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# a-2 - CpSc 421 Homework 10 Solutions Attempt any three of...

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CpSc 421 Homework 10 Solutions Attempt any three of the six problems below. The homework is graded on a scale of 100 points, even though you can attempt fewer or more points than that. Your recorded grade will be the total score on the problems that you attempt. 1. ( 20 points , Sipser problems 5.17 and 5.18) (a) ( 10 points ) Prove that the Post Correspondence Problem is decidable if the alphabet is unary, i.e., Σ = { 1 } . Solution (sketch): if the PCP instance has a tile of the form 1 i 1 i , then that tile solves the problem. Other- wise, if there is at least one tile of the form 1 i 1 j with i > j and one with i < j , then it is straightforward to solve the problem. Finally, if all tiles have more 1 ’s on the top than on the bottom (or all have more on the bottom than on the top), then the problem is not solvable. (b) ( 10 points ) Prove that the Post Correspondence Problem is undecidable if the alphabet is binary, i.e., Σ = { 0 , 1 } . Solution (sketch): Consider any instance of PCP with some alphabet, Σ 1 . Let k = log 2 | Σ 1 | . We can encode each symbol of Σ 1 with k symbols of { 0 , 1 } . With this encoding, every tile will a top string whose length is a multple of k and likewise for the bottom string. This ensures that the strings of { 0 , 1 } that encode symbols of Σ 1 will stay properly aligned, and the new, binary, problem is solvable iff the original problem is solvable. 2. ( 30 points , Sipser problem 5.21) Let AMBIG CFG = { G | G describes an ambiguous CFG } . Show that AMBIG CFG is undecidable. (Hint: Use a reduction from PCP . Given a PCP instance P = t 1 b 1 , t 2 b 2 , . . . , t k b k , construct a CFG G with the rules S T | B T t 1 T a 1 | . . . | t k T a k | t 1 a 1 | . . . | t k a k B b 1 B a 1 | . . . | b k B a k | b 1 a 1 | . . . | b k a k , where a 1 . . . a k are new terminal symbols. Prove that this reduction works.) Solution (sketch): If the PCP problem has a solution, i 1 , i 2 , . . . i m such that t i 1 t i 2 . . . t i m = b i 1 b i 2 . . . b i m , then S T * t i 1 t i 2 . . . t i k a i k . . . a i 2 a i 1 and S B * b i 1 b i 2 . . . b i k a i k . . . a i 2 a i 1 . Because t i 1 t i 2 . . . t i m = b i 1 b i 2 . . . b i m , these two derivations produce the same string. Thus, G is ambiguous. Now, I’ll show that if G is ambiguous, then the PCP problem has a solution. The critical observation is that a grammar with starts symbol T and has the rules given above is unambiguous, as there is only one way to get any suffix of a i symbols. Likewise, a grammar that starts with B is unambiguous. So, if G is ambiguous, there must be a string, x L ( G ) such that S T * x and S B * x . The derivation gives a solution to the PCP problem (in fact, the solution is the reverse of the string of a i ’s at the end of x ).

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a-2 - CpSc 421 Homework 10 Solutions Attempt any three of...

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