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## PowerPoint Slideshow about ' Undecidable Problems From Language Theory' - clementine-ball

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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.

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.

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

- 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 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 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.

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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