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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)
examples of decidable problems
As we are primarily concerned with questions of what is and what is not computable relative to some particular model of computation, we will usually base our explorations of langua
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This close relationship between the SL2 languages and the recognizable languages lets us use some of what we know about SL 2 to discover properties of the recognizable languages.
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
1. Simulate a TM with infinite tape on both ends using a two-track TM with finite storage 2. Prove the following language is non-Turing recognizable using the diagnolization
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
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
One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included
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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