SURVEY OF APPROXIMATION ALGORITHMS FOR SET COVER PROBLEM. Himanshu Shekhar Dutta. Thesis Prepared for the Degree of MASTER OF SCIENCE

SURVEY OF APPROXIMATION ALGORITHMS FOR SET COVER PROBLEM Hmanshu Shekha Dutta Thess Pepaed fo the Degee of MASTER OF SCIENCE UNIVERSITY OF NORTH TEXAS Decembe 2009 APPROVED: Fahad Shahokh, Mao Pofesso Jose Peez, Commttee Membe Yan Huang, Commttee Membe Ian Pabey, Cha of the Depatment Compute Scence and Engneeng Costas Tsatsouls, Dean of the College of Engneeng Mchael Montcno, Dean of the Robet B. Toulouse School of Gaduate Studes

Dutta, H manshu S hekha. Suvey o f A ppoxmaton Algothms fo S et C ove Poblem. Maste of Scence (Compute Scence), Decembe 2009, 76 pages, 10 fgues, 1 table, efeences, 29 ttles. In t hs t hess, I s uvey 11 a ppoxmaton algothms fo u nweghted s et cove poblem. I have also mplemented the thee algothms and ceated a softwae lbay that stoes the code I have wtten. The a lgothms I s uvey a e: 1. J ohnson s s tandad g eedy 2. f-fequency geedy 3. Goldsmdt, Hochbaum and Yu s modfed geedy 4. Halldosson s local optmzaton 5. Du and Fue sem local optmzaton 6. Asaf Levn s mpovement to Du and Fue 7. S mple oundng 8. R andomzed oundng 9. L P dua lty 10. P mal-dual schema 11. Netwok flow technque. Most of the algothms suveyed ae efnements of standad geedy algothm. Tll 196 now the best known algothm has a pefomance ato of Hk ( ). 390

Copyght 2009 by Hmanshu Shekha Dutta

ACKNOWLEDGEMENTS My sncee thanks and egads to my advso D. Fahad Shahokh fo hs constant suppot and gudance.

TABLE OF CONTENTS ACKNOWLEDGEMENTS LIST OF FIGURES...v LIST OF TABLES...v CHAPTER 1 INTRODUCTION.1 1.1 Motvaton....1 1.2 Pelmnaes....3 2 APPROXIMATION ALGORITHMS BASED ON STANDARD GREEDY...8 2.1 Johnson s Standad Geedy Algothm fo Set Cove.....8 2.1.1 Standad Geedy Algothm...9 2.1.2 Standad Geedy Algothm s Rato Bound....11 2.1.3 Analyss of the Standad Geedy Algothm....13 2.1.4 A Tght Analyss of the Standad Geedy Algothm by Pet Slavk..14 2.2 Httng Set..16 2.2.1 Algothm fo Solvng Httng Set Poblem.....16 2.3 f-appoxmaton Geedy Algothm fo Set Cove....17 2.3.1 Geedy Algothm fo f-appoxmaton...18 2.4 Modfed Geedy (MG) Algothm by Goldschmdt et al.. 21 2.4.1 MG (Modfed Geedy) Algothm.. 22 2.5 Set Cove va Local Impovement by Halldosson.... 32 v

2.5.1 The Local Impovement Famewok...33 2.5.2 Appoxmate k-set Cove Usng Set Packng fo (S, C, k).34 2.5.2.1 AppoxSetCove(S, C, k) Algothm...35 2.5.3 Halldosson s Dect Applcaton of Local Impovement...37 2.5.3.1 Algothm fo 3-SC Poblem...38 2.5.3.2 Algothm fo k-sc Poblem...40 2.6 Duh and Fue Algothm Usng Sem Local Optmzaton....43 2.6.1 3-Set Cove Usng Sem Local Optmzaton..43 2.6.2 k-set Cove Usng Sem Local Optmzaton Algothm.44 2.7 Modfed Duh and Fue Algothm by Asaf Levn....45 2.7.1 Asaf Levn s Algothm...45 3 LINEAR PROGRAMMING TECHNIQUE FOR SET COVER PROBLEM.47 3.1 Lnea Pogammng Intoducton.....47 3.2 Set Cove Usng LP... 51 3.2.1 Set Cove va Roundng...52 3.2.2 Set Cove va LP Dualty. 54 3.2.3 Pmal-Dual Schema....56 3.2.3.1 Set Cove va Pmal-Dual Schema...57 4 SET COVER SOLUTION USING NETWORK FLOW APPROACH...60 4.1 Tansfomaton of Set Cove nto Flow Poblem..60 4.2 Mnmum Flow Algothm fo Set Cove Poblem.... 64 4.3 Pefomance Analyss of Mnmum Flow Algothm...68 v

5 CONCLUSION...70 REFERENCES..73 v

LIST OF FIGURES 1.1 Set Cove n Geomety. 2 1.2 Httng Set n Geomety...3 1.3 Example Illustatng Edge Cove and Maxmum Matchng....6 1.4 Example Illustatng Vetex Cove..7 2.1 Gaph Constucted fo Example 2.4...30 4.1 Bpatte gaph B fo Set Cove Poblem n Example 4.1.63 4.2 Layeed Netwok N fo Bpatte Gaph n Fg-4.1...63 4.3 Netwok Constucted fo Example 4.1..67 4.4 Suvvng Netwok Afte Deletng Vetex S 2 and Its Neghbos.67 4.5 Recalculated Flow Values fo Suvvng Netwok....68 v

LIST OF TABLES 5.1 Pefomance Rato fo Dffeent Set Cove Algothms 72 v

CHAPTER 1 INTRODUCTION 1.1 Motvaton Set cove s an optmzaton poblem whch seves as a model to many eal wold poblems. Its applcaton ncludes weless netwoks, semconducto ndusty, flexble manufactung, schedulng, outng etc.. One of the mao applcaton of set cove poblem s n weless netwok [1,2]. Take an example, whee a set of custome locatons have been povded. Assumng unt dsk s geometc modelng of coveage aea of base statons. The am s to fnd the best locatons fo placng base statons, so that all the customes ae sevced. Snce placng base staton also costs money, attempt s made to mnmze the numbe of base statons, ensung all the customes ae sevced by the mnmum collecton of base staton. Ideally, each base staton should povde sevce to as many customes as possble. Ths s essentally a set cove poblem n geometc settngs. Hee n custome locatons coespond to n ponts of the plane o elements of a unvesal set. m n 2 possble base staton o dsk placements n the plane coesponds to a famly of m sets [1,2]. The am s to fnd mnmum numbe of dsks (o sets) n famly of possble m dsks (o sets), so that all the custome locatons (o all the elements of unvesal set) ae coveed. 1

In the example shown n Fg-1.1, thee ae 22 ponts n the plane. The am s to cove all these ponts wth mnmum numbe of dsks. Consde applyng a smple geedy heustcs whch at each teaton selects a dsk that coves maxmum numbe of uncoveed ponts. Ths algothm selects the 3 dsks shown n the boldface. At teaton-1, teaton-2 and teaton-3 t selected dsks that cove 12, 6, 4 ponts espectvely. Ths s not an optmal soluton. The optmal soluton s shown by two dashed dsks. Each of the dashed dsks coves 11 ponts each. Fg-1.1 Set Cove n Geomety Fg-1.2 shows anothe applcaton of set cove poblem. Hee a set of unt ectangles ae gven and the am s to ht all the ectangles wth mnmum numbe of ponts. Ths poblem s elevant n settng up emegency facltes so that all the potental customes ae wthn the coveage aea of emegency faclty [2]. Unt ectangles epesent a set of customes (o ctes) whch need emegency facltes. Ths poblem s 2

known as httng set poblem whch can be solved by tansfomng the httng set poblem to set cove poblem and then applyng any of the avalable algothms fo set cove poblem. Clealy 4 such facltes should be enough fo ths example. Fg-1.2 Httng Set n Geomety 1.2 Pelmnaes Many combnatoal optmzaton poblems ae NP-had,.e., t s hghly unlkely that a polynomal-tme algothm exsts fo solvng the poblem, unless P = NP [3,4]. Hence when one faces solvng an NP-had poblem one should focus on fndng nea optmal soluton n polynomal tme [3,4,5,6]. Appoxmaton algothm s an algothm that etuns nea optmal soluton [3]. Fo maxmzaton poblems, the pefomance ato of an appoxmate algothm s the wost case ato of the sze of the optmal soluton to the sze of the appoxmate soluton. Fo mnmzaton poblems, pefomance ato s the nvese of ths ato,.e., wost-case ato of the sze of the appoxmate soluton to the sze of the optmal soluton. 3

In set cove poblem, a unvesal set and collecton of subsets ae povded, such that each element of unvesal set s ncdent to at least one of the subsets [8,9,10,11,12]. The am s to fnd mnmum numbe of subsets, whose unon s unvesal set. Bascally, attempt s to cove all the elements of the unvesal set. Fo example, a company eques at least a pogamme, to cove ts pogammng need of dffeent languages, example C, C++, JAVA, SQL etc.. Thee ae people who know multple languages. The am s to fnd mnmum numbe of people so that fo evey language thee s at least one peson who knows t,.e., fnd mnmum numbe of people so that all the languages ae coveed. Moe pecsely, Let U, U = m be a unvesal set and a famly of subsets S 2 U A sub collecton C (C S) coves the unvesal set U, f U C S C Mnmum set cove s sub collecton C of mnmum cadnalty. Example-1.1: Fnd the set cove fo followng collecton of sets. U = {1, 2, 3, 4} S 1 = {1, 2} S 2 = {2, 3} S 3 = {3, 4} S 4 = {4, 1} 4

