HARDNESS OFHARDNESS OFAPPROXIMATIONSAPPROXIMATIONS

Gap Introducing ReductionGap Introducing ReductionFor simplicity we assume that we are always reducing from

SAT(or any other NP- hard problem).

Let Π be a minimization problem.

A gap introducing reduction from SAT to Π comes with two parameters f and α.

• f is a function of the instance.

• α is a function of the size of the instance.

)()()( esatisfiablnot is

)()( esatisfiabl is

xfxxOPT

xfxOPT

Given an instance Φ of SAT it outputs, in polynomial time, an instance x of Π such that:

Gap α(|x|) represents the hardness factor established by the gap-introducing reduction for the NP-hard optimization problem.

Gap Preserving ReductionGap Preserving Reduction

)()()()( 21 yfyOPTxfxOPT

Once we have obtained a gap-introducing reduction from SAT to an optimization problem, say Π1, a hardness result for another optimization problem, say Π2 can be proved by a gap preserving reduction from Π1 to Π2.

Composed reduction shows that there is no β(|y|) factor approximation algorithm for Π2 assuming

We assume

• Π1 is a minimization problem, and

• Π2 is a maximization problem.A gap-preserving reduction Γ from Π1 το Π2 comes with 4 parameters (functions), f1, α, f2, β.

Given an instance x of Π1, it computes in polynomial time, an instance y of Π2 such that•

• )()()()()()( 21 yfyyOPTxfxxOPT

NPP

Since Π1 is a minimization problem

Since Π2 is a maximization problem

1)( x

1)( x

Remarks on reductionsRemarks on reductions• The “gap” β can, in general, be bigger or smaller than α. In

this sense, “gap-preserving” is a slight misnomer.

• An approximation algorithm for Π2 together with a gap reduction Γ from Π1 to Π2 does not necessarily yield an approximation algorithm for Π1.

• Gap preserving reduction Γ together with an appropriate gap-introducing reduction from SAT to Π1 does suffice for proving a hardness of approximation for results for Π2.

Probabilistically Checkable Probabilistically Checkable Proof SystemsProof Systems

An input tape.

A work tape.

A tape that contains a random string.

A tape called the proof string and denoted as Π. Proof system should be thought of as an array of bits; out of which the verifier will examine a few.

A verifier is a polynomial-time probabilistic Turing Machine containing

DefinitionsDefinitions on PCP Ion PCP IA verifier is O(r(n),q(n)) – restricted if on each input size n

it

• uses at most O(r(n)) random bits for its computation, and

• queries at most O(q(n)) bits of the proof.In other words, an (r(n),q(n)) – restricted verifier has two

associated integers c, k

• Random string has length cr(n).

• Verifier reads the random string R, computes a sequence of kq(n) locations, and queries these locations in Π

otherwise0

bits random of string a using Π, proof the toaccess with ,input accepts 1),(

xMrxM

DefinitionsDefinitions on PCP IIon PCP IIA verifier M probabilistically checks membership proof for

language L, if

• For every input x in L, there is a proof Πx that causes M to accept for every random string, i.e., with probability 1, .

• For any input x not in L, every proof is rejected with probability at least 0.5, i.e.,

11),(Pr RxMR

2

11),(Pr RxMR

11),(Pr such that RxMLx R

2

11),(Pr , RxMLx R

The probability of accepting in case is called the error probability.

Lx

Observation: The choice of probability ½ in the second part is arbitrary. By repeating the verifier’s program O(1) times, and rejecting if the verifier rejects once, the error probability can be reduced to any arbitrary positive constant

0

)( where,))(,0(

k

knnpolynpolyPCPNP

))(),(( nqnrPCPL

Note that

A language L is in PCP(r(n),q(n)), written , if there is an (r(n),q(n)) - verifier M that probabilistically checks membership proof for L.

