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| ''Not to be confused with the "[[squeeze theorem]]" (sometimes called the "sandwich theorem").''
| | "Why does my computer keep freezing up?" I was asked by a lot of persons the cause of their computer freeze problems. And I am fed up with spending much time inside answering the query time plus time again. This article is to tell we the real cause of the PC Freezes.<br><br>Most of the reliable businesses might provide a full cash back guarantee. This signifies which we have the chance to receive your cash back in the event you find the registry cleaning has not delivered what we expected.<br><br>It doesn't matter whether you're not very obvious regarding what rundll32.exe is. However remember which it plays an important character in maintaining the stability of our computers plus the integrity of the program. Whenever some software or hardware couldn't answer usually to a program surgery, comes the rundll32 exe error, that can be caused by corrupted files or missing data inside registry. Usually, error message can shows up at booting or the beginning of running a system.<br><br>The issue with nearly all of the persons is the fact that they do not like to invest income. In the damaged variation 1 does not have to pay anything plus could download it from internet surprisingly easily. It is easy to install because well. However, the problem comes whenever it is very not able to detect all possible viruses, spyware and malware inside the program. This really is because it is actually obsolete in nature plus does not receive any normal updates within the site downloaded. Thus, the system is accessible to problems like hacking.<br><br>Besides, should you will get a [http://bestregistrycleanerfix.com/tune-up-utilities tuneup utilities 2014] which may do the job for we effectively plus immediately, then why not? There is one such program, RegCure that is surprisingly good plus complete. It has qualities which different products do not have. It is the most recommended registry cleaner today.<br><br>Why this problem occurs frequently? What are the causes of it? In truth, there are 3 major causes that can cause the PC freezing issue. To solve the issue, we have to take 3 procedures in the following paragraphs.<br><br>Reboot PC - Simply reboot the PC to see when the error is gone. Frequently, rebooting the PC readjusts the internal settings plus software and therefore fixes the issue. If it doesn't then move on to follow the instructions below.<br><br>Fortunately, there's a simple technique to fix most a computer mistakes. You just should be able to fix corrupt registry files on the computer. And to do that, you could just utilize a tool recognised as a registry cleaner. These simple pieces of software actually scan by your PC and fix every corrupt file that might result a issue to Windows. This enables your computer to utilize all the files it wants, that not only speeds it up - however, also stops all the errors on the system also. |
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| In [[measure theory]], a branch of [[mathematics]], the '''ham sandwich theorem''', also called the '''Stone–Tukey theorem''' after [[Arthur Harold Stone|Arthur H. Stone]] and [[John Tukey]], states that given {{mvar|n}} [[measurable]] "objects" in {{mvar|n}}-[[dimension]]al space, it is possible to divide all of them in half (with respect to their [[Measure (mathematics)|measure]], i.e. volume) with a single {{math|(''n'' − 1)}}-dimensional [[hyperplane]]. Here the "objects" should be [[Set (mathematics)|sets]] of finite [[Measure (mathematics)|measure]] (or, in fact, just of finite [[outer measure]]) for the notion of "dividing the volume in half" to make sense.
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| ==Naming==
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| [[File:Ham sandwich.jpg|thumb|right|A ham sandwich]]
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| The ham sandwich theorem takes its name from the case when {{math|1=''n'' = 3}} and the three objects of any shape are a chunk of [[ham (meat)|ham]] and two chunks of [[bread]] — notionally, a [[sandwich]] — which can then all be simultaneously bisected with a single cut (i.e., a [[plane (mathematics)|plane]]). In two dimensions, the theorem is known as the '''pancake theorem''' of having to cut two infinitesimally thin [[pancake]]s on a plate each in half with a single cut (i.e., a straight [[line (mathematics)|line]]).