S 1 and S 3 fom the mnmum set cove of sze 2. Mnmum set cove s not unque as S 2 and S 4, too, fom the set cove of sze 2. Dual of set cove poblem s known as httng set poblem [12]. In httng set poblem, a famly of sets S, S = m ae gven. The am s to ceate a subset H such that H has at least one element n each of the sets,.e., evey set has non-empty ntesecton wth H. Mnmum httng set s ceatng a set wth mnmum cadnalty. H s U S Fo collecton of sets povded n Example-1.1, esultng httng set would be: H = {1, 3} Ths s non-unque, as {2, 4} s anothe httng set of sze 2. Set packng s a sub collecton P ' ( P' S), whee all the membes of ths sub collecton ae mutually dsont [12]. Maxmum set packng s the sub collecton of maxmum cadnalty. In the above example, both S 1 and S 3 o S 2 and S 4, fom the maxmum set packng of sze 2. An nteestng concept elated to set cove s that of an edge cove [7]. Edge cove of an undected gaph G = (V,E) s a subset E ' E of edges such that evey vetex of V s ncdent to some edge of E '. Mnmum edge cove s one wth mnmum cadnalty. It s mpotant to note that mnmum edge cove poblem s specal nstance of the set cove poblem, whee each subset has maxmum cadnalty of two. In fact some of the algothms suveyed n ths thess use the concept of edge cove togethe wth geedy appoach, to get bette pefomance guaantee. 5

In the Fg-1.3, edges ab and cd fom the mnmum edge cove, of cadnalty 2, whee G has a path length of 4. b d a c Fg-1.3 Example Illustatng Edge Cove and Maxmum Matchng A matchng n a gaph G s a set of non loop edges wth no shaed ponts [7]. Maxmal matchng s a matchng whch can not be extended by addng any edge. Maxmum matchng s a matchng of maxmum cadnalty. Pefect matchng coves all the nodes of the gaph [assumpton - thee ae even numbe of vetces]. In the Fg-1.3, edge bc fom the maxmal matchng. If ths edge s selected n the matchng set, then no othe edge can be added. But clealy ths s not the maxmum matchng. Edges ab and cd, fom the maxmum matchng of cadnalty 2. Incdentally ths s also the edge cove and theefoe t also epesents pefect matchng. Maxmum matchng s specal nstance of the set packng poblem, whee each subset has maxmum cadnalty of two. Fo a gaph G wth n vetces havng no solated vetex, mnmum edge cove and maxmum matchng, sum to the total numbe of vetces. 6

In a gven undected gaph G = (V, E) a vetex cove s a subset V ' V such that fo evey edge uv E, at least one of the end pont o vetex belongs to V '. The mnmum vetex cove s subset V ' of mnmum cadnalty [3]. In the Fg-1.4, vetex a and d fom the mnmum vetex cove. All the edges ae ncdent to these 2 vetces. Fg-1.4 Example Illustatng Vetex Cove Mnmum vetex cove poblem s a specal nstance of the dual set cove poblem, whee each subset has maxmum cadnalty of two. 7

CHAPTER 2 APPROXIMATION ALGORITHMS BASED ON STANDARD GREEDY As descbed eale, n set cove poblem, a unvesal set and collecton of subsets ae povded, such that each element of unvesal set s ncdent to at least one of the subsets [8,9,10,11,12]. The am s to fnd mnmum numbe of subsets, whose unon s unvesal set. Bascally, the attempt s to cove all the elements of the unvesal set. Set cove poblem has pactcal applcatons n many aeas, such as logc desgn, sem conducto ndusty, fault testng etc.. of subsets Moe pecsely, let U, U = m be the gound set (o unvesal set) and S be a famly S 2 U. A sub collecton C (C S) coves the unvesal set U, f U C S C In ths chapte, seveal appoxmate solutons to set cove poblem wll be evewed whch ae based on Johnson s standad geedy algothm. 2.1 Johnson s Standad Geedy Algothm fo Set Cove [8] Geedy appoxmaton of SC was ognally pesented by Johnson n hs pape, Appoxmaton Algothms fo Combnatoal Poblems. The standad geedy algothm fo set cove woks by pckng a subset that coves the most numbe of emanng uncoveed elements, at each stage of algothm. 8

2.1.1 Standad Geedy Algothm [8] 1. Set C =, UNCOV = U. SET() = S, 1 N. 2. If UNCOV =, halt and etun C. 3. Choose N such that SET() s maxmzed. 4. Set C = C { S }, UNCOV = UNCOV - SET(), SET() = SET() - SET(), 1 N. 5. Go to 2. Example-2.1: Ths example fnds the set cove fo followng collecton of sets, usng standad geedy algothm. S 1 = {1, 2, 3, 4, 5, 6} S 2 = {5, 6, 8, 9} S 3 = {1, 4, 7, 10} S 4 = {2, 5, 7, 8, 11} S 5 = {3, 6, 9, 12} S 6 = {10, 11} U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} ITERATION I: C = 9

UNCOV = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} UNCOV, theefoe contnue Fo = 1, SET(1) = 6 has maxmum cadnalty C = {S 1 } UNCOV = {7, 8, 9, 10, 11, 12} S 2 = {8, 9} S 3 = {7, 10} S 4 = {7, 8, 11} S 5 = {9, 12} S 6 = {10, 11} ITERATION II: UNCOV, theefoe contnue Fo = 4, SET(4) = 3 has maxmum cadnalty C = {S 1, S 4 } UNCOV = {9, 10, 12} S 2 = {9} S 3 = {10} S 5 = {9, 12} S 6 = {10} 10

ITERATION III: UNCOV, theefoe contnue Fo = 5, SET(5) = 2 has maxmum cadnalty C = {S 1, S 4, S 5 } UNCOV = {10} S 3 = {10} S 6 = {10} ITERATION IV: UNCOV, theefoe contnue Fo = 3, SET(3) = 1 has maxmum cadnalty C = {S 1, S 4, S 5, S 6 } UNCOV = ITERATION V: UNCOV, theefoe stop. C = {S 1, S 4, S 5, S 6 } Theefoe, Johnson s standad geedy algothm poduces a set cove {S 1, S 4, S 5, S 6 } of sze 4. Ths s not the optmum set cove. Caeful analyss shows that optmum set cove s {S 3, S 4, S 5 }. 2.1.2 Standad Geedy Algothm s Rato Bound [3] The complexty of standad geedy algothm descbed above s of O( mn ). But, by 11

mantanng a poty queue, standad geedy algothm can be mplemented to un n tme O ( S ). S S The standad geedy algothm has a ato bound of Hk ( ) = k 1 1/, whee k s the sze of the lagest set. Hks ( ) also known as hamonc sees. Lemma 1 [3]: Johnson s standad geedy set cove has a ato bound Hk ( ), whee k s the sze of the lagest set. Poof [3]: Assgn the cost to each subset selected. Fo unweghted set cove poblem the cost of selectng a subset s 1. Let C denote the sze of an optmal set cove C, C the sze of the set cove C etuned by the standad geedy algothm and S the th subset selected. When algothm adds S to set cove, t ncus a cost of 1. Ths cost of selectng S s evenly spead among the elements coveed fo fst tme by S. Let cu denote the cost/weght allocated to element u, fo each element u U. Each element s assgned a cost only fo the fst tme t s coveed. If u s coveed fo the fst tme by S then c u 1 S ( S S... S ) 1 2 1 The algothm fnds a set cove soluton C whch has a total cost C as each set ncus a cost of 1. Snce the optmal cove C also coves U. Theefoe, C = uu c u 12

