The language SAT - Tel Aviv Universitybchor/CM/Compute13.pdf · Lecture 13 The language SAT Cook-Levin: SAT is NP-Complete. Outline of the proof to Cook-Levin theorem. Additional

Lecture 13The language SAT

Cook-Levin: SAT is NP-Complete.

Outline of the proof to Cook-Levin theorem.

Additional NP-Complete problemsby chains of reductions.

Slides modified by Benny Chor, based on original slides by Maurice Herlihy, Brown University. – p.1





















NP-Completeness (reminder)

A language B is NP-complete if it satisfies

B ∈ NP , and

For every A in NP, A ≤P B


3SAT (reminder)

Definition: A Boolean formula is in 3CNF form if itis a CNF formula, and all terms have three literals.

(x1 ∨ x2 ∨ x3) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6 ∨ x4)

Define

3SAT = {〈φ〉 | φ is satisfiable 3CNF formula}

Clearly, if φ is a satisfiable 3CNF formula, then forany satisfying assignment of φ, every clause mustcontain at least one literal assigned 1.


3SAT (reminder)


(x1 ∨ x2 ∨ x3) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6 ∨ x4)

Define




3SAT (reminder)


(x1 ∨ x2 ∨ x3) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6 ∨ x4)

Define




The Language SAT

Definition: A Boolean formula is in conjunctivenormal form (CNF) if it consists of terms, connectedwith ∧s.

For example(x1 ∨ x2 ∨ x3 ∨ x4) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6)

Definition:SAT = {〈φ〉 | φ is satisfiable CNF formula}


The Language SAT

Definition: A Boolean formula is in conjunctivenormal form (CNF) if it consists of terms, connectedwith ∧s.

For example(x1 ∨ x2 ∨ x3 ∨ x4) ∧ (x3 ∨ x5 ∨ x6) ∧ (x3 ∨ x6)

Definition:SAT = {〈φ〉 | φ is satisfiable CNF formula}


Strategy

Once we have one structured NP-completeproblem, we can generate more by poly-timereductions.

Getting the first one requires some work.


Strategy

Once we have one structured NP-completeproblem, we can generate more by poly-timereductions.

Getting the first one requires some work.


Cook-Levin (early 70s)

Theorem: SAT is NP complete.

Must show that every NP problem reduces toSAT in poly-time.

Proof Idea: Suppose L ∈ NP , and M is anNTM that accepts L.

On input w of length n, M runs in timet(n) = nc.

We consider the nc-by-nc tableau that describesthe computation of M on input w.






























The Tableau

cell[1,1]

cell[1,t(n)]

q 0 0 10 0 . . .1 2 3 . . . t(n)

0 00

1


The Tableau

cell[1,1]

cell[1,t(n)]

q 0 0 10 0 . . .1 2 3 . . . t(n)

0 00

1

Row 1 in tableau represents initial configurationof M on input w.

Row i in tableau represents i-th configuration in acomputation of M on input w.


The Tableau

cell[1,1]

cell[1,t(n)]

q 0 0 10 0 . . .1 2 3 . . . t(n)

0 00

1




The Tableau

cell[1,1]

cell[1,t(n)]

q 0 0 10 0 . . .1 2 3 . . . t(n)

0 00

1




A Formula Simulating the Tableau

We construct a Boolean CNF formula φw that“mimics” the tableau.

Given the string w, it takes O(n2c) steps toconstruct φw.

The following property holds:φw ∈ SAT iff M accepts w.

So the mapping w 7→ φw is a poly time reductionfrom L to SAT , establishing L ≤P SAT .

We still got a few small details to take care off...





































Details of Formula (Partial List)

We construct a Boolean CNF formula φw that“mimics” the tableau:

φw uses Boolean variables of three types.bi,j,σ is true iff the j-th cell in i-thconfiguration contains the letter σ ∈ Γ.si,q is true iff in i-th configuration, M is instate q ∈ Q.hi,j is true iff in i-th configuration M , has ishead in cell j on tape.

The formula φw consists of four parts:φw = φunique(M) ∧ φstart(w) ∧ φaccept(M) ∧

φcompute(M)




φw uses Boolean variables of three types.

bi,j,σ is true iff the j-th cell in i-thconfiguration contains the letter σ ∈ Γ.si,q is true iff in i-th configuration, M is instate q ∈ Q.hi,j is true iff in i-th configuration M , has ishead in cell j on tape.


