De nitions Basic techniques Relax and rounddiaz/slides8-19.pdf · Basic techniques Relax and round...

DefinitionsBasic techniquesRelax and round

Approximation Algorithms

AlCo-GMat

Fall 2019

AlCo-GMat Approximation Algorithms


Optimization ProblemComplexity classesAlgorithmic techniques for hard problemsDefinition and exampleComplexity classes for approximation

1 Definitions

2 Basic techniques

3 Relax and round




References:Approximation

(S. = Springer)

Garey, Johnson. Computers and intractability a Theory of theNP-completessa, Freeman, 1979

Ausiello, et al. Complexity and Approximation S. 1999

Vazirani. Approximation Algorithms, S. 2001.

Willliamsom, Shmoys, The Design of ApproximationAlgorithms. Cambridge University Press, 2011.




Problems

Example:

Given a graph and two vertices, is there a path joining them?decison problem

Given a graph and two vertices, obtain a path joining themwith minimum length.optimization problem

Given a set of vertices and two of them, obtain a path joiningthem with minimum length




Decision Problems: Complexity classes

P polynomial timethere is a polynomial time algorithm providing the correctanswer. for any input.

EXP exponential time optimizationthere is an exponential time algorithm providing the correctanswer. for any input.

NP non-deterministic polynomial timeSyntactic definition!




Optimization Problems

An optimization problem is a structure P = (I, sol,m, goal), where

I is the input set to P;

sol(x) is the set of feasible solutions for an input x .

The objective function m is an integer (rational) measuredefined over pairs (x , y), for x ∈ I and y ∈ sol(x).

goal is the optimization criterium max or min.

That is the function problem whose goal, with respect to aninstance x , is to find an optimum solution, that is, a feasiblesolution y such that

y = goal{(m(x , y ′) | y ′ ∈ sol(x)}.




Optimization Problems: Complexity classes

PO polynomial time optimizationthere is a polynomial time algorithm computing an optimalsolution for any input.

EXPO exponential time optimizationthere is an exponential time algorithm computing an optimalsolution for any input.

NPO NP optimizationSyntactic definition:




The NPO class

An optimization problem P = (I, sol,m, goal) belongs to NPO iff

I is recognizable in polynomial time.

The feasible solutions are short:a polynomial p exists such that, for y ∈ sol(x), |y | ≤ p(|x |) .Moreover, it is decidable in polynomial time whethery ∈ sol(x), for x , y with |y | ≤ p(|x |),

The objective function m is computable in polynomial time.




The NPO class: Justification

The bounded version of an optimization problem is thedecision problem

minimization P = (I, sol,m,min) is

Given x ∈ I and an integer kIs there a solution y ∈ sol(x) such that m(x , y) 6 k?

maximization P = (I, sol,m,max) is

Given x ∈ I and an integer kIs there a solution y ∈ sol(x) such that m(x , y) > k?

The bounded version of a NPO problem belongs to NP.

For a NPO problem the bounded version and the optimizationproblem are polynomially equivalent, via Turing reductions.




Some NPO problems

min-bin packingGiven n objects, object i has volume vi , 0 ≤ vi ≤ 1, compute theminimum number of unit bins needed to pack all the objects.

max-satGiven a cnf formula F , compute an assignment that satisfies themaximum number of clauses.

max-w-satGiven a cnf formula F , in which each clause has an assigned weight.Define the value of an assignment as the sum of the weights of thesatisfied clauses. The problem consists in computing an assignmentwith maximum value.

And the bounded literal per clause families max-k-sat.




How to “solve” NP-hard optimization problems in practice

We know that an optimization problem, whose decision version isNP-hard, cannot be solved in polynomial time unless P =NP.

Exact algorithms: for instances of small size or for restrictedclasses of instances.

Randomization: algorithms that use random bits, either withhigh probability of succes or with high probability of poly time.

Heuristic method: a feasible solution with empirical guarantee.

Approximation algorithm: a feasible solution with aperformance guarantee.

Parameterization: solve efficiently some slices of the problem.




Approximation algorithms

Let P be an optimization problem.

