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	<updated>2026-09-21T11:43:54Z</updated>
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	<entry>
		<id>https://en.formulasearchengine.com/w/index.php?title=Epistemic_modal_logic&amp;diff=246271</id>
		<title>Epistemic modal logic</title>
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		<updated>2014-12-31T07:26:28Z</updated>

		<summary type="html">&lt;p&gt;24.19.233.213: /* Semantics */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Nice to satisfy you, my name is Refugia. Years ago we moved to North Dakota. Doing ceramics is what my family members and I enjoy. Since she was 18 she&#039;s been operating as a meter reader but she&#039;s always wanted her personal company.&amp;lt;br&amp;gt;&amp;lt;br&amp;gt;Feel free to surf to my web blog [http://www.biogids.nl/biobank/2014-06-13/how-can-1-especially-get-around-todays-diseases std testing at home]&lt;/div&gt;</summary>
		<author><name>24.19.233.213</name></author>
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	<entry>
		<id>https://en.formulasearchengine.com/w/index.php?title=Lagrange%27s_theorem_(number_theory)&amp;diff=245059</id>
		<title>Lagrange&#039;s theorem (number theory)</title>
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		<updated>2014-02-17T08:48:58Z</updated>

		<summary type="html">&lt;p&gt;24.19.12.242: Undid revision 595835057 by Wcherowi (talk) Reverted revert; see talk page for rationale&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;== Hollister London  och det bör vara en prioriterad . ==&lt;br /&gt;
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== Air Max 2014  samt ställa boskap lös. Nicholas Dorsett ==&lt;br /&gt;
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== Nike Free 5.0  perfekt orienterade ==&lt;br /&gt;
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== Nike Free Run Dam  Patrick Duffy ==&lt;br /&gt;
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== Aflevering een Ray Ban Wayfarer ==&lt;br /&gt;
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== ze naar een vuilnisbelt Ray Ban Aviator ==&lt;br /&gt;
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Zorg ervoor dat je gefocust blijven op de titel voor een artikel. Het probleem met dat is dat het bevat slechts 200 comments als je meer dan dat, zult u het overschot verliezen. Plannen voor de toekomst is een langdurig proces. Sommige zijn te gevoelig dat zelfs de verlichting van mobiele telefoons kunt ze wakker. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;Wat kunnen we anders doen? Zijn we medeverantwoordelijk? Als we enige verantwoordelijkheid en we zijn in staat om die verantwoordelijkheid te nemen, het maakt het een beetje makkelijker om deze volgende suggestie te overwegen.. Elke keer dat deze items hebben hun doel gediend en worden vervolgens gedumpt, ze naar een vuilnisbelt, waar ze een enorme hoeveelheid ruimte innemen vervoerd. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;Denk na over bedrijven die u hoog in het vaandel ondersteunen, of wie je echt hebben gevoeld veel waardering voor. Uiteraard al deze fasen zal een bepaalde periode van cursus. Borstvoeding is een effectieve manier om gewichtstoename te beheersen na de zwangerschap. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;Gezien het feit dat je in staat om moeilijke situaties in het dagelijks leven omgaan, zou het ook eenvoudig voor u om te gaan met uw dolende tieners. Dus, moet u de combinatietherapie van de Bluze Capsules en Mast Mood olie gebruiken. Ontwikkelen van de behoeften van uw bedrijf dan beslissen over een tool baseding op hen. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;Bijvoorbeeld, vloer zorg experts gebruiken neutrale pH (7) Reiniger voor het dagelijks onderhoud en alkalische pH (12 13) voor afwerking verwijderen. Ja, u. Gewoon niet roken is de eerste en het makkelijke gedeelte, maar om permanent te stoppen met roken vergt een diepere verandering. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;&#039;We hebben hun langdurige vrijheidsstraffen ondervraging aan de samenzwering ontgraven en naar de bron die dit idee van de vermeende afpersing van geld van de Jindal Group [http://www.studiodeprez.be/studioverhuur/images/reservatie.asp?r=36-Ray-Ban-Aviator Ray Ban Aviator] had gekookt want het is niet een enkele persoon beslissing weten.. Een voordeel van uw kant is het feit dat je [http://www.campagnesurmer.be/includes/kalender.asp?p=17-Hollister-Belgie-Online Hollister Belgie Online] niet hoeft te uw letselschade advocaat te betalen, tenzij je je nederzetting winnen. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;Helaas voor veel te meerdere het is een van de laatste kwesties die wordt gemaakt als het gaat om de opleidingen van onze kinderen. [http://www.dekringledegem.be/backoffice/nvt/reserveren.asp?id=49-Longchamp-Tassen-Online Longchamp Tassen Online] Tijdens het weekend, Fielder deed het weer, de opening van een nep koffie winkel genaamd Dumb Starbucks in Los Feliz, Californië, dat [http://www.studiodeprez.be/studioverhuur/images/reservatie.asp?r=3-Ray-Ban-Justin-Rb4165 Ray Ban Justin Rb4165] de lijnen trok rond het blok voor de gemiddelde koffie en supermarkt gebak.. &amp;lt;br&amp;gt;&amp;lt;br&amp;gt;Blijf op de mogelijkheden van een tegenbod na elk interview te brengen. Als gevolg van de vicieuze cirkel van alle do&#039;s en don&#039;ts voor het uitvoeren van Super Power Yajnas en sinds de yajna uitvoerder moet [http://www.daelprinting.be/en/cms/inc/categorie.asp?page=105-Woolrich-Jas-Kopen Woolrich Jas Kopen] de last van de enorme kosten van het betalen van Dakshina (tegen) tot priesters dragen / pandits mensen genegeerd Super Power Yajnas.&amp;lt;ul&amp;gt;&lt;br /&gt;
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&amp;lt;/ul&amp;gt;&lt;/div&gt;</summary>
		<author><name>24.19.12.242</name></author>
	</entry>
	<entry>
		<id>https://en.formulasearchengine.com/w/index.php?title=129_(number)&amp;diff=7356</id>
		<title>129 (number)</title>
		<link rel="alternate" type="text/html" href="https://en.formulasearchengine.com/w/index.php?title=129_(number)&amp;diff=7356"/>
		<updated>2014-01-13T08:09:59Z</updated>

		<summary type="html">&lt;p&gt;24.19.91.7: /* In other fields */&lt;/p&gt;
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&lt;div&gt;&#039;&#039;&#039;Granular computing&#039;&#039;&#039; (GrC) is an emerging computing paradigm of [[information processing]].  It concerns the processing of complex information entities called information granules, which arise in the process of data abstraction and derivation of knowledge from information or data.  Generally speaking, information granules are collections of entities that usually originate at the numeric level and  are arranged together due to their similarity, functional or physical adjacency, indistinguishability, coherency, or the like.  &lt;br /&gt;
&lt;br /&gt;
At present, granular computing is more a &#039;&#039;theoretical perspective&#039;&#039; than a coherent set of methods or principles.  As a theoretical perspective, it encourages an approach to data that recognizes and exploits the knowledge present in data at various levels of resolution or scales.  In this sense, it encompasses all methods which provide flexibility and adaptability in the resolution at which knowledge or information is extracted and represented.&lt;br /&gt;
&lt;br /&gt;
== Types of granulation ==&lt;br /&gt;
[[Image:Catarina 26 mar 2004 1310Z.jpg|thumb|upright=1.10|Satellite view of cyclone.]] &lt;br /&gt;
[[Image:NASA Manhattan.jpg|thumb|upright=1.10|Satellite view of Manhattan.]] &lt;br /&gt;
As mentioned above, &#039;&#039;granular computing&#039;&#039; is not an algorithm or process; there is not a particular method that is called &amp;quot;granular computing&amp;quot;.  It is rather an approach  to looking at data that recognizes how different and interesting regularities in the data can appear at different levels of granularity, much as different features become salient in [[satellite images]] of greater or lesser resolution.  On a low-resolution satellite image, for example, one might notice interesting cloud patterns representing [[cyclones]] or other large-scale weather phenomena, while in a higher-resolution image, one misses these large-scale atmospheric phenomena but instead notices smaller-scale phenomena, such as the interesting pattern that is the streets of [[Manhattan]].  The same is generally true of all data: At different resolutions or granularities, different features and relationships emerge.  The aim of granular computing is ultimately simply to try to take advantage of this fact in designing more-effective machine-learning and reasoning systems. &lt;br /&gt;
&lt;br /&gt;
There are several types of granularity that are often encountered in [[data mining]] and [[machine learning]], and we review them below:&lt;br /&gt;
&lt;br /&gt;
=== Value granulation (discretization/quantization) ===&lt;br /&gt;
One type of granulation is the [[Quantization (signal processing)|quantization]] of variables.  It is very common that in data mining or machine-learning applications that the resolution of variables needs to be &#039;&#039;decreased&#039;&#039; in order to extract meaningful regularities.  An example of this would be a variable such as &amp;quot;outside temperature&amp;quot; (&amp;lt;math&amp;gt;temp&amp;lt;/math&amp;gt;), which in a given application might be recorded to several decimal places of [[Arithmetic precision|precision]] (depending on the sensing apparatus).  However, for purposes of extracting relationships between &amp;quot;outside temperature&amp;quot; and, say, &amp;quot;number of health-club applications&amp;quot; (&amp;lt;math&amp;gt;club &amp;lt;/math&amp;gt;), it will generally be advantageous to quantize &amp;quot;outside temperature&amp;quot; into a smaller number of intervals.&lt;br /&gt;
&lt;br /&gt;
==== Motivations ====&lt;br /&gt;
There are several interrelated reasons for granulating variables in this fashion:&lt;br /&gt;
* Based on prior domain knowledge, there is no expectation that minute variations in temperature (e.g., the difference between {{convert|80|-|80.7|°F|C|1}}) could have an influence on behaviors driving the number of health-club applications.  For this reason, any &amp;quot;regularity&amp;quot; which our learning algorithms might detect at this level of resolution would have to be &#039;&#039;spurious&#039;&#039;, as an artifact of overfitting.  By coarsening the temperature variable into intervals the difference between which we &#039;&#039;do&#039;&#039; anticipate (based on prior domain knowledge) might influence  number of health-club applications, we eliminate the possibility of detecting these spurious  patterns.  Thus, in this case, reducing resolution is a method of controlling [[overfitting]].&lt;br /&gt;
* By reducing the number of intervals in the temperature variable (i.e., increasing its &#039;&#039;grain size&#039;&#039;), we increase the amount of sample data indexed by each interval designation.  Thus, by coarsening the variable, we increase sample sizes and achieve better statistical estimation.  In this sense, increasing granularity provides an antidote to the so-called &#039;&#039;[[curse of dimensionality]]&#039;&#039;, which relates to the exponential decrease in statistical power with increase in number of dimensions or variable cardinality.&lt;br /&gt;