))(),(( nqnrPCPL

since NP is the set of languages for which membership proofs can be checked in deterministic polynomial time

PCP TheoremPCP Theorem

)1,(lognPCPNP

NPP

The PCP theorem directly gives an optimization problem in particular a maximization problem for which there is no factor ½ approximation algorithm, assuming

Gives an alternative characterization of NP in terms of PCPTheorem: Proof is divided into two parts• (easy to prove)

• (difficult to prove))1,(lognPCPNP

NPnPCP )1,(log

In terms of the 3SAT problem, the interesting and difficult part of the PCP theorem is decreasing the error probability to <1/2 (i.e., maximizing the acceptance probability of M), even though the verifier M is allowed to read only a constant number of bits.Use of PCP Theorem:

Maximizing Accept Maximizing Accept ProbabilityProbability

NPP

2

1

2

11),(Pr RMR

Maximization Problem: Let M be a PCP(logn,1) verifier for SAT. On input Φ, a SAT problem, find a proof that maximizes the probability of acceptance of V. Theorem: Assuming P is not NP, there is no factor ½ approximation algorithm for the above problem. Proof: Suppose there is a factor ½ approximation algorithm.

If Φ is satisfiable, then this algorithm must provide a proof on which M’s acceptance probability is

But the acceptance probability can be computed in polynomial time, by simply simulating M for all random strings of length O(logn).

Thus this polynomial time can be computed in polynomial time, contradicting the assumption that

Reductions Using PCP-Reductions Using PCP-theoremtheorem

Recent inapproximability results divide into four broad classes based on the approximation ratio that is provably hard to achieve

ClassClass Factor of approximation Factor of approximation that is hardthat is hard

Represenative Represenative problemsproblems

II MAX-3SATMAX-3SAT

IIII SET COVERSET COVER

IIIIII LABELCOVERLABELCOVER

IVIV CLIQUECLIQUE

1

)(lognO

n

n1log2

Inapproximability ResultsInapproximability Results

Hardness of MAX-3SATHardness of MAX-3SATMAX-3SAT is the restriction to of MAX-SAT to instances in which each clause has at most 3 literals.

In MAX-3SAT optimization problem, feasible solutions are truth assignments and objective function is the number or fraction of satisfied clauses.

MAX-3SAT plays a similar role in hardness of approximation as 3SAT plays in the theory of NP – Hardness.We will prove the next theoremTheorem: There is a constant for which there is a gap introducing reduction from SAT to MAX-3SAT that transforms a boolean formula Φ το Ψ such that

0

• If Φ is satisfiable, OPT(Ψ)=m, and

• If Φ is not satisfiable, mOPT )1()(

m denotes the number of clauses in Ψ

StrategyStrategyProof will be accomplished in two stages.

Definition of MAX k-function ProblemGiven

• n boolean variables x1,x2,…,xn

• m functions f1,f2,…,fm each of which is a function of k of the boolean variables.

Find

A truth assignment to x1,…,xn that maximizes the number of functions satisfied. Here k is assumed to be a fixed constant.

From SAT to MAX k-From SAT to MAX k-functions SATfunctions SATLemma: There is a constant k for which there is a gap-

introducing reduction from SAT to MAX k-FUNCTION SAT such that transforms a boolean formula Φ to an instance Ι of MAX k-FUNCTION SAT such that• If Φ is

satisfiable, , and

• If Φ is not satisfiable, mIOPT

2

1)(

mIOPT )(where m is the number of

formulae in IProof• Φ: instance of SAT of length n

• M: PCP(logn,1) verifier for SAT, with associated parameters c and q. Corresponding to each random string r of length c*log(n), M reads q bits of the proof. Thus M reads a total of at most bits of the proof

• B: is the set of boolean variables corresponding to each of these bits

• fr: A boolean function of q variables from B corresponding to each string r. There is a polynomial algorithm which given input Φ, outputs the m=nc functions fr.

cnc nqq log2

If Φ is satisfiable, there is a proof Π that makes M accepts with probability 1. The corresponding truth assignment to B satisfies all nc functions fr.If Φ is not satisfiable, then on every proof Π, M accepts with probability <1/2. Thus every truth assignment satisfies ½ nc functions fr.