For any instance x of P let opt(x) be the cost of an optimalsolution.

Let A be an algorithm such that for any instance x of Pcomputes a feasible solution with cost A(x).

A is an r -approximation for P (r ≥ 1) if for any instance x of P

1

r≤ opt(x)

A(x)≤ r

P is r -approximable in polynomial time if there is a polynomialtime computable r -approximation for P.




Nex Fit: An approximation algorithm for BinPacking

min-bin packingGiven n objects, object i has volume vi , 0 ≤ vi ≤ 1, compute theminimum number of unit bins needed to pack all the objects.





Let us assume that the bins are labeled B1,B2, . . . ,Bn and thatinitially all are closed and during the execution of the algorithmonly one bin will be open.

Next Fit places the n objects, one after the other, as follows:

Opens B1 and places the first object in B1.

If the i-th object fits in the open box, we put it inside.Otherwise, we close the bin, open the next one and place theobject in it.

For the case of v1 = 0.3, v2 = 0.8, and v3 = 0.7, Next Fit solutionneeds three bins. But there is a solution with uses only two bins.Not optimal!





Theorem

Let x be an input to the min-bin packing problem and let opt(x)be the minimum number of bins needed to pack de objects in x . IfNF(x) is the number of bins in the solution computed by Next Fit,then opt(x) ≤ NF(x) ≤ 2opt(x).





Theorem

Let x = (v1, . . . , vn be an input to the min-bin packing. LetNF(x) be the number of bins in the solution computed by NextFit, we have opt(x) ≤ NF(x) ≤ 2opt(x).

Proof.

The first inequality is always true:

min-bin packing is a minimization problem

Next Fit provides a feasible solution

Let V =∑n

i=1 vi , we have opt(x) ≥ dV e.





Proof.

Let us look to two consecutive bins in the Next Fit solution Thetotal packet size in the two bins must be bigger than 1, otherwisewe will never have opened the second bin. So,

NF(x) ≤ 2dV e.

But we have seen opt(x) ≥ dV e, so

NF(x) ≤ 2dV e ≤ 2opt(x).




NPO: approximation classes

Classification of NPO problems as approximablewithin a constant r in polynomial time:

APX exists r .

PTAS Polynomial Time Approximation Schemafor any r , but time may depend exponentially in 1/(r − 1).

FPTAS Fully Polynomial Time Approximation Schemafor any r in time polynomial in the input size and 1/(r − 1).

Further classification can be obtained by considering non-constantr , for example log n or log log n . . . .

Negative results through hardness APX-hard etc.




Hard to approximate problems

Hardness levels:

APX-hard: unless P =NP no PTASunless P =NP no constant approximation...Non-approximableunless P =NP, for any r at most a polynomial function of n, thereis no polynomial time r -approximation algorithm



GreedyLocal SearchScalingCombinatorial algorithms

1 Definitions

2 Basic techniques

3 Relax and round




Load Balancing problem




Processing scenario

We have m identical machines;n jobs, job j has processing time tj

Job j must run contiguously on one machine.

A machine can process at most one job at a time.

We want to assign jobs to machines optimizing the makespan.

Let J(i) be the subset of jobs assigned to machine i .

The load of machine i is Li =∑

j∈J(i) tj

The makespan is the maximum load on any machine,L = maxi Li .




Load Balancing problem

lbalGiven n jobs, job j has processing time tj , assign jobs to m identicalmachines as to minimize the makespan.

The problem is NP-hard and belongs to NPO.Load balancing is hard even if m = 2 machines(reduction from Partition).

The approximation algorithm we propose is a greedy algorithmcalled list-scheduling.




List scheduling

List schedulingFor j = 1, . . . , n:

Assign job j to the machine having smallest load so far.




List scheduling: Implementation

function List Scheduling(m, n,T )for i = 1, dots,m do

L[i ] = 0 load on machineJ[i ] = ∅ jobs assigned to

end forfor j = 1, . . . , n do

i = argminkLk machine with smallest loadJ[i ] = J[i ] ∪ {j}L[i ] = L[i ] + T [j ]

end forend function

Cost: Using a priority queue to maintain L, the cost is O(n log m)




List scheduling: Approximation rate

Theorem

List scheduling is a polynomial 2-approximation algorithm forlbal.