* S C u S c (2.1) u Also, fo any subset S belongng to the famly of subsets S u S c H ( S ) (2.2) u Fo nequaltes (2.1) and (2.2), t follows C H( S ) * S C * C H(max( S ) * = C H ( k ) (2.3) Hee k s the sze of lagest set,.e., max( S ). Rato bound fo standad geedy set cove s O(1) ln k. Ths s tue fo nonweghted as well as weghted set cove. Ths can be poved usng the fact that hamonc sees has ato bound O(1) ln k. The above esult mples standad geedy algothm soluton s not too lage than optmal soluton. If the maxmum set sze s 2 then ths ato bound s = 1 + ½ = 3/2. Fo maxmum set sze of 3, ths ato bound becomes 1 + 1/2 + 1/3 = 11/6. 2.1.3 Analyss of the Standad Geedy Algothm Lund and Yanakaks [13] establshed that set cove poblem cannot be appoxmated wth ato c log 2 m fo any c < l/4 unless all NP poblems ae solvable n DTIME m poly log m ( ). The Lund and Yanakaks esult, bascally ules out any dastc 13

mpovement to standad geedy algothm. Theefoe, mao effots have been made to apply mpovements to geedy algothm athe developng a new algothm fom scatch. Even fo maxmum set sze of 3, set cove poblem s NP-had. Ths can be poved usng the fact that when the edges of a gaph ae coveed by tangles the poblem becomes a set cove poblem [17]. Hee each tangle epesents a set. Coveng the edges of a gaph by tangles s NP-had poblem. Fo the tangle coveng poblem, the standad geedy algothm s pefomance ato s 11/6. Standad geedy set cove algothm can be used to solve vetex cove poblem fo a gaph of degee of 3 at most [3]. As mentoned above, the soluton found by standad geedy set cove algothm s bound by H(3) = 1 + 1/2 +1/3 = 11/6 tmes the optmal soluton. Ths appoxmaton ato s bette than that of appoxmate vetex cove. 2.1.4 A Tght Analyss of the Standad Geedy Algothm by Pet Slavk [15] Johnson showed that the pefomance ato fo set cove standad geedy algothm s Hm ( ), whch les between ln m and ln m + 1. Fo 20 yeas ths emaned the bound fo standad geedy set cove algothm. Pet Slavk n 1996 [15] poved that the pefomance ato of standad geedy algothm s exactly ln m ln ln m + (1). Hee m s the sze of gound set o unvese. Also the lowe and the uppe bound dffe only by less than 1.1. Slavk s analyss was fst tght analyss of standad geedy algothm. Fo a functon gong to nfnty wth m, hs analyss povded the fst uppe bound fo standad geedy algothm that les below H(m). Slavk used the hadness esults shown by Fege. Fegh poved a vey stong esult 14

showng that fo any > 0, set cove cannot be appoxmated wthn (1 - ) ln m by any polynomal tme algothm, unless NPTIME m O(loglog m) [ ]. Theefoe, fo any polynomal tme algothm maxmum achevable mpovement on pefomance ato Hm ( ), s at most a functon f(m) = o(ln m). All the appoxmatons tll ths pont wee geneally based on assumpton of some po knowledge of m and c. mn Then vaous algothms ae used to obtan the bounds on c geedy. But Pet Slavk stated wth c mn and c geedy, and obtaned bounds on m. Slavk s analyss showed that fo any set U wth U = m 2 c geedy c mn (ln m ln c mn + 1) Also c geedy ( c mn - ½) (ln m ln c mn + 1) + c mn Above two esults gve uppe and lowe bound fo standad geedy algothm. The uppe bound comes as: c geedy c mn < ln m ln ln m + 0.78 and The lowe bound c geedy c mn > ln m ln ln m - 0.31 It can be clealy seen that uppe and lowe bound dffe by only 1.11. The uppe and lowe bound povded above by Pet Slavk ae tght up to a constant. 15

2.2 Httng Set Dual of set cove poblem s httng set poblem. In httng set poblem, a famly of sets S, S = m ae gven. The am s to ceate a subset H such that H has at least one element n each of the sets,.e., evey set has non-empty ntesecton wth H. Mnmum httng set s ceatng a set wth mnmum cadnalty [12]. H s U S Httng set poblem can be conveted nto set cove poblem. To acheve that, all the elements of U ae eplaced by names of subset that contan t. Now ths s set cove poblem wth oles of U and S evesed. 2.2.1 Algothm fo Solvng Httng Set Poblem 1. Replace all the elements of U by names of subset that contan t. Ths foms the famly of subsets. 2. Famly of subsets ognally povded, now fom the unvesal set U. 3. Apply any set cove algothm, of new set cove poblem fomed by evesng oles of U and S. The set cove found s the httng set o dual of set cove. Example-2.2: Bulds httng set, fo followng collecton of subsets and unvesal set. S 1 = {1, 2} S 2 = {2, 3} S 3 = {3, 4} 16

S 4 = {4, 1} U = {1, 2, 3, 4} Replace the elements of U by names of subset that contan t. 1 = {S 1, S 4 } 2 = {S 1, S 2 } 3 = {S 2, S 3 } 4 = {S 3, S 4 } U = {S 1, S 2, S 3, S 4 } Now the httng set poblem has been tansfomed nto set cove poblem. Hee 1, 2, 3 and 4 fom the famly of subsets and the famly of subsets ognally povded, fom the unvesal set. Clealy 1 and 3 s the soluton set fo new set cove poblem as they cove the unvesal set U. It can be easly vefed that 1 and 3 s also the mnmum httng set soluton, fo ognal httng set poblem. Ths s non-unque as 2 and 4, too, fom the httng set of cadnalty 2. 2.3 f-appoxmaton Geedy Algothm fo Set Cove [16] f-appoxmaton geedy algothm povdes a soluton whch s bounded by the maxmum numbe of sets that an element belong. If each element occus n at most f sets, then ths algothm fnds a soluton whch has a pefomance ato of f. Ths algothm s vey useful n low fequency systems. 17

Let s assume [16], Unvesal set U = { u 1, u 2,..., u m }. Subsets S1, S2,..., Sn U. The cost c 0 fo each set S. The cost functon s 1 fo unweghted set cove. The am s to fnd I {1,..., n } that mnmzes subect to I S Ic U. In othe wods, the goal s to select the mnmum cost collecton of sets that cove the unvesal set. Fequency of an element n a set cove nstance, s defned as the numbe of sets t s n,.e., f max { : u S }. 2.3.1 Geedy Algothm fo f-appoxmaton [16] Intalze I Whle U Pck u wth maxmum fequency. u U I I { : u S } U U S I Example-2.3: Ths example fnds the set cove fo collecton of sets n Fg-2.1, usng f-appoxmaton geedy algothm. S 1 = {1, 2, 3, 4, 5, 6} S 2 = {5, 6, 8, 9} 18

S 3 = {1, 4, 7, 10} S 4 = {2, 5, 7, 8, 11} S 5 = {3, 6, 9, 12} S 6 = {10, 11} U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} ITERATION I: f 5 = 3 s the maxmum fequency element. Theefoe, coespondng set n soluton s S 1. Hence, C = {S 1 } UNCOV = {7, 8, 9, 10, 11, 12} S 2 = {8, 9} S 3 = {7, 10} S 4 = {7, 8, 11} S 5 = {9, 12} S 6 = {10, 11} ITERATION II: f 8 = 2 s the maxmum fequency, n the emanng elements. The coespondng set n soluton s S 2. 19

Hence, C = {S 1, S 2 } UNCOV = {7, 10, 11, 12} S 3 = {7, 10} S 4 = {7, 11} S 5 = {12} S 6 = {10, 11} ITERATION III: f 7 = 2 s the maxmum fequency, n the emanng elements. The coespondng set n soluton s S 3. Hence, C = {S 1, S 2, S 3 } UNCOV = {11, 12} S 4 = {11} S 5 = {12} S 6 = {11} ITERATION IV: f 11 = 1 s the maxmum fequency, n the emanng elements. Theefoe, coespondng set n soluton s S 4. 20

Hence, C = {S 1, S 3, S 2, S 4 } UNCOV = {12} S 5 = {12} ITERATION V: 12 s the only uncoveed element. Theefoe, set cove soluton must nclude S 5. Hence, C = {S 1, S 3, S 2, S 4, S 5 } Theefoe, the fnal set cove s, C = {S 1, S 3, S 2, S 4, S 5 } 2.4 Modfed Geedy (MG) Algothm by Goldschmdt et al. [17] Olve Goldschmdt, Dot S. Hochbaum and Gang Yu poposed a soluton wth ato bound of (ln k 5 / 6), whee k s the maxmum set sze. Ths algothm woks by applyng the combnaton of moe than one algothm, appoxmate and exact, to obtan mpovement ove standad geedy algothm. The fnal set cove soluton s the unon of the outputs fo dffeent algothms appled. Ths algothm uses the fact that optmal soluton fo set cove poblem can be obtaned n polynomal tme, fo maxmum set sze of 2. Fo maxmum set sze 2, the set cove poblem educes to edge cove poblem. 21