φcompute(M)




φw uses Boolean variables of three types.bi,j,σ is true iff the j-th cell in i-thconfiguration contains the letter σ ∈ Γ.

si,q is true iff in i-th configuration, M is instate q ∈ Q.hi,j is true iff in i-th configuration M , has ishead in cell j on tape.


φcompute(M)




φw uses Boolean variables of three types.bi,j,σ is true iff the j-th cell in i-thconfiguration contains the letter σ ∈ Γ.si,q is true iff in i-th configuration, M is instate q ∈ Q.

hi,j is true iff in i-th configuration M , has ishead in cell j on tape.


φcompute(M)






φcompute(M)






φcompute(M)






φcompute(M)


Details of Formula (cont.)

φunique(M) guarantees that the variables encode

legal configurations. For example, at most one ofbi,j,0 and bi,j,1 is true.

φstart(w) guarantees that the variablescorresponding to the first row (i = 1) encode theinitial configuration of M on w.

φaccept(M) guarantees that M reached anaccepting configuration.

φcompute(M) guarantees that the configurationdescribed by the i + 1-st row is a legal successionof the configuration described by the i-th row.













































φcompute(M) is the “heart” of φw. To constructit, employ locality of computations.

To determine contents of tableau entry (i, j) (cellj in configuration i), only the contents of threetableau entries (from configuration i− 1),(i− 1, j − 1), (i− 1, j), (i− 1, j + 1), and M ’stable, are needed.

If head not in area, nothing changes. And and if itis, changes are local and determined using M .

0 00

1






0 00

1






0 00

1


The Tableau in Perspective

cell[1,1]

cell[1,t(n)]

q 0 0 10 0 . . .1 2 3 . . . t(n)

0 00

1


Correctness of ReductionAll four components of φw can be put in CNF, soφw itself (∧ of the four) is also in CNF.

The transformation w 7→ φw is computable intime O(n2c).An assignment satisfyingφunique(M) ∧ φstart(w)∧ φcompute(M)

corresponds to a valid computation of M on w.An assignment satisfying, in addition φaccept(M),corresponds to an accepting computation of Mon w.Therefore M accepts w iff φw ∈ SAT .

For complete details, consult Sipser or take theComplexity course. ♣


Correctness of ReductionAll four components of φw can be put in CNF, soφw itself (∧ of the four) is also in CNF.The transformation w 7→ φw is computable intime O(n2c).

An assignment satisfyingφunique(M) ∧ φstart(w)∧ φcompute(M)




Correctness of ReductionAll four components of φw can be put in CNF, soφw itself (∧ of the four) is also in CNF.The transformation w 7→ φw is computable intime O(n2c).An assignment satisfyingφunique(M) ∧ φstart(w)∧ φcompute(M)

corresponds to a valid computation of M on w.

An assignment satisfying, in addition φaccept(M),corresponds to an accepting computation of Mon w.Therefore M accepts w iff φw ∈ SAT .




corresponds to a valid computation of M on w.An assignment satisfying, in addition φaccept(M),corresponds to an accepting computation of Mon w.

Therefore M accepts w iff φw ∈ SAT .















Strategy

We have seen that SAT is NP-complete.

We now reduce SAT to 3SAT.

And then will reduce 3SAT to a bunch of otherproblems in NP.

In class and recitation will give in detail just afew examples.

Full list contains hundreds or thousands of knownNP-complete problems (from combinatorics,operation research, VLSI design, computationalgeometry, bioinformatics, . . . ).

NP-completeness of new and of old problems isstill established these days.


Strategy








Strategy








Strategy








Strategy








Strategy








SAT and 3SATRecallSAT = {〈φ〉 | φ is a satisfiable CNF formula}


The reduction maps CNF formulae to 3CNF ones“clause by clause”. A clause with ` literals is mappedto ` clauses, built on the original literals together with`− 1 new ones.

For example:

(x1 ∨ x2 ∨ x3 ∨ x4 ∨ x8)7−→(x1 ∨ y1) ∧ (y1 ∨ x2 ∨ y2) ∧ (y2 ∨ x3 ∨ y3) ∧(y3 ∨ x4∨y4) ∧ (y4 ∨ x8)





For example:






For example:



SAT ≤P 3SATConsider mapping φ 7→ φ3, e.g. (x1 ∨ x2 ∨ x3 ∨ x4 ∨ x8) 7−→(x1∨y1)∧ (y1∨x2∨y2)∧ (y2∨x3∨y3)∧ (y3∨x4∨y4)∧ (y4∨x8)

Claim: φ has a satisfying assignment iff φ3 does.

