# costas busch - rpi1 undecidable problems for recursively enumerable languages continued…

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- Slide 1
- Costas Busch - RPI1 Undecidable problems for Recursively enumerable languages continued
- Slide 2
- Costas Busch - RPI2 is empty? is finite? contains two different strings of the same length? Take a recursively enumerable language Decision problems: All these problems are undecidable
- Slide 3
- Costas Busch - RPI3 Theorem: For a recursively enumerable language it is undecidable to determine whether is finite Proof: We will reduce the halting problem to this problem
- Slide 4
- Costas Busch - RPI4 finite language problem decider YES NO Suppose we have a decider for the finite language problem: Let be the TM with finite not finite
- Slide 5
- Costas Busch - RPI5 Halting problem decider YES NO halts on We will build a decider for the halting problem: doesnt halt on
- Slide 6
- Costas Busch - RPI6 YES NO YES Halting problem decider finite language problem decider We want to reduce the halting problem to the finite language problem
- Slide 7
- Costas Busch - RPI7 YES NO YES Halting problem decider finite language problem decider We need to convert one problem instance to the other problem instance convert input ?
- Slide 8
- Costas Busch - RPI8 Construct machine : If enters a halt state, accept ( inifinite language) Initially, simulates on input Otherwise, reject ( finite language) On arbitrary input string
- Slide 9
- Costas Busch - RPI9 halts on is infinite if and only if
- Slide 10
- Costas Busch - RPI10 construct YES NO YES halting problem decider finite language problem decider
- Slide 11
- Costas Busch - RPI11 is empty? is finite? contains two different strings of the same length? Take a recursively enumerable language Decision problems: All these problems are undecidable
- Slide 12
- Costas Busch - RPI12 Theorem: For a recursively enumerable language it is undecidable to determine whether contains two different strings of same length Proof: We will reduce the halting problem to this problem
- Slide 13
- Costas Busch - RPI13 Two-strings problem decider YES NO Suppose we have the decider for the two-strings problem: Let be the TM with contains Doesnt contain two equal length strings
- Slide 14
- Costas Busch - RPI14 Halting problem decider YES NO halts on We will build a decider for the halting problem: doesnt halt on
- Slide 15
- Costas Busch - RPI15 YES NO YES NO Halting problem decider Two-strings problem decider We want to reduce the halting problem to the empty language problem
- Slide 16
- Costas Busch - RPI16 YES NO YES NO Halting problem decider Two-strings problem decider We need to convert one problem instance to the other problem instance convert inputs ?
- Slide 17
- Costas Busch - RPI17 Construct machine : When enters a halt state, accept if or Initially, simulate on input (two equal length strings ) On arbitrary input string Otherwise, reject ( )
- Slide 18
- Costas Busch - RPI18 halts on if and only if accepts two equal length strings accepts and
- Slide 19
- Costas Busch - RPI19 construct YES NO YES NO Halting problem decider Two-strings problem decider
- Slide 20
- Costas Busch - RPI20 Rices Theorem
- Slide 21
- Costas Busch - RPI21 Non-trivial properties of recursively enumerable languages: any property possessed by some (not all) recursively enumerable languages Definition:
- Slide 22
- Costas Busch - RPI22 Some non-trivial properties of recursively enumerable languages: is empty is finite contains two different strings of the same length
- Slide 23
- Costas Busch - RPI23 Rices Theorem: Any non-trivial property of a recursively enumerable language is undecidable
- Slide 24
- Costas Busch - RPI24 The Post Correspondence Problem
- Slide 25
- Costas Busch - RPI25 Some undecidable problems for context-free languages: Is context-free grammar ambiguous? Is ? are context-free grammars
- Slide 26
- Costas Busch - RPI26 We need a tool to prove that the previous problems for context-free languages are undecidable: The Post Correspondence Problem
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- Costas Busch - RPI27 The Post Correspondence Problem Input: Two sequences of strings
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- Costas Busch - RPI28 There is a Post Correspondence Solution if there is a sequence such that: PC-solution: Indeces may be repeated or ommited
- Slide 29
- Costas Busch - RPI29 Example: PC-solution:
- Slide 30