Proof of Hardness of MAX-Proof of Hardness of MAX-3SAT3SATWe show how to obtain a 3SAT formula from the

nc functions• Ψ: Boolean formula defined by

• Ψr: fr boolean function written as a SAT formula containing at most 2q clauses. Each clause contains at most q literals.

• : 3SAT formula, obtained by using the standard trick of introducing new variables to every clause of Ψ containing more than 3 literals. The resultant formula contains at most clauses

rr

If a truth assignment satisfies formula fr, then it must satisfies all clauses of Ψr

On the other hand if it does not satisfy fr, then it must leave at least one clause of Ψr un-satisfied.

Thus if Φ is not satisfiable, any truth assignment must leave >1/2*nc clauses of Ψ unsatisfied

)2(2 qn qc

If Φ is satisfiable, then there is a truth assignment satisfying all clause of

If Φ is not satisfiable >1/2*nc remain unsatifiable, under any truth assignment.

MAX-3SAT with bounded occurrences of MAX-3SAT with bounded occurrences of variablesvariables

VS

),min(, SSSSE

Useful Notions and some NotationsExpander Graph G(V, E):

• Every vertex has the same degree

• For any nonempty subset

• Gx: A degree 14 expander graph on k vertices. Label the vertices with distinct boolean variables x1,x2,…,xk

• ψx: A CNF formula

• B: The set of boolean variables occuring in Φ

• Consistent truth assignments to x1,…,xk. All the variables are set to true or all are set to false

where

SSE , is the set of edges in the cut

SS ,

The purpose of the Expander graph is to ensure that in any optimal truth assignment, a given set of Boolean must have consistent assignment, i.e., all true or all false.

),( ji xx )( ji xx )( ij xx Corresponding to each edge of Gx we will include the clauses and

Description of reductionDescription of reduction

• W.l.o.g it is assumed that every variable occurs in Φ at least Νο times. If not we can replicate each clause No times.

• For each variable x in B which occurs times in Φ

• Each variable in of Vx occurs exactly 29 times

},,{ 1 kx xxV oNk

Bx

xVV

is a set of completely new variables.

Let Gx be a 14 degree expander on k-vertices. Label its vertices with variables from Vx and obtain formula ψx.

Replace each occurrence of variable x in Φ by a distinct variable from Vx

After the end of the above for loop every occurrence of a variable in Φ is replaced by a distinct variable from the set of new variables.

denotes the new formula after the replacement in Φ

xBx

Type I clauses: Clause of

Type II clauses: Remaining clauses of ψ

Proof (continued)Proof (continued)An inconsistent truth assignment partition the vertices of Gx into two sets S and

ijji xxxx

SCorresponding to each edge in the cut ψx will have an unsatisfied clause (see example).

Since by definition the number of unsatisfied clause in ψx is at least assuming w.l.og that S is the smallest subset

),min(, SSSSE

1S

Claim: An optimal truth assignment for ψ must satisfy all type II clauses, and therefore must be consistent for each set Vx.Proof: By contradiction.

τ is an optimal assignment for ψ that is not consistent for some Vx, with x in B. Thus τ partitions the edges of Gx into two disjoint sets.

Flip the truth assignment to variables in S, keeping the rest of the assignment the same as τ. As a result, some type I clauses that were satisfied under τ may now be unsatisfied.

Each of these must contain a variable of S so their number is at most |S|. On the other hand we get at least |S|+1 new satisfied clauses corresponding to the edges in the cut.

Consequently the flipped assignment satisfies more clauses than τ.(Contradiction)

Gap analysisGap analysis m: Number of clauses in Φ

: Number of clauses in ψ.

3m: Total number of occurences of all variables in Φ is at most 3m

Each occurrence participates in 28 type II two literal clauses giving a total cost of at most 42m clauses.

In addition, ψ has m type I clauses. Therefore m+42m=43m.

