计算理论20110421y

计算理论导引与算法复杂性
Introduction to the Theory of Computation and Algorithm Complexities
主讲： 主讲：刘国华 （学院楼①225室，ghliu@） 学院楼①225室 ） 助教： 助教：辛婷婷 学院楼①153室 （学院楼①153室，xtt.moon@） ） FTP:222.204.215.3
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• THE HAITING PROBLEM(ACCEPTANCE PROBLEM) 1 THE DIAGONALIZATION METHOD 1)Correspondence:a function that is both one-to-one and onto. 2)Countable set: a set A is countable if either it is finite or it has the same size as N. THEOREM 3.8 R is incountable. PROOF IDEA. To show that no correspondence exists between N and R. METHOD:contradiction To construct a real number x whose ith fractional digit is different from f(i)’s,1≤i≤n.
On input <D>: 1. Run H on input <D, <D>>. 2. D rejects <D>, if D accepts <D>; D accepts <D>,if D rejects <D>.”
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The universe of languages Turing-recognizable languages
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COROLARY 3.2 ATM is not Turing-recognizable. • REDUCIBILITY A way of converting one problem to another problem in such a way that a solution to the second problem can be used to solve the first problem. If problem A can reduce to problem B. We can know: 1.Solving problem A can not be harder than problem B. 2.If B is decidable, then A also is decidable. 3.If A is undecidable, then B also is undecidable.
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• THE HAITING PROBLEM(ACCEPTANCE PROBLEM) ATM={<M,w>|M is a TM and M accepts w} U=“On input<M,w>,where M is a TM and w is a string: 1.Simulate M on input w. 2.If M ever enters its accept state, accept; if M ever enter its reject state, reject.” If M loops, what will we do?
The complement of Turing-recognizable languages
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3 A TURING-UNRECOGNIZABLE LANGUAGE Co-Turing-recognizable language: a language is co-Turingrecognizable if it is the complement of a Turingrecognizable language and Turing-recognizable. THEOREM 3.10 A language is decidable iff it is Turing-recognizable and coTuring-recognizable. PROOF IDEA M=“On input w: 1.Run both M1 and M2 on input w in parallel. 2.If M1 accepts, accept; if M2 accept, reject.”
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•UNDECIDABLE PROBLEMS FROM LANGUEAGE THEORY 1. HALTTM={<M, w>|M is a TM and M halts on input w}. THEOREM 3.11 HALTTM is undecidable. PROOF IDEA METHOD: contradiction Problem A is ATM, problem B is HALTTM, if A reduces to B, then we can know: If B is decidable, then A also is decidable. To assume that we have a TM R that decides HALTTM S=“On input <M, w>, an encoding of a TM M and a string w: 1.Run TM R on input<M, w>. 2.If R rejects, reject. 3.If R accepts, simulate M on w until it halts. 4.If M has accepted, accept; if M has rejected, reject.”
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• MAPPING REDUCIBILITY
* putable function: A function f:Σ→Σ*is a computable Σ→Σ function if some Turing machine M, on every input w, halts with just f(w) on its tape. 2.Mapping reducible language: Language A is mapping reducible to language B, written A≤m B ,if there is a ≤ * computable function f: Σ→Σ* , where for every w, w∈A⇔f(w)∈B. ∈ ⇔ ∈ The function f is called the reduction of A to B. THEOREM 3.14 If A≤ mB and B is decidable, then A is decidable. ≤
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n 1 2 3 4 f(n) 3.14159… 55.55555… 0.12345… 0.50000…
…
…
x=0. 2111 …
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COROLLARY 3.1 Some language are not Turing-recognizaTuring machine is countable; 2)The set of all language is uncountable; (1)The set of all infinite binary sequences is uncountable; (2) The set of all language and the set of all infinite binary sequences are the same size.
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3. REGULARTM={<M>|M is a TM and L(M) is a regular language}. THEOREM 3.13 EREGULARTM is undecidable. 4.EQTM={<M1,M2>|M1 and M2 are TMs and L(M1) =L(M2)}. THEOREM 3.13 EQTM is undecidable.
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PROOF IDEA M for B. f is the reduction of A to B. N=“ On input w: pute f(w). 2.Run M on input f(w) and output whatever M ouputs.” COROLLARY 3.3 If A≤ mB and is A undecidable, then B is undecidable. ≤ THEOREM 3.15 If A≤ mB and B is Turing-recognizable, then A is Turing≤ recognizable. COROLLARY 3.4 If A≤ B and A is not Turing-recognizable, then B is not ≤ Turing-recognizable.
CHAPTER 3 ALGORITHEMS(2)
2 THE HAITING PROBLEM(ACCEPTANCE PROBLEM) ATM={<M, w>|M is a TM and M accepts w} THEORM 3.9 ATM is undecidable. PROOF IDEA accept if M accepts w H(<M, w>)= reject if M does not accept w D=“On input <M>, where M is a TM: 1. Run H on input <M, <M>>. 2.Output the opposite of what H output; that is, if H accepts, reject and if H rejects, accept.”