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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?
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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 give
We will assume that the string has been augmented by marking the beginning and the end with the symbols ‘?' and ‘?' respectively and that these symbols do not occur in the input al
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
Computer has a single LIFO stack containing ?xed precision unsigned integers (so each integer is subject to over?ow problems) but which has unbounded depth (so the stack itself nev
So we have that every language that can be constructed from SL languages using Boolean operations and concatenation (that is, every language in LTO) is recognizable but there are r
The generalization of the interpretation of strictly local automata as generators is similar, in some respects, to the generalization of Myhill graphs. Again, the set of possible s
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We developed the idea of FSA by generalizing LTk transition graphs. Not surprisingly, then, every LTk transition graph is also the transition graph of a FSA (in fact a DFA)-the one
Paths leading to regions B, C and E are paths which have not yet seen aa. Those leading to region B and E end in a, with those leading to E having seen ba and those leading to B no
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