*Independent of prior domain knowledge, it is often the case that meaningful regularities (i.e., which can be detected by a given learning methodology, representational language, etc.) may exist at one level of resolution and not at another.&lt;br /&gt;
&lt;br /&gt;
[[Image:Value granulation.png|thumb|200 px|Benefits of value granulation: Implications here exist at the resolution of &amp;lt;math&amp;gt;\{X_i,Y_j\}&amp;lt;/math&amp;gt; that do not exist at the higher resolution of &amp;lt;math&amp;gt;\{x_i,y_j\}&amp;lt;/math&amp;gt;; in particular, &amp;lt;math&amp;gt;\forall x_i,y_j: x_i \not\to y_j&amp;lt;/math&amp;gt;, while at the same time, &amp;lt;math&amp;gt;\forall X_i \exists Y_j: X_i \leftrightarrow Y_j&amp;lt;/math&amp;gt;.]]&lt;br /&gt;
For example, a simple learner or pattern recognition system may seek to extract regularities satisfying a [[conditional probability]] threshold such as &amp;lt;math&amp;gt;p(Y=y_j|X=x_i) \ge \alpha &amp;lt;/math&amp;gt;.  In the special case where &amp;lt;math&amp;gt;\alpha = 1 &amp;lt;/math&amp;gt;, this recognition system is essentially detecting &#039;&#039;[[logical implication]]&#039;&#039; of the form &amp;lt;math&amp;gt;X=x_i \rightarrow Y=y_j &amp;lt;/math&amp;gt; or, in words, &amp;quot;if &amp;lt;math&amp;gt;X=x_i&amp;lt;/math&amp;gt;, then &amp;lt;math&amp;gt;Y=y_j &amp;lt;/math&amp;gt;&amp;quot;.  The system&#039;s ability to recognize such implications (or, in general, conditional probabilities exceeding threshold)  is partially contingent on the resolution with which the system analyzes the variables.&lt;br /&gt;
&lt;br /&gt;
As an example of this last point, consider the feature space shown to the right.  The variables may each be regarded at two different resolutions.  Variable &amp;lt;math&amp;gt;X&amp;lt;/math&amp;gt; may be regarded at a high (quaternary) resolution wherein it takes on the four values &amp;lt;math&amp;gt;\{x_1, x_2, x_3, x_4\}&amp;lt;/math&amp;gt; or at a lower (binary) resolution wherein it takes on the two values &amp;lt;math&amp;gt;\{X_1, X_2\}&amp;lt;/math&amp;gt;.  Similarly, variable &amp;lt;math&amp;gt;Y&amp;lt;/math&amp;gt; may be regarded at a high (quaternary) resolution or at a lower (binary) resolution, where it takes on the values &amp;lt;math&amp;gt;\{y_1, y_2, y_3, y_4\}&amp;lt;/math&amp;gt; or &amp;lt;math&amp;gt;\{Y_1, Y_2\}&amp;lt;/math&amp;gt;, respectively.  It will be noted that at the high resolution, there are &#039;&#039;&#039;no&#039;&#039;&#039; detectable implications of the form &amp;lt;math&amp;gt;X=x_i \rightarrow Y=y_j &amp;lt;/math&amp;gt;, since every &amp;lt;math&amp;gt;x_i&amp;lt;/math&amp;gt; is associated with more than one &amp;lt;math&amp;gt;y_j&amp;lt;/math&amp;gt;, and thus, for all &amp;lt;math&amp;gt;x_i&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;p(Y=y_j|X=x_i) &amp;lt; 1 &amp;lt;/math&amp;gt;.  However, at the low (binary) variable resolution, two bilateral implications become detectable:    &amp;lt;math&amp;gt;X=X_1 \leftrightarrow Y=Y_1 &amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;X=X_2 \leftrightarrow Y=Y_2 &amp;lt;/math&amp;gt;, since every &amp;lt;math&amp;gt;X_1&amp;lt;/math&amp;gt; occurs &#039;&#039;iff&#039;&#039; &amp;lt;math&amp;gt;Y_1&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;X_2&amp;lt;/math&amp;gt; occurs &#039;&#039;iff&#039;&#039; &amp;lt;math&amp;gt;Y_2&amp;lt;/math&amp;gt;.  Thus, a pattern recognition system scanning for implications of this kind would find them at the binary variable resolution, but would fail to find them at the   higher quaternary variable resolution.&lt;br /&gt;
&lt;br /&gt;
====Issues and methods====&lt;br /&gt;
It is not feasible to exhaustively test all possible discretization resolutions on all variables in order to see which combination of resolutions yields interesting or significant results.  Instead, the feature space must be preprocessed (often by an [[information entropy|entropy]] analysis of some kind) so that some guidance can be given as to how the discretization process should proceed.  Moreover, one cannot generally achieve good results by naively analyzing and discretizing each variable independently, since this may obliterate the very interactions that we had hoped to discover. &lt;br /&gt;
&lt;br /&gt;
A sample of papers that address the problem of variable discretization in general, and multiple-variable discretization in particular, is as follows: {{Harvtxt|Chiu|Wong|Cheung|1991}}, {{Harvtxt|Bay|2001}},  {{Harvtxt|Liu|Hussain|Tan|Dasii|2002}}, {{Harvtxt|Wang|Liu|1998}}, {{Harvtxt|Zighed|Rabaséda|Rakotomalala|1998}}, {{Harvtxt|Catlett|1991}}, {{Harvtxt|Dougherty|Kohavi|Sahami|1995}}, {{Harvtxt|Monti|Cooper|1999}}, {{Harvtxt|Fayyad|Irani|1993}}, {{Harvtxt|Chiu|Cheung|Wong|1990}}, {{Harvtxt|Nguyen|Nguyen|1998}}, {{Harvtxt|Grzymala-Busse|Stefanowski|2001}}, {{Harvtxt|Ting|1994}}, {{Harvtxt|Ludl|Widmer|2000}}, {{Harvtxt|Pfahringer|1995}}, {{Harvtxt|An|Cercone|1999}}, &lt;br /&gt;
{{Harvtxt|Chiu|Cheung|1989}}, {{Harvtxt|Chmielewski|Grzymala-Busse|1996}}, {{Harvtxt|Lee|Shin|1994}}, {{Harvtxt|Liu|Wellman|2002}}, {{Harvtxt|Liu|Wellman|2004}}.&lt;br /&gt;
&lt;br /&gt;
=== Variable granulation (clustering/aggregation/transformation) ===&lt;br /&gt;
Variable granulation is a term that could describe a variety of techniques, most of which are aimed at reducing dimensionality, redundancy, and storage requirements.  We briefly describe some of the ideas here, and present pointers to the literature.&lt;br /&gt;
&lt;br /&gt;
====Variable transformation====&lt;br /&gt;
A number of classical methods, such as [[principal component analysis]], [[multidimensional scaling]], [[factor analysis]], and [[structural equation modeling]], and their relatives, fall under the genus of &amp;quot;variable transformation.&amp;quot;  Also in this category are more modern areas of study such as [[dimensionality reduction]], [[projection pursuit]], and [[independent component analysis]].  The common goal of these methods in general is to find a representation of the data in terms of new variables, which are a linear or nonlinear transformation of the original variables, and in which important statistical relationships emerge. The resulting variable sets are almost always smaller than the original variable set, and hence these methods can be loosely said to impose a granulation on the feature space.  These dimensionality reduction methods are all reviewed in the standard texts, such as {{Harvtxt|Duda|Hart|Stork|2001}}, {{Harvtxt|Witten|Frank|2005}}, and {{Harvtxt|Hastie|Tibshirani|Friedman|2001}}.&lt;br /&gt;
&lt;br /&gt;
====Variable aggregation====&lt;br /&gt;
A different class of variable granulation methods derive more from [[data clustering]] methodologies than from the linear systems theory informing the above methods.  It was noted fairly early that one may consider &amp;quot;clustering&amp;quot; related variables in just  the same way that one considers clustering related data.  In data clustering, one identifies a group of similar entities (using a measure of &amp;quot;similarity&amp;quot; suitable to the domain), and then in some sense &#039;&#039;replaces&#039;&#039; those entities with a prototype of some kind.  The prototype may be the simple average of the data in the identified cluster, or some other representative measure.  But the key idea is that in subsequent operations, we may be able to use the single prototype for the data cluster (along with perhaps a statistical model describing how exemplars are derived from the prototype) to &#039;&#039;stand in&#039;&#039; for the much larger set of exemplars.   These prototypes are generally such as to capture most of the information of interest concerning the entities.&lt;br /&gt;
&lt;br /&gt;
[[Image:Kraskov tree.png|thumb|400 px|A Watanabe-Kraskov variable agglomeration tree. Variables are agglomerated (or &amp;quot;unitized&amp;quot;) from the bottom-up, with each merge-node representing a (constructed) variable having entropy equal to the joint entropy of the agglomerating variables. Thus, the agglomeration of two m-ary variables &amp;lt;math&amp;gt;X_1&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;X_2&amp;lt;/math&amp;gt; having individual entropies &amp;lt;math&amp;gt;H(X_1)&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;H(X_2)&amp;lt;/math&amp;gt; yields a single &amp;lt;math&amp;gt;m^2&amp;lt;/math&amp;gt;-ary variable &amp;lt;math&amp;gt;X_{1,2}&amp;lt;/math&amp;gt; with entropy &amp;lt;math&amp;gt;H(X_{1,2})=H(X_1,X_2)&amp;lt;/math&amp;gt;. When &amp;lt;math&amp;gt;X_1&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;X_2&amp;lt;/math&amp;gt; are highly dependent (i.e., redundant) and have large mutual information &amp;lt;math&amp;gt;I(X_1;X_2)&amp;lt;/math&amp;gt;, then &amp;lt;math&amp;gt;H(X_{1,2})&amp;lt;/math&amp;gt; &amp;amp;#x226A; &amp;lt;math&amp;gt;H(X_1)+H(X_2)&amp;lt;/math&amp;gt; because &amp;lt;math&amp;gt;H(X_1,X_2)=H(X_1)+H(X_2)-I(X_1;X_2)&amp;lt;/math&amp;gt;, and this would be considered a parsimonious unitization or aggregation.]]&lt;br /&gt;
Similarly, it is reasonable to ask whether a large set of variables might be aggregated into a smaller set of &#039;&#039;prototype&#039;&#039; variables that capture the most salient relationships between the variables.    Although variable clustering methods based on [[linear correlation]] have been proposed ({{Harvnb|Duda|Hart|Stork|2001}};{{Harvnb|Rencher|2002}}), more powerful methods of variable clustering are based on the [[mutual information]] between variables. Watanabe has shown ({{Harvnb|Watanabe|1960}};{{Harvnb|Watanabe|1969}}) that for any set of variables one can construct a &#039;&#039;polytomic&#039;&#039; (i.e., n-ary) tree representing a series of variable agglomerations in which the ultimate &amp;quot;total&amp;quot; correlation  among the complete variable set is the sum of the &amp;quot;partial&amp;quot; correlations exhibited by each agglomerating subset (see figure). Watanabe suggests that an observer might seek to thus partition a system in such a way as to minimize the interdependence between the parts &amp;quot;... as if they were looking for a natural division or a hidden crack.&amp;quot; &lt;br /&gt;
&lt;br /&gt;
One practical approach to building such a tree is to successively choose for agglomeration the two variables (either atomic variables or previously agglomerated variables) which have the highest pairwise mutual information {{Harv|Kraskov|Stögbauer|Andrzejak|Grassberger|2003}}. The product of each agglomeration is a new (constructed) variable that reflects the local [[joint distribution]] of the two agglomerating variables, and thus possesses an entropy equal to their [[joint entropy]].&lt;br /&gt;
(From a procedural standpoint, this agglomeration step involves replacing two columns in the attribute-value table—representing the two agglomerating variables—with a single column that has a unique value for every unique combination of values in the replaced columns {{Harv|Kraskov|Stögbauer|Andrzejak|Grassberger|2003}}.  No information is lost by such an operation; however, it should be noted that if one is exploring the data for inter-variable relationships, it would generally &#039;&#039;not&#039;&#039; be desirable to merge redundant variables in this way, since in such a context it is likely to be precisely the redundancy or &#039;&#039;dependency&#039;&#039; between variables that is of interest;  and once redundant variables are merged, their relationship to one another can no longer be studied.&lt;br /&gt;
