# 1 Approximation Algorithms

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1 CME 305: Dscrete Mathematcs and Algorthms 1 Approxmaton Algorthms In lght of the apparent ntractablty of the problems we beleve not to le n P, t makes sense to pursue deas other than complete solutons to these problems. Three standard approaches nclude: Explotng specal problem structure: perhaps we do not need to solve the general case of the problem but rather a tractable specal verson; Heurstcs: procedures that tend to gve reasonable estmates but for whch no proven guarantees exst; Approxmaton algorthms: procedures whch are proven to gve solutons wthn a factor of optmum. Of these approaches, approxmaton algorthms are arguably the most mathematcally satsfyng, and wll be the subject of dscusson for ths secton. An algorthm s a factor α approxmaton α-approxmaton algorthm) for a problem ff for every nstance of the problem t can fnd a soluton wthn a factor α of the optmum soluton. If the problem at hand s a mnmzaton then α > 1 and ths defnton mples that the soluton found by the algorthm s at most α tmes the optmum soluton. If the problem s a maxmzaton, α < 1 and ths defnton guarantees that the approxmate soluton s at least α tmes the optmum. 1.1 Mnmum Vertex Cover Gven a graph GV, E), the mnmum vertex cover problem s to fnd a subset S V wth mnmum cardnalty such that every edge n E has at least one endpont n S. Algorthm 1 2-Approxmaton Algorthm for Mnmum Vertex Cover Fnd a maxmal matchng M n G. Output the endponts of edges n M: S = e M e. Clam 1 The output of algorthm 1 s feasble. Proof: We prove ths by contradcton: suppose there exsts an edge e = u, v) such that u, v / S. Snce e does not share an endpont wth any of the vertces n M, M {e} s a larger matchng, whch contradcts M beng a maxmal matchng. Lemma 1 Algorthm 1 gves a 2-approxmaton for mnmum vertex cover regardless of the choce of M. Proof: The edges of M are ndependent; thus any feasble cover must take at least one vertex from every edge n M. Ths means that M OP T and then we have: M OP T S = 2 M 2OP T Ths technque of lower boundng the optmum s often key n provng approxmaton factors, as we are usually unable to compute the value of OPT.

2 2 CME 305: Dscrete Mathematcs and Algorthms - Lecture Mnmum Weght Vertex Cover Gven a graph GV, E), and a weght functon W : V R +, the mnmum weght vertex cover problem s to fnd a subset S V that covers all the edges and has mnmum weght W S) = v S wv). We may formulate ths problem as an nteger program as follows: assocate varable x v wth node v, and solve: mnmze: wv)x v s.t. v V x u + x v 1, u, v) E x v {0, 1} v V Solvng ths nteger program s equvalent to solvng mn weght vertex cover, an NP-complete problem. We nstead attempt to solve a smpler problem for whch polynomal-tme algorthms exsts. We modfy the second constrant to read: 0 x v 1 v V. Ths s known as a lnear programmng LP) relaxaton of the nteger program. It s well known that lnear programs can be solved n polynomal tme. Let {x v, v V } be the soluton of the LP. We need to convert ths fractonal soluton to an ntegral one; we use the followng roundng polcy: f x v < 0 then we set x v = 0, otherwse we set x v = 1. Note that { x v, v V } s a feasble soluton. Because {x v, v V } s a feasble soluton of the LP, x v + x u 1, and thus x v 1/2 or x u 1/2 whch mples that x v + x u 1. Lemma 2 Let S = {v V : x v = 1}. Then S gves a 2-approxmaton to mn weght vertex cover,.e. wv) 2wS ) where S s the optmum soluton. v S Proof: Snce the feasble regon of the IP s a subset of the feasble regon of the LP, the optmum of the LP s a lower bound for the optmum of the IP. Moreover, note that our roundng procedure ensures that x v 2x v for all v V, thus: OP T IP ws) = v V wv) x v 2 v V wv)x v = 2OP T LP 2OP T IP. 1.3 Job Schedulng Suppose we have n jobs each of whch take tme t to process and m dentcal machnes on whch to schedule ther completon. Jobs cannot be splt between machnes. For a gven schedulng, let A j be the set of jobs assgned to machne j. Let L j = A j t be the load of machne j. The mnmum makespan schedulng problem s to fnd an assgnment of jobs to machnes that mnmzes the makespan, defned as the maxmum load over all machnes.e. max j L j ). We consder the followng greedy algorthm for ths problem whch sorts the jobs so that t 1 t 2... t n, and teratvely allocates the next job to the machne wth the least load. We note that algorthm 2 need not return an optmal soluton. Consderng jobs of szes {3, 3, 2, 2, 2} to be assgned to two machnes we see that t s no better than a 7/6-approxmaton algorthm. We prove n these

