In particular, the pumping lemma states that for any regular language L {displaystyle L} there exists a constant p {displaystyle p}, so that every string w {displaystyle w} in L {displaystyle L} with a length of at least p {displaystyle p} in three substrings x {displaystyle x} , y {displaystyle y} and z {displaystyle z} ( w = x y z {displaystyle w=xyz} , where y {displaystyle y} is not empty), so the strings x z , x y z , x y z , x y y z ,. {displaystyle xz,xyz,xyyz,xyyyz,…} built by repeating y {displaystyle y} zero or more times are always in L {displaystyle L}. This repetition process is called pumping. In addition, the pump lemma ensures that the length of x y {displaystyle xy} is at most p {displaystyle p}, which imposes a limit on how w {displaystyle w} can be divided. Finite languages satisfy the empty pumping lemma by having p {displaystyle p} equal to the maximum string length in L {displaystyle L} plus one. The pumping lemma is useful for refuting the regularity of a particular language. It was first detected in 1959 by Michael Rabin and Dana Scott,[1] and shortly thereafter, rediscovered by Yehoshua Bar-Hillel, Micha A. Perles and Eli Shamir in 1961, as a simplification of their pumping lemma for contextless languages. [2] [3] The pumping lemma is often used to prove that a particular language is not regular: a contradictory proof may consist of showing a string (of the required length) in the language that does not have the property described in the pump lemma. The FSA accepts the string: abcd. Since this string has a length at least as large as the number of states, which is four, the drawer principle states that there must be at least one repeated state between the start state and the next four states visited. In this example, only q 1 {displaystyle q_{1}} is a repeating state. Since the substring bc passes the machine through transitions starting at state q 1 {displaystyle q_{1}} and ending at state q 1 {displaystyle q_{1}}, this part could be repeated and the FSA would still accept what the string abcbcd gives.
Alternatively, part bc could be removed and the FSA would still agree to give the channel display. In terms of pumping lemma, the string abcd is divided into part b {displaystyle x} x, part b of y {displaystyle y} and part b of z {displaystyle z}. If a language L {displaystyle L} is regular, then there is a number p ≥ 1 {displaystyle pgeq 1} (the length of the pump) so that each string u w v {displaystyle uwv} in L {displaystyle L} with | w | ≥ p {displaystyle |w|geq p} can be written as For example, prove that L01={0n1n | n ≥ 0} is irregular. If we assume that L is regular, then when pumping the lemma, the above rules follow. Now leave x L and |x ∈| ≥ n. Thus, by pumping the lemma, u, v, w exists, such that (1) – (3) hold. We show that for all u, v, w, (1) – (3) does not apply. If (1) and (2) are maintained, then x = 0n1n = uvw with |uv| ≤ n and |v| ≥ 1. Thus, u = 0a, v = 0b, w = 0c1n where: a + b ≤ n, b ≥ 1, c ≥ 0, a + b + c = n But then (3) fails for i = 0 uv0w = uw = 0a0c1n = 0a + c1n ∉ L, since a + c ≠ n. Pumping lemma for contextless languages (CFL) Pumping lemma for CFL states that for any language without context L, it is possible to find two substrings that can be “pumped” an unlimited number of times and are always in the same language. For each L language, we break down their strings into five parts and pump the second and fourth substrings. The pumping lemma is also used here as a tool to prove that a language is not CFL.
Because if a string doesn`t meet its requirements, then the language is not CFL. Thus, if L is a CFL, there exists an integer n such that for all x ∈ L with |x| ≥ n u, v, w, x, y ∈ Σ∗ such that x = uvwxy and (1) |vwx| ≤ n (2) |vx| ≥ 1 (3) for all i ≥ 0: uviwxiy ∈ L Use the pump lemma to obtain a contradiction – There are two pump lemmas defined for 1. regular languages and 2. Background – Free languages Pumping lemma for regular languages For every regular language L, there exists an integer n, so for all x ∈ L with |x| ≥ n there exist u, v, w ∈ Σ∗ such that x = uvw and (1) |uv| ≤ n (2) |v| ≥ 1 (3) for all i ≥ 0: uviw ∈ L Simply put, this means that if a chain v is “pumped”, that is, if v is inserted any number of times, the resulting chain always remains in L. So if a language is regular, it always satisfies the lemma of the pump. If there is at least one pump chain that is not L-shaped, then L is definitely not regular. It is not always the other way around. That is, if the pumping contains a lemma, it does not mean that the tongue is regular. For the example above, 0n1n is CFL because each string can be the result of pumping in two places, one for 0 and another for 1. Let`s prove that L012={0n1n2n | n ≥ 0} is not without context.
If we assume that L is contextless, then the pumping lemma follows the above rules. Now leave x L and |x ∈| ≥ n. Thus, by pumping the lemma, u, v, w, x, y exists such that (1) – (3) holds. We show that for all u, v, w, x, y (1) – (3) do not apply. If (1) and (2) are maintained, then x = 0n1n2n = uvwxy with |vwx| ≤ n and |vx| ≥ 1. (1) tells us that VWX does not contain both 0 and 2. Thus, vwx does not have a 0 or vwx no 2. So we have two cases to consider. Suppose vwx has no 0. Par (2) vx contains a 1 or a 2. So uwy has `n` 0 and uwy has less than `n` 1` or has less than `n` 2`.
But (3) tells us that uwy = uv0wx0y ∈ L. So uwy has an equal number of 0, 1 and 2 gives us a contradiction. The case where vwx has no 2 is similar and also gives us a contradiction. L is therefore not without context. While the pumping lemma states that all regular languages meet the conditions described above, the opposite of this statement is not true: a language that satisfies these conditions can still be irregular. In other words, the original and general versions of the pumping lemma give a necessary but not sufficient condition for a language to be regular. In formal language theory, the pump lemma for regular languages is a lemma that describes an essential property of all regular languages. Informally, it indicates that all sufficiently long strings can be pumped into a regular language, that is, a central section of the chain is repeated as many times as desired to produce a new string that is also part of the language. Let L {displaystyle L} be a regular language. Then there exists an integer p ≥ 1 {displaystyle pgeq 1} that depends only on L {displaystyle L}, so that any string w {displaystyle w} in L {displaystyle L} of length at least p {displaystyle p} (p {displaystyle p} is called “pump length”)[4] can be written as w = x y z {displaystyle w=xyz} (i.e.
w {displaystyle w} can be divided into three substrings), that meet the following conditions: Let w , x , y , z , p {displaystyle w,x,y,z,p} and n {displaystyle n} as used in the formal statement for the pump lemma above. For example, suppose there exists a constant p {displaystyle p}, as required by the lemma. Let w {displaystyle w} in L {displaystyle L} given by w = a p b p {displaystyle w=a^{p}b^{p}} , which is a longer string than p {displaystyle p}. Due to the pump lemma, a w = x y z {displaystyle w=xyz} decomposition must be decomposed with | x y | ≤ p {displaystyle |xy|leq p} and | There | ≥ 1 {displaystyle |y|geq 1} so that x y i z {displaystyle xy^{i}z} in L {displaystyle L} for each i ≥ 0 {displaystyle igeq 0}.
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