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Suppose G = (N, Σ, P, S) is a reduced grammar (we can certainly reduce G if we haven't already). Our algorithm is as follows:
1. Define maxrhs(G) to be the maximum length of the right hand side of any production.
2. While maxrhs 3 we convert G to an equivalent reduced grammar G' with smaller maxrhs.
3. a) Choose a production A → α where is of maximal length in G.
b) Rewrite α as α1α2 where |α1| = |α1|/2 (largest integer ≤ |α1|/2) and |α2| = |α2|/2 (smallest integer ≥ |α2|/2)
c) Replace A -> α in P by A -> α1B and B -> α2
If we repeat step 3 for all productions of maximal length we create a grammar G' all of whose productions are of smaller length than maxrhs.
We can then apply the algorithm to G' and continue until we reach a grammar that has maxrhs ≤ 2.
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design a turing machine that accepts the language which consists of even number of zero''s and even number of one''s?
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
Ask question #Minimum 100 words accepte
LTO was the closure of LT under concatenation and Boolean operations which turned out to be identical to SF, the closure of the ?nite languages under union, concatenation and compl
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
The SL 2 languages are speci?ed with a set of 2-factors in Σ 2 (plus some factors in {?}Σ and some factors in Σ{?} distinguishing symbols that may occur at the beginning and en
s-> AACD A-> aAb/e C->aC/a D-> aDa/bDb/e
Intuitively, closure of SL 2 under intersection is reasonably easy to see, particularly if one considers the Myhill graphs of the automata. Any path through both graphs will be a
We'll close our consideration of regular languages by looking at whether (certain) problems about regular languages are algorithmically decidable.
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