hw1slns - CS6505 Computability, Algorithms, and Complexity...

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Unformatted text preview: CS6505 Computability, Algorithms, and Complexity Homework 1 solutions sketch. 1. (15 points) Consider the simulation of a multitape Turing machine M by a single-tape Turing machine S as described in the proof of Theorem 3.13 in the text. Let k = 2. (a) Give a formal definition of the simulating single-tape Turing machine S without specifying the transi- tions. (b) Show the transitions of the machine S for implementing a left move of the machine M . See the attachment. 2. (15 points) A verifier is a deterministic Turing machine V that is a decider that takes two arguments x (the input) and y (the proof). Show that a language L is Turing-Recognizable if and only if there exists a verifier V such that for all x * : x L : y * such that V ( x,y ) accepts x 6 L : y * , V ( x,y ) rejects Solution: Only If Direction : The crucial idea here is that y (as a function of x ) can be chosen to encode the number of steps required by M to halt and accept x . The verifier V is the one that on input h x,y i simulates the recognizer M on input x for y number of steps, and then halts. If L is Turing-recognizable, then for all x L , there is a number y such that M halts and accepts in y steps. When V simulates M on x for y steps, it can verify that x L . If x / L , then no matter what number y we provide, V will always halt after y steps, but will never accept x . Therefore V is a halting verifier that satisfies the conditions of the problem. Source of a common mistake: note that the verifier V is NOT accepting the language L itself - L need not be decidable, while V must be a decider. V is accepting {h x,y i| x L and y a proof that x L } . If Direction : If there exists a verifier V satisfying the conditions of the problem, M is a recognizer that uses V as follows. M on input x does the following: y { , 1 } * in lexicographic order, M simulates V on h x,y i and accepts whenever V accepts. For all x L , V ( h x,y i ) will accept for some y by hypothesis, therefore M will halt and accept x . For all x / L , V ( h x,y i ) will never accept for any y , therefore M will loop on all x . Thus M is a recognizer for L . Note: In this case, we cannot assume that the proof string y is the number of steps as before, it can be any arbitrary proof string that satisfies the condition of the problem. 3. (15 points) A 2-dimensional Turing machine has a finite control with its tape being a 2-dimensional array of cells infinite in all the four directions. Based on the current state and the symbol under the read/write head, the machine changes state, writes a new symbol, and moves in one of the four directions (up, down, left, right)....
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This note was uploaded on 09/20/2009 for the course CS 6505 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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hw1slns - CS6505 Computability, Algorithms, and Complexity...

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