\"WARC-Block-Digest\":\"sha1:EG6EJ3VEVBJ5RSLENBHAQY336TOFHXLE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107876500.43_warc_CC-MAIN-20201021122208-20201021152208-00699.warc.gz\"}"}
https://www.hackmath.net/en/example/6871	[ "# Proportion 2\n\nA car is able to travel 210 km in 3 hours. How far can it travel in 5 hours?\nPut what kind of proportion is this and show your solution.\n\nResult\n\nx = 350 km\n\n#### Solution:", null, "Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):\n\nShowing 0 comments:", null, "Be the first to comment!", null, "#### To solve this example are needed these knowledge from mathematics:\n\nNeed help calculate sum, simplify or multiply fractions? Try our fraction calculator. Do you want to convert length units? Most natural application of trigonometry and trigonometric functions is a calculation of the triangles. Common and less common calculations of different types of triangles offers our triangle calculator. Word trigonometry comes from Greek and literally means triangle calculation.\n\n## Next similar examples:\n\n1. Candies", null, "There are red, blue and green candies in bad. Red to green is in 6:11 ratio and blue to red in a 7: 5 ratio. In what proportion are blue to green candies?\n2. Good swimmer", null, "Good swimmer swims 23 m distance with ten shots. With how many shots he swim to an island located 81 m if still swims at the same speed?\n3. Land area", null, "A land area of Asia and Africa are in a 3: 2 ratio, the European and African are is 1:3. What are the proportions of Asia, Africa, and Europe?\n4. Masons", null, "1 mason casts 30.8 meters square in 8 hours. How long casts 4 masons 178 meters square?\n5. Medians 2:1", null, "Median to side b (tb) in triangle ABC is 12 cm long. a. What is the distance of the center of gravity T from the vertex B? b, Find the distance between T and the side b.\n6. Three monks", null, "Three medieval monks has task to copy 600 pages of the Bible. One rewrites in three days 1 page, second in 2 days 3 pages and a third in 4 days 2 sides. Calculate for how many days and what day the monks will have copied whole Bible when they begin Wednesd\n7. Scale of the map", null, "Determine the scale of the map if the actual distance between A and B is 720 km and distance on the map is 20 cm.\n8. Liters od milk", null, "The cylinder-shaped container contains 80 liters of milk. Milk level is 45 cm. How much milk will in the container, if level raise to height 72 cm?\n9. Simple equation 6", null, "Solve equation with one variable: X/2+X/3+X/4=X+4\n10. Poplar", null, "How tall is a poplar by the river, if we know that 1/5 of its total height is a trunk, 1/10th of the height is the root and 35m from the trunk to the top of the poplar?\n11. Seamstress", null, "The seamstress cut the fabric into 3 parts. The first part was the eighth fabric, the second part was three-fifths of the fabric and the third part had a length of 66 cm. Calculate the original length of the fabric.\n12. Equation with x", null, "Solve the following equation: 2x- (8x + 1) - (x + 2) / 5 = 9\n13. Inquality", null, "Solve inequality: 3x + 6 > 14\n14. Wiring 2", null, "Willie cut a piece of wire that was 3/8 of the total length of the wire. He cut another piece that was 9m long. The two pieces together were one half of the total length of the wire. How long was the wire before he cut it?\n15. Tailor", null, "Tailor bought 2 3/4 meters of textile and paid 638 CZK. Determine the price per 1 m of the textile.\n16. UN 1", null, "If we add to an unknown number his quarter, we get 210. Identify unknown number.\n17. Lengths of the pool", null, "Miguel swam 6 lengths of the pool. Mat swam 3 times as far as Miguel. Lionel swam 1/3 as far as Miguel. How many lengths did mat swim?" ]	[ null, "https://www.hackmath.net/tex/2fa/2fae670ab34de.png", null, "https://www.hackmath.net/hashover/images/first-comment.png", null, "https://www.hackmath.net/hashover/images/avatar.png", null, "https://www.hackmath.net/thumb/56/t_4656.jpg", null, "https://www.hackmath.net/thumb/30/t_3630.jpg", null, "https://www.hackmath.net/thumb/97/t_2697.jpg", null, "https://www.hackmath.net/thumb/96/t_1896.jpg", null, "https://www.hackmath.net/thumb/9/t_8209.jpg", null, "https://www.hackmath.net/thumb/66/t_2466.jpg", null, "https://www.hackmath.net/thumb/27/t_2427.jpg", null, "https://www.hackmath.net/thumb/9/t_1809.jpg", null, "https://www.hackmath.net/thumb/94/t_7594.jpg", null, "https://www.hackmath.net/thumb/46/t_7946.jpg", null, "https://www.hackmath.net/thumb/36/t_5636.jpg", null, "https://www.hackmath.net/thumb/64/t_3564.jpg", null, "https://www.hackmath.net/thumb/5/t_7605.jpg", null, "https://www.hackmath.net/thumb/7/t_6807.jpg", null, "https://www.hackmath.net/thumb/57/t_2257.jpg", null, "https://www.hackmath.net/thumb/40/t_3640.jpg", null, "https://www.hackmath.net/thumb/94/t_6994.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9370961,"math_prob":0.98291874,"size":2956,"snap":"2019-13-2019-22","text_gpt3_token_len":827,"char_repetition_ratio":0.104674794,"word_repetition_ratio":0.03697479,"special_character_ratio":0.27401894,"punctuation_ratio":0.1040724,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9854818,"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],"im_url_duplicate_count":[null,1,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,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-19T17:25:36Z\",\"WARC-Record-ID\":\"<urn:uuid:bcba2d5f-1e2c-450f-95f6-9d8116b02045>\",\"Content-Length\":\"20044\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1162d5fe-2aef-44f9-8f7f-6a6f615b7708>\",\"WARC-Concurrent-To\":\"<urn:uuid:0b7d07cc-41a9-4af2-967f-edc763166af9>\",\"WARC-IP-Address\":\"104.24.105.91\",\"WARC-Target-URI\":\"https://www.hackmath.net/en/example/6871\",\"WARC-Payload-Digest\":\"sha1:UNEIR5FKI5WXBYJQVLMQCOZLJMSSDWY2\",\"WARC-Block-Digest\":\"sha1:PCJGH2BSLQPTTTK2434CZJ742EDWNSLH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232255071.27_warc_CC-MAIN-20190519161546-20190519183546-00487.warc.gz\"}"}
http://mathandmultimedia.com/tag/decimal-number-system/	[ "## How to Change Number Bases Part 2\n\nIn the previous post, we have learned how to change numbers form one base to other. In this post, we are going to discuss more examples of number bases particularly the two number systems used in computers: the binary and the hexadecimal system.\n\nThe Binary Number System\n\nThe binary number system has base 2 and only uses 1 and 0 as digits. The binary number 1101 in expanded form is", null, "$1 \\times 2^3 + 1 \\times 2^2 + 0 \\times 2^1 + 1 \\times 2^0$ or » Read more\n\n## How to Change Number Bases Part 1\n\nI have already discussed clock arithmetic, modulo division, and number bases. We further our discussion in this post by learning how to change numbers from one base to another.\n\nThe number system that we are using everyday is called the decimal number system or the base 10 number system (deci means 10). It is believed that this system was developed because we have 10 fingers.\n\nIn the base 10 system, the digits are composed of 0 up to 9. Adding 1 to 9, the largest digit in this system, will give us 10. That is, we replace 9 in the ones place with 0, and add 1 to the tens place which is the next larger place value.\n\nAnother way to write a number in base 10 is by multiplying its digits by powers of 10 and adding them. For example, the number 2578 can be rewritten in expanded form as", null, "$2(10^3) + 5(10^2) + 7(10^1) + 8(10^0)$» Read more\n\n## What If We Have 12 Fingers?\n\nOur number system is called the decimal system (deci means 10) because we count in groups of 10’s. This is probably because we have 10 fingers. What do I mean when I said when we count in groups of 10?", null, "Our number system has the digits 0 to 9, and then we when we reach the 10th number, we place 1 in the tens place 0 in the ones digit. In the decimal number system, 23 means that we have 2 tens and 3 ones. Similarly, the number 452 means that we have 4 groups hundreds (10 tens), 5 groups of tens and 8 ones. In fact, if we use the expanded notation, 452 is equal to", null, "$4 \\times 10^2 + 5 \\times 10^1 + 2 \\times 10^0$.\n\nNotice that each number is multiplied by powers of 10. » Read more" ]	[ null, "http://s0.wp.com/latex.php", null, "http://s0.wp.com/latex.php", null, "http://upload.wikimedia.org/wikipedia/commons/thumb/3/32/Human-Hands-Front-Back.jpg/320px-Human-Hands-Front-Back.jpg", null, "http://s0.wp.com/latex.php", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92682403,"math_prob":0.9968373,"size":1836,"snap":"2022-05-2022-21","text_gpt3_token_len":449,"char_repetition_ratio":0.15829694,"word_repetition_ratio":0.011235955,"special_character_ratio":0.25599128,"punctuation_ratio":0.0964467,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99759984,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-26T10:20:37Z\",\"WARC-Record-ID\":\"<urn:uuid:5b08300d-91ed-48ee-9aeb-44582ee20a94>\",\"Content-Length\":\"38658\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c37fdb37-becd-442e-a1e1-2e3e781b1204>\",\"WARC-Concurrent-To\":\"<urn:uuid:2e9df73c-753e-4f08-b7eb-df594d31c6a5>\",\"WARC-IP-Address\":\"166.62.28.131\",\"WARC-Target-URI\":\"http://mathandmultimedia.com/tag/decimal-number-system/\",\"WARC-Payload-Digest\":\"sha1:SOCG2D2VCA55QJAZIQO5XJNN4JPNO5DE\",\"WARC-Block-Digest\":\"sha1:NBAJBIABGDMGTITO7TYVRNHMMZTLI3GC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662604794.68_warc_CC-MAIN-20220526100301-20220526130301-00000.warc.gz\"}"}
https://mathematica.stackexchange.com/questions/265260/rectangular-image-segmentation/268469	[ "# Rectangular Image Segmentation?\n\nI'd like to segment an image such that the components are rectangles. Is this possible out of the box? For example:\n\nimg = CloudGet[CloudObject[\"https://www.wolframcloud.com/obj/1e937fa7-80d2-4db2-8e47-80fe376c0e8f\"]]\nColorize @ WatershedComponents[ ColorConvert[img, \"Grayscale\"] ]", null, "You can imagine the output being something like the above but with rectangles instead of polygons, the fewer the better, could look something like this (a poor approximation drawn using Canvas[]):", null, "A follow-up would be to specify thresholds and control min and max size/aspects of the rectangles.\n\nUpdate\n\nHere are a few additional examples to test solutions:\n\nmoreTests = CloudGet[\"https://www.wolframcloud.com/obj/\"<>#]&/@\n{\"04b63372-f4cb-4c4f-9161-a8ca581b01fa\",\"68b066cb-9775-44de-aba4-717449293713\",\n\"cb413bbb-56ac-40b3-92dd-2029c0c40b2e\",\"3f4b3af3-ce6e-4512-b917-17b0efa80fc9\",\n\n• The principal singular vectors make an interesting first approximation, but I don't where to go from here: {uu, ss, vv} = SingularValueDecomposition[ImageData@RemoveAlphaChannel@ColorConvert[img0, \"Grayscale\"]]; Colorize@ WatershedComponents[Image[uu[[All, {1}]] . Transpose@vv[[All, {1}]]], Method -> {\"MinimumSaliency\", 0.07}]. Needs something like a follow-up routine to combine some adjacent rectangles. But image-processing is not my thing. Mar 19 at 3:07\n• It looks like a superpixels problem, I haven't seen a square superpixel algorithm yet, a similar effect is: epfl.ch/labs/ivrl/research/slic-superpixels May 17 at 2:35\n• Do you have any guidance on the criteria you want for your segmentation? As stated, the problem is too unconstrained to really come up with a meaningful solution. Let's say your image is a rectangle, split along the diagonal such that on one side of the diagonal it's solid red and the other side of the diagonal is solid blue. What would you want this algorithm to do with that? May 17 at 15:17\n\nModification 2\n\nI have significantly improved the code to get even closer to what M.R. was looking for. Rectangles now grow in width and height, reducing the number of narrow slices. Also, they grow from the four corners of the image. The function has two options:\n\n1. \"RectCount\" allows you to specify the number of rectangles drawn, so you can get a partial result superimposed on the original image.\n2. \"Collage\" (true or false) indicates if a collage of the image covered with rectangles with the original image is desired (default is false).\n(* ===== The function ======= )\ngrowRects[img_, OptionsPattern[]] :=\nModule[{image, pixelBlockSize, colorDiffLimit, subimages,\nsubColorList, pixelBlockCoordinates, pixelBlockCoordinatesLinear,\nblocksInRect, initialBlockCoordinate, initialBlockColor, maxWidth,\nmaxHeight, lastColumn, lastRow, newColumn,\ninitialBlockCoordinateLinear, linearIndex, available, colorMatched,\nwithinBounds, newColumnAllowed, newRowAllowed, rectsList, lb,\nnewRow, rectColor, t, rt, initialBlockCoordinateLinear2,\ninitialBlockCoordinateLinear3, initialBlockCoordinateLinear4,\nlinearIndex2, linearIndex3, linearIndex4},\n\n( Users may modify the following to see the effect )\n\npixelBlockSize = 10;\ncolorDiffLimit = 0.15;\n\n( Create block size images and a corresponding list of mean color \\\nfor each subimage )\nsubimages = ImagePartition[img, pixelBlockSize];\nsubColorList =\nTable[\nImageMeasurements[subimages[[i]], \"Mean\"], {i, 1,\nLength[subimages]}];\n\n( Reassemble the image -\nimage partition may have introduced some cropping )\n\nimage = ImageAssemble[subimages];\n\n( get the coordinates in terms of pixel blocks -\nwe will grow rectangles by adding rows or columns of blocks )\n(\nwe need four versions to facilitate dealing with four coordinate \\\nsystems (one to draw from each corner) )\npixelBlockCoordinates = {\nTable[{i, j}, {i, 1, Length[subimages]}, {j, 1,\nLength[subimages[]]}],\nTable[{i, j}, {i, Length[subimages], 1, -1}, {j, 1,\nLength[subimages[]]}],\nTable[{i, j}, {i, 1, Length[subimages]}, {j,\nLength[subimages[]], 1, -1}],\nTable[{i, j}, {i, Length[subimages], 1, -1}, {j,\nLength[subimages[]], 1, -1}]\n};\n\n( The equivalent linear list of coordinates for each of the four \\\nsystems )\n\npixelBlockCoordinatesLinear =\nFlatten[#, 1] & /@ pixelBlockCoordinates;\n\n( area to fill in pixelBlocksSize )\n{maxWidth, maxHeight} = Dimensions[subimages];\nareaPoints = maxWidthmaxHeight;\n\n(* number of blocks filled )\nblocksInRect = 0;\n\n( Using a rectangle counter )\n\nIf[IntegerQ[count = OptionValue[\"RectCount\"]], useCounter = True;\ncounter = 1, useCounter = False];\n\n( refers to our four modes of filling rectangles from each corner \\\n- we will rotate later)\nseedArray = {1, 2, 3, 4};\n\n( Start the rectangle loop -\neach rectangle starts at the first point of \\\npixelBlockCoordinatesLinear that is not 0 )\n\nWhile[areaPoints > blocksInRect,\nIf[useCounter && (counter++ > count), Break[]];\n\n( Each mode has different directions to grow rectangles )\n\ns = First[seedArray];\nWhich[\n];\nseedArray = RotateLeft[seedArray];\n\n( color assigned to the next rectangle )\n\nrectColor = RandomColor[];\nim = Image[\nRandomChoice[{rectColor}, {pixelBlockSize, pixelBlockSize}]];\n\n( the algorithm ensure that the first non zero element of each \\\nlist of coordinates is the one we want )\ninitialBlockCoordinate =\nFirstCase[pixelBlockCoordinatesLinear[[s]], Except];\n\n( find the linear position in each list of coordinate \\\ncorresponding to our starting point )\n\ninitialBlockCoordinateLinear =\nmaxHeight(initialBlockCoordinate[] -\n1) + (initialBlockCoordinate[]);\ninitialBlockCoordinateLinear2 =\nmaxHeight(maxWidth -\ninitialBlockCoordinate[]) + (initialBlockCoordinate[]);\ninitialBlockCoordinateLinear3 =\nmaxHeight(initialBlockCoordinate[] - 1) + {maxHeight -\ninitialBlockCoordinate[] + 1};\ninitialBlockCoordinateLinear4 =\nmaxHeight(maxWidth - initialBlockCoordinate[]) + {maxHeight -\ninitialBlockCoordinate[] + 1};\n\n( remove used blocks from availability in all list )\n\npixelBlockCoordinatesLinear[][[initialBlockCoordinateLinear]] =\n0;\npixelBlockCoordinatesLinear[][[initialBlockCoordinateLinear2]] =\n0;\npixelBlockCoordinatesLinear[][[initialBlockCoordinateLinear3]] =\n0;\npixelBlockCoordinatesLinear[][[initialBlockCoordinateLinear4]] =\n0;\n\n( color the initial block with the random color )\n\nsubimages[[initialBlockCoordinate[]]][[\ninitialBlockCoordinate[]]] = im;\n\n( this is the color of the subimage at this location )\n\ninitialBlockColor =\nsubColorList[[initialBlockCoordinate[]]][[\\\ninitialBlockCoordinate[]]];\n\n( the first block in our rectangle )\n\nlastColumn = {initialBlockCoordinate};\nlastRow = {initialBlockCoordinate};\nblocksInRect = blocksInRect + 1;\n\n( Now we can alternate between adding rows and columns )\nnewColumnAllowed = True;\nnewRowAllowed = True;\n\n( Start the rectangle growing loop )\n\nWhile[newColumnAllowed \|\| newRowAllowed,\n\n( Attempt to add a column - extend the X coordinate -\ntest validity (available blocks, color) )\nIf[newColumnAllowed,\n( create new column from existing last column )\n\nnewColumn = {#[] + cadd, #[]} & /@ lastColumn;\n\n( is the new column within bounds )\n\nwithinBounds = AllTrue[newColumn, 1 <= #[] <= maxWidth &];\nIf[withinBounds,\n( get position ou the points in our four list of coordinates )\n\nlinearIndex = maxHeight(#[] - 1) + (#[]) & /@ newColumn;\nlinearIndex2 =\nmaxHeight(maxWidth - #[]) + (#[]) & /@ newColumn;\nlinearIndex3 =\nmaxHeight(#[] - 1) + {maxHeight - #[] + 1} & /@\nnewColumn;\nlinearIndex4 =\nmaxHeight(maxWidth - #[]) + {maxHeight - #[] + 1} & /@\nnewColumn;\n\n( verifiy that coordinates have not been used before )\n\navailable =\nNoneTrue[\npixelBlockCoordinatesLinear[][[#]] & /@ linearIndex,\nIntegerQ];\nIf[available,\n( check color )\n\ncolorMatched =\nAllTrue[subColorList[[#[], #[]]] & /@ newColumn,\nMax@Abs[Take[#, 3] - Take[initialBlockColor, 3]] <\ncolorDiffLimit &];\nIf[colorMatched,\n( keep the new column,\nreplacing the last column and adding a point to last row )\n\nblocksInRect = blocksInRect + Length[newColumn];\nlastColumn = newColumn;\nlastRow = AppendTo[lastRow, Last[lastRow] + {cadd, 0}];\n\n(\nMark column coordinates as not available in all four lists by \\\nsetting coordinate to 0 )\n(pixelBlockCoordinatesLinear[[\n1]][[#]] = 0) & /@ linearIndex;\n(pixelBlockCoordinatesLinear[][[#]] = 0) & /@\nlinearIndex2;\n(pixelBlockCoordinatesLinear[][[#]] = 0) & /@\nlinearIndex3;\n(pixelBlockCoordinatesLinear[][[#]] = 0) & /@\nlinearIndex4;\n\n( convert new column to full size colored blocks )\n\nFor[i = 1, i <= Length[newColumn], i++,\n\nsubimages[[newColumn[[i]][]]][[newColumn[[i]][]]] =\nim;\n]\n\n, newColumnAllowed = False ( not color matched )]\n, newColumnAllowed = False ( not available )]\n, newColumnAllowed = False ( not within bounds )]\n];\n\n( Attempt to add a row - extend the Y coordinate -\ntest validity (available blocks, color) )\nIf[newRowAllowed,\n( create new row from existing last row )\n\nnewRow = {#[], #[] + radd} & /@ lastRow;\n\n( is the new row within bounds )\n\nwithinBounds = AllTrue[newRow, 1 <= #[] <= maxHeight &];\nIf[withinBounds ,\n( get position ou the points in our four list of coordinates )\n\nlinearIndex = maxHeight(#[] - 1) + (#[]) & /@ newRow;\nlinearIndex2 =\nmaxHeight(maxWidth - #[]) + (#[]) & /@ newRow;\nlinearIndex3 =\nmaxHeight(#[] - 1) + {maxHeight - #[] + 1} & /@ newRow;\nlinearIndex4 =\nmaxHeight(maxWidth - #[]) + {maxHeight - #[] + 1} & /@\nnewRow;\n\n( verifiy that coordinates have not been used before )\n\navailable =\nNoneTrue[\npixelBlockCoordinatesLinear[][[#]] & /@ linearIndex,\nIntegerQ];\nIf[available,\n( check color )\n\ncolorMatched =\nAllTrue[subColorList[[#[], #[]]] & /@ newRow,\nMax@Abs[Take[#, 3] - Take[initialBlockColor, 3]] <\ncolorDiffLimit &];\nIf[colorMatched,\n( keep the new row,\nreplacing the last row and adding a point to last column )\n\nblocksInRect = blocksInRect + Length[newRow];\nlastRow = newRow;\nlastColumn =\n\n(\nMark row coordinates as not available in all four lists by \\\nsetting coordinate to 0 )\n(pixelBlockCoordinatesLinear[[\n1]][[#]] = 0) & /@ linearIndex;\n(pixelBlockCoordinatesLinear[][[#]] = 0) & /@\nlinearIndex2;\n(pixelBlockCoordinatesLinear[][[#]] = 0) & /@\nlinearIndex3;\n(pixelBlockCoordinatesLinear[][[#]] = 0) & /@\nlinearIndex4;\n\n( convert new row to full size colored blocks )\n\nFor[i = 1, i <= Length[newRow], i++,\nsubimages[[newRow[[i]][]]][[newRow[[i]][]]] = im;\n];\n, newRowAllowed = False ( not color matched )]\n, newRowAllowed = False ( not available )]\n, newRowAllowed = False ( not within bounds )]\n]\n];\n\n];\nassembly = ImageAssemble[subimages];\nIf[OptionValue[\"Collage\"], ImageCollage[{image, assembly}], assembly]\n]\nOptions[growRects] = {\"RectCount\" -> Infinity, \"Collage\" -> False};\n\n\nHere are function calls with examples of results:\n\ngrowRects[img1]\ngrowRects[img1,\"RectCount\"->10]\ngrowRects[img1,\"Collage\"->True]\n... and a few more collages", null, "", null, "", null, "", null, "", null, "", null, "Modification (original post, now replaced by the above)\n\nIn the various examples provided, some have 4 channels, some 3 channels. The code now tests for this, so all the examples work. Also, the variable subsize was set at 16 pixels before, but the optimal value depends on the size of the file. It is now set to 1/100 of the width of the image, but you can play with it to optimize a particular image. Unfortunately, this code produces narrow rectangles, which is not exactly what was required.\n\n( Get the image )\n\nimg = CloudGet[\nCloudObject[\n\"https://www.wolframcloud.com/obj/1e937fa7-80d2-4db2-8e47-\\\n80fe376c0e8f\"]];\n\n( Partition the image )\nsubsize = 1/100ImageDimensions[img][];\nsubimages = ImagePartition[img, subsize];\n\n(* Measure average color of subimages )\n\nt = Table[\nImageMeasurements[subimages[[i]], \"Mean\"], {i, 1,\nLength[subimages], 1}];\n\n( Get positions where color change greater than 0.1 in one direction \\\n)\n\nchannels = ImageMeasurements[img, \"Channels\"];\nWhich[channels == 4, ch1 = {0., 0., 0., _}; ch2 = {_, _, _, _},\nchannels == 3, ch1 = {0., 0., 0.}; ch2 = {_, _, _}];\n\npv = Reverse[\nPosition[\nTable[\nThreshold[Differences[t[[i]]], {\"Hard\", 0.1}] /. ch1 -> \" \" /.\nch2 -> \"\|\", {i, 1, Length[t], 1}], \"\|\"], 2];\n\n( Get positions where color change greater than 0.1 in the other \\\ndirection )\n\ntrt = Transpose[t];\nph = Position[\nTable[\nThreshold[Differences[trt[[i]]], {\"Hard\", 0.1}] /. ch1 -> \" \" /.\nch2 -> \"_\", {i, 1, Length[trt], 1}], \"_\"];\n\n( Find intersection )\ninter = Intersection[pv, ph];\n\n( Convert the list to image coordinates and add missing points \\\n(where y=0) )\n\ninterS = {#[], ImageDimensions[img][] - #[]} & /@ (subsize\ninter);\nfinal = {};\nAppendTo[final, {#[], 0}] & /@ interS;\np = SortBy[DeleteDuplicates@Join[interS, final], {First, Greater}];\n\n(* Function to process list p of coordinates and draw rectangles. h \\\nis the height )\n\nrectangleSegments[h_, p_List] :=\nModule[{rects, anchor, prevCol, prevRow},\nrects = {};\nanchor = {0, h};\nprevCol = 0;\nprevRow = 0;\n\ni = 1;\nWhile[i <= Length[p],\nIf[\np[[i]][] != prevCol,\nanchor = {prevCol, h},\nanchor = {anchor[], prevRow};\n];\nAppendTo[rects, {anchor, p[[i]]}];\nprevCol = p[[i]][];\nprevRow = p[[i]][];\ni++;\n];\nrects\n]\n\n( Calling function *)\n\nrseg = rectangleSegments[ImageDimensions[img][], p];\nShow[Graphics[{EdgeForm[Black], FaceForm[RandomColor[]],\nRectangle[#[], #[]]}] & /@ rseg]", null, "• I like this approach, if you can make it do the right thing for the additional few test images I posted above, I will accept!\n– M.R.\nMay 20 at 21:23\n• In the six examples I added, whenever there is a big clear space (constant or perhaps a smooth gradient) the algo should produce a larger rectangle that \"fills that space\", not just lots of tiny ones...\n– M.R.\nMay 20 at 21:27\n• Also, the big yellow rect on the left in your example should not have three short ones below it (because the image is all white down to the bottom edge)\n– M.R.\nMay 20 at 21:32\n• I made some modifications (see main window), but I am afraid that the problem filling big clear spaces is not solved. May 20 at 22:59\n• Have you considered using something like ColorQuantize or DominantColors in preprocessing May 20 at 23:07" ]	[ null, "https://i.stack.imgur.com/mF4i8.png", null, "https://i.stack.imgur.com/7bxKx.png", null, "https://i.stack.imgur.com/pQb08.jpg", null, "https://i.stack.imgur.com/l451n.jpg", null, "https://i.stack.imgur.com/ODsic.jpg", null, "https://i.stack.imgur.com/MRwOP.jpg", null, "https://i.stack.imgur.com/6y8DE.jpg", null, "https://i.stack.imgur.com/2ddYi.jpg", null, "https://i.stack.imgur.com/4kTak.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.57389724,"math_prob":0.9776522,"size":11891,"snap":"2022-27-2022-33","text_gpt3_token_len":3326,"char_repetition_ratio":0.18103811,"word_repetition_ratio":0.22093023,"special_character_ratio":0.32301742,"punctuation_ratio":0.20586608,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9751273,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-07-03T08:08:38Z\",\"WARC-Record-ID\":\"<urn:uuid:b2e09082-8824-46df-ac87-62e30e28df00>\",\"Content-Length\":\"249389\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6317342e-a7fb-4cff-9eab-b2a9ed6c9a21>\",\"WARC-Concurrent-To\":\"<urn:uuid:08385d08-4f5b-4509-b199-be5b41240cd1>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://mathematica.stackexchange.com/questions/265260/rectangular-image-segmentation/268469\",\"WARC-Payload-Digest\":\"sha1:X7AXFN7XBBUWZLEN2DZPBJSYZZGJKINM\",\"WARC-Block-Digest\":\"sha1:AFWCDYM2UFWCYXZM7YB5SHC6V6EGZAHU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656104215805.66_warc_CC-MAIN-20220703073750-20220703103750-00232.warc.gz\"}"}
