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CSC 3130: Automata theory and formal languages

Andrej Bogdanov

http://www.cse.cuhk.edu.hk/~andrejb/csc3130

NP-completeness

Fall 2008

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P and NP

languages that can be decided on a TMwith polynomial running time

P =

(problems that admit efficient algorithms)

NP = languages decidable on a nondeterministic

TM with polynomial running time

(solutions can be checked efficiently)

decidable

NP

P

We believe that NP is bigger than P,

but we are not 100% sure

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Evidence that NP is bigger than P

These (and many others) are in NP

Their solutions, once found, are easy to verify

But no efficient algorithms are known for any of

them

The fastest known algorithms take time 2n

CLIQUE = {(G, k): G is a graph with a clique ofk vertices}

IS= {(G, k): G is a graph with an independent set ofk vertices}

VC= {(G, k): G is a graph with a vertex cover ofk vertices}

SAT= {f:fis a satisfiable Boolean formula

}

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Equivalence of certain NP languages

We strongly suspect that problems like CLIQUE,SAT, etc. require time 2nto solve

We do not know how to prove this, but what we

can prove is that

If any one of them is tractable (by a polynomial-

time algorithm), then all of them are tractable

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Equivalence of certain NP languages

All these problems are as hard as one another

Moreover, they are at the frontier ofNP

They are at least as hard as any problem in NPNP

P

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Polynomial-time reductions

What do we mean when we say, for example,

We mean that

Or, we can convert an imaginary poly-timealgorithm forIS into one forCLIQUE

CLIQUE is at least as hard as IS

IfCLIQUE has a polynomial-time algorithm,

so does IS

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Polynomial-time reductions

Theorem

IfCLIQUE has a

polynomial-time

algorithm, so does IS

CLIQUE = {(G, k): G is a graph with a clique ofk vertices}

IS= {(G, k): G is a graph with an independent set ofk vertices}

1 2

3 4

{1, 4}, {2, 3, 4}, {1} are cliques

{1, 2}, {1, 3}, {4} are

independent sets

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Polynomial-time reductions

Proof: Suppose CLIQUE has an algorithmA

We want to use it to solve IS

IfCLIQUE has a polynomial-timealgorithm, so does IS

G, k

reject

if not

accept

ifG has IS of size k

A

G, k

reject

if not

accept

ifGhas clique of size k

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Reducing IS to CLIQUE

We look for a polynomial-time transformation s.t.:

RG, k G, k

IfG has a clique of size k,

then G has an IS of size k

IfG does not have a clique of size k,then G has no IS of size k

1 2

3 4

G

ISs of size k cliques of size k

1 2

3 4

Gflip all edges

set k = k

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Reducing IS to CLIQUE

RG, k G, kOn input (G, k)Construct G by flipping all edges ofG

Set k = k

Output (G, k)

independent sets in G cliques in G

IfG has a clique of size k,

then G has an IS of size k

IfG does not have a clique of size k,

then G has no IS of size k

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Reduction recap

We showed that

by converting an imaginary algorithm forCLIQUE

into one for IS

To do this, we came up with a reduction thattransforms instances ofIS into ones forCLIQUE

IfCLIQUE has a polynomial-time algorithm,

so does IS

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Polynomial-time reductions

Language Lpolynomial-time reduces to Lif

there exists a polynomial-time computable map R

that takes an instance xofLinto instanceyofL

s.t.xL if and only if yL

L L(IS) (CLIQUE)

x y(G, k) (G,k)

xL yL(G has IS of size k) (Ghas clique of size k)

R

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Meaning of poly-time reductions

Saying Lreduces to L means Lis easier than L In other words, if we can solve L, then we can also

solve L

Theorem

Proof

IfLreduces toLand LP, then LP

x y

xL yL

R algorithm forLacc

rej

algorithm

accepts

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Notes about reductions

The direction of the reduction is very important This makes sense, because A is easier than B and

B is easier than A mean different things

However, it is possible that Lreduces to Land L

reduces to L

This means that Land Lare as hard as one another

For example, IS and CLIQUE reduce to one another

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The Cook-Levin Theorem

So every problem in NP is easier than SAT

But SAT itself is in NP, so SAT must be the

hardest problem in NP:

Every LNP reduces to SAT

IfSATP, then P = NP

SAT= {f:f is a satisfiable Boolean formula}

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Interpretation of Cook-Levin Theorem

Optimistic view:

Pessimistic view:

If we manage to solve SAT, then wecan also solve

CLIQUE, scheduling, and almost anything

Since we do not believe P = NP, it is unlikely that

we will ever have a fast algorithm forSAT