# prof. busch - lsu1 reductions. prof. busch - lsu2 problem is reduced to problem if we can solve...

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• Slide 1
• Prof. Busch - LSU1 Reductions
• Slide 2
• Prof. Busch - LSU2 Problem is reduced to problem If we can solve problem then we can solve problem
• Slide 3
• Prof. Busch - LSU3 Language is reduced to language There is a computable function (reduction) such that: Definition:
• Slide 4
• Prof. Busch - LSU4 Computable function : which for any string computes There is a deterministic Turing machine Recall:
• Slide 5
• Prof. Busch - LSU5 If: a: Language is reduced to b: Language is decidable Then: is decidable Theorem: Proof: Basic idea: Build the decider for using the decider for
• Slide 6
• Prof. Busch - LSU6 Decider for Decider for compute accept reject accept reject (halt) Input string END OF PROOF Reduction YES NO
• Slide 7
• Prof. Busch - LSU7 Example: is reduced to:
• Slide 8
• Prof. Busch - LSU8 Turing Machine for reduction DFA We only need to construct:
• Slide 9
• Prof. Busch - LSU9 Let be the language of DFA construct DFA by combining and so that: DFA Turing Machine for reduction
• Slide 10
• Prof. Busch - LSU10
• Slide 11
• Prof. Busch - LSU11 Decider Decider for compute Input string YES NO Reduction
• Slide 12
• Prof. Busch - LSU12 If: a: Language is reduced to b: Language is undecidable Then: is undecidable Theorem (version 1): (this is the negation of the previous theorem) Proof: Using the decider for build the decider for Suppose is decidable Contradiction!
• Slide 13
• Prof. Busch - LSU13 Decider for Decider for compute accept reject accept reject (halt) Input string Reduction END OF PROOF If is decidable then we can build: CONTRADICTION! YES NO
• Slide 14
• Prof. Busch - LSU14 Observation: In order to prove that some language is undecidable we only need to reduce a known undecidable language to
• Slide 15
• Prof. Busch - LSU15 State-entry problem Input: Turing Machine State Question:Does String enter state while processing input string ? Corresponding language:
• Slide 16
• Prof. Busch - LSU16 Theorem: (state-entry problem is unsolvable) Proof: Reduce (halting problem) to (state-entry problem) is undecidable
• Slide 17
• Prof. Busch - LSU17 Decider for YES NO state-entry problem decider DeciderCompute Reduction YES NO Given the reduction, if is decidable, then is decidable A contradiction! since is undecidable Halting Problem Decider
• Slide 18
• Prof. Busch - LSU18 Compute Reduction We only need to build the reduction: So that:
• Slide 19
• Prof. Busch - LSU19 halting states special halt state Construct from : A transition for every unused tape symbol of
• Slide 20
• Prof. Busch - LSU20 halts on statehalts halting states special halt state
• Slide 21
• Prof. Busch - LSU21 halts on state on input halts on inputTherefore: Equivalently: END OF PROOF
• Slide 22
• Prof. Busch - LSU22 Blank-tape halting problem Input:Turing Machine Question:Doeshalt when started with a blank tape? Corresponding language:
• Slide 23
• Prof. Busch - LSU23 Theorem: (blank-tape halting problem is unsolvable) Proof: Reduce (halting problem) to (blank-tape problem) is undecidable
• Slide 24
• Prof. Busch - LSU24 Decider for YES NO blank-tape problem decider DeciderCompute Reduction YES NO Given the reduction, If is decidable, then is decidable A contradiction! since is undecidable Halting Problem Decider
• Slide 25
• Prof. Busch - LSU25 Compute Reduction We only need to build the reduction: So that:
• Slide 26
• Prof. Busch - LSU26 no yes Write on tape Tape is blank? Run with input Construct from : If halts then halt Accept and halt
• Slide 27
• Prof. Busch - LSU27 halts when started on blank tape halts on input no yes Write on tape Tape is blank? Run with input Accept and halt
• Slide 28
• Prof. Busch - LSU28 END OF PROOF halts when started on blank tape halts on input Equivalently:
• Slide 29
• Prof. Busch - LSU29 If: a: Language is reduced to b: Language is undecidable Then: is undecidable Theorem (version 2): Proof: Using the decider for build the decider for Suppose is decidable Contradiction! Then is decidable
• Slide 30
• Prof. Busch - LSU30 Suppose is decidable Decider for accept reject (halt)
• Slide 31