https://5dok.net/document/yr38rpko-evolution-how-long-does-speciation-take.html	[ "# evolution: how long does speciation take?\n\nN/A\nN/A\nProtected\n\nShare \"evolution: how long does speciation take?\"\n\nCopied!\n79\n0\n0\nLaat meer zien ( pagina)\n\nHele tekst\n\n(1)\n\nfaculteit Wiskunde en Natuurwetenschappen\n\n## evolution: how long does speciation take?\n\n### Bachelor thesis Mathematics\n\nJuly 2014\n\nStudent: Gerard Hekkelman\n\nFirst Supervisor Mathematics: Prof. Dr. E.C. Wit Second Supervisor Mathematics : Dr. W.P. Krijnen\n\nSupervisor Centre for Evological and Evolutionary Studies: Prof. Dr. R.S.\n\nEtienne\n\n(2)\n\nAbstract\n\nIn this paper, we study the constant birth-death process and the protracted birth-death process which describes macro-evolution in biology. We derive some properties of these models, and an- alytically derive the likelihood of the constant birth-death model. For the protracted birth-death model, we derive an appromation of the likelihood using the concept of Gaussian processes. Using the exact likelihood of the constant birth-death process and the approximated likelihood of the protracted birth-death model, we infer a given phylogenetic tree and discuss the results.\n\n(3)\n\n### Contents\n\nIntroduction 4\n\n1 Birth-death models 6\n\n1.1 Pure Birth Model . . . 6\n\n1.2 Pure Death model . . . 7\n\n1.3 Birth-death model . . . 8\n\n1.4 Protracted Pure Birth Model . . . 9\n\n1.5 Protracted Birth-Death Model . . . 10\n\n2 Inference 11 2.1 Inferences of Constant Birth-Death Models . . . 11\n\n2.2 Inference of the Yule Process . . . 12\n\n2.2.1 Confidence intervals for λ . . . 14\n\n2.3 Likelihood of the birth-death process . . . 17\n\n2.3.1 Probability functions of simulated trees. . . 18\n\n2.3.2 Probability Density Function of a Simulated Tree . . . 21\n\n2.3.3 Maximum Likelihood Estimation . . . 22\n\n2.4 Approximating the Likelihood of the Protracted Birth-Death Model . . . 23\n\n2.4.1 Approximation of the Process . . . 23\n\n2.4.2 Approximation of the likelihood . . . 26\n\n3 Simulation studies 27 3.1 Simulation of a Birth-Death Model . . . 27\n\n3.1.1 Basic Properties . . . 27\n\n3.1.2 Results of simulation . . . 32\n\n3.1.3 Inferences . . . 33\n\n3.2 Simulation of a Protracted Birth-Death Model . . . 40\n\n3.2.1 Basic Properties . . . 40\n\n3.2.2 Approximation of the Process . . . 44\n\n3.2.3 Inferences . . . 44\n\n4 Conclusion and Discussion 50 4.1 Conclusion . . . 50\n\n4.2 Discussion and Impossible Improvements . . . 51\n\nA R codes 54 A.1 Review: Other Models and Biological Properties . . . 54\n\nA.1.1 Moran Process . . . 54\n\nA.1.2 Random Walk . . . 54\n\nA.1.3 Shapes of clades . . . 55\n\nA.1.4 Sampling and paraphyly . . . 55\n\nA.1.5 Diversification models . . . 56\n\n(4)\n\nA.1.6 Diversification slowdowns . . . 58 A.2\n\nBirth-Death Simulation Functions . . . 60 A.3\n\nCode for Constant Birth-Death Simulations . . . 62 A.4\n\nCode for Protracted Birth-Death Simulations . . . 72\n\n(5)\n\n### Introduction\n\nA phylogenetic tree (that is, the ancestry, predigree, genealogy) of a group of species represents the evolutionary relatedness between the species, and their common ancestry, for example between human and chimpansee. Currently, DNA data are routinely used to infer the phylogenetic tree.\n\nAn example of an actual phylogenetic tree is given in figure 1.\n\nFigure 1: An example of a phylogenetic tree.\n\nIn this example, we can see that the group of virusses is represented in a tree form. Every branch represents a species, and each node represents a common ancestor for two species in the tree. For example, ”virus2” and ”virus5” have one common ancestor, which is ”virus1”. For these kind of phylogenies, we are interested in estimating the parameters which have their influence on the evolutionary process. However, the phylogenetic trees used as data only contain the genetic information of extant species. By this, we don’t have information of the extant species which existed in the past, but are extinct today. Consider figure 1 again, the data we observe is given in figure 2.\n\nThese trees are called reconstructed phylogenies. We infere these phylogenies to obtain the parameters which cause the growth of a phylogeny. Sophisticated software, often using Bayesian approaches, has been developed for this. One component of this software is the probability of the phylogenetic tree given a model of species diversification (speciation and extinction). There is a lot of debate about the proper model of species diversification.\n\nOne example is a model which assumes that speciation and extinction are instantaneous events,\n\n(6)\n\nFigure 2: An example of a phylogenetic tree.\n\nwith occur with a constant rate through time. This model is called the constant birth-death model.\n\nAnother example is a model that assumes that speciation takes time rather than being instanta- neous (which all other models assume). This model, called the protracted birth-death model, allows us to estimate how long a speciation process takes to complete. However, the current math- ematical representation of the protracted speciation model seems incoherent.\n\nIn this paper we first represent the constant birth-death model. We give the properties of the constant birth-death model and make simulation studies. Thereafter, we make inferences of this reconstructed phylogeny and obtain a maximum likelihood estimate of the diversification rate. Sec- ondly, we introduce the protracted model and search for an approximation of the representation of the protracted model and a way to infer its parameters from a given phylogenetic tree.\n\n(7)\n\n### Birth-death models\n\nThe birth-death models are mathematical models of the dynamical process of speciation and ex- tinction, which are used for informative thinking about macroevolutionary patterns. The pure birth-model and the pure death model are submodels of the birth-death models. We can simply model the growth of a clade through speciation to deduce the rate of speciation, discovering any irregularities. Also less obvious questions can be answered using the birth-death models.\n\nThe use of the birth-death models started back in 1924 to the work of the statistician Yule (Yule 1924), who modelled a clade growing according the pure birth process. Here, extinction does not occur.\n\nThe modern usage of the birth-death models started with the work of Raup and colleagues in 1973 (Raup et al 1973), when computers were starting to become a readily used tool (Nee 2006).\n\nThey modelled a scenario where the probabilities of a species is either speciating or coming to ex- tinct are equal, and the clade size was roughly constant. They discovered that this purely random process could produce trends and patterns resembling those in the fossil records.\n\nWe first investigate several models of macro-evolution, where the constant birth-death model and the protracted birth-death model are the most important ones in this paper.\n\n### 1.1 Pure Birth Model\n\nIn this model, we assume that each species in a clade has a constant rate λ of producing a new species at any point in time and that extinction never occurs. We assume that the duration of speciation Ti is exponentially distributed for each species in the clade:\n\nTb(i) ∼ exp(λ), ∀i = 1, 2, . . . (1.1) Let Ng denote the number of species in the clade, we are interested in the probability density function of the number of species at a time t. The probability that at time t there are Ng species is given by the following stochastic differential equation (Yule 1924):\n\nd\n\ndtPr(Ng; t) = λ(Ng− 1) Pr(Ng− 1; t) − λNgPr(Ng; t) (1.2) This equation is known as the master equation of the pure birth model, or Yule process named after its discoverer George Udny Yule. This equation will be solved analytically later in this paper.\n\nInstead of looking at the probability density function, we can also focus on the expected number of species in the clade at a time t:\n\n(8)\n\nFigure 1.1: George Udny Yule (1871-1951) .\n\nLet E [N (t)] denote the expected number of species in the clade at time t and set E [N0] = E [N (0)] as the expected initial number of species. The expected number of species E [N (t)] grows exponentially over time with a rate λ, hence we have the ODE:\n\nd\n\ndtE [N (t)] = λE [N (t)] , E [N (0)] = E [N0] (1.3) The solution is straightforward:\n\nE [N (t)] = E [N0] eλt (1.4)\n\nAs a consequence, a plot of log N (t) against the time should be linear and the slope of this plot provides an estimate of b, the per-capita birth rate.\n\n### 1.2 Pure Death model\n\nIn this model, we assume that each species in a clade has a constant rate µ of going extinct in time t and that speciation never occurs. We assume that the duration of extinction Ti is exponentially distributed for each species in the clade:\n\nTd(i) ∼ exp(µ), ∀i = 1, 2, . . . (1.5) Let Ng denote the number of species, we are interested in the probability density function of the number of species at a time t. The probability that at time t there are Ngspecies is similiar to the probability density function given in equation 1.2, and it is given implicitely by the following stochastic differential equation:\n\nd\n\ndtPr(Ng; t) = µ(Ng+ 1) Pr(Ng+ 1; t) − µNgPr(Ng; t) (1.6)\n\n(9)\n\nWith initial condition:\n\nPr(Ng= Ng(0); t = 0) = 1 (1.7)\n\nThe equation 1.6 is the master equation of the pure death model. Again, this equation can be solved analytically. First, we look at the expected number of species in the clade at a time t:\n\nLet E [N (t)] denote the expected number of species in the clade at time t and set E [N0] = E [N (0)] as the expected initial number of species. The expected number of species that survive N (t) decays exponentially over time with a rate µ, hence we have the following ODE:\n\nd\n\ndtE [N (t)] = −µE [N (t)] , E [N (0)] = E [N0] (1.8) The solution is straightforward:\n\nE [N (t)] = E [N0] e−µt (1.9)\n\n### 1.3 Birth-death model\n\nIn this model, we assume that each species in a clade speciate with a constant rate λ and that extinction occurs with a constant rate µ in time. We assume that the duration of speciation Tb\n\nand the duration of extinction Td are exponetially distributed for each species:\n\nTb(i) ∼ exp(λ), ∀i = 1, 2, . . . (1.10) Td(i) ∼ exp(µ), ∀i = 1, 2, . . . (1.11) Let Ng denote the number of species, we are interested in the probability density function of the number of species at a time t. The probability that at time t there are Ng species is obtained by Kendall in 1948 (Kendall 1948):\n\nd\n\ndtPr(Ng; t) = λ(Ng− 1) Pr(Ng− 1; t) + µ(Ng+ 1) Pr(Ng+ 1; t)\n\n− (λ + µ)NgPr(Ng; t) (1.12)\n\nWith initial condition:\n\nPr(Ng= Ng(0); t = 0) = 1 (1.13)\n\nThis equation will be solved analytically later in this paper. Again, we look first to the expected number of species in the clade at a time t. We define the net rate of diversification as r := λ − µ.\n\nThe clades grow exponentially at a rate r, we therefore have the following ODE:\n\nd\n\ndtE [N (t)] = (λ − µ)E [N (t)] , E [N (0)] = E [N0] (1.14) The solution is straight forward:\n\nE [N (t)] = E [N0] e(λ−µ)t (1.15)\n\nThe semilog representation of the growth of a molecular phylogeny was expected to be linear with a constant slope r, however when the extinction rate is nonzero it is not. But, over much over the history the plot is expected to be linear with slope r (Nee 2006). Molecular phylogenies are completely based on data of extant species, and species that originated more recently in the past had less time to go extinct. Therefore, the extinction rate µ gets smaller as we approach the present. So as we approach the present, the slope of the semilog presentation is expected to\n\n(10)\n\nFigure 1.2: David George Kendall (1918-2007) .\n\nincrease and assymptotically approach λ. By this, we can estimate λ and µ seperately on intuition (Nee 2006). Notice that in molecular phylogenies, we can only see the history of extant species.\n\nSpecies which have become extinct are not represented in these phylogenies.\n\nWe have two surprises from the birth-death models (Nee 2006):\n\n• We can estimate speciation and extinction rates from molecular phylogenies, even though they do not contain information from extant species.\n\n• We can estimate per-species speciation and extinction rates even from fossil data that are not resolved to a level below that of the genus.\n\n### 1.4 Protracted Pure Birth Model\n\nIn this model we make speciation a protracted process. That is, that each species in a clade pro- duce new species with a rate λ1, but these species are not fully completed. These incipient species become good species with a rate λ2, and give rise to new incipient species by rate λ3. This means that speciation in the protracted birth model takes one extra step with respect to the regular birth model.\n\nTo be more precise: in this model, we assume that each species in a clade speciate with a constant rate λ1 and that speciation is completed by a constant rate λ2. We assume that the duration of speciation Tb and the duration of completion Tc are exponetially distributed for each species:\n\nTb(i) ∼ exp(λ1), ∀i = 1, 2, . . . (1.16) Tc(i) ∼ exp(λ2), ∀i = 1, 2, . . . (1.17)\n\n(11)\n\nLet Ng denote the number of good species (the completed species), and let Ni denote the incipient species (the incomplete species). Now, we are interested in the probability density function of the number of good species Ngand incipient species Niat a time t. The probability that at time t there are Ng good species and Ni incipient species is implicitely obtained (Etienne, Rosindell 2012 ):\n\nd\n\ndtPr(Ng, Ni; t) = λ1NgPr(Ng, Ni− 1; t) + λ3(Ni− 1) Pr(Ng, Ni− 1; t)\n\n+ λ2(Ni+ 1) Pr(Ng− 1, Ni+ 1; t) − (λ1Ng+ (λ2+ λ3)Ni) Pr(Ng, Ni; t) (1.18) With initial condition:\n\nPr(Ng= Ng(0), Ni= 0; t = 0) = 1 (1.19) This model cannot be solved analytically, so we investigate the expected number of good and incipient species first. We obtain the following system of ODE’s:\n\nd dt\n\n\u0014E[Ng; t]\n\nE[Ni; t]\n\n\u0015\n\n=\u0014 0 λ2\n\nλ1 λ3− λ2\n\n\u0015 \u0014E[Ng; t]\n\nE[Ni; t]\n\n\u0015\n\n(1.20) With initial condition:\n\n\u0014E[Ng; t = 0]\n\nE[Ni; t = 0]\n\n\u0015\n\n=\u0014Ng(0) 0\n\n\u0015\n\n(1.21) The general solution of this system is given in (Etienne and Rosindell 2012). The case that λ1 = λ3 is much easier to solve, the solution of this system in the case is straight forward the solution of a linear system ODE:\n\nE[Ng; t] = Ng(0) 1 +λλ1\n\n2\n\nexp(λ1t) + Ng(0) 1 +λλ2\n\n1\n\nexp(−λ2t) (1.22)\n\nE[Ni; t] = Ng(0) 1 +λλ2\n\n1\n\n(exp(λ1t) − exp(λ2t)) (1.23)\n\n### 1.5 Protracted Birth-Death Model\n\nIf we make speciation in the constant birth-death model protracted, analogously to the protracted birth model, we assume that species give birth to incipient species by a rate λ1, incipient species reach completion by a rate λ2, give birth to new incipient species by a rate λ3 and extinction occurs for both species with a rate µ.1\n\nd\n\ndtPr(Ng, Ni; t) = λ1NgPr(Ng, Ni− 1; t) + λ1(Ni− 1) Pr(Ng, Ni− 1; t)\n\n+ λ2(Ni+ 1) Pr(Ng− 1, Ni+ 1; t) + µ(Ng+ 1) Pr(Ng+ 1, Ni; t) + µ(Ni+ 1) Pr(Ng, Ni+ 1; t)\n\n((λ1+ µ)Ng+ (λ1+ λ2+ µ)Ni) Pr(Ng, Ni; t) (1.24) In this paper, we focus on obtaining an approximation of the the probability density function implicitely given in equation 1.24.\n\nThere are many other models describing evolutionary processes, However, they lie beyond our scope in this paper. A review of these models, and some biological properties are given in appendix A.1.\n\n1Note that in the paper of Etienne and Rosindell, they assume that extinction rates may differ for good species with rate µ1and incipient species with rate µ2. Here we assume µ = µ1= µ2.\n\n(12)\n\n### 2.1 Inferences of Constant Birth-Death Models\n\nMolecular systematics produces phylogenies that may have a temporal dimension, thus containing information about the tempo of the clade’s evolution as well as the relationships among taxa (Nee 2001). We are particulary interested in extracting this information. We are able to study the rates of diversification in the clade. Bladwin and Sanderson (1998) used the simple Yule process to study the rate of diversification . The Yule Process is a fairly simple process, but we can use nice statistical approaches for obtaining results under this model.\n\nFor inference, we first must distinguish between actual and reconstructed phylogenies. We note four points (Nee et al. 1994):\n\n• Both phylogenies have the same number of taxa at the present day\n\n• At any point in the past, the number of lineages in the reconstructed phylogeny is less or equal than the number of lineages of the actual phylogeny.\n\n• The number of lineages in the reconstructed phylogeny cannot decrease towards the present, this can happen in the actual phylogeny.\n\n• The reconstructed phylogeny provides timings for when each pair of species has last shared a common ancestor and commences at that point in the past when all present-day species shared their most recent common ancestor.\n\nFor making the decision whether we have to investigate the causes of an apparently high di- versification should be invesigated, it is desirable to whether or not the diversification really is remarkable in reference to some null model.\n\nIn a broad class of models, the number of progreny lineages of any particular ancestral lineages in a reconstructed phylogeny has a geometric distribution (Nee et al. 1994). The derivation of this will be given later in section 2.3.1. We are interested in how many lineages each ancestral lineage gives rise to and if the distribution of progeny lineages fits our geometric expectation.\n\nUnder the constant rate birth-death model, there are interesting properties for both the actual and the reconstructed phylogeny:\n\n• If there was no extinction, the curves representing the number of lineages through time for the actual phylogeny and the reconstructed phylogeny are the same.\n\n• The push of the past is observed as an apparently higher diversification rate at the beginning of the growth of the actual phylogeny. It is a result from the fact that we consider clades which have survived to the present day, and these are the ones which got a ”flying start”\n\nmost of the times.\n\n(13)\n\n• The pull of the present is the observed increase in the diversification rate in the recent past of the reconstructed phylogeny. It is a result of the fact that lineages who arose more recent in the past have had less time to go extinct.\n\n• The slope of both phylogenies is λ − µ most of the time.\n\n• The slope of the reconstructed phylogeny assymptotically approaches the birth rate.\n\n• The pull of the present and the pull of the past increases as the fraction µ/λ approaches one.\n\nUsing reconstructed phylogenies, it is tempting to take a pure birth process intuitively, so d = 0, as a model for the data since there are no extinct species in the phylogeny. However, using likelihood plots in (Nee et al. 1994) show that we cannot exclude the possibility that it is actu- ally nonzero. Using a likelihood surface apporach, we check in section 3.1.3 if this is indeed the case.\n\nWhen only a sample of a clade has been used, so not the whole clade, creates an effect of slowdown in diversification. This effect becomes more pronounced the smaller the sample is (Nee et al. 1994). Lineages that have arisen in the recent past are likely to have fewer progeny than lineages which arose in a more distant past. So, they are less likely to have any progeny represented in the sample which causes the oberserved diversification slowdown.\n\n### 2.2 Inference of the Yule Process\n\nWe first make the following assumptions, before we make inferences with the Yule Process:\n\n• From the time of its origin with two lineages time t ago, the tree has grown according to a Yule Process with parameter λ\n\n• the age of the clade, t, is a fixed variable.\n\nNote that in this way, at each point in time each lineage has the same probability of giving birth to a new lineage. This probability is proportional to the parameter λ, controlling the rate of growth of the tree. Note also that the clade size N is not predetermined, it is a random variable.\n\nWe assume that our data consists if the length of time from each node to the present day , denoted by xi. For example, a clade with four lineages at the end has 3 nodes. We let x2 denote the time between the present and the last ancestor, which is also the age of this monophyletic clade. So we have x2= t. Moreover, we consider in general the following quantities:\n\nsr=\n\nn\n\nX\n\ni=3\n\nxi (2.1)\n\ns = 2x2+\n\nn\n\nX\n\ni=3\n\nxi= 2x2+ sr (2.2)\n\nWe immediately that s represents the sum of all branch lengths in the tree, and sr only repre- sents the sum of the branch lengths, except the two basal branches.\n\nTo obtain the probability density function of the data, we have the following:\n\n• We have n branches, from which the 2 base branches have the same length and are therefore the same. We therefore have (n − 1)! different permuations for the clade.\n\n• We assume the branch lengths are independent exponential random variables\n\n(14)\n\nThe probability for a branch i giving birth after a time xi is:\n\nPr(Xi≥ xi) = Z xi\n\n0\n\nλ exp[−λxi]dxi = exp[−λxi] (2.3) The probability of j lineages in the tree at the time of a birth event is proportional to λj. Hence, for a tree which has 2 birth events after its first node and thus has 3 species at this moment. These events contribute therefore the term 2λ ∗ 3λ tot the likelihood expression. In general: for n species we have n − 1 birth events. So, we have a contribution of the term (n − 1)!λn−2 to the likelihood expression.\n\nThe n branches x, of which the two basal branches are the same, contribute a term exp[−λs]\n\nto the likelihood by the combined probabilities of equation 2.3. Combining this and the latter, we obtain the likelihood for the data:\n\nPr(x, n; λ, t) = (n − 1)!λn−2e−λs (2.4) Actually, we are only interested in the probability density function of x only. The probability of n lineages, given λ and t is obtained by the following:\n\n• A clade starting with two species, has given birth to n − 2 new species.