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Computer has a single unbounded precision counter which you can only increment, decrement and test for zero. (You may assume that it is initially zero or you may include an explicit instruction to clear.) Limit your program to a single unsigned integer variable, and limit your methods of accessing it to something like inc(i), dec(i) and a predicate zero?(i) which returns true i? i = 0. This integer has unbounded precision-it can range over the entire set of natural numbers-so you never have to worry about your counter over?owing. It is, however, restricted to only the natural numbers-it cannot go negative, so you cannot decrement past zero.
(a) Sketch an algorithm to recognize the language: {aibi| i ≥ 0}. This is the set of strings consisting of zero or more ‘a's followed by exactly the same number of ‘b's.
(b) Can you do this within the ?rst model of computation? Either sketch an algorithm to do it, or make an informal argument thatit can't be done.
(c) Give an informal argument that one can't recognize the language: {aibici| i ≥ 0} within this second model of computation (i.e, witha single counter)
Let G be a graph with n > 2 vertices with (n2 - 3n + 4)/2 edges. Prove that G is connected.
examples of decidable problems
write short notes on decidable and solvable problem
We saw earlier that LT is not closed under concatenation. If we think in terms of the LT graphs, recognizing the concatenation of LT languages would seem to require knowing, while
We'll close our consideration of regular languages by looking at whether (certain) problems about regular languages are algorithmically decidable.
The key thing about the Suffx Substitution Closure property is that it does not make any explicit reference to the automaton that recognizes the language. While the argument tha
The Equivalence Problem is the question of whether two languages are equal (in the sense of being the same set of strings). An instance is a pair of ?nite speci?cations of regular
how to find whether the language is cfl or not?
As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′ 0 . Since they cannot be reached from Q′ 0 there is no path from Q′ 0 to a sta
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
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