3 CME 305: Dscrete Mathematcs and Algorthms - Lecture 10 3 Algorthm 2 Greedy Approxmaton Algorthm for Job Schedulng j, A j, T j 0 for = 1 to n do j argmn k T k A j = A j {} T j = T j + t end for notes that algorthm 2 has an approxmaton factor of no worse than 3/2; we leave as an exercse to the reader to prove that t s actually a 4/3-approxmaton algorthm. Let T denote the optmal makespan. The followng two facts are self-evdent: T max T 1 m t n t. Clam 2 The soluton of the greedy makespan algorthm s at most 1 m n t + max t Proof: Consder machne j wth maxmum load T j. Let be the last job scheduled on machne j. When was scheduled, j had the smallest load, so j must have had load smaller than the average load. Then, T j = T j t j ) + t j 1 m n t + max t. Ths clam shows mmedately that algorthm 2 s a 2-approxmaton algorthm. Slghtly more careful analyss proves α = 3/2. Lemma 3 The approxmaton factor of the greedy makespan algorthm s at most 3/2. Proof: If there are at most m jobs, the schedulng s optmal snce we put each job on ts own machne. If there are more than m jobs, then by the pgeonhole prncple at least one processor n the optmal schedulng must get 2 of the frst m + 1 jobs. Each of these jobs s at least as bg as t m+1. Thus, T 2t m+1. Take the output of algorthm 2 and consder machne j assgned maxmum load T j. Let be the last job assgned to j. We may assume > m or else the algorthm s output s optmal. Snce the jobs are sorted, t t m+1 T /2 and we have T j = T j t ) + t 1 m n t + t T + T /2 = 3 2 T. =1

