Combnatoral Auctons wth Structured Item Graphs Vncent Contzer and Jonathan Derryberry and Tuomas Sandholm Carnege Mellon Unversty 5000 Forbes Avenue Pttsburgh, PA 15213 {contzer, jonderry, sandholm}@cs.cmu.edu Abstract Combnatoral auctons (CAs) are mportant mechansms for allocatng nterrelated tems. Unfortunately, wnner determnaton s NP-complete unless there s specal structure. We study the settng where there s a graph (wth some desred property), wth the tems as vertces, and every bd bds on a connected set of tems. Two computatonal problems arse: 1) clearng the aucton when gven the tem graph, and 2) constructng an tem graph (f one exsts) wth the desred property. 1 was prevously solved for the case of a tree or a cycle, and 2 for the case of a lne graph or a cycle. We generalze the frst result by showng that gven an tem graph wth bounded treewdth, the clearng problem can be solved n polynomal tme (and every CA nstance has some treewdth; the complexty s exponental n only that parameter). We then gve an algorthm for constructng an tem tree (treewdth 1) f such a tree exsts, thus closng a recognzed open problem. We show why ths algorthm does not work for treewdth greater than 1, but leave open whether tem graphs of (say) treewdth 2 can be constructed n polynomal tme. We show that fndng the tem graph wth the fewest edges s NP-complete (even when a graph of treewdth 2 exsts). Fnally, we study how the results change f a bd s allowed to have more than one connected component. Even for lne graphs, we show that clearng s hard even wth 2 components, and constructng the lne graph s hard even wth 5. Introducton Combnatoral auctons (CAs), where bdders can bd on bundles of tems, have emerged as key mechansms for allocatng goods, tasks, resources, etc., both n computer scence and economcs. They are desrable when a bdder s valuaton of a bundle mght not equal the sum of hs valuatons of the ndvdual tems contaned n the bundle. Unfortunately, determnng the wnners n a general CA s NP-complete (Rothkopf, Pekeč, & Harstad 1998). Three man approaches have been pursued to address ths: 1) desgnng optmal search algorthms that are often fast (e.g., (Sandholm 2002; Sandholm et al. 2001; Fujshma, Leyton-Brown, & Shoham 1999; Boutler 2002)), but requre exponental tme n the worst case (unless P = NP), 2) desgnng approxmaton algorthms (e.g., (Hoos Copyrght c 2004, Amercan Assocaton for Artfcal Intellgence (www.aaa.org). All rghts reserved. & Boutler 2000; Zurel & Nsan 2001; van Hoesel & Müller 2001)) but unfortunately no polytme algorthm can guarantee an approxmaton (unless ZPP = NP) (Sandholm 2002), and 3) desgnng optmal polytme algorthms for restrcted classes of CAs (e.g., (Rothkopf, Pekeč, & Harstad 1998; Tennenholtz 2000; Penn & Tennenholtz 2000; Sandholm & Sur 2003)). Ths paper falls roughly wthn the thrd approach: we present hardness and easness results for natural classes of CAs. However, we also develop a problem nstance parameter (treewdth of the tem graph, descrbed below), such that any CA nstance falls wthn our framework, and wnner determnaton complexty s exponental n the parameter only. Lke almost all of the work on polynomal-tme solvable combnatoral auctons, we restrct our attenton to the case where any two bds on dsjont subsets can be smultaneously accepted (that s, no XOR-constrants between bds are allowed). Consder graphs wth the aucton s tems as vertces, whch have the property that for any bd, the tems occurrng n t consttute a connected set n the graph. (For nstance, the fully connected graph (wth an edge between every par of tems) always has ths property.) Such graphs can have potentally useful structure (for example, the graph may be a tree). For any type of structure, one can ask the followng two questons: 1) how hard s the clearng problem when we are gven a vald tem graph wth the desred structure? 2) f the graph s not gven beforehand, how hard s t to construct a vald tem graph wth ths structure (f t exsts)? We wll nvestgate both questons. 1 has been solved for the specal case where the graph s a tree or a cycle (Sandholm & Sur 2003); 2 has been solved for the specal case where the graph s a lne (Korte & Mohrng 1989) or a cycle (Eschen & Spnrad 1993). In each of these cases, a low-order polynomal algorthm was presented. One practcal use of such polynomally detectable and solvable specal cases s to ncorporate them nto optmal search algorthms (Sandholm & Sur 2003). At every node of the search tree, we can detect whether the remanng problem s polynomally solvable, and f t s, we use the polynomal specal-purpose algorthm for solvng t. Otherwse the search wll contnue nto the subtree. Also, there are two pure uses for graph structures whch make questons 1 and 2 easy. Frst, the auctoneer can de-