\n\n• Two species did not speciate before time t.\n\n• There occured n − 1 birth events.\n\nBecause we have n lineages and n−1 birth events, there are n−1 different possibilities of getting n lineages. This contributes a (n − 1) term to the likelihood. In the same reasoning as before, we have two branches which did not give birth, contributing the exp(−λt)2= exp(−2λt) term to the likelihood. Finally, the n − 2 branches who gave birth, contributed the (1 − exp[−λt])n−2term to the likelihood. Combining this we get the probability of n lineages in the clade:\n\nPr(n; λ, t) = (n − 1)e−2λt(1 − e−λt)n−2 (2.5) Where the probability density functions obtained are the same as in (Nee 2001). Thus, the probability of x given n, λ and t is:\n\nPr(x; n, λ, t) = Pr(x, n; λ, t)\n\nPr(n; λ, t) =(n − 2)!λn−2e−λsr\n\n(1 − e−λt)n−2) (2.6)\n\nRemark Observe that:\n\nPr(x; n, λ, t) =(n − 2)!λn−2e−λsr\n\n(1 − e−λt)n−2) (2.7)\n\n= (n − 2)!\n\nn\n\nY\n\ni=3\n\nλe−λxi\n\n1 − e−λt (2.8)\n\nTherefore, this is also the probability density function of the order statistics of n − 2 inde- pendent and identically distributed random variables, where the random variables are truncated exponentially distributed: i.e. they have the density:\n\nPr(Xi = xi; λ, t) = λe−λxi\n\n1 − e−λt (2.9)\n\n(15)\n\n### 2.2.1 Confidence intervals for λ\n\nMoran’s Maximum Likelihood Estimation\n\nConsider the likelihood function given in equation 2.4 again. For maximum likelihood estimation, we need both the outcome of our random variables n and x. Taking the natural logarithm of equation 2.4, we obtain the loglikelihood:\n\n`(x, n; λ, t) = log(n − 1)! + (n − 2) log(λ) − λs (2.10) Taking the partial derivative of 2.10 to λ yields us\n\n∂λ`(x, n; λ, t) = n − 2\n\ns − λ (2.11)\n\nWhere setting 2.11 equal to zero yields our maximum likelihood estimate of the parameter λ:\n\nλˆKM=n − 2\n\ns (2.12)\n\nThis estimator has been called the Kendall-Moran estimator, after those who have derived it first. To obtain the variance of this estimator, we look to the inverse of the Fisher information matrix J , which is a scalar in this case:\n\nJ (λ, n) = −E\u0014 ∂2\n\n∂λ2log (Pr(x, n; λ, t)))\n\n\u0015\n\n= n − 2 λ2\n\n⇒ Var(ˆλKM) = λ2\n\nn − 2 (2.13)\n\nTherefore, we can i see by the results in (Dobson and Barnett 2008):\n\nλˆKM∼ N\n\n\u0012 λ, λ2\n\nn − 2\n\n\u0013\n\n(2.14) By this result, we can obtain a two sided 95% confidence interval simply. By properties of the normal distribution, we know that z0.975 = 1.96 = −z0.025. Hence we obtain:\n\n−1.96 <\n\nˆλKM− λ J12 =\n\nλˆKM− λ\n\nλ n−2\n\n< 1.96 (2.15)\n\nFrom this last equation 2.15, we obtain the 95% confidence interval for λ by basic calculus:\n\nˆλKM\n\n1 − 1.96n−2 < λ <\n\nˆλKM\n\n1 + 1.96n−2 (2.16)\n\nKendall’s Maximum Likelihood Estimation\n\nKendall obtained in 1949 a different variance from equation 2.13, λ2\n\n2(eλt− 1) (2.17)\n\nHowever, it is easy to see the relationship between the variances given in equations 2.13 and 2.17, since we can rewrite equation 2.17 as:\n\n(16)\n\nλ2\n\n2(eλt− 1) = λ2\n\n2eλt− 2) = λ2\n\n2E[n] − 2 (2.18)\n\nWhere we treat n now a random variable representing the population, because of the pure- birth assumption. Note that it bega with two ancestral lineages time t ago. To obtain a 95%\n\nconfidence intervalm we proceed similiarly to the Kendall-Moran case. However, we obtain after some calculations:\n\nλˆK= λ\n\n\u0012\n\n1 ± 1.96\n\n2eλt− 2\n\n\u0013\n\n(2.19) This equation can’t be solved to λ analytically. So we can’t derive any explicit confidence interval in this case. However, in any particular case we can obtain these intervals numerically (Nee 2001).\n\nThe variances of Kendall-Moran (equation 2.13) and Kendall (equation 2.17) differ by the as- sumptions they made for both situations. Moran was considering a population of processes that grew untill they reach exactly n lineages. In that case, n is a predetermined variable in the like- lihood (2.4.) and the age of the clade t is a random variable. In this model, the branch lengths, the elements of x, are exponentially distributed. In the Kendall model, the time t was fixed and the number of lineages n was a random variable. In this model the branch lengths are truncated exponentially distributed. Allthough the maximum likelihood estimates are the same for both models, the variances differ. The Kendall model seems to be the appropiated one for inference in this context, corresponding to our original specifications (Nee 2001).\n\nHowever, if we want to take the Moran model then it is not necessary to make use of any approximations because we obtain an exact confidence interval. Notice that an exponential random variable with λ = 0.5 has a chi-squared distribution with two degrees of freedom. Let χ2n,α be the upper α point of the chi-squared distribution with 2n degress of freedom. Then the exact 95%\n\nconfidence interval for λ under the Moran model is (Nee 2001):\n\nχ2(n−2),0.025\n\n2s < λ < χ2(n−2),0.975\n\n2s (2.20)\n\nParadis (1997) suggested a third choice for the variance, where he uses the observed Fisher infor- mation instead of the expected information:\n\nˆλ2P\n\nn − 2 (2.21)\n\nThis variance yields another 95% confidence interval, which is straight forward:\n\nλˆP\n\n\u0012\n\n1 − 1.96\n\n√n − 2\n\n\u0013\n\n< λ < ˆλP\n\n\u0012\n\n1 + 1.96\n\n√n − 2\n\n\u0013\n\n(2.22) The analysis of Paradis differs from all the others. He assumes the branch lengths are expo- nentially distributed, so he is studying the same hypothetical population of processes as Moran.\n\nHowever, Paradis’ maximum likelihood estimate of λ differs:\n\nˆλP = n − 1 P i = 2nxi\n\n(2.23) The numerator is larger by one, and the denumerator smaller by x2 in comparison with the maximum likelihood estimate of Kendall and Moran. This difference is the result of the use of a different likelihood than equation (2.4).\n\n(17)\n\nHey’s Maximum Likelihood Estimation\n\nHey (1992) ignores the length of time between the last node in the tree and the present. That is equivalent to substracting nxn from s. For equation (2.4), this is only a appropiate likelihood corresponding to the Moran model if we assume that a speciation event occured at the present day, such that xn = 0. If this is not suitable but one wished to use Moran’s model, one has to use Hey’s form. From now, we assume that xn= 0 when discussing the Moran model.\n\nLikelihood ratio analysis\n\nObserve the likelihood ratio statistic, which is chi-squared distributed with one degree of freedom (Dobson and Barnett 2008):\n\nW (λ0) = 2[`(ˆλ) − `(λ0)] ∼ χ2(1) (2.24) Where `(λ) is the log likelihood function given in equation (2.10). A 95% confidence interval is given by the set of points λ ∈ Λ such that:\n\nW (λ) < χ20.95(1) = 3.841 (2.25)\n\nWhich can also be solved numerically (Nee 2001).\n\nSummarizing the last sections:\n\n• Kendall’s model correspond to our original specifications of the correct probability model for inference, but the confidence interval it provids is a numerical approximation whose accuracy is unknown. We can get an exact interval, when we discard the information about λ in the clade size.\n\n• Moran’s model provides an exact confidence interval, but the model assumes a fixed clade size and a randomly varying clade age. This does not seem appropiate in the present context.\n\n• The likelihood ratio test analysis and the Paradis variant falls outside the natural development of this topic, because we base our analysis on models in which clade size, age or both are fixed.\n\nBecause non of the confidence intervals presents an overwhelming case for itself, we compare their performances in simulations (Nee 2001). The following was observed:\n\n• The Paradis variant was the least precise variant, since it delivered the largest confidence interval for an equivalent precision as the Moran and Kendall model.\n\n• The truncated exponential model was also dropped for the same reason as the Paradis variant.\n\n• By making the ratio of the variances of Moran and Kendall’s models as a new parameter, we can naturally compare the results of both models for different values of this ratio. Due to the better performances, Moran’s model was the best model.\n\n(18)\n\n### 2.3 Likelihood of the birth-death process\n\nWe set t0= 0 as the time where our phylogeny begins. Also, we set the time T as the time of the present day, and the time t as some arbitrary time between 0 and T . For the birth death process, we can distinguish four related processes (Nee et al. 1994). These processes are plotted in figure 2.1, where the blue line represents a time t > 0 and the red line represents a time T > t:\n\nFigure 2.1: Four related processes within the constant birth-death process.\n\n1. A simple birth death process which may or may not survive to time t.\n\n2. A subset of the realizations of the first process and consists of those realizations which survive to time t between times 0 and T , but may or may not go extinct before the present day.\n\n3. A subset of the second process which do survive to the present.\n\n4. The reconstructed process, derived from the third by pruning the historical record of those lineages which do not have contemporary descendants. This process corresponds to a perfect phylogeny.\n\n(19)\n\n### 2.3.1 Probability functions of simulated trees.\n\nWe define Pr (i; t) as the probability that a process has i lineages at time t. For each of the four processes, we subsript the probabilities by 1 to 4 to clearify on which process we describe.\n\nA crucial probability relevant to both paleontological and molecular phylogenetic data is the probability that a lineage that arose at some time t in the past, still has some descendants at the time T later (Kendall 1948):\n\nP (t, T ) = 1 −µλ\n\n1 −µλe−(λ−µ)(T −t) (2.26)\n\nThe probability equation (2.26) is a crucial probabilty, because a lineage will not appear in a molecular phylogeny if it has no extant descendants.\n\nWe can estimate both composite parameters a = µλ and r = b − d, but in general the estimations of r are much more precise than the estimations of a (Foote 1988, Nee et al 1995a, ). Together with the probability:\n\nut= 1 − e−(λ−µ)t\n\n1 − µλe−(λ−µ)t (2.27)\n\nWe can express Prk(i; t) in terms of ut and P (t, T ) for all four processes. We will show that these process all have a geometric distribution, except process 3, which has the distribution of two independent geometric random variables.\n\nFor the first process, we obtain that the probability of no descendants is equal to 1 minus the probability that a single lineage alive at time 0 has still some descendants at time t. Hence:\n\nPr\n\n1(i, t) =\n\n\u001a 1 − P (0, t) if i = 0\n\nP (0, t)(1 − ut)ui−1t if i > 0 (2.28) From the probability of the first process in equation 2.28, we immediately obtain the probability for process 2 by conditioning the probability on the event that the process survives untill time t:\n\nPr\n\n2(i, t) = Pr(i lineages\|no extinction untill t)\n\n=Pr(i lineages ∧ no extinction untill t) Pr(no extinction untill t)\n\n=Pr1(i, t\|i > 0) P (0, t)\n\n= (1 − ut)ui−1t (2.29)\n\nFrom the last probability equation 2.29, we obtain the conditional probability Pr3(i, t; T ) for a birth-death process that survives to T. To do this, we compound distribution 2.29 with the probability that at least one of the i lineages existing at time t has some descendants at time T (Nee et al. 1994):\n\nPr\n\n3(i, t; T ) = Pr2(i, t)(1 − (1 − P (t, T ))i) P\n\nj=1Pr2(j, t)(1 − (1 − P (t, T ))i)\n\n= (1 − ut)ui−1t (1 − (1 − P (t, T ))i) P\n\nj=1(1 − ut)ui−1t (1 − (1 − P (t, T ))i) (2.30) This function is an ugly expression, but there is a simple underlying structure. To show this, we use the moment generating function (mgf) of random variables. To obtain our desired result, we first need the moment generating function of geometric random variables.\n\n(20)\n\nProperty 1 The moment generating function of a random variable X which is geometric dis- tributed with parameter p, is given by:\n\nmX(t) = pet\n\n1 − et(1 − p) (2.31)\n\nProof\n\nmX(t) = E\u0002eXt\u0003\n\n=\n\nX\n\nx=1\n\nextPr(X = x)\n\n=\n\nX\n\nx=1\n\n(et)xp(1 − p)x−1\n\n= etp\n\nX\n\nx=1\n\n(et)x−1(1 − p)x−1\n\n= etp\n\nX\n\nx=1\n\n((1 − p)et)x−1\n\n= etp\n\nX\n\ny=0\n\n((1 − p)et)y\n\n= et p\n\n1 − (1 − p)et As desired.\n\nUsing the properties of the moment generating function, we observe the following for the prob- ability density given for the third process:\n\nProperty 2 The moment generating function of the number of lineages i in the third process, equals the moment generating function of the sum of two independent geometric variables G1 and G2, where\n\nG1∼ geo(ut)\n\n(G2+ 1) ∼ geo(ut(1 − P (t, T )) That is, that the pdf of G2 is given by:\n\nPr(G2= i) = (1 − ut(1 − P (t, T ))(ut(1 − P (t, T )))i, i ≥ 0 Proof We begin the proof with the moment generating function of G2:\n\nmG2(t) =\n\nX\n\ni=0\n\neitPr(G2= i) (2.32)\n\n=\n\nX\n\ni=0\n\nsi(1 − ut(1 − P (t, T ))(ut(1 − P (t, T ))i (2.33)\n\n= (1 − ut(1 − P (t, T ))\n\nX\n\ni=0\n\n(ut(1 − P (t, T )s))i (2.34)\n\n= 1 − ut(1 − P (t, T ))\n\n1 − ut(1 − P (t, T )s (2.35)\n\n(21)\n\nThe moment generating of the the third process is given by:\n\nmi(t) =\n\nX\n\ni=1\n\neitPr\n\n3(i, t; T )\n\n=\n\nX\n\ni=1\n\nsi (1 − ut)ui−1t (1 − (1 − P (t, T ))i) P\n\nj=1(1 − ut)ui−1t (1 − (1 − P (t, T ))i)\n\n= s(1 − ut)\n\nP\n\nj=1(1 − ut)ui−1t (1 − (1 − P (t, T ))i)\n\nX\n\ni=1\n\n(sut)i−1(1 − (1 − P (t, T ))i)\n\n= s(1 − ut) κ\n\n\" X\n\ni=1\n\n(sut)i−1\n\nX\n\ni=1\n\n(sut)i−1(1 − P (t, T ))i\n\n#\n\n= s(1 − ut) κ\n\n\"\n\n1\n\n1 − sut− (1 − P (t, T )\n\nX\n\ni=1\n\n(sut(1 − P (t, T )))i−1\n\n#\n\n= s(1 − ut) κ\n\n\u0014 1\n\n1 − sut\n\n− (1 − P (t, T ) sut(1 − P (t, T ))\n\n\u0015\n\n= s(1 − ut) κ\n\n\u0014 P (t, T )\n\n(1 − sut)(1 − (1 − P (t, T ))sut)\n\n\u0015\n\n=\u0014 s(1 − ut) 1 − uts\n\n\u0015 \u0014 1 κ\n\nP (t, T ) 1 − ut(1 − P (t, T ))s\n\n\u0015\n\nWhere:\n\nκ =\n\nX\n\nj=1\n\n(1 − ut)uj−1t [1 − (1 − P (t, T ))i]\n\n=\n\nX\n\nj=1\n\n(1 − ut)uj−1t\n\nX\n\nj=1\n\n(1 − ut)uj−1t [1 − (1 − P (t, T ))i]\n\n= 1 − (1 − ut)(1 − P (t, T ))\n\nX\n\nj=1\n\nuj−1t (1 − P (t, T ))i−1\n\n= 1 − (1 − ut)(1 − P (t, T )) 1\n\n1 − ut(1 − P (t, T ))\n\n= P (t, T ) 1 − ut(1 − P (t, T )) In that case:\n\nmi(t) =\u0014 s(1 − ut) 1 − uts\n\n\u0015 \u0014 1 − ut(1 − P (t, T )) 1 − ut(1 − P (t, T ))s\n\n\u0015\n\n(2.36) We recognize that the moment generating function is the product of two moment generating functions of variables G1and G2. The first term is the moment generating function of G1∼ geo(ut) and the second term is the moment generating function of (G2+ 1) ∼ geo(ut(1 − P (t, T )).\n\nThus, the number of lineages existing at time t for a birth-death process which will survive to a time T later can be treated as the sum of two independent variables, with a geometric distribution.\n\nNow we will see that process 4, which is reconstructed from this one, is again geometric distributed for Pr(i, t).\n\nLet zi be the probabilities of the third process, given by equation 2.30. Of the j lineages existing at time t, i will have some descendants at the time T . Here, i is binomial distributed with parameter P (t, T ) and n > 0, since at least one survives to T . So, we obtain for the reconstructed process:\n\n(22)\n\nPr4(i, t; T ) =\n\nX\n\nj=1\n\nzj\n\n1 − (1 − P (t, T ))j\n\n\u0012j i\n\n\u0013\n\nP (t, T )i(1 − P (t, T ))j−i (2.37)\n\nWhich simplifies to (Nee et al. 1994):\n\nPr4(i, t; T ) =\n\n\u0012\n\n1 − utP (0, T ) P (0, t)\n\n\u0013 \u0012\n\nutP (0, T ) P (0, t)\n\n\u0013i−1\n\n, i > 0 (2.38)\n\nWhich is a geometric distribution with parameter utP (0,T )\n\nP (0,t). Therefore, the first, second and fourth process have a geometric distributed number of lineages, the third process’ number of lineages is distributed as the sum of two geometric variables.\n\n### 2.3.2 Probability Density Function of a Simulated Tree\n\nWe now look more closely to a birth-death process which generated the distribution given in equation 2.38. We let n(t) be the number of lineages at time t. We suppose that we grow a reconstructed phylogenetic tree, starting at time t = 0, thus n(0) = 1. Each lineage a time t give rise to a daughter lineage at a rate λ Pr(t, T ). So, after a small amount of time dt, we have:\n\nn(t) =\n\n\u001a n(t) + 1 with probability n(t)λP (t, T )dt\n\nn(t) with probability 1 − n(t)λP (t, T )dt (2.39) We extend the probability model given in equation 2.39 by the following. Given that we have n lineages at a time tn, we denote the time untill the next lineage as τ . We then have:\n\nPr(τ > t + dt; 0 < t < T − tn) = Pr(τ > t; 0 < t < T − tn) Pr(no lineage in dt)\n\n= Pr(τ > t; 0 < t < T − tn)(1 − λn(t)P (t, T )dt) (2.40) Which is equivalent to:\n\nd\n\ndtPr(τ > t; 0 < t < T − tn) = −λn(t)P (t, T ) Pr(τ > t; 0 < t < T − tn) (2.41) Since we know the number of lineages n(t) at time t, we can treat it as a fixed variable. The solution of this ODE is then straight forward:\n\nPr(τ > t; 0 < t < T − tn) = exp\n\n\u0012\n\n−λn Z tn+t\n\ntn\n\nP (s, T )ds\n\n\u0013\n\n(2.42) Where the last integral needs some computations to get the solution:\n\n(23)\n\nZ tn+t tn\n\nP (s, T )ds = Z tn+t\n\ntn\n\n1 −µλ\n\n1 − µλe−(λ−µ)(T −s)ds\n\n= Z tn+t\n\ntn\n\nλ − µ\n\nλ − µe−(λ−µ)(T −s)ds substitute u(t) = (λ − µ)t, to obtain:\n\n=\n\nZ u(tn+t) u(tn)\n\n1\n\nλ − µeu−(λ−µ)Tdu\n\n=\n\nZ u(tn+t) u(tn)\n\n1 λ − κeudu\n\nsubstitute v(u) = λ − κeu to obtain\n\n=\n\nZ v(u(tn+t)) v(u(tn))\n\n1 w\n\n1 w − λdw\n\n=\n\nZ v(u(tn+t)) v(u(tn))\n\n\u0012\n\n− 1\n\nλw + 1\n\nλ(w − λ)\n\n\u0013 dv\n\n= −1\n\nλlog(w) + 1\n\nλlog(λ(w − λ))\|v(u(tv(u(tn+t))\n\nn))\n\n= t −µ λt − 1\n\nλlog\u0012 1 −µλexp(−(λ − µ)(T − tn− t)) 1 −µλexp(−(λ − µ)(T − tn))\n\n\u0013\n\n(2.43) Which yields us the solution of the ODE:\n\nPr(τ > t; 0 < t < T − tn) = exp(−n(λ − µ)t)\u0012 1 −µλexp(−(λ − µ)(T − tn− t)) 1 − µλexp(−(λ − µ)(T − tn))\n\n\u0013n\n\n(2.44) From the last equation 2.45, we can obtain the probability density function of the waiting time for a birth, t (Nee et al. 1994):\n\nPr(t; tn, T, λ, µ) = n(λ − µ)e−n(λ−µ)t 1 − µλe−(λ−µ)(T −tn−t)\u0001n−1\n\n1 −µλe−(λ−µ)(T −tn)\u0001n (2.45) From this probability density function in equation 2.45, we obtain the likelihood of a tree which has grown according to a birth-death process. Let xi denote the time between the i-th birth event and the present day T . Then, the time between two lineages is ti = xi− xi+1 and xi = T − tn. Suppose we have a tree which has N lineages at the present day. Let the vector x contain all the times between lineages, which starts from the time that the first species gave birth to two new species. Therefore, x = (x2, x3, . . . , xN). We obtain the likelihood of the tree by multiplying all independent probabilities of the times between lineages:\n\nL(x\|λ, µ) =\n\nN −1\n\nY\n\ni=1\n\nPr(ti; tn, T, λ, µ)\n\n=\n\nN −1\n\nY\n\ni=1\n\nn(λ − µ)e−n(λ−µ)(xi−xi+1) 1 −µλe−(λ−µ)xi+1\u0001n−1\n\n1 −µλe−(λ−µ)xi\u0001n (2.46)\n\n### 2.3.3 Maximum Likelihood Estimation\n\nTo find the maximum likelihood estimator of the parameters of the birth-death process, we need the likelihood obtain in equation 2.46. Instead of maximizing the likelihood, we maximize the log-likelihood:\n\n(24)\n\n`(x\|λ, µ) = log\n\nN −1\n\nY\n\ni=1\n\nn(λ − µ)e−n(λ−µ)(xi−xi+1) 1 −µλe−(λ−µ)xi+1\u0001n−1 1 −µλe−(λ−µ)xi\u0001n\n\n!\n\n=\n\nN −1\n\nX\n\ni=2\n\n{log(n) + log(λ − µ) − n(λ − µ)(xi− xi+1)}\n\n+\n\nN −1\n\nX\n\ni=2\n\n\u001a\n\n(n − 1) log(1 − λ\n\nµe−(λ−µ)xi+1) − n log(1 −λ\n\nµe−(λ−µ)xi)\n\n\u001b\n\n(2.47)\n\nThe normal approach, i.e. taking partial derrivatives and setting equal to zero is not applicable in this case, since this equation can’t be solved analytically. Therefore, we obtain maximum likelihood estimates of the parameters λ and µ numerically. This will be done in chapter 3.\n\n### 2.4 Approximating the Likelihood of the Protracted Birth- Death Model\n\nInstead of finding the likelihood of a tree which has grown according to a protracted birth-death model, we make an approximation using Gaussian processes. A protracted birth-death tree which has grown according to a Gaussian process is not a useful tree for the estimation, since the tree its lineages might not be a natural number in this process. However, treating a tree which has grown according to a protracted birth-death process, can be inferred by an approximation with a birth-death process. We first give the definition of a Gaussian process:\n\nDefinition A real-valued stochastic process, {Xt; t ∈ T }, where T is an index set, is a Gaussian process if all the finite-dimensional distributions have a multivariate normal distribution. That is, for any choice of distinct values t1, . . . , tk ∈ T , the random vector X = (Xt1, . . . , Xtk)0 has a multivariate normal distribution with mean µ = E[X] and covariance matrix Σ = cov(X, X), which has the probability density function:\n\nfX(X; µ, Σ) = 1 (2π)k2\|Σ\|12\n\nexp\n\n\u0014\n\n−1\n\n2(X − µ)0Σ−1(X − µ)\n\n\u0015\n\n(2.48) Which is denoted by:\n\nX ∼ N (µ, Σ) (2.49)\n\nThe mean and covariance of a Gaussian process are defined by:\n\nµ(t) = E[Xt], γ(s, t) = cov(Xs, Xt), t, s ∈ T (2.50)\n\n### 2.4.1 Approximation of the Process\n\nWe let Ntdenote the state at time t, which consists of the number of good species Ng(t) and the number of incipient species Ni(t) at time t. Suppose that we take a step in time dt, such that the next event occurs. For the protracted birth-death process, five different event may occur:\n\n1. A good species giving birth to an incipient species, with a rate λ1Ng(t)dt.\n\n2. A good species going extinct, with a rate µNg(t)dt.\n\n3. An incipient species reaching completion, with a rate λ2Ni(t)dt\n\n4. An incipient species giving birth to a new incipient species, with a rate λ3Ni(t)dt 5. An incipient species going extinct, with a rate µNi(t)dt.\n\n(25)\n\nThe Mean and Covariance of the Process\n\nWe look to the difference between the state of today Nt= [Ng(t), Ni(t)] and the state of the next event Nt+dt, which is denoted by ∆Nt= Nt+dt− Nt. To avoid confusion with the mean µ and the death rate µ, we let the vector Λ = (λ1, λ2, λ3, µ) denote the parameters. Doing so, we have the following expectation:\n\nE[∆Nt\|Nt] =\n\n\u0014 −µNg(t)dt + λ2Ni(t)dt\n\n−µNi(t)dt − λ2Ni(t)dt + λ1Ng(t)dt + λ3Ni(t)dt\n\n\u0015\n\n=\n\n\u0014 −µNg(t) + λ2Ni(t) +λ1Ng(t) − (λ2+ λ3+ µ)Ni(t)\n\n\u0015 dt\n\n= µ {Nt, Λ} dt (2.51)\n\nWe again look to the difference difference between the state of today ∆Nt. As we have earlier defined the expectation of the development in the protracted birth-death process, given the last state of the process, we do exactly the same for the conditional covariance. In a similiar way as for the expectation, we obtain the following conditional covariance:\n\ncov(∆Nt\|Nt) =\n\n\u0014 Var (∆Ng(t + dt)\|Nt) cov(∆Ng(t + dt), ∆Ni(, t + dt)\|Nt) cov(∆Ng(t + dt), ∆Ni(, t + dt)\|Nt) Var (∆Ni(t + dt)\|Nt)\n\n\u0015 (2.52) We obtain the entries of the matrix part by part. We start with Σ1,1 :\n\nVar (∆Ng(t + dt)\|Nt) = E[(∆Ng(t + dt))2\|Nt] − (E[∆Ng(t + dt)\|Nt])2\n\n= µNg(t)dt + λ2Ni(t)dt − (µNg(t)dt + λ2Ni(t)dt)2\n\n= µNg(t)dt + λ2Ni(t)dt − (µNg(t) + λ2Ni(t))2dt2\n\n∝ µNg(t)dt + λ2Ni(t)dt (2.53)\n\nΣ1,2 = Σ2,1 is given by:\n\ncov(∆Ng(t + dt), ∆Ni(t + dt)\|Nt) = E[∆Ng(t + dt) · ∆Ni(t + dt)\|Nt]\n\n− E[∆Ng(t + dt)\|Nt] · E[∆Ni(t + dt)\|Nt]\n\n= −λ2Ni(t)dt − (−µNg(t)dt + λ2Ni(t)dt)\n\n× (−µNi(t)dt − λ2Ni(t)dt + λ1Ng(t)dt + λ3Ni(t)dt)\n\n= −λ2Ni(t)dt − (−µNg(t) + λ2Ni(t))\n\n× (−µNi(t) − λ2Ni(t) + λ1Ng(t) + λ3Ni(t))dt2\n\n∝ −λ2Ni(t)dt (2.54)\n\nΣ2,2 is given by:\n\nVar (∆Ni(t + dt)\|Nt) = E[(∆Ni(t + dt))2\|Nt] − (E[∆Ni(t + dt)\|Nt])2\n\n= λ1Ng(t)dt + (λ2+ λ3+ µ)Ni(t)dt\n\n− (−µNi(t)dt − λ2Ni(t)dt + λ1Ng(t)dt + λ3Ni(t)dt)2\n\n= λ1Ng(t)dt + (λ2+ λ3+ µ)Ni(t)dt\n\n− (−µNi(t) − λ2Ni(t) + λ1Ng(t) + λ3Ni(t))2dt2\n\n∝ λ1Ng(t)dt + (λ2+ λ3+ µ)Ni(t)dt (2.55) We therefore obtain the covariance matrix Σ as:\n\n(26)\n\ncov(∆Nt\|Nt) =\u0014µNg(t) + λ2Ni(t) −λ2Ni(t)\n\n−λ2Ni(t) λ1Ng(t) + (λ2+ λ3+ µ)Ni(t)\n\n\u0015\n\ndt = Σ{Nt, Λ} · dt (2.56) If we assume that the number good species and incipient species is nonnegative, the matrix Σ{Nt, Λ} is nonsingular in most of the situations because the determinant is nonzero. Observe that the determinant of Σ is given by the following expression:\n\ndet (Σ) =\n\nµNg(t) + λ2Ni(t) −λ2Ni(t)\n\n−λ2Ni(t) λ1Ng(t) + (λ2+ λ3+ µ)Ni(t)\n\n= µNg(t) (λ1Ng(t) + [λ2+ λ3+ µ]Ni(t)) + λ2Ni(t) (λ1Ng(t) + (λ3+ µ)Ni(t)) (2.57) The determinant is nonzero, except if:\n\n1. All species are extinct.