Fo modfed geedy algothm, Goldschmdt et al. used followng temnology: Gven a famly { S 1, S 2,..., S n } of fnte sets, defne J = {1,2,.,n} and I J S. J J s called a cove f I S = I. They assumed that no set s embedded n J anothe set,.e., fo all, J and. If ths assumpton s not tue then t eques a pe-pocessng step, to emove all the nstances of sets beng subsets of othes. A k-set element s called set of sze k. k S denotes the elements stll uncoveed afte k teaton of the algothm. Index set fo the fst k sets added to the set cove s denoted k J and the k elements stll uncoveed afte k ounds s denoted by I. The modfed geedy algothm executes the Johnson s standad geedy algothm, tll the maxmum set sze becomes 2. Once the maxmum set sze becomes 2, t fnds the optmal cove and gets the mpovement ove standad geedy algothm. 2.4.1 MG (Modfed Geedy) Algothm [17] Step 0: Set 0 J = ; S = S, 0 0 0 J; I = U S I; k 0. Step 1: k = k + 1. Select k = agmax k 1 J S. Step 2: If S k 1 k 3, set k k 1 J J k and S S \ S, J, k k 1 k 1 k I I \ S and go to step 1. Else, contnue. k k 1 k 1 k k Step 3: If I, stop the output cove * k J J, othewse, fom the undected gaph G(V,E) wth V= U J k k k k S I and : 2, E S S J. Afte that, fnd the matchng of maxmum cadnalty on gaph G. Theeafte, nclude sets that coespond to the 22

matchng edges of G to J k. Exclude the elements coveed by the edges of the matchng fom S, k J. Include the emanng sngle element set n k J. End pocessng and etun J * J k as set cove. Bascally, algothm apples the standad geedy algothm tll maxmum set sze becomes 2. At maxmum set sze of 2, the poblem s equvalent to the edge cove poblem. Edge cove poblem can be solved n polynomal tme. Ths can be done by fndng maxmum matchng. Then the algothm adds all the solated vetces (sngle element set) not coveed by maxmum matchng. Note: Fo a gaph G wth n vetces havng no solated vetex, mnmum edge cove and maxmum matchng, sum to the total numbe of vetces. Lemma 2 [17]: Fo set sze 2 o less, MG algothm povdes an optmal set cove. Poof [17]: If S 2, J, step 1 and 2 can be gnoed and algothm can dectly poceed to step 3. It can be easly obseved that to obtan set cove each 1-element set needs to be ncluded n the output. Theefoe, algothm eques only mnmum numbe of 2-element sets fo coveng emanng elements. Bascally, the am s to solve the set cove poblem wth exact 2 elements pe set. Ths poblem s same as solvng edge cove poblem fo gaph. Edge cove of an undected gaph G = (V,E) s a subset E ' E of edges such that evey vetex of V s ncdent to some edge of E '. Mnmum edge cove s edge cove wth mnmum cadnalty. 23

Fo modfed geedy, a famly of fnte sets { S 1, S 2,..., S n } = E and I = V ae gven. Each set s of sze 2 snce the edges ae fomed by onng the end ponts. Thee ae known polynomal tme algothms to fnd soluton of edge cove poblem. These algothms wok by fndng maxmum matchng and set of adacent nodes uncoveed by matchng [18]. Let EC denote the set of edges fo optmal edge cove and M ' denote the set of edges n the maxmum matchng fo the gaph fomed by EC, G(EC) = (V, EC). Clealy, M ' M, as M s maxmum matchng. The gaph G(EC) also satsfes followng elatonshp: V = 2 M ' + EC - M ' Hence EC = V - M '. Edge cove EC(M), can be obtaned fom maxmum matchng by addng the nodes (o vetces) that ae not yet coveed. Ths cove, too, satsfes the elatonshp V = 2 M + EC(M) - M Hence EC(M) = V - M. Ths mples EC EC(M) = M - M ' 0, whch s possble only f EC = EC(M), as EC s mnmum edge cove. The untme of modfed geedy algothm s domnated by the tme spent fo solvng the maxmum matchng poblem. The maxmum matchng poblem n a gaph G = (V, E) can be solved n unnng tme of O( E V ). In a gaph of ode n, maxmum matchng can be found n 2.5 On ( )[19,20,21]. 24

Lemma 3 [17]: Let J * be a cove found by MG and let J be any othe cove,.e., Let S S S I J * '. J J ' n S, J, then * ( ), ' J J Q n (2.4) whee, 0 n 0 Q ( n ) 1 n 1 n 1 1/ 1/ 6 n 2 Poof [17]: The above can be poved by nducton on lagest sze set. When maxmum set sze s 1 o 2, then above holds tue as step 3 of MG povdes the optmal cove, and the ght hand sde of nequalty (2.4) s geate than o equal to ' J. As * J s optmal, * J cannot exceed the numbe of sets equed n any cove ' J. Assume that the esult holds tue fo maxmum set sze = q - 1, then t needs to be poved that the esult also holds tue fo q 3. Let denote the smallest ndex k such that S k q, fo all, whch means afte teatons of modfed geedy algothm s appled, thee s no set that contans moe than q 2 elements. Let U = I \ I denote the coveed elements tll teaton. As U s coveed by dsont sets of sze geate than o equal to q, J U / q (2.5) 25

Let us now defne n = S. Afte - 1 teatons of MG algothm, uncoveed elements I I \ U and unused sets S S \ U, 1 J - satsfes assumpton fo nducton, as n q 1, J. Hence, J Q ( n ), (2.6) * 1 ' J whee * J 1 s defned as the ndex set fo the emanng cove,.e., Fo all combnatons of values of n and Q( n ), Q ( n ) Q ( n ) S \ S / q (2.7) J J J. * * 1 Now fo any gven, consde the possble values of n and n, and n evey scenao use the fact that n n q. Case n 1: Q( n ) 1. If n = 0, then Q ( n ) = 0 and S \ S = n n = 1. If n = 1, then Q ( n ) = 1 and S \ S = 0. Theefoe, n both cases Q ( n ) Q ( n ) S \ S / q s tue. Case n 2 : Q( n ) = Hn ( ) 1. Thee ae 3 possble condton that coespond to the 6 sze of set S, n, afte the fst teatons. () n = 0: Q ( n ) = 0 and Q ( n ) = 1 Hn ( ) = 6 1 1 1 1 1... 3 3 4 n 26

4 3 Q ( n ) ( n n ) / q = Q ( n ) S \ S / q A stong nequalty follows fom Q ( n ) = 0 and q n. () n = 1: Q ( n ) = 1 and Q ( n ) = 1 1 1 1 Q( n)... 3 3 4 n Q ( n ) ( n 1) / n Q ( n ) ( n n ) / q because n q Q ( n ) S \ S / q One thng to note hee s that when q = 3, the nequalty changes to equalty () n 1: 1 1 Q ( n ) H( n ) 1 1/ 6 3 n 2 1 1 Qn( n) 1 3 n 2 n n 1 1 1 = 1 3 2 n 1 Q ( n ) ( n n ) / q snce 1 1 n q Theefoe, Q ( n ) Q ( n ) S \ S / q. Fo q 3, fom (), () and () t follows 27

Q ( n ) Q ( n ) S \ S / q (2.8) Applyng nequalty (2.8) and because S \ \ =, ' S J J ' S S U followng esult s obtaned: J ' Q ( n ) ( Q ( n ) S \ S / q ) J ' J U / q * 1 J J * 1 Lemma 3 gves a elatonshp between a set cove found by modfed geedy algothm and set cove povded by any othe algothm. If ' J J opt, algothm etuns the wost case pefomance ato. 1 Theoem [17]: The sze of cove found by MG algothm s no moe than Hk ( ) tmes the sze of optmal cove. Poof [17]: The wost case pefomance ato fo MG algothm s acheved, by settng 6 ' J J opt,, n k J,.e., opt J k 1 opt * J 1 ( H ( k) ) Jopt k 2 6 Theefoe, the sze of cove found by modfed geedy algothm by Goldschmdt el 1 al. s Hk ( ) tmes the sze of an optmal cove, whee k s the maxmum set sze. 6 Theefoe, modfed geedy algothm mpoves the pefomance ato of standad geedy algothm by 1/6. The uppe bound s tght fo k 3. 28

Fo example, bound fo standad geedy algothm fo k = 3 s 11/6, whee as the MG algothm has a ato bound of 10/6. Theefoe, standad geedy algothm s bound fo coveng the edges of a gaph by tangles, s 11/6. Modfed geedy algothm s bound fo the same poblem s 10/6. It can be clealy obseved that modfed geedy by Goldschmdt el al., s manly suted to the set cove poblem fo small k, whee k s the maxmum numbe of elements pe set. The mpovement comes fom exact soluton when set sze s 2 o less. Modfed geedy algothm s mpotant n the feld of semconducto ndusty o flexble manufactung. Example-2.4: Fnd the set cove fo followng collecton of sets, usng modfed geedy algothm. S 1 = {1, 2} S 2 = {1, 3} S 3 = {1, 4} S 4 = {2, 3} S 5 = {2, 4} S 6 = {3, 4} The gaph fo poblem, usng MG algothm s shown n Fg-2.1. The gaph has edges: 1-2, 1-3, 1-4, 2-3, 2-4 and 3-4. 29

Fg-2.1 Gaph Constucted fo Example 2.4 1-2 and 3-4 fom the maxmum matchng. Ths s also the pefect matchng as t coves all the nodes of the gaph. Ths means {S 1, S 6 } foms the edge cove. Clealy, {S 1, S 6 } s also a set cove. Ths s optmal soluton, at the same tme non-unque as well. {S 2, S 5 } and {S 3, S 4 } ae othe exact set cove soluton fo ths poblem. Example-2.5: Fnd the set cove fo followng collecton of sets. Hee maxmum set sze s geate than 2. S 1 = {1, 2, 3, 4, 5, 6} S 2 = {5, 6, 8, 9} S 3 = {1, 4, 7, 10} S 4 = {2, 5, 7, 8, 11} S 5 = {3, 6, 9, 12} S 6 = {10, 11} 30

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} STEP I: Apply standad geedy algothm, tll set sze becomes 2. Ths selects S 1 and S 4. Note: efe example 1. Hence, C = {S 1, S 4 } UNCOV = {9, 10, 12} S 2 = {9} S 3 = {10} S 5 = {9, 12} S 6 = {10} STEP II: Fnd edge cove on emanng elements. S 5 s the only edge. S 3 coves the emanng solated vetex 10. Hence, C = {S 1, S 4, S 3, S 5 } Theefoe, the fnal set cove s C = {S 1, S 4, S 3, S 5 } 31

