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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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