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CS4026Formal Models of Computation

Part IIIComputability & Complexity

Part III-A – Computability Theory

formal models of computation 2

Reducibility: Introduction• Chapter 5 of Sipser’s Book

• Overview:– 2 Examples of proofs using reduction

– Mapping Reducibility


Reducibility: Introduction• Reducibility: method to prove that problems are

computationally unsolvable

• Reduction: convert problem A into problem B– so that B’s solution applies to A OR

– so that A’s lack of solution applies to B

• Intuitive example: travelling from ABZ to New York– Reduces to: buying a ticket WHICH

– Reduces to: getting money to buy ticket WHICH

– Reduces to: getting a Summer job WHICH

– Reduces to: …

• Reducibility in maths:– Measuring the area of a rectangle reduces to measuring

its height and width


Reducibility: Introduction (Cont’d)• Reducibility helps us classifying problems

• When A is reducible to B– Solving A cannot be harder than solving B, because a

solution to B gives a solution to A

• In computability theory:– If A is reducible to B and B is decidable, A is also

decidable

– If A is undecidable and reducible to B, B is also undecidable

• Method to prove that a problem B is undecidable:– Show that some other problem A already known to be

undecidable reduces to B

– “If B was decidable then A would also be decidable, and we know that that’s not the case”


Problem No. 1• Problem: determine if a Turing machine halts

(accepting or rejecting) on a given input

• Let

HALTTM = {⟨M,w ⟩|M is a TM and M halts on input w }

And remember

ATM = {⟨M,w ⟩|M is a TM that accepts input stringw }

• HALTTM is the real halting problem– ATM is called the acceptance problem

• Theorem: HALTTM is undecidable

• Proof idea:– Reduce ATM to HALTTM then

– Use undecidability of ATM to prove undecidability of HALTTM


Problem No. 1 (Cont’d)Proof:• Assume (to obtain contradiction) that TM R decides

HALTTM

• We can then construct TM S to decide ATM:

S = “On input ⟨M,w ⟩ where M is a TM and w is a string:

1. Run TM R on input ⟨M,w ⟩.

2. If R rejects, reject. {<M,w> does not halt}

3. If R accepts {<M,w> halts}

then simulate M on w until it halts.

4. If M accepts w, accept;

If M rejects w, reject.”

• If R decides HALTTM then S decides ATM– Because ATM is undecidable, HALTTM must also be

undecidable!!


Problem No. 1 (Cont’d)

ATM consists of two problems: For given <M,w>,

1. Decide whether <M,w> halts.Given R, we can do this!

2. If yes then determine whether M accepts w. (We know a TM exists that recognises ATM , so we can do this too, by constructing a suitable TM.)

3. If no then reject.


S⟨M,w ⟩Yes

No

M accepts w

M rejects or loops on w

S

⟨M,w ⟩

Yes M accepts w

NoM rejects or loops on w

Problem No. 1 (Cont’d)Diagrammatically:

R⟨M,w ⟩Yes

No

M halts on w

M loops on w

R⟨M,w ⟩Yes

No

M accepts w

M rejects w

simulateM on w


Problem No. 1 (Cont’d)In summary:

• ATM reduces to HALTTM

– A TM S would require a TM R if it could be built

– The added functionality is straightforward

• Since ATM is undecidable, HALTTM is also undecidable

S

⟨M,w ⟩

Yes M accepts w


R⟨M,w ⟩Yes

No

M accepts w

M rejects w

simulateM on w


THE REST OF THIS LECTURE IS OPTIONAL


Problem No. 2• Problem: determine if a Turing machine does not

accept any input, that is, its language is empty

• Let

ETM = {⟨M ⟩|M is a TM and L(M ) = ∅ }

• Theorem: ETM is undecidable

• Proof idea: same as before– Assume that ETM is decidable and

– Show that ATM is decidable – a contradiction

• Let R be a TM that decides ETM

• We use R to build S that decides ATM


Problem No. 2 (Cont’d)• However, we run R on a modification of ⟨M ⟩:

– We modify ⟨M ⟩ to ensure M rejects all strings except w

– On input w, M works as usual

– In our notation, the modified machine is

– The only string M1 may now accept is w

– M1 language is non-empty if, and only if, it accepts w

M1 = “On input x :

1. If x ≠ w, reject.

2. If x =w, run M on input w and accept if M does.”


Problem No. 2 (Cont’d)Proof:

