Costas Busch - RPI1 Undecidable problems for Recursively enumerable languages continued…

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

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

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

Costas Busch - RPI5 Halting problem decider YES NO halts on We will build a decider for the halting problem: doesn’t halt on

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

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 ?

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

Costas Busch - RPI9 halts on is infinite if and only if

Costas Busch - RPI10 construct YES NO YES halting problem decider finite language problem decider

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

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

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 Doesn’t contain two equal length strings

Costas Busch - RPI14 Halting problem decider YES NO halts on We will build a decider for the halting problem: doesn’t halt on

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

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 ?

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 ( )

Costas Busch - RPI18 halts on if and only if accepts two equal length strings accepts and

Costas Busch - RPI19 construct YES NO YES NO Halting problem decider Two-strings problem decider

Costas Busch - RPI20 Rice’s Theorem

Costas Busch - RPI21 Non-trivial properties of recursively enumerable languages: any property possessed by some (not all) recursively enumerable languages Definition:

Costas Busch - RPI22 Some non-trivial properties of recursively enumerable languages: is empty is finite contains two different strings of the same length

Costas Busch - RPI23 Rice’s Theorem: Any non-trivial property of a recursively enumerable language is undecidable

Costas Busch - RPI24 The Post Correspondence Problem

Costas Busch - RPI25 Some undecidable problems for context-free languages: Is context-free grammar ambiguous? Is ? are context-free grammars

Costas Busch - RPI26 We need a tool to prove that the previous problems for context-free languages are undecidable: The Post Correspondence Problem

Costas Busch - RPI27 The Post Correspondence Problem Input: Two sequences of strings

Costas Busch - RPI28 There is a Post Correspondence Solution if there is a sequence such that: PC-solution: Indeces may be repeated or ommited

Costas Busch - RPI29 Example: PC-solution:

Costas Busch - RPI30 Example: There is no solution Because total length of strings from is smaller than total length of strings from

Costas Busch - RPI31 The Modified Post Correspondence Problem Inputs: MPC-solution:

Costas Busch - RPI32 Example: MPC-solution:

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:

Costas Busch - RPI34 Theorem: The MPC problem is undecidable Proof: We will reduce the membership problem to the MPC problem

Costas Busch - RPI35 Membership problem Input: recursive language string Question: Undecidable

Costas Busch - RPI36 Membership problem Input: unrestricted grammar string Question: Undecidable

Costas Busch - RPI37 Suppose we have a decider for the MPC problem MPC solution? YES NO String Sequences MPC problem decider

Costas Busch - RPI38 We will build a decider for the membership problem YES NO Membership problem decider

Costas Busch - RPI39 MPC problem decider Membership problem decider The reduction of the membership problem to the MPC problem: yes no

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 ?

Costas Busch - RPI41 : special symbol For every symbol Grammar : start variable For every variable

Costas Busch - RPI42 Grammar For every production : special symbol string

Costas Busch - RPI43 Example: Grammar : String

Costas Busch - RPI44

Costas Busch - RPI45

Costas Busch - RPI46 Grammar :

Costas Busch - RPI47

Costas Busch - RPI48

Costas Busch - RPI49

Costas Busch - RPI50 has an MPC-solution if and only if

Costas Busch - RPI51 MPC problem decider Membership problem decider Construct yes no

Costas Busch - RPI52 Since the membership problem is undecidable, The MPC problem is uncedecidable END OF PROOF

Costas Busch - RPI53 Theorem: The PC problem is undecidable Proof: We will reduce the MPC problem to the PC problem

Costas Busch - RPI54 Suppose we have a decider for the PC problem PC solution? YES NO String Sequences PC problem decider

Costas Busch - RPI55 We will build a decider for the MPC problem MPC solution? YES NO String Sequences MPC problem decider

Costas Busch - RPI56 PC problem decider MPC problem decider The reduction of the MPC problem to the PC problem: yes no

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

Costas Busch - RPI58 : input to the MPC problem

Costas Busch - RPI59 We construct new sequences

Costas Busch - RPI60 We insert a special symbol between any two symbols

Costas Busch - RPI61

Costas Busch - RPI62 Special Cases

Costas Busch - RPI63 has a PC solution has an MPC solution if and only if

Costas Busch - RPI64 PC-solution MPC-solution

Costas Busch - RPI65 PC problem decider MPC problem decider yes no Construct

Costas Busch - RPI66 Since the MPC problem is undecidable, The PC problem is undecidable END OF PROOF

Costas Busch - RPI67 Some undecidable problems for context-free languages: Is context-free grammar ambiguous? Is ? are context-free grammars We reduce the PC problem to these problems

Costas Busch - RPI68 Theorem: Proof: Let be context-free grammars. It is undecidable to determine if Rdeduce the PC problem to this problem

Costas Busch - RPI69 Suppose we have a decider for the empty-intersection problem Empty- interection problem decider YES NO Context-free grammars

Costas Busch - RPI70 We will build a decider for the PC problem PC solution? YES NO String Sequences PC problem decider

Costas Busch - RPI71 PC problem decider The reduction of the PC problem to the empty-intersection problem: yes no Empty- interection problem decider

Costas Busch - RPI72 PC problem decider noyes no Empty- interection problem decider convert inputs ? We need to convert the input instance of one problem to the other

Costas Busch - RPI73 : input to the PC problem

Costas Busch - RPI74 Introduce new unique symbols:

Costas Busch - RPI75 Context-free grammar :

Costas Busch - RPI76 Context-free grammar :

Costas Busch - RPI77 if and only if has a PC solution

Costas Busch - RPI78 Because are unique There is a PC solution:

Costas Busch - RPI79 PC problem decider noyes no Empty- interection problem decider Construct Context-Free Grammars

Costas Busch - RPI80 Since PC is undecidable, the empty-intersection problem is undecidable END OF PROOF

Costas Busch - RPI81 For a context-free grammar, Theorem: it is undecidable to determine if G is ambiguous Proof: Reduce the PC problem to this problem

Costas Busch - RPI82 PC problem decider noyes no Ambiguous- grammar problem decider Construct Context-Free Grammar

Costas Busch - RPI83 start variable of

Costas Busch - RPI84 if and only if is ambiguous if and only if has a PC solution