Lecture9x

# Lecture9x - 15-453 FORMAL LANGUAGES AUTOMATA AND...

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FORMAL LANGUAGES, AUTOMATA, AND COMPUTABILITY 15-453 * Read chapter 4 of the book for next time * Lecture9x.ppt

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A Turing Machine is represented by a 7-tuple T = (Q, Σ, Γ, δ , q 0 , q accept , q reject ): Q is a finite set of states Γ is the tape alphabet, a superset of Σ ; Γ q 0 Q is the start state Σ is the input alphabet, where Σ δ : Q × Γ → Q × Γ × {L, R} is the transition func q accept Q is the accept state q reject Q is the reject state, and q reject q accept

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CONFIGURATION: (1) tape contents, (2) current state, (3) location of read/write head. 11010 q 7 00110 q 7 1 0 0 0 0 0 1 1 1 1
A TM recognizes a language iff it accepts all and only those strings in the language. A TM decides a language L iff it accepts all strings in L and rejects all strings not in L. A language L is called Turing-recognizable or recursively enumerable iff some TM recognizes L. A language L is called decidable or recursive iff some TM decides L.

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A language is called Turing-recognizable or recursively enumerable (r.e.) if some TM recognizes it. A language is called decidable or recursive if some TM decides it. recursive languages r.e. languages
Theorem: If A and ¬ A are r.e. then A is decidable. Given: a Turing Machine TM A that recognizes A and a Turing Machine TM R that recognizes ¬ A, we can build a new machine that decides A. How can we prove this? Run TM A and TM R in parallel. (Or more precisely, interleave them.) One of them will eventually recognize the input string. If TM A recognizes it, then accept. If TM R recognizes it, then reject.

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A TM that decides { 0 | n ≥ 0 } High-Level Idea. Repeatedly divide the number of zeros in half until it becomes an odd number. If we are left with a single zero, then accept. Otherwise, reject. 2 n We want to accept iff: the input string consists entirely of zeros, and the number of zeros is a power of 2.
A TM that decides { 0 | n ≥ 0 } 1. Sweep from left to right, cross out every other 0 . (Divides number in half.) 2. If in step 1, the tape had only one 0 , accept . 3. Else if the tape had an odd number of 0 ’s, reject . 4. Move the head back to the first input symbol. 5. Go to step 1. PSEUDOCODE: 2 n

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0 → , R , R q accept q reject 0 → x, R x → x, R , R x → x, R 0 → 0, L x → x, L x → x, R , L , R 0 → x, R 0 → 0, R , R x → x, R { 0 | n ≥ 0 } 2 n q 0 q 1 q 2 q 3 q 4
0 → , R , R q accept

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