Undecidable problems for Recursively enumerable languages

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Undecidable problemsfor Recursively enumerable
languages

• continued

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Take a recursively enumerable language
Decision problems

• is empty?

• is finite?

• contains two different strings
• of the same length?

All these problems are undecidable
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Theorem
For a recursively enumerable language
it is undecidable to determine whether is
finite
Proof
We will reduce the halting problem to this problem
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Let be the TM with
Suppose we have a decider for the finite
language problem
YES
finite
finite language problem decider
NO
not finite
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We will build a decider for the halting problem
YES
halts on
Halting problem decider
doesnt halt on
NO
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We want to reduce the halting problem to the
finite language problem
Halting problem decider
NO
YES
finite language problem decider
YES
NO
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We need to convert one problem instance to the
other problem instance
Halting problem decider
NO
YES
convert input ?
finite language problem decider
YES
NO
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Construct machine
On arbitrary input string
Initially, simulates on input
If enters a halt state, accept (
inifinite language)
Otherwise, reject ( finite language)
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halts on
if and only if
is infinite
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halting problem decider
NO
YES
finite language problem decider
construct
YES
NO
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Take a recursively enumerable language
Decision problems

• is empty?

• is finite?

• contains two different strings
• of the same length?

All these problems are undecidable
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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
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Let be the TM with
Suppose we have the decider for the two-strings
problem
YES
contains
Two-strings problem decider
Doesnt contain
NO
two equal length strings
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We will build a decider for the halting problem
YES
halts on
Halting problem decider
doesnt halt on
NO
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We want to reduce the halting problem to the
empty language problem
Halting problem decider
YES
YES
Two-strings problem decider
NO
NO
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We need to convert one problem instance to the
other problem instance
Halting problem decider
YES
YES
Two-strings problem decider
convert inputs ?
NO
NO
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Construct machine
On arbitrary input string
Initially, simulate on input
When enters a halt state, accept if
or
(two equal length strings
)
Otherwise, reject (
)
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halts on
if and only if
accepts two equal length strings
accepts and
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Halting problem decider
YES
YES
Two-strings problem decider
construct
NO
NO
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Rices Theorem

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Definition
Non-trivial properties of recursively enumerable
languages
any property possessed by some (not
all) recursively enumerable languages
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Some non-trivial properties of recursively
enumerable languages

• is empty

• is finite

• contains two different strings
• of the same length

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Rices Theorem
Any non-trivial property of a recursively
enumerable language is undecidable
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The Post Correspondence Problem

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Some undecidable problems for context-free
languages

• Is ?

are context-free grammars

• Is context-free grammar ambiguous?

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We need a tool to prove that the
previous problems for context-free languages are
undecidable
The Post Correspondence Problem
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The Post Correspondence Problem
Input
Two sequences of strings
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There is a Post Correspondence Solution if there
is a sequence such that
PC-solution
Indeces may be repeated or ommited
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Example
PC-solution
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Example
There is no solution
Because total length of strings from is smaller
than total length of strings from
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The Modified Post Correspondence Problem
Inputs
MPC-solution
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Example
MPC-solution
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We will show
1. The MPC problem is undecidable
(by reducing the membership to MPC)
2. The PC problem is undecidable
(by reducing MPC to PC)
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Theorem The MPC problem is undecidable
Proof We will reduce the membership
problem to the MPC problem
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Membership problem
Input recursive language string
Question
Undecidable
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Membership problem
Input unrestricted grammar string
Question
Undecidable
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Suppose we have a decider for the MPC problem
MPC solution?
String Sequences
YES
MPC problem decider
NO
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We will build a decider for the membership
problem
YES
Membership problem decider
NO
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The reduction of the membership problem to the
MPC problem
Membership problem decider
yes
yes
MPC problem decider
no
no
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We need to convert the input instance of one
problem to the other
Membership problem decider
yes
yes
convert inputs ?
MPC problem decider
no
no
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Grammar
start variable
special symbol
For every symbol
For every variable
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Grammar
string
special symbol
For every production
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Example
Grammar
String
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(No Transcript)
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(No Transcript)
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Grammar
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(No Transcript)
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(No Transcript)
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(No Transcript)
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has an MPC-solution
if and only if
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Membership problem decider
yes
yes
Construct
MPC problem decider
no
no
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Since the membership problem is undecidable, The
MPC problem is uncedecidable
END OF PROOF
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Theorem The PC problem is undecidable
Proof We will reduce the MPC problem
to the PC problem
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Suppose we have a decider for the PC problem
PC solution?
String Sequences
YES
PC problem decider
NO
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We will build a decider for the MPC problem
MPC solution?
String Sequences
YES
MPC problem decider
NO
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The reduction of the MPC problem to the PC
problem
MPC problem decider
yes
yes
PC problem decider
no
no
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We need to convert the input instance of one
problem to the other
MPC problem decider
yes
yes
convert inputs ?
PC problem decider
no
no
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input to the MPC problem
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We construct new sequences
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We insert a special symbol between any two
symbols
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(No Transcript)
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Special Cases
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has a PC solution
if and only if
has an MPC solution
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PC-solution
MPC-solution
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MPC problem decider
yes
yes
Construct
PC problem decider
no
no
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Since the MPC problem is undecidable, The PC
problem is undecidable
END OF PROOF
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Some undecidable problems for context-free
languages

• Is ?

are context-free grammars

• Is context-free grammar
• ambiguous?

We reduce the PC problem to these problems
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Theorem
Let be context-free grammars. It
is undecidable to determine if
Rdeduce the PC problem to this problem
Proof
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Suppose we have a decider for the empty-intersecti
on problem
Context-free grammars
Empty- interection problem decider
YES
NO
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We will build a decider for the PC problem
PC solution?
String Sequences
YES
PC problem decider
NO
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The reduction of the PC problem to the
empty-intersection problem
PC problem decider
yes
yes
Empty- interection problem decider
no
no
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We need to convert the input instance of one
problem to the other
PC problem decider
no
yes
Empty- interection problem decider
convert inputs ?
no
yes
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input to the PC problem
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Introduce new unique symbols
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Context-free grammar
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Context-free grammar
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has a PC solution
if and only if
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Because are unique
There is a PC solution
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PC problem decider
no
yes
Empty- interection problem decider
Construct Context-Free Grammars
no
yes
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Since PC is undecidable, the empty-intersection
problem is undecidable
END OF PROOF
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For a context-free grammar ,
Theorem
it is undecidable to determine if G is ambiguous
Reduce the PC problem to this problem
Proof
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PC problem decider
no
yes
Ambiguous- grammar problem decider
Construct Context-Free Grammar
no
yes
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start variable of
start variable of
start variable of
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has a PC solution
if and only if
if and only if
is ambiguous