– We assume TM R decides ETM

– We then build TM S (using R ) that decides ATM:

• N.B.: S must be able to compute M1 from ⟨M,w ⟩

– Just add extra states to M to perform x =w test

• The reasoning:– If R were a decider for ETM, S would be a decider for ATM

– A decider for ATM cannot exist

– Hence, ETM must be undecidable

S = “On input ⟨M,w ⟩ where M is a TM and w is a string:

1. Use the description of M andw to built M1 as explained

2. Run R on input ⟨M1⟩.

3. If R accepts, reject; if R rejects, accept.”


S⟨M,w ⟩Yes

No

M accepts w

M rejects or loops on w

R⟨M ⟩No

Yes

L(M ) ≠ ∅

L(M ) = ∅

S

⟨M,w ⟩Yes M accepts w


Problem No. 2 (Cont’d)Diagrammatically:

RNo

Yes

⟨M1⟩modify

M onto M1⟨M,w ⟩


Mapping Reducibility• Let’s now formalise the notion of reducibility

– There are various alternatives

• Ours is called mapping reducibility– Also called “many-one reducibility”

• In a nutshell– If we can reduce problem A to problem B and

– We have a solution to problem B

– Then we have a solution to problem A

• Diagrammatically:

aReduction

f(a) = bb

Solver for B

output


Mapping Reducibility (Cont’d)• To reduce problem A to problem B via mapping

reducibility:– We must find a computable function to convert instances

of problem A to instances of problem B

– If we have such function (called a reduction) we can solve A with a solver for B


Computable Functions• A function f : ∑* → ∑* is a computable function if

some Turing machine M, on every input w, halts with just f (w ) on its tape

• Example:– A TM that takes input ⟨m,n⟩ and returns m +n


Definition of Mapping Reducibility• Language A is mapping reducible to language B if

there is a computable function f : ∑* → ∑* such that for every w,

w ∈A ↔ f (w )∈ B

• This is denoted as A ≤m B

• Function f is called the reduction of A to B

• Diagrammatically:

A Bf


Mapping Reducibility (Cont’d)• If problem A is mapping reducible to problem B

– If problem B has been previously solved, then

– We can obtain a solution to problem A

• Theorem:– if A ≤m B and B is decidable, then A is decidable

• Proof: – Let M be the decider for B

– Let f be the reduction from A to B

– We describe a decider N for A asN = “On input w :

1. Compute f (w ).

2. Run M on input f (w ) and output whatever M outputs.”

– Clearly, if w ∈ A then f (w ) ∈ B as f is a reduction from A to B.– Thus, M accepts f (w ) whenever w ∈ A.

– Therefore M works as desired.


Revisiting an Example• Let’s revisit our example and prove that

HALTTM = {⟨M,w ⟩|M is a TM and M halts on input w }

is undecidable.

– We shall do this using mapping reducibility

– We give a reduction f from ATM to HALTTM

– A computable function (i.e., a TM!) that takes as input ⟨M,w ⟩ and returns as output ⟨M′,w′ ⟩ where

⟨M,w ⟩∈ATM if and only if ⟨M′,w′ ⟩∈HALTTM

– The following machine F computes a reduction fS = “On input ⟨M,w ⟩:

1. Construct the following machine M′ :M′ = “On input x :

1. Run M on x2. If M accepts, accept.3. If M rejects, enter a loop.”

2. Output ⟨M′,w ⟩”


Mapping Reducibility (Cont’d)• Corollary:

– if A ≤m B and A is undecidable, then B is undecidable

• Theorem: – if A ≤m B and B is Turing-recognisable, then A is Turing-

recognisable

• Corollary: – if A ≤m B and A is not Turing-recognisable, then B is not

Turing-recognisable


The end ...

• In the course 2007-2008, this is the end of the material for the CS4026 exam

• Given more time, we would have shown that the method of reduction is also applicable to proofs concerning the complexity of a problem. A typical reasoning pattern:– “If problem A was solvable within time limitations X then

B would also be solvable within X. We know (or believe) that B is not solvable within X, therefore A is not either”


Reading List

• Introduction to the Theory of Computation. Michael Sipser. PWS Publishing Co., USA, 1997. (A 2nd Edition has recently been published). Chapter 5.

• Algorithmics: The Spirit of Computing. 3rd

Edition. David Harel with Yishai Feldman. Addison-Wesley, USA, 2004. Chapter 8.

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