Thus

m

mm 43

If Φ is satisfiable, then by construction ψ is satisfiable, i.e .

If Φ is unsatisfiable, then i.e., clauses of Φ remain unsatisfied

mOPT )1()( m

Thus

mm

mOPT MM

431

43)(

mOPT )(

Hardness of Vertex CoverHardness of Vertex CoverInput

• An undirected graph G=(V,E)

• A cost function on vertices

Find

A minimum cost vertex cover, i.e., a subset such that every edge has at least on endpoint incident at

QVc :

VV V

Cardinality vertex cover is a special case in which all vertices are of unit cost.

VC(d): Restriction of the cardinality vertex cover problem to instances in which each vertex has degree at most d.Theorem: There is a gap-preserving reduction from MAX-3SAT(29) to VC(30) that transforms a boolean formula Φ to a graph such that•

•

where m is the number of clauses in Φ.

VGOPTmOPT 3

2)( then )( if

VGOPTOPT ub 3

21)( then m1)( if

Description of reductionDescription of reduction• W.l.o.g it is assumed that each clause of Φ has exactly three literals.

• Corresponding to each clause of Φ, graph G has 3 vertices.

• Each of these vertices is labeled with one literal of the clause. Thus

• Graph G has two types of edges

mV 3

1. For each clause of Φ, G has 3 edges connecting its vertices, and

2. For each u, v in V, if the literals u and v are negations of each other, then (u,v) is an edge in V.

By construction each vertex of G has two edges of the first type and at most 28 edges of the second type. Hence G has at most degree (28+2)=30.

Proof (continued)Proof (continued)Claim: The size of a maximum independent set in G is precisely OPT(Φ)Proof:• Consider an optimal truth assignment for clause Φ

• Pick one vertex, corresponding to a satisfied clause, from each satisfied clause. Clearly the picked vertices form an independent set.Conversely

:• Consider an independent set I in G

• Set literals corresponding to its vertices to be true. Any extension of this truth setting to all variables must satisfy at least |I| clauses in Φ.

Gap Analysis: Note that the complement of a maximum independent set in G is a minimum vertex cover.

mmmmVGOPTmOPT 23)( then )( if •

• mmm

mVGOPTOPT

bb

b

213

1)( then m1)( if

Hardness of Steiner TreeHardness of Steiner TreeInput

• An undirected graph G=(V,E)

• Nonnegative edge costs

• Vertices are partitioned into two sets, Required and Steiner

Find

A minimum cost tree in G that contains all the required vertices and any subset of the Steiner verticesTheorem: There is a gap-preserving reduction from VC(30) to

the Steiner tree problem. It transforms an instance G=(V,E) of VC(30) to an instance H(R,S,cost) of Steiner tree, where R and S are the required and steiner vertices of H, and cost is a metric on . It satisfies •

•

13

2HOPT then

3

2)( if SRVGOPT

uSSu SRHOPTGOPT

97

4 where,1

3

21)( then V

3

21)( if

SR

Description of reduction (Construction Description of reduction (Construction of graph H)of graph H)

Construct a graph H(R, S, cost) such that G(V,E) has a vertex cover of size c iff H has a Steiner tree of cost |R|+c-1.

• H will have a required vertex re corresponding to each edge e in E.

• H will have a Steiner vertex su corresponding to each vertex u in V.

• Assigned edge costs on graph H

Gin u at vertex incident not is e edge If2)s,(r

Gin u at vertex incident is e edge If1)s,(r

2Vertices RequiredBetween

1VerticesSteiner Between

Costs Edge

ue

ue

ProofProofClaim: G(V,E) has a vertex cover of size c iff H has a Steiner tree of cost |R|+c-1Proof:

• Let G has a vertex cover of size c.

• Let Sc be the set of Steiner vertices in H corresponding to the c vertices in the cover.

There is a steiner tree in H covering using cost 1 edges only, since every edge e in E must be incident at a vertex in the cover.

