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	Pumping lemma - Wikipedia, the free encyclopedia
		Thus, if there is a pumping lemma for a given language class, any nonempty language in the class will contain an infinite set of finite strings all produced by a simple rule given by the lemma.
		The two most important examples are the pumping lemma for regular languages and the pumping lemma for context-free languages.
		Ogden's lemma is a second, stronger pumping lemma for context-free languages.
	en.wikipedia.org /wiki/Pumping_lemma (222 words)

	Pumping lemma for regular languages - Wikipedia, the free encyclopedia
		In the theory of formal languages, the pumping lemma for regular languages is a lemma that gives a property that all regular languages have.
		It is obviously in L, and the pumping lemma therefore yields a decomposition of w = xyz with
		In terms of the pumping lemma, the string abcbcd is broken into a x portion a, a y portion bc and a z portion bcd.
	en.wikipedia.org /wiki/Pumping_lemma_for_regular_languages (1001 words)

	CS 3133, A Term 1998, Pumping Lemma
		Then L is accepted by some deterministic finite automaton (DFA), and by the pumping lemma every sufficiently long string z of L may be written in the form stated in the lemma.
		The aim is to choose a string for which pumping will be guaranteed to produce strings that can't possibly be in L. The choice of the string z is related to the number k that appears in the statement of the pumping lemma.
		The number of a's in the pumped string is i + jn + k - i - j = j(n-1) + k, and the number of b's is 2k.
	web.cs.wpi.edu /~alvarez/CS3133/pumping.html (692 words)

	[No title]
		Use the contradiction in (c) to conclude that the pumping lemma does not apply for L. Use the conclusion in (d) to imply that the assumption in (a), that L is regular, is false.
		It should be emphasized that in the previous schema the pumping lemma implies only the existence of a constant m for the assumed regular language L, and the existence of a decomposition xyz for the chosen string w.
		The proof of the pumping lemma is based on the observation that a state is repeated in each computation on a "long" input, with a portion of the input being consumed between the repetition.
	www.cse.ohio-state.edu /~gurari/theory-bk/theory-bk-twose4.html (1530 words)

	An Example of Pumping Context-Free Languages, CS 532/051, 092, 251, Spring 2006, IIT
		CFL Pumping Lemma II The pumping lemma given in the textbook (let’s call it Pumping Lemma I) is an awfully weak one; let’s use this slightly stronger one:
		As usual, we’ll actually use the contrapositive of the lemma, to prove that a language is not context-free.
		Pumping Lemma II, on the other hand, is powerful enough.
	www.cs.iit.edu /~cs532/Notes/CFL_Pumping.html (908 words)

	[No title]
		The integer p associated with the pumping lemma is just the number of states a DFA that recognizes the regular language in question.
		Assume L is regular and let p be the constant given by the pumping lemma.
		Proof is by contradiction again, using the pumping lemma to get the contradiction, but has to work slightly differently.
	condor.depaul.edu /~glancast/444class/docs/lecOct09.html (1845 words)

	[No title]
		The pumping lemma for context-free languages can be used to show that a language is not context-free.
		The method is similar to that for using the pumping lemma for regular languages to show that a language is not regular.
		As in the case of the pumping lemma for regular languages the choice of the string w is of critical importance when trying to show that a language is not context-free.
	www.cse.ohio-state.edu /~gurari/theory-bk/theory-bk-threese4.html (1116 words)

	The Pumping Lemma for Regular Languages
		Application of the Pumping Lemma to prove the set of palindromes is not regular (long commented version)
		Application of the Pumping Lemma to prove the set of palindromes is not regular (short uncommented version)
		In particular, this pumping lemma will be the main method we use to prove specific languages are not regular.
	www.cse.msu.edu /~torng/360Book/RegLang/Pumping (1886 words)

	Example 1: (Site not responding. Last check: 2007-11-03)
		By the pumping lemma, z can be written as uvwxy, s.t.
		Therefore, z’ is not in L, in violation of the pumping lemma.
		In some circumstances, it may be possible to avoid needing to go through such cases by coming up with some necessary condition for membership in the language and showing that applying the pumping lemma to a candidate string results in a string that no longer satisfies the condition.
	boole.stanford.edu /~rvg/154/handouts/section3.html (646 words)

	Tests, CS 532/051, 092, 251, Spring 2006, IIT
		Bad use of pumping lemma: You need to show that the pumped/unpumped string must be ∉ L, not “may” be ∉ L. The language of balanced strings is not regular.
		Bad use of pumping lemma: You want the pumped/unpumped string to be ∉ L; yours is ∈ L. Finite automata don’t have stacks.
		Bad use of pumping lemma: You should start with a string w ∈ L, not w ∉ L. Bad use of pumping lemma: Your statement of the lemma is incorrect.
	www.cs.iit.edu /~cs532/Misc/exam1_comments.html (1178 words)

