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The path function δ : Q × Σ* → P(Q) is the extension of δ to strings: This just says that the path labeled ε from any given state q goes only to q itself (or rather never l
It is not hard to see that ε-transitions do not add to the accepting power of the model. The underlying idea is that whenever an ID (q, σ v) directly computes another (p, v) via
how to find whether the language is cfl or not?
shell script to print table in given range
The fact that SL 2 is closed under intersection but not under union implies that it is not closed under complement since, by DeMorgan's Theorem L 1 ∩ L 2 = We know that
State & prove pumping lemma for regular set. Show that for the language L={ap |p is a prime} is not regular
Find the Regular Grammar for the following Regular Expression: a(a+b)*(ab*+ba*)b.
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(
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
The fact that the Recognition Problem is decidable gives us another algorithm for deciding Emptiness. The pumping lemma tells us that if every string x ∈ L(A) which has length grea
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