\n\n2. Only one of the parameters is nonzero.\n\n3. All good species have become extinct, and λ2= 0 or λ3= µ = 0.\n\n4. There are no incipient species left, µ = 0 or λ1= 0.\n\nBecause the matrix Σ is in general nonsingular, we can find the square root of the matrix by diagonalization. That is:\n\nB(Xt\|Λ) = V D12V−1 (2.58)\n\nWhere D is the matrix consisting of the eigenvalues of Σ, and V is the matrix consisting of the eigenvalues of Σ.\n\nDefining the New Process\n\nNow we have obtained the expressions of µ(Nt, Λ) and Σ(Nt, Λ), we can obtain a Gaussian process for our protracted birth-death model. We first need to define the concept of a Wiener Process:\n\nDefinition A statistical process {Wt; t ∈ T } in continuous time is called a Wienerprocess or Brow- nian Motion if it has the following properties:\n\n1. W0= 0\n\n2. The function t → Wt is everywhere continuous almost surely.\n\n3. Wt has independent increments with Wt− Ws∼ N (0, I(t − s)) for 0 ≤ s < t.\n\nWe define the new process Ytby means of a Wiener Process {Wt; t ∈ T }:\n\ndYt= µ(Yt; Λ)dt + B(Yt; Λ)dWt (2.59) When we look more closely to equation 2.59, we see that the conditional expectation and the conditional variance of dYtgiven Ytare the same for the protracted birth-death process. Therefore, we conclude that this newly defined process is an approximation of the original protracted birth- death process. However, notice that is it not exactly the same process. We can even see that this construction of the approximation yields us a Gaussian process:\n\nE[dYt\|Yt] = µ(Yt; Λ)dt\n\ncov[dYt\|Yt] = B(Yt; Λ)dtB(Yt; Λ)0= Σ · dt dYt∼ N (µdt, Σdt)\n\nTherefore, the process given by Yk+1= Yk+ dYk is a Gaussian process. This means that we can approximate the protracted birth-death model, which is originally a Poisson process, by means of a Gaussian process, given by {Yt, t ∈ T }.\n\n(27)\n\n### 2.4.2 Approximation of the likelihood\n\nSuppose that we have a tree which has grown according to a protracted birth-death process. For this model, we assume that we use actual phylogenies for inferences, instead of the reconstructed phylogenies. Doing so, we can list the number of good species and incipient species on a time t in a vector, called Nt. At a time t + ∆t later, the next event occurs in our protracted birth-death process. We are interested in the difference between Ntand Nt+∆t:\n\n∆Nt= Nt+∆t− Nt (2.60)\n\nThis difference can be seen as a development in the process. Normally, the difference is obvi- ously a vector containing whole numbers. We can view these differences however, as an outcome in a Gaussian process. Intuitively, the set {∆Nt\|t ∈ T } can be treated a Gaussian process. Fur- thermore, by the construction of our approximation dYt, we can even claim that all these variables are indepent multivariate normal random variables. This is the case, because the approximation dYt is constructed by means of a Wienerprocess which has independent increments.\n\nSince the likelihood of a development ∆Ntin the protracted process is now approximated by the probability density of the multivariate normal distribution, we have that for each event time t ∈ T :\n\nf∆Nt(∆Nt; µt, Σt) ≈ 1 (2π)k2t\|12\n\nexp\n\n\u0014\n\n−1\n\n2(∆Nt− µt)0Σ−1t (∆Nt− µt)\n\n\u0015\n\n= dmnorm(∆Nt; Σt, µt) (2.61)\n\nBy the independent increments of the Wienerprocess, we therefore obtain the following loglike- lihood of the actual phylogeny:\n\n`({Nt, t ∈ T }; Λ) ≈X\n\nt∈T\n\nlog (dmnorm(∆Nt; Σt, µt)) (2.62)\n\nThismaximum likelihood estimate can’t be solved analytically for Λ = (λ1, λ2, λ3, µ). The inferences will therefore be computed numerically in chapter 3.\n\n(28)\n\n### Simulation studies\n\nIn this chapter, we investigate the simulations of the constant birth-death model and the protracted birth-death model. We first review and proof basic properties of the models.\n\n### 3.1 Simulation of a Birth-Death Model\n\nThe simulation of the birth-death model has been done numerically in R. In order to do so, we use a recursive algorithm. First, we give an outline of important analytical properties of the algorithm.\n\n### 3.1.1 Basic Properties\n\nThe first event\n\nIn a birth-death model, we suppose that speciation and extinction are processes that proceed simulataneously. The process that finished first, is the process that we observe. Let Tb(1) denote the time that speciation of the first species in a phylogeny would occur, and Td(1) the time that extinction of this species would occure. Furthermore, we assume that Tb(1) ∼ exp(λ) and Td(1) ∼ exp(µ) are independent random variables. We now consider the time that the first of these two events occurs: T (1) = min(Tb(1), Td(1)). First, we state an useful theorem.\n\nProperty 3 Suppose X1, . . . , Xn are independent exponential random variables with parameters λ1, . . . , λn. Then the minimum of the random variables, X = min{X1, . . . , Xn}, is exponentially distributed with parameter λ =Pn\n\ni=1λi.\n\nProof We make use of unicity of the surivival function for random variables. Recall that the survival function of an exponentially distributed random variable Y with parameter θ, is defined by:\n\nS(y; θ) = Pr(Y ≥ y) = exp(−θy) (3.1)\n\nNow we are going to make use of the fact, that for a certain a > 0: X = min{X1, . . . , Xn} > a if and only if Xi > a for all i = 1 . . . n. By independency of the random variables, we obtain the survival function of X:\n\nPr(X = min{X1, . . . , Xn} > a) =\n\nn\n\nY\n\ni=1\n\nPr(Xi > a)\n\n=\n\nn\n\nY\n\ni=1\n\nexp(−aλi)\n\n= exp(−a\n\nn\n\nX\n\ni=1\n\nλi)\n\nSo by the unicity of the survival function, X ∼ exp(Pn i=1λi).\n\nReferenties\n\nGERELATEERDE DOCUMENTEN\n\nEconomische Zaken presenteerden in november 2014, had als ondertitel: ‘Keuzes voor de toekomst’. Het visiedocument kiest voor excellente wetenschap en onderzoek dat de\n\nOur main results are representations for the decay parameter under four different scenarios, derived from a unified perspective involving the orthogonal polynomials appearing in\n\nFor estimating the parameters and calculating the values of the log-likelihoods of the unrestricted and restricted Lee-Carter models, we considered the pop- ulations of the\n\nA wider VFoV combined with a longer scanning range, high output rate of points and better range accuracy will always be of significant importance when selecting the most suitable\n\nToday, people demand up‐to‐date, or (near‐) real-time information, about virtually anything, anywhere, and\n\nIs het een engel met een bazuin (beschermer van de kerk of bedoeld voor het kerkhof ), een mannelijke of vrouwelijke schutheilige met een attribuut, of een begijn (wat op die\n\nThe study has aimed to fill a gap in the current literature on the relationship between South Africa and the PRC by looking at it as a continuum and using asymmetry\n\n(*): To what extent were the firm’s marketing research resources, people, knowledge and skills adequate for the gathering of market information needed for this application.. O\n\n derive a minimum distance estimator based on the likelihood of the first-di fference transform of a dynamic panel data model utilizing assumptions regarding the\n\nA Monte Carlo comparison with the HLIM, HFUL and SJEF estimators shows that the BLIM estimator gives the smallest median bias only in case of small number of instruments\n\nIn this paper we propose a method for forecasting emerging technologies in their take-off phase by identifying mature technologies with a similar behavior in that phase,\n\nChanges in the extent of recorded crime can therefore also be the result of changes in the population's willingness to report crime, in the policy of the police towards\n\n(nr. 1, 2 en 3) maar niet bij de bedrijven met recirculatie bij paprika (nr. 23 t/m 27), waar door het overlopen van de goten aan- zienlijke hoeveelheden water en kunstmest\n\nSelecta Klemm rood West-Stek geel West-Stek roze Van Staaveren roze Selecta Klemm roze Van Staaveren lila West-Stek roze Van Staaveren geel Vergelijkingsras roze... Kooij\n\nOmdat de concentratie fosfor in dit water (veel) groter kan zijn dan in deze studie opgelegd, zijn de berekende concentraties fosfor in het uittredend grondwater aan de lage kant\n\nWhile it is assumed that the code is outdated, it does not mean that its objectives are different from the objectives of the HR department to „construct a\n\nTo assess the effect of culture on extreme response style and midpoint response style, four of Hofstede’s cultural dimensions are examined: individualism,\n\nThis is partly connected to the im- pression management theory, whose final aim is to control and influence the perceptions other people have about the giver of the\n\nTegelijkertijd daalt de vaccinatiegraad zozeer, dat sommige landen voor specifieke infecties, zoals bij- voorbeeld mazelen, het vaccineren van\n\nThese peculiar properties can be entirely attributed to uncorrected contamination by the galaxy’s spheroidal component, and when this element is properly modelled, the galaxy\n\n• Instead of counting how “close” the verifying category is to the one with the highest probability, we should count how often we observe the category with the highest probability," ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8650027,"math_prob":0.99345255,"size":49555,"snap":"2023-40-2023-50","text_gpt3_token_len":14940,"char_repetition_ratio":0.16399266,"word_repetition_ratio":0.11671233,"special_character_ratio":0.294158,"punctuation_ratio":0.12761799,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99818575,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T00:40:40Z\",\"WARC-Record-ID\":\"<urn:uuid:59982faa-6396-4234-9443-e5b2931d7a4a>\",\"Content-Length\":\"253197\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f420c5f4-85c5-4a70-b479-567308d5fcae>\",\"WARC-Concurrent-To\":\"<urn:uuid:7259d21c-6722-47c5-98c8-15d1e5abd1a6>\",\"WARC-IP-Address\":\"142.93.139.232\",\"WARC-Target-URI\":\"https://5dok.net/document/yr38rpko-evolution-how-long-does-speciation-take.html\",\"WARC-Payload-Digest\":\"sha1:PNP6DIB4H4N55K446VL6ATB6DDABHY22\",\"WARC-Block-Digest\":\"sha1:2MFVBNN6V3GGL3WIXP4OZ6JAFS7V3H25\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510942.97_warc_CC-MAIN-20231002001302-20231002031302-00236.warc.gz\"}"}
https://open.library.ubc.ca/soa/cIRcle/collections/ubctheses/24/items/1.0074375	[ "# Open Collections\n\n## UBC Theses and Dissertations", null, "## UBC Theses and Dissertations\n\n### CQ algorithms : theory, computations and nonconvex extensions Guo, Yipin\n\n#### Abstract\n\nThe split feasibility problem (SFP) is important due to its occurrence in signal processing and image reconstruction, with particular progress in intensity-modulated radiation therapy. Mathematically, it can be formulated as ﬁnding a point x∗ such that x∗ ∈ C and Ax∗ ∈ Q, where A is a bounded linear operator, C and Q are subsets of two Hilbert spaces H₁ and H₂ respectively. One particular algorithm for solving this problem is the CQ algorithm. In this thesis, previous work on CQ algorithm is presented and a new proof of convergence of the relaxed CQ algorithm is given. The CQ algorithm is shown to be a special case of the subgradient projection algorithm. The SFP is extended into two nonconvex cases. The ﬁrst one is on S-subdiﬀerentiable functions, and the other one is on prox-regular functions. The subgradient projection algorithm and CQ algorithm are proved to converge to a solution of the ﬁrst and second case respectively." ]	[ null, "https://open.library.ubc.ca/staticfile/img/featured/icon-ubctheses.svg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88449293,"math_prob":0.9702414,"size":1167,"snap":"2022-40-2023-06","text_gpt3_token_len":277,"char_repetition_ratio":0.14015478,"word_repetition_ratio":0.0,"special_character_ratio":0.18251929,"punctuation_ratio":0.08737864,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97192556,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-10-01T22:09:59Z\",\"WARC-Record-ID\":\"<urn:uuid:2f3951ba-6c56-40b1-a622-dab80731b916>\",\"Content-Length\":\"106337\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:efe04069-a02b-4c4a-ab6e-bd36955fe29c>\",\"WARC-Concurrent-To\":\"<urn:uuid:0db8609c-32dc-4e54-a1ee-4a60b23a90d3>\",\"WARC-IP-Address\":\"142.103.96.89\",\"WARC-Target-URI\":\"https://open.library.ubc.ca/soa/cIRcle/collections/ubctheses/24/items/1.0074375\",\"WARC-Payload-Digest\":\"sha1:IOWFJE4MLTGPMQL5IQNMKL5RRUHPAETN\",\"WARC-Block-Digest\":\"sha1:ERA4ANJUPTDRWW3T25C7YQXULNOW26YJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030336921.76_warc_CC-MAIN-20221001195125-20221001225125-00609.warc.gz\"}"}
http://15462.courses.cs.cmu.edu/spring2022/newsfeed	[ "What's Going On", null, "How exactly does the triangle mesh become a grid again?", null, "bt1 commented on slide_022 of Variance Reduction ()\n\n@pbhuang In this case, I think \"pathological\" just means a worst-case scenario for rendering - naive path tracing works really badly with a small directional light source + near-perfect mirror, since the probability of tracing back to the light source is small", null, "pbhuang commented on slide_022 of Variance Reduction ()\n\nWhat does \"pathological\" mean in this context?", null, "nsp commented on slide_010 of Dynamics and Time Integration ()\n\nForce should really be mg for this slide, meaning that acceleration would just be g, and we would not need mass in any of these equations. Motion of an object in flight should not depend on its mass.", null, "nsp commented on slide_032 of Introduction to Animation ()\n\nLet me add a comment here -- in practice, it is easiest to define our polynomials p(u) for 0 <= u <= 1. What that means for this slide is that we can keep this notion of knots occurring at different times. However, it might be easier to reparameterize our f(t) where t_i <= t <= t_{i+1} in terms of parameter u, such that 0 <= u <= 1. This is pretty simple -- u is just (t-t_i) / (t_{i+1} - t_i). This reparameterization makes our polynomials easier to work with, derive, and think about. We will make the assumption of this parameterization in the class.", null, "I'm a bit confused with this slide here. For the premultiplied (from the previous slide), isn't $$B' = \\alpha_B B$$? So the non-premultiplied equation = premultiplied equation?", null, "coconut commented on slide_023 of Introduction to Geometry ()\n\nCould you elaborate more on why it would be harder to sample an implicit surface like the one above?", null, "Question! (i,j) looks like it's in the middle of the pixel ($$f_{00}$$) but i and j are the result of a floor so should be an int (and so shouldn't be in the middle of a pixel), - is the diagram not quite right? Or am I missing something here?", null, "AugustSZ commented on slide_038 of 3D Rotations ()\n\nDon't understand what's q bar", null, "pbhuang commented on slide_050 of Math Review Part I (Linear Algebra) ()\n\nA and C", null, "pbhuang commented on slide_029 of Math Review Part I (Linear Algebra) ()\n\nYes it does", null, "AugustSZ commented on slide_010 of Math Review Part II (Vector Calculus) ()\n\n<u,v>=<v,u>", null, "AugustSZ commented on slide_034 of Math Review Part I (Linear Algebra) ()\n\nit's symmetric", null, "AugustSZ commented on slide_030 of Math Review Part I (Linear Algebra) ()\n\nyes again", null, "AugustSZ commented on slide_029 of Math Review Part I (Linear Algebra) ()\n\nYes. It satisfies the 4 properties of a norm on Page 28", null, "nsp commented on slide_038 of Introduction ()\n\nBefore I replaced these slides with a few fixes, AugustSZ commented — There're 2 kinds of projections mentioned in Fundamentals of Computer Graphics: Orthographic Projection and Perspective projection. Camera Definition: position, orientation(up/look-at)\n\nI replied — Yes, that's right. We'll have a closer look when we get to the Perspective Projection lecture on Feb 9th. We've made an assumption here that we have Perspective projection, with camera position at c (really, c=(0,0,0) given the math on the slide), up vector is y, and look-at vector is z. If we want a general camera, we'll need to loosen those restrictions later to allow arbitrary camera positions and orientations. However, you can always fall back on the rule of similar triangles. They may be oriented oddly in 3D space to begin with, but we'll see how to transform them to look like the picture above.\n\nOrthographic projection is very simple. For orthographic projection in the picture above, v=y and u=x. Nothing more to do." ]	[ null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/ohno.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/bt1.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/pbhuang.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/nsp.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/nsp.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/hannahhe.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/coconut.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/hannahhe.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/AugustSZ.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/pbhuang.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/pbhuang.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/AugustSZ.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/AugustSZ.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/AugustSZ.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/AugustSZ.jpg", null, "http://15462.courses.cs.cmu.edu/spring2022content/profile_pictures/nsp.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.93611634,"math_prob":0.80671537,"size":2785,"snap":"2022-27-2022-33","text_gpt3_token_len":662,"char_repetition_ratio":0.094210714,"word_repetition_ratio":0.0041067763,"special_character_ratio":0.24021544,"punctuation_ratio":0.10326087,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.992689,"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,1,null,1,null,3,null,3,null,3,null,2,null,1,null,2,null,5,null,3,null,3,null,5,null,5,null,5,null,5,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-16T00:27:24Z\",\"WARC-Record-ID\":\"<urn:uuid:863eac66-7641-4081-8c5b-46494789fa5a>\",\"Content-Length\":\"18374\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1905d5f0-9588-4d9d-98bf-4961dd9fbfd0>\",\"WARC-Concurrent-To\":\"<urn:uuid:5a4b9b0c-b184-4190-9957-253a632f27bf>\",\"WARC-IP-Address\":\"128.2.220.105\",\"WARC-Target-URI\":\"http://15462.courses.cs.cmu.edu/spring2022/newsfeed\",\"WARC-Payload-Digest\":\"sha1:SBL4USOKCOMPSRWXKVBWCE22M5EI52TW\",\"WARC-Block-Digest\":\"sha1:WCK7JMNZC3MWRDMRQTCUVDT32BG4J52Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882572215.27_warc_CC-MAIN-20220815235954-20220816025954-00036.warc.gz\"}"}
https://codinginterviewpro.com/interview-questions-and-answers/all-new-60-oops-interview-questions/	[ "# All New 60 OOPs Interview Questions\n\n## Introduction\n\nWelcome to the world of Object-Oriented Programming (OOP)! In this interview, we will explore some fundamental concepts and principles of OOP, which is a programming paradigm widely used in modern software development. OOP aims to structure code in a way that mirrors the real world, making it easier to design, understand, and maintain software systems.\n\nRemember that understanding OOP goes beyond memorizing syntax; it’s about grasping the underlying principles and applying them to solve real-world problems. So, let’s embark on this learning journey together, and I’m excited to see your knowledge and passion for Object-Oriented Programming in action! Good luck!\n\n## Basic Questions\n\n### 1. What is Object-Oriented Programming (OOP)?\n\nObject-Oriented Programming (OOP) is a programming paradigm that organizes data and behaviors into objects. It focuses on modeling real-world entities as objects, encapsulating data and methods within them. OOP promotes code reusability, maintainability, and modularity.\n\n### 2. What are the main principles of OOP?\n\nThe main principles of OOP are:\n\n1. Encapsulation: Bundling data and methods that operate on the data within a single unit (object).\n2. Abstraction: Hiding unnecessary implementation details and exposing only relevant features of an object.\n3. Inheritance: Creating new classes (subclasses) based on existing classes (superclasses) to inherit their attributes and behaviors.\n4. Polymorphism: The ability of different objects to be treated as instances of their common superclass, allowing methods to be called on them without knowing their specific types.\n\n### 3. What is a class in OOP?\n\nA class in OOP is a blueprint that defines the structure and behavior of objects. It acts as a template for creating objects with specific attributes and methods.\n\nPython\n``````class Car:\ndef __init__(self, make, model):\nself.make = make\nself.model = model\n\ndef start_engine(self):\nreturn f\"{self.make} {self.model}'s engine started.\"``````\n\n### 4. What is an object in OOP?\n\nAn object in OOP is an instance of a class. It represents a real-world entity or concept and contains data and methods defined in the class.\n\nPython\n``````# Creating objects (instances) of the Car class\ncar1 = Car(\"Toyota\", \"Corolla\")\ncar2 = Car(\"Honda\", \"Civic\")\n\n# Accessing object's attributes and methods\nprint(car1.make) # Output: Toyota\nprint(car2.start_engine()) # Output: Honda Civic's engine started.``````\n\n### 5. What is inheritance in OOP?\n\nInheritance in OOP allows a new class (subclass) to inherit attributes and methods from an existing class (superclass). It promotes code reuse and supports the “is-a” relationship.\n\nPython\n``````# Base class (Superclass)\nclass Animal:\ndef speak(self):\npass\n\n# Derived class (Subclass)\nclass Dog(Animal):\ndef speak(self):\nreturn \"Woof!\"\n\nclass Cat(Animal):\ndef speak(self):\nreturn \"Meow!\"\n\ndog = Dog()\ncat = Cat()\nprint(dog.speak()) # Output: Woof!\nprint(cat.speak()) # Output: Meow!``````\n\n### 6. What is polymorphism in OOP?\n\nPolymorphism allows objects of different classes to be treated as instances of their common superclass. It enables the same method to be called on different objects, producing different results based on the actual object’s type.\n\nPython\n``````# Using the Animal superclass from the previous example\ndef animal_sound(animal):\nreturn animal.speak()\n\ndog = Dog()\ncat = Cat()\n\nprint(animal_sound(dog)) # Output: Woof!\nprint(animal_sound(cat)) # Output: Meow!``````\n\n### 7. What is abstraction in OOP?\n\nAbstraction in OOP involves hiding implementation details and exposing only essential features to the outside world.\n\nPython\n``````from abc import ABC, abstractmethod\n\nclass Shape(ABC):\n@abstractmethod\ndef area(self):\npass\n\nclass Circle(Shape):\n\ndef area(self):\n\ncircle = Circle(5)\nprint(circle.area()) # Output: 78.5``````\n\n### 8. What is encapsulation in OOP?\n\nEncapsulation in OOP involves bundling data (attributes) and methods that operate on that data within a class, providing control over access to the data.\n\nPython\n``````class BankAccount:\ndef __init__(self):\nself.