cde on the graph beforehand, and allow only bds on connected sets of tems. (In ths case, 1 s most mportant, but 2 may also be useful f the auctoneer wants to make sure that bds on certan bundles are allowed.) Second, the auctoneer can allow bds on any bundle; then, once the bds have been submtted, attempt to construct an tem graph that s vald for these bds; and fnally, clear the aucton usng ths graph. Clearly, the second approach s only practcal f real nstances are lkely to have tem graphs wth some structure. To a lesser extent, ths s also mportant for the frst approach: f bdders must bd on bundles too dfferent from ther desred bundles, economc value wll be lost. Fortunately, real-world nstances are lkely to have graphs that are not fully connected. For nstance, consder a combnatoral aucton for tourst actvtes n the Bay Area. One tem for sale may be a tcket to Alcatraz (n San Francsco). Another may be a tcket to the Chldren s Dscovery Museum (n San Jose). A thrd tem may be a Caltran tcket to get back and forth between the two ctes. Supposng (for now) that there s no alternatve transportaton between the two ctes, the only bundle that s unlkely to receve a bd s {Alcatraz, Chldren s Dscovery Museum}, because the bdder wll need transportaton (no matter whch cty she s based out of for the duraton of her vst). Thus, a vald graph for ths aucton s the followng lne graph (because ts only dsconnected set s the one we just ruled out). Alcatraz Caltran Chldren s Museum To extend the example, suppose that there are alternatve modes of transportaton whch are also ncluded n the aucton: a Rental Car, and a Bus tcket. Now the followng bundles are unlkely to receve a bd: {Alcatraz, Chldren s Dscovery Museum} (because the bdder requres a form of transportaton) and any bundle contanng more than one mode of transportaton. Thus, the followng s a vald tem graph (because we just ruled out all ts dsconnected sets). Alcatraz Rental Car Caltran Bus Chldren s Museum Ths graph does not fall under any of the prevously studed structures (t s not a lne, tree, or cycle). Stll, t has nterestng structure: for nstance, t has a treewdth of 2. The rest of the paper s organzed as follows. We frst show that gven an tem graph wth bounded treewdth, the clearng problem can be solved n polynomal tme. Next, we show how to construct an tem tree (treewdth = 1), f t exsts, n polynomal tme. Ths answers the proposed open queston of whether ths can be done (Sandholm & Sur 2003). (We leave open the queston of whether an tem graph wth small treewdth (say, 2) can be constructed f t exsts.) We show that constructng the tem graph wth the fewest edges s NP-complete (even when a graph of treewdth 2 s easy to construct). Fnally, we study a varant where a bd s allowed to consst of k connected sets, rather than just one. We show that the clearng problem s NP-complete even for lne graphs wth k =2, and the graph constructon problem s NP-complete for lne graphs wth k =5. Item graphs We frst formally defne tem graphs. Defnton 1 Gven a combnatoral aucton clearng problem nstance, the graph G =(I,E), whose vertces correspond to the tems n the nstance, s a (vald) tem graph f for every bd, the set of tems n that bd consttutes a connected set n G. We emphasze that an tem graph, n our defnton, does not need to have an edge connectng every par of tems that occurs n a bd. Rather, each par only needs to be connected va a path consstng only of tems n the bd. In other words, the subgraph consstng of each bd must form only one connected component, but t needs not be a clque. Clearng wth bounded treewdth tem graphs In ths secton, we show that combnatoral auctons can be cleared n polynomal tme when an tem graph wth bounded treewdth s gven. Ths generalzes a result by Sandholm and Sur (Sandholm & Sur 2003) whch shows polynomal tme clearablty when the tem graph s a tree (treewdth = 1). 