5.17 - 6.045J/18.400J: Automata, Computability and...

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Unformatted text preview: 6.045J/18.400J: Automata, Computability and Complexity Prof. Nancy Lynch Recitation 6: Decidability and Undecidability March 15, 2007 Problem 1: These are the key concepts from lecture this week: 1. Undecidability - p. 172-176 have great example proofs. Elena Grigorescu 3. Computation history - p. 176, 179, 185 give the definition and some examples. 4. Diagonalization method - p. 160-168. This concept is both elegant and difficult; make sure you understand it. Problem 2: Show that the following languages are undecidable: is a power of . Problem 3: Show that the following language is undecidable: Reduce from both and . Recall that was decidable. 2. Run on . =“On input , 1. reject.” 6: Decidability and Undecidability-1 7 p ¦£ Notice that is the TM that we constructed when we proved (i.e., ). =“On input , i q E rWH e QdF' fd  2. If accepts, accept” was undecidable by reducing from b  1. Run on . i ¢ 3. If accepts, reject; otherwise, accept.” f gd b 1. Construct TM from and and TM as detailed below. b  ! f dB9 e h ¢ d  eUd ! b9  c@a =“On input , an encoding of a TM and a string , ` 7  ¦£ that Solution 3: In class we saw how to reduce to is undecidable. Let be a TM that decides as follows. . Here we will reduce from to prove . We could then construct a decider for   &H E '  H E SRQAFPI$GF' ¢   7¦£  7¦£  £  XYVWU¦£ T7     £ A # ! A9      7 D@CB@ ©$80¦£ and are TMs such that 3 42  2. is a Turing machine and accepts exactly the strings in & & 65 ) !   10" ©$(' # !     ¢§ ¥ £ %$" ©©¨¦¤¢ 1. halts on any input in accept or reject ¡ 2. Reductions - p. 171-172 will help with the terminology (e.g., “reduce from ”, etc.) whose length `   f gd   £ ed Thus, we contrive that if and only if rejects , while always. Since, by assumption, we have a decider that tells us if these two machines recognize the same language, we know that if rejects and , then this implies that accepts . Problem 4: (From Sipser, problems 5.17 and 5.18) Consider the Post Correspondence Problem over small alphabets. Solution 4: 1. Sketch of Proof: We prove it is decidable, by giving an algorithm that decides it. Each 3. If 4. If We know that the number of symbols counted in step 1 is finite, since the number and content of each domino is finite. By giving unique binary encodings of equal length to each domino, the problem reduces nicely. Observe that this would not necessarily be true if our encoding for each unique symbol was allowed to be of different lengths. ¢ 5. If ¢ 4. Run on accepts, accept; otherwise, reject.” ! „r B™%€€9 rf B9 er h 9kkk   ! „ƒB%€%€9 ƒB9 e6h r  9 k k k rf  r  3. Construct new dominos using the binary encoding. . 6: Decidability and Undecidability-2 H En qpovQml  2. Assign to each unique symbol a unique (iterative) value. Front-pad with zeros. -bit (or you could also reduce this to a # ¨ #2 1. Count the number of different symbols on the dominos: ‚ =“On input , where each is a domino, . -bit) i¥ ¨Ui ` ! „ e€%€€9 f e9 e “ 9kkk   i¥i ¨U™j ¢ Assume a TM that decides . Build a TM to decide . 2 i¥ ¨Ui 2. Sketch of Proof: We prove it is undecidable by reducing from h g (4f ud “0e” u ! ‚” u  ‚” u ! ‚” 2. If ‚ ™” 1. If for some domino, accept. [That single domino is a match.] for all dominos, reject. for all dominos, reject. and for , then accept. [You can even these out.] (over an arbitrary alphabet ). u ’ ‚ 9‚ “I‘cB©‰ …‡ ˆƒu …„ †6u domino in the set has a top portion of the values : and a bottom portion of &y9u 1€xv 2. Show that the problem is undecidable over the binary alphabet &u wv 1. Show that the problem is decidable over the unary alphabet . (bPCP). for some q E CtH f Gd4' b b   ¢ q E CsH e Qd4' fd ‚  –‚‰ ‚ ‘˜—©P•€” ed ‚ ƒ ¢ . Lets consider ` ...
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This note was uploaded on 04/13/2010 for the course CS 235 taught by Professor Sabnis during the Spring '10 term at Punjab Engineering College.

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