# 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