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| ==History==
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| According to {{harvtxt|Beyer|Zardecki|2004}}, the earliest known paper about the ham sandwich theorem, specifically the {{math|1=''n'' = 3}} case of bisecting three solids with a plane, is by {{harvtxt|Steinhaus|1938}}. Beyer and Zardecki's paper includes a translation of the 1938 paper. It attributes the posing of the problem to [[Hugo Steinhaus]], and credits [[Stefan Banach]] as the first to solve the problem, by a reduction to the [[Borsuk–Ulam theorem]]. The paper poses the problem in two ways: first, formally, as "Is it always possible to bisect three solids, arbitrarily located, with the aid of an appropriate plane?" and second, informally, as "Can we place a piece of ham under a meat cutter so that meat, bone, and fat are cut in halves?" Later, the paper offers a proof of the theorem.
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| A more modern reference is {{harvtxt|Stone|Tukey|1942}}, which is the basis of the name "Stone–Tukey theorem". This paper proves the {{mvar|n}}-dimensional version of the theorem in a more general setting involving measures. The paper attributes the {{math|1=''n'' = 3}} case to [[Stanislaw Ulam]], based on information from a referee; but {{harvtxt|Beyer|Zardecki|2004}} claim that this is incorrect, given Steinhaus's paper, although "Ulam did make a fundamental contribution in proposing" the [[Borsuk–Ulam theorem]].
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| ==Reduction to the Borsuk–Ulam theorem==
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| The ham sandwich theorem can be proved as follows using the [[Borsuk–Ulam theorem]]. This proof follows the one described by Steinhaus and others (1938), attributed there to [[Stefan Banach]], for the {{math|1=''n'' = 3}} case.
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| Let {{math|''A''<sub>1</sub>, ''A''<sub>2</sub>, …, ''A''<sub>''n''</sub>}} denote the {{mvar|n}} objects that we wish to simultaneously bisect. Let {{mvar|S}} be the [[unit sphere|unit]] [[n-sphere|{{math|(''n'' − 1)}}-sphere]] embedded in {{mvar|n}}-dimensional [[Euclidean space]] <math>\mathbb{R}^n</math>, centered at the [[Origin (mathematics)|origin]]. For each point {{mvar|p}} on the surface of the sphere {{mvar|S}}, we can define a [[Linear continuum|continuum]] of oriented affine [[hyperplane]]s (not necessarily centred at 0) perpendicular to the ([[Surface normal|normal]]) [[vector (geometric)|vector]] from the origin to {{mvar|p}}, with the "positive side" of each hyperplane defined as the side pointed to by that vector. By the [[intermediate value theorem]], every family of such hyperplanes contains at least one hyperplane that bisects the bounded object {{math|''A''<sub>''n''</sub>}}: at one extreme translation, no volume of {{math|''A''<sub>''n''</sub>}} is on the positive side, and at the other extreme translation, all of {{math|''A''<sub>''n''</sub>}}'s volume is on the positive side, so in between there must be a translation that has half of {{math|''A''<sub>''n''</sub>}}'s volume on the positive side. If there is more than one such hyperplane in the family, we can pick one canonically by choosing the midpoint of the interval of translations for which {{math|''A''<sub>''n''</sub>}} is bisected. Thus we obtain, for each point {{mvar|p}} on the sphere {{mvar|S}}, a hyperplane {{math|π(''p'')}} that is perpendicular to the vector from the origin to {{mvar|p}} and that bisects {{math|''A''<sub>''n''</sub>}}.
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| Now we define a function {{mvar|f}} from the {{math|(''n'' − 1)}}-sphere {{mvar|S}} to {{math|(''n'' − 1)}}-dimensional Euclidean space <math>\mathbb{R}^{n-1}</math> as follows:
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| :{{math|1=''f''(''p'') = (}}vol of {{math|''A''<sub>1</sub>}} on the positive side of {{math|π(''p'')}}, vol of {{math|''A''<sub>2</sub>}} on the positive side of {{math|π(''p'')}}, ..., vol of {{math|''A''<sub>''n''−1</sub>}} on the positive side of {{math|π(''p''))}}.