1 Lnear-tme approxmaton algorthms for clearng when the tem graph has bounded treewdth have also been gven, where the approxmaton rato s the treewdth, plus one (Akcoglu et al. 2002). In contrast, we wll clear the aucton optmally. Frst we wll gve a very bref revew of treewdth. Defnton 2 A tree decomposton T of a graph G =(I,E) s a tree wth the followng propertes. 1. Each v T has an assocated set I v of vertces n G. 2. v T I v = I (each vertex of G occurs somewhere n T ). 3. For each ( 1, 2 ) E, there s some v T wth 1, 2 I v (each edge of G s contaned wthn some vertex of T ). 4. For each I, {v T : I v } s connected n T. We say that the wdth of the tree s max v T I v 1. Whle the general problem of fndng a tree decomposton of a graph wth mnmum wdth s NP-complete, when the treewdth s bounded, the tree decomposton can be constructed n polynomal tme (Areborg, Cornel, & Proskurowsk 1987). Because we are only nterested n the case where the treewdth of the tem graph s bounded, we may assume that the tree decomposton s gven to us as well as the graph tself. The followng (known) lemma wll be useful n our proof. Lemma 1 If X I s a connected set n G, then {v T : I v X {}} s connected n T. 1 The specal case of a tree can also be solved n polynomal tme usng algorthms for perfect constrant matrces (de Vres & Vohra 2003), but those algorthms are slower n practce.
We are now ready to present our frst result. Theorem 1 Suppose we are gven a combnatoral aucton problem nstance, together wth a tree decomposton T wth wdth tw of an tem graph G. Then the optmal allocaton can be determned n O( T 2 ( B +1) tw+1 ) usng dynamc programmng. (Both wth and wthout free dsposal.) Proof: Fx a root n T (the top of the tree). At every vertex v T wth tems I v, consder all functons f : I v B {0}, ndcatng possble assgnments of the tems to the bds. (f() =0ndcates no commtment as to whch bd tem s assgned to.) Ths set has sze ( B +1) Iv. Consder the subset F Iv of these functons satsfyng: 1. If f() =b, b must bd on ; 2. All bds n the mage f(i v ) nclude tems that occur hgher up n T than v; 3. Iff() =b and b also bds on tem j I v, then f(j) =b also. The nterpretaton s that each functon n F Iv corresponds to a constrant from hgher up n the tree as to whch bds should be accepted. We now compute, for every node v (startng from the leaves and gong up to the root), for every functon f F Iv, the maxmum value that can be obtaned from tems that occur n v and ts descendant vertces (but not n any other vertces), and that do not occur n bds n the mage of f. (We observe that f v s the root node, there can be no constrants from hgher up n the tree (that s, there s only one f functon), and the correspondng value s the maxmum value that can be obtaned n the aucton.) Denotng ths value by r(v, f), we can compute t usng dynamc programmng from the leaves up n the followng manner: Consder all assgnments g : { I v : f() =0} B {0}, 2 wth the propertes that: 1. If g() =b, b must bd on ; 2. The mage of g does not nclude any bds that nclude tems that occur hgher n T than v. 3.Ifg() =b and b also bds on tem j I v, then f(j) =0and g(j) =b also. (Thus, g ndcates whch bds concernng the unallocated tems n I v we are consderng acceptng, but only bds that we have not consdered hgher n the tree.) The value of such an assgnment s b g(i a(b) + v) w T :p(w)=v r(w, q w(f,g)), where g(i v ) s the mage of g, a(b) s the value of bd b, p(w) s the parent of w, and q w (f,g) :I w B maps tems occurrng n a bd n the mage of ether f or g to that bd, and everythng else to 0. The maxmum such value over all g s r(v, f). Because we need to do ths computaton once for each vertex v n T, the number of assgnments g s at most ( B + 1) Iv where I v tw+1, and for each assgnment we need to do a lookup for each of the chldren of v, ths algorthm has runnng tme O( T 2 ( B +1) tw+1 ). The allocaton that the algorthm produces can be obtaned gong back down the tree, as follows: at the root node, there s only one constrant functon f mappng everythng to 0 (because no bd has tems hgher up the tree). Thus, consder the r-value maxmzng assgnment g root for the root; all the bds n ts mage are accepted. Then, for each of ts chldren, consder the r-value maxmzng assgnment under the constrant mposed