&lt;br /&gt;
=== System granulation (aggregation) ===&lt;br /&gt;
&lt;br /&gt;
In [[database systems]], aggregations (see e.g. [[OLAP|OLAP aggregation]] and [[Business intelligence]] systems) result in transforming original data tables (often called information systems) into the tables with different semantics of rows and columns, wherein the rows correspond to the groups (granules) of original tuples and the columns express aggregated information about original values within each of the groups. Such aggregations are usually based on SQL and its extensions. The resulting granules usually correspond to the groups of original tuples with the same values (or ranges) over some pre-selected original columns. &lt;br /&gt;
&lt;br /&gt;
There are also other approaches wherein the groups are defined basing on, e.g., physical adjacency of rows. For example, [[Infobright]] implements a database engine wherein data is partitioned onto &#039;&#039;rough rows&#039;&#039;, each consisting of 64K of physically consecutive (or almost consecutive) rows. Rough rows are automatically labeled with compact information about their values on data columns, often involving multi-column and multi-table relationships. It results in a higher layer of granulated information systems where objects correspond to rough rows and attributes - to various flavors of rough information. Database operations can be efficiently supported within such a new framework, with an access to the original data pieces still available.&lt;br /&gt;
&lt;br /&gt;
=== Concept granulation (component analysis) ===&lt;br /&gt;
The origins of the &#039;&#039;granular computing&#039;&#039; ideology are to be found in the [[rough sets]] and [[fuzzy sets]] literatures.  One of the key insights of rough set research—although by no means unique to it—is that, in general, the selection of different sets of features or variables will yield different &#039;&#039;concept&#039;&#039; granulations.  Here, as in elementary rough set theory, by &amp;quot;concept&amp;quot; we mean a set of entities that are &#039;&#039;indistinguishable&#039;&#039; or &#039;&#039;indiscernible&#039;&#039; to the observer (i.e., a simple concept), or a set of entities that is composed from such simple concepts (i.e., a complex concept).  To put it in other words, by projecting a data set ([[value-attribute system]]) onto different sets of variables, we recognize alternative sets of equivalence-class &amp;quot;concepts&amp;quot; in the data, and these different sets of concepts will in general be conducive to the extraction of different relationships and regularities.&lt;br /&gt;
&lt;br /&gt;
====Equivalence class granulation====&lt;br /&gt;
We illustrate with an example.  Consider the attribute-value system below:&lt;br /&gt;
&lt;br /&gt;
:{| class=&amp;quot;wikitable&amp;quot; style=&amp;quot;text-align:center; width:30%&amp;quot; border=&amp;quot;1&amp;quot;&lt;br /&gt;
|+ Sample Information System&lt;br /&gt;
! Object !! &amp;lt;math&amp;gt;P_{1}&amp;lt;/math&amp;gt; !! &amp;lt;math&amp;gt;P_{2}&amp;lt;/math&amp;gt; !! &amp;lt;math&amp;gt;P_{3}&amp;lt;/math&amp;gt; !! &amp;lt;math&amp;gt;P_{4}&amp;lt;/math&amp;gt; !! &amp;lt;math&amp;gt;P_{5}&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{1}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 1 || 2 || 0 || 1 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 1 || 2 || 0 || 1 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{3}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 0 || 0 || 1 || 0&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{4}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 0 || 0 || 1 || 2 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{5}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 1 || 0 || 2 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{6}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 0 || 0 || 1 || 2 || 2&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{7}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 0 || 0 || 1 || 0&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{8}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 0 || 1 || 2 || 2 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{9}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 1 || 0 || 2 || 2&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 0 || 0 || 1 || 0&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
When the full set of attributes &amp;lt;math&amp;gt;P = \{P_{1},P_{2},P_{3},P_{4},P_{5}\}&amp;lt;/math&amp;gt; is considered, we see that we have the following seven equivalence classes or primitive (simple) concepts:&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;&lt;br /&gt;
\begin{cases} &lt;br /&gt;
\{O_{1},O_{2}\} \\ &lt;br /&gt;
\{O_{3},O_{7},O_{10}\} \\ &lt;br /&gt;
\{O_{4}\} \\ &lt;br /&gt;
\{O_{5}\} \\&lt;br /&gt;
\{O_{6}\} \\&lt;br /&gt;
\{O_{8}\} \\&lt;br /&gt;
\{O_{9}\} \end{cases}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Thus, the two objects within the first equivalence class, &amp;lt;math&amp;gt;\{O_{1},O_{2}\}&amp;lt;/math&amp;gt;,  cannot be distinguished from one another based on the available attributes, and the three objects within the second equivalence class, &amp;lt;math&amp;gt;\{O_{3},O_{7},O_{10}\}&amp;lt;/math&amp;gt;, cannot be distinguished from one another based on the available attributes.  The remaining five objects are each discernible from all other objects.  Now, let us imagine a projection of the attribute value system onto attribute &amp;lt;math&amp;gt;P_{1}&amp;lt;/math&amp;gt; alone, which would represent, for example, the  view from an observer which is only capable of detecting this single attribute. Then we obtain the following much coarser equivalence class structure.  &lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;&lt;br /&gt;
\begin{cases} &lt;br /&gt;
\{O_{1},O_{2}\} \\ &lt;br /&gt;
\{O_{3},O_{5},O_{7},O_{9},O_{10}\} \\ &lt;br /&gt;
\{O_{4},O_{6},O_{8}\} \end{cases}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This is in a certain regard the same structure as before, but at a lower degree of resolution (larger grain size).  Just as in the case of [[#Value granulation (discretization/quantization)|value granulation (discretization/quantization)]], it is possible that relationships (dependencies) may emerge at one level of granularity that are not present at another.  As an example of this, we can consider the effect of concept granulation on the measure known as &#039;&#039;attribute dependency&#039;&#039; (a simpler relative of the [[mutual information]]).&lt;br /&gt;
&lt;br /&gt;
To establish this notion of dependency (see also [[rough sets]]), let &amp;lt;math&amp;gt;[x]_Q = \{Q_1, Q_2, Q_3, \dots, Q_N \}&amp;lt;/math&amp;gt; represent a particular concept granulation, where each &amp;lt;math&amp;gt;Q_i&amp;lt;/math&amp;gt; is an equivalence class from the concept structure induced by attribute set &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt;.  For example, if the  attribute set &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt; consists of attribute &amp;lt;math&amp;gt;P_{1}&amp;lt;/math&amp;gt; alone, as above,  then the concept structure &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; will be composed of    &amp;lt;math&amp;gt;Q_1 = \{O_{1},O_{2}\}&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;Q_2 = \{O_{3},O_{5},O_{7},O_{9},O_{10}\}&amp;lt;/math&amp;gt;, and &amp;lt;math&amp;gt;Q_3 = \{O_{4},O_{6},O_{8}\}&amp;lt;/math&amp;gt;.  The &#039;&#039;&#039;dependency&#039;&#039;&#039; of attribute set &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt; on another attribute set &amp;lt;math&amp;gt;P&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;\gamma_{P}(Q)&amp;lt;/math&amp;gt;, is given by&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;&lt;br /&gt;
\gamma_{P}(Q) =  \frac{\left | \sum_{i=1}^N {\underline P}Q_i \right |} {\left | \mathbb{U} \right |} \leq 1&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
That is, for each equivalence class &amp;lt;math&amp;gt;Q_i&amp;lt;/math&amp;gt; in &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt;, we add up the size of its &amp;quot;lower approximation&amp;quot; (see [[rough sets]]) by the attributes in &amp;lt;math&amp;gt;P&amp;lt;/math&amp;gt;, i.e., &amp;lt;math&amp;gt;{\underline P}Q_i&amp;lt;/math&amp;gt;.  More simply, this approximation  is the number of objects which on attribute set &amp;lt;math&amp;gt;P&amp;lt;/math&amp;gt; can be positively identified as belonging to target set &amp;lt;math&amp;gt;Q_i&amp;lt;/math&amp;gt;.  Added across all equivalence classes in &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt;, the numerator above represents the total number of objects which—based on attribute set &amp;lt;math&amp;gt;P&amp;lt;/math&amp;gt;—can be positively categorized according to the classification induced by  attributes &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt;.  The dependency ratio therefore expresses the proportion (within the entire universe) of such classifiable objects, in a sense capturing the &amp;quot;synchronization&amp;quot; of the two concept structures &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt;.  The dependency &amp;lt;math&amp;gt;\gamma_{P}(Q)&amp;lt;/math&amp;gt; &amp;quot;can be interpreted as a proportion of such objects in the information system for which it suffices to know the values of attributes in &amp;lt;math&amp;gt;P&amp;lt;/math&amp;gt; to determine the values of attributes in &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt;&amp;quot; (Ziarko &amp;amp; Shan 1995).&lt;br /&gt;
&lt;br /&gt;
Having gotten definitions now out of the way, we can make the simple observation that the choice of concept granularity (i.e., choice of attributes) will influence the detected dependencies among attributes.  Consider again the attribute value table from above:&lt;br /&gt;
&lt;br /&gt;
:{| class=&amp;quot;wikitable&amp;quot; style=&amp;quot;text-align:center; width:30%&amp;quot; border=&amp;quot;1&amp;quot;&lt;br /&gt;
|+ Sample Information System&lt;br /&gt;
! Object !! &amp;lt;math&amp;gt;P_{1}&amp;lt;/math&amp;gt; !! &amp;lt;math&amp;gt;P_{2}&amp;lt;/math&amp;gt; !! &amp;lt;math&amp;gt;P_{3}&amp;lt;/math&amp;gt; !! &amp;lt;math&amp;gt;P_{4}&amp;lt;/math&amp;gt; !! &amp;lt;math&amp;gt;P_{5}&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{1}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 1 || 2 || 0 || 1 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{2}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 1 || 2 || 0 || 1 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{3}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 0 || 0 || 1 || 0&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{4}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 0 || 0 || 1 || 2 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{5}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 1 || 0 || 2 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{6}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 0 || 0 || 1 || 2 || 2&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{7}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 0 || 0 || 1 || 0&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{8}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 0 || 1 || 2 || 2 || 1&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{9}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 1 || 0 || 2 || 2&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;O_{10}&amp;lt;/math&amp;gt;&lt;br /&gt;