Proof.

Let L∗ be the optimum makespan and L the makespan of thesolution computed by list scheduling.

L∗ ≥ maxj tj and L∗ ≥ 1m

∑j tj .

Assume that L = Li . Let j be the last job scheduled inmachine i .

When job j was assigned, all the other machines have higherload, so Li − tj ≤ Lk , for k 6= i .




List scheduling: Approximation rate

Theorem

List scheduling is a polynomial 2-approximation algorithm forlbal.

Proof.

When job j was assigned, all the other machines have higherload, so Li − tj ≤ Lk , for 1 ≤ k ≤ m. Summing up for all kand dividing by m we get

Li − tj ≤ 1m

∑k Lk ≤ 1

m

∑j tj ≤ L∗

We have, L = Li = (Li − tj) + tj ≤ L∗ + L∗ = 2L∗




List scheduling: Tightness?

m machines, m(m − 1) length 1 jobs and 1 job of length m.

List scheduling assigns the first m length 1 jobs, to differentmachines, and

after scheduling the unit length jobs, all machines have loadLi = m − 1.

The last job is assigned to one machine, giving a makespanL = m + m − 1 = 2m − 1.

An optimal solution assigns the big job to one machine and munit jobs to the other machines, so L∗ = m.

The approximation rate is tight




Longest processing first

Analizamos una variante,

Longest processing first:Sort jobs in decreasing order of processing time.Run list scheduling.




Longest processing first: Approximation rate

Theorem

Longest processing first is a polynomial 3/2-approximationalgorithm for lbal.

Proof.

If n ≤ m, L = t1 = L∗ ≤ 3/2L∗

If n > m, since there are more jobs than machines, a machinemust take two jobs in (t1, . . . , tm+1).

Therefore, L∗ ≥ 2tm+1 ≥ 2tj , being j the last job assigned tothe machine i that gives the makespan.

So, Li = (Li − tj) + tj ≤ L∗ + 12L∗ ≤ 3

2L∗.




Longest processing first: Tightness?

The 3/2 bound on the approximation rate is not tight

In fact Longest processing first is a 4/3-approximationalgorithm [Graham 1969]

4/3 is tight:

m machinesn = 2m + 1 jobs2 jobs of length m,m + 1, . . . , 2m − 1 and one more job oflenght m.L∗ = 3m and L = 4m − 1, which gives a ratio tending to 4/3.4/3 is tight




Max Cut

max-cutGiven a graph G = (V ,A) we want to find a partition of V into V1,V2 in such a way that

cut(V1,V2) = ||{(u, v) | u ∈ V1 v ∈ V2}||

is maximum.

The problem is NP-hard and belongs to NPO.

Let us analyze a local search algorithm using the HillClimbingparadigm.




Local search

A neighborhood structure is defined on the solution set.

The algorithm performs an exploration of the neighborhoodgraph.

It starts at one feasible solution moves to a better one andfinishes at a local optimum

A local optimum is a solution such that all its neighbors have equalor worse cost.




Neighborhood for MaxCut

Given a graph G = (V ,E ) define N (G ):

The neighbors of a solution (V1,V2) are all those partitionsthat can be obtained by moving either one element fromV1 to V2 or one element from V2 to V1.

Using N (G ) we consider the following algorithm:

HillClimbing Max-Cut (G:graph, n: integer)V1,V2: set of [1 . . . n];V1 := ∅; V2 := V(G );while not local-optimum(V1,V2) do(V1,V 2):= a neighboring partition of (V1,V2)with improved cost




HillClimbing MaxCut

Observe that the algorithm needs only polyomial time as

we can compute cut(X ,Y ) in O(|E(G )|) teps.

The number of neighbors of a solution is n.

The cost of the initial solution is 0.

The maximum partition cut is upper bounded by |E(G )|.At each step the cut is increased in one unit.