4 4 CME 305: Dscrete Mathematcs and Algorthms - Lecture Non-Unform Job Schedulng Consder a more general verson of the mnmum makespan schedulng problem n whch job can have dfferent processng tme on dfferent machnes. Let t j be the tme t takes machne j to process job. We want to mnmze the makespan T = max j x jt j where the varable x j {0, 1} ndcates whether job s assgned to machne j. We may wrte ths as an nteger program wth constrants ensurng that each job s assgned to exactly one machne and that the load of each machne does not exceed T, the makespan. mnmze: s.t. T x j = 1 j x j t j T x j {0, 1} j To approxmate a soluton to ths nteger program, one mght frst try to relax varable x j and let x j [0, 1], however, the soluton of the correspondng LP may be too far from the soluton of the IP. Thus soluton of LP does not serve as a tght lower bound for OPT. For nstance, f we have only 1 job, m machnes, and t 1j = m for every machne j, then OPT= m but the soluton of the LP s 1 by assgnng x 1j = 1 m. The maxmum rato of the optmal IP and LP solutons s called the ntegralty gap. We need to defne the relaxed problem n such a way that the ntegralty gap s small. The dea s to ensure that f t j > T we assgn x j = 0. We do ths by defnng a seres of feasblty LPs wth a makespan parameter T as follows. x j = 1 j x j t j T j 0 x j 1, j x j = 0, j s.t. t j > T Usng bnary search we can obtan the smallest value of T, T, for whch the above feasblty lnear program FLP) has a soluton. We outlne a procedure for roundng the soluton of the FLP wth T = T. Let GJ, M, E) be a bpartte graph defned on the set of jobs and machnes where edge {, j} between job J and machne j M has weght x j. Our clam s that ths allocaton graph can be transformed nto a forest n such a way that the load of every machne remans the same or decreases. Suppose G = J, M, E) s not a tree, thus t has cycle c = j 1, 1, j 2, 2,..., j r, r, j 1 ; suppose we update x 1j 1 to x 1j 1 ɛ 1, we proceed around cycle c and update the weght of edges n the followng way: Snce the total weght of edges ncdent to 1 must add up to one, f we decrease x 1j 1 by ɛ 1, we need to ncrease x 1j 2 by ɛ 1, thus we update x 1j 2 to x 1j 2 + ɛ 1. Now to keep the load of machne 2 less than or equal to T, we decrease x 2j 2 by ɛ 2 = t 1 j 2 t 2 ɛ 1, repeatng ths procedure, we modfy the weghts keepng the j 2 constrants satsfed. The only constrant that may become unsatsfed s the load of machne j 1 ; we decrease the load of j 1 by ɛ 1 va edge { 1, j 1 } and at the end, we may ncrease the load by t 1 j 2 t 2 j 3 t 2 j 2 t 3... t r 1 jr j 3 t r jr ɛ. If t 1 j 2 t 2 j 3 t 2 j 2 t 3... t r 1 jr j 3 t r jr > 1 we use the smple observaton that f we start from {j 1, r } and go around the cycle ) t1 j n that drecton, then we would need to ncrease the weght of { 1, j 1 } by 2 t 1 2 j 3 t 2 j 2 t 3... t r 1 jr j 3 t r jr < 1 thus the total load of 1 would become less that T.

5 CME 305: Dscrete Mathematcs and Algorthms - Lecture 10 5 Usng the above scheme, we are able to decrease the weght of { 1, j 1 } by ɛ keepng the soluton feasble. Repeatng ths reducton, we can make the weght of one of the edges zero, thus we can remove cycle c. We obtan our forest by breakng all the cycles. Suppose now that we have a soluton of the FLP wth T = T whose allocaton graph s a forest, call t x. Note that f job J s a leaf of the tree wth parent j M, then x j = 1, thus there s no j leaf wth fractonal weght. However, we can have fractonal edge {, j} between leaf j M and ts parent J. Suppose has k leaves j 1, j 2,..., j k, we choose one of the leaf machnes unformly at random and assgn job to t wth probablty x j. Then we remove from the graph. We repeat ths procedure of assgnng fractonal load of a parent to one of ts chldren and removng the job from the graph untl all the jobs are assgned. Clam 3 The above roundng procedure produces a factor 2 approxmaton. Proof: Let OP T be the makespan of the optmal assgnment and let T be the mnmum value of T found usng bnary search on FLP. Then, T OP T snce the FLP s clearly feasble usng the optmal assgnment. x s a feasble soluton therefore load of each machne s at most T. Durng the roundng procedure, we add the load of at most one job to each machne because a node can only have one parent n the forest G. Any machne j s load before ths addton s J x jt j T, so the new load of machne j s less than 2T. Hence, the fnal makespan s at most 2T. 1.5 Maxmum Satsfablty We return to the settng of boolean formulas and consder a problem related to satsfablty: for a gven formula n CNF, what s the maxmum number of ts clauses that can be satsfed by assgnng truth values to ts varables? More concretely, suppose we have n varables x 1,..., x n and m clauses C 1,..., C m where C = x x. S + S The maxmum satsfablty problem s to fnd the maxmum number of clauses that may be satsfed by an assgnment x. We frst propose a smple randomzed algorthm to approxmate a soluton to ths problem. Set each x ndependently to be 0 or 1 wth probablty 1/2. Then the probablty that any C s satsfed s 1 2 C. If we let Z denote the event that clause C s satsfed by ths random assgnment and Z = m =1 Z be the total number of satsfed clauses, we may compute: m m E[Z] = E[Z ] = 1 2 C ). =1 In the case that all of our clauses are large,.e. C K for each, then ths randomzed algorthm has an approxmaton rato of 1 2 K n expectaton: =1 m1 2 K ) E[Z] OP T m. An expected approxmaton rato s sometmes undesrable because t does not shed much lght on the probablty that the randomzed algorthm returns a good soluton. Concentraton nequaltes can help us estmate such probabltes, but n some cases we may do even better. Ths algorthm may be derandomzed usng condtonal expectaton as follows.