| 2 || 0 || 0 || 1 || 0&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
Let us consider the dependency of attribute set  &amp;lt;math&amp;gt;Q = \{P_4, P_5\}&amp;lt;/math&amp;gt;&lt;br /&gt;
on attribute set &amp;lt;math&amp;gt;P = \{P_2, P_3\}&amp;lt;/math&amp;gt;.  That is, we wish to know what proportion of objects can be correctly classified into classes of &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; based on knowledge of &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt;.  The equivalence classes of &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; and of &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt; are shown below.&lt;br /&gt;
&lt;br /&gt;
:{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt;&lt;br /&gt;
! &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
| &amp;lt;math&amp;gt;&lt;br /&gt;
\begin{cases} &lt;br /&gt;
\{O_{1},O_{2}\} \\ &lt;br /&gt;
\{O_{3},O_{7},O_{10}\} \\ &lt;br /&gt;
\{O_{4},O_{5},O_{8}\} \\&lt;br /&gt;
\{O_{6},O_{9}\}\end{cases}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
| &amp;lt;math&amp;gt;&lt;br /&gt;
\begin{cases} &lt;br /&gt;
\{O_{1},O_{2}\} \\ &lt;br /&gt;
\{O_{3},O_{7},O_{10}\} \\ &lt;br /&gt;
\{O_{4},O_{6}\} \\&lt;br /&gt;
\{O_{5},O_{9}\} \\&lt;br /&gt;
\{O_{8}\}\end{cases}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
The objects that can be &#039;&#039;definitively&#039;&#039; categorized according to concept structure &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; based on &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt; are those in the set &amp;lt;math&amp;gt;\{O_{1},O_{2},O_{3},O_{7},O_{8},O_{10}\}&amp;lt;/math&amp;gt;, and since there are six of these, the dependency of &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt; on &amp;lt;math&amp;gt;P&amp;lt;/math&amp;gt;,  &amp;lt;math&amp;gt;\gamma_{P}(Q) = 6/10&amp;lt;/math&amp;gt;.  This might be considered an interesting dependency in its own right, but perhaps in a particular data mining application only stronger dependencies are desired.  &lt;br /&gt;
&lt;br /&gt;
We might then consider the dependency of the smaller attribute set  &amp;lt;math&amp;gt;Q = \{P_4\}&amp;lt;/math&amp;gt;&lt;br /&gt;
on the attribute set &amp;lt;math&amp;gt;P = \{P_2, P_3\}&amp;lt;/math&amp;gt;.  The move from &amp;lt;math&amp;gt;Q = \{P_4, P_5\}&amp;lt;/math&amp;gt; to &amp;lt;math&amp;gt;Q = \{P_4\}&amp;lt;/math&amp;gt; induces a coarsening of the class structure &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt;, as will be seen shortly.  We wish again to know what proportion of objects can be correctly classified into the (now larger) classes of &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; based on knowledge of &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt;.  The equivalence classes of the new &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; and of &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt; are shown below.&lt;br /&gt;
&lt;br /&gt;
:{| class=&amp;quot;wikitable&amp;quot;&lt;br /&gt;
|-&lt;br /&gt;
! &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt;&lt;br /&gt;
! &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt;&lt;br /&gt;
|-&lt;br /&gt;
| &amp;lt;math&amp;gt;&lt;br /&gt;
\begin{cases} &lt;br /&gt;
\{O_{1},O_{2},O_{3},O_{7},O_{10}\} \\ &lt;br /&gt;
\{O_{4},O_{5},O_{6},O_{8},O_{9}\} \end{cases}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
| &amp;lt;math&amp;gt;&lt;br /&gt;
\begin{cases} &lt;br /&gt;
\{O_{1},O_{2}\} \\ &lt;br /&gt;
\{O_{3},O_{7},O_{10}\} \\ &lt;br /&gt;
\{O_{4},O_{6}\} \\&lt;br /&gt;
\{O_{5},O_{9}\} \\&lt;br /&gt;
\{O_{8}\}\end{cases}&lt;br /&gt;
&amp;lt;/math&amp;gt;&lt;br /&gt;
|}&lt;br /&gt;
&lt;br /&gt;
Clearly, &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; has a coarser granularity than it did earlier.  The objects that can now be &#039;&#039;definitively&#039;&#039; categorized according to the concept structure &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; based on &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt; constitute the complete universe &amp;lt;math&amp;gt;\{O_{1},O_{2},\ldots,O_{10}\}&amp;lt;/math&amp;gt;, and thus  the dependency of &amp;lt;math&amp;gt;Q&amp;lt;/math&amp;gt; on &amp;lt;math&amp;gt;P&amp;lt;/math&amp;gt;,  &amp;lt;math&amp;gt;\gamma_{P}(Q) = 1&amp;lt;/math&amp;gt;.  That is, knowledge of membership according to category set  &amp;lt;math&amp;gt;[x]_P&amp;lt;/math&amp;gt; is adequate to determine category membership in &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; with complete certainty; In this case we might say that &amp;lt;math&amp;gt;P \rightarrow Q&amp;lt;/math&amp;gt;.  Thus, by coarsening the concept structure, we were able to find a stronger (deterministic) dependency.  However, we also note that the classes induced in &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt; from the reduction in resolution necessary to obtain this deterministic dependency are now themselves large and few in number; as a result, the dependency we found, while strong, may be less valuable to us than the weaker dependency found earlier under the higher resolution view of &amp;lt;math&amp;gt;[x]_Q&amp;lt;/math&amp;gt;. &lt;br /&gt;
&lt;br /&gt;
In general it is not possible to test all sets of attributes to see which induced concept structures yield the strongest dependencies, and this search must be therefore be guided with some intelligence.  Papers which discuss this issue, and others relating to intelligent use of granulation, are those by Y.Y. Yao and [[Lotfi Zadeh]] listed in the [[#References]] below.&lt;br /&gt;
&lt;br /&gt;
====Component granulation====&lt;br /&gt;
Another perspective on concept granulation may be obtained from work on parametric models of categories.  In [[mixture model]] learning, for example, a set of data is explained as a mixture of distinct [[Gaussian distribution|Gaussian]] (or other) distributions.  Thus, a large amount of data is &amp;quot;replaced&amp;quot; by a small number of distributions.  The choice of the number of these distributions, and their size, can again be viewed as a problem of &#039;&#039;concept granulation&#039;&#039;.  In general, a better fit to the data is obtained by a larger number of distributions or parameters, but in order to extract meaningful patterns, it is necessary to constrain the number of distributions, thus deliberately  &#039;&#039;coarsening&#039;&#039; the concept resolution.  Finding the &amp;quot;right&amp;quot; concept resolution is a tricky problem for which many methods have been proposed (e.g., [[Akaike information criterion|AIC]], [[Bayesian information criterion|BIC]], [[Minimum description length|MDL]], etc.), and these are frequently  considered under the rubric of &amp;quot;[[model regularization]]&amp;quot;.&lt;br /&gt;
&lt;br /&gt;
==Different interpretations of granular computing==&lt;br /&gt;
Granular computing can be conceived as a framework of theories, methodologies, techniques, and tools that make use of information granules in the process of problem solving.  In this sense, granular computing is used as an umbrella term to cover topics that have been studied in various fields in isolation.  By examining all of these existing studies in light of the unified framework of granular computing and extracting their commonalities, it may be possible to develop a general theory for problem solving. &lt;br /&gt;
&lt;br /&gt;
In a more philosophical sense, granular computing can describe a way of thinking that relies on the human ability to perceive the real world under various levels of granularity (i.e., abstraction) in order to abstract and consider only those things that serve a specific interest and to switch among different granularities. By focusing on different levels of granularity, one can obtain different levels of knowledge, as well as a greater understanding of the inherent knowledge structure.  Granular computing is thus essential in human problem solving and hence has a very significant impact on the design and implementation of intelligent systems.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
* [[Rough set|Rough Sets]], [[Discretization]]&lt;br /&gt;
* [[Type-2 Fuzzy Sets and Systems]]&lt;br /&gt;
&lt;br /&gt;
== References ==&lt;br /&gt;
&amp;lt;div class=&amp;quot;references-small&amp;quot; style=&amp;quot;-moz-column-count:2; column-count:2;&amp;quot;&amp;gt;&lt;br /&gt;
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*Bargiela, A. and Pedrycz, W. (2003) &#039;&#039;Granular Computing. An introduction&#039;&#039;, Kluwer Academic Publishers&lt;br /&gt;
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*{{cite conference&lt;br /&gt;
  | first = Y. Y. | last = Yao&lt;br /&gt;
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*{{cite conference&lt;br /&gt;
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&amp;lt;/div&amp;gt;&lt;br /&gt;
&lt;br /&gt;
[[Category:Theoretical computer science]]&lt;br /&gt;
[[Category:Machine learning]]&lt;/div&gt;</summary>
		<author><name>24.19.91.7</name></author>
	</entry>
	<entry>
		<id>https://en.formulasearchengine.com/w/index.php?title=Hyperbolic_function&amp;diff=1760</id>
		<title>Hyperbolic function</title>
		<link rel="alternate" type="text/html" href="https://en.formulasearchengine.com/w/index.php?title=Hyperbolic_function&amp;diff=1760"/>
		<updated>2013-11-27T08:04:17Z</updated>

		<summary type="html">&lt;p&gt;24.19.55.181: /* Inverse functions as logarithms */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;[[Image:Tower of Hanoi.jpeg|300px|thumb|A model set of the Towers of Hanoi (with 8 disks)]]&lt;br /&gt;
[[Image:Tower of Hanoi 4.gif|300px|thumb|An animated solution of the &#039;&#039;&#039;Tower of Hanoi&#039;&#039;&#039; puzzle for &#039;&#039;T(4,3)&#039;&#039;.]]&lt;br /&gt;
[[File:UniversumUNAM34.JPG|thumb|300px|Tower of Hanoi interactive display at the [[Universum (UNAM)|Universum museum]] in Mexico City]]&lt;br /&gt;
The &#039;&#039;&#039;Tower of Hanoi&#039;&#039;&#039; (also called the &#039;&#039;&#039;Tower of Brahma&#039;&#039;&#039; or &#039;&#039;&#039;Lucas&#039; Tower&#039;&#039;&#039;,&amp;lt;ref&amp;gt;{{cite book |last=Hofstadter |first=Douglas R. |title=Metamagical Themas : Questing for the Essence of Mind and Pattern |year=1985 |publisher=Basic Books |location=New York |isbn=0-465-04540-5}}&amp;lt;/ref&amp;gt; and sometimes pluralised) is a [[mathematical game]] or [[puzzle]]. It consists of three rods, and a number of disks of different sizes which can slide onto any rod. The puzzle starts with the disks in a neat stack in ascending order of size on one rod, the smallest at the top, thus making a conical shape.&lt;br /&gt;
&lt;br /&gt;
The objective of the puzzle is to move the entire stack to another rod, obeying the following simple rules:&lt;br /&gt;
&lt;br /&gt;
# Only one disk can be moved at a time.&lt;br /&gt;
# Each move consists of taking the upper disk from one of the stacks and placing it on top of another stack i.e. a disk can only be moved if it is the uppermost disk on a stack. &lt;br /&gt;
# No disk may be placed on top of a smaller disk.&lt;br /&gt;
&lt;br /&gt;
