Theory of Computer Science E3. Proving NP-Completeness Gabriele R¨oger

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Theory of Computer Science

E3. Proving NP-Completeness

Gabriele R¨oger

University of Basel

May 6, 2019

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Theory of Computer Science

May 6, 2019 — E3. Proving NP-Completeness

E3.1 Overview

E3.2 Cook-Levin Theorem E3.3 3SAT

E3.4 Summary

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Course Overview

Theory

Background Logic Automata Theory Turing Computability

Complexity

Nondeterminism P, NP

Polynomial Reductions Cook-Levin Theorem NP-complete Problems More Computability

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E3. Proving NP-Completeness Overview

E3.1 Overview

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Proving NP-Completeness by Reduction

I Suppose we know one NP-complete problem

(we will use satisfiability of propositional logic formulas).

I With its help, we can then prove quite easily that further problems are NP-complete.

Theorem (Proving NP-Completeness by Reduction) Let A and B be problems such that:

I A is NP-hard, and

I A≤_pB.

Then B is also NP-hard.

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Proving NP-Completeness by Reduction: Proof

Proof.

First part: We must show X ≤_pB for all X ∈NP.

FromX ≤_pA(becauseA is NP-hard) andA≤_pB

(by prerequisite), this follows due to the transitivity of≤_p. Second part: follows directly by definition of NP-completeness.

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NP-Complete Problems

I There are thousands of known NP-complete problems.

I An extensive catalog of NP-complete problems from many areas of computer science is contained in:

Michael R. Garey and David S. Johnson:

Computers and Intractability —

A Guide to the Theory of NP-Completeness W. H. Freeman, 1979.

I In the remaining chapters, we get to know some of these problems.

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Overview of the Reductions

SAT

3SAT

Clique

IndSet

VertexCover

DirHamiltonCycle

HamiltonCycle

TSP

SubsetSum

Partition

BinPacking

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What Do We Have to Do?

I We want to show the NP-completeness of these 11 problems.

I We first show that SATis NP-complete.

I Then it is sufficient to show

I thatpolynomial reductionsexist for all edges in the figure (and thus all problems are NP-hard)

I and that the problems are all in NP.

(It would be sufficient to show membership in NP only for the leaves in the figure. But membership is so easy to show that this would not save any work.)

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E3. Proving NP-Completeness Cook-Levin Theorem

E3.2 Cook-Levin Theorem

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Course Overview

Theory

Complexity

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SAT is NP-complete

Definition (SAT)

The problemSAT(satisfiability) is defined as follows:

Given: a propositional logic formulaϕ Question: Isϕsatisfiable?

Theorem (Cook, 1971; Levin, 1973) SATis NP-complete.

Proof.

SAT∈NP: guess and check.

SATis NP-hard: somewhat more complicated (to be continued) . . .

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NP-hardness of SAT (1)

Proof (continued).

We must show: A≤_pSATfor all A∈NP.

LetAbe an arbitrary problem in NP.

We have to find a polynomial reduction ofAto SAT, i. e., a functionf computable in polynomial time

such that for every input wordw over the alphabet of A:

w ∈A iff f(w) is a satisfiable propositional formula. . . .

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NP-hardness of SAT (2)

Proof (continued).

BecauseA∈NP, there is an NTM M and a polynomial p such thatM accepts the problem Ain time p.

Idea: construct a formula that encodes the possible configurations whichM can reach in time p(|w|) on input w

and that issatisfiable if and only if

an end configuration can be reachedin this time. . . .

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NP-hardness of SAT (3)

Proof (continued).

LetM =hQ,Σ,Γ, δ,q₀,,Ei be an NTM for A,

and letp be a polynomial bounding the computation time ofM. Without loss of generality,p(n)≥n for all n.

Letw =w₁. . .w_n∈Σ^∗ be the input forM.

We number the tape positions with integers (positive and negative) such that the TM head initially is on position 1.

Observation: withinp(n) computation steps the TM head can only reach positions in the set

Pos={−p(n) + 1,−p(n) + 2, . . . ,−1,0,1, . . . ,p(n) + 1}.

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NP-hardness of SAT (4)

Proof (continued).

We can encode configurations ofM by specifying:

I what the currentstate of M is

I on which position in Pos theTM head is located

I which symbols from Γ thetape contains at positionsPos can be encoded by propositional variables

To encode a fullcomputation(rather than just one configuration), we needcopies of these variables for each computation step.

We only need to consider the computation steps Steps={0,1, . . . ,p(n)} becauseM should accept

withinp(n) steps. . . .

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NP-hardness of SAT (5)

Proof (continued).

Use the following propositional variables in formulaf(w):

I statet,q (t ∈Steps,q ∈Q)

encodes the state of the NTM in thet-th configuration

I headt,i (t ∈Steps,i ∈Pos)

encodes the head position in the t-th configuration

I tape_t,i,a (t∈Steps,i ∈Pos,a∈Γ)

encodes the tape content in thet-th configuration Constructf(w) such that every satisfying interpretation

I describes asequence of TM configurations

I that begins with the start configuration,

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NP-hardness of SAT (6)

Proof (continued).

Auxiliary formula:

oneof X := _

x∈X

x

!

∧ ¬



 _

x∈X

_

y∈X\{x}

(x∧y)





Auxiliary notation:

The symbol⊥stands for an arbitrary unsatisfiable formula

(e.g., (A∧ ¬A), whereA is an arbitrary proposition). . . .

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NP-hardness of SAT (7)

Proof (continued).

1. describe the configurations of the TM:

Valid:= ^

t∈Steps

oneof {state_t,q|q∈Q} ∧

oneof {head_t,i |i ∈Pos} ∧

^

i∈Pos

oneof {tape_t,i,a|a∈Γ}

. . .

