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Claim Under the assumptions above, if there is an algorithm for checking a problem then there is an algorithm for solving the problem. Before going on, you should think a bit about how to do this. For this claim the assumption that the solution of each instance is unique is not necessary; but both of the others are. If you had a program that checks whether a proposed solution to an instance of a problem is correct and another that systematically generates every instance of the problem along with every possible solution, how could you use them (as subroutines) to build a program that, when given an instance, was guaranteed to ?nd a correct solution to that problem under the assumption that such a solution always exists?
Normal forms are important because they give us a 'standard' way of rewriting and allow us to compare two apparently different grammars G1 and G2. The two grammars can be shown to
DEGENERATE OF THE INITIAL SOLUTION
Theorem (Myhill-Nerode) A language L ⊆ Σ is recognizable iff ≡L partitions Σ* into ?nitely many Nerode equivalence classes. Proof: For the "only if" direction (that every recogn
One of the first issues to resolve, when exploring any mechanism for defining languages is the question of how to go about constructing instances of the mechanism which define part
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Computation of a DFA or NFA without ε-transitions An ID (q 1 ,w 1 ) computes (qn,wn) in A = (Q,Σ, T, q 0 , F) (in zero or more steps) if there is a sequence of IDs (q 1
prove following function is turing computable? f(m)={m-2,if m>2, {1,if
For every regular language there is a constant n depending only on L such that, for all strings x ∈ L if |x| ≥ n then there are strings u, v and w such that 1. x = uvw, 2. |u
Given any NFA A, we will construct a regular expression denoting L(A) by means of an expression graph, a generalization of NFA transition graphs in which the edges are labeled with
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(
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