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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.
20*2
a) Let n be the pumping lemma constant. Then if L is regular, PL implies that s can be decomposed into xyz, |y| > 0, |xy| ≤n, such that xy i z is in L for all i ≥0. Since the le
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
To see this, note that if there are any cycles in the Myhill graph of A then L(A) will be infinite, since any such cycle can be repeated arbitrarily many times. Conversely, if the
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
S-->AAA|B A-->aA|B B-->epsilon
A common approach in solving problems is to transform them to different problems, solve the new ones, and derive the solutions for the original problems from those for the new ones
DEGENERATE OF THE INITIAL SOLUTION
The fact that the Recognition Problem is decidable gives us another algorithm for deciding Emptiness. The pumping lemma tells us that if every string x ∈ L(A) which has length grea
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