HillClimbing MaxCut

Lemma

Let G = (V ,A) be a graph, if (V1,V2) is a local optimum ofN (G ) then cut((V1,V2)) ≤ 2opt(G ).

Proof.

Let

(V1,V2) be a local optimum

m1 be the number of edges in G joining two vertices in V1

m2 be the number of edges in G joining two vertices in V2

m = |E(G )|Therefore, m = m1 + m2 + cut(V1,V2).




HillClimbing MaxCut

Proof.

For any vertex u, set

mu1 = |{v | v ∈ V1, (v , u) ∈ A}|

mu2 = |{v | v ∈ V2, (v , u) ∈ A}|.




HillClimbing MaxCut

Proof.

As (V1,V2) is a local optimum, for any u, moving only u providesa new partition with equal or smaller cost.

for u ∈ V1,cut((V1 − u,V2 + u))− cut((V1,V2)) = mu

1 −mu2 ≤ 0

for u ∈ V1,cut((V1 + u,V2 − u))− cut((V1,V2) = mu

2 −mu1 ≤ 0

So, ∑u∈V1

(mu1 −mu

2) = 2m1 − cut(V1,V2) ≤ 0

∑u∈V2

(mu2 −mu

1) = 2m2 − cut(V1,V2) ≤ 0




HillClimbing MaxCut

Proof.

Adding the las two quantities we have

2m1 + 2m2 − 2cut(V1,V2) ≤ 0

As m = m1 + m2 + cut(V1,V2) we get

cut(V1,V2) ≥ m/2.

As we always have opt(G ) ≤ m, then we have that

2cut(V1,V2) ≥ opt(G ).




HillClimbing MaxCut

Taking into account that the proposed algorithm always finds alocal optimum in polynomial time, we have

Theorem

HillClimbing Max-Cut is a polynomial 2-approximationalgorithm for max-cut.




A problem in FPTAS: 0-1 Knapsack

0-1 knapsackGiven an integer b and a set of n objects, object i has weight wi andvalue vi , compute a selection of objects with total size less than orequal to b and maximum profit.

The problem is NP-hard and belongs to NPO. There is a dynamic

programming algorithm that solves 0-1 knapsack in time

∼ nn∑

i=1

vi .

The algorithm is polynomial for poly(n) values.




Consider the following Scaledown algorithm which has r asinput:

Scaledown(w , v , b, r)

vmax := max vi ;

t := blog[ r−1rvmaxn ]c;

z := instance obtained by changing profits to v ′i = vi/2t ;

y := optimal solution for z ;

return y

Is Scaledown a polynomial time approximation schema? time?rate of approximation?




TimeThe most difficult part is the computation of the optimal solutionthat takes time

nn∑

i=1

v ′i = n

∑ni=1 vi2t

= n2 vmax

2t

But, t = blog r−1r

vmaxn c

Thus, for r → 1, = n2 vmaxr−1r

vmaxn

= rn3

r−1 = O( n3

r−1).

polynomial in input size and 1/(r − 1)




Quality of the solutionopt(x)− A(x , r) ≤ n2t , and nvmax ≥ opt(x) ≥ vmax. Thus

n2t

vmax≥ opt(x)− A(x , r)

opt(x)= 1− A(x , r)

opt(x)

A(x , r)

opt(x)≥ 1− n2t

vmax=

vmax − n2t

vmax

But t = blog r−1r

vmaxn c and n2t = n r−1

rvmaxn = vmax

r−1r .

opt(x) ≤ vmax

vmax − n2tA(x , r) =

vmax

vmax − Vmax(r−1)r

A(x , r) ≤ rA(x , r).

Thus, Scaledown is an r -approximation, we have a FPTAS forKnapsack.




Min-TSP with triangle inequality

min-m-tspGiven a set of n cities together with distances among any pair of cities,under the assumption that distances verify the triangle inequality, finda shortest tour.

We model the instance by a weighted graph G = (V ,E , d).




Algorithm TSP-ST

Compute a minimum spanning tree T of G .

Find the directed graph T ′ obtained from T by replacing eachedge with two arcs in oposite directions.