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| This function {{mvar|f}} is [[continuous function|continuous]]. By the [[Borsuk–Ulam theorem]], there are [[antipodal points]] {{mvar|p}} and {{mvar|q}} on the sphere {{mvar|S}} such that {{math|1=''f''(''p'') = ''f''(''q'')}}. Antipodal points {{mvar|p}} and {{mvar|q}} correspond to hyperplanes {{math|π(''p'')}} and {{math|π(''q'')}} that are equal except that they have opposite positive sides. Thus, {{math|1=''f''(''p'') = ''f''(''q'')}} means that the volume of {{math|''A''<sub>''i''</sub>}} is the same on the positive and negative side of {{math|π(''p'')}} (or {{math|π(''q'')}}), for {{math|1=''i'' = 1, 2, ..., ''n'' − 1}}. Thus, {{math|π(''p'')}} (or {{math|π(''q'')}}) is the desired ham sandwich cut that simultaneously bisects the volumes of {{math|''A''<sub>1</sub>, ''A''<sub>2</sub>, …, ''A''<sub>''n''</sub>}}.
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| ==Measure theoretic versions==
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| In [[measure theory]], {{harvtxt|Stone|Tukey|1942}} proved two more general forms of the ham sandwich theorem. Both versions concern the bisection of {{mvar|n}} [[subset]]s {{math|''X''<sub>1</sub>, ''X''<sub>2</sub>, …, ''X''<sub>''n''</sub>}} of a common set {{mvar|X}}, where {{mvar|X}} has a [[Carathéodory]] [[outer measure]] and each {{math|''X''<sub>''i''</sub>}} has finite outer measure.
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| Their first general formulation is as follows: for any suitably restricted real [[function (mathematics)|function]] <math>f \colon S^n \times X \to \mathbb{R}</math>, there is a point {{mvar|p}} of the {{mvar|n}}-[[sphere]] {{math|''S<sup>n</sup>''}} such that the surface {{math|''f''(''s'',''x'') = 0}}, dividing {{mvar|X}} into {{math|''f''(''s'',''x'') < 0}} and {{math|''f''(''s'',''x'') > 0}}, simultaneously bisects the outer measure of {{math|''X''<sub>1</sub>, ''X''<sub>2</sub>, …, ''X''<sub>''n''</sub>}}. The proof is again a reduction to the Borsuk-Ulam theorem. This theorem generalizes the standard ham sandwich theorem by letting {{math|1=''f''(''s'',''x'') = ''s''<sub>0</sub> + ''s''<sub>1</sub>''x''<sub>1</sub> + ... + ''s<sub>n</sub>x<sub>n</sub>''}}.
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| Their second formulation is as follows: for any {{math|''n''+1}} measurable functions {{math|''f''<sub>0</sub>, ''f''<sub>1</sub>, …, ''f''<sub>''n''</sub>}} over {{mvar|X}} that are [[linearly independent]] over any subset of {{mvar|X}} of positive measure, there is a [[linear combination]] {{math|1=''f'' = ''a''<sub>0</sub>''f''<sub>0</sub> + ''a''<sub>1</sub>''f''<sub>1</sub> + ... + ''a<sub>n</sub>f<sub>n</sub>''}} such that the surface {{math|1=''f''(''x'') = 0}}, dividing {{mvar|X}} into {{math|''f''(''x'') < 0}} and {{math|''f''(''x'') > 0}}, simultaneously bisects the outer measure of {{math|''X''<sub>1</sub>, ''X''<sub>2</sub>, …, ''X''<sub>''n''</sub>}}. This theorem generalizes the standard ham sandwich theorem by letting {{math|1=''f''<sub>0</sub>(''x'') = 1}} and letting {{math|''f<sub>i</sub>''(''x'')}}, for {{math|''i'' > 0}}, be the {{mvar|i}}th coordinate of {{mvar|x}}.