RSc

We will show that

• T can be transformed into a Steiner tree of the same cost that uses edged of cost 1 only. If so the latter must contain exactly c Steiner Vertices.

• Every required vertex of H must have a unit cost edge to one of these Steiner Vertices. (Therefore, the corresponding c vertices of G form a cover).

=>

The cost of the Steiner tree is .11 cRSR c

<=Let T be a Steiner tree in H of cost .1 cR

Proof (continued)Proof (continued)Let (u,v) be an edge of cost 2 in T. (W.l.og vertices u and v are both required) • Let eu be the edge in G corresponding to the required vertex

u in T.

• Let ev be the edge in G corresponding to the required vertex v in T.

Since G is connected there is a path, p, from one of the endpoints of eu to one of the endpoints of ev in GRemoving edge (u,v) from T gives two connected components• Let R1 be the set of required vertices in the first connected

component.

• Let R2 be the set of required vertices in the second connected component.

Vertices u,v lie in different sets, so path p in G must have two adjacent edges, say (a,b) and (b,c) such that their corresponding vertices in H, say w and lie in R1 and R2 respectively.

w

Proof (continued)Proof (continued)Let the Steiner vertex in H, corresponding to b be sb.

Now throwing the edges and must connect the two components . These two edges are of unit cost.

wsb , wsb ,

Gap Analysis

•

•

13

2HOPT then

3

2)( if SRVGOPT

13

21)( then V

3

21)( if SRHOPTGOPT uu

HARDNESS OFHARDNESS OFCLIQUE PROBLEMCLIQUE PROBLEM

A First Approach On Hardness of Clique A First Approach On Hardness of Clique ProblemProblemInput

• An undirected graph G=(V,E)

• Nonnegative weights on vertices. Cardinality version: all weights are equal to 1.

Find

A clique of maximum weight. A clique in G is a subset of vertices such that for each pair u, v in S, (u,v) is in E. Its weight is the sum of weights of its vertices Theorem: For fixed constants b and q, there is a gap

introducing reduction from SAT to Clique that transforms a boolean formula of size n to a graph G=(V,E), where such that •

•

bnGOPT )( e,satisfiabl is Φ if

bqnV 2

VS

bnGOPT 2

1)( e,satisfiablnot is Φ if

Preparation for the proofPreparation for the proof• F: A PCP(logn,1) verifier for SAT. F requires blogn random

bits and queries q bits of the proof

• r: Binary string of blogn bits.

• τ: truth assignment to q boolean variables.

• Q(r): The q positions in the proof that F queries when it is given string r as the random string.

• p(r): Truth setting assigned by proof p to positions Q(r)

• Ur,τ: A vertex in G for each choice of the random string, r, of blogn bits, and each truth assignment, τ, of q boolean variables

otherwiserejecting

proof theof positions Q(r) in the τreadsit when

andr string randomgiven isit when accepts F Ifaccepting

is vertex ,ru

ProofProof

• Vertices and are consistent if τ1 and τ2 agree at each position at which Q(r1) and Q(r2) overlap.

• Two distinct vertices and are connected by an edge in G iff they are consistent and they are both accepting.

11 ,ru

22 ,ru

11 ,ru

22 ,ru

• If Φ is satisfiable, there is a proof, p, on which F accepts for each choice of r, of the random string. There are 2blogn=nb possible random strings .

• If Φ is not satisfiable, suppose that C is a clique in G.

For each random string r, let p(r) be the truth setting assigned by proof p to positions Q(r). Thus, the vertices form a clique of size nb.

nbru rpr log ,

Since the vertices of C are pairwise consistent, there is a proof p such that the Q(r) positions of p contain the truth assignment τ for each vertex of clique. The number of vertices in clique is 2q|C|.

The total number of vetices in G is 2qnb.

Thus the probability of acceptance is at least bbq

q

n

C

n

C

2

2

Generalization of definitionGeneralization of definition of of PCPPCP

• If x is in L, there is a proof y that makes V accept with probability >=c.

• If x is not in L, for every proof y, V accepts with probability <s.

