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Undecidable Problems From Language Theory. Lecture 33 Section 5.1 Wed, Nov 7, 2007. Reducibility. A problem A is reducible to a problem B if a solution to B gives us a solution to A . More specifically, an instance of problem A can be restated as an instance of problem B .

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Undecidable problems from language theory

Undecidable Problems From Language Theory

Lecture 33

Section 5.1

Wed, Nov 7, 2007


Reducibility
Reducibility

  • A problem A is reducible to a problem B if a solution to B gives us a solution to A.

  • More specifically, an instance of problem A can be restated as an instance of problem B.

  • We will have a precise definition of reducibility later.


Simple examples
Simple Examples

  • Sorting a list of numbers reduces to the pair of problems

    • Find the smallest number in a list.

    • Swapping two numbers in a list.


Simple examples1
Simple Examples

  • How could we reduce the problem of finding the area of the hexagon to simpler problems?

6

10

2

3

10

5

11

5


Simple examples2
Simple Examples

  • How could we reduce the problem of finding the area of the hexagon to simpler problems?

6

10

2

3

10

5

11

5


Simple examples3
Simple Examples

  • How could we reduce the problem of finding the area of the hexagon to simpler problems?

6

10

2

10

3

Area = 137.5

30

10

60

12.5

5

25

11

5


Simple examples4
Simple Examples

  • In a very similar way, OpenGL reduces the problem of shading polygons to the problem of shading triangles.


Reducibility and decidability
Reducibility and Decidability

  • Theorem: If A is reducible to B and B is decidable, then A is decidable.

  • Proof:

    • Let R be a Turing machine that reduces A to B.

    • Let DB be a decider of B.


Reducibility and decidability1

DA

acc

acc

IB

IA

R

DB

rej

rej

Reducibility and Decidability


Reducibility and decidability2
Reducibility and Decidability

  • Theorem: If A is reducible to B and A is undecidable, then B is undecidable.

  • Proof:

    • This is the contrapositive of the previous theorem.


The halting problem for turing machines
The Halting Problem for Turing Machines

  • Define the language HALTTM to be

    HALTTM = {M, w | M halts on w}.


The halting problem for turing machines1
The Halting Problem for Turing Machines

  • Theorem: HALTTM is undecidable.

  • Proof:

    • We will show that ATM is reducible to HALTTM.

    • Suppose that HALTTM is decidable.

    • Let DH be a decider for HALTTM.

    • We build a decider DA for ATM.


The halting problem for turing machines2

acc

acc

DA

acc

M, w

U

M, w

DH

rej

rej

rej

The Halting Problem for Turing Machines


The halting problem for turing machines3
The Halting Problem for Turing Machines

  • Thus, we could build a decider for ATM, which we know to be impossible.

  • Therefore, HALTTM is undecidable.


The emptiness problem for turing machines
The Emptiness Problem for Turing Machines

  • Define the language ETM to be

    ETM = {M | L(M ) = }.


The emptiness problem for turing machines1
The Emptiness Problem for Turing Machines

  • Theorem: ETM is undecidable.

  • Proof:

    • Suppose ETM is decidable.

    • Let DE be a decider for ETM.

    • Let COMP be a Turing machine that compares two strings.

    • Given M, w, build the Turing machine Mw as follows.


The emptiness problem for turing machines2

Mw

acc

acc

x = w

M, w

U

x

COMP

rej

rej

xw

The Emptiness Problem for Turing Machines


The emptiness problem for turing machines3
The Emptiness Problem for Turing Machines

  • What is the language of Mw?

  • Let CONVERT be a Turing Machine that will read the M, w pair and construct the Turing machine Mw.


The emptiness problem for turing machines4

DA

acc

rej

M, w

Mw

CONVERT

DE

acc

rej

The Emptiness Problem for Turing Machines


The emptiness problem for turing machines5
The Emptiness Problem for Turing Machines

  • Thus, we could build a decider for ATM, which we know to be impossible.

  • Therefore, ETM is undecidable.


The equivalence problem for turing machines
The Equivalence Problem for Turing Machines

  • Define the language EQTM to be

    EQTM = {A, B | L(A ) = L(B)}.


The equivalence problem for turing machines1
The Equivalence Problem for Turing Machines

  • Theorem: EQTM is undecidable.

  • Proof:

    • Suppose EQTM is decidable.

    • Let DEQ be a decider for EQTM.

    • Let M be a Turing machine that accepts no input.


The emptiness problem for turing machines6

DE

acc

acc

M

M, M

DEQ

rej

rej

The Emptiness Problem for Turing Machines


The emptiness problem for turing machines7
The Emptiness Problem for Turing Machines

  • What is the language of DE?

  • Thus, we could build a decider for ETM, which we know to be impossible.

  • Therefore, EQTM is undecidable.


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