	CSC 4170 The Pumping Lemma for Regular Languages (Site not responding. Last check: 2007-11-03)
		The pumping lemma for regular languages is another way of proving that a given (infinite) language is not regular.
		(The pumping lemma cannot be used to prove that a given language is regular.)
		Show that repeating the cycle some number of times ("pumping" the cycle) yields a string that is not in L. Conclude that L is not regular.
	www.seas.upenn.edu /~cit596/notes/dave/pumping2.html (234 words)

	Solutions to Practice Homework 10 (Site not responding. Last check: 2007-11-03)
		Part of the pumping lemma statement is a requirement that the string x have length at least n.
		Since EQUAL does not satisfy the properties of the pumping lemma for regular languages, we conclude that EQUAL is not regular.
		Since MOREA does not satisfy the properties of the pumping lemma for regular languages, we conclude that MOREA is not regular.
	www.cse.msu.edu /~torng/460/Homework/sphw10.html (1516 words)

	Geneseo CSci 342 Fall 2001 Pumping 3
		Pumping lemma to prove that L is regular?
		The thinking behind this choice: Key characteristic of language is that the number of a's is bigger than the number of b's, see if we can attack that by pumping to produce a string not in the language.
		But then to be in the language, we need enough a's to exceed the number of b's, so follow the b's with p+1 a's.
	www.cs.geneseo.edu /~baldwin/csci342/fall2001/pumping3.html (373 words)

	Title page for ETD etd-04282003-101336
		The pumping lemma for regular languages and its application are among the more dicult concepts students encounter in an introductory theory of computing course.
		The pumping lemma is used to prove that particular languages are not regular.
		The pumping lemma animator will be included in an ongoing project designed to create animations and interactive tools for a complete course on theory of computing.
	www.montana.edu /etd/available/cogliati_0805.html (443 words)

	Applying the pumping lemma
		such that the properties given by the pumping lemma hold.
		Using the intuition that finite automata can only use finite memory it should be clear that this language is not regular, becasue one has to remember the first half of the word to check whetherthe 2nd half is the same word read backwards.
		From the pumping lemma we know that there is a splitting of the word
	www.cs.nott.ac.uk /~txa/g51mal.2001/notes/node22.html (270 words)

	JFLAP: Regular Pumping Lemmas
		This game approach to the pumping lemma is based on the approach in Peter Linz's An Introduction to Formal Languages and Automata.
		We have structured the pumping lemma proof in JFLAP such that JFLAP takes the role of player A, and the user takes the role of player B, and they play the game with a few examples we have included.
		Resetting the pumping lemma clears all the information the user has entered, allowing us to restart the example.
	www.jflap.org /tutorial/pumpinglemma/regPL.html (973 words)

	Pumping Lemma Practice Answers (Site not responding. Last check: 2007-11-03)
		Note: because y is not the empty string, y contains at least one "a" if this is the case that applies.
		Pumping twice results in a string not in the language because it is not of the form a
		= 2 * k + 1 >= k, the pumping lemmas applies.
	csr.uvic.ca /~wendym/courses/320/03summer/ans_pump.html (548 words)

	Pumping Lemma (Site not responding. Last check: 2007-11-03)
		This is a variation on the pumping lemma, described above.
		Since z is near the top of the derivation, rather than the bottom, it may generate a verry long substring.
		There is no bound on the length of vwx, as there was in the pumping lemma.
	www.mathreference.com /lan-cfl,pump.html (552 words)

	Pumping Lemma for CFLs - Java
		pumping lemma => contradiction, thus L cannot be context free.
		pumping lemma*, don't bother with the closure properties.
		You say your pumping lemma proof has "too many cases," but it is usually
	www.thescripts.com /forum/thread16448.html (264 words)

	The Pumping Lemma: Examples
		be the constant associated with this grammar by the Pumping Lemma.
		By the Pumping Lemma this must be representable as
		By the same argument as for the previous lemma neither
	www.cs.may.ie /~jpower/Courses/parsing/node41.html (169 words)

	CMSC 451 Lecture 8, Pumping Lemma for Regular Languages (Site not responding. Last check: 2007-11-03)
		If a regular grammar can be constructed to exactly generate the strings in a language, then the language is regular.
		To prove a language is not regular requires a specific definition of the language and the use of the Pumping Lemma for Regular Languages.
		For the Pumping Lemma, the statement "A" is "L is a Regular Language", The statement "B" is a statement from the Predicate Calculus.
	www.cs.umbc.edu /~squire/s01-451/cs451_l8.html (280 words)