by g root ; all the bds n the mage of that assgnment are also accepted, etc. 2 In the case of no free dsposal, g cannot map to 0. To show that the algorthm works correctly, we need to show that no bds that are accepted hgh up n the tree are forgotten about lower n the tree (and ts tems lower n the tree awarded to other bds). Because the tems n a bd consttute a connected set n the tem graph G, by Lemma 1, the vertces n T contanng tems from such a bd are also connected. Now, f a bd b s accepted at the hghest vertex v T contanng an tem n b (that s, the tems n that vertex occurrng n b are awarded to b), each of v s chldren must also award all ts tems occurrng n b to b; and by the connectedness ponted out above, for each chld, ether there s at least one such an tem n that chld, or none of ts descendants have any tems occurrng n b. In the former case, b s also n the mage of the chld s allocaton functon, and the same reasonng apples to ts chldren, etc.; n the latter case the fact that b has been accepted s rrelevant to ths part of the tree. So, ether an accepted bd forces a constrant n a chld, and the fact that the bd was accepted s propagated down the tree; or the bd s rrelevant to all that chld s descendants as well, and can be safely forgotten about. Constructng a vald tem tree So far we dscussed queston 1: how to clear the aucton gven a vald tem graph. In ths secton, we move on to the second queston of constructng the graph. We present a polynomal-tme algorthm that constructs an tem tree (that s, an tem graph wthout cycles), f one exsts for the bds. Ths closes a recognzed open research problem (Sandholm & Sur 2003), and s necessary f one wants to use the polynomal tem tree clearng algorthm as a subroutne of a search algorthm, as dscussed n the ntroducton. Frst, we ntroduce some notaton. In a combnatoral aucton wth bd set B and tem set I, defne tems(b) I to be the set of tems n bd b. Also, let T b refer to the subgraph of a tree contanng only vertces represented by tems(b) and all edges among elements of tems(b). Wth these defntons n hand, we are now ready to present the man theorem of ths secton. Ths theorem shows how to gve a tree that mnmally volates the requrement of an tem tree that each bd bds on only one component. Thus, f t s actually possble to gve a vald tem tree, such a tree wll be produced by the algorthm. Theorem 2 Gven an arbtrary set of bds B for tems I, a correspondng tree T that mnmzes the number of connected components n T b b B can be found n O( B I 2 ) tme. Proof: Consder the algorthm MAKETREE(B,I) shown below, whch returns the maxmum spannng tree of the complete undrected weghted graph over vertces I n whch each edge (, j) has a weght equal to the number of bds b such that, j tems(b).
MAKETREE(B,I) 1 A An I I matrx of 0s 2 for each b n B 3 do for each n tems(b) 4 do for each j n tems(b) 5 do A(, j) A(, j)+1 6 return the maxmum spannng tree of the graph A The runnng tme of MAKETREE(B,I) s O( B I 2 ) from the trply nested for loops, plus the tme needed to fnd the maxmum spannng tree. The maxmum spannng tree can be found n O( I 2 ) tme (Cormen, Leserson, & Rvest 1990), so the runnng tme of the algorthm as a whole s O( B I 2 )+O( I 2 )=O( B I 2 ). To see that MAKETREE(B,I) returns the tree T wth the mnmum sum of connected components across all T b, note that the total weght of T can be wrtten as the number of edges n T among tems(b). b B Because T s a tree, each subgraph T b s a forest, and the number of edges n any forest equals the number of vertces n the forest, mnus the number of components n the forest. It follows that we can rewrte the above expresson as s = b B tems(b) the number of components n T b. Because b B tems(b) s a constant, maxmzng s s the same as mnmzng the sum of the number of connected components across all T b. In partcular, f an tem tree exsts for the gven bds, then n that tree, each bd bds on only one connected component, so the summaton n Theorem 2 s equal to the number of bds. Because each term n the summaton must always be greater than or equal to 1, ths tree mnmzes the summaton. Thus, MAKETREE wll return a tree for whch the summaton n Theorem 2 s equal to the number of bds as well. But ths can only happen f each bd bds on only one connected component. So, MAKETREE wll return an