With three disks, the puzzle can be solved in seven moves. The minimum number of moves required to solve a Tower of Hanoi puzzle is 2&amp;lt;sup&amp;gt;&#039;&#039;n&#039;&#039;&amp;lt;/sup&amp;gt; - 1, where &#039;&#039;n&#039;&#039; is the number of disks.&lt;br /&gt;
&lt;br /&gt;
== Origins ==&lt;br /&gt;
&lt;br /&gt;
The puzzle was first publicized in [[Western world|the West]] by the [[French people|French]] [[mathematician]] [[Édouard Lucas]] in 1883. There is a story about an [[India]]n temple in [[Kashi Vishwanath Temple|Kashi Vishwanath]] which contains a large room with three time-worn posts in it surrounded by 64 golden disks. [[Brahmin]] priests, acting out the command of an ancient prophecy, have been moving these disks, in accordance with the immutable rules of the Brahma, since that time. The puzzle is therefore also known as the Tower of [[Brahma]] puzzle. According to the legend, when the last move of the puzzle will be completed, the world will end.&amp;lt;ref&amp;gt;{{cite book |last=Spitznagel |first=Edward L. |title=Selected topics in mathematics |year=1971 |publisher=Holt, Rinehart and Winston |page=137 |isbn=0-03-084693-5}}&amp;lt;/ref&amp;gt; It is not clear whether Lucas invented this legend or was inspired by it.&lt;br /&gt;
&lt;br /&gt;
If the legend were true, and if the priests were able to move disks at a rate of one per second, using the smallest number of moves, it would take them 2&amp;lt;sup&amp;gt;64&amp;lt;/sup&amp;gt;−1 seconds or roughly 585 [[1,000,000,000 (number)|billion]] years&amp;lt;ref&amp;gt;{{cite book |last=Moscovich |first=Ivan |authorlink=Ivan Moscovich |title=1000 playthinks: puzzles, paradoxes, illusions &amp;amp; games |publisher=Workman |year=2001 |isbn=0-7611-1826-8 }}&amp;lt;/ref&amp;gt; or 18,446,744,073,709,551,615 turns to finish, or about 127 times the current age of the sun.&lt;br /&gt;
&lt;br /&gt;
There are many variations on this legend. For instance, in some tellings, the temple is a [[monastery]] and the priests are [[monk]]s. The temple or monastery may be said to be in different parts of the world — including [[Hanoi]], [[Vietnam]], and may be associated with any [[religion]]. In some versions, other elements are introduced, such as the fact that the tower was created at the beginning of the world, or that the priests or monks may make only one move per day.&lt;br /&gt;
&lt;br /&gt;
== Solution ==&lt;br /&gt;
The puzzle can be played with any number of disks, although many toy versions have around seven to nine of them. The minimum number of moves required to solve a Tower of Hanoi puzzle is 2&amp;lt;sup&amp;gt;&#039;&#039;n&#039;&#039;&amp;lt;/sup&amp;gt; - 1, where &#039;&#039;n&#039;&#039; is the number of disks.&amp;lt;ref&amp;gt;{{cite book |last=Petković |first=Miodrag |title=Famous Puzzles of Great Mathematicians |year=2009 |publisher=AMS Bookstore |isbn=0-8218-4814-3 |page=197}}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Iterative solution ===&lt;br /&gt;
A simple solution for the toy puzzle: Alternate moves between the smallest piece and a non-smallest piece. When moving the smallest piece, always move it to the next position in the same direction (to the right if the starting number of pieces is even, to the left if the starting number of pieces is odd). If there is no tower position in the chosen direction, move the piece to the opposite end, but then continue to move in the correct direction. For example, if you started with three pieces, you would move the smallest piece to the opposite end, then continue in the left direction after that. When the turn is to move the non-smallest piece, there is only one legal move. Doing this will complete the puzzle using the fewest number of moves to do so.&amp;lt;ref&amp;gt;{{cite journal |last=Troshkin |first=M. |title=Doomsday Comes: A Nonrecursive Analysis of the Recursive Towers-of-Hanoi Problem |journal=Focus |volume=95 |issue=2 |pages=10–14 |year= |language=Russian }}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==== Simpler statement of iterative solution ====&lt;br /&gt;
Alternating between the smallest and the next-smallest disks, follow the steps for the appropriate case:&lt;br /&gt;
&lt;br /&gt;
For an even number of disks:&lt;br /&gt;
* make the legal move between pegs A and B&lt;br /&gt;
&lt;br /&gt;
* make the legal move between pegs A and C&lt;br /&gt;
* make the legal move between pegs B and C&lt;br /&gt;
* repeat until complete&lt;br /&gt;
&lt;br /&gt;
For an odd number of disks:&lt;br /&gt;
* make the legal move between pegs A and C&lt;br /&gt;
* make the legal move between pegs A and B&lt;br /&gt;
* make the legal move between pegs C and B&lt;br /&gt;
* repeat until complete&lt;br /&gt;
&lt;br /&gt;
In each case, a total of 2ⁿ-1 moves are made.&lt;br /&gt;
&lt;br /&gt;
==== Equivalent iterative solution ====&lt;br /&gt;
&lt;br /&gt;
Another way to generate the unique optimal iterative solution:&lt;br /&gt;
&lt;br /&gt;
Number the disks 1 through n (largest to smallest).&lt;br /&gt;
* If n is odd, the first move is from the Start to the Finish peg.&lt;br /&gt;
* If n is even, the first move is from the Start to the Using peg.&lt;br /&gt;
&lt;br /&gt;
Now, add these constraints:&lt;br /&gt;
* No odd disk may be placed directly on an odd disk.&lt;br /&gt;
* No even disk may be placed directly on an even disk.&lt;br /&gt;
* Never undo your previous move (that is, do not move a disk back to its immediate last peg).&lt;br /&gt;
Considering those constraints after the first move, there is only one legal move at every subsequent turn.&lt;br /&gt;
&lt;br /&gt;
The sequence of these unique moves is an optimal solution to the problem equivalent to the iterative solution described above.&amp;lt;ref&amp;gt;{{cite journal |first1=Herbert |last1=Mayer |first2=Don |last2=Perkins |title=Towers of Hanoi Revisited |pages=80–84 |journal=SIGPLAN Notices |year=1984 |doi=10.1145/948566.948573}}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Recursive solution ===&lt;br /&gt;
&lt;br /&gt;
A key to solving this puzzle is to recognize that it can be solved by breaking the problem down into a collection of smaller problems and further breaking those problems down into even smaller problems until a solution is reached. For example:&lt;br /&gt;
&lt;br /&gt;
* label the pegs A, B, C — these labels may move at different steps&lt;br /&gt;
* let &#039;&#039;n&#039;&#039; be the total number of discs&lt;br /&gt;
* number the discs from 1 (smallest, topmost) to &#039;&#039;n&#039;&#039; (largest, bottommost)&lt;br /&gt;
&lt;br /&gt;
To move &#039;&#039;n&#039;&#039; discs from peg A to peg C:&lt;br /&gt;
# move &#039;&#039;n&#039;&#039;−1 discs from A to B. This leaves disc &#039;&#039;n&#039;&#039; alone on peg A&lt;br /&gt;
# move disc &#039;&#039;n&#039;&#039; from A to C&lt;br /&gt;
# move &#039;&#039;n&#039;&#039;−1 discs from B to C so they sit on disc &#039;&#039;n&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
The above is a recursive algorithm, to carry out steps 1 and 3, apply the same algorithm again for &#039;&#039;n&#039;&#039;−1. The entire procedure is a finite number of steps, since at some point the algorithm will be required for &#039;&#039;n&#039;&#039; = 1. This step, moving a single disc from peg A to peg B, is trivial. This approach can be given a rigorous mathematical formalism with the theory of [[dynamic programming]],&amp;lt;ref&amp;gt;{{cite journal |first=Moshe |last=Sniedovich |title= OR/MS Games: 2. The Towers of Hanoi Problem, |journal=INFORMS Transactions on Education |volume=3 |issue=1 |year=2002 |pages=34–51 |url=http://archive.ite.journal.informs.org/Vol3No1/Sniedovich/}}&amp;lt;/ref&amp;gt;&amp;lt;ref&amp;gt;{{cite book | last = Sniedovich | first = Moshe | title = Dynamic Programming: Foundations and Principles | publisher = Taylor &amp;amp; Francis | year = 2010 | isbn = 978-0-8247-4099-3 |url= http://www.taylorandfrancis.com/books/details/9780824740993/}}&amp;lt;/ref&amp;gt; and is often used as an example of recursion when teaching programming.&lt;br /&gt;
&lt;br /&gt;
==== Logical analysis of the recursive solution ====&lt;br /&gt;
&lt;br /&gt;
&amp;lt;!-- TODO:  This section needs copy-editing for tone, professionalism and clarity: edited/bolded .--&amp;gt;&lt;br /&gt;
As in many mathematical puzzles, finding a solution is made easier by solving a slightly more general problem: how to move a tower of h (h=height) disks from a starting peg &#039;&#039;&#039;A&#039;&#039;&#039; (f=from) onto a destination peg &#039;&#039;&#039;C&#039;&#039;&#039; (t=to), &#039;&#039;&#039;B&#039;&#039;&#039; being the remaining third peg and assuming &#039;&#039;&#039;t&#039;&#039;&#039;≠&#039;&#039;&#039;f&#039;&#039;&#039;. First, observe that the problem is symmetric for permutations of the names of the pegs ([[Symmetric group|symmetric group S&amp;lt;sub&amp;gt;&#039;&#039;3&#039;&#039;&amp;lt;/sub&amp;gt;]]). If a solution is known moving from peg &#039;&#039;&#039;A&#039;&#039;&#039; to peg &#039;&#039;&#039;C&#039;&#039;&#039;, then, by renaming the pegs, the same solution can be used for every other choice of starting and destination peg. If there is only one disk (or even none at all), the problem is trivial. If h=1, then simply move the disk from peg &#039;&#039;&#039;A&#039;&#039;&#039; to peg &#039;&#039;&#039;C&#039;&#039;&#039;. If h&amp;gt;1, then somewhere along the sequence of moves, the largest disk must be moved from peg &#039;&#039;&#039;A&#039;&#039;&#039; to another peg, preferably to peg &#039;&#039;&#039;C&#039;&#039;&#039;. The only situation that allows this move is when all smaller h-1 disks are on peg &#039;&#039;&#039;B&#039;&#039;&#039;. Hence, first all h-1 smaller disks must go from &#039;&#039;&#039;A&#039;&#039;&#039; to &#039;&#039;&#039;B&#039;&#039;&#039;. Subsequently move the largest disk and finally move the h-1 smaller disks from peg &#039;&#039;&#039;B&#039;&#039;&#039; to peg &#039;&#039;&#039;C&#039;&#039;&#039;. The presence of the largest disk does not impede any move of the h-1 smaller disks and can temporarily be ignored. Now the problem is reduced to moving h-1 disks from one peg to another one, first from &#039;&#039;&#039;A&#039;&#039;&#039; to &#039;&#039;&#039;B&#039;&#039;&#039; and subsequently from &#039;&#039;&#039;B&#039;&#039;&#039; to &#039;&#039;&#039;C&#039;&#039;&#039;, but the same method can be used both times by renaming the pegs. The same strategy can be used to reduce the h-1 problem to h-2, h-3, and so on until only one disk is left. This is called recursion. This algorithm can be schematized as follows. Identify the disks in order of increasing size by the natural numbers from 0 up to but not including h. Hence disk 0 is the smallest one and disk h-1 the largest one.&lt;br /&gt;
&lt;br /&gt;
The following is a procedure for moving a tower of h disks from a peg &#039;&#039;&#039;A&#039;&#039;&#039; onto a peg &#039;&#039;&#039;C&#039;&#039;&#039;, with &#039;&#039;&#039;B&#039;&#039;&#039; being the remaining third peg:&lt;br /&gt;
*Step 1: If h&amp;gt;1 then first use this procedure to move the h-1 smaller disks from peg &#039;&#039;&#039;A&#039;&#039;&#039; to peg &#039;&#039;&#039;B&#039;&#039;&#039;.&lt;br /&gt;
*Step 2: Now the largest disk, i.e. disk h can be moved from peg &#039;&#039;&#039;A&#039;&#039;&#039; to peg &#039;&#039;&#039;C&#039;&#039;&#039;.&lt;br /&gt;
*Step 3: If h&amp;gt;1 then again use this procedure to move the h-1 smaller disks from peg &#039;&#039;&#039;B&#039;&#039;&#039; to peg &#039;&#039;&#039;C&#039;&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