Proof sketch: ⇐ An assignment satisfying φ3 cannot“rely” on new literals alone – at least one originalliteral must be satisfied.⇐ An assignment satisfying φ makes at least oneliteral per clause happy. In the “φ3 clause” of thisliteral the new veriable is under no constraints. Thisenables propagation to a satisfying assignment that“relies” on new vars alone in rest of φ3 clauses.

This establishes validity of the reduction. Since it is inpolynomial time (why?), we get SAT ≤P 3SAT. ♣.









Proof sketch: ⇐ An assignment satisfying φ3 cannot“rely” on new literals alone – at least one originalliteral must be satisfied.

⇐ An assignment satisfying φ makes at least oneliteral per clause happy. In the “φ3 clause” of thisliteral the new veriable is under no constraints. Thisenables propagation to a satisfying assignment that“relies” on new vars alone in rest of φ3 clauses.













3SAT – Cousins and CambriansWe now know that SAT ≤P 3SAT. Since SAT isNP-complete and 3SAT∈ NP, this proves that 3SAT isitself NP-complete.

What about the 3SAT ≤P SAT direction?

We now want to examine what happens if we furtherreduce the number of literals per clause in CNFformulae.

Definition: A Boolean formula is in 2CNF if it is aCNF formula, and all terms have at most two literals.For example

(x1 ∨ x2) ∧ (x5 ∨ x6) ∧ (x6 ∨ x4)






(x1 ∨ x2) ∧ (x5 ∨ x6) ∧ (x6 ∨ x4)






(x1 ∨ x2) ∧ (x5 ∨ x6) ∧ (x6 ∨ x4)






(x1 ∨ x2) ∧ (x5 ∨ x6) ∧ (x6 ∨ x4)


3SAT – Cousins and CambriansDefinition:


Betting time: Is 2SAT NP-complete? Is it in P?Or maybe we do not know? . . .

Well, turns out 2SAT is in P. For details, though,you’ll have to refer to the algorithms, ahhhm,efficiency of computations, course.












Chains of Reductions: NPC Problems

SAT

IntegerProg 3SAT

Clique

IndepSet

VertexCover

SetCover

3ExactCover

Knapsack

Scheduling

3Color HamPath

HamCircuit

TSP


Hamiltonian Path (reminder)

HAMPATH = {〈G, s, t〉 | G has Hamiltonian pathfrom s to t}

Hamiltonian path in directed G visits each node once.


Hamiltonian CircuitHAMCIRCUIT = {〈G〉 | G has Hamiltonian circuit}


Hamiltonian CircuitHAMCIRCUIT = {〈G〉 | G has Hamiltonian circuit}


ReductionTheorem: HAMPATH is polynomial-time reducibleto HAMCIRCUIT ,

HAMPATH ≤P HAMCIRCUIT .


Traveling Salesman

10 9

36

9

a

b

c

d

5

Roger WilliamsZoo

Brown UniversityAl FornoRestaurant

StateCapitol

(not drawn to scale)

Parameters:

set of cities C

set of inter-city distances D

goal k


Traveling Salesman

Define TSP = {〈C,D, k〉 | (C,D) has a TS tourof total distance ≤ k}

Remark: Can consider two versions –undirected and directed.

RecallHAMCIRCUIT = {〈G〉 | G has Hamiltonian circuit}

Theorem: HAMCIRCUIT is polynomial-timereducible to TSP,

HAMCIRCUIT ≤P TSP .


Traveling Salesman







Traveling Salesman







HAMCIRCUIT≤P TSP

The reduction: Given a directed graph G = (V,E) weconstruct a directed traveling salesman instance.

The cities are identical to the nodes of theoriginal graph, C = V .

The distance of going from v1 to v2 is 1 if(v1, v2) ∈ E, and 2 otherwise.

The bound on the total distance of a tour isk = |V |.