We introduce two parameters

• c parameter stands for completeness, and

• s parameter stands for soundness.

Question: Why the need for generalizing the definition of PCPAnswer: Error probability needs to be made inverse polynomial )(),(, nqnrPCP scDefinition

of

Definition: if there is a verifier V, which on input x of length n, obtains a random string of length O(r(n)), queries O(q(n)) bits of the proof and satisfies.

According to the previous definition

)(),(, nqnrPCPL sc

)(),()(),(2

1,1

nqnrPCPnqnrPCP

In the previous proof the hardness factor established is precisely the bound on the error probability of the PCP verifier for SAT.

How to reduce parameter sHow to reduce parameter s

• Obvious way: Simulate a PCP[logn,1] verifier multiple number of times and accepts iff the verifier accepts each time

• Clever way: Use a constand degree expander graph to generate O(logn) strings of blogn bits each, using only O(logn) truly random bits.

Two ways

Simulating k times will reduce soundness to

However this will increase the number of random bits needed to O(klogn) and the number of query bits to O(k)

k2

1

Verifier will be simulated using these O(logn) strings as the random strings. Clearly these are not truly random strings. Properties of expanders show that they are almost random.

Probability of error still drops exponentially in the number of times the verifier is simulated

A useful theoremA useful theoremGraph H with the following properties

• A constant degree expander on nb vertices.

• Each vetrex has a unique blogn bitsTheorem: Let S be any set of vertices of H and

2

bnS

There is a constant k such that

2

1Sin lie walk randomlength log a of verticesallPr nk

Question 1: Why we introduce graph H

Answer: We will use it to generate O(logn) strings of blogn bits, using only O(logn) truly random strings. The verifier will be simulated using these O(logn) strings as ‘random’ strings.Question 2: How we construct a random walk on graph H of length O(logn)Answer: We use only O(logn) random bits.

1. blogn bits to pick the starting vertex, and

2. a constant number of bits to pick successive vertex

Theorem Theorem

Proof is divided into two parts

•

•

Let which means there is a verifier F for L which requires blogn random bits and queries q bits of the proof, where b and q are constants We give a verifier for language L, which.

• constructs an expander graph H

• constructs a random walk of length klogn using only O(logn) random bits. By construction of H, the label of each vertex on this path specifies a blogn bit string.

• It uses these these klogn+1 strings as the ‘random’ string on which it simulates verifier F.

nnPCPNPn

log,log1,1

1,loglog,log2

1,1

1,1

nPCPnnPCPn

nnPCPnPCPn

log,log1,log 1,1

2

1,1

1,log2

1,1

nPCPL

nnPCPFn

log,log 1,1

Proof Proof

• Consider x is in L. Let p be a proof that makes verifier F accepts with probability 1.

• Consider x is not in L. Let p be an arbitrary proof supplied to

runs 1log allon accepts iff accepts nkFF

2

bn

Given proof p, also accepts x with probability 1. Hence the completeness parameter is 1 (c=1)

F

F

Given proof p, verifier F accepts on random strings of length blogn

Let S denote the corresponding set of vertices of H, .

Since accepts x iff F accepts x on all klogn+1 random strings, accepts x if the random walk remains entirely in S. But the probability of this event is <1/n.

Thus the soundness of is 1/n.

2

bnS

FF

F

Observe that requires only O(logn) random bits and queries O(logn) bits of the query.