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| ==Discrete and computational geometry versions==
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| [[Image:Discrete ham sandwich cut.svg|thumb|right|A ham-sandwich cut of eight red points and seven blue points in the plane.]]
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| In [[discrete geometry]] and [[computational geometry]], the ham sandwich theorem usually refers to the special case in which each of the sets being divided is a [[finite set]] of [[point (geometry)|point]]s. Here the relevant measure is the [[counting measure]], which simply counts the number of points on either side of the hyperplane. In two dimensions, the theorem can be stated as follows:
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| :For a finite set of points in the plane, each colored "red" or "blue", there is a [[line (mathematics)|line]] that simultaneously bisects the red points and bisects the blue points, that is, the number of red points on either side of the line is equal and the number of blue points on either side of the line is equal.
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| There is an exceptional case when points lie on the line. In this situation, we count each of these points as either being on one side, on the other, or on neither side of the line (possibly depending on the point), i.e. "bisecting" in fact means that each side contains less than half of the total number of points. This exceptional case is actually required for the theorem to hold, of course when the number of red points or the number of blue is odd, but also in specific configurations with even numbers of points, for instance when all the points lie on the same line and the two colors are separated from each other (i.e. colors don't alternate along the line). A situation where the numbers of points on each side cannot match each other is provided by adding an extra point out of the line in the previous configuration.
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| In computational geometry, this ham sandwich theorem leads to a computational problem, the '''ham sandwich problem'''. In two dimensions, the problem is this: given a finite set of {{mvar|n}} points in the plane, each colored "red" or "blue", find a ham sandwich cut for them. First, {{harvtxt|Megiddo|1985}} described an algorithm for the special, separated case. Here all red points are on one side of some line and all blue points are on the other side, a situation where there is a unique ham sandwich cut, which Megiddo could find in linear time. Later, {{harvtxt|Edelsbrunner|Waupotitsch|1986}} gave an algorithm for the general two-dimensional case; the running time of their algorithm is {{math|''O''(''n'' log ''n'')}}, where the symbol {{mvar|O}} indicates the use of [[Big O notation]]. Finally, {{harvtxt|Lo|Steiger|1990}} found an optimal {{math|''O''(''n'')}}-time [[algorithm]]. This algorithm was extended to higher dimensions by {{harvtxt|Lo|Matoušek|Steiger|1994}}. Given {{mvar|d}} sets of points in general position in {{mvar|d}}-dimensional space, the algorithm computes a {{math|(''d''−1)}}-dimensional hyperplane that has equal numbers of points of each of the sets in each of its half-spaces, i.e., a ham-sandwich cut for the given points.
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| ==Generalizations==
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| The original theorem works for at most n collections, where n is the number of dimensions. If we want to bisect a larger number of collections, we can use, instead of a hyperplane, an algebraic surface of degree k, i.e., an n-1 dimensional surface defined by a polynomial function of degree k:
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| Given <math>\binom{k+n}{n}-1</math> measures in an n-dimensional space, there exists an algebraic surface of degree k which bisects them all. ({{harvtxt|Smith|Wormald|1998}}).
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| This generalization is proved by mapping the n-dimensional plane into a <math>\binom{k+n}{n}-1</math> dimensional plane, and then applying the original theorem. For example, for n=2 and k=2, the 2 dimensional plane is mapped to a 5 dimensional plane via:
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| <math> (x,y) \to (x,y,x^2,y^2,xy) </math>.
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| ==References==
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| *{{citation
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| | doi = 10.2307/4145019
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| | last1 = Beyer | first1 = W. A.
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| | last2 = Zardecki | first2 = Andrew
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| | issue = 1
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| | journal = [[American Mathematical Monthly]]
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| | pages = 58–61
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| | title = The early history of the ham sandwich theorem
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| | url = http://proquest.umi.com/pqdweb?did=526216421&Fmt=3&clientId=5482&RQT=309&VName=PQD
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| | volume = 111
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| | year = 2004
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| | jstor = 4145019}}.