tem tree. Corollary 1 MAKETREE wll return a vald tem tree f and only f one exsts, n O( B I 2 ) tme. (And whether a tree s a vald tem tree can be checked n O( B I ) tme.) Implcatons for bd sets wthout an tem tree The above result presents an algorthm for constructng a tree T from a set of bds that mnmzes the sum of the number of connected components across all T b. Even when the tree returned s not a vald tem tree, we can stll use t to help us clear the aucton, as follows. Suppose MAKETREE was close to beng able to construct an tem tree, n the sense that only a few bds were splt nto multple components. Then, we could use brute force to determne whch of these splt bds to accept, and solve the rest of the problem usng dynamc programmng as n (Sandholm & Sur 2003). If the number of splt bds s k, ths algorthm takes O(2 k B I ) tme (so t s effcent f k s small). We note, however, that MAKETREE(B,I) does not mnmze the number of splt bds (k), as would be desrable for the proposed search technque. Rather, t mnmzes the total number of components (summed over bds). Thus, t may prefer splttng many bds nto few components each, over splttng few bds nto many components each. So there may exst trees that have fewer splt bds than the tree returned by MAKETREE. Also, MAKETREE does not solve the general problem of constructng an tem graph of small treewdth f one exsts. The straghtforward adaptaton of the MAKETREE algorthm to fndng an tem graph of treewdth 2 (where we fnd the maxmum spannng graph of treewdth 2 n the last step) does not always provde a vald tem graph, even when a vald tem graph of treewdth 2 exsts. To see why, consder an aucton nstance for whch the followng graph s the unque tem graph of treewdth 2 (for example, because for each edge, there s a bd on only ts two endponts). B A D If there are many bds on the bundle {A, B, D}, and few other bds, the adapted algorthm wll draw the edge (A, D). As a result, t wll fal to draw one of the other edges (because otherwse the graph would have treewdth 3), and thus the graph wll not be a vald tem graph. For now, we leave open the queston of how to construct a vald tem graph wth treewdth 2 (or 3,or4,...) f one exsts. However, n the next secton, we solve a related queston. Constructng the tem graph wth the fewest edges s hard The more edges an tem graph has, the less structure there s n the nstance. A natural queston s therefore to construct the vald tem graph wth the fewest edges. It should be ponted out that ths s not necessarly the best graph to work on. For example, gven our algorthm, a graph of treewdth 2 may be more desrable to work on than a graph wth fewer edges of hgh treewdth. On the other hand, assumng that the tems cannot be dsjont nto two separate components (whch s easy to check), a tree s always a graph wth the mnmum number of edges (and f a tree exsts, then only trees have the mnmum number of edges). So n ths case, generatng a graph of mnmum treewdth s the same as generatng a graph wth the mnmum number of edges. We next show that constructng the graph wth the fewest edges s hard. Interestngly, the queston s hard already for nstances wth treewdth 2. (For nstances of treewdth 1 (forests) t s easy: dvde the tems nto as many separate components (wth no bds across more than one component) as possble, and run our MAKETREE algorthm on each.) Thus, f a graph of treewdth 2 can be constructed n polyno- C
mal tme (and P NP), the algorthm for dong so cannot be used to get the fewest edges unlke the case of treewdth 1. Theorem 3 Determnng whether an tem tree wth fewer than q edges exsts s NP-complete, even when an tem graph of treewdth 2 s guaranteed to exst and each bd s on at most 5 tems, and whether or not the tem tree we construct s requred to be of treewdth 2. Proof: The problem s n NP because we can nondetermnstcally generate a graph wth the tems as vertces and at most k edges, and check whether t s vald tem graph. To show that the problem s NP-complete, we reduce an arbtrary 3SAT nstance to the followng set of tems and bds. For every varable v V, let there be two tems +v, v. Furthermore, let there be two more tems, 0 and 1. Let the set of bds be as follows. For every v V, let there be bds on the followng sets: { 0, +v }, { 0, v }, { +v, v }, { +v, v, 1 }. Fnally, for every clause