F

Attack On Hardness of Clique ProblemAttack On Hardness of Clique ProblemTheorem: For fixed constants b and q, there is a gap introducing reduction from SAT to Clique that transforms a boolean formula of size n to a graph G=(V,E), where such that •

•

bnGOPT )( e,satisfiabl is Φ if

bqbq nnnV

11)( e,satisfiablnot is Φ if bb nnn

GOPT

Proof: Let F be a verifier for SAT that requires blogn random bits and queries qlogn bits of the proof

nnPCPn

log,log1,1

• If Φ is satisfiable and p is a good proof, choose the nb vertices of G such that the klogn positions of p associated with each chosen vertex contains assignment τ. These vertices form a clique

• If Φ is not satisfiable, suppose that C is a clique in G

We have shown that any clique C in G gives rise to a proof that is accepted by F with probability

Since the soundness of F is 1/n, the largest clique in G is of size <nb-1.

bbq

q

n

C

nn

Cn

Attack On Hardness of Clique Problem Attack On Hardness of Clique Problem IIII

Corollary: There is no factor approximation algorithm for the cardinality

clique problem assuming , where

qn1

NPP bqq

1

HARDNESS OFHARDNESS OFSET-COVER PROBLEMSET-COVER PROBLEM

Input

• A universe U of n elements,

• a collection of subsets of U, , and

• a cost function

Find

A minimum cost subcollection of S that covers all elements of U.

A known approximation algorithmA known approximation algorithm

QSc :

kssS ,,1

C=0

while ( )

Let the cost-effectiveness of S

find the set whose cost-effectiveness is smallest, say S

Pick S, and for each e in S-C, set

end while

A greedy approximation algorithm with factor

UC

CS

Sc

)(

)(eprice

SCC

nH n

1

2

11

Two-prover one-round proof Two-prover one-round proof System System

(Introduction)(Introduction)Since now for the purpose of showing hardness of MAX-3SAT and Clique we did not require a detailed description of the kinds of queries made by the verifier.

The only restriction was that we only required a bound on the number of queries made on the proof.Question: Which is the notion behind a Two-prover One-round proof systemAnswer: Think of the proof system as a game between the prover and the verifier.

Prover is trying to cheat in the sense that it is trying to convince the verifier that a “no” instance for Language L is actually in L.Question: Is there a verifier that can ensure that the probability of getting cheated is <1/2 fro every “no” instance?

Two-prover one-round proof Two-prover one-round proof SystemSystem

Two-prover model • Verifier is allowed to query two non-communicating

provers, denoted P1 and P2.

• Verifier can cross check the prover’s answer. Thus the prover’s ability to cheat gets restricted in this model.

One-round proof system • Verifier is allowed one round of communication with each

prover. The simplest way formalizing this is as follows

o Proof P1 is written in alphabet Σ1. The size of the alphabet, |Σ1| may be unbounded.

o Proof P2 is written in alphabet Σ2. The size of the alphabet, |Σ2| may be unbounded.

• Verifier is allowed to query one position of the two proofs.

Two-prover one-round proof Two-prover one-round proof SystemSystem

Two-prover One-round model defines the class

• for every input , there is a pair of proofs and that makes M accepts with probability .

• for every input , and every pair of proofs and that makes M accepts with probability <s.

Comes with 3 parameters

• Completeness (c),

• Soundness (s), and

• # of random bits provided to the verifier (r(n)) nrRP sc,12

nrRPL sc,12

There is a polynomial time bounded verifier M that receives O(r(n)) truly random bits and satisfies

*11 y *

22 ys

Lx

Lx *11 y *

22 y

TheoremTheoremTheorem: There is a constant εp>0 such that

nRPNPp

log12 1,1 Proof: Divided into two parts

•

•

NPnRPp

log12 1,1

nRPSATnRPNPpp

log12or log12 1,11,1

Proving the second part. We know that

• If Φ is satisfiable, OPT(Ψ)=m

• If Φ is not satisfiable, OPT(Ψ)<(1-ε5)m

Proof (Continued)Proof (Continued)The Two-prover one-round verifier, M, for SAT works as follows

1. Given a SAT formula Φ, it uses the aforementioned reduction to obtain a MAX-3SAT(5) instance of Ψ.

2. M assumes that P1 contains an optimal truth assignment, τ, for Ψ. (|Σ1|=2). It assumes that P2 contains for each clause, the assignment to its three boolean variables under τ (|Σ2|=23).

3. It uses the O(logn) random bits to pick

• A random clause C from Ψ.