__balance = 0\n\ndef deposit(self, amount):\nself.__balance += amount\n\ndef withdraw(self, amount):\nif amount <= self.__balance:\nself.__balance -= amount\nelse:\nprint(\"Insufficient balance.\")\n\ndef get_balance(self):\nreturn self.__balance\n\naccount = BankAccount()\naccount.deposit(1000)\naccount.withdraw(500)\nprint(account.get_balance()) # Output: 500``````\n\n### 9. Explain the concept of a constructor.\n\nA constructor in OOP is a special method that gets automatically called when an object is created from a class. It is used to initialize the object’s attributes and perform any necessary setup.\n\nPython\n``````class Person:\ndef __init__(self, name, age):\nself.name = name\nself.age = age\n\nperson1 = Person(\"Alice\", 25)\nperson2 = Person(\"Bob\", 30)\n\nprint(person1.name, person1.age) # Output: Alice 25\nprint(person2.name, person2.age) # Output: Bob 30``````\n\n### 10. What is a destructor?\n\nA destructor in OOP is a special method that gets automatically called when an object is about to be destroyed (e.g., goes out of scope or is explicitly deleted). In Python, the `__del__()` method serves as a destructor.\n\nPython\n``````class MyClass:\ndef __init__(self, name):\nself.name = name\n\ndef __del__(self):\nprint(f\"{self.name} object is being destroyed.\")\n\nobj1 = MyClass(\"Object 1\")\nobj2 = MyClass(\"Object 2\")\n\ndel obj1 # Output: Object 1 object is being destroyed.``````\n\n### 11. What is an instance in OOP?\n\nAn instance in OOP refers to a specific object created from a class. It has its own unique data and can access the methods defined in the class.\n\nPython\n``````class Person:\ndef __init__(self, name):\nself.name = name\n\nperson1 = Person(\"Alice\")\nperson2 = Person(\"Bob\")\n\nprint(person1.name) # Output: Alice\nprint(person2.name) # Output: Bob``````\n\n### 12. What is a method in OOP?\n\nA method in OOP is a function defined within a class, which operates on the class’s data and provides behavior to objects of that class.\n\nPython\n``````class Calculator:\nreturn a + b\n\ndef subtract(self, a, b):\nreturn a - b\n\ncalc = Calculator()\nresult2 = calc.subtract(10, 4)\n\nprint(result1) # Output: 8\nprint(result2) # Output: 6``````\n\nMethod overloading in OOP allows a class to define multiple methods with the same name but different parameter lists. The correct method is chosen based on the number or types of arguments passed during the method call.\n\nPython\n``````class OverloadingExample:\nreturn a + b\n\nreturn a + b + c\n\nresult1 = overload.add(2, 3) # This will raise an error since the first 'add' method is overwritten.\n\n### 14. What does ‘method overriding’ mean?\n\nMethod overriding in OOP occurs when a subclass provides a specific implementation for a method that is already defined in its superclass. The overridden method in the subclass takes precedence over the method in the superclass.\n\n### 15. What is a superclass and a subclass?\n\n• A superclass is a class that is extended or inherited by other classes. It provides common attributes and behaviors that its subclasses can utilize.\n• A subclass is a class that inherits from a superclass. It extends the superclass by adding specific attributes and behaviors or overriding existing ones.\nPython\n``````# Superclass\nclass Animal:\ndef speak(self):\nreturn \"Generic animal sound.\"\n\n# Subclass\nclass Dog(Animal):\ndef speak(self):\nreturn \"Woof!\"\n\ndog = Dog()\nprint(dog.speak()) # Output: Woof!``````\n\n### 16. What is an interface?\n\nIn Python, there is no explicit interface keyword as in some other languages. Instead, interfaces are achieved through abstract base classes (ABCs) using the `ABC` module.\n\nPython\n``````from abc import ABC, abstractmethod\n\nclass Shape(ABC):\n@abstractmethod\ndef area(self):\npass\n\nclass Circle(Shape):\n\ndef area(self):\n\ncircle = Circle(5)\nprint(circle.area()) # Output: 78.5``````\n\n### 18. What is a static method in OOP?\n\nA static method in OOP is a method that belongs to the class rather than an instance of the class. It does not have access to instance-specific data.\n\nPython\n``````class MathOperations:\n@staticmethod\ndef multiply(a, b):\nreturn a * b\n\nresult = MathOperations.multiply(3, 4)\nprint(result) # Output: 12``````\n\n### 19. What are access modifiers in OOP?\n\nAccess modifiers in OOP define the level of visibility and access to attributes and methods within a class.\n\n• `public`: No restriction, can be accessed from anywhere.\n• `protected`: Accessible within the class and its subclasses.\n• `private`: Accessible only within the class.\nPython\n``````class MyClass:\ndef __init__(self):\nself.public_attr = \"I am public.\"\nself._protected_attr = \"I am protected.\"\nself.__private_attr = \"I am private.\"\n\nobj = MyClass()\n\nprint(obj.public_attr) # Output: I am public.\nprint(obj._protected_attr) # Output: I am protected.\nprint(obj.__private_attr) # This will raise an AttributeError.``````\n\n## Intermediate Questions\n\n### 1. What are the different types of inheritance in OOP?\n\nThere are five types of inheritance in Object-Oriented Programming:\n\n1. Single Inheritance\n2. Multiple Inheritance\n3. Multilevel Inheritance\n4. Hierarchical Inheritance\n5. Hybrid Inheritance\nPython\n``````# Example of Single Inheritance\nclass Animal:\ndef speak(self):\npass\n\nclass Dog(Animal):\ndef speak(self):\nreturn \"Woof!\"\n\n# Example of Multiple Inheritance\nclass A:\ndef method_a(self):\npass\n\nclass B:\ndef method_b(self):\npass\n\nclass C(A, B):\ndef method_c(self):\npass\n\n# Other types of inheritance (Multilevel, Hierarchical, Hybrid) can also be shown similarly.``````\n\n### 2. Explain the concept of multiple inheritances. How is it handled in languages that do not directly support it, like Java?\n\nMultiple Inheritance is the ability of a class to inherit properties and behaviors from more than one parent class. Java does not directly support multiple inheritance for classes, but it can be achieved using interfaces.\n\nJava\n``````interface A {\nvoid methodA();\n}\n\ninterface B {\nvoid methodB();\n}\n\nclass MyClass implements A, B {\n@Override\npublic void methodA() {\n// Implementation\n}\n\n@Override\npublic void methodB() {\n// Implementation\n}\n}``````\n\n### 3. What is the diamond problem in multiple inheritance and how can it be resolved?\n\nThe Diamond Problem occurs when a class inherits from two classes, both of which have a common ancestor. It leads to ambiguity in the inheritance hierarchy. In some languages, like C++, this is an issue.\n\nC++\n``````class A {\npublic:\nvoid foo() {}\n};\n\nclass B : public A {};\n\nclass C : public A {};\n\nclass D : public B, public C {};\n// Here, class D inherits from both B and C, which both inherit from A.\n\nint main() {\nD d;\nd.foo(); // Ambiguity: Which 'foo' to call, B's or C's?\n\nreturn 0;\n}``````\n\nTo resolve the Diamond Problem, virtual inheritance can be used in C++.\n\nC++\n``````class A {\npublic:\nvoid foo() {}\n};\n\nclass B : virtual public A {};\n\nclass C : virtual public A {};\n\nclass D : public B, public C {};\n// Virtual inheritance ensures only one instance of A in D.\n\nint main() {\nD d;\nd.foo(); // No ambiguity now.\n\nreturn 0;\n}``````\n\n### 4. Explain the Liskov Substitution Principle (LSP).\n\nLSP states that objects of a superclass should be replaceable with objects of its subclasses without affecting the correctness of the program.\n\n### 5. What is the Open Closed Principle (OCP)?\n\nOCP states that software entities (classes, modules, functions, etc.) should be open for extension but closed for modification. It encourages the use of abstraction and polymorphism to allow easy extension without changing existing code.\n\n### 6. What is the Single Responsibility Principle (SRP)?\n\nSRP states that a class should have only one reason to change, i.e., it should have only one responsibility.\n\n### 7. What is the Interface Segregation Principle (ISP)?\n\nISP states that a client should not be forced to implement interfaces it does not use. It promotes smaller, specific interfaces rather than large, general interfaces.\n\n### 8. What is the Dependency Inversion Principle (DIP)?\n\nDIP states that high-level modules should not depend on low-level modules, but both should depend on abstractions. It encourages the use of dependency injection.\n\n### 10. What is a virtual function?\n\nA virtual function allows dynamic binding or late binding, which enables the correct function to be called based on the object at runtime.\n\nC++\n``````class Base {\npublic:\nvirtual void show() {\ncout << \"Base class\" << endl;\n}\n};\n\nclass Derived : public Base {\npublic:\nvoid show() override {\ncout << \"Derived class\" << endl;\n}\n};\n\nint main() {\nBase* ptr;\nDerived d;\nptr = &d;\nptr->show(); // Output: \"Derived class\"\n\nreturn 0;\n}``````\n\n### 11. What are pure virtual functions?\n\nA pure virtual function is a virtual function with no implementation in the base class, making it an abstract function. Classes containing pure virtual functions are abstract and cannot be instantiated.\n\nC++\n``````class Shape {\npublic:\nvirtual void draw() = 0; // Pure virtual function\n};\n\nclass Circle : public Shape {\npublic:\nvoid draw() override {\ncout << \"Drawing a circle\" << endl;\n}\n};\n\nint main() {\nCircle c;\nc.draw(); // Output: \"Drawing a circle\"\n\n// Shape s; // Error: Cannot instantiate abstract class\n\nreturn 0;\n}``````\n\n### 13. Explain what friend classes and friend functions are.\n\nA friend class or function can access private and protected members of another class.\n\nC++\n``````class A {\nprivate:\nint x;\nfriend class B; // B is a friend class of A\n\npublic:\nA() : x(10) {}\n};\n\nclass B {\npublic:\nvoid display(A obj) {\ncout << \"Value of x in A: \" << obj.x << endl;\n}\n};\n\nint main() {\nA a;\nB b;\nb.display(a); // Output: \"Value of x in A: 10\"\n\nreturn 0;\n}``````\n\nC++\n``````class Complex {\nprivate:\ndouble real;\ndouble imag;\n\npublic:\nComplex(double r, double i) : real(r), imag(i) {}\n\nComplex operator+(const Complex& other) {\nreturn Complex(real + other.real, imag + other.imag);\n}\n};\n\nint main() {\nComplex a(2.0, 3.5);\nComplex b(1.5, 2.0);\n// c\n\n.real = 3.5, c.imag = 5.5\n\nreturn 0;\n}``````\n\n### 15. What is a copy constructor?\n\nA copy constructor creates a new object by copying the content of an existing object.\n\nC++\n``````class MyClass {\nprivate:\nint data;\n\npublic:\nMyClass(int d) : data(d) {}\n\n// Copy constructor\nMyClass(const MyClass& other) : data(other.data) {}\n};\n\nint main() {\nMyClass obj1(42);\nMyClass obj2 = obj1; // Uses the copy constructor\n\nreturn 0;\n}``````\n\n### 16. What is the Difference Between ‘Pass by Value’ and ‘Pass by Reference’\n\nC++\n``````// Pass by Value\nvoid modifyValue(int val) {\nval = val * 2; // Changes only the local copy\n}\n\nint main() {\nint num = 5;\nmodifyValue(num);\n// num still remains 5\n\nreturn 0;\n}\n\n// Pass by Reference\nvoid modifyReference(int& val) {\nval = val * 2; // Changes the original value\n}\n\nint main() {\nint num = 5;\nmodifyReference(num);\n// num is now 10\n\nreturn 0;\n}``````\n\n### 17. What is a template in OOP?\n\nTemplates allow the creation of generic classes or functions that can work with different data types.\n\nC++\n``````template <typename T>\nT add(T a, T b) {\nreturn a + b;\n}\n\nint main() {\nint sum1 = add(3, 5); // sum1 = 8\ndouble sum2 = add(1.5, 2.3); // sum2 = 3.8\n\nreturn 0;\n}``````\n\n### 18. Explain the term ‘aggregation’.\n\nAggregation represents a “has-a” relationship between classes, where one class contains another as a part.\n\nC++\n``````class Address {\n};\n\nclass Person {\nprivate:\n\npublic:\n};``````\n\n### 19. Explain the term ‘composition’.\n\nComposition represents a stronger form of aggregation, where one class is a part of another and has a lifecycle dependent on the containing class.\n\nC++\n``````class Engine {\n// Engine details\n};\n\nclass Car {\nprivate:\nEngine engine; // Composition: Car has an Engine\n\npublic:\nCar() : engine(Engine()) {}\n};``````\n\n### 20. What is a ‘this’ pointer? What is its use?\n\n‘this’ pointer is a pointer that refers to the current instance of the class. It is used to access members of the class when they have the same names as function parameters or to return the current object from member functions.\n\nC++\n``````class MyClass {\nprivate:\nint value;\n\npublic:\nvoid setValue(int value) {\nthis->value = value; // 'this' refers to the object's member\n}\n\nint getValue() {\nreturn this->value;\n}\n};\n\nint main() {\nMyClass obj;\nobj.setValue(42);\ncout << obj.getValue(); // Output: 42\n\nreturn 0;\n}``````\n\n### 1. Can you describe a situation where it would be appropriate to use a “friend” class or function in C++?\n\nUsing the “friend” keyword in C++ allows a class or function to access private members of another class. It should be used sparingly and with caution, as it breaks encapsulation. An appropriate situation to use “friend” is when you have a utility function or class that needs access to the private data of another class for efficiency reasons.\n\nExample:\n\nC++\n``````class MySecretClass {\nprivate:\nint secretValue;\n\npublic:\nMySecretClass(int value) : secretValue(value) {}\n\nfriend int getSecretValue(const MySecretClass& obj);\n};\n\nint getSecretValue(const MySecretClass& obj) {\nreturn obj.secretValue;\n}\n\nint main() {\nMySecretClass secretObj(42);\nint value = getSecretValue(secretObj);\n// value will be 42, accessed the private member using a friend function.\nreturn 0;\n}``````\n\n### 2. How does polymorphism promote extensibility?\n\nPolymorphism allows objects of different classes to be treated as objects of a common base class, enabling dynamic dispatch of function calls based on the actual type of the object. This promotes extensibility, as new derived classes can be added without modifying existing code, making it easier to introduce new behaviors.\n\nExample:\n\nC++\n``````class Shape {\npublic:\nvirtual void draw() const = 0;\n};\n\nclass Circle : public Shape {\npublic:\nvoid draw() const override {\n// Draw circle here.\n}\n};\n\nclass Square : public Shape {\npublic:\nvoid draw() const override {\n// Draw square here.\n}\n};\n\nvoid drawShape(const Shape& shape) {\nshape.draw(); // Calls the appropriate draw() based on the actual object type.\n}\n\nint main() {\nCircle circle;\nSquare square;\n\ndrawShape(circle); // Draws a circle.\ndrawShape(square); // Draws a square.\n\nreturn 0;\n}``````\n\n### 3. What are virtual destructors, and when would you use one?\n\nA virtual destructor is a destructor declared with the “virtual” keyword in the base class. When you have a class hierarchy and intend to delete derived class objects through a pointer to the base class, the base class’s destructor should be made virtual to ensure proper cleanup of resources.\n\nExample:\n\nC++\n``````class Base {\npublic:\nvirtual ~Base() {\n// Virtual destructor allows proper cleanup when deleting through base pointer.\n}\n};\n\nclass Derived : public Base {\npublic:\n~Derived() {\n// Derived destructor code here.\n}\n};\n\nint main() {\nBase* ptr = new Derived();\n// Use the object through the base pointer.\n\ndelete ptr; // Calls the proper destructors (Derived and then Base).\nreturn 0;\n}``````\n\n### 4. Explain the concept of pure virtual functions. What is an abstract class, and when should it be used?\n\nA pure virtual function is a virtual function in the base class that is declared with “= 0” and has no implementation. A class containing at least one pure virtual function becomes an abstract class, and objects cannot be created directly from it. Instead, it serves as a base for derived classes to implement the pure virtual functions.\n\nExample:\n\nC++\n``````class Shape {\npublic:\nvirtual void draw() const = 0; // Pure virtual function.\n\n// Non-pure virtual functions are optional in an abstract class.\nvirtual void printArea() const {\nstd::cout << \"Area: \" << area() << std::endl;\n}\n\n// Abstract classes can have regular functions with implementation.\ndouble area() const {\nreturn 0.0; // Base implementation; will be overridden in derived classes.\n}\n};\n\nclass Circle : public Shape {\npublic:\nvoid draw() const override {\n// Draw circle here.\n}\n\ndouble area() const override {\n// Calculate circle area here.\n}\n};\n\nclass Square : public Shape {\npublic:\nvoid draw() const override {\n// Draw square here.\n}\n\ndouble area() const override {\n// Calculate square area here.\n}\n};\n\nint main() {\n// Shape* shape = new Shape(); // Error, cannot create an instance of an abstract class.\nShape* circle = new Circle();\nShape* square = new Square();\n\ncircle->draw();\ncircle->printArea();\n\nsquare->draw();\nsquare->printArea();\n\ndelete circle;\ndelete square;\nreturn 0;\n}``````\n\n### 5. What is the difference between composition and aggregation?\n\nComposition and aggregation are both forms of association between classes, but they differ in ownership and lifetime relationships.\n\nComposition implies a strong ownership relationship where the lifetime of the part (member object) is controlled by the whole (containing object). In aggregation, the containing object may reference or use the aggregated object, but it doesn’t have sole responsibility for the aggregated object’s lifetime.\n\nExample of Composition:\n\nC++\n``````class Engine {\n// Engine class implementation.\n};\n\nclass Car {\nprivate:\nEngine engine; // Car owns an Engine object.\n\npublic:\n// Car class implementation.\n};``````\n\nExample of Aggregation:\n\nC++\n``````class Person {\n// Person class implementation.\n};\n\nclass Company {\nprivate:\nstd::vector<Person> employees; // Company references Person objects, but doesn't own them.\n\npublic:\n// Company class implementation.\n};``````\n\n### 6. What is the use of the “volatile” keyword in C++? What implications does it have on compiler optimizations?\n\nThe “volatile” keyword is used to indicate that a variable can be modified by external sources not explicitly known to the compiler. It informs the compiler not to perform any optimizations (e.g., caching) that could interfere with the variable’s actual value.\n\nExample:\n\nC++\n``````volatile int sensorValue; // Declaring a volatile variable.\n\n// Assume some external hardware updates the sensorValue.\n// Without volatile, the compiler might optimize this loop as the value doesn't change in the loop.\nwhile (sensorValue < 100) {\n// Do something based on sensorValue.\n}\n}``````\n\n### 7. Explain the diamond problem in multiple inheritance. How is it addressed in C++ and in other OOP languages like Java?\n\nThe diamond problem occurs when a class inherits from two classes, and both those classes inherit from a common base class. This creates an ambiguity in the hierarchy, as the derived class has two copies of the base class, leading to conflicts in member access.\n\nExample:\n\nC++\n``````class A {\npublic:\nvoid doSomething() { / A's implementation / }\n};\n\nclass B : public A {\n};\n\nclass C : public A {\n};\n\nclass D : public B, public C {\n};\n\nint main() {\nD obj;\n// obj.doSomething(); // Error, ambiguity due to diamond problem.\nreturn 0;\n}``````\n\nIn C++, this problem is addressed by using virtual inheritance.\n\nThe virtual keyword is used to specify that there should be only one instance of the base class shared by all the derived classes.\n\nC++\n``````class A {\npublic:\nvoid doSomething() { / A's implementation / }\n};\n\nclass B : virtual public A {\n};\n\nclass C : virtual public A {\n};\n\nclass D : public B, public C {\n};\n\nint main() {\nD obj;\nobj.doSomething(); // No ambiguity due to virtual inheritance.\nreturn 0;\n}``````\n\nJava automatically addresses the diamond problem by allowing a class to inherit from only one concrete class (single inheritance) but supporting multiple interface implementations.\n\n### 8. In what scenario would you use multiple inheritance instead of interfaces?\n\nMultiple inheritance is used when a class needs to inherit functionality from multiple base classes, and those base classes provide implementation along with data. If the goal is to share the implementation code among multiple classes, multiple inheritance is preferred.\n\nExample:\n\nC++\n``````class Shape {\npublic:\nvoid draw() const {\n// Common draw code for all shapes.\n}\n};\n\nclass Color {\npublic:\nvoid setColor(int r, int g, int b) {\n// Set color implementation.\n}\n};\n\nclass ColoredShape : public Shape, public Color {\n// ColoredShape inherits both draw() from Shape and setColor() from Color.\n};\n\nint main() {\nColoredShape cShape;\ncShape.draw();\ncShape.setColor(255, 0, 0);\nreturn 0;\n}``````\n\n### 9. What is a covariant return type in Java?\n\nCovariant return types allow a method in a subclass to return a more specific type than the method in the superclass. This feature was introduced in Java 5 to improve readability and make the code more concise.\n\nExample:\n\nJava\n``````class Fruit {\n// Fruit class implementation.\n}\n\nclass Apple extends Fruit {\n// Apple class implementation.\n}\n\npublic Fruit getFruit() {\nreturn new Fruit();\n}\n}\n\n@Override\npublic Apple getFruit() {\nreturn new Apple(); // Covariant return type allows returning a more specific type.\n}\n}``````\n\n### 10. How does the Law of Demeter apply to Object-Oriented design?\n\nThe Law of Demeter (LoD) states that an object should only interact with its immediate connections and not with the objects further away in the class hierarchy. In simpler terms, it encourages limiting the number of direct dependencies between classes to reduce coupling and make the code easier to maintain and understand.\n\nExample:\n\nC++\n``````class Engine {\npublic:\nvoid start() {\n// Engine start implementation.\n}\n};\n\nclass Car {\nprivate:\nEngine engine;\n\npublic:\nvoid startCar() {\nengine.start(); // Following the Law of Demeter, Car should interact with its immediate component (Engine).\n}\n};``````\n\n### 11. How can you implement Object-Oriented Programming in languages that are not inherently object-oriented?\n\nIn languages that are not inherently object-oriented, you can still simulate object-oriented programming using structures or records to represent objects and functions to model methods. You’ll need to manually manage memory and dispatch function calls.\n\nExample in C:\n\nC++\n``````typedef struct {\nint length;\nint width;\n} Rectangle;\n\nvoid drawRectangle(Rectangle rect) {\n// Drawing code for the rectangle.\n}\n\nint calculateArea(Rectangle* rect) {\nreturn rect->length * rect->width;\n}\n\nint main() {\nRectangle myRect = {10, 5};\ndrawRectangle(&myRect);\nint area = calculateArea(&myRect);\n// Use area and perform other operations.\nreturn 0;\n}``````\n\n### 12. In Java, is it possible to inherit from two classes? If not, how would you work around this?\n\nNo, in Java, a class can only inherit from a single direct superclass (single inheritance). However, you can achieve similar functionality by using interfaces, which allow a class to implement multiple interfaces and provide implementation for their methods.\n\nExample:\n\nJava\n``````interface Shape {\nvoid draw();\n}\n\ninterface Color {\nvoid setColor(int r, int g, int b);\n}\n\nclass ColoredShape implements Shape, Color {\n@Override\npublic void draw() {\n// Draw the colored shape.\n}\n\n@Override\npublic void setColor(int r, int g, int b) {\n// Set color implementation.\n}\n}``````\n\n### 13. What is operator overloading, and what are some good practices and pitfalls when using it?\n\nOperator overloading allows you to define custom behaviors for operators when used with user-defined types. It can improve code readability and make the class interface more intuitive. However, it should be used with caution to avoid confusing code and ensure consistency with the standard meaning of operators.