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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)
S-->AAA|B A-->aA|B B-->epsilon
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
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Let L1 and L2 be CGF. We show that L1 ∩ L2 is CFG too. Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the sec
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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
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