c C, let there be a bd on { +v : +v c} { v : v c} { 0, 1 } (the set of all tems correspondng to lterals n the clause, plus the two extra tems we note that because we are reducng from 3SAT, these are at most 5 tems). Let the target number of edges be q =4 V. We proceed to show that the two nstances are equvalent. Frst, suppose there exsts a soluton to the 3SAT nstance. Then, let there be an edge between any two tems whch consttute a bd by themselves; addtonally, let there be an edge between +v and 1 whenever v s set to true n the SAT soluton, and an edge between v and 1 whenever v s set to false n the SAT soluton (for a total of 4 V edges). We observe that all the bds of the form { +v, v, 1 } are now connected. Also, for any c C, because the 3SAT soluton satsfed c, ether 1 s connected to some +v wth +v c, or 1 s connected to some v wth v c. (And all the tems besdes 1 n the bd correspondng to c are clearly connected.) So all the bds consttute connected subsets, and there exsts a vald tem graph wth at most 4 V edges. (Also, ths s a seres parallel graph, and such graphs have treewdth 2.) Now, suppose there exsts a vald tem graph wth at most 4 V edges. Of course, there must be an edge between any two tems whch consttute a bd by themselves; and because of the bds on three tems, for every v V, there must be an edge ether between +v and 1, or between v and 1. Ths already requres 4 V edges, so there cannot be any more edges. Because each bd correspondng to a clause c must be connected, there must be ether an edge between some +v wth +v c and 1, or between some v wth v c and 1. But then, t follows that f we set v to true f there s an edge between +v and 1, and to false f there s an edge between v and 1, we have a soluton to the SAT nstance. All that remans to show s that n these nstances, a vald tem graph of treewdth 2 always exsts. Consder the graph that has an edge between any two tems whch consttute a bd by themselves; an edge between +v and 1 for any v V ; and an edge between 0 and 1 (for a total of 4 V +1edges). Ths s a seres parallel graph, and such graphs have treewdth 2. Bds on multple connected sets In ths secton, we nvestgate what happens f we reduce the requrements on tem graphs somewhat. Specfcally, let the requrement be that for each bd, the tems form at most k connected components n the graph. (The case where k =1 s the one we have studed up to ths pont.) So, to see f a bd s vald gven the graph, consder how many connected components the tems n the bd consttute n the graph; f (and only f) there are at most k components, the bd s vald. The followng fgure shows an example tem tree. One bd bds on all the tems encapsulated by rectangles (2 connected components); the other, on all the tems encapsulated by ellpses (3 connected components). Thus, f k =2, then the frst bd s vald, but the second one s not. (Equvalently, the graph s not vald for the second bd.) As we wll see, both clearng when a smple graph s known, and detectng whether a smple graph exsts, become hard for k>1. Clearng s hard even wth 2 connected sets on a lne graph Even f the tem graph s a lne, t s hard to clear auctons n whch bdders may bundle two ntervals together. To show ths, we prove the followng slghtly stronger theorem. Theorem 4 If an tem graph s created that conssts of two dsconnected lne graphs, and bdders are permtted to bundle one connected component from each lne, then determnng f the aucton can generate revenue r s NP-complete. Proof: The problem s n NP because the general clearng problem s n NP. To show that t s NP-complete, an arbtrary nstance of VERTEXCOVER wll be reduced to a correspondng aucton problem. In VERTEXCOVER, the goal s to determne whether there exsts a set of vertces C of sze at most k n a graph G =(V,E) such that for each edge (x, y) E, ether x C or y C. To perform the reducton, gven G =(V,E), create an tem u for each edge e. Place all of the u tems nto the upper lne. In addton, for each vertex v V, create the tems lv ej for each edge e j that v s part of. For each v, algn all of ts correspondng lv ej tems nto a contguous nterval on the lower lne. Now, create the followng bds: 1. A bd of prce 2 for each edge tem par (lv ej,u j ). 2. A bd of prce 1 for each vertex nterval bundle, {lv ej for all e j for whch v s part of e j }. We now show that there exsts a vertex cover of sze at most k exactly when the optmal revenue of the correspondng aucton s at least 2 E + V k.
