4-solutions

4-solutions - CS 360 Introduction to the Theory of...

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Unformatted text preview: CS 360 Introduction to the Theory of Computing Spring 2008 Assignment 4 Solutions 1. [4 points] Prove that the following two languages are decidable: INF DFA = {( D ) : D is a DFA for which L ( D ) is infinite } INF CFG = {( G ) : G is a CFG for which L ( G ) is infinite } Solution 1. Let Σ ≥ n denote the language given by Σ ≥ n = { x ∈ Σ ∗ : | x | ≥ n } . The DTMs that decide INF DFA and INF CFG are almost exactly the same. They are both described as follows: Given input ( D ) (or ( G ) ); Compute the pumping length n of L ( D ) (or L ( G ) ); Compute a DFA E such that L ( E ) = L ( D ) ∩ Σ ≥ n ; (or, Compute a CFG H such that L ( H ) = L ( G ) ∩ Σ ≥ n ); Decide whether L ( E ) (or L ( H ) ) is empty; if L ( E ) negationslash = ∅ (or L ( H ) negationslash = ∅ ) then Accept else Reject; Let us argue that these DTMs decide INF DFA and INF CFG . First, we point out that the pumping lengths for both L ( D ) and L ( G ) are easily computed: • According to Section 4.1.1 of our text, the pumping length of L ( D ) for any DFA D is merely the number of states in D . • According to Section 7.2.2 of our text, the pumping length of L ( G ) for any CFG G is equal to 2 m where m is the number of variables in the Chomsky normal form of G . In either case, this pumping length is clearly computable. Next, is clear that Σ ≥ n is regular. We saw in our lecture that regular languages are closed under intersection and that context-free languages are closed under intersection with a regular language. Hence, L ( D ) ∩ Σ ≥ n is regular and L ( G ) ∩ Σ ≥ n is context-free. Moreover, the proofs of these closure properties are constructive. Hence, we may compute the DFA E and CFG H satisfying L ( E ) = L ( D ) ∩ Σ ≥ n and L ( H ) = L ( G ) ∩ Σ ≥ n . We also saw in our lecture that testing regular and context-free languages for emptiness are both decidable problems. In particular, it is possible to decide whether L ( E ) = ∅ and whether L ( H ) = ∅ . Finally, we argue that L ( D ) is infinite ⇔ L ( E ) negationslash = ∅ , and L ( G ) is infinite ⇔ L ( H ) negationslash = ∅ . Clearly, L ( E ) negationslash = ∅ if and only if L ( D ) contains a string of length n . Similarly, L ( H ) negationslash = ∅ if and only if L ( G ) contains a string of length at least n . So it remains only to prove that L ( D ) and L ( G ) are infinite if and only if they contain a string of length at least n . One direction is easy: if L ( D ) is infinite then it surely contains a string of length at least n , as otherwise L ( D ) would be finite. (The same holds for L ( G ) .) For the other direction, suppose that L ( D ) contains a string x of length at least n . By the pumping lemma for regular languages, there exist strings u,v,w with x = uvw such that: 1 • v negationslash = ε , • | uv | ≤ n , • uv i w ∈ L ( D ) for each i ≥ ....
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4-solutions - CS 360 Introduction to the Theory of...

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