\n\nExample:\n\nC++\n``````class ComplexNumber {\npublic:\ndouble real;\ndouble imag;\n\nComplexNumber(double r, double i) : real(r), imag(i) {}\n\nComplexNumber operator+(const ComplexNumber& other) const {\nreturn ComplexNumber(real + other.real, imag + other.imag);\n}\n\n// Other operator overloads, such as -, , /, etc., can also be implemented.\n};\n\nint main() {\nComplexNumber num1(3.0, 4.0);\nComplexNumber num2(1.5, 2.5);\nComplexNumber result = num1 + num2; // Uses the overloaded + operator.\n\nreturn 0;\n}``````\n\nGood practices:\n\n• Overload operators to maintain their intuitive meaning.\n• Provide consistent behavior with built-in types.\n\nPitfalls:\n\n• Overusing operator overloading, which can make the code harder to understand.\n\n### 14. Can you explain a practical example of when and why you would use a nested or inner class?\n\nNested or inner classes are used when one class should be closely related to and used only within another class. It provides better encapsulation and helps to organize the code better.\n\nExample:\n\nJava\n``````class Outer {\nprivate int outerData;\n\nclass Inner {\nprivate int innerData;\n\nInner(int innerData) {\nthis.innerData = innerData;\n}\n\nint getSum() {\nreturn outerData + innerData; // Inner class can access outer class members.\n}\n}\n\nOuter(int outerData) {\nthis.outerData = outerData;\n}\n\nInner createInner(int innerData) {\nreturn new Inner(innerData);\n}\n}\n\npublic class Main {\npublic static void main(String[] args) {\nOuter outerObj = new Outer(10);\nOuter.Inner innerObj = outerObj.createInner(5);\nint sum = innerObj.getSum(); // Accessing inner class method.\n\nSystem.out.println(\"Sum: \" + sum); // Output: Sum: 15\n}\n}``````\n\nIn this example, the `Inner` class is closely related to the `Outer` class and requires access to its members. It’s used to perform a specific calculation involving both `Inner` and `Outer` data. By using an inner class, we keep the code more organized and encapsulate the `Inner` class’s functionality within the context of the `Outer` class.\n\n### 15. What is the problem with multiple inheritance, and how does the “interface” concept help resolve this issue?\n\nThe problem with multiple inheritance is the potential for the diamond problem, where a class inherits from two or more classes, which in turn inherit from the same base class. This creates ambiguity and can lead to conflicts in member access and function calls.\n\nThe “interface” concept, like in Java, helps to resolve this issue by providing a way to achieve multiple inheritance of type without the inheritance of implementation. An interface specifies a set of methods that a class implementing the interface must provide, ensuring a contract for behavior without introducing conflicts.\n\nExample:\n\nJava\n``````interface Animal {\nvoid makeSound();\n}\n\ninterface Flyable {\nvoid fly();\n}\n\nclass Bird implements Animal, Flyable {\n@Override\npublic void makeSound() {\nSystem.out.println(\"Chirp Chirp\");\n}\n\n@Override\npublic void fly() {\nSystem.out.println(\"Bird is flying\");\n}\n}\n\nclass Bat implements Animal, Flyable {\n@Override\npublic void makeSound() {\nSystem.out.println(\"Screech\");\n}\n\n@Override\npublic void fly() {\nSystem.out.println(\"Bat is flying\");\n}\n}\n\npublic class Main {\npublic static void main(String[] args) {\nBird bird = new Bird();\nbird.makeSound();\nbird.fly();\n\nBat bat = new Bat();\nbat.makeSound();\nbat.fly();\n}\n}``````\n\nIn this example, both `Bird` and `Bat` classes implement both `Animal` and `Flyable` interfaces, which specify the behavior they must provide. This allows us to achieve multiple inheritance of type without running into the diamond problem.\n\n### 16. What is the use of a private constructor? When would you use it?\n\nA private constructor is used to prevent the creation of objects of a class directly by users. It is often used in the singleton design pattern to ensure that only one instance of the class can be created.\n\nExample of Singleton using a private constructor:\n\nC++\n``````class Singleton {\nprivate:\nstatic Singleton instance;\nSingleton() {} // Private constructor.\n\npublic:\nstatic Singleton* getInstance() {\nif (!instance) {\ninstance = new Singleton();\n}\nreturn instance;\n}\n};\n\nSingleton* Singleton::instance = nullptr;\n\nint main() {\nSingleton* singleton1 = Singleton::getInstance();\nSingleton* singleton2 = Singleton::getInstance();\n\n// Both pointers will point to the same instance.\nif (singleton1 == singleton2) {\nstd::cout << \"Same instance.\" << std::endl;\n}\n\nreturn 0;\n}``````\n\nIn this example, the `Singleton` class has a private constructor, and the only way to get an instance of it is through the `getInstance()` static method, which ensures that only one instance is created and returned.\n\n### 17. What is a virtual function, and why is it important in a base class?\n\nA virtual function is a function declared in the base class with the `virtual` keyword. It allows the function to be overridden in derived classes, enabling dynamic dispatch of function calls based on the actual object type.\n\nExample:\n\nC++\n``````class Shape {\npublic:\nvirtual void draw() const {\n// Base implementation for draw().\n}\n};\n\nclass Circle : public Shape {\npublic:\nvoid draw() const override {\n// Circle-specific draw implementation.\n}\n};\n\nclass Square : public Shape {\npublic:\nvoid draw() const override {\n// Square-specific draw implementation.\n}\n};\n\nint main() {\nCircle circle;\nSquare square;\n\nShape* shape1 = &circle;\nShape* shape2 = &square\n\nshape1->draw(); // Calls the draw() function of the Circle class.\nshape2->draw(); // Calls the draw() function of the Square class.\n\nreturn 0;\n}``````\n\nIn this example, the `draw()` function in the base class `Shape` is virtual, allowing it to be overridden in the derived classes `Circle` and `Square`. The function call is resolved dynamically at runtime based on the actual object type, facilitating polymorphism.\n\n### 18. In the context of Object-Oriented design, what does SOLID stand for?\n\nSOLID is an acronym for five principles of Object-Oriented Design, aimed at creating maintainable and scalable software.\n\n1. Single Responsibility Principle (SRP): A class should have only one reason to change, meaning it should have only one responsibility.\n\nExample:\n\nC++\n``````class Shape {\npublic:\nvoid draw() const {\n// Draw the shape.\n}\n\ndouble calculateArea() const {\n// Calculate the area of the shape.\n}\n};``````\n1. Open/Closed Principle (OCP): Software entities should be open for extension but closed for modification. New functionality should be added without altering existing code.\n\nExample:\n\nC++\n``````class Shape {\npublic:\nvirtual double calculateArea() const = 0; // Open for extension through derived classes.\n};``````\n1. Liskov Substitution Principle (LSP): Subtypes should be substitutable for their base types without affecting the program’s correctness.\n\nExample:\n\nC++\n``````class Rectangle {\npublic:\nvirtual void setWidth(int width) { /* Implementation / }\nvirtual void setHeight(int height) { / Implementation / }\n};\n\nclass Square : public Rectangle {\npublic:\nvoid setWidth(int width) override { / Implementation / }\nvoid setHeight(int height) override { / Implementation / }\n};``````\n1. Interface Segregation Principle (ISP): Clients should not be forced to depend on interfaces they do not use. It’s better to have specific interfaces rather than one large interface.\n\nExample:\n\nC++\n``````class Printer {\npublic:\nvirtual void print() = 0;\n};\n\nclass Scanner {\npublic:\nvirtual void scan() = 0;\n};\n\nclass MultiFunctionDevice : public Printer, public Scanner {\npublic:\nvoid print() override { / Implementation / }\nvoid scan() override { / Implementation / }\n};``````\n1. Dependency Inversion Principle (DIP): High-level modules should not depend on low-level modules; both should depend on abstractions. Abstractions should not depend on details; details should depend on abstractions.\n\nExample:\n\nC++\n``````class Database {\n// Implementation details.\n};\n\nclass UserRepository {\nprivate:\nDatabase db; // Dependency on the Database implementation.\n\npublic:\nUserRepository(Database* db) : db(db) {}\n\n// Repository methods using the Database.\n};``````\n\nIn this example, the `UserRepository` is directly dependent on the `Database` implementation, violating the DIP. To follow the principle, we can introduce an abstraction, such as an interface or a base class, to decouple the high-level module from the low-level details.\n\n### 19. Explain the concept of Reflection in Java and C#.\n\nReflection is a feature in Java and C# that allows a program to examine its own structure at runtime. It enables the inspection and manipulation of classes, methods, fields, and interfaces, even if they were not known at compile-time.\n\nExample in Java:\n\nJava\n``````import java.lang.reflect.*;\n\nclass MyClass {\nprivate int value;\npublic void setValue(int value) { this.value = value; }\npublic int getValue() { return value; }\n}\n\npublic class Main {\npublic static void main(String[] args) {\nMyClass obj = new MyClass();\n\n// Using Reflection to get and set the private field \"value\".\ntry {\nField field = obj.getClass().getDeclaredField(\"value\");\nfield.setAccessible(true);\n\nfield.setInt(obj, 42);\nSystem.out.println(\"Value: \" + field.getInt(obj));\n} catch (NoSuchFieldException \| IllegalAccessException e) {\ne.printStackTrace();\n}\n}\n}``````\n\nIn this example, we use Reflection to access and modify the private field “value” in the `MyClass` instance, which we wouldn’t normally be able to do directly.\n\n### 20. What are some issues with the OOP paradigm, and how might they be resolved?\n\nSome issues with the OOP paradigm include:\n\n• Overuse of inheritance: Deep inheritance hierarchies can lead to tight coupling and difficult-to-maintain code. Consider favoring composition and interfaces over inheritance when possible.\n• Lack of flexibility in some languages: Some languages don’t support multiple inheritance or don’t provide enough features for modern OOP. In such cases, using design patterns and interfaces can help work around these limitations.\n• Performance overhead: Object-oriented designs can sometimes introduce performance overhead due to dynamic dispatch and object creation. In performance-critical scenarios, consider using more optimized data structures and algorithms.\n• Complexity and overengineering: Overusing design patterns and creating excessively abstract class hierarchies can lead to unnecessary complexity. Follow the KISS (Keep It Simple, Stupid) principle and YAGNI (You Ain’t Gonna Need It) to avoid overengineering.\n• Encapsulation violations: Improper use of access modifiers can lead to encapsulation violations, breaking the principle of information hiding. Ensure that classes expose only necessary interfaces and keep their internal details hidden.\n\n## MCQ Questions\n\n### 1. What is OOPs?\n\na) Object-Oriented Programming System\nb) Object-Oriented Programming Structure\nc) Object-Oriented Programming Style\nd) Object-Oriented Programming Syntax\n\na) Encapsulation\nb) Inheritance\nc) Polymorphism\nd) Compilation\n\n### 3. What is the main purpose of encapsulation in OOPs?\n\na) Data hiding\nb) Code reusability\nc) Code organization\nd) Code optimization\n\na) “is a”\nb) “has a”\nc) “uses a”\nd) “contains a”\n\na) this\nb) new\nc) abstract\nd) final\n\n### 6. What is the purpose of the super keyword in Java?\n\na) Accessing the superclass methods and variables\nb) Initializing the superclass object\nc) Calling the constructor of the superclass\nd) Modifying the superclass behavior\n\nAnswer: a) Accessing the superclass methods and variables\n\na) Inheritance\nb) Encapsulation\nc) Polymorphism\nd) Abstraction\n\na) private\nb) protected\nc) public\nd) default\n\n### 9. What is the purpose of the final keyword in Java?\n\na) It is used to define a constant value.\nb) It is used to prevent inheritance.\nc) It is used to prevent method overriding.\nd) All of the above.\n\nAnswer: d) All of the above.\n\n### 10. Which of the following is an example of runtime polymorphism in Java?\n\nb) Method overriding\nc) Method hiding\nd) Method abstracting\n\na) Inheritance\nb) Encapsulation\nc) Polymorphism\nd) Interface\n\n### 12. What is the output of the following code?\n\nJava\n``````class A {\nvoid display() {\nSystem.out.println(\"Class A\");\n}\n}\n\nclass B extends A {\nvoid display() {\nSystem.out.println(\"Class B\");\n}\n}\n\npublic class Main {\npublic static void main(String[] args) {\nA obj = new B();\nobj.display();\n}\n}``````\n\na) Class A\nb) Class B\nc) Compile-time error\nd) Runtime error\n\na) static\nb) final\nc) abstract\nd) private\n\na) Inheritance\nb) Encapsulation\n\nc) Polymorphism\nd) Abstraction\n\na) Aggregation\nb) Composition\nc) Inheritance\nd) Association\n\n### 16. What is the output of the following code?\n\nJava\n``````interface MyInterface {\ndefault void display() {\nSystem.out.println(\"Interface\");\n}\n}\n\nclass MyClass implements MyInterface {\npublic static void main(String[] args) {\nMyClass obj = new MyClass();\nobj.display();\n}\n}``````\n\na) Interface\nb) MyClass\nc) Compile-time error\nd) Runtime error\n\na) Abstraction\nb) Encapsulation\nc) Polymorphism\nd) Inheritance\n\na) _variable\nb) 123abc\nc) \\$name\nd) firstName\n\na) this\nb) new\nc) class\nd) void\n\n### 20. Which of the following is not a type of exception in Java?\n\na) Checked exception\nb) Unchecked exception\nc) Runtime exception\nd) Custom exception\n\nCheck Also\nClose" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7385756,"math_prob":0.68974626,"size":42773,"snap":"2023-40-2023-50","text_gpt3_token_len":9686,"char_repetition_ratio":0.14564288,"word_repetition_ratio":0.115185566,"special_character_ratio":0.24424286,"punctuation_ratio":0.16281185,"nsfw_num_words":4,"has_unicode_error":false,"math_prob_llama3":0.9526269,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-21T18:44:11Z\",\"WARC-Record-ID\":\"<urn:uuid:6616c747-b49f-48c7-8d6a-f4593cea0973>\",\"Content-Length\":\"968680\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c4ef37fa-35ce-4fa3-85f5-9d2e7348802e>\",\"WARC-Concurrent-To\":\"<urn:uuid:8729e7b9-bf86-4bc9-91cc-6b1544fdb386>\",\"WARC-IP-Address\":\"172.67.212.88\",\"WARC-Target-URI\":\"https://codinginterviewpro.com/interview-questions-and-answers/all-new-60-oops-interview-questions/\",\"WARC-Payload-Digest\":\"sha1:AWFY4YLJ5I5PZ5FL25H3BZ2TLDNUKZQV\",\"WARC-Block-Digest\":\"sha1:47P7TYZJOKIAGEX6PB4JQXJKG65PASB6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233506029.42_warc_CC-MAIN-20230921174008-20230921204008-00444.warc.gz\"}"}
https://www.openml.org/d/42897	[ "Data\nbias-correction\n\n# bias-correction\n\nactive ARFF CC-BY Visibility: public Uploaded 19-05-2021 by Meilina Reksoprodjo\nIssue #Downvotes for this reason By\n\nAuthor: Dongjin Cho,Cheolhee Yoo Source: [UCI](https://archive.ics.uci.edu/ml/datasets/Bias+correction+of+numerical+prediction+model+temperature+forecast) - 2020 Please cite: [Paper](https://www.semanticscholar.org/paper/Comparative-Assessment-of-Various-Machine-Bias-for-Cho-Yoo/a24c181f054e35e8caf4fff78949f20569cb11b2 Bias correction of numerical prediction model temperature forecast dataset This data is for the purpose of bias correction of next-day maximum and minimum air temperatures forecast of the LDAPS model operated by the Korea Meteorological Administration over Seoul, South Korea. This data consists of summer data from 2013 to 2017. The input data is largely composed of the LDAPS model's next-day forecast data, in-situ maximum and minimum temperatures of present-day, and geographic auxiliary variables. There are two outputs (i.e. next-day maximum and minimum air temperatures) in this data. Hindcast validation was conducted for the period from 2015 to 2017. ### Attribute information 1. station - used weather station number: 1 to 25 2. Date - Present day: yyyy-mm-dd ('2013-06-30' to '2017-08-30') 3. Present_Tmax - Maximum air temperature between 0 and 21 h on the present day (degrees Celsius): 20 to 37.6 4. Present_Tmin - Minimum air temperature between 0 and 21 h on the present day (degrees Celsius): 11.3 to 29.9 5. LDAPS_RHmin - LDAPS model forecast of next-day minimum relative humidity (%): 19.8 to 98.5 6. LDAPS_RHmax - LDAPS model forecast of next-day maximum relative humidity (%): 58.9 to 100 7. LDAPS_Tmax_lapse - LDAPS model forecast of next-day maximum air temperature applied lapse rate (degrees Celsius): 17.6 to 38.5 8. LDAPS_Tmin_lapse - LDAPS model forecast of next-day minimum air temperature applied lapse rate (degrees Celsius): 14.3 to 29.6 9. LDAPS_WS - LDAPS model forecast of next-day average wind speed (m/s): 2.9 to 21.9 10. LDAPS_LH - LDAPS model forecast of next-day average latent heat flux (W/m2): -13.6 to 213.4 11. LDAPS_CC1 - LDAPS model forecast of next-day 1st 6-hour split average cloud cover (0-5 h) (%): 0 to 0.97 12. LDAPS_CC2 - LDAPS model forecast of next-day 2nd 6-hour split average cloud cover (6-11 h) (%): 0 to 0.97 13. LDAPS_CC3 - LDAPS model forecast of next-day 3rd 6-hour split average cloud cover (12-17 h) (%): 0 to 0.98 14. LDAPS_CC4 - LDAPS model forecast of next-day 4th 6-hour split average cloud cover (18-23 h) (%): 0 to 0.97 15. LDAPS_PPT1 - LDAPS model forecast of next-day 1st 6-hour split average precipitation (0-5 h) (%): 0 to 23.7 16. LDAPS_PPT2 - LDAPS model forecast of next-day 2nd 6-hour split average precipitation (6-11 h) (%): 0 to 21.6 17. LDAPS_PPT3 - LDAPS model forecast of next-day 3rd 6-hour split average precipitation (12-17 h) (%): 0 to 15.8 18. LDAPS_PPT4 - LDAPS model forecast of next-day 4th 6-hour split average precipitation (18-23 h) (%): 0 to 16.7 19. lat - Latitude (degrees): 37.456 to 37.645 20. lon - Longitude (degrees): 126.826 to 127.135 21. DEM - Elevation (m): 12.4 to 212.3 22. Slope - Slope (degrees): 0.1 to 5.2 23. Solar radiation - Daily incoming solar radiation (wh/m2): 4329.5 to 5992.9 24. Next_Tmax - The next-day maximum air temperature (degrees Celsius): 17.4 to 38.9 25. Next_Tmin - The next-day minimum air temperature (degrees Celsius): 11.3 to 29.8\n\n### 25 features\n\n station numeric 25 unique values 2 missing Date string 310 unique values 2 missing Present_Tmax numeric 167 unique values 70 missing Present_Tmin numeric 155 unique values 70 missing LDAPS_RHmin numeric 7672 unique values 75 missing LDAPS_RHmax numeric 7664 unique values 75 missing LDAPS_Tmax_lapse numeric 7675 unique values 75 missing LDAPS_Tmin_lapse numeric 7675 unique values 75 missing LDAPS_WS numeric 7675 unique values 75 missing LDAPS_LH numeric 7675 unique values 75 missing LDAPS_CC1 numeric 7569 unique values 75 missing LDAPS_CC2 numeric 7582 unique values 75 missing LDAPS_CC3 numeric 7599 unique values 75 missing LDAPS_CC4 numeric 7524 unique values 75 missing LDAPS_PPT1 numeric 2812 unique values 75 missing LDAPS_PPT2 numeric 2510 unique values 75 missing LDAPS_PPT3 numeric 2356 unique values 75 missing LDAPS_PPT4 numeric 1918 unique values 75 missing lat numeric 12 unique values 0 missing lon numeric 25 unique values 0 missing DEM numeric 25 unique values 0 missing Slope numeric 27 unique values 0 missing Solar radiation numeric 1575 unique values 0 missing Next_Tmax numeric 183 unique values 27 missing Next_Tmin numeric 157 unique values 27 missing\n\n### 19 properties\n\n7752\nNumber of instances (rows) of the dataset.\n25\nNumber of attributes (columns) of the dataset.\nNumber of distinct values of the target attribute (if it is nominal).\n1248\nNumber of missing values in the dataset.\n164\nNumber of instances with at least one value missing.\n24\nNumber of numeric attributes.\n0\nNumber of nominal attributes.\n0\nPercentage of binary attributes.\n2.12\nPercentage of instances having missing values.\n0.64\nPercentage of missing values.\nAverage class difference between consecutive instances.\n96\nPercentage of numeric attributes.\n0\nNumber of attributes divided by the number of instances.\n0\nPercentage of nominal attributes.\nPercentage of instances belonging to the most frequent class.\nNumber of instances belonging to the most frequent class.\nPercentage of instances belonging to the least frequent class.\nNumber of instances belonging to the least frequent class.\n0\nNumber of binary attributes." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7910959,"math_prob":0.83946127,"size":3290,"snap":"2021-43-2021-49","text_gpt3_token_len":1017,"char_repetition_ratio":0.2008521,"word_repetition_ratio":0.21703854,"special_character_ratio":0.33708206,"punctuation_ratio":0.15712188,"nsfw_num_words":1,"has_unicode_error":false,"math_prob_llama3":0.9506652,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-24T07:15:32Z\",\"WARC-Record-ID\":\"<urn:uuid:7a4e870d-5489-4da7-9803-84b5d763c307>\",\"Content-Length\":\"34750\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f2285786-3c5c-4c0d-a7f2-329021ac5421>\",\"WARC-Concurrent-To\":\"<urn:uuid:4801b970-2d47-48d5-afc8-d54396f7cd63>\",\"WARC-IP-Address\":\"131.155.11.11\",\"WARC-Target-URI\":\"https://www.openml.org/d/42897\",\"WARC-Payload-Digest\":\"sha1:XHL575BSD2ZUPRAF6ZOFMDLXZA37EGRB\",\"WARC-Block-Digest\":\"sha1:URN2SBO3LMLEZYI6D3OSTT35UTE6YALJ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585911.17_warc_CC-MAIN-20211024050128-20211024080128-00020.warc.gz\"}"}
https://lists.freebsd.org/pipermail/freebsd-bugs/2007-June/024874.html	[ "# kern/109024: [msdosfs] mount_msdosfs: msdosfs_iconv: Operation not permitted\n\nAlexander Pyhalov alp at rsu.ru\nThu Jun 21 13:10:07 UTC 2007\n\n```The following reply was made to PR kern/109024; it has been noted by GNATS.\n\nFrom: Alexander Pyhalov <alp at rsu.ru>\nTo: bug-followup at FreeBSD.org, ggg_mail at inbox.ru\nCc:\nSubject: Re: kern/109024: [msdosfs] mount_msdosfs: msdosfs_iconv: Operation\nnot permitted\nDate: Thu, 21 Jun 2007 16:49:06 +0400\n\n<!DOCTYPE html PUBLIC \"-//W3C//DTD HTML 4.01 Transitional//EN\">\n<html>\n<body bgcolor=3D\"#ffffff\" text=3D\"#000000\">\n<font face=3D\"Courier New, Courier, monospace\">Bug was wrongly closed,\nbecause with loaded module msdosfs_iconv.ko mount says mount_msdosfs:\nmsdosfs_iconv: Operation not permitted. <br>\nThe reason, as I understand, is in kiconv. If you compile and run the\nfollowing code at sturtup=9A ( you should run it as root, for example,\nusing rc.d), everything works correctly.<br>\n<br>\n#include <sys/stat.h><br>\n#include <stdio.h><br>\n#include <sys/iconv.h><br>\n<br>\nint main()<br>\n{<br>\n=9A=9A=9A=9A=9A=9A=9A int er;<br>\n;<br>\n=9A=9A=9A=9A=9A=9A=9A if(er)<br>\n=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A printf(\"Er=3D%d\\n\",er);<br>=\n\n>\n=9A=9A=9A=9A=9A=9A=9A if(er)<br>\n=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A=9A printf(\"Er2=3D%d\\n\",er);<br=\n>\n=9A=9A=9A=9A=9A=9A=9A return 0;<br>\n}<br>\n<br>\n<br>\n</font>\n<pre class=3D\"moz-signature\" cols=3D\"72\">--=20\n=F3 =D5=D7=C1=D6=C5=CE=C9=C5=CD,=20\n=E1=CC=C5=CB=D3=C1=CE=C4=D2 =F0=D9=C8=C1=CC=CF=D7,\n=D3=C9=D3=D4=C5=CD=CE=D9=CA =C1=C4=CD=C9=CE=C9=D3=D4=D2=C1=D4=CF=D2 =E0=E7=\n=E9=EE=E6=EF =E0=E6=F5.\n</pre>\n</body>\n</html>\n\n```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5765892,"math_prob":0.7547852,"size":1828,"snap":"2019-43-2019-47","text_gpt3_token_len":840,"char_repetition_ratio":0.19188596,"word_repetition_ratio":0.032786883,"special_character_ratio":0.4119256,"punctuation_ratio":0.13475177,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99967575,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-17T07:07:12Z\",\"WARC-Record-ID\":\"<urn:uuid:b846baea-7d67-4d1d-aa52-52778696f74e>\",\"Content-Length\":\"5280\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ccca1415-7a44-4e71-8212-f128867f6712>\",\"WARC-Concurrent-To\":\"<urn:uuid:a2bdf3b6-09e9-44a7-bb62-ad27db6d3534>\",\"WARC-IP-Address\":\"96.47.72.84\",\"WARC-Target-URI\":\"https://lists.freebsd.org/pipermail/freebsd-bugs/2007-June/024874.html\",\"WARC-Payload-Digest\":\"sha1:J6ZSSJTUPGZTET57J5ONJVWBLLESGUN6\",\"WARC-Block-Digest\":\"sha1:C3P6E3NWSUQLSQ75OVUTAYSMHOAYPK5U\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496668896.47_warc_CC-MAIN-20191117064703-20191117092703-00182.warc.gz\"}"}
