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Theorem The class of recognizable languages is closed under Boolean operations.
The construction of the proof of Lemma 3 gives us a DFA that keeps track of whether or not a given string is in either or both of any pair of recognizable languages. We can modify the construction for other Boolean operations simply by selecting the appropriate set of accepting states:
• Union: Let F′
= {(q, p) | q ∈ F1 or p ∈ F2}. Then L(A′ ) = L1 ∪ L2.
• Relative complement: Let F′ = F1 × (Q2 - F2). Then L(A′ ) = L1 -L2.
• Complement: Let L1 = Σ* and use the construction for relative complement.
A problem is said to be unsolvable if no algorithm can solve it. The problem is said to be undecidable if it is a decision problem and no algorithm can decide it. It should be note
De?nition (Instantaneous Description) (for both DFAs and NFAs) An instantaneous description of A = (Q,Σ, δ, q 0 , F) , either a DFA or an NFA, is a pair h q ,w i ∈ Q×Σ*, where
Find a regular expression for the regular language L={w | w is decimal notation for an integer that is a multiple of 4}
We saw earlier that LT is not closed under concatenation. If we think in terms of the LT graphs, recognizing the concatenation of LT languages would seem to require knowing, while
dfa for (00)*(11)*
a) Let n be the pumping lemma constant. Then if L is regular, PL implies that s can be decomposed into xyz, |y| > 0, |xy| ≤n, such that xy i z is in L for all i ≥0. Since the le
De?nition Instantaneous Description of an FSA: An instantaneous description (ID) of a FSA A = (Q,Σ, T, q 0 , F) is a pair (q,w) ∈ Q×Σ* , where q the current state and w is the p
draw pda for l={an,bm,an/m,n>=0} n is in superscript
Proof (sketch): Suppose L 1 and L 2 are recognizable. Then there are DFAs A 1 = (Q,Σ, T 1 , q 0 , F 1 ) and A 2 = (P,Σ, T 2 , p 0 , F 2 ) such that L 1 = L(A 1 ) and L 2 = L(
This was one of the ?rst substantial theorems of Formal Language Theory. It's maybe not too surprising to us, as we have already seen a similar equivalence between LTO and SF. But
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