By means of [[mathematical induction]], it is easily proven that the above procedure requires the minimal number of moves possible, and that the produced solution is the only one with this minimal number of moves. Using [[recurrence relation]]s, the exact number of moves that this solution requires can be calculated by: &amp;lt;math&amp;gt;2^h - 1&amp;lt;/math&amp;gt;. This result is obtained by noting that steps 1 and 3 take &amp;lt;math&amp;gt;T_{h-1}&amp;lt;/math&amp;gt; moves, and step 2 takes one move, giving &amp;lt;math&amp;gt;T_h = 2T_{h-1} + 1&amp;lt;/math&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
=== Non-recursive solution ===&lt;br /&gt;
The list of moves for a tower being carried from one peg onto another one, as produced by the recursive algorithm has many regularities. When counting the moves starting from 1, the ordinal of the disk to be moved during move &#039;&#039;m&#039;&#039; is the number of times &#039;&#039;m&#039;&#039; can be divided by 2. Hence every odd move involves the smallest disk. It can also be observed that the smallest disk traverses the pegs f, t, r, f, t, r, etc. for odd height of the tower and traverses the pegs f, r, t, f, r, t, etc. for even height of the tower. This provides the following algorithm, which is easier, carried out by hand, than the recursive algorithm.&lt;br /&gt;
&lt;br /&gt;
In alternate moves:&lt;br /&gt;
* move the smallest disk to the peg it has not recently come from.&lt;br /&gt;
* move another disk legally (there will be one possibility only)&lt;br /&gt;
For the very first move, the smallest disk goes to peg t if h is odd and to peg r if h is even.&lt;br /&gt;
&lt;br /&gt;
Also observe that:&lt;br /&gt;
* Disks whose ordinals have even parity move in the same sense as the smallest disk.&lt;br /&gt;
* Disks whose ordinals have odd parity move in opposite sense.&lt;br /&gt;
* If h is even, the remaining third peg during successive moves is t, r, f, t, r, f, etc.&lt;br /&gt;
* If h is odd, the remaining third peg during successive moves is r, t, f, r, t, f, etc.&lt;br /&gt;
&lt;br /&gt;
With this knowledge, a set of disks in the middle of an optimal solution can be recovered with no more state information than the positions of each disk:&lt;br /&gt;
* Call the moves detailed above a disk&#039;s &#039;natural&#039; move.&lt;br /&gt;
* Examine the smallest top disk that is not disk 0, and note what its only (legal) move would be: (if there is no such disc, then we are either at the first or last move).&lt;br /&gt;
* If that move is the disk&#039;s &#039;natural&#039; move, then the disc has not been moved since the last disc 0 move, and that move should be taken.&lt;br /&gt;
* If that move is not the disk&#039;s &#039;natural&#039; move, then move disk 0.&lt;br /&gt;
&lt;br /&gt;
=== Binary solution ===&lt;br /&gt;
Disk positions may be determined more directly from the [[binary numeral system|binary]] (base 2) representation of the move number (the initial state being move #0, with all digits 0, and the final state being #2&amp;lt;sup&amp;gt;&#039;&#039;n&#039;&#039;&amp;lt;/sup&amp;gt;−1, with all digits 1), using the following rules:&lt;br /&gt;
* There is one binary digit ([[bit]]) for each disk&lt;br /&gt;
* The most significant (leftmost) bit represents the largest disk. A value of 0 indicates that the largest disk is on the initial peg, while a 1 indicates that it&#039;s on the final peg.&lt;br /&gt;
* The bitstring is read from left to right, and each bit can be used to determine the location of the corresponding disk.&lt;br /&gt;
* A bit with the same value as the previous one means that the corresponding disk is stacked on top the previous disk on the same peg.&lt;br /&gt;
** (That is to say: a straight sequence of 1&#039;s or 0&#039;s means that the corresponding disks are all on the same peg).&lt;br /&gt;
* A bit with a different value to the previous one means that the corresponding disk is one position to the left or right of the previous one. Whether it is left or right is determined by this rule:&lt;br /&gt;
** Assume that the initial peg is on the left and the final peg is on the right.&lt;br /&gt;
** Also assume &amp;quot;wrapping&amp;quot; - so the right peg counts as one peg &amp;quot;left&amp;quot; of the left peg, and vice versa.&lt;br /&gt;
** Let n be the number of greater disks that are located on the same peg as their first greater disk and add 1 if the largest disk is on the left peg. If n is even, the disk is located one peg to the left, if n is odd, the disk located one peg to the right.&lt;br /&gt;
&lt;br /&gt;
For example, in an 8-disk Hanoi:&lt;br /&gt;
* Move 0 = 00000000&lt;br /&gt;
** The largest disk is 0, so it is on the left (initial) peg.&lt;br /&gt;
** All other disks are 0 as well, so they are stacked on top of it. Hence all disks are on the initial peg.&lt;br /&gt;
* Move 2&amp;lt;sup&amp;gt;8&amp;lt;/sup&amp;gt;-1 = 11111111&lt;br /&gt;
** The largest disk is 1, so it is on the right (final) peg.&lt;br /&gt;
** All other disks are 1 as well, so they are stacked on top of it. Hence all disks are on the final peg and the puzzle is complete.&lt;br /&gt;
* Move 216&amp;lt;sub&amp;gt;10&amp;lt;/sub&amp;gt; = 11011000&lt;br /&gt;
** The largest disk is 1, so it is on the right (final) peg.&lt;br /&gt;
** Disk two is also 1, so it is stacked on top of it, on the right peg.&lt;br /&gt;
** Disk three is 0, so it is on another peg. Since n is odd(n=3), it is one peg to the right, i.e. on the left peg.&lt;br /&gt;
** Disk four is 1, so it is on another peg. Since n is even(n=2), it is one peg to the left, i.e. on the right peg.&lt;br /&gt;
** Disk five is also 1, so it is stacked on top of it, on the right peg.&lt;br /&gt;
** Disk six is 0, so it is on another peg. Since n is odd(n=5), the disk is one peg to the right, i.e. on the left peg.&lt;br /&gt;
** Disks seven and eight are also 0, so they are stacked on top of it, on the left peg.&lt;br /&gt;
&lt;br /&gt;
The source and destination pegs for the &#039;&#039;m&#039;&#039;th move can also be found elegantly from the binary representation of &#039;&#039;m&#039;&#039; using [[bitwise operation]]s. To use the syntax of the [[C (programming language)|C programming language]], move &#039;&#039;m&#039;&#039; is from peg &amp;lt;code&amp;gt;(m&amp;amp;m-1)%3&amp;lt;/code&amp;gt; to peg &amp;lt;code&amp;gt;((m|m-1)+1)%3&amp;lt;/code&amp;gt;, where the disks begin on peg 0 and finish on peg 1 or 2 according as whether the number of disks is even or odd. Another formulation is from peg &amp;lt;code&amp;gt;(m-(m&amp;amp;-m))%3&amp;lt;/code&amp;gt; to peg &amp;lt;code&amp;gt;(m+(m&amp;amp;-m))%3&amp;lt;/code&amp;gt;.&lt;br /&gt;
&lt;br /&gt;
Furthermore the disk to be moved is determined by the number of times the move count (m) can be divided by 2 (i.e. the number of zero bits at the right), counting the first move as 1 and identifying the disks by the numbers 0, 1, 2 etc. in order of increasing size. This permits a very fast non-recursive computer implementation to find the positions of the disks after m moves without reference to any previous move or distribution of disks.&lt;br /&gt;
&lt;br /&gt;
The [[count trailing zeros]] (ctz) operation, which counts the number of consecutive zeros at the end of a binary number, gives a simple solution to the problem: the disks are numbered from zero, and at move &#039;&#039;m&#039;&#039;, disk number ctz(&#039;&#039;m&#039;&#039;) is moved the minimum possible distance to the right (circling back around to the left as needed).&amp;lt;ref&amp;gt;{{cite book |last=Warren |first=Henry S. |title=Hacker&#039;s delight |year=2003 |publisher=Addison-Wesley |location=Boston MA |isbn=0-201-91465-4 |edition=1st |chapter=Section 5-4: Counting Trailing 0&#039;s.}}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
=== Gray code solution ===&lt;br /&gt;
The binary numeral system of [[Gray code]]s gives an alternative way of solving the puzzle. In the Gray system, numbers are expressed in a binary combination of 0s and 1s, but rather than being a standard [[numeral system|positional numeral system]], Gray code operates on the premise that each value differs from its predecessor by only one (and exactly one) bit changed. The number of bits present in Gray code is important, and leading zeros are not optional, unlike in positional systems.&lt;br /&gt;
&lt;br /&gt;
If one counts in Gray code of a bit size equal to the number of disks in a particular Tower of Hanoi, begins at zero, and counts up, then the bit changed each move corresponds to the disk to move, where the least-significant-bit is the smallest disk and the most-significant-bit is the largest.&lt;br /&gt;
&lt;br /&gt;
:Counting moves from 1 and identifying the disks by numbers starting from 0 in order of increasing size, the ordinal of the disk to be moved during move m is the number of times m can be divided by 2.&lt;br /&gt;
&lt;br /&gt;
This technique identifies which disk to move, but not where to move it to. For the smallest disk there are always two possibilities. For the other disks there is always one possibility, except when all disks are on the same peg, but in that case either it is the smallest disk that must be moved or the objective has already been achieved. Luckily, there is a rule which does say where to move the smallest disk to. Let f be the starting peg, t the destination peg and r the remaining third peg. If the number of disks is odd, the smallest disk cycles along the pegs in the order f→t→r→f→t→r, etc. If the number of disks is even, this must be reversed: f→r→t→f→r→t etc.&amp;lt;ref&amp;gt;{{cite book |first=Charles D. |last=Miller |chapter=Ch. 4: Binary Numbers and the Standard Gray Code |chapterurl=https://web.archive.org/web/20040821062630/http://occawlonline.pearsoned.com/bookbind/pubbooks/miller2_awl/chapter4/essay1/deluxe-content.html#tower |title=Mathematical Ideas |edition=9 |publisher=Addison Wesley Longman |year=2000 |isbn=0321076079 }}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== Graphical representation ==&lt;br /&gt;
The game can be represented by an undirected [[graph (mathematics)|graph]], the nodes representing distributions of disks and the edges representing moves. For one disk, the graph is a triangle:&lt;br /&gt;
&lt;br /&gt;
[[Image:Tower of Hanoi 1-disk graph.svg|center]]&lt;br /&gt;
&lt;br /&gt;
The graph for two disks is three triangles arranged in a larger triangle:&lt;br /&gt;
&lt;br /&gt;
[[Image:Tower of Hanoi-2.svg|center]]&lt;br /&gt;
&lt;br /&gt;
The nodes at the vertices of the outermost triangle represent distributions with all disks on the same peg.&lt;br /&gt;
&lt;br /&gt;
For h+1 disks, take the graph of h disks and replace each small triangle with the graph for two disks.&lt;br /&gt;
&lt;br /&gt;
For three disks the graph is:&lt;br /&gt;
&lt;br /&gt;
[[Image:Tower of Hanoi-3.svg|440px|center]]&lt;br /&gt;
* call the pegs a, b and c&lt;br /&gt;
* list disk positions from left to right in order of increasing size&lt;br /&gt;
&lt;br /&gt;
The sides of the outermost triangle represent the shortest ways of moving a tower from one peg to another one. The edge in the middle of the sides of the largest triangle represents a move of the largest disk. The edge in the middle of the sides of each next smaller triangle represents a move of each next smaller disk. The sides of the smallest triangles represent moves of the smallest disk.&lt;br /&gt;
