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Another striking aspect of LTk transition graphs is that they are generally extremely ine?cient. All we really care about is whether a path through the graph leads to an accepting node or not. From that perspective, there is surely no reason to distinguish the nodes in the region marked H in Figure 2. Every one of these is an accepting node and every path from any one of them leads only to others in the same region. Every string with an initial segment which reaches one of these nodes will be accepted regardless of what the rest of the string looks like.
With a little more thought, it should become clear that the nodes in each of the other regions marked out in the ?gure are equivalent in a similar way. Any path which, when appended to a path leading to any one of the nodes, extends it to a path leading to an accepting state will do the same for paths leading to any node in the same region.
We can characterize the paths leading to the nodes in each region in terms of the components of aa ∧ (¬bb ∨ ba) they satisfy. Paths leading to region H satisfy aa ∧ ba. Strings starting this way will be accepting no matter what occurs in the remainder of the string. Regions D, F and G all satisfy aa. D and F also satisfy ¬bb and, so, are accepting. Paths reaching region G have seen bb and no longer accept until they have been extended with an a, thus satisfying aa ∧ ba and entering region H. We need to distinguish the nodes inregions D and F because paths leading to D end in a and, therefore, can be extended with b harmlessly, while if a path leading to F is extended with b we will no longer accept it.
Can v find the given number is palindrome or not using turing machine
Since the signi?cance of the states represented by the nodes of these transition graphs is arbitrary, we will allow ourselves to use any ?nite set (such as {A,B,C,D,E, F,G,H} or ev
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
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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(
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
Both L 1 and L 2 are SL 2 . (You should verify this by thinking about what the automata look like.) We claim that L 1 ∪ L 2 ∈ SL 2 . To see this, suppose, by way of con
Find a regular expression for the regular language L={w | w is decimal notation for an integer that is a multiple of 4}
Prove xy+yz+ýz=xy+z
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