Suppose there s a vertex cover of sze k for a graph G =(V,E). Then t s possble to sell all E of the edge tems breakng up only k of the vertex nterval bundles. The resultng revenue f only those k ntervals are rendered unsellable by matchng one or more of ther members wth edge tems s r =2 E + V k as requred; the proft s 2 for each of the edge pars of the form (lv ej,u j ) that are sold, plus 1 for each of the vertex nterval bundles that have not had any of ther tems matched to edges. Conversely, suppose there s a way to acheve revenue r = 2 E + V k. Suppose all of the edge tem pars were sold n ths case. Then, because V k vertex ntervals were sold, only k were spoled by sellng some of ther tems wth edge pars. These k vertces can be used as a vertex cover. On the other hand, suppose not all edge pars were sold. Then the revenue could be ncreased by sellng the rest of the edge pars even f t means spolng addtonal vertex bundles because edge pars carry a prce of 2 whle vertex nterval bundles only have prce 1. Ths mples that t s possble, by sellng all of the edge tem pars, to acheve revenue r =2 E + V k >rso that k <k, where k represents the sze of a possble vertex cover. Corollary 2 The problem of optmally clearng bds that bundle together at most two connected components of a lne tem graph s NP-hard because jonng the two lnes of tem graphs of the form dscussed n Theorem 4 consttutes a trvally correct reducton. Constructng a lne graph s hard even wth 5 connected sets We now move on to the task of constructng a smple tem graph that s vald when bds are allowed to consst of multple components. The graph constructon queston s perhaps less nterestng here because, as we just showed, clearng remans hard even f we are gven an tem lne graph. Nevertheless, the graph may stll be helpful n reducng the clearng tme: maybe the clearng tme bound can be reduced to a smaller exponental functon, or maybe t can reduce the clearng tme n practce. In ths secton, we show that unfortunately, detectng whether a vald lne graph exsts wth multple (5) components s also NP-complete. Lemma 2 Suppose a bd s allowed to contan up to k connected components of tems. Then, f there are m 2k +5 tems, there exsts a set of O(m k ) bds such that there s exactly one lne graph (one orderng of the tems, up to symmetry) consstent wth these bds. Proof: Label the tems 1 through m. For any subset of k +1 tems of whch at least two tems are successors ( and +1), let there be a bd on that set of tems. We observe that there are at most (m 1) ( m k 1) such bds (choosng the successve par of tems frst, and then the remanng k 1 of course we are double-countng some combnatons ths way, but we only want an upper bound), whch s O(m k ). Orderng the tems 1, 2,..., m (or equvalently m, m 1,...,1), we get a vald tem graph (because any two successve tems are adjacent n ths graph, there are at most k components n every bd). What remans to show s that f the tems are ordered dfferently, there s at least one bd wth k +1 components. If the tems are ordered dfferently, there s at least one par of successve (accordng to the orgnal labelng) tems, +1whch are not adjacent n the graph. Consder the set of these two tems, plus every tem that has an odd ndex n the orderng of ths graph (besdes the ones that concde wth, or are adjacent to, the frst two). Ths set has at least 2+(k +3) 4=k +1tems, two of whch are adjacent n the orgnal labelng, and each of whch s a separate component. It follows that there exsts a subset of ths set whch consttuted one of the bds, and now has k +1components. Lemma 3 Suppose each bd s allowed to contan at most k connected components, and we have a set of bds that forces a unque orderng of the tems (up to symmetry). Then suppose we replace one tem r wth two new tems n 1 and n 2, and let every bd bddng on the orgnal tem bd on both the new tems. Then the only vald orderngs of the new set of tems are the vald orderngs for the orgnal set, where r s replaced by n 1,n 2 (where these two can be placed n any order). Ths extends to replacng multple tems by pars. Proof: We omt the proof because of space constrant. Theorem 5 Gven the bds, detectng