6 6 CME 305: Dscrete Mathematcs and Algorthms - Lecture 10 Algorthm 3 Derandomzed Approxmaton Algorthm for Maxmum Satsfablty for = 1 to n do Compute 1 2 E[Z x = 1, x 1,..., x 1 ] and 1 2 E[Z x = 0, x 1,..., x 1 ]. Set x = 1 f the frst expresson s larger than the second, set x = 0 otherwse. end for return x For motvaton, we consder the frst step of the algorthm. Note that E[Z] = 1 2 E[Z x 1 = 1] E[Z x 1 = 0]. Both terms on the rght hand sde may be computed smply by summng over all clauses the probablty that C s satsfed gven the nformaton on x 1. The equaton above mples that E[Z x 1 = 1] E[Z] or E[Z x 1 = 0] E[Z]. Thus f we choose the greater expectaton n each step of the algorthm, we wll determnstcally buld up an assgnment x such that E[Z x] E[Z] m1 2 K ) where E[Z x] s no longer an expectaton, but merely an evaluaton. The approxmaton rato of algorthm 3 depends on all of the clauses havng K or more varables. We present a dfferent algorthm for dealng wth the possblty that some of the clauses may be small. It s based on the now famlar concept of LP relaxaton. We wrte down an nteger program for maxmum satsfablty. maxmze: s.t. m =1 q q y j + 1 y j ) j S + q, y j {0, 1} j S, j The varables q correspond to the truth value of each clause C, and the varables y j correspond to the values of each boolean varable x. We relax the last condton to be 0 q, y j 1 n order to get a lnear program. Algorthm 4 LP Relaxaton Algorthm for Maxmum Satsfablty Solve the LP gven above. for j = 1 to n do Independently set x j = { 1 : wth probablty y j 0 : wth probablty 1 y j end for In order to analyze algorthm 4 we consder the probablty that a partcular clause s satsfed; WLOG we

7 CME 305: Dscrete Mathematcs and Algorthms - Lecture 10 7 consder C 1 = x 1... x k. We have q1 = mn{y yk, 1} and by the AM-GM nequalty: P r[c 1 ] = 1 k 1 yj ) j=1 1 1 k k 1 yj ) j=1 ) k 1 1 q 1 k q 1 1 q 1 1 1/e) 1 1 k Ths last lne mples, through the lnearty of expectaton, that ths roundng procedure gves a 1 1/e)- approxmaton for maxmum satsfablty regardless of the sze of the smallest clause. Algorthm 4 may also be derandomzed by the method of condtonal expectatons. These two derandomzed algorthms may be combned to gve a factor 3/4 approxmaton algorthm for maxmum satsfablty. We smply run both algorthms on a gven problem nstance and output the better of the two assgnments. Ths procedure tself may be vewed as a derandomzaton of an algorthm that flps a far con to decde whch randomzed sub-algorthm to run. That algorthm has approxmaton factor 3/4: m m 1 E[Z] = E[Z ] 1 2 C ) =1 =1 ) k ) k 1 1 ) )) C 3/4)m. C

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