Find an Eulerian circuit R of T ′.

Let S be the walk in G directed by R.

Transform S in a tour C , by removing (in order) all thevertices that have been already visited in S .




Theorem

Algorithm TSP-ST is a polynomial 2-aproximation algorithm formin-m-tsp.




Theorem

Algorithm TSP-ST is a polynomial 2-aproximation algorithm formin-m-tsp.

Proof.

Observe that:

opt(G ) ≤ c(C ) ≤ c(S) = c(R) due to triangle inequality.

c(R) = 2c(T ) as we use each edge twice.

Furthermore, any circuit provides a spanning tree, just byremoving one of their edges, amb total distance below thecircuit’s distance:c(T ) ≤ opt(G ).




Christofides’ algorithm TSP-CH

Handshaking lemma: every finite undirected graph has an evennumber of vertices with odd degree

Compute a minimum spanning tree T of G .

Let O be the set of vertices with odd degree in T . By thehandshaking lemma, O has an even number of vertices.

Find a minimum-weight perfect matching M in the inducedsubgraph given by the vertices from O.

Combine the edges of M and T to form a connectedmultigraph H in which each vertex has even degree.

Find an Eulerian circuit R of H.

Let S be the walk in G directed by R.

Transform S in a tour C , by removing (in order) all thevertices that have been already visited in S .




Lemma

G = (V ,E ) is a graph. Let M be a minimum-weight perfectmatching for O. Then c(M) ≤ opt(G )/2.

Proof.

Take any optimal tour of G , take the shortcuts to make a tourK for O. By the triangle inequality c(K ) ≤ opt(G )

As O has even number of vertices K can be decomposed intotwo (alternating) perfect matchings S and S ′.M is a minimum weight perfect matching and2c(M) ≤ c(S) + c(S ′) = c(K ) ≤ opt(G )




Theorem

Algorithm TSP-CH is a polynomial 3/2-aproximation algorithmfor min-m-tsp.

Proof.

By the previous lemma c(M) ≤ opt(G )/2.

We already know that c(T ) ≤ opt(G )

Also, opt(G ) ≤ c(C ) ≤ c(S) = c(R) due to triangleinequality.

By construction, c(R) ≤ c(T ) + w(M) ≤ opt(G ) + opt(G)2




Shortest superstring

shsstrGiven an alphabet Σ and a set of n strings, S = {s1, ..., sn} ⊆ Σ∗,we want to find the shortest string s that, for i = 1, . . . , n, containssi as a substring.

Input

CATGCCTAAGTGCTATTCAATGCATC

Output GCTAAGTTCATGCATC

CATGCCTAAGT

GCTATTCA

ATGCATC

We assume that no si ∈ S is a substring of sj ∈ S .ShSStr belongs to NPO and it is NP-hard, by a reductionfrom TSP.




An expression for the optimal cost

For two strings s and t define

overlap(s, t) = the longest suffix of s that is a prefix of t.

prefix(s, t) = s after deleting overlap(s, t).

Assume that in an optimal solution s∗ the strings appear in theorder s1, . . . , sn.

The overlap among two consecutive strings must bemaximum.

opt(S) = |s∗| =|prefix(s1, s2)|+ · · ·+ |prefix(sn, s1)|+ |overlap(sn, s1)|.




Minimum weight cycle cover

minwcyclecoverGiven a weighted graph G find a minimum weight collection of disjointcycles covering all the vertices in G .

MinWCycleCover can be solved as a minimum weightperfect matching in a bipartite graph

G ′ = (V1,V2,E′) if (u, v) ∈ E we have an edge (u1, v2) with

weight w(u, v).

Cycle covers in G are in a one to one correspondence withperfect matchings in G ′.

A minimum weight perfect matching in G ′ can be computedusing a MaxFlow algorithm in poly time.




From cycles to strings

Assume that c is a cycle in the prefix graph of a set of stringsS .

c = i1 → i2 → · · · → i` → i1.

Consider the stringσ(c) = prefix(si1 , si2) · · · · · prefix(si` , si1) · si1 .