	Pumping Lemma argument template (Site not responding. Last check: 2007-11-03)
		To prove that a language L is not regular, we show that the pumping lemma fails, that is, we show that
		is in L. By the pumping lemma, there must be some choice of x,y,z satisfying the numbered constraints above.
		Thus, L fails to satisfy the pumping lemma.
	www.cs.uiuc.edu /class/sp06/cs273/Misc/pumping-lemma.html (225 words)

	Pumping for contradictions! (Site not responding. Last check: 2007-11-03)
		Here is a short description of how to use the pumping lemma to show a language L is not regular.
		They must reveal the n that is guaranteed to exist by the pumping lemma.
		All you know about x, y, and z is the restrictions on their lengths (xy isn't too long and y is non-empty).
	www.cs.williams.edu /~kim/cs361/F02/Pumping.html (229 words)

	Computational Complexity: A New-To-Me Pumping Lemma for Regular Languages
		Here is an example from the Winter 1982 SIGACT News, a variation of the regular language pumping lemma due to Donald Stanat and Stephen Weiss.
		The rest follows like the usual pumping lemma.
		Using a result of Jaffe, Stanat and Weiss show that this condition is not only necessary but also sufficient to characterize the regular languages.
	weblog.fortnow.com /2003/08/new-to-me-pumping-lemma-for-regular.html (273 words)

	Limitations of the Pumping Lemma (Site not responding. Last check: 2007-11-03)
		The Pumping Lemma is very unspecific as to where in the generated string the pumping is to occur.
		is a prime number} is impossible by direct application of the Pumping Lemma.
		In such cases, specialised versions, such as Parikh's Theorem, are required.
	www.cs.may.ie /~jpower/Courses/parsing/node42.html (42 words)

	CMSC 451 Lecture 8, Pumping Lemma for Regular Languages (Site not responding. Last check: 2007-11-03)
		The Pumping Lemma is generally used to prove a language is not regular.
		If a DFA, NFA or NFA-epsilon machine can be constructed to exactly accept a language, then the language is a Regular Language.
		A note about proofs using the Pumping Lemma: Given: Formal statements A and B. A implies B. If you can prove B is false, then you have proved A is false.
	www.cs.umbc.edu /~squire/s04-451/cs451_l8.html (280 words)

	[No title] (Site not responding. Last check: 2007-11-03)
		¨Pumping a Parse TreeóQ'¨A Tree Tall Enoughó? uvxyz ÎðL¡$ ªóI uv2xy2z ÎðL¡HóJ! uxz Îð L¡$ªóS¨Pumping Lemma* for CFLó@¨Formal Proof of Pumping LemmaóC¨Pumping anbncn (Ex.
		¨Pumping a Parse TreeóQ'¨A Tree Tall Enoughó? uvxyz ÎðL¡$ ªóI uv2xy2z ÎðL¡HóJ! uxz Îð L¡$ªók@óS¨Pumping Lemma* for CFLólA¨Exampleó@¨Formal Proof of Pumping LemmaóC¨Pumping anbncn (Ex.
		¨Pumping a Parse TreeóQ'¨A Tree Tall Enoughó? uvxyz ÎðL¡$ ªóI uv2xy2z ÎðL¡HóJ! uxz Îð L¡$ªóS¨Pumping Lemma* for CFLólA¨Exampleó@¨Formal Proof of Pumping LemmaóC¨Pumping anbncn (Ex.
	zippy.sonoma.edu /~ravi/cs454sp03/Lectures/pumpingCFL.ppt (471 words)

	Geneseo CSci 342 Fall 2001 Pumping 1
		How does the proof of the pumping lemma work?
		Mini-Assignment: Figure out how using "p" as the number of times substrings repeat helps.
		Look at Sipser's third condition in the Pumping Lemma, xy
	www.cs.geneseo.edu /~baldwin/csci342/fall2001/pumping1.html (271 words)

	Pumping Lemma for Regular Sets (Site not responding. Last check: 2007-11-03)
		Next: Applying Pumping Lemma Up: Properties of Regular Sets Previous: Properties of Regular Sets
		Pumping Lemma: Let L be a regular set.
		Furthermore, n is no greater than the number of states of the smallest FA accepting L.
	www.cs.gsu.edu /~cscskp/Automata/propReg/node2.html (54 words)

	The Pumping Lemma (Site not responding. Last check: 2007-11-03)
		By contradiction you have shown that language L is not
		So there is a pumping lemma for all languages context-free,
		Although we do not have the same for those that are r.e.
	www.cs.brandeis.edu /~mairson/poems/node1.html (179 words)

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