https://cstheory.stackexchange.com/questions/27460/approximation-algorithms-for-min-vector-subset-sum-over-gf2	[ "# Approximation algorithms for min vector subset-sum over GF(2)\n\nIn this question vzn asked about the following problem, which I'll call Vector-Subset-Sum.\n\nGiven a set of vectors $v_i$ over GF(2) and a target vector $y$, is there a subset of the $v_i$ summing to $y$?\n\n1. The decision problem is in P.\n2. The natural minimization problem (finding the smallest subset) is NP-hard.\n3. It's related to the problem of finding sparse codewords in a linear code.\n\nMy question is: are there any known polynomial-time approximation algorithms for the minimization problem in (2)? I know very little about coding theory, so there may be some obvious or well-known approximation-preserving reductions to (3) that I have not seen.\n\n• By the way, in a paper with Indyk, Woodruff and Xie (drona.csa.iisc.ernet.in/~arnabb/boolcycles.pdf), we showed that the problem of finding whether there are $k$ $v_i$'s that sum to $y$ requires at least $\\min(n^{\\Omega(k)}, 2^{\\Omega(d)})$ time assuming ETH, if $v_1, \\dots, v_n$ are $n$ vectors of dimension $d$. This bound is also tight. Nov 16, 2014 at 5:42\n\nDumer, Miccancio and Sudan showed that the minimum distance of a linear code is not approximable to any constant factor in randomized polynomial time, unless $\\mathsf{NP} = \\mathsf{RP}$. The minimum distance problem is the same as your problem if the $v_i$'s form the columns of the parity check matrix and $y = 0$.\n• “The minimum distance problem is the same as your problem if the v_i's form the columns of the parity check matrix and y=0.” As is stated, this is not true. The special case of the problem in question with y=0 is easy to solve: the zero vector is always the optional solution. The minimum distance problem is equivalent to finding the solution with the second smallest weight. I think that you need a slightly more delicate randomized reduction (by adding one constraint $c^T x=1$ with a randomized chosen vector c). Nov 16, 2014 at 19:17" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9273737,"math_prob":0.9972413,"size":705,"snap":"2023-40-2023-50","text_gpt3_token_len":165,"char_repetition_ratio":0.115549214,"word_repetition_ratio":0.0,"special_character_ratio":0.22836879,"punctuation_ratio":0.09701493,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99984217,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-08T02:47:54Z\",\"WARC-Record-ID\":\"<urn:uuid:bc9048ff-9b18-4727-8429-bc030fe29e03>\",\"Content-Length\":\"163314\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:78c250c0-351f-4235-85e4-f732c639ec8e>\",\"WARC-Concurrent-To\":\"<urn:uuid:a01b2828-670f-4d14-8c6a-007ebe92dd80>\",\"WARC-IP-Address\":\"172.64.144.30\",\"WARC-Target-URI\":\"https://cstheory.stackexchange.com/questions/27460/approximation-algorithms-for-min-vector-subset-sum-over-gf2\",\"WARC-Payload-Digest\":\"sha1:CHYIFYCKXUYWE4YKBWWDVBKM7KHXLR4Y\",\"WARC-Block-Digest\":\"sha1:IA6DI74LFNTOFXTNIQEA7N5OFT4HNAIK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100710.22_warc_CC-MAIN-20231208013411-20231208043411-00233.warc.gz\"}"}
https://islamtarihikaynaklari.com/piercing/140-pounds-in-kg.php	[ "# Bikini 140 Pounds In Kg Foton\n\nNya Inlägg\n\n• ## Proon Sex", null, "", null, "", null, "", null, "### Erotisk Convert lbs to kg - Conversion of Measurement Units Pictures\n\nPlease enable Javascript to use the Rainbow Spånga converter. How many lbs in 1 kg? The answer is 2. We assume you are converting between pound and kilogram. Use this page to learn how to convert between pounds and kilograms.\n\nType in your own numbers in the form to convert the units! You can do the reverse unit conversion from kg to lbs Fluktare, or enter Kilted Bros two units below:.\n\nThe Pounsd abbreviation: lb is a unit of mass or weight in a number of different systems, including English units, Imperial units, and United States customary units. Its size can vary 140 Pounds In Kg system to system.\n\nThe international avoirdupois pound is equal to exactly The definition of the international pound was agreed by the United States and countries of the 140 Pounds In Kg of Nations in In the United Kingdom, the use of the international pound was implemented in the Weights and Measures Act An avoirdupois pound is equal to 16 avoirdupois ounces and to exactly 7, grains.\n\nThe kilogram or kilogramme, symbol: kg is the SI base unit of mass. A gram is defined as one thousandth of a kilogram. Conversion of units describes equivalent units of mass in other systems. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type Poounds unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types.", null, "", null, "", null, "", null, "Please enable Javascript to use the unit converter. How many lbs in 1 kg? The answer is 2.", null, "Use this to learn how to convert between pounds and kilograms. Type in your own numbers in the form to convert the units! ›› Quick conversion chart of lbs to kg. 1 lbs to kg = kg. 5 lbs to kg = kg. 10 lbs to kg = kg. 20 lbs to kg = kg. 30 lbs to kg = kg. 40 lbs to kg = kg. 50 lbs to kg.", null, "", null, "", null, "Pounds (lb) = Kilograms (kg) Visit Kilograms to Pounds Conversion. Pounds: The pound or pound-mass (abbreviations: lb, lbm, lbm, ℔) is a unit of mass with several definitions. Nowadays, the common is the international avoirdupois pound which is legally defined as exactly kilograms. A pound is equal to\n\n2021 islamtarihikaynaklari.com" ]	[ null, "https://lbs-to-kg.appspot.com/image/140.png", null, "https://qph.fs.quoracdn.net/main-qimg-8e802a0e9d7fb708ec431fab5d51a2a0", null, "https://pbs.twimg.com/media/CG_OXmfW0AAdD_n.jpg", null, "https://convertermaniacs.com/images/lbs-to-kg.png", null, "https://www.thecalculatorsite.com/images/charts/pounds-kg-84-to-195.png", null, "https://i2.wp.com/tim.blog/wp-content/uploads/2013/05/baseline-before-experiment-190-pounds.jpg", null, "https://i.pinimg.com/736x/8a/ab/f5/8aabf5b61c9cd950cc85e70cbd5c852a--affirmations-inspirational.jpg", null, "https://images-na.ssl-images-amazon.com/images/I/6192VP2LQtL.__AC_SY445_SX342_QL70_ML2_.jpg", null, "https://www.sycor.com/pub/media/wysiwyg/technical_information/KG_and_LB.png", null, "https://converter.ninja/images/140_lb_in_kg.jpg", null, "https://i.dietdoctor.com/wp-content/uploads/2015/12/Suzz2-800x804.png", null, "https://i2.wp.com/tim.blog/wp-content/uploads/2013/05/baseline-before-experiment-190-pounds.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.89066344,"math_prob":0.9467914,"size":2141,"snap":"2021-43-2021-49","text_gpt3_token_len":518,"char_repetition_ratio":0.1483388,"word_repetition_ratio":0.1658031,"special_character_ratio":0.22886501,"punctuation_ratio":0.12785389,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9537574,"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],"im_url_duplicate_count":[null,6,null,1,null,1,null,null,null,4,null,3,null,1,null,1,null,10,null,2,null,1,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T00:38:08Z\",\"WARC-Record-ID\":\"<urn:uuid:c4f2f4eb-63b2-4b28-9cf9-2307ed2c2f36>\",\"Content-Length\":\"14105\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:817d681e-4ddf-4ef2-8dac-678f96b6c099>\",\"WARC-Concurrent-To\":\"<urn:uuid:8556f65e-c906-4f4f-aba7-e0f6f1d0d881>\",\"WARC-IP-Address\":\"172.67.182.59\",\"WARC-Target-URI\":\"https://islamtarihikaynaklari.com/piercing/140-pounds-in-kg.php\",\"WARC-Payload-Digest\":\"sha1:DLSNZ2BCTBLM2TSEDZS4XSRTMQ2NS55F\",\"WARC-Block-Digest\":\"sha1:2J7U3Y5X2QUX426B7DW3O62CKJI6DV3G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585186.33_warc_CC-MAIN-20211018000838-20211018030838-00106.warc.gz\"}"}
https://six-sigma-online.com/index.php/component/content/article/105-six-sigma-heretic/157-performing-a-long-term-msa?Itemid=437	[ "## Testing through time stability\n\nAhh, measurement system analysis—the basis for all our jobs because, as Lord Kelvin said, “… When you cannot measure it, when you cannot express it in numbers, your knowledge is of a meager and unsatisfactory kind.” How interesting it is then, that we who thrive on data so frequently don't have any proof that the numbers we're using relate to the event we are measuring—hence my past few articles about the basics of measurement system analysis in “Letting You In on a Little Secret,” on how to do a potential study in “The Mystery Measurement Theatre, and on how to do a short-term study in “Performing a Short-Term MSA Study.” The only (and most important) topic remaining is how to perform a long-term study, which is the problem I left with you last month.\n\nSo read on to see how.\n\nThe potential study told us if we even have a hope of being able to use a measurement system to generate data (numbers related to an event) as opposed to numbers (symbols manipulated by mathematicians for no apparent reason other than entertainment). The short-term study allowed us to test the system performance a little more rigorously, perhaps in preparation for using it in our process. A measurement system’s performance is quantified by:\n\n• Repeatability—the amount of variability the same system (operator, device) exhibits when measuring exactly the same thing multiple times\n• Reproducibility—the amount of variability due to different operators using the same device, or maybe the same operators using different devices\n• %R&R—a combination of the repeatability and reproducibility that tells us how easy it is for this measurement system to correctly classify product as conforming or nonconforming to our specification\n• Bias—the amount that the average measurement is off of the “true” value\n\nHowever, none of these things have any meaning if the measurement system changes through time, and neither the potential nor short-term study really tests for through-time stability. With an unstable gauge, I might convene a Six Sigma team to work on a problem that doesn’t exist, put in systems to control a process based on a random number generator, or scrap product that is conforming. Simply put, my ability to understand and control my process is completely compromised. A measurement system that is out of control is even worse than a process that is out of control, since we don’t even have a hint of what is really going on.\n\nThus the need for the long-term study, which allows us to assess in detail exactly how our measurement system is performing through time. The long-term study is the Holy Grail (not the Money Python kind) of measurement system analysis, and with it we can state with confidence that the system is producing data, and not just numbers.\n\nAs before, I gave you a link to a totally free and awesome spreadsheet that will help you with your MSA work, and I gave you some data for the following scenario:\n\nA statistical facilitator and an engineer wish to conduct a gauge capability analysis (long-term) for a particular ignition signal processing test on engine control modules. The test selected for study measures voltage which has the following specifications:\n\nIGGND = 1.4100 ± 0.0984 Volts (Specification)\n\nEight control modules are randomly selected from the production line at the plant, and run (in random order, of course) through the tester at one hour intervals (but randomly within each hour). This sequence is repeated until 25 sample measures (j = 25) of size eight (n = 8) have been collected. The assumption is that these voltages are constant and the only variation we see in remeasuring a module is gauge variation.\n\nRight off the bat, we know that we only measure this engine control module; if we measured others or to different target voltages, we would include them in the study as well. We select these eight parts and keep remeasuring the same ones each hour. We also are assuming that different operators have no effect, since we are only using one.\n\nTo start off the spreadsheet calculates the mean and standard deviation across each hour’s measurements. Regardless of the actual voltages the eight modules have, the average of those voltages must be the same, right? One way we will eventually look at the measurement error over time will be by looking at how that average moves around. Because we have eight modules, we can also calculate a standard deviation across these eight. But be careful that you understand what this is. There are two components of variability in this standard deviation, only one of which relates to gauge variability. There is some measurement error as I take a reading for each one, but there is also the fact that the eight modules are producing somewhat different voltages from each other as well. Even if there were no measurement error, we still would calculate a standard deviation due to the part differences. This second variance component is of no interest to us for the MSA, but we had better not forget about it as we go forward.", null, "Figure 1: Worksheet calculations\n\nFirst, we need some validation that we can use the numbers. If the measurement process is in control, then the measurements for each part over time should be in control. So we take a look at each part on an individuals chart (the limit comes from the moving ranges, which I leave out of sight for this example). In order for the moving range to relate to the dispersion, we want to check normality for all the parts:", null, "Figure 2: Normality tests for the eight modules—output from MVP stats\n\nWith a sample size of 25 we probably are going to rely on the skewness and kurtosis indices, and they allow us to assume the variability is distributed normally. So let’s take a look at those individual charts on all the parts.", null, "", null, "Figure 3: Individuals charts for each part through time—limits from the moving range\n\nWe do see a point outside the lower control limit on part 4 and a larger than expected moving range on part 7 (both of which we would have investigated once the long-term study was in place). But two out of 200 observations is well within what we would expect with Type I error rate of 0.0027 (the rate you get a point out of the ±3σ control limits due to chance and chance alone) so I am comfortable saying that, so far, the gauge looks stable with respect to location. If a particular part became damaged at some point, or was a lot more difficult to read than another part, it should show up on these charts.\n\nWhile individual charts are really useful, they have their weaknesses, one of which is a lack of sensitivity to pretty big shifts in the mean. Thankfully, we have those means we calculated back in the worksheet calcul figure 1 to increase the sensitivity to a global shift in the average. Remember how the standard deviation across the parts has two components of variability? That is why we can’t just do an X-bar and an s chart using the usual limits—that “s” is inflated due to the part-to-part differences and will give us limits on the mean and the s chart that are too large. We get around this by recalling that we can plot any statistic on an individuals chart—though we may have to adjust the limits for the shape of the distribution. We are in good shape (pun intended) to use the individuals chart for our means, because due to the central limit theorem, those 25 averages of eight modules will tend to be distributed normally, and therefore the moving range of the means will relate to the dispersion of the means.", null, "Figure 4: Means as individuals control chart—limits from the moving range\n\nFigure 4 further supports our notion that the location measurements are stable through time. If there were some sort of a shift across many or all the parts, it would show up here, so actions such as a recalibration, retaring, or dropping the gauge on the floor would show up on this chart. Due to the central limit theorem, this chart will be far more sensitive to shifts in the average than the individuals charts on each part.\n\nWe also know that control for continuous data isn't assessed by just looking at the average—we need to look at that dispersion as well. Using the same trick as with the means, we will plot the standard deviations (extra component of variability and all) on an individuals chart. If the measurement error is normal and the same for every part, then regardless of the actual voltages the standard deviations across the parts ought to be distributed close to (though not exactly) normal. (Again, this is different than the random sampling distribution of the standard deviation, which would be very clearly positively skewed.)\n\nThe upshot is that we plot the standard deviations on an individuals chart with limits from the moving ranges as well.", null, "Figure 5: Standard deviations across eight parts plotted as individuals—limits from the moving range\n\nHere we have one point outside of the limits, which if we were up and running with the chart, we would have reacted to by investigating. Because the measurements are already done, I am going to continue watching that control chart very closely and any statement about control is going to be provisional. The types of events that would show up here would be if there was a global change in the dispersion of the measurements—perhaps a control circuit going bad or a change in the standard operating procedure.\n\nAs with the short-term study, we also want to keep an eye on the relationship between the average measurement and the variation of the readings. We want to see no correlation between them—if we do, then the error of the measurement changes with the magnitude of what you are measuring, and so your ability to correctly classify as conforming or not changes too. We will check that with a correlation between the mean and standard deviation.", null, "Figure 6: Correlation between magnitude (mean) and variation (standard deviation)\n\nWe only have eight points because we only have eight different parts. Normally, we won’t do correlations on such few points, but we would have already tested this with the short-term study. (Which you DID do, right?) This is just to make sure nothing really big changed. This correlation is not significant, so we can check that off our list.\n\nAt this point we are (provisionally) saying that there is nothing terribly strange going on in our measurement system through time—it seems to be stable sliced a couple of ways, and the magnitude and dispersion are independent. We can now, finally, begin to answer our question about the repeatability and reproducibility. In our case, we only have one operator and system, so measurement error due to reproducibility is assumed to be zero. If we had tested multiple operators, the estimate of the variability due to operator would be:", null, "Which is pretty similar to that formula you remember from SPC. The range across the operator averages divided by our old friend d2 on the expanded d2 table with g = 1 and m = the number of appraisers. The spreadsheet is set up to handle up to two operators.\n\nAll we are looking at (in this case, to estimate) is the variability due to repeatability, which is:", null, "You recognize that from SPC as well I bet. The only difference is that we are taking the average of the standard deviations for each of the j = 8 modules and dividing by c4 for the number of measurements (25 here). The spreadsheet cleverly does all this for you.", null, "Again, we are interested in the ability of this device to correctly categorize whether a given module is in or out of spec. Our spec width was 0.1968V, and so we put that into the %R&R formula:", null, "We find that the measurement error alone takes up 134.74 percent of the spec.\n\nUh-oh. What is it with these measurement devices?\n\nIf I measure a part that is smack dab in the middle of the spec over time, I would see this distribution:", null, "Figure 8: Measurement error\n\nThat means that on any individual measurement (how we have been using this gauge up to this point) of a module that is exactly on target at 1.41, I could reasonably see a reading somewhere between 1.26 and 1.56—in or out of spec at the whim of the gauge variability.\n\nDo you remember that crazy graph that showed the probability of incorrectly classifying a part as conforming or nonconforming on a single measure? Here it is for our voltage measurement system:", null, "Figure 9: Probability of incorrectly classifying module conformance on a single measurement\n\nOnce again, we have a measurement system that is probably no good for making conformance decisions on a single measurement. It is stable through time, yes, but so highly variable that we stand a pretty good chance of calling good stuff bad and bad stuff good.\n\nBecause it's stable, we could conceive of taking multiple measurements to use the average of those readings to determine conformance:\n\nSay we are looking for a %R&R of 10 percent:", null, "Leaving us measuring the voltage 182 times to get that %R&R.\n\nAhh, that’s not gonna happen.\n\nWe need to figure out what is causing all the variability in this measurement device, or replace it with something that is capable of determining if a module is in conformance with the spec. It is also possible that the modules themselves are contributing noise to the measurements—maybe our assumption that they give the same voltage time after time is wrong. That would be a good thing to find out, too.\n\nWe have some more work to do, it seems.\n\nNote that we did not assess bias—with this gauge, it would be a waste of time since it is unusable for this specification. If we wanted to, all we have to do is get the “true” voltages of those eight modules, get the average of the true values, and see how far off that average is from our measured average. If they are statistically different, we just add or subtract the amount of bias to get an accurate (but not precise with this system) measurement. You would want to track this bias with time on an control chart as well, to make sure it was stable too.\n\nIf you have been reading these mini-dissertations on MSA, you will know that a common assumption of them all is that the measurement is not destructive—that what you are measuring remains constant with time. What do you do if that is not the case?" ]	[ null, "https://six-sigma-online.com/images/stories/heretic_longterm.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm01.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm02.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm13.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm03.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm04.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm05.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm05.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm07.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm08.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm09.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm10.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm11.jpg", null, "https://six-sigma-online.com/images/stories/heretic_longterm12.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9568768,"math_prob":0.9449278,"size":14246,"snap":"2021-43-2021-49","text_gpt3_token_len":2964,"char_repetition_ratio":0.14176379,"word_repetition_ratio":0.002426203,"special_character_ratio":0.2037765,"punctuation_ratio":0.07919708,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97168946,"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,8,null,8,null,4,null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-22T03:28:21Z\",\"WARC-Record-ID\":\"<urn:uuid:0c93d2de-4932-4fd9-8f9c-bf03be5bed71>\",\"Content-Length\":\"43860\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4bcc18f8-7709-4be1-92b5-26baeb7cc922>\",\"WARC-Concurrent-To\":\"<urn:uuid:44f32347-2400-4a3f-8aa4-dfc1cdf9ea1b>\",\"WARC-IP-Address\":\"35.208.100.176\",\"WARC-Target-URI\":\"https://six-sigma-online.com/index.php/component/content/article/105-six-sigma-heretic/157-performing-a-long-term-msa?Itemid=437\",\"WARC-Payload-Digest\":\"sha1:VPAM2GXW5H4SLGKOMXHM7B2AQ7O4XJMN\",\"WARC-Block-Digest\":\"sha1:LLSIUH5QC6HUGDTAHYH45JZYELYIKCLR\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585450.39_warc_CC-MAIN-20211022021705-20211022051705-00214.warc.gz\"}"}