σ(c) is a superstring of si1 · · · · · si`σ(c) starts and ends with si1 .




SSS-CycleCover

Algorithm SSS-CycleCover(S)

Construct the prefix graph associated to S .

Find a minimum weight cycle cover C = {c1, . . . , ck}.Break each cycle i at a position i1.

Output σ(c1) · · · · · σ(ck)

Observe that, for two consecutive cycles c and c ′ with breakingstrings t and t ′, we have |overlap(t, t ′)| ≤ w(c) + w(c ′).




SSS-CycleCover

Theorem

Algorithm SSS-CycleCover is a 4-approximation algorithm forShSString.

Proof.

opt(S) = w(C) =∑k

i=1 w(ci ).

A(S) =∑k

i=1 |σ(ci )| = w(C ) +∑k

i=1 |ti |.




Proof.

Assume that ti (and si ) are numbered in order of theirleftmost occurrence in a shortest superstring s∗ for S .

opt(S) =∑n−1

i=1 |prefix(si , si+1)|+ |prefix(sn, s1)|+|overlap(sn, s1)| ≤

∑n−1i=1 |prefix(si , si+1)|

|ti | − overlap(ti , ti+1) = |prefix(si , si+1)|, so

opt(S) ≥k∑

i=1

|ti | ≥k∑

i=1

|ti | −k−1∑i=1

|overlap(ti , ti+1)|

k∑i=1

|ti | − 2k∑

i=1

w(ci ) =k∑

i=1

|ti | − 2opt(S)

Therefore,∑k

i=1 |ti | ≤ 3opt(S), and A(S) ≤ 4opt(S).



The frameworkVertex CoverMin Weighted Vertex COverMin 2-SATMax Satisfiability

1 Definitions

2 Basic techniques

3 Relax and round




Relaxation and rounding

Many real-life problems can be modeled as Integer LinearPrograms (IP).The IP can be relaxed to a linear program (LP).We can solve the LP in polynomial timeSimplex method: Dantzig in 1947 (exponential Klee andMinty 1972), Ellipsoid method: Khachiyan 1979 (O(n6)),Interior-point method: Karmarkar 1984 (O(n3)).

LP optimal solution might not be integral, transform it to geta feasible solution not far from opt.




Vertex cover

min-vcGiven a graph G = (V ,A) we want to find a set S ⊂ V with minimumcardinality, so that every edge in G has at least one end point in S .

VC-IP

minn∑

i=1

cixi

s.t. xi + xj ≥ 1 for all (i , j) ∈ E

xi ∈ {0, 1} for all i ∈ V

VC-LP

minn∑

i=1

cixi

s.t. xi + xj ≥ 1 for all(i , j) ∈ E

xi ≥ 0 for all i ∈ V




Min Vertex cover: relaxed solutions

Lemma

VC-LP has an optimal solution x∗ such that xi ∈ {0, 1, 1/2}.Furthermore, such a solution can be computed in polynomial time.

Proof.

Let y be an optimal solution s.t. not all its coordinates are in {0, 1, 1/2}.Set ε = minyi /∈{0,1,1/2}{yi , |yi − 1/2|, 1− yi}. Consider

y ′i =

yi − ε 0 < yi < 1/2

yi + ε 1/2 < yi < 1

yi otherwise

y ′′i =

yi + ε 0 < yi < 1/2

yi − ε 1/2 < yi < 1

yi otherwise∑yi = (

∑y ′i +

∑y ′′i )/2, so both are optimal solutions. One of them

has more {0, 1, 1/2} coordinates than y .




Vertex cover: Relax + Round algorithm

function Relax+Round VC(G )Construct the LP-VC associated GLet y be an optimal relaxed solution (of the LP instance)Using the previous lemma, construct an optimal relaxed

solution y ′ such that y ′i ∈ {0, 1, 1/2}Let x defined as xi = 0 if y ′i = 0, xi = 1 otherwise.return (x)

end function

Relax+Round VC

runs in polynomial time

x defines a vertex cover∑ni=1 xi ≤ 2

∑ni=1 y ′i ≤ 2opt

is a 2-approximation for VC.




