Pumping lemma for regular languages

Formal statement

Let L be a regular language. Then there exists an integer p ≥ 1 depending only on L such that every string w in L of length at least p (p is called the "pumping length") can be written as w = xyz (i.e., w can be divided into three substrings), satisfying the following conditions:

1. |y| ≥ 1
2. |xy| ≤ p
3. for all i ≥ 0, xyⁱz ∈ L

y is the substring that can be pumped (removed or repeated any number of times, and the resulting string is always in L). (1) means the loop y to be pumped must be of length at least one; (2) means the loop must occur within the first p characters. |x| must be smaller than p (conclusion of (1) and (2)), but apart from that, there is no restriction on x and z.

In simple words, for any regular language L, any sufficiently long word w (in L) can be split into 3 parts. i.e. w = xyz , such that all the strings xy^kz for k≥0 are also in L.

Below is a formal expression of the Pumping Lemma.

( ∀ L ⊆ Σ ∗ ) ( regular ( L ) ⇒ ( ( ∃ p ≥ 1 ) ( ( ∀ w ∈ L ) ( ( | w | ≥ p ) ⇒ ( ( ∃ x , y , z ∈ Σ ∗ ) ( w = x y z ∧ ( | y | ≥ 1 ∧ | x y | ≤ p ∧ ( ∀ i ≥ 0 ) ( x y i z ∈ L ) ) ) ) ) ) ) )

Use of the lemma

The pumping lemma is often used to prove that a particular language is non-regular: a proof by contradiction (of the language's regularity) may consist of exhibiting a word (of the required length) in the language that lacks the property outlined in the pumping lemma.

For example, the language L = {aⁿbⁿ : n ≥ 0} over the alphabet Σ = {a, b} can be shown to be non-regular as follows. Let w, x, y, z, p, and i be as used in the formal statement for the pumping lemma above. Let w in L be given by w = a^pb^p. By the pumping lemma, there must be some decomposition w = xyz with |xy| ≤ p and |y| ≥ 1 such that xyⁱz in L for every i ≥ 0. Using |xy| ≤ p, we know y only consists of instances of a. Moreover, because |y| ≥ 1, it contains at least one instance of the letter a. We now pump y up: xy²z has more instances of the letter a than the letter b, since we have added some instances of a without adding instances of b. Therefore, xy²z is not in L. We have reached a contradiction. Therefore, the assumption that L is regular must be incorrect. Hence L is not regular.

The proof that the language of balanced (i.e., properly nested) parentheses is not regular follows the same idea. Given p, there is a string of balanced parentheses that begins with more than p left parentheses, so that y will consist entirely of left parentheses. By repeating y, we can produce a string that does not contain the same number of left and right parentheses, and so they cannot be balanced.

Proof of the pumping lemma

For every regular language there is a finite state automaton (FSA) that accepts the language. The number of states in such an FSA are counted and that count is used as the pumping length p. For a string of length at least p, let q₀ be the start state and let q₁, ..., q_p be the sequence of the next p states visited as the string is emitted. Because the FSA has only p states, within this sequence of p + 1 visited states there must be at least one state that is repeated. Write q_s for such a state. The transitions that take the machine from the first encounter of state q_s to the second encounter of state q_s match some string. This string is called y in the lemma, and since the machine will match a string without the y portion, or with the string y repeated any number of times, the conditions of the lemma are satisfied.

For example, the following image shows an FSA.

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 pigeonhole principle indicates that there must be at least one repeated state among the start state and the next four visited states. In this example, only q₁ is a repeated state. Since the substring bc takes the machine through transitions that start at state q₁ and end at state q₁, that portion could be repeated and the FSA would still accept, giving the string abcbcd. Alternatively, the bc portion could be removed and the FSA would still accept giving the string ad. In terms of the pumping lemma, the string abcd is broken into an x portion a, a y portion bc and a z portion d.

General version of pumping lemma for regular languages

If a language L is regular, then there exists a number p ≥ 1 (the pumping length) such that every string uwv in L with |w| ≥ p can be written in the form

with strings x, y and z such that |xy| ≤ p, |y| ≥ 1 and

From this, the above standard version follows a special case, with both u and v being the empty string.

Since the general version imposes stricter requirements on the language, it can be used to prove the non-regularity of many more languages, such as { a^mbⁿcⁿ : m≥1 and n≥1 }.

Converse of lemma not true

While the pumping lemma states that all regular languages satisfy the conditions described above, the converse of this statement is not true: a language that satisfies these conditions may still be non-regular. In other words, both the original and the general version of the pumping lemma give a necessary but not sufficient condition for a language to be regular.

For example, consider the following language L:

In other words, L contains all strings over the alphabet {0,1,2,3} with a substring of length 3 including a duplicate character, as well as all strings over this alphabet where precisely 1/7 of the string's characters are 3's. This language is not regular but can still be "pumped" with p = 5. Suppose some string s has length at least 5. Then, since the alphabet has only four characters, at least two of the first five characters in the string must be duplicates. They are separated by at most three characters.

For a practical test that exactly characterizes regular languages, see the Myhill-Nerode theorem. The typical method for proving that a language is regular is to construct either a finite state machine or a regular expression for the language.