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FIT2014 Theory of computation - MUM S2 2025
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To use the pumping lemma for CFL, we assume that a language L has a CFG in CNF. Next, we choose a string W in L that is long enough. We then divide the string W into 5 parts, i.e., W = uvxyz. Which of the following is (are) true?
When we do a proof by contradiction using the pumping lemma for CFL, as long as we can reach a contradiction (i.e., the resulting string is NOT in the language assumed to be context-free) for just one way to divide the string W = uvxyz, then we are done.
When we prove that a language L is not context free, we use the pumping lemma. Part of the proof involves the parameter K, which refers to
We appeal to the pumping lemma for context-free language (CFL) to prove that a language L is CFL.
Convert the following Nondeterministic Finite Automaton (NFA) into an equivalent Deterministic Finite Automaton (FA).
Your FA must be presented by filling in some rows in the table further down. You may not need all the rows available.
This question is also about the language ANTIPALINDROME, defined in the previous question.
Give a Context-Free Grammar for the language ANTIPALINDROME.
Let ANTIPALINDROME be the language of all even-length strings over the alphabet {a,b} such that, for all i, the i-th letter from the start is different to the i-th letter from the end. Examples of strings in this language include:
ε, ab, ba, aabb, abab, baba, bbaa, aaabbb, ...
Using the Pumping Lemma for Regular Languages, prove that ANTIPALINDROME is not regular.
Consider the five-state Finite Automaton represented by the following table.
| state | a | b |
| Start 1 | 3 | 1 |
| 2 | 4 | 2 |
| Final 3 | 1 | 3 |
| Final 4 | 2 | 4 |
| Final 5 | 3 | 5 |
Find an equivalent FA with the minimum number of states.
Enter your simplified FA in the table below. You may not need all rows of the table.
Convert the following Nondeterministic Finite Automaton (NFA) into an equivalent Deterministic Finite Automaton (FA).
Your FA must be presented by filling in some rows in the table further down. You may not need all the rows available.
Let x be a string, and let M be a Finite Automaton with just one Final State that accepts the strings x and xx.
(a) Prove, by induction on n, that M accepts the string xn for every n ≥ 1.
(b) Would the same statement hold if M is a Nondeterministic Finite Automaton, also with just one Final State, instead? Why or why not?