Min Weighted vertex cover: Relax+Round approximation

LP WVC

minn∑

i=1

cixi

s.t. xi + xj ≥ 1 for all(i , j) ∈ E

xi ≥ 0 for all i ∈ V

function WVC(G , c)Construct the LP WVC, Iy = LP.solve(I )for i = 1, . . . , n do

if yi < 1/2 thenxi = 0

elsexi = 1

end ifend forreturn (x)

end functionRelax+Round WVC

runs in polynomial timex defines a vertex cover∑n

i=1 cixi ≤ 2∑n

i=1 ciyi ≤ 2optis a 2-approximation for WVC.




Minimum 2-Satisfiability

min 2-satGiven a Boolean formula in 2-CNF, determine whether it is satisfi-able and, in such a case, find a satisfying assignment with minimumnumber of true variables.

2-SAT can be solved in polynomial time

Min 2-SAT IP formulation?




Minimum 2-Satisfiability: IP formulation

Suppose that F has n variables x1, . . . xn and m clauses with 2literals per clause

IP Min 2-SAT

minn∑

i=1

xi

s.t. xi + xj ≥ 1 for all clauses (xi ∨ xj) ∈ F

(1− xi ) + xj ≥ 1 for all clauses (x i ∨ xj) ∈ F

(1− xi ) + (1− xj) ≥ 1 for all clauses (x i ∨ x j) ∈ F

xi ∈ {0, 1} 1 ≤ i ≤ n

LP Min 2-SAT is obtaining replacing xi ∈ {0, 1} by xi ≥ 0.




Minimum 2-Satisfiability: LP relaxation

LP Min 2-SAT

minn∑

i=1

xi

s.t. xi + xj ≥ 1 for all clauses (xi ∨ xj) ∈ F

(1− xi ) + xj ≥ 1 for all clauses (x i ∨ xj) ∈ F

(1− xi ) + (1− xj) ≥ 1 for all clauses (x i ∨ x j) ∈ F

xi ≥ 0 1 ≤ i ≤ n

Let y be an optimal solution to LP Min 2-SAT.Can we use the same rounding scheme as for WVC?Setting xi = 1 if yi > 1/2 and xi = 0 if yi < 1/2 is safe, allclauses with at least one literal with value > 1/2 will besatisfied.When yi = 1/2?




Minimum 2-Satisfiability: LP relaxation

Let y be an optimal solution to IP Min 2-SAT.

What to do when yi = 1/2? 1? 0?

If F contains the clauses (xi ∨ xj) and (x i ∨ x j) andyi = yj = 1/2, neither xi = xj = 1 nor xi = xj = 0 satisfy theformula.

F1 = clauses whose two variables have y value = 1/2.

Rounding those values to 1 or 0 would keep the approximationratio to 2, provided the constructed solution x to Min 2-SATis still a satisfying assignment.

Any satisfying assignment for the clauses in F1 and get a2-approximation




Minimum 2-Satisfiability: Relax+Round approximation

function Relax+Round Min 2-SAT(F )if F is not satisfiable then return falseend ifConstruct the LP Min 2-SAT, Iy = LP.solve(I )for i = 1, . . . , n do

if y ′i < 1/2 then xi = 0end ifif y ′i > 1/2 then xi = 1end if

end forF1 = clauses with both y values = 1/2.Let J = {j | xj ∈ F1}for i=1,. . . , n do

if yi = 1/2 and i 6∈ J then xi = 1end if

end forComplete x with a satisfying assignment for F1

return (x)end function




Minimum 2-Satisfiability: Relax+Round approximation

Theorem

Relax+Round Min 2-SAT is a 2-approximation for Min2-SAT.




Max Satisfiability

max satGiven a Boolean formula in CNF and weights for each clause, find aBoolean assignment to maximize the weight of the satisfied clauses.

Suppose that F has n variables x1, . . . xn and m clauses C1, . . . ,Cm.