&lt;br /&gt;
[[Image:Hanoi-Graph-7.svg|thumb|right|The game graph of level 7 shows the relatedness to the [[Sierpiński triangle]].]]&lt;br /&gt;
In general, for a puzzle with &#039;&#039;n&#039;&#039; disks, there are 3&amp;lt;sup&amp;gt;&#039;&#039;n&#039;&#039;&amp;lt;/sup&amp;gt; nodes in the graph; every node has three edges to other nodes, except the three corner nodes, which have two: it is always possible to move the smallest disk to one of the two other pegs; and it is possible to move one disk between those two pegs &#039;&#039;except&#039;&#039; in the situation where all disks are stacked on one peg. The corner nodes represent the three cases where all the disks are stacked on one peg. The diagram for &#039;&#039;n&#039;&#039;&amp;amp;nbsp;+&amp;amp;nbsp;1 disks is obtained by taking three copies of the &#039;&#039;n&#039;&#039;-disk diagram—each one representing all the states and moves of the smaller disks for one particular position of the new largest disk—and joining them at the corners with three new edges, representing the only three opportunities to move the largest disk. The resulting figure thus has 3&amp;lt;sup&amp;gt;&#039;&#039;n&#039;&#039;+1&amp;lt;/sup&amp;gt; nodes and still has three corners remaining with only two edges.&lt;br /&gt;
&lt;br /&gt;
As more disks are added, the graph representation of the game will resemble a [[fractal]] figure, the [[Sierpiński triangle]]. It is clear that the great majority of positions in the puzzle will never be reached when using the shortest possible solution; indeed, if the priests of the legend are using the longest possible solution (without re-visiting any position), it will take them 3&amp;lt;sup&amp;gt;64&amp;lt;/sup&amp;gt;&amp;amp;nbsp;&amp;amp;minus;&amp;amp;nbsp;1 moves, or more than 10&amp;lt;sup&amp;gt;23&amp;lt;/sup&amp;gt; years.&lt;br /&gt;
&lt;br /&gt;
The longest non-repetitive way for three disks can be visualized by erasing the unused edges:&lt;br /&gt;
&lt;br /&gt;
[[Image:Tower of Hanoi-3 Longest Path.svg|440px|center]]&lt;br /&gt;
&lt;br /&gt;
Incidentally, this longest non-repetitive path can be obtained by forbidding all moves from &#039;&#039;a&#039;&#039; to &#039;&#039;b&#039;&#039;.&lt;br /&gt;
&lt;br /&gt;
The circular [[Hamiltonian path]] for three disks is:&lt;br /&gt;
&lt;br /&gt;
[[Image:Tower of Hanoi-4 Longest Cycle.svg|440px|center]]&lt;br /&gt;
&lt;br /&gt;
The graphs clearly show that:&lt;br /&gt;
* From every arbitrary distribution of disks, there is exactly one shortest way to move all disks onto one of the three pegs.&lt;br /&gt;
* Between every pair of arbitrary distributions of disks there are one or two different shortest paths.&lt;br /&gt;
* From every arbitrary distribution of disks, there are one or two different longest non selfcrossing paths to move all disks to one of the three pegs.&lt;br /&gt;
* Between every pair of arbitrary distributions of disks there are one or two different longest non self-crossing paths.&lt;br /&gt;
* Let &#039;&#039;N&#039;&#039;&amp;lt;sub&amp;gt;&#039;&#039;h&#039;&#039;&amp;lt;/sub&amp;gt; be the number of non selfcrossing paths for moving a tower of &#039;&#039;h&#039;&#039; disks from one peg to another one. Then:&lt;br /&gt;
** &#039;&#039;N&#039;&#039;&amp;lt;sub&amp;gt;1&amp;lt;/sub&amp;gt; = 2&lt;br /&gt;
** &#039;&#039;N&#039;&#039;&amp;lt;sub&amp;gt;&#039;&#039;h&#039;&#039;+1&amp;lt;/sub&amp;gt; = (&#039;&#039;N&#039;&#039;&amp;lt;sub&amp;gt;&#039;&#039;h&#039;&#039;&amp;lt;/sub&amp;gt;)&amp;lt;sup&amp;gt;2&amp;lt;/sup&amp;gt; + (&#039;&#039;N&#039;&#039;&amp;lt;sub&amp;gt;&#039;&#039;h&#039;&#039;&amp;lt;/sub&amp;gt;)&amp;lt;sup&amp;gt;3&amp;lt;/sup&amp;gt;.&lt;br /&gt;
** For example: &#039;&#039;N&#039;&#039;&amp;lt;sub&amp;gt;8&amp;lt;/sub&amp;gt; ≈ 1.5456&amp;amp;times;10&amp;lt;sup&amp;gt;795&amp;lt;/sup&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== Applications ==&lt;br /&gt;
The Tower of Hanoi is frequently used in psychological research on [[problem solving]]. There also exists a variant of this task called [[Tower of London Test|Tower of London]] for neuropsychological diagnosis and treatment of executive functions.&lt;br /&gt;
&lt;br /&gt;
The Tower of Hanoi is also used as a [[Backup rotation scheme]] when performing computer data [[Backups]] where multiple tapes/media are involved.&lt;br /&gt;
&lt;br /&gt;
As mentioned above, the Tower of Hanoi is popular for teaching recursive algorithms to beginning programming students. A pictorial version of this puzzle is programmed into the [[emacs]] editor, accessed by typing M-x hanoi. There is also a sample algorithm written in [[Prolog]].&lt;br /&gt;
&lt;br /&gt;
The Tower of Hanoi is also used as a test by neuropsychologists trying to evaluate [[frontal lobe]] deficits.&lt;br /&gt;
&lt;br /&gt;
In 2010, researchers published the results of an experiment that found that the ant species [[linepithema humile]] were successfully able to solve the Tower of Hanoi problem through non-linear dynamics and pheromone signals.&amp;lt;ref&amp;gt;{{cite journal |last1=Reid |first1=C.R. |last2=Sumpter |first2=D.J. |last3=Beekman |first3=M. |title=Optimisation in a natural system: Argentine ants solve the Towers of Hanoi |journal=J. Exp. Biol. |volume=214 |issue=Pt 1 |pages=50–8 |date=January 2011 |pmid=21147968 |doi=10.1242/jeb.048173 |url=http://jeb.biologists.org/cgi/pmidlookup?view=long&amp;amp;pmid=21147968}}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== General shortest paths and the number 466/885 ==&lt;br /&gt;
A curious generalization of the original goal of the puzzle is to start from a given configuration of the disks where all disks are not necessarily on the same peg, and to arrive in a minimal number of moves at another given configuration. In general it can be quite difficult to compute a shortest sequence of moves to solve this problem. A solution was proposed by Andreas Hinz, and is based on the observation that in a shortest sequence of moves, the largest disk that needs to be moved (obviously one may ignore all of the largest disks that will occupy the same peg in both the initial and final configurations) will move either exactly once or exactly twice.&lt;br /&gt;
&amp;lt;br/&amp;gt;&lt;br /&gt;
&amp;lt;br/&amp;gt;The mathematics related to this generalized problem becomes even more interesting when one considers the &#039;&#039;&#039;average&#039;&#039;&#039; number of moves in a shortest sequence of moves between two initial and final disk configurations that are chosen at random. Hinz and Chan Hat-Tung independently discovered&lt;br /&gt;
&amp;lt;ref&amp;gt;{{cite journal |first=A. |last=Hinz |title=The Tower of Hanoi |journal=L&#039;Enseignement Mathematique |volume=35 |year=1989 |pages=289–321 |doi=10.5169/seals-57378}}&amp;lt;/ref&amp;gt;&amp;lt;ref&amp;gt;{{cite journal |first=T. |last=Chan |title=A statistical analysis of the towers of Hanoi problem |journal=Internat. J. Comput. Math. |volume=28 |year=1988 |pages=57–65 |doi=10.1080/00207168908803728}}&amp;lt;/ref&amp;gt; (see also,&lt;br /&gt;
&amp;lt;ref&amp;gt;{{cite book |last=Stewart |first=Ian |title=Another Fine Math You&#039;ve Got Me Into... |publisher=Courier Dover |year=2004 |isbn=0-7167-2342-5 |chapter=}}&amp;lt;/ref&amp;gt; Chapter 1, p.&amp;amp;nbsp;14)&lt;br /&gt;
that the average number of moves in an n-disk Tower is given by the following exact formula:&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt; \frac{466}{885}\cdot 2^n - \frac{1}{3} - \frac{3}{5}\cdot \left(\frac{1}{3}\right)^n +&lt;br /&gt;
\left(\frac{12}{29} + \frac{18}{1003}\sqrt{17}\right)\left(\frac{5+\sqrt{17}}{18}\right)^n +&lt;br /&gt;
\left(\frac{12}{29} - \frac{18}{1003}\sqrt{17}\right)\left(\frac{5-\sqrt{17}}{18}\right)^n.&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Note that for large enough &#039;&#039;n&#039;&#039;, only the first and second terms do not converge to zero, so we get an [[asymptotic analysis|asymptotic expression]]: &amp;lt;math&amp;gt;466/885\cdot 2^n - 1/3 + o(1)&amp;lt;/math&amp;gt;, as &amp;lt;math&amp;gt;n \to \infty&amp;lt;/math&amp;gt;. Thus intuitively, we could interpret the fraction of &amp;lt;math&amp;gt;466/885\approx 52.6\%&amp;lt;/math&amp;gt; as representing the ratio of the labor one has to perform when going from a randomly chosen configuration to another randomly chosen configuration, relative to the difficulty of having to cross the &amp;quot;most difficult&amp;quot; path of length &amp;lt;math&amp;gt;2^n-1&amp;lt;/math&amp;gt; which involves moving all the disks from one peg to another. An alternative explanation for the appearance of the constant 466/885, as well as a new and somewhat improved algorithm for computing the shortest path, was given by Romik.&amp;lt;ref&amp;gt;{{cite journal |first=D. |last=Romik |title=Shortest paths in the Tower of Hanoi graph and finite automata |journal=[[SIAM Journal on Discrete Mathematics]] |volume=20 |issue=3 |year=2006 |pages=610–622 |doi=10.1137/050628660}}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
== Variations ==&lt;br /&gt;
&lt;br /&gt;
=== Cyclic Hanoi ===&lt;br /&gt;
Cyclic Hanoi is a variation of the Hanoi in which each disk must be moved in the same cyclic direction, in most cases, clockwise.&amp;lt;ref&amp;gt;{{cite journal |first=T. D. |last=Gedeon |title=The Cyclic Towers of Hanoi: An Iterative Solution Produced by Transformation |journal=The Computer Journal |volume=39 |issue=4 |year=1996  |doi=10.1093/comjnl/39.4.353 |url=http://comjnl.oxfordjournals.org/content/39/4/353.short}}&amp;lt;/ref&amp;gt; For example, given a standard three peg set-up, a given disk can be moved from peg A to peg B, then from B to C, C to A, etc. This can be solved using two mutually recursive procedures:&lt;br /&gt;
&lt;br /&gt;
To move &#039;&#039;n&#039;&#039; discs &#039;&#039;&#039;clockwise&#039;&#039;&#039; from peg A to peg C:&lt;br /&gt;
# move &#039;&#039;n&#039;&#039; − 1 discs &#039;&#039;&#039;clockwise&#039;&#039;&#039; from A to C&lt;br /&gt;
# move disc #&#039;&#039;n&#039;&#039; from A to B&lt;br /&gt;
# move &#039;&#039;n&#039;&#039; − 1 discs &#039;&#039;&#039;counterclockwise&#039;&#039;&#039; from C to A&lt;br /&gt;
# move disc #&#039;&#039;n&#039;&#039; from B to C&lt;br /&gt;
# move &#039;&#039;n&#039;&#039; − 1 discs &#039;&#039;&#039;clockwise&#039;&#039;&#039; from A to C&lt;br /&gt;
&lt;br /&gt;
To move &#039;&#039;n&#039;&#039; discs &#039;&#039;&#039;counterclockwise&#039;&#039;&#039; from peg A to peg C:&lt;br /&gt;
&lt;br /&gt;
# move &#039;&#039;n&#039;&#039; − 1 discs &#039;&#039;&#039;clockwise&#039;&#039;&#039; from A to B&lt;br /&gt;
# move disc #&#039;&#039;n&#039;&#039; from A to C&lt;br /&gt;
# move &#039;&#039;n&#039;&#039; − 1 discs &#039;&#039;&#039;clockwise&#039;&#039;&#039; from B to C&lt;br /&gt;
&lt;br /&gt;
=== With four pegs and beyond ===&lt;br /&gt;
Although the three-peg version has a simple recursive solution as outlined above, the &#039;&#039;optimal&#039;&#039; solution for the Tower of Hanoi problem with four pegs (called &#039;&#039;&#039;Reve&#039;s puzzle&#039;&#039;&#039;), let alone more pegs, is still an [[open problem]]. This is a good example of how a simple, solvable problem can be made dramatically more difficult by slightly loosening one of the problem constraints.&lt;br /&gt;
&lt;br /&gt;