HAMCIRCUIT≤P TSP






HAMCIRCUIT≤P TSP






HAMCIRCUIT≤P TSPWe analyse the reduction

Validity

=⇒ Suppose G has a Hamiltonian circuit.The distance assigned by the reduction to alledges in this circuit is 1. Thus in (C,D) thereis a traveling salesman tour of total distance|V | = k, namely (C,D, k) ∈ TSP.⇐= Suppose (C,D) has a traveling salesmantour of total distance |V | = k. Tour cannotcontain any edge of distance 2. Therefore itgives a Hamiltonian circuit in G.

Efficiency: Reduction in quadratic time (filling updistances for all edges of the complete graph). ♣



Validity=⇒ Suppose G has a Hamiltonian circuit.The distance assigned by the reduction to alledges in this circuit is 1. Thus in (C,D) thereis a traveling salesman tour of total distance|V | = k, namely (C,D, k) ∈ TSP.

⇐= Suppose (C,D) has a traveling salesmantour of total distance |V | = k. Tour cannotcontain any edge of distance 2. Therefore itgives a Hamiltonian circuit in G.




Validity=⇒ Suppose G has a Hamiltonian circuit.The distance assigned by the reduction to alledges in this circuit is 1. Thus in (C,D) thereis a traveling salesman tour of total distance|V | = k, namely (C,D, k) ∈ TSP.⇐= Suppose (C,D) has a traveling salesmantour of total distance |V | = k. Tour cannotcontain any edge of distance 2. Therefore itgives a Hamiltonian circuit in G.




Validity=⇒ Suppose G has a Hamiltonian circuit.The distance assigned by the reduction to alledges in this circuit is 1. Thus in (C,D) thereis a traveling salesman tour of total distance|V | = k, namely (C,D, k) ∈ TSP.⇐= Suppose (C,D) has a traveling salesmantour of total distance |V | = k. Tour cannotcontain any edge of distance 2. Therefore itgives a Hamiltonian circuit in G.



Hamiltonian PathTheorem: HAMPATH is NP-complete.

Must demonstrate

HAMPATH ∈ NP

every A ∈ NP poly-time reducible toHAMPATH.

Note thatwe have already seen the firstwill now show that 3SAT ≤P HAMPATH



Must demonstrate

HAMPATH ∈ NP





Must demonstrate

HAMPATH ∈ NP


Note that

we have already seen the firstwill now show that 3SAT ≤P HAMPATH



Must demonstrate

HAMPATH ∈ NP


Note thatwe have already seen the first

will now show that 3SAT ≤P HAMPATH



Must demonstrate

HAMPATH ∈ NP




Hamiltonian PathFor any 3CNF formula φ,

we construct a graph G

with vertices s and t

such that φ is satisfiable iff there is a Hamiltonianpath from s to t.












Hamiltonian PathHere is a 3CNF formula φ:

(a1 ∨ b1 ∨ c1) ∧ (a2 ∨ b2 ∨ c2) ∧ · · · (ak ∨ bk ∨ ck)∧

where

each ai, bi, ci is xi or xi

the ` clauses are C1, . . . , C`,

the k variables are x1, . . . , xk.


Hamiltonian PathHere is a 3CNF formula φ:

(a1 ∨ b1 ∨ c1) ∧ (a2 ∨ b2 ∨ c2) ∧ · · · (ak ∨ bk ∨ ck)∧

where

each ai, bi, ci is xi or xi

the ` clauses are C1, . . . , C`,

the k variables are x1, . . . , xk.


Chains of Reductions: NPC Problems

SAT

IntegerProg 3SAT

Clique

IndepSet

VertexCover

SetCover

3ExactCover

Knapsack

Scheduling

3Color HamPath

HamCircuit

TSP


On Search, Decision, and Optimization

Let R(·, ·) be a poly time computable predicate.

Decision Problem: Given input x, decide if thereis some y satisfying R(x, y)?

Using the “certificate” characterization oflanguages in NP, the decision problem is the sameas deciding membership x ∈ L for L ∈ NP .

Search Problem: Given input x, find some ysatisfying R(x, y), or declare that none exist.

The search problem seems harder to solve thanthe decision problem.


























Turns out that for NP complete languages, searchand decision have the same difficulty.

Specifically, given access to an oracle for L (thedecision problem), we can solve the searchproblem in poly time.

Examples: SAT and Clique (on board).






























Documents

The language SAT - Tel Aviv Universitybchor/CM/Compute13.pdf · Lecture 13 The language SAT Cook-Levin: SAT is NP-Complete. Outline of the proof to Cook-Levin theorem. Additional