2.5 Set Cove va Local Impovement by Halldosson [22] Halldosson poposed an old optmzaton technque and appled t to the appoxmaton of collectons of dscete tems. He called ths technque as local mpovement. The appoach conssted of extendng a collecton, by addng some tems whle emovng othes. Essentally, Halldosson poposed a combnaton of dffeent algothms n seveal subnstances of the ntal set cove nstance. Halldosson descbed local seach as tool to solve had combnatoal optmzaton poblems. Local seach can be appled to geedy seach o hll clmbng, smulated annealng o andomzed hll clmbng, augmentng paths, and local changes. Local seach technque has been successful n pactcal nstances. But the technque fals many tmes as t doesn t have good wost case esults. In ths pape, Halldosson pesented seveal postve esults on the qualty of locally optmal solutons. The goal of local mpovement s to mpove the best pefomance atos known fo vaous optmzaton poblems. At the same tme, t also hghlghts the effectveness of local mpovement heustcs as appoxmaton algothms. Local seach appoach stats wth a maxmal/mnmal collecton dependng on maxmzaton/mnmzaton poblem, and expands/shnks the collecton untl no bette soluton can be found. Usng ths technque, the algothm fo 3-set cove,.e., set cove wth maxmum set sze of at most 3 gves a soluton of pefomance ato 11/7 1.57. 11 Ths genealzes to a Hk ( ) ato fo k-set cove, whee k s the sze of the lagest set. 42 32

2.5.1 The Local Impovement Famewok [22] Halldosson consdeed the poblem of maxmzaton. A soluton s maxmal f the addton of any tem destoys the feasblty of cuent soluton. A non-maxmal soluton can be extended by ust addng an tem. Suppose A s a feasble soluton and thee ae sets x, y, z whee x A and y, z A such that A ' = (A - {x}) U {y, z} s also a soluton. Then, A ' s an extenson of A, and {x, y, z} a 2-mpovement of A. Bascally, a set of tems I such that A I (the unon of A and I ) s a soluton and A I > A, s an mpovement of A. Halldosson used the temnology t-mpovements. A t-mpovement adds n new tems and emoves n-1 tems, fo some n t. Bascally, mpovement adds one moe tem than the numbe of tems t emoves. A soluton s sad to be t-locally optmal f no t-mpovements s possble. Local seach tes to mpove ntal soluton by successon of mpovements, untl no mpovement exsts. Moe pecsely, t-optmal soluton: A maxmal soluton. Repeat I n-mpovement of A, whee n t. A A I (the symmetc unon of A and I). untl no futhe mpovement exsts. Retun A. 33

Mnmzaton poblem s vey much lke maxmzaton poblem, but t-mpovement poceeds n opposte decton. Fo mnmzaton poblem, local seach stat wth a mnmal soluton and t attempts to mpove t by shnkng the soluton set. Impovement emoves one moe tem than the numbe of tems t adds. A soluton s sad to be t-locally optmal f no t-mpovement s possble. 2.5.2 Appoxmate k-set Cove Usng Set Packng fo (S, C, k) [22] Halldosson povded a genec algothm usng set packng. Halldosson used s to denote the numbe of elements n the base set(o gound set o unvesal set o supe set) S, A to denote the set cove output of the algothm and B as the set cove output by any othe algothm. Hee C s collecton of sets and k s maxmum sze of sets n ths collecton. Fo smplcty Halldosson assumed that the nput set collecton C s monotone,.e., wheneve X C, so s the evey subset ' X of X. Set cove s a patton of the base set S nto sets n C. Ths can be vefed fom the fact that eplacng sets by an appopate supeset n the ognal nput does not ncease the sze of the set cove soluton,.e., total numbe of sets. The Halldosson s appoach can be vewed as a sequence of maxmal solutons to set packng poblems. It stats wth k-set packng poblem and ends wth 1- set packng poblem, fo whch the soluton s tval. Halldosson s AppoxSetCove algothm uses an appoxmate set packng algothm to poduce a set cove soluton. If geedy algothm s used to solve the set packng poblem then the esultng soluton of AppoxSetCove algothm s also geedy. 34

Fo a collecton C to be the collectons of subsets of S ', algothm uses the notaton C ' S S ', whose supesets ae contaned n C. to denote estcton of C to S ', 2.5.2.1 AppoxSetCove(S, C, k) Algothm [22] If k =1 etun C. C k The sets n C of sze k. A k AppoxSetPackng( C k, k). S S ( ). k C c Ak Retun Ak AppoxSetCove( S k, C Sk, k-1). 1 Halldosson compaed the above algothm to the mpoved bound of Hk ( ) fo k-sc obtaned by modfed geedy algothm by Goldschmdt, Hochbaum and Yu. MG algothm woks on fndng maxmum matchng on maxmum set sze of 2, whch s essentally fndng set packng soluton fo maxmum set sze of 2. Theefoe, 2-opt on 2- SP obtans the same ato. 6 Lemma 4 [22]: AppoxSetCove, usng 2-opt on the 2- SP sub-poblem, fnds an appoxmate 3-set cove wth at most (s + 2b)/3 sets. Hee s denotes the numbe of elements n the base set S, a be the numbe of sets n the set cove soluton by cuent Halldosson algothm A and b be the numbe of sets n any othe cove set cove Soluton B. 35

Poof [22]: Let A ( C ), 1,2,3, denote the sets n A ( C) of sze, espectvely. Let S denote the elements contaned n sets n A. Consde the estcton B of B to the elements contaned n A 1 A 2. Let S 2 ' B 1 and ' B2 be the sets n Ths mples s2 s 3 a 3, so B S 2 wth 1 and 2 elements espectvely. b 2b 3a s ' ' 1 2 3 Theefoe, the numbe of 2-sets n B s 2 s at least b s 3 a ( b b ) ' ' ' 2 3 1 2 s 3 a b 3 Snce the set packng algothm dscussed s 3/2 optmal, t wll fnd at least twothds of the sets. Thus, the soluton povded conssts of a 3 3-set, at least 2( s 3 a3 b ) / 3 2-sets, emande beng 1-set. Theefoe, the cost of the soluton wll be at most 2 4 a3 [ s 3 a3 b] [ s 3 a3 ( s 3 a3 b )] 3 3 s 2b 3 36

Ths mples that 3-SC poblem, can be solved usng 2-opt on the 2-set packng sub-poblem. The above algothm fnds appoxmate set cove whch s 5/3 optmal. Ths 1 esult can be extended to k-set cove whose soluton s Hk ( ) optmal. The assumpton fo k-set cove soluton s, 2-SP poblem s equvalent to maxmum matchng poblem and the exact maxmum matchng can be found n polynomal tme. 6 2.5.3 Halldosson s Dect Applcaton of Local Impovement [22] Halldosson povded the algothm fo 3-set cove poblem,.e., set cove poblem wth maxmum set sze of 3. He used the esult of 3-SC, along wth the modfed geedy algothm suggested by Goldschmdt et al. to get local mpovements. Halldosson suggested the local mpovement method dectly on the set cove soluton. The method employed local shnkng of soluton to acheve mpovement. Fo example, say x, x..., x x A, y, y..., y C Sets 1 2, k, k 1 1 2, k such that A { x} { y} s a set cove. Ths would eplace k + 1 sets n ognal soluton A by only k sets, whle the esultant soluton stll emanng a set cove. The local mpovement method seeks mpovements usng patcula type of shnkng. The am s to ty to mege a set wth some unt sze sets, poducng a numbe of sets of sze two. The appoach seeks: 1) a set n A 3 and thee sets n A 1 fo whch thee ae thee sets n C 2 who cove the same sx elements, o 2) a set n A 2 and two sets n A 1 37

fo whch thee ae two sets n C 2 who cove the same fou elements. 2.5.3.1 Algothm fo 3-SC Poblem [22] Fnd a maxmal 3-set packng (3-SP). Fnd an optmal 2-set cove (2-SC). Repeat shnkng steps untl obtaned set cove s locally mnmum. The above algothm fnds a cove wth at most (2s + 5b)/7 sets. 3-SC algothm fo set cove wth maxmum set sze of at most 3, has a soluton of pefomance ato 11/7 = Hk ( ) 1.57. 11 42 Halldosson poposed a method of fndng a maxmal collecton of ndependent shkng mpovement n lnea tme. It assumes collecton of sets ' C 2 n C 2 whch have one element n S 1. Then t maks each element n S1 S2wth at most thee sets n contanng that element. Then test f fo each set X n A2, A3, thee can be chosen a goup of X dsont sets, one fom the goup of sets maked to each element. If answe s yes, ' C 2 then mpovement s possble. Theefoe, update mpovements wll be ndependent. S, C accodngly to ensue that the ' 2 Halldosson obseved that unde these shnkng mpovements A 1 s monotone deceasng. Theefoe, once a maxmum collecton of mpovements have been obtaned, futhe local mpovements ae not possble. Theefoe, the ente shnkng pocess uns n the lnea tme. Paallel mplementaton of shnkng pocedue s also possble, by fndng a maxmal ndependent set n the gaph of all possble mpovements. 38