• A random boolean variable x occurring in clause C.

4. M obtains the truth assignment to x and the three variables in C by querying P1 and P2, respectively.

5. M accepts if C is satisfied and the two proofs agree on their assignment for variable x.

Proof (Continued)Proof (Continued)• If Φ is satisfiable, then so is Ψ. Clearly there are proofs y1 and

y2 such that M accepts with probability 1.

• If Φ is unsatisfiable, any truth assignment must leave strictly more than ε5 clauses unsatisfiedConsider any pair of proofs (y1,y2), and assume y1 as a truth assignment, say, τ.

The random clause C picked by M, is not satisfied by τ with probability

If so, and if the assignment for C contained in y2 is satisfying then y1 and y2 must be inconsistent.

55

m

m

Let A, B be two events

A: Random Clause C is not satisfied by τ.

Β: Random clause C is satisfied by the assignment contained in y2

3

1ntinconsiste are 2 and 1 5 BPAPBAPyyP

Hence overall, verifier M must reject with probability 35

Main ReductionMain ReductionTheorem: There is a constant c>0 for which there is a randomized gap-introducing reduction Γ, requiring time n(O(loglogn)), from SAT to the cardinality set cover problem that transforms a Boolean formula Φ to a set system S over a universal set of size n(O(loglogn)) such that•

•

knSOPT 2)( e,satisfiabl is Φ if

2

1log)( e,satisfiablnot is Φ if nkncSOPTPR k

where

• n: the length of each of the two proofs for SAT under the two-prover one-round model (polynomial in the size of Φ).

• k=O(loglogn).Observation: A slight abuse of notation, since gap introducing reductions were defined to run in polynomial time

Each boolean variable occurs in exactly 5 clauses.

Each clause contains 3 distinct variables (negated or unnegated).

As a result of the uniformity conditions, if Ψ has n variables, then it has clauses. Therefore, the two proofs are of length n and

This modification changes the constant ε5 to some other constant.

What we need for the proof IWhat we need for the proof I• Uniformity Conditions for MAX-3SAT(5) formula.

3

5n

3

5n

• The two proofs have equal lenght.

The two proofs are of length n and

Equality of length can be easily achieved by repeating the first proof 5 times and the second proof 3 times.

Verifier will query a random copy of each proof.

3

5n

A set system where,

U is the universal subset.

C1,…,Cm are subsets of U.

What we need for the proof IIWhat we need for the proof II• A gadget (set system)

mm CCCCU ,,,,,, 11

Good Cover: U is covered by picking a set Ci and its complement.

Bad Cover: A cover that does not include a set and its complement.• A useful theorem for constructing efficiently set

systems as the above one Theorem: There exists a polynomial p(., .) such that there is a randomized algorithm which generates, for each m and l, a set system mm CCCCU ,,,,,, 11

With |U|=p(m,2l). With probability >1/2 the gadget produced satisfies that every bad cover is of size l. Moreover the running time is polynomial in |U|.

What we need for the proof IIWhat we need for the proof II• Reduce error probability of two-prover one-round proof

systemTwo ways Parallel repetition (usual way): Verifier picks k

clauses randomly and inpependently, and a random boolean variable from each of the clauses Verifier queries P1 on the k-variables.

Verifier queries P2 on the k-clauses.

Accepts if all the answers are acceptingUnder this schema, the probability that the provers manage to cheat drops to <(1-εp)k. This is true only if it is assumed that the provers are required to answer each question before being given the next questionTwo major drawbacks1. Each prover is allowed to look at all k questions

before providing its k answers. Able to coordinate its answers.

2. If the provers are required to answer each question before being given the next question, probability error drops in the usual fashion. However this required k-round of communiocations.

What we need for the proof IIIWhat we need for the proof III

Parallel repetition (proposed by the following theorem)Theorem: Let the error probability of a two-prover one-round proof system be δ<1.

Then the error probability on k parallel is at kost δkd, where d is a constant that depends only on the length of the answers of the original proof system.