Undecidable Problems From Language Theory

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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

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.
• We will have a precise definition of reducibility later.
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 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 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 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 Examples
• In a very similar way, OpenGL reduces the problem of shading polygons to the problem of shading triangles.
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.

DA

acc

acc

IB

IA

R

DB

rej

rej

Reducibility and Decidability
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
• Define the language HALTTM to be

HALTTM = {M, w | M halts on w}.

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.

acc

acc

DA

acc

M, w

U

M, w

DH

rej

rej

rej

The Halting Problem for Turing Machines
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
• Define the language ETM to be

ETM = {M | L(M ) = }.

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.

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 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.

DA

acc

rej

M, w

Mw

CONVERT

DE

acc

rej

The Emptiness Problem for Turing Machines
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
• Define the language EQTM to be

EQTM = {A, B | L(A ) = L(B)}.

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.

DE

acc

acc

M

M, M

DEQ

rej

rej

The Emptiness Problem for Turing Machines
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.