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Unformatted text preview: Problem Set 5 Fall 09 Due: Thursday Dec 3 at 11:00 AM in class (i.e., Room 103 Talbot Lab) Please follow the homework format guidelines posted on the class web page: http://www.cs.uiuc.edu/class/fa09/cs373/ 1. [ Points : 15] (a) For each of the following PCP problems, either nd a solution or prove that a solution does not exist. i. 11 101 11 11011 110 1 Solution: 110 1 11 101 110 1 11 11011 ii. 10 1 10 01 01 10 10 Solution: This one has no solution: Since all the strings here have length at least one, the two string of the starting piece should start with the same character, therefore the only possibility for the starting piece is (10 , 1) . Now the bottom string of the next domino should start with a zero, the only such piece is (10 , 01) . Consider the sequence (10 , 1)(10 , 01) we observe that again the next piece should start with a zero in its bottom string, that is we should play (10 , 01) again. Inspecting (10 , 1)(10 , 01)(10 , 01) we see that again the next piece should be (10 , 01) which means that the game has entered a loop and never nishes. iii. 1110 1 1 0111 Solution: 1110 1 1110 1 1110 1 1 0111 1 0111 1 0111 (b) Prove that PCP is undecidable even if we restrict its alphabet to two symbols, for example Σ = { , 1 } . 1 Solution: Let's name this restricted version of PCP, PCP . We reduce PCP to PCP (which proves that PCP is undecidable since PCP is undecidable). Let Σ be the alphabet of PCP. Our reduction function f replaces the ith character in Σ with i 1 . Note that the reduction is reversible, that is if x,y ∈ Σ * and f ( x ) = f ( y ) , then x = y . To reduce our instance of PCP to PCP , we just apply function f to the strings of all domino pieces. Since f is reversible, some ordering of domino pieces is a solution for the reduced instance i the same ordering is a solution for the original instance. (c) Prove that PCP is decidable if we restrict its alphabet to one symbol, for example Σ = { } . Solution: Note that the goal here is to pick some dominos such that the total number of zero's on the top of them equals the total number of zero's on the bottom of those selected dominos. If there is a domino with the same number of zeros on the top and bottom, then that domino itself is a solution, therefore we may assume that no domino has equal number of zeros on its sides. If all dominos have more zeros on top (bottom), then obviously any subset of dominos has more zeros on top (bottom) and therefore there is no solution. Finally assume that there is at least one domino ( a,b ) such that a < b and on domino ( c,d ) such that c > d . Note that if we pick b a instances of ( c,d ) together with c d instances of ( a,b ) , then we have a solution....
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This note was uploaded on 04/18/2010 for the course CS 373 taught by Professor Viswanathan,m during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Viswanathan,M

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