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NP-hardness of SAT (8)

Proof (continued).

2. begin in the start configuration Init:=state_0,q₀∧head_0,1∧

n

^

i=1

tape_0,i_,w_i ∧ ^

i∈Pos\{1,...,n}

tape_0,i, . . .

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NP-hardness of SAT (9)

Proof (continued).

3. reach an accepting configuration

Accept:= _

t∈Steps

_

qe∈E

statet,qe

. . .

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NP-hardness of SAT (10)

Proof (continued).

4. follow the rules inδ:

Trans:= ^

t∈Steps



 _

qe∈E

statet,qe ∨ _

R∈δ

Rule_t,R





where. . . .

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NP-hardness of SAT (11)

Proof (continued).

4. follow the rules inδ (continued):

Rulet,hhq,ai,hq⁰,a⁰,Dii:=

statet,q∧statet+1,q⁰ ∧

^

i∈Pos

head_t,i →tape_t,i,a∧head_t+1,i+D∧tape_t+1,i,a⁰

∧

^

i∈Pos

^

a⁰⁰∈Γ

¬head_t,i∧tape_t,i,a00→tape_t+1,i_,a00

I For D, interpret L −1, N 0, R +1.

I special case: tape andhead variables with a tape index i+D

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NP-hardness of SAT (12)

Proof (continued).

Putting the pieces together:

Setf(w) :=Valid∧Init∧Accept∧Trans.

I f(w) can be constructed in time polynomial in |w|.

I w ∈A iffM acceptsw in p(|w|) steps w ∈A ifff(w) is satisfiable

w ∈A ifff(w)∈SAT A≤_pSAT

SinceA∈NP was arbitrary, this is true for every A∈NP.

HenceSATis NP-hard and thus also NP-complete.

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E3. Proving NP-Completeness 3SAT

E3.3 3SAT

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Course Overview

Theory

Complexity

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SAT ≤

_p

3SAT

SAT

3SAT

Clique

IndSet

DirHamiltonCycle

HamiltonCycle

SubsetSum

Partition

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SAT and 3SAT

Definition (Reminder: SAT)

The problemSAT(satisfiability) is defined as follows:

Given: a propositional logic formulaϕ Question: Isϕsatisfiable?

Definition (3SAT)

The problem3SATis defined as follows:

Given: a propositional logic formulaϕ in conjunctive normal form with at most three literals per clause

Question: Isϕsatisfiable?

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3SAT is NP-Complete (1)

Theorem (3SATis NP-Complete) 3SATis NP-complete.

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3SAT is NP-Complete (2)

Proof.

3SAT∈NP: guess and check.

3SATis NP-hard: We showSAT≤_p3SAT.

I Let ϕbe the given input for SAT. LetSub(ϕ) denote the set of subformulas of ϕ, includingϕitself.

I For all ψ∈Sub(ϕ), we introduce a new proposition X_ψ.

I For each new proposition Xψ, define the following auxiliary formula χψ:

I Ifψ=Afor an atomA: χ_ψ =(X_ψ↔A)

I Ifψ=¬ψ⁰: χ_ψ =(X_ψ↔ ¬Xψ⁰)

I Ifψ= (ψ⁰∧ψ⁰⁰): χ_ψ=(X_ψ ↔(X_ψ⁰∧X_ψ⁰⁰))

I Ifψ= (ψ⁰∨ψ⁰⁰): χ_ψ=(X_ψ ↔(X_ψ⁰∨X_ψ⁰⁰))

. . .

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3SAT is NP-Complete (3)

Proof (continued).

I Consider the conjunction of all these auxiliary formulas, χ_all :=V

ψ∈Sub(ϕ)χ_ψ.

I Every interpretation I of the original variables can be extended to a model I⁰ ofχ_all in exactly one way:

for eachψ∈Sub(ϕ), setI⁰(Xψ) = 1 iff I |=ψ.

I It follows thatϕ is satisfiable iff (χ_all∧X_ϕ) is satisfiable.

I This formula can be computed in linear time.

I It can also be converted to 3-CNF in linear time because it is the conjunction of constant-size parts involving at most three variables each.

(Each part can be converted to 3-CNF independently.)

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Restricted 3SAT

Note: 3SAT remains NP-complete if we also require that

I every clause containsexactly three literals and

I a clause may not contain the same literal twice Idea:

I remove duplicated literals from each clause.

I add new variables: X,Y,Z

I add new clauses: (X∨Y ∨Z), (X∨Y ∨ ¬Z), (X ∨ ¬Y ∨Z), (¬X∨Y ∨Z), (X ∨ ¬Y ∨ ¬Z), (¬X ∨Y ∨ ¬Z),

(¬X∨ ¬Y ∨Z)

satisfied if and only if X,Y,Z are all true

I fill up clauses with fewer than three literals with ¬X and if necessary additionally with¬Y

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E3. Proving NP-Completeness Summary

E3.4 Summary

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E3. Proving NP-Completeness Summary

Summary

I Thousands of important problems are NP-complete.

I The satisfiability problem of propositional logic (SAT) is NP-complete.

I Proof idea forNP-hardness:

I Every problem in NP can be solved by an NTM in polynomial timep(|w|) for inputw.

I Given a wordw, construct a propositional logic formulaϕ that encodes the computation steps of the NTM on inputw.

I Constructϕso that it is satisfiable if and only if there is an accepting computation of lengthp(|w|).

I Usually (as seen for3SAT), the easiest way to show that another problem is NP-complete is to

I show that it is in NP with a guess-and-check algorithm, and

I polynomially reduce a known NP-complete to it.