whether an orderng of the tems (a lne graph) exsts such that each bdder bds on at most 5 connected components s NP-complete. Proof: The problem s n NP because we can nondetermnstcally generate an orderng of the tems, and check whether any bd s bddng on more than 5 components. To show that the problem s NP-hard, we reduce an arbtrary 3SAT nstance to the followng sets of tems and bds. For every varable v V, let there be four tems, v, v, +v, v. Let the set of bds be as follows. Frst, usng Lemma 2 and Lemma 3, let there be O(m 5 ) bds such that the only remanng vald orderngs are v 1, v1, { +v1, v1 }, v 2, v2, { +v2, v2 },..., v n, vn, { +vn, vn }. (Here, two tems are n set notaton f ther relatve order s not yet determned.) Fnally, for every clause c C, let there be a bd bddng on any v wth v occurrng n c (whether t s +v or v), and on any +v wth +v occurrng n c, and on any v wth v occurrng n c. (So, 6 tems n total.) We show the nstances are equvalent. Frst suppose there exsts a soluton to the 3SAT nstance. Then, whenever a varable v s set to true, let +v be ordered to the left of v ; otherwse, let +v be ordered to the rght of v. Then, for every clause, for some lteral +v (or v) occurrng n that clause, +v (or v ) s adjacent to v, and t follows that the bd correspondng to the clause has at most 5 connected components. So, there s a vald orderng. Now suppose there exsts a vald orderng. Because of the v tems, the only tems n a bd correspondng to a clause that can possbly be adjacent are an +v and the
correspondng v,oran v and the correspondng v. Ths must happen at least once for every bd correspondng to a clause (or the bd would have 6 components. But then, f we set a varable v to true f +v and v are adjacent, and to false otherwse, every clause must have at least one +v n t where v s set to true, or at least one v n t where v s set to false. It follows that there s a soluton to the 3SAT nstance. Conclusons and future research Combnatoral auctons (CAs) are mportant mechansms for allocatng nterrelated tems. Unfortunately, (even approxmate) wnner determnaton s NP-complete unless there s specal structure. In ths paper, we studed the settng where there s a graph (wth some desred property), wth the tems as vertces, and every bd bds on a connected set of tems. Two computatonal problems arse: 1) clearng the aucton when gven the tem graph, and 2) constructng an tem graph (f one exsts) wth the desred property. 1 was prevously solved for the case of a tree or a cycle, and 2 for the case of a lne graph or a cycle. We generalzed the frst result by showng that gven an tem graph wth bounded treewdth, the clearng problem can be solved n polynomal tme (and every CA nstance has some treewdth; the complexty s exponental n only that parameter). We then gave an algorthm for constructng an tem tree (treewdth 1) f such a tree exsts, thus closng a recognzed open problem. We showed why ths algorthm does not work for treewdth greater than 1, but leave open whether tem graphs of (say) treewdth 2 can be constructed n polynomal tme. We showed that fndng the tem graph wth the fewest edges s NP-complete (even when a graph of treewdth 2 exsts). Fnally, we studed how the results change f a bd s allowed to have a few connected components (rather than just one). Even for lne graphs, we showed that clearng s hard even wth 2 components, and constructng the lne graph s hard even wth 5 components. For future research, the most mportant specfc queston left open s whether tem graphs of bounded treewdth w can be constructed n polynomal tme (when they exst), for w 2. Another queston s whether a lne graph can be constructed n polynomal tme n the varant where a bd s allowed to have k components, when k < 5. Also, we can ask whether we can reduce the runtme bound of the tree detecton algorthm. More open-ended future research ncludes dong expermental work wth the algorthms presented here (for nstance, comparng ther runnng tme to solvers such as CPLEX), and ntegratng these algorthms nto search algorthms as subroutnes, as dscussed n the ntroducton. Acknowledgements Ths materal s based upon work supported by the Natonal Scence Foundaton under CAREER Award IRI- 9703122, Grant IIS-9800994, ITR IIS-0081246, and ITR IIS-0121678. 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