Lemma 5 [22]: 3-SC algothm fnds a cove wth at most (2s +5b)/7 sets. Poof [22]: Fo two set collectons ' A, A : let ' ' A Adenote { X A: X ( ' ' X ) X A } o the collecton of sets n A contanng elements n some set n ' A. Patton B nto B1 B A1, B2 ( B B1 ) A 2, and B B B B. Let 3 1 2 b1, b2, b 3 denote the szes of these collectons espectvely. Usng the optmalty of 2-set packng algothm a1 b 1 (2.9) a1 a2 b1 b 2 (2.10) Snce the soluton s locally mnmal. Theefoe, at least one element of each set n A 2 must belong to a set n ethe B2 o B 3. Each set n B 2 contans an element n A 2. Theefoe, slots avalable fo elements fom A 3 ae at most 2, n such a set. Ths mples a3 2b2 3b 3 (2.11) The total numbe of set n the soluton s pecsely s. a1 2a2 3a3 s (2.12) Now f add (2.9) twce, (2.10) thee tmes, (2.11) once and (2.12) twce, 7( a1 a2 a3) 2s 5( b1 b2 b3 ) 2b 3 ( a1 a2 a3) (2s 5( b1 b2 b3 )) / 7 2 b 3 / 7 whch mples 3-SC algothm fnds a set cove wth (2s +5b)/7 sets at most. 39

Snce s 3, b ths mples pefomance ato of 11/7 = 2b, whch mples that 3-SC algothm has pefomance ato of 11/7. 2.5.3.2 Algothm fo k-sc Poblem [22] Execute standad geedy algothm, untl set sze becomes at most 3. Let A 1 be the obtaned SC soluton. On the suvvng set cove nstance, apply 3-SC algothm, to get soluton A 2. A = A 1 U A 2 s the fnal SC-soluton. 11 The k-sc algothm povdes a soluton whch s Hk ( ) optmal. 42 Example 2.6: Fnd set cove fo followng collecton of sets usng k-sc algothm. S 1 = {1, 2, 3, 4, 5, 6} S 2 = {5, 6, 8, 9} S 3 = {1, 4, 7, 10} S 4 = {2, 5, 7, 8, 11} S 5 = {3, 6, 9, 12} S 6 = {10, 11} U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} STEP I: Apply standad geedy algothm, tll set sze becomes 3. 40

S 1 = 6 has maxmum cadnalty. Theefoe, select S 1. Now emanng sets have maxmum set sze of 3. Afte elements of S 1 s emoved fom emanng sets, C = {S 1 } UNCOV = {7, 8, 9, 10, 11, 12} S 2 = {8, 9} S 3 = {7, 10} S 4 = {7, 8, 11} S 5 = {9, 12} S 6 = {10, 11} STEP II: Now fnd maxmum set packng. S 2 and S 3 fom the maxmum set packng. Note: ths s non-unque. Afte emovng elements of S 2 and S 3 fom emanng sets, C = {S 1, S 2, S 3 } UNCOV = {11, 12} S 4 = {11} S 5 = {12} S 6 = {11} 41

STEP III: Now fnd the mnmum edge cove. But n ths case, thee ae only solated edges emanng. Remanng element of S 4 and S 6 ae essentally the same. Theefoe, S 4 and S 5 complete the set cove. C = {S 1, S 2, S 3, S 4, S 5 } STEP IV: Now apply local mpovements to sets S 2, S 3, S 4, S 5. S 2 = {8, 9} S 3 = {7, 10} S 4 = {7, 8, 11} S 5 = {9, 12} S 2 and S 3 fom the maxmum set packng. Note: ths s non-unque. Afte emovng elements of S 2 and S 3 fom emanng sets, C = {S 1, S 2, S 3 } UNCOV = {11, 12} S 4 = {11} S 5 = {12} In ths case no futhe local optmzaton s possble. Theefoe, the fnal set cove s C = {S 1, S 2, S 3, S 4, S 5 } 42

2.6 Duh and Fue Algothm Usng Sem Local Optmzaton [23] Duh and Fue offeed a new appoach, sem-local optmzaton to solve set cove poblem. Ths appoach s based on local optmzaton. Howeve ths appoach s moe poweful. The advantage of ths appoach comes fom global changes made to appoxmate soluton. Ths appoach apples standad geedy algothm tll the maxmum set sze becomes 3. Fo maxmum set sze of 5 and maxmum set sze of 4 t apples standad geedy algothm wth some estcton. In the end, fo maxmum set sze 3, t uses sem local optmzaton. Sem local optmzaton fo 3-set cove has geatly mpoved pefomance ato of 4/3. Based on esult of 3-set cove and a estctve phase, 1 pefomance ato of k-set cove poblem tuns out to be H( k ). 2 2.6.1 3-Set Cove Usng Sem Local Optmzaton [23] In pue local optmzaton technque, cuent appoxmate soluton gets bette by eplacng a fxed numbe of sets n the cuent set cove wth a lesse numbe of sets to get a new set cove. Duh and Fue suggested followng appoach to solve 3-set cove poblem and called t sem local optmzaton. In sem local optmzaton, once sets of sze 3 has been selected fo patal cove, the emanng sets wth maxmum set sze of 2 ae coveed optmally n polynomal tme, by fndng maxmum matchng. A sem-local(n,t) mpovement step fo 3-set cove conssts of nsetng up to n 3- sets and deletng up to t sets. Also, an abtay numbe of sets wth maxmum set sze of 2 ae optmally eplaced. The algothm looks fo numbe of sets wth set sze equal 3, nstead of all the smalle subsets. It allows global changes to the sets wth maxmum set 43

sze of 2. Fo qualty of soluton, algothm fst looks fo numbe of sets n the set cove and second the numbe of sets wth set sze equal 1. Set cove wth smallest sze s pefeed and among set coves of same sze the one wth fewe 1 element sets s pefeed. Sem local-optmzaton algothm fo 3-set cove poduces a set cove wth pefomance ato 4/3. 2.6.2 k-set Cove Usng Sem Local Optmzaton Algothm [23] 1. The Geedy Phase: Fo = k, down to 6 do Apply standad geedy algothm to select maxmum collecton of -sets. -sets denote the sets coveng exactly new element. 2. Restcted Phase: Fo = 5, down to 4 do Select maxmal collecton of -sets wth the condton that the choce of these -sets would not ncease the numbe of 1-sets. 3. Sem Local Impovement Phase fo 3-Set Cove: Apply sem local optmzaton on the elements that ae stll not coveed. Sem local-optmzaton algothm fo k-set cove poduces a set cove wth a 1 pefomance ato of H( k ). 2 44

2.7 Modfed Duh and Fue Algothm by Asaf Levn [24] In 2006, Asaf pesented an mpovement to Duh and Fue algothm whch has 196 pefomance ato H( k ). Ths s the best known mpovement fo standad geedy 390 algothm. The pevous best known mpovement of standad geedy algothm was Duh 1 and Fue algothm whch had a pefomance ato of H( k ). 2 2.7.1 Asaf Levn s Algothm [24] 1. The Geedy Phase: Fo = k, down to 6 do Apply standad geedy algothm to select maxmum collecton of -sets. -sets denote the sets coveng exactly new element. 2. Restcted Phase: Select a maxmal collecton of dsont 5-sets wth the condton that 5-sets would not ncease the numbe of 1- sets. 3. Restcted Local Phase: (a) Select a maxmal collecton of dsont 4-sets wth the condton that these 4- sets would not ncease the numbe of 1-sets. (b) Whle thee ae 4-sets C C (cove) and C1, C 2 C such that C ' = (C \{ C}) { C1, C2} s a collecton of dsont 4-sets, so that t nceases 45

the numbe of 1-sets, eplace C by C '. 4. Sem Local Impovement Phase fo 3-Set Cove: Run sem local optmzaton on stll uncoveed elements. Asaf Levn algothm eplaces the geedy constucton fo = 4, n Duh and Fue algothm by local seach appoach. The mpovement of ths algothm, compaed to Duh and Fue algothm, comes fom local seach phase. Above algothm by Asaf Levn, fo k-set cove poduces a set cove wth 196 pefomance ato H( k ). Ths s best known mpovement fo standad geedy algothm. The tme complexty of ths algothm s 390 5 6 O( m n ). 46

CHAPTER 3 LINEAR PROGRAMMING TECHNIQUE FOR SET COVER PROBLEM 3.1 Lnea Pogammng Intoducton Lnea pogammng s the poblem of optmzng a lnea obectve functon, subect to lnea nequalty constants [3]. Bascally, the am s to acheve best outcome,.e., maxmze o mnmze lnea functon. In lnea pogammng poblem fo maxmzaton [3], a m n matx A, a m-vecto b and a n-vecto c ae gven. The am s to ty and acheve a vecto x of n elements that n maxmzes the obectve functon 1 cx, subect to the m constants gven Ax b. Feasble soluton fo lnea pogammng poblem s any vecto x that satsfes Ax b. Fo mnmzaton poblem, constants ae of knd. Many pactcal poblems can be vey easly expessed as lnea pogams [3]. Because of ths a lots of eseach wok has been done n the aea of lnea pogammng. Geneal lnea pogams can be solved vey quckly usng the smplex algothm. If a poblem can be defned as polynomal szed lnea pogammng poblem [3] then t means that thee exsts a polynomal tme algothm fo the poblem. Also n many specal cases faste algothms exst fo the lnea pogammng poblem. Sometmes the obectve functon s gnoed and an attempt s made to fnd whethe any feasble soluton exsts o establsh that no feasble soluton exsts. 47