- Costas Busch - RPI30 Example: There is no solution Because total length of strings from is smaller than total length of strings from
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- Costas Busch - RPI31 The Modified Post Correspondence Problem Inputs: MPC-solution:
- Slide 32
- Costas Busch - RPI32 Example: MPC-solution:
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- Costas Busch - RPI33 1. The MPC problem is undecidable 2. The PC problem is undecidable (by reducing MPC to PC) (by reducing the membership to MPC) We will show:
- Slide 34
- Costas Busch - RPI34 Theorem: The MPC problem is undecidable Proof: We will reduce the membership problem to the MPC problem
- Slide 35
- Costas Busch - RPI35 Membership problem Input: recursive language string Question: Undecidable
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- Costas Busch - RPI36 Membership problem Input: unrestricted grammar string Question: Undecidable
- Slide 37
- Costas Busch - RPI37 Suppose we have a decider for the MPC problem MPC solution? YES NO String Sequences MPC problem decider
- Slide 38
- Costas Busch - RPI38 We will build a decider for the membership problem YES NO Membership problem decider
- Slide 39
- Costas Busch - RPI39 MPC problem decider Membership problem decider The reduction of the membership problem to the MPC problem: yes no
- Slide 40
- Costas Busch - RPI40 MPC problem decider Membership problem decider We need to convert the input instance of one problem to the other yes no convert inputs ?
- Slide 41
- Costas Busch - RPI41 : special symbol For every symbol Grammar : start variable For every variable
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- Costas Busch - RPI42 Grammar For every production : special symbol string
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- Costas Busch - RPI43 Example: Grammar : String
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- Costas Busch - RPI44
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- Costas Busch - RPI45
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- Costas Busch - RPI46 Grammar :
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- Costas Busch - RPI47
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- Costas Busch - RPI48
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- Costas Busch - RPI49
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- Costas Busch - RPI50 has an MPC-solution if and only if
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- Costas Busch - RPI51 MPC problem decider Membership problem decider Construct yes no
- Slide 52
- Costas Busch - RPI52 Since the membership problem is undecidable, The MPC problem is uncedecidable END OF PROOF
- Slide 53
- Costas Busch - RPI53 Theorem: The PC problem is undecidable Proof: We will reduce the MPC problem to the PC problem
- Slide 54
- Costas Busch - RPI54 Suppose we have a decider for the PC problem PC solution? YES NO String Sequences PC problem decider
- Slide 55
- Costas Busch - RPI55 We will build a decider for the MPC problem MPC solution? YES NO String Sequences MPC problem decider
- Slide 56
- Costas Busch - RPI56 PC problem decider MPC problem decider The reduction of the MPC problem to the PC problem: yes no
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- Costas Busch - RPI57 PC problem decider MPC problem decider yes no convert inputs ? We need to convert the input instance of one problem to the other
- Slide 58
- Costas Busch - RPI58 : input to the MPC problem
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- Costas Busch - RPI59 We construct new sequences
- Slide 60
- Costas Busch - RPI60 We insert a special symbol between any two symbols
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- Costas Busch - RPI61
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- Costas Busch - RPI62 Special Cases
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- Costas Busch - RPI63 has a PC solution has an MPC solution if and only if
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- Costas Busch - RPI64 PC-solution MPC-solution
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- Costas Busch - RPI65 PC problem decider MPC problem decider yes no Construct
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- Costas Busch - RPI66 Since the MPC problem is undecidable, The PC problem is undecidable END OF PROOF
- Slide 67
- Costas Busch - RPI67 Some undecidable problems for context-free languages: Is context-free grammar ambiguous? Is ? are context-free grammars We reduce the P