• Prof. Busch - LSU31 Suppose is decidable Decider for accept reject (halt) Then is decidable (we have proven this in previous class) reject accept (halt) Decider for NO YES NO
• Slide 32
• Prof. Busch - LSU32 Decider for Decider for compute accept reject accept reject (halt) Input string Reduction If is decidable then we can build: CONTRADICTION! YES NO
• Slide 33
• Prof. Busch - LSU33 Decider for Decider for compute accept reject accept reject (halt) Input string Reduction END OF PROOF CONTRADICTION! Alternatively: NO YES NO
• Slide 34
• Prof. Busch - LSU34 Observation: In order to prove that some language is undecidable we only need to reduce some known undecidable language to or to (theorem version 1) (theorem version 2)
• Slide 35
• Prof. Busch - LSU35 Undecidable Problems for Turing Recognizable languages is empty? is regular? has size 2? Let be a Turing-acceptable language All these are undecidable problems
• Slide 36
• Prof. Busch - LSU36 is empty? is regular? has size 2? Let be a Turing-acceptable language
• Slide 37
• Prof. Busch - LSU37 Empty language problem Input: Turing Machine Question:Isempty? Corresponding language:
• Slide 38
• Prof. Busch - LSU38 Theorem: (empty-language problem is unsolvable) is undecidable Proof: Reduce (membership problem) to (empty language problem)
• Slide 39
• Prof. Busch - LSU39 Decider for YES NO empty problem decider DeciderCompute Reduction YES NO Given the reduction, if is decidable, then is decidable membership problem decider A contradiction! since is undecidable
• Slide 40
• Prof. Busch - LSU40 Compute Reduction We only need to build the reduction: So that:
• Slide 41
• Prof. Busch - LSU41 Write on tape, and Simulate on input Tape of input string accepts ? Construct from : yes Turing Machine Accept yes
• Slide 42
• Prof. Busch - LSU42 The only possible accepted string Louisiana Prof. Busch - LSU42Prof. Busch - LSU42 Write on tape, and Simulate on input accepts ? yes Turing Machine Accept yes
• Slide 43
• Prof. Busch - LSU43 accepts does not accept Prof. Busch - LSU43Prof. Busch - LSU43Prof. Busch - LSU43 Write on tape, and Simulate on input accepts ? yes Turing Machine Accept yes
• Slide 44
• Prof. Busch - LSU44 Therefore: accepts Equivalently: END OF PROOF
• Slide 45
• Prof. Busch - LSU45 is empty? is regular? has size 2? Let be a Turing-acceptable language
• Slide 46
• Prof. Busch - LSU46 Regular language problem Input: Turing Machine Question:Isa regular language? Corresponding language:
• Slide 47
• Prof. Busch - LSU47 Theorem: (regular language problem is unsolvable) is undecidable Proof: Reduce (membership problem) to (regular language problem)
• Slide 48
• Prof. Busch - LSU48 Decider for YES NO regular problem decider DeciderCompute Reduction YES NO Given the reduction, If is decidable, then is decidable membership problem decider A contradiction! since is undecidable
• Slide 49
• Prof. Busch - LSU49 Compute Reduction We only need to build the reduction: So that:
• Slide 50
• Prof. Busch - LSU50 Tape of input string Construct from : Prof. Busch - LSU50Prof. Busch - LSU50Prof. Busch - LSU50Prof. Busch - LSU50 Write on tape, and Simulate on input Accept accepts ? yes Turing Machine
• Slide 51
• Prof. Busch - LSU51 accepts does not accept not regular regular Prof. Busch - LSU51Prof. Busch - LSU51Prof. Busch - LSU51Prof. Busch - LSU51Prof. Busch - LSU51 Write on tape, and Simulate on input Accept accepts ? yes Turing Machine
• Slide 52
• Prof. Busch - LSU52 Therefore: accepts Equivalently: END OF PROOF is not regular
• Slide 53
• Prof. Busch - LSU53 is empty? is regular? has size 2? Let be a Turing-acceptable language
• Slide 54
• Prof. Busch - LSU54 Does have size 2 (two strings)? Size2 language problem Input: Turing Machine Question: Corresponding language:
• Slide 55
• Prof. Busch - LSU55 Theorem: (size2 language problem is unsolvable) is undecidable Proof: Reduce (membership problem) to (size 2 language problem)
• Slide 56
• Prof. Busch - LSU56 Decider for YES NO size2 problem decider DeciderCompute Reduction YES NO Given the reduction, If is decidable, then is decidable membership problem decider A contradiction! since is undecidable
• Slide

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