MathChic68

26.11.2019 • 

Social Studies

When word about the recursion theorem reached lem e. hackett, he immediately implemented it in code. his associate, ben bitdiddle, who insists the recursion theorem is too bizarre to be true, used lem's implementation to create the following program: tm b: on input x 1. construct own description (b) 2. run b(x): if it accepts then reject else if it rejects then accept "a program like this that contradicts itself cannot exist," claimed ben. therefore, your code cannot possibly be right, and so the recursion theorem is false. "let's run the program and see what happens," said lem who prefers running code to proofs. what is the result of running ben's program? what was the mistake in ben's reasoning (other than thinking that the recursion theorem is false)? problem 4. (10 points) we have seen several examples of undecidable languages of the form l = {(m): m is a tm whose language l(m) satisfies property p} examples of p include: "l(m) = $", "i̇z(m)]22", and vw: w e l(m) wr e l(m) notice that in each example two conditions are satisfied: a) if l(m) = l(m2) then (m) and (m2) are both in l or are both not in l, and b) there are tms in l and tms that are not in l. in this problem we will prove a powerful result: if p is any property that satisfies conditions a and b, then l is undecidable. in other words, every interesting property of tms that depends on the language of the tm is undecidable. fill in the blanks in the following outline to complete the proof of this result: 1. let p be any property of the language of a tm that satisfies conditions a and b. 2. assume that l is decidable and let tm d decide l. 3. since condition bis satisfied by p, let (m) el and (n) e l 4. now consider the following tm x: on input w: 1. compute own description (x) 2. if d accepts (x) then 3. if drejects (x) then since, in both cases x contradicts d, we conclude that

Ответ:

kealinwiley

08.03.2021

Social Studies

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