Fo example, gven a poblem [26]: mnmze 7 x 1 + x 2 + 5 x 3 subect to x 1 - x 2 + 3 x 3 10 5 x 1 + 2 x 2 - x 3 6 x 1, x 2, x 3 0 Ths s the standad fom of mnmzaton poblem [26]. Hee all the constants ae of knd and all the vaables ae constant to be non-negatve. The mpotance of nonnegatve s that t doesn t evese the decton of constant nequalty. As stated eale, any soluton fo the vaables n the lnea pogam whch satsfes all the constant s known as feasble soluton. Let z epesent the optmal value of ths lnea pogam and be a gven atonal numbe. The queston asked s f z s at most. If answe to ths queston s yes, then t means that thee s a feasble soluton, whose obectve functon value s at most. Ths povdes the uppe bound fo z. In the above example, consde f obectve functon value of at most 30,.e., z 30. x = (2,1,3) consttutes one soluton as 7 * 2 + 1 + 5 * 3 = 30. Now the othe obectve s to fnd good lowe bound on z. In the above example, one such bound can be obtaned by the fst constant, usng coeffcent compason tem by tem. Snce the x s ae estcted to be non-negatve, clealy 7 x 1 + x 2 + 5 x 3 x 1 - x 2 + 3 x 3 10. Theefoe, obectve functon s at least 10 fo any feasble soluton. A bette lowe bound s obtaned by summng up the two constants 7 x 1 + x 2 + 5 x 3 ( x 1 - x 2 + 3 x 3 ) + (5 x 1 + 2 48

x 2 - x 3 ) 16. A lowe bound s placed to the obectve functon so that sutable non-negatve multples can be found fo the constants so that when the sum s taken, the coeffcent of each x s domnated by the coeffcent n the obectve functon [26]. Snce x s ae estcted to be non-negatve, ght-hand sde of ths sum s the lowe bound on z. Bascally, the am s to choose the multples n such a way that ght hand sde of the sum s as lage as possble. The poblem of fndng lowe bound as lnea pogam can be expessed as [26]: maxmze 10 y 1 + 6 y 2 subect to y 1 + 5 y 2 7 - y 1 + 2 y 2 1 3 y 1 - y 2 5 y 1, y 2 0 Hee y 1 and y2 ae chosen to be the non negatve multples fo the fst and second constant, espectvely. Fst lnea pogam s called the pmal lnea pogam and the second lnea pogam s called dual lnea pogam [26]. Dual of dual lnea pogam s pmal lnea pogam tself. Hee t can be obseved fom dual of lnea pogam s poblem maxmzaton, whee pmal lnea pogam s mnmzaton poblem. By constuctng, evey feasble soluton to the dual pogam gves a lowe bound on the optmum value of the pmal. The evese also hold tue,.e., evey feasble soluton on the pmal pogam 49

povdes the uppe bound on the optmal value of the dual. Theefoe, f the feasble solutons fo the dual and the pmal wth matchng value functon ae known, then both solutons must be optmal. In the example above, x = (7/4, 0, 11/4) and y = (2,1 ), both acheve the obectve functon value of 26. Theefoe, both ae optmal solutons. L-P Dualty Theoem [26]: The pmal lnea pogam has fnte optmum f and only f ts dual lnea pogam has fnte optmum. Moeove, f x ( x,... x ) and * * * 1 n y ( y,... y ) ae optmal solutons fo pmal lnea pogam and the dual lnea * * * 1 m pogam espectvely, then n m * * c x b y 1 1 In standad fom, mn-max dual theoem can be stated n the followng way: mnmze subect to n 1 n 1 cx a x b = 1,...,m x 0 = 1,...,n whee a, b and c ae gven atonal numbes. Then the dual lnea pogam s: maxmze m 1 by = 1,...,n subect to m 1 a y c = 1,...,m 50

Weak L-P Dualty Theoem [26]: If x ( x1,... xn ) s a feasble soluton fo pmal lnea pogam and y ( y1,... y m) s a feasble soluton fo the dual lnea pogam, then n c x m b y 1 1 Complementay Slackness Condtons [26]: Let x and y denote the soluton fo pmal lnea pogam and dual lnea pogam espectvely. Then x and y ae both optmal f and only f each of the condtons stated below ae satsfed: Pmal complementay slackness condtons fo each1 n; ethe x = 0 o m 1 a y c ; and Dual complementay slackness condtons fo each 1 m; ethe y = 0 o n 1 a x b 3.2 Set Cove Usng LP Fo solvng mnmzaton poblem fo an NP-had poblem, usng appoxmaton lnea pogammng algothm, key step s to establsh a good lowe bound at the cost of the optmal soluton [26]. Fo ths, set cove poblem s expessed as ntege pogammng(ip). The cost of an optmal soluton to LP-elaxaton povdes the desed lowe bound. The geneal appoach fo solvng set cove poblem usng LP s [16]: 1. Fomulate the set cove poblem as an IP. 2. Relax IP to a LP. 51

3. Use the LP soluton to get a soluton set cove fo IP. 3.2.1 Set Cove va Roundng [16,26] In the oundng stategy, fst the lnea pogam s solved and then the soluton obtaned fo lnea pogam s tansfomed to ntegal soluton. At the same tme, attempt s made to make sue that n the pocess the cost of the soluton does not ncease much. The ntegal and factonal solutons cost s compaed to obtan the pefomance ato. Ths stategy s known as oundng. The set cove poblem s defned as Gound set o unvese U = { u 1, u 2,..., u m } Subsets S1, S2,..., Sn U weght w 0 fo each set S. The cost functon s 1 fo unweghted set cove. The am s to fnd I {1,..., n } that mnmzes c subect to I S U. I Bascally, the goal s to select the mnmum weght collecton of sets that cove the unvese. The fequency of an element n a set cove nstance s defned to be the numbe of sets t s n,.e., f max { : u S } Defne a vaable x fo each subset x wll be set to 1 ff coespondng subset S, whch can have the values of ethe 0 o 1. S s selected n the set cove. Clealy, the 52

estcton beng that, fo each element u U, whee the am s that at least one of the set contanng t should be pcked. Fo weghted SC poblem the IP fomulaton s Mn n 1 wx Subect to: : u S x 1 u U x {0,1} The LP elaxaton s obtaned by changng the last constant to1 x 0. Snce the uppe bound on x s edundant, Mn n 1 wx Subect to: : u n S x 1 u U n 1 x 0 Note: fo unweghted set cove w 1 1 n 53

One way of convetng a soluton to lnea pogam nto an ntege soluton s to ound up all non-zeo vaables to 1. Ths algothm acheves the pefomance guaantee of f. Followng s the slghtly modfed algothm whch pcks fewe sets n geneal. Smple Roundng Algothm fo Set Cove [26]: 1. Solve LP-elaxaton to get an optmal soluton x *. 2. Pck all sets S fo whch x * 1 f n ths soluton. The pefomance ato acheved by oundng algothm s f. Randomzed Roundng Appoach to Set Cove [26]: Each element selected by oundng pocess has a constant pobablty. Repeatng a pocess O(ln n) tmes set cove of hgh pobablty s obtaned. To get complete set cove, c ln n sub collectons ae pcked ndependently and the unon s computed. Hee c s a constant. The pobablty that set cove etuned by ths appoach s not a vald set cove s 1/4. Wth pobablty 1/2 ths appoach etuns a set cove that has a pefomance ato of 4ln(4 n ). The above pocedue s epeated untl a vald set cove s found. The expected numbe of epettons equed s at most 2. Clealy, the expected cost of algothm s of O(ln n) optmal. 3.2.2 Set Cove va LP Dualty [26] As aleady stated, any feasble soluton to the dual gves a lowe bound on the pmal pogam. Theefoe, t gves lowe bound on the ognal ntege pogam also. The algothm pesented n ths secton fnds an ntegal soluton to the pmal and smultaneously a soluton to the dual. The pefomance ato s obtaned by compang the 54

cost of these two solutons,.e., soluton to the pmal and the dual. The man advantage of ths algothm ove oundng algothm s that nstead of havng to wok wth an abtay optmal soluton to LP-elaxaton, the two solutons can be caefully pcked so that they have nce combnatoal popetes. Also snce the algothm doesn t have to fst solve LP-elaxaton optmally, ths algothm can be made moe effcent. Integalty gap of a mnmzed ntege pogam s defned as the maxmum ato of an optmal ntege pogam and optmal factonal soluton. The am s to mnmze ntegalty gap. Usng the LP-dualty appoach, s essentally fomulaton of ntege pogammng of the ntegalty gap. Intoduce a vaable y coespondng to each element u U. Now consde the pmal lnea pogam Mn m 1 wx Subect to: : u S x 1 u U x 0 The dual lnea pogam to the above LP-elaxaton fo set cove s: Max u U y Subect to: 55

