Pumping lemma Flashcards
(34 cards)
What is the main purpose of the Pumping Lemma?
To prove that certain languages are not regular.
When can the pumping lemma be applied?
Only to prove that a language is not regular — not to prove that it is regular.
What must be shown to prove a language is regular?
That it is accepted by a DFA, NFA, or described by a regular expression.
What does the Pumping Lemma state?
If L is regular, then there exists an n ≥ 1 such that every w ∈ L with |w| ≥ n can be split as w = xyz with |xy| ≤ n, y ≠ ε, and xyⁱz ∈ L for all i ≥ 0.
What is the significance of the condition |xy| ≤ n in the pumping lemma?
It ensures the loop y appears early — within the first n letters of the word.
Why must y ≠ ε in the Pumping Lemma?
So that something is actually being repeated — y must be a non-empty loop.
What does it mean to ‘pump’ the loop y in the Pumping Lemma?
To repeat the substring y any number of times and remain in the language: xyⁱz ∈ L for all i ≥ 0.
Why does the pumping phenomenon occur in DFAs?
Because if the input word is longer than the number of states, the DFA must revisit a state — creating a loop (by the pigeonhole principle).
What is the intuition behind the loop y in DFA processing?
It’s a repeatable sequence of transitions that brings the DFA back to a previous state.
Why is the pumping lemma not used to prove regularity?
Because it can only show that a language must have the pumping property if it’s regular — it can’t confirm regularity on its own.
Give an example of a language the pumping lemma shows is not regular.
L = { w ∈ {a,b}* | w has equal numbers of a’s and b’s }
Why is the language with equal numbers of a’s and b’s not regular?
Because tracking an arbitrary difference between a’s and b’s would require infinite states, which a finite automaton can’t provide.
What does it mean if a word w violates the pumping lemma conditions?
It means the language L is not regular.
What is the first step in applying the pumping lemma to prove non-regularity?
Assume for contradiction that the language is regular and let n be the pumping length.
What is the typical strategy in using the pumping lemma?
Choose a word w ∈ L with |w| ≥ n, show that all decompositions into xyz fail the pumping condition.
What is the main use of the Pumping Lemma?
To prove that certain languages are not regular.
Can the Pumping Lemma be used to prove regularity?
No — it can only be used to show a language is not regular.
What is the form of the Pumping Lemma?
For a regular language L, there exists n ≥ 1 such that every word w ∈ L with |w| ≥ n can be split as w = xyz with |xy| ≤ n, y ≠ ε, and xyⁱz ∈ L for all i ≥ 0.
What are the three conditions in the Pumping Lemma?
(1) |xy| ≤ n, (2) y ≠ ε, (3) xyⁱz ∈ L for all i ≥ 0.
Why must the loop y appear early in the word?
Because |xy| ≤ n — pumping must occur within the first n characters.
What is the proof strategy for using the Pumping Lemma?
Assume the language is regular, then find a word that violates the pumping conditions — leading to a contradiction.
Why is a contradiction important in Pumping Lemma proofs?
It shows that the assumption of regularity must be false, so the language is not regular.
What is a common mistake when applying the Pumping Lemma?
Not specifying n or not defining the decomposition xyz.
Why must you carefully choose the word w in a Pumping Lemma proof?
Choosing a word that can be pumped will not create a contradiction — and won’t prove the language is non-regular.