http://www.percentagecal.com/answer/120-is-what-percent-of-1000	[ "#### Solution for 120 is what percent of 1000:\n\n120:1000100 =\n\n(120100):1000 =\n\n12000:1000 = 12\n\nNow we have: 120 is what percent of 1000 = 12\n\nQuestion: 120 is what percent of 1000?\n\nPercentage solution with steps:\n\nStep 1: We make the assumption that 1000 is 100% since it is our output value.\n\nStep 2: We next represent the value we seek with {x}.\n\nStep 3: From step 1, it follows that {100\\%}={1000}.\n\nStep 4: In the same vein, {x\\%}={120}.\n\nStep 5: This gives us a pair of simple equations:\n\n{100\\%}={1000}(1).\n\n{x\\%}={120}(2).\n\nStep 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS\n(left hand side) of both equations have the same unit (%); we have\n\n\\frac{100\\%}{x\\%}=\\frac{1000}{120}\n\nStep 7: Taking the inverse (or reciprocal) of both sides yields\n\n\\frac{x\\%}{100\\%}=\\frac{120}{1000}\n\n\\Rightarrow{x} = {12\\%}\n\nTherefore, {120} is {12\\%} of {1000}.\n\n#### Solution for 1000 is what percent of 120:\n\n1000:120100 =\n\n(1000100):120 =\n\n100000:120 = 833.33\n\nNow we have: 1000 is what percent of 120 = 833.33\n\nQuestion: 1000 is what percent of 120?\n\nPercentage solution with steps:\n\nStep 1: We make the assumption that 120 is 100% since it is our output value.\n\nStep 2: We next represent the value we seek with {x}.\n\nStep 3: From step 1, it follows that {100\\%}={120}.\n\nStep 4: In the same vein, {x\\%}={1000}.\n\nStep 5: This gives us a pair of simple equations:\n\n{100\\%}={120}(1).\n\n{x\\%}={1000}(2).\n\nStep 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS\n(left hand side) of both equations have the same unit (%); we have\n\n\\frac{100\\%}{x\\%}=\\frac{120}{1000}\n\nStep 7: Taking the inverse (or reciprocal) of both sides yields\n\n\\frac{x\\%}{100\\%}=\\frac{1000}{120}\n\n\\Rightarrow{x} = {833.33\\%}\n\nTherefore, {1000} is {833.33\\%} of {120}.\n\nCalculation Samples" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8230369,"math_prob":0.9997551,"size":1821,"snap":"2019-35-2019-39","text_gpt3_token_len":618,"char_repetition_ratio":0.13593836,"word_repetition_ratio":0.52684563,"special_character_ratio":0.44920373,"punctuation_ratio":0.14861462,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999949,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-23T11:29:42Z\",\"WARC-Record-ID\":\"<urn:uuid:81057a49-b9fd-4151-a871-c5d117c7e330>\",\"Content-Length\":\"10097\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:126457b3-a374-4d7c-a586-08abdbe700cf>\",\"WARC-Concurrent-To\":\"<urn:uuid:7c2cea3c-2f49-4455-a37f-4d9744af121a>\",\"WARC-IP-Address\":\"217.23.5.136\",\"WARC-Target-URI\":\"http://www.percentagecal.com/answer/120-is-what-percent-of-1000\",\"WARC-Payload-Digest\":\"sha1:5DGMU5UZ4DJXKA2NS2KKHQ3EQRJ2EYLP\",\"WARC-Block-Digest\":\"sha1:REOFQIHSVWUY4QNY7OVLGBA3KMXWXACM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027318375.80_warc_CC-MAIN-20190823104239-20190823130239-00168.warc.gz\"}"}
https://my.homecampus.com.sg/Learn/Primary-Grade-4/Decimal/Decimals-Multiplication	[ "## Multiplying Decimals with Whole Number\n\nPractice Unlimited Questions\n\n#### How to Multiply Decimals with Whole Number?\n\nStep 1: Multiply the hundredths.\n\nStep 2: Multiply and add the tenths.\n\nStep 3: Multiply and add the ones.\n\nSee working examples below.\n\n#### 1. Bob had 3 packets of butter each containing 0.4 kg butter. How much butter did Bob have altogether?\n\nMethod 1:\n0.4 + 0.4 + 0.4 = 1.2 kg\nBob had 1.2 kg of butter altogether.\n\nMethod 2:\nMultiply 0.4 kg by 3.\n0.4 × 3 = ?", null, "", null, "Bob had 1.2 kg of butter altogether.\n\n#### 2. Bob had 3 cartons of milk each containing 0.45 L of milk. How much milk did Bob have altogether?\n\nMultiply 0.45 L by 3.\n0.45 × 3 = ?", null, "", null, "Bob had 1.35 L of milk altogether.\n\n#### 3. Bob had 7 bags of flour each containing 1.85 kg flour. How much flour did Bob have altogether?\n\nMultiply 1.85 kg by 7.\n1.85 × 7 = ?", null, "Bob had 12.95 kg of flour altogether.\n\n#### 4. Serena bought 5 drawing books for \\$4.25 each. How much did she spend on the drawing books altogether?\n\nCost of 1 drawing book = \\$4.25\nCost of 5 drawing books = \\$4.25 × 5\n= \\$21.25\n\nShe spent \\$21.25 on the drawing books altogether.", null, "" ]	[ null, "https://my.homecampus.com.sg/images/notes/P4_Decimals_Multiplication_Tenths_1a.png", null, "https://my.homecampus.com.sg/images/notes/P4_Decimals_Multiplication_Tenths_1b.png", null, "https://my.homecampus.com.sg/images/notes/P4_Decimals_Multiplication_Tenths_2a.png", null, "https://my.homecampus.com.sg/images/notes/P4_Decimals_Multiplication_Tenths_2b.png", null, "https://my.homecampus.com.sg/images/notes/P4_Decimals_Multiplication_Tenths_3a.png", null, "https://my.homecampus.com.sg/images/notes/P4_Decimals_Multiplication_Problem_Sum.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9506132,"math_prob":0.94482887,"size":1445,"snap":"2023-40-2023-50","text_gpt3_token_len":475,"char_repetition_ratio":0.18459404,"word_repetition_ratio":0.46206897,"special_character_ratio":0.36885813,"punctuation_ratio":0.21818182,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99590796,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12],"im_url_duplicate_count":[null,4,null,4,null,4,null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-04T16:05:41Z\",\"WARC-Record-ID\":\"<urn:uuid:1404d46e-af9c-419e-8f9c-d2933a3f6213>\",\"Content-Length\":\"36907\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:4d6fbe67-68ef-49e9-b9f2-ed550758df1b>\",\"WARC-Concurrent-To\":\"<urn:uuid:e8e3bd7a-c116-4ff9-adea-3cb361f7f1cc>\",\"WARC-IP-Address\":\"18.213.98.197\",\"WARC-Target-URI\":\"https://my.homecampus.com.sg/Learn/Primary-Grade-4/Decimal/Decimals-Multiplication\",\"WARC-Payload-Digest\":\"sha1:I2GBICCYOWWTPX74KYUK3V2VLHLOZYV3\",\"WARC-Block-Digest\":\"sha1:VMTMLNYBQZ6P6HBW34RER5YTX236I56S\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511386.54_warc_CC-MAIN-20231004152134-20231004182134-00132.warc.gz\"}"}
https://www.luxist.com/content?q=descartes+cube+route+login+kraft+road&nojs=1&ei=UTF-8&nocache=1&s_pt=source7&s_chn=1&s_it=rs-rhr1&s_it=rs-rhr2	[ "# Luxist Web Search\n\n2. ### Descartes Systems Group - Wikipedia\n\nen.wikipedia.org/wiki/Descartes_Systems_Group\n\nThe Descartes Systems Group Inc. (commonly referred to as Descartes) is a Canadian multinational technology company specializing in logistics software, supply chain management software, and cloud -based services for logistics businesses. Descartes is perhaps best known for its abrupt and unexpected turnaround in the mid-2000s after coming close ...\n\n3. ### René Descartes - Wikipedia\n\nen.wikipedia.org/wiki/René_Descartes\n\nRené Descartes ( / deɪˈkɑːrt / or UK: / ˈdeɪkɑːrt /; French: [ʁəne dekaʁt] ( listen); Latinized: Renatus Cartesius; [note 3] 31 March 1596 – 11 February 1650 : 58 ) was a French philosopher, mathematician, scientist and lay Catholic who invented analytic geometry, linking the previously separate fields of ...\n\n4. ### Dimension - Wikipedia\n\nen.wikipedia.org/wiki/Dimension\n\nv. t. e. In physics and mathematics, the dimension of a mathematical space (or object) is informally defined as the minimum number of coordinates needed to specify any point within it. Thus, a line has a dimension of one (1D) because only one coordinate is needed to specify a point on it – for example, the point at 5 on a number line.\n\n5. ### Think the California DMV’s driver licensing exam is tough to ...\n\nwww.aol.com/news/think-california-dmv-driver...\n\nTry taking mine. Jack Ohman. July 22, 2022, 5:00 AM. The California Department of Motor Vehicles’ written driver licensing exam has come under scrutiny for its chronically high failure rate ...\n\n6. ### Cube (algebra) - Wikipedia\n\nen.wikipedia.org/wiki/Cube_(algebra)\n\nThe cube of a number or any other mathematical expression is denoted by a superscript 3, for example 2 3 = 8 or (x + 1) 3. The cube is also the number multiplied by its square: n 3 = n × n 2 = n × n × n. The cube function is the function x ↦ x 3 (often denoted y = x 3) that maps a number to its cube. It is an odd function, as\n\n7. ### Games on AOL.com: Free online games, chat with others in real ...\n\nwww.aol.com/games/play/masque-publishing/just-words\n\nDiscover the best games on AOL.com - Free online games and chat with others in real-time.\n\n8. ### Euclidean geometry - Wikipedia\n\nen.wikipedia.org/wiki/Euclidean_geometry\n\nEuclidean geometry is a mathematical system attributed to ancient Greek mathematician Euclid, which he described in his textbook on geometry: the Elements. Euclid's approach consists in assuming a small set of intuitively appealing axioms (postulates), and deducing many other propositions ( theorems) from these." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9289494,"math_prob":0.84083575,"size":1993,"snap":"2022-27-2022-33","text_gpt3_token_len":509,"char_repetition_ratio":0.08697838,"word_repetition_ratio":0.0,"special_character_ratio":0.25338686,"punctuation_ratio":0.13527851,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96078575,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-08-09T11:31:17Z\",\"WARC-Record-ID\":\"<urn:uuid:4b679beb-479a-4487-9d43-13138e298de4>\",\"Content-Length\":\"54318\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5b0821a3-013f-4c6f-8611-bf76275d704a>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b850699-6f5a-4415-956a-f950193d1d4f>\",\"WARC-IP-Address\":\"66.218.84.137\",\"WARC-Target-URI\":\"https://www.luxist.com/content?q=descartes+cube+route+login+kraft+road&nojs=1&ei=UTF-8&nocache=1&s_pt=source7&s_chn=1&s_it=rs-rhr1&s_it=rs-rhr2\",\"WARC-Payload-Digest\":\"sha1:ARIRTZ7XVSW53UG4UZDZXSIQAONHKUCU\",\"WARC-Block-Digest\":\"sha1:YFWE2QFHRCLS5W37ZFVKARQW2NSUMZFB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-33/CC-MAIN-2022-33_segments_1659882570921.9_warc_CC-MAIN-20220809094531-20220809124531-00241.warc.gz\"}"}
https://tools.carboncollective.co/future-value/1014-in-26-years/	[ "# Future Value of $1,014 in 26 Years Calculating the future value of$1,014 over the next 26 years allows you to see how much your principal will grow based on the compounding interest.\n\nSo if you want to save $1,014 for 26 years, you would want to know approximately how much that investment would be worth at the end of the period. To do this, we can use the future value formula below: $$FV = PV \\times (1 + r)^{n}$$ We already have two of the three required variables to calculate this: • Present Value (FV): This is the original$1,014 to be invested\n• n: This is the number of periods, which is 26 years\n\nThe final variable we need to do this calculation is r, which is the rate of return for the investment. With some investments, the interest rate might be given up front, while others could depend on performance (at which point you might want to look at a range of future values to assess whether the investment is a good option).\n\nIn the table below, we have calculated the future value (FV) of $1,014 over 26 years for expected rates of return from 2% to 30%. The table below shows the present value (PV) of$1,014 in 26 years for interest rates from 2% to 30%.\n\nAs you will see, the future value of $1,014 over 26 years can range from$1,696.85 to $930,175.97. Discount Rate Present Value Future Value 2%$1,014 $1,696.85 3%$1,014 $2,186.78 4%$1,014 $2,811.28 5%$1,014 $3,605.45 6%$1,014 $4,613.07 7%$1,014 $5,888.66 8%$1,014 $7,499.90 9%$1,014 $9,530.75 10%$1,014 $12,085.03 11%$1,014 $15,290.98 12%$1,014 $19,306.63 13%$1,014 $24,326.38 14%$1,014 $30,588.92 15%$1,014 $38,386.79 16%$1,014 $48,077.92 17%$1,014 $60,099.43 18%$1,014 $74,984.27 19%$1,014 $93,381.09 20%$1,014 $116,078.12 21%$1,014 $144,031.53 22%$1,014 $178,399.47 23%$1,014 $220,582.40 24%$1,014 $272,271.24 25%$1,014 $335,504.46 26%$1,014 $412,735.79 27%$1,014 $506,914.48 28%$1,014 $621,580.19 29%$1,014 $760,975.06 30%$1,014 \\$930,175.97\n\nThis is the most commonly used FV formula which calculates the compound interest on the new balance at the end of the period. Some investments will add interest at the beginning of the new period, while some might have continuous compounding, which again would require a slightly different formula.\n\nHopefully this article has helped you to understand how to make future value calculations yourself. You can also use our quick future value calculator for specific numbers." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87645936,"math_prob":0.9991121,"size":2468,"snap":"2023-14-2023-23","text_gpt3_token_len":846,"char_repetition_ratio":0.1911526,"word_repetition_ratio":0.025188917,"special_character_ratio":0.47082657,"punctuation_ratio":0.19773096,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9992984,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-03T21:10:26Z\",\"WARC-Record-ID\":\"<urn:uuid:b8a9c308-32f0-4812-a028-4bcb0edbf9eb>\",\"Content-Length\":\"21989\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1ee7dba5-f014-44e7-96db-42404193b10e>\",\"WARC-Concurrent-To\":\"<urn:uuid:ee9b2e89-01ea-456f-b62a-9bf230a87f51>\",\"WARC-IP-Address\":\"138.197.3.89\",\"WARC-Target-URI\":\"https://tools.carboncollective.co/future-value/1014-in-26-years/\",\"WARC-Payload-Digest\":\"sha1:BS5DNK3OH4RFQVSB3OQAMDRYXMOKOPTO\",\"WARC-Block-Digest\":\"sha1:JXB65QESHOJUAXPQOQ3J5LW365O3QIHZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224649343.34_warc_CC-MAIN-20230603201228-20230603231228-00016.warc.gz\"}"}
https://mollysmtview.com/recipes/how-many-ounces-is-6-cups	[ "# How Many Ounces is 6 Cups? \| Get The Helpful Answer Now\n\nDo you often find yourself in the kitchen with recipes that call for measurements such as cups, tablespoons, and ounces but you just don’t know how to convert them? Don’t worry—you’re not alone. Many people ask the same question: How many ounces is 6 cups? It is a simple unit conversion and this blog post will explain exactly how to solve the equation. After reading this article, you’ll never need another conversion calculator; it’s easy to understand all of your measuring needs.\n\n## Understanding Measuring Cups and Ounces\n\nWhen it comes to cooking or baking, measuring cups are an essential kitchen tool. But what exactly do they measure? A cup is a unit of measurement used in the United States and other countries for measuring dry ingredients (such as flour, sugar, rice, etc.) by volume. It is equivalent to 8 fluid ounces (which is approximately 23 cl). One cup is equal to approximately 0.23 liters or 8 Imperial fluid ounces.\n\nOn the other hand, ounces are a unit of measurement that typically measure weight. Ounces can be used to measure both dry and liquid ingredients; however, it’s usually preferred to use grams for dry ingredients and milliliters for liquids. One ounce is equal to 28.35 grams or 1/16th of a pound (0.0625 lbs).\n\n## Calculating from Ounces to Cups\n\nTo convert from ounces to cups, just divide the number of ounces by 8. Here is the formula:\n\nNumber of cup = number of ounces / 8\n\nFor example, if you have 24 ounces and want to know how many cups that is, you would divide 24 by 8 and get 3 cups.\n\n## Converting from Cups to Ounces\n\nConverting from cups to ounces is just as easy. The formula for this conversion is:\n\nNumber of ounces = number of cups x 8\n\nIf you have 10 cups and want to know how many ounces that is, you would multiply 10 by 8 and get 80 ounces.\n\n## How Many Ounces is 6 Cups?\n\nNow that you know how cups and ounces are measured, let’s answer the question: How many ounces is 6 cups? The answer is simple; one cup is equal to 8 fluid ounces, so 6 cups is equal to 48 fluid ounces. To make the equation easier to remember, just remember that one cup is 8 fluid ounces and multiply that number by 6.\n\n## Why Should Know How Many Ounces is 6 Cups?\n\nKnowing how many ounces is 6 cups can be helpful in a number of ways. It helps you make accurate conversions between different measuring systems and save time when cooking. In addition, it ensures that the ingredients in your recipes are measured correctly so you can get consistent results every time. This knowledge will also help ensure that your baked goods turn out exactly as they should, every time.\n\n## How to Convert 6 Cups to Ounces?\n\nNow that we understand why this conversion is important, let’s talk about how to make it happen. The equation for converting 6 cups to ounces is simple: one cup equals 8 ounces. Therefore, 6 cups is equal to 48 ounces.\n\nIf you want to convert a different number of cups into ounces, simply multiply the number of cups by 8. For instance, if a recipe calls for 12 cups of liquid, you would need 96 ounces (12 x 8 = 96).\n\n## Cup to Ounce Conversion Chart\n\nIf you’re looking for a quick reference guide to convert from cups to ounces, then look no further. Here’s a handy chart that will help make your measuring needs much easier.\n\n\| Cups \| Ounces \|\n\n\| — \| —– \|\n\n\| 1 Cup \| 8 Ounces \|\n\n\| 2 Cups \| 16 Ounces \|\n\n\| 3 Cups \| 24 Ounces \|\n\n\| 4 Cups \| 32 Ounces \|\n\n\| 5 Cups \| 40 Ounces \|\n\n\| 6 Cups \| 48 Ounces \|\n\n## Common Conversions of 6 Cups to Ounces for Dry Ingredients\n\nWhen it comes to measuring dry ingredients, the conversion isn’t as straightforward. Dry ingredients are typically measured in grams or milliliters instead of ounces. To convert 6 cups to grams, you would multiply 6 by 16 to get 96 g; and to convert 6 cups to milliliters, you would multiply 6 by 237 (which is the number of milliliters in 1 cup) to get 1422 ml.\n\n## Common Applications of 6 cups to ounces Conversion\n\nKnowing how to convert 6 cups to ounces is essential for many kitchen tasks. Here are a few examples of when this conversion is useful:\n\n• Baking: If you’re baking a cake or other recipe, it’s important to get the measurements right. Many popular recipes call for measuring ingredients in cups rather than ounces, so understanding this conversion can help you make sure your baking comes out perfectly.\n• Cooking: If a recipe calls for 6 cups of liquid, knowing how many ounces are in 6 cups can help you calculate the amount of liquid needed to make the dish. This is especially important if you’re making large, multi-ingredient dishes like soup or stew.\n• Measuring drinks: If you’re making a pitcher of iced tea or other beverage, understanding the 6 cups to ounces conversion can help you make sure that you have enough to serve your guests.\n\n## Conclusion: How Many Ounces is 6 Cups?\n\nSo if you’re ever wondering “how many ounces is 6 cups?” the answer is 48. Now that you understand the basics of measuring cups and ounces, converting between them will be a breeze. Whether you are baking a cake or making soup, you can now easily convert between different measurement units with expert accuracy.\n\n## FAQ: ounces is 6 cups\n\n### Is 32 Oz 6 cups?\n\nRest assured, 32 ounces equates to 4 cups. No need for further concern.\n\n### Is 12 oz 6 cups?\n\nLooking to convert 12 oz to cups? Simplify the process with the easiest conversion guide – 12 US fluid ounces equals 1.5 cups. Make accurate measurements a breeze with this straightforward conversion.\n\n### What is 6 cups in ounces on a digital scale?\n\nIf you’re looking for an even more precise measurement, you may be wondering what 6 cups in ounces on a digital scale would be. To get the most accurate reading, make sure to tare your scale—this will ensure that any container or plate you place on the scale is taken into account when weighing your ingredients. Once you’ve done this, you can simply put 6 cups of the ingredient onto the scale and read the measurement in ounces.\n\n### Does 6 cups in ounces of liquid equal 6 cups in ounces of dry ingredients?\n\nNo, 6 cups in ounces of liquid does not equal 6 cups in ounces of dry ingredients. This is because different ingredients have different densities—liquid ingredients are generally much more dense than dry ones. For example, 1 cup of water is equal to 8 ounces while 1 cup of flour is only 4.5 ounces. Therefore, if a recipe calls for 6 cups of liquid, you would need 48 ounces while 6 cups of flour would only require 27 ounces.\n\n### How many ounces is 6 cups of sugar?\n\nThe same equation applies for other ingredients as well. For example, how many ounces is 6 cups of sugar? The answer is the same: 6 x 8 = 48. So, 6 cups of sugar is equal to 48 ounces.\n\n### What is 6 cups in ounces of liquid?\n\nWhen measuring liquids, you may find yourself wondering what 6 cups in ounces of liquid would be. As we’ve seen in the examples above, 6 cups is equal to 48 ounces.\n\n### Is 16 Oz 6 cups?\n\nTo find out the number of cups in 16 fluid ounces, simply divide 16 by 8. Therefore, 16 ounces equals 2 cups.\n\n### How many ounces is 6 cups of water?\n\nNow that we’ve discussed the general conversion equation, let’s look at a specific example. How many ounces is 6 cups of water? To calculate this, simply multiply 6 by 8: 6 x 8 = 48. So, 6 cups of water is equal to 48 ounces.\n\n### Does 6 cups in ounces of liquid equal 6 cups in grams?\n\nNo, 6 cups in ounces of liquid does not equal 6 cups in grams. This is because the conversion between ounces and grams varies depending on the ingredient being measured. For instance, 1 cup of water is equal to 8 ounces and 236 grams while 1 cup of flour is 4.5 ounces and 120 grams. So, 6 cups of water would be equal to 48 ounces and 1392 grams while 6 cups of flour would be 27 ounces and 720 grams.\n\n### Is 6 cups in ounces more or less than 6 cups in milliliters?\n\n6 cups in ounces is typically much more than 6 cups in milliliters—this is because one cup of liquid is equal to 8 ounces and 236 milliliters. Therefore, 6 cups of liquid would be equal to 48 ounces and 1416 milliliters." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.92404455,"math_prob":0.9840871,"size":9232,"snap":"2023-40-2023-50","text_gpt3_token_len":2316,"char_repetition_ratio":0.19375813,"word_repetition_ratio":0.1632537,"special_character_ratio":0.25433275,"punctuation_ratio":0.102564104,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98951316,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-09-27T21:25:01Z\",\"WARC-Record-ID\":\"<urn:uuid:11b0f9af-1581-4b71-a8d6-936f6146b0d9>\",\"Content-Length\":\"212258\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:93e7e8f9-cc2f-496b-a292-a29a099c21fb>\",\"WARC-Concurrent-To\":\"<urn:uuid:77b505f1-d638-49a5-b026-c1b7602eee6e>\",\"WARC-IP-Address\":\"67.202.92.13\",\"WARC-Target-URI\":\"https://mollysmtview.com/recipes/how-many-ounces-is-6-cups\",\"WARC-Payload-Digest\":\"sha1:UZDTXF5BAEPHD4JOJPDOKF4L3QZ6M5HB\",\"WARC-Block-Digest\":\"sha1:7IGVHOGOTU7WU537F5JB7ACJNYRLT4YT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510326.82_warc_CC-MAIN-20230927203115-20230927233115-00078.warc.gz\"}"}