The fact that the problem with four or more pegs is an open problem does not imply that no algorithm exists for finding (all of) the optimal solutions. Simply represent the game by an undirected graph, the nodes being distributions of disks and the edges being moves and use [[breadth first search]] to find one (or all) shortest path(s) moving a tower from one peg onto another one. However, even smartly implemented on the fastest computer now available, this algorithm provides no way of effectively computing solutions for large numbers of disks; the program would require more time and memory than available. Hence, even having an algorithm, it remains unknown how many moves an optimal solution requires and how many optimal solutions exist for 1000 disks and 10 pegs.&lt;br /&gt;
&lt;br /&gt;
Though it is not known exactly how many moves must be made, there are some asymptotic results. There is also a &amp;quot;presumed-optimal solution&amp;quot; given by the &#039;&#039;&#039;Frame-Stewart algorithm&#039;&#039;&#039;, discovered independently by Frame and Stewart in 1941.&amp;lt;ref&amp;gt;{{cite journal |first1=B.M. |last1=Stewart |first2=J.S. |last2=Frame |title=Solution to advanced problem 3819 |journal=American Mathematical Monthly |volume=48 |issue=3 |pages=216–9 |date=March 1941 |jstor=2304268}}&amp;lt;/ref&amp;gt; The related open Frame-Stewart conjecture claims that the Frame-Stewart algorithm always gives an optimal solution. The optimality of the Frame-Stewart algorithm has been computationally verified for 4 pegs with up to 30 disks.&amp;lt;ref&amp;gt;{{cite journal |last1=Korf |first1=Richard E. |first2=Ariel |last2=Felner |title=Recent Progress in Heuristic Search: a Case Study of the Four-Peg Towers of Hanoi Problem |journal=[[IJCAI]] |year=2007 |pages=2324–9 |url=http://www.ijcai.org/papers07/Papers/IJCAI07-374.pdf |format=PDF}}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
For other variants of the four-peg Tower of Hanoi problem, see Paul Stockmeyer&#039;s survey paper.&amp;lt;ref&amp;gt;{{cite journal |first=Paul |last=Stockmeyer |title=Variations on the Four-Post Tower of Hanoi Puzzle |journal=Congressus Numerantium |volume=102 |year=1994 |pages=3–12 |url=http://www.cs.wm.edu/~pkstoc/boca.ps |format=Postscript}}&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==== Frame–Stewart algorithm ====&lt;br /&gt;
The Frame–Stewart algorithm, giving a &#039;&#039;presumably optimal solution&#039;&#039; for four (or even more) pegs, is described below:&lt;br /&gt;
* Let &amp;lt;math&amp;gt;n&amp;lt;/math&amp;gt; be the number of disks.&lt;br /&gt;
* Let &amp;lt;math&amp;gt;r&amp;lt;/math&amp;gt; be the number of pegs.&lt;br /&gt;
* Define &amp;lt;math&amp;gt;T(n,r)&amp;lt;/math&amp;gt; to be the minimum number of moves required to transfer n disks using r pegs&lt;br /&gt;
The algorithm can be described recursively:&lt;br /&gt;
# For some &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt;, &amp;lt;math&amp;gt;1 \leq k &amp;lt; n&amp;lt;/math&amp;gt;, transfer the top &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; disks to a single peg other than the start or destination pegs, taking &amp;lt;math&amp;gt;T(k,r)&amp;lt;/math&amp;gt; moves.&lt;br /&gt;
# Without disturbing the peg that now contains the top &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; disks, transfer the remaining &amp;lt;math&amp;gt;n-k&amp;lt;/math&amp;gt; disks to the destination peg, using only the remaining &amp;lt;math&amp;gt;r-1&amp;lt;/math&amp;gt; pegs, taking &amp;lt;math&amp;gt;T(n-k,r-1)&amp;lt;/math&amp;gt; moves.&lt;br /&gt;
# Finally, transfer the top &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; disks to the destination peg, taking &amp;lt;math&amp;gt;T(k,r)&amp;lt;/math&amp;gt; moves.&lt;br /&gt;
The entire process takes &amp;lt;math&amp;gt;2T(k,r)+T(n-k,r-1)&amp;lt;/math&amp;gt; moves. Therefore, the count &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt; should be picked for which this quantity is minimum.&lt;br /&gt;
&lt;br /&gt;
This algorithm (with the above choice for &amp;lt;math&amp;gt;k&amp;lt;/math&amp;gt;) is presumed to be optimal, and no counterexamples are known.&lt;br /&gt;
&lt;br /&gt;
=== Multistack Tower of Hanoi ===&lt;br /&gt;
{{unreferenced section|date=November 2013}}&lt;br /&gt;
U.S. patent number 7,566,057 issued to Victor Mascolo discloses multistack Tower of Hanoi puzzles with two or more stacks and twice as many pegs as stacks. After beginning on a particular peg, each stack displaces and is displaced by a different colored stack on another peg when the puzzle is solved. Disks of one color also have another peg that excludes all other colors, so that there are three pegs available for each color disk, two that are shared with other colors, and one that is not shared. On the shared pegs, a disk may not be placed on a different colored disk of the same size, a possibility that does not arise in the standard puzzle.&lt;br /&gt;
&lt;br /&gt;
The simplest multistack game, Tower of Hanoi (2&amp;amp;nbsp;&amp;amp;times;&amp;amp;nbsp;4), has two stacks and four pegs, and it requires 3[&#039;&#039;T&#039;&#039;(&#039;&#039;n&#039;&#039;)] moves to solve where &#039;&#039;T&#039;&#039;(&#039;&#039;n&#039;&#039;) is the number of moves needed to solve a single stack classic of &#039;&#039;n&#039;&#039; disks. The game proceeds in seesaw fashion with longer and longer series of moves that alternate between colors. It concludes in reverse seesaw fashion with shorter and shorter such series of moves. Starting with the second series of three moves, these alternate series of moves double in length for the first half of the game, and the lengths are halved as the game concludes. The solution involves nesting an algorithm suitable for Tower of Hanoi into an algorithm that indicates when to switch between colors. When there are k stacks of n disks apiece in a game, and &#039;&#039;k&#039;&#039;&amp;amp;nbsp;&amp;gt;&amp;amp;nbsp;2, it requires &#039;&#039;k&#039;&#039;[&#039;&#039;T&#039;&#039;(&#039;&#039;n&#039;&#039;)]&amp;amp;nbsp;+&amp;amp;nbsp;&#039;&#039;T&#039;&#039;(&#039;&#039;n&#039;&#039;&amp;amp;nbsp;&amp;amp;minus;&amp;amp;nbsp;1)&amp;amp;nbsp;+&amp;amp;nbsp;1 moves to relocate them.&lt;br /&gt;
&lt;br /&gt;
The addition of a centrally located universal peg open to disks from all stacks converts these multistack Tower of Hanoi puzzles to multistack Reve&#039;s puzzles as described in the preceding section. In these games each stack may move among four pegs, the same combination of three in the 2&amp;amp;nbsp;&amp;amp;times;&amp;amp;nbsp;4 game plus the central universal peg. The simplest game of this kind (2&amp;amp;nbsp;&amp;amp;times;&amp;amp;nbsp;5) has two stacks and five pegs. A solution conjectured to be optimal interlocks the optimal solution of the 2&amp;amp;nbsp;&amp;amp;times;&amp;amp;nbsp;4 puzzle with the presumed optimal solution to Reve&#039;s puzzle. It takes &#039;&#039;R&#039;&#039;(&#039;&#039;n&#039;&#039;)&amp;amp;nbsp;+&amp;amp;nbsp;2&#039;&#039;R&#039;&#039;(&#039;&#039;n&#039;&#039;&amp;amp;nbsp;&amp;amp;minus;&amp;amp;nbsp;1)&amp;amp;nbsp;+&amp;amp;nbsp;2 moves, where &#039;&#039;R&#039;&#039;(&#039;&#039;n&#039;&#039;) is the number of moves in the presumed optimal Reve&#039;s solution for a stack of &#039;&#039;n&#039;&#039; disks.&lt;br /&gt;
&lt;br /&gt;
== In popular culture ==&lt;br /&gt;
In the science fiction story &amp;quot;Now Inhale&amp;quot;, by [[Eric Frank Russell]],&amp;lt;ref&amp;gt;{{cite journal |journal=Astounding Science Fiction |date=April 1959 |first=Eric Frank |last=Russell |title=Now Inhale}}&amp;lt;/ref&amp;gt; the human is a prisoner on a planet where the local custom is to make the prisoner play a game until it is won or lost, and then his execution will be immediate. The protagonist knows that a rescue ship might take a year or more to arrive, so he chooses to play Towers of Hanoi with 64 disks. (This story makes reference to the legend about the Buddhist monks playing the game until the end of the world.)&lt;br /&gt;
&lt;br /&gt;
In the 1966 &#039;&#039;[[Doctor Who]]&#039;&#039; story &#039;&#039;[[The Celestial Toymaker]]&#039;&#039;, the [[eponym]]ous villain forces [[The Doctor (Doctor Who)|the Doctor]] to play a ten-piece 1,023-move Tower of Hanoi game entitled [[List of Doctor Who items#T|The Trilogic Game]] with the pieces forming a pyramid shape when stacked.&lt;br /&gt;
&lt;br /&gt;
In 2007, the concept of the Towers Of Hanoi problem was used in &#039;&#039;[[Professor Layton and the Diabolical Box]]&#039;&#039; in puzzles 6, 83, and 84, but the discs had been changed to pancakes. The puzzle was based around a dilemma where the chef of a restaurant had to move a pile of pancakes from one plate to the other with the basic principles of the original puzzle (i.e. three plates that the pancakes could be moved onto, not being able to put a larger pancake onto a smaller one, etc.)&lt;br /&gt;
&lt;br /&gt;
In the film &#039;&#039;[[Rise of the Planet of the Apes]]&#039;&#039; (2011), this puzzle, called in the film the &amp;quot;Lucas Tower&amp;quot;, is used as a test to study the intelligence of apes.&lt;br /&gt;
&lt;br /&gt;
The puzzle is featured regularly in [[adventure game|adventure]] and [[computer puzzle game|puzzle]] games. Since it is easy to implement, and easily recognised, it is well-suited to use as a puzzle in a larger graphical game (e.g. &#039;&#039;[[Star Wars: Knights of the Old Republic]]&#039;&#039; and &#039;&#039;[[Mass Effect (video game)|Mass Effect]]&#039;&#039;&amp;lt;ref&amp;gt;{{cite web |url=http://www.giantbomb.com/tower-of-hanoi/92-5744/ |title=Tower of Hanoi (video game concept)  |publisher=giantbomb.com |accessdate=2010-12-05}}&amp;lt;/ref&amp;gt;). Some implementations use straight disks, but others disguise the puzzle in some other form. There is an arcade version by [[Sega]]/[[Andamiro]].&amp;lt;ref&amp;gt;http://www.segaarcade.com/towerofhanoi&amp;lt;/ref&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The problem is featured as part of a reward challenge in a [[Survivor: South Pacific#Episode 1: &amp;quot;I Need Redemption&amp;quot;|2011 episode of the American version of the &#039;&#039;Survivor&#039;&#039; TV series]]. Both players ([[Ozzy Lusth]] and [[Benjamin Wade (Survivor contestant)|Benjamin &amp;quot;Coach&amp;quot; Wade]]) struggle to understand how to solve the puzzle and are aided by their fellow tribe members.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
*[[Backup rotation scheme]], a TOH application&lt;br /&gt;
*[[Baguenaudier]]&lt;br /&gt;
*[[Recursion (computer science)]]&lt;br /&gt;
&lt;br /&gt;
== Notes ==&lt;br /&gt;
{{reflist|30em}}&lt;br /&gt;
&lt;br /&gt;
== External links ==&lt;br /&gt;
{{Commons category|Tower of Hanoi}}&lt;br /&gt;
*{{mathworld|title=Tower of Hanoi|urlname=TowerofHanoi}}&lt;br /&gt;
*{{dmoz|Science/Math/Recreations/Famous_Problems/Tower_of_Hanoi|Tower of Hanoi}}&lt;br /&gt;
*[http://www.codeminima.com/Lisp:_Tower_of_Hanoi Tower of Hanoi: Code in Lisp]&lt;br /&gt;
*[https://play.google.com/store/apps/details?id=com.ADP.pyramidmovers Tower of Hanoi Variant Game: Pyramid Mover]&lt;br /&gt;
&lt;br /&gt;
{{DEFAULTSORT:Tower Of Hanoi}}&lt;br /&gt;
[[Category:Mechanical puzzles]]&lt;br /&gt;
&lt;br /&gt;
{{Link GA|de}}&lt;br /&gt;
{{Link FA|he}}&lt;/div&gt;</summary>
		<author><name>24.19.55.181</name></author>
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		<updated>2012-07-25T20:13:06Z</updated>

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