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| *{{citation
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| | doi = 10.1016/S0747-7171(86)80020-7
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| | last1 = Edelsbrunner | first1 = H. | author1-link = Herbert Edelsbrunner
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| | last2 = Waupotitsch | first2 = R.
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| | journal = J. Symbolic Comput.
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| | pages = 171–178
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| | title = Computing a ham sandwich cut in two dimensions
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| | volume = 2
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| | year = 1986}}.
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| *{{citation
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| | last1 = Lo | first1 = Chi-Yuan
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| | last2 = Steiger | first2 = W. L.
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| | contribution = An optimal time algorithm for ham-sandwich cuts in the plane
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| | pages = 5–9
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| | title = Proceedings of the Second Canadian Conference on Computational Geometry
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| | year = 1990}}.
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| *{{citation
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| | last1 = Lo | first1 = Chi-Yuan
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| | last2 = Matoušek | first2 = Jiří | author2-link = Jiří Matoušek (mathematician)
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| | last3 = Steiger | first3 = William L.
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| | doi = 10.1007/BF02574017
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| | journal = [[Discrete and Computational Geometry]]
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| | pages = 433–452
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| | title = Algorithms for Ham-Sandwich Cuts
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| | volume = 11
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| | year = 1994}}.
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| *{{citation
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| | doi = 10.1016/0196-6774(85)90011-2
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| | last = Megiddo | first = Nimrod
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| |authorlink= Nimrod Megiddo
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| | journal = Journal of Algorithms
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| | pages = 430–433
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| | title = Partitioning with two lines in the plane
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| | volume = 6
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| | year = 1985}}.
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| *{{citation
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| | last1 = Smith | first1 = W. D.
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| | last2 = Wormald | first2 = N. C.
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| | doi = 10.1109/sfcs.1998.743449
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| | chapter = Geometric separator theorems and applications
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| | title = Proceedings 39th Annual Symposium on Foundations of Computer Science (Cat. No.98CB36280)
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| | pages = 232
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| | year = 1998
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| | isbn = 0-8186-9172-7
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| | pmid =
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| | pmc =
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| }}
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| *{{citation
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| | last = Steinhaus | first = Hugo | author-link = Hugo Steinhaus
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| | journal = Mathesis Polska
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| | pages = 26–28
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| | title = A note on the ham sandwich theorem
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| | volume = 9
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| | year = 1938}}.
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| *{{citation
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| | doi = 10.1215/S0012-7094-42-00925-6
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| | last1 = Stone | first1 = A. H.
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| | last2 = Tukey | first2 = J. W. | author2-link = John Tukey
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| | journal = [[Duke Mathematical Journal]]
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| | pages = 356–359
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| | title = Generalized "sandwich" theorems
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| | url = http://projecteuclid.org/euclid.dmj/1077493229
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| | volume = 9
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| | year = 1942}}.
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| ==External links==
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| * {{MathWorld|title=Ham Sandwich Theorem|urlname=HamSandwichTheorem}}
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| *[http://jeff560.tripod.com/h.html ham sandwich theorem] on the [http://jeff560.tripod.com/mathword.html Earliest known uses of some of the words of mathematics]
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| *[http://cgm.cs.mcgill.ca/~athens/cs507/Projects/2002/DanielleMacNevin/index.htm Ham Sandwich Cuts] by Danielle MacNevin
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| *[http://gfredericks.com/sandbox/ham_sandwich An interactive 2D demonstration]
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| [[Category:Theorems in measure theory]]
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| [[Category:Articles containing proofs]]
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| [[Category:Theorems in topology]]
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"Why does my computer keep freezing up?" I was asked by a lot of persons the cause of their computer freeze problems. And I am fed up with spending much time inside answering the query time plus time again. This article is to tell we the real cause of the PC Freezes.
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