IP Max SAT

maxm∑j=1

wjzj

s.t.∑xi∈Cj

yi +∑x i∈Cj

(1− yi ) ≥ zj = 1, . . . ,m

yi ∈ {0, 1} 1 ≤ i ≤ n

zj ∈ {0, 1} 1 ≤ j ≤ m

LP Max SAT is obtaining replacing a ∈ {0, 1} by 0 ≤ a ≤ 1.




Max Satisfiability: Relax+RRound

function Relax+RRound(F )Construct the LP Max SAT, I(y , z) = LP.solve(I )for i=1,. . . , n do

Set xi = 1 with probability yiend forreturn (x)

end function

The optimal LP solution is used as an indicator of theprobability that the variable has to been set to 1.

The performance of a randomized algorithm is the expectednumber of satisfiable clause.

This expectation has to be compared with opt.





Let (y∗, z∗) be an optimal solution of LP Max SAT

Let Zj be the indicator random variable for the event thatclause Cj is satisfied.

Assume that Cj has k-literals and that ` of them are negatedvariables.

Lemma

For any 1 ≤ j ≤ m, E [Zj ] ≥ z∗j (1− 1/e).

Recall (a1 . . . ak)1/k ≤ (a1 + · · ·+ ak)/k .





Proof.Zj is an indicator random variable, and soE [Zj ] = Pr [Zj = 1] = 1− Pr [Zj = 0]

Pr [Zj = 0] =∏xi∈Cj

(1− y∗i ) ·∏x i∈Cj

y∗i ≥

((k − `)−

∑xi∈Cj

y∗i +∑

x i∈Cjy∗i

k

)k

≥

((k −

∑xi∈Cj

y∗i −∑

x i∈Cj(1− y∗i )

k

)k

≥(

(k − z∗j )

k

)k

≥(

1−z∗jk

)k

E [Zj ] ≤ 1−(

1−z∗jk

)k

≥ z∗j

(1− 1

k

)k

≥ z∗j (1− 1/e)




Max Satisfiability: Relax+RRound approximation

Theorem

Relax+RRound is a e/(e − 1)-approximation for Max SAT.

Proof.

Let (y∗, z∗) be an optimal solution of LP Max SAT

Let Zj be the indicator r.v.a for clause Cj is satisfied.

Let W be the r.v. weight of satisfied clauses:W =

∑mj=1 wjZj .

E [W ] =∑m

j=1 wjE [Zj ] ≥ (1− 1/e)∑m

j=1 wjz∗j ≥ (1− 1/e)opt




Max Satisfiability:RandAssign

function RandAssign(F )for i=1,. . . , n do

Set xi = 1 with probability 1/2end forreturn (x)

end function

Theorem

RandAssign is a 2-approximation for Max SAT.

Proof.

E [W ] =∑m

j=1 wjE [Zj ] =∑m

j=1 wj

(1− (12)kj

)≥ 1

2

∑mj=1 wj ≥

12opt.

We move from r = 2 (RandAssign) to r = 1.581977(Relax+RRound).




Max Satisfiability:Best2

function Best2(F )x1,W1 = RandAssign(F )x2,W2 = Relax+RRound(F )if W1 ≥W2 then

return (x1)else

return (x2)end if

end function

Theorem

Best2 is a 4/3 (1.33333)-approximation for Max SAT.





Proof.

E [W ] = E [max{W1,W2}] ≥ E [(W1 + W2)/2}].

E [W ] ≥m∑j=1

wj

[1

2

(1−

(1

2

)kj)

+1

2z∗j

(1−

(1

kj

)kj)]

≥m∑j=1

wj3

4z∗j ≥

3

4

m∑j=1

wjz∗j ≥

3

4opt.





Proof.

Is

[12

(1−

(12

)kj)+ 12z∗j

(1−

(1kj

)kj)]≥ 3

4z∗j ?

kj = 1: 1212 + 1

2z∗j ≥34z∗j .

kj = 2: 1234 + 1

234z∗j ≥

34z∗j .

kj ≥ 3: the minimum possible of each term is

1

2

7

8+

1

2

(1− 1

e

)z∗j ≥

3

4z∗j


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