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CS 461 – Nov. 9 Chomsky hierarchy of language classes –Review –Let’s find a language outside the TM world! –Hints: languages and TM are countable, but the set of all languages is uncountable Undecidability –We have seen that ATM is undecidable. –Let’s look at some more undecidable problems.

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TM properties 2 kinds of TMs 2 kinds of languages. –Turing-recognizable (a.k.a. recursively enumerable) Example: A TM –Decidable (a.k.a. recursive) Example: 0* If L is decidable, then L’ is decidable. If L and L’ are both Turing-recognizable, then L is decidable. (since either L or L’ must accept) Therefore, the complement of A TM is not even Turing recognizable.

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Decidable? for n = 3.. maxint: for each combo of (x, y, z) starting at 1: if (x n + y n == z n ) print (“hello, world”); Are we able to tell if this program will print “ hello, world ” or not? –Only since 1993. For 300 years we didn’t know. –Solving this problem in general (for any program) must be hard… but what if it could be done?

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“hello, world” tester Suppose H can decide the question. Then we can create similar algorithm H2. –Returns “hello, world” instead of “no”. –Let’s run H2 on itself! H program & input Print “yes” Print “no” H2 Print “yes” Print “hello, world”

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Regular? REG = “Is there an algorithm to determine if a TM recognizes a regular language?” Proof by contradiction, p. 191 Assume REG is decidable. –Then, TM “R” that can tell if an input recognizes a regular language. –We’ll use R to solve the A TM problem by creating a decider S. –In other words: Machine R solves REG -- which implies existence of …. Machine S which solves A TM. –Do we understand the I/O of R and S?

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Regular, continued Here is how to build machine S, which will “solve” A TM. Input to S is. Create a new TM “M2” –M2 sees if its input, x, is of the form 0 n 1 n. If so, M2 accepts x. –If x is not of the form 0 n 1 n, M2 runs T on w. If T accepts w, M2 accepts x. –Thus, L(M2) is either Σ*, if T accepts w, or 0 n 1 n. –The only way for M2 to accept Σ* is for T to accept w. Feed M2 into R. –If R says that M2 accepts a regular language, we can conclude T accepts w, so S can say “yes”, i.e. accept. Otherwise S rejects. –But S can’t exist. So we have a contradiction. REG undecidable.

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Rice’s Theorem Any question Q about a TM language is undecidable. Proof by contradiction: Assume Q is decidable. Then TM “QD” that –always halts, and –answers Q for any input TM it receives. Using QD, we can build a machine AD to decide A TM. (But we know that AD can’t exist, therefore QD can’t exist. That is the contradiction.)

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Building TM “AD” AD will do the following: –Its input is. –Build a TM “Q1” that satisfies Q. –Build a TM “M2” whose input is x. M2 runs T on w. If T loops/rejects, M2 will loop/reject. If T accepts w, next run Q1 on x. M2 accepts if Q1 accepts. Note that L(M2) = L(Q1) if T accepts w. L(M2) = if T does not accept w. In other words, if T accepts, then M2 satisfies Q. –Feed M2 into QD. If QD accepts, AD accepts. Otherwise AD rejects. Thus: QD is a decider AD is a decider.

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Glossary to proof Q = some question or property about a TM language QD = a TM that can decide Q Q1 = a TM for which Q is true, built inside AD A TM = the acceptance question for TM’s, already known to be undecidable AD = a hypothetical decider for A TM T = the TM input to AD w = the corresponding word also input to AD M2 = a second TM built inside AD Rice’s theorem is a little complex because 5 TM’s!

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Examples Let Q = “Is the language infinite?” or …. “Does the language include 001?” “Is the language finite?” “Is the language nonregular?” Can also prove by contradiction: –Is it decidable to determine if one TM language is the subset of another?

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Recursively Enumerable and Recursive Languages

Recursively Enumerable and Recursive Languages

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