y w S U : u S y 0 u U Smple Low Cost Dual-LP Algothm fo Set Cove [26]: 1. Fnd an optmal soluton 2. Pck all sets S fo whch * y to dual lnea pogam. y * : u S w n ths soluton. The pefomance ato acheved by dual LP algothm s f. 3.2.3 Pmal-Dual Schema LP-dualty also povdes a geneal schema fo obtanng the appoxmaton lnea pogammng algothm: the pmal dual schema [16]. Pmal dual schema behaves much lke dual LP. But athe than fndng the optmal dual soluton, t constucts ts own dual soluton. Now consde the pmal pogam wtten n standad fom [26]: mnmze subect to n 1 n 1 cx a x b = 1,...,m x 0 = 1,...,n whee a, b and c ae specfed n the nput. Then the dual pogam s: 56

maxmze subect to m 1 m 1 by a y c = 1,...,n y 0 = 1,...,m Algothms usng pmal-dual schema un by ensung the pmal complementay slackness condtons and elaxng the dual condtons. Pmay Complementay Slackness Condtons [26]: Fo each 1 n: ethe x 0 o a y 1 m Relaxed Dual Complementay Slackness Condtons [26]: Fo each 1 m: ethe y 0 o a x 1 m c and b whee > 1 s an constant; f becomes 1 then t s usual condton. Poposton [26]: Let x and y be the pmal and dual feasble solutons satsfyng the condtons mentoned above the n c x m b 1 1 3.2.3.1 Set Cove va Pmal-Dual Schema [26] Pmal-dual schema stats by mentonng the pmal complementay slackness condton and then the dual condtons ae elaxed appopately. Pmal condton can be wtten as 57

S U : x 0 y w : u S Set S s called tght f y w. Snce the pmal vaables ae ntegally : u S ncemented, pmal complementay slackness condton can be stated as: Select only tght sets n the set cove. To mantan the feasblty of dual, ovepackng of any set s not allowed. A set s called ovepacked f the total amount packed nto ts elements exceeds ts cost. Pmal and dual can be thought as coveng and packng lnea pogam espectvely. Dual condtons ae elaxed wth f. u y : y 0 y f : u S Snce x has 0/1 soluton, these condtons ae equvalent to: Cove each element wth non-zeo dual at most f tmes. Snce each element occus n at most f sets, theefoe, ths condton s satsfed fo all elements tvally. Usng the above two condtons: Pmal-Dual Algothm fo Set Cove [26]: 1. Intalze: x 0; y 0 2. Tll all the elements ae coveed pefom: Select an uncoveed element, say I, and ase y tll some tght sets ae found. Select each tght set n the cove and update x. Mak each of the elements occung n these sets as coveed. 58

3. Retun the obtaned set cove x. The above algothm acheves a pefomance ato of f. Geedy heustcs can be used to pesent set cove lnea pogammng and n that case appoxmaton facto wll be Hk ( ). 59

CHAPTER 4 SET COVER SOLUTION USING NETWORK FLOW APPROACH Mohamed Aff, Mhand Hf, Vangels Th. Paschos and Vassls Zssmopoulos pesented the set cove soluton, whch was based on netwok flow algothm of Fod and Fulkeson. Ths polynomal tme algothm was based on tansfomng set cove poblem nto a patcula netwok flow poblem and poposng a flow algothm whch was based on Fod and Fulkeson algothm. 4.1 Tansfomaton of Set Cove nto Flow Poblem [27] Mohamed et. al used the followng defnton fo tansfomng set cove nto a flow poblem. Defnton 1 [27]: A bpatte gaph B = (S, C, E) s constucted fom set cove poblem defnton. In ths bpatte gaph B = (S, C, E), vetex set S coesponds to the famly of sets, vetex set C coesponds to the gound set (o base set o unvesal set) C, and set of edges E { sc : c S }. Once bpatte gaph s ceated, a layeed netwok N = (X, A, c, b) s constucted wth vetex X S C s0 c0 whee c 0 epesents the souce of the netwok and s 0 epesents the snk of the netwok and A E Es Ec s the ac set of the netwok. 60

E { s s : s S}, E { c c : c C } and the acs ae oented fom c 0 to s 0. The s 0 c 0 vectos c ( ca ) a A, b ( b a) b A epesent capactes and lowe bounds espectvely, on the acs of N. Theefoe, 1 a 1 ca a c, 0 0 ( s ) ' a Ec c a c c, 1,.., m E { c s : s ( c )} ' a E 0 s c b a 0 a E E ' mn{ ' : { : ( )}} 0, 1,..., a c a c s s c a c c m s whee ( x) s the set of neghbos of x, ( x ) and ( x ) ae the oute and nne degee of vetex x. Set of ncomng acs of vetex v ae denoted by I ( v ). Set of outgong acs of vetex v ae denoted by I ( v ). Usng constucton mpled by above defnton set cove poblem educes to a mnmum flow poblem on N. mn a Es a a a a I ( v) a I ( v) b c a A a a a a a {0,1} a v X \{ s, c } E s a E E 0 0 c 61

The set cove poblem can be educed to mnmum flow poblem n polynomal tme O( mn ). Also the (optmal) obectve functon values fo the soluton of set cove and ae equal. Gven an nstance of I of set cove t takes O( mn) steps fo the constucton of bpatte gaph B. The constucton of layeed netwok N, too, takes O( E ) = O( mn ) steps. Snce computaton of uppe and lowe capactes of ac can be pefomed n constant tme, theefoe, the ente tansfomaton fom set cove to netwok flow poblem can be pefomed n O( mn ). Example 4.1: Ceate the chaactestc bpatte gaph B and the layeed netwok N, fo followng set cove poblem. { S1, S2, S3, S 4} C { a, b, c, d, e } S1 { a, b } S2 { b, c, d } S3 { d, e } S4 { a, e }. Usng Defnton 1, above set cove poblem can be conveted nto netwok flow poblem. Fg-4.1 shows the chaactestc bpatte gaph B fo ths and Fg-4.2 epesents the layeed netwok N. 62

Fg-4.1 Bpatte Gaph B fo Set Cove Poblem n Example 4.1 Fg-4.2 Layeed Netwok N fo Bpatte Gaph n Fg-4.1 63

4.2 Mnmum Flow Algothm fo Set Cove Poblem [27] Mohamed et. al gave followng algothm fo solvng set cove poblem. Algothm uses t(a) to denote the extemty of ac a and (a) to denote ntal extemty. begn * S epeat CONSTRUCT(N); epeat { ca : a A } stop false; ' ( c ) ; X { c }; A ' 0 0 whle ( s X stop do ' 0 ) f a xy A x X y X X b then ' ', \ a a ' ' ' X X y A y a { }; ( ) ; ' ' ( y) mn{ ( y), a b}; A A { a} else stop tue end-f end-whle f s0 ' X then end-f ( s 0) else 0 64

f 0 then end-f x s ; 0 untl 0 whle x c0 do end-whle a A ' ( x); ; x ( a ) a a * * S S s ss0 a a Es { : ag max{ : }}; A { s s } { a : t( a) s } { c c : c ( s )}; 0 0 C C \ ( s ); S S \{ s }; A A\ A untl C end Pocedue CONSTRUCT uses the defnton 1 to tansfom ethe the ognal set cove poblem o ts suvvng nstance afte deletng of an S-vetex and ts C- neghbohoods, on oute epeat loop. The nne epeat loop gves the mnmum flow algothm. It computes a flow educton and a educng path along whch the flow wll be educed by. Algothm stats wth a feasble soluton, t satuates all acs of N, and tes to educe flow fom c 0 to s. The educton step along a path 0 [ c0c ss 0] s defned ecusvely as follows: ( c 0), fo each ac a = xy A, ( y) mn{ ( x), a ba} and fnally ( s 0). 65

Once a mnmum flow has been obtaned, algothm checks f the flows on the acs of Es ae ntegal o not. If the flows ae not ntegal and gven that flows ae atonal numbes, algothm chooses the S-vetex though whch the maxmum flow closest to 1 s sent. Algothm then deletes ths vetex and ts neghbos n C. Then uppe and lowe capactes n the suvvng netwok ae e-calculated. Thus at the end of each teaton, a numbe of C-vetces of N ae deleted. Theefoe, afte a numbe of steps all C-vetces wll be emoved, whch mples they ae coveed by S-neghbos. The algothm teats evey path of algothm only once because the way a and b ae computed fo a E. Example 4.2: Fnd set cove soluton, fo the poblem defned n Example 4.1, usng mnmum flow netwok algothm. Fst the flow values need to be calculated. Fg-4.3 shows the flow values, calculated by the netwok flow algothm, fo the netwok shown n Fg-4.2. Algothm selects the vetex S 2, because of the maxmum flow value on the ac SS, 2 0 on the acs of E. s Afte selectng S 2, ths vetex and all ts neghbos ae deleted. Fg-4.4 shows the suvvng netwok afte deletng vetex S 2 and ts neghbohood. Thee afte flow values ae ecalculated on the suvvng netwok. Fg-4.5 shows ecalculated flow value fo suvvng netwok. 66

Fg-4.3 Netwok Constucted fo Example 4.1 Fg-4.4 Suvvng Netwok Afte Deletng Vetex S 2 and Its Neghbos 67