College of Computer & Informaton Scence Fall 2009 Northeastern Unversty 20 October 2009 CS7880: Algorthmc Power Tools Scrbe: Jan Wen and Laura Poplawsk Lecture Outlne: Prmal-dual schema Network Desgn: Stener Tree Stener Forest (prmal-dual algorthm) 1 Prmal-dual schema In a prevous lecture we have ntroduced the concept of dualty for LP problems. A common form of the LP problem wth ts dualty s showed below: P rmal Dual s.t. mn c x max b y a x b s.t. a y c x 0 y 0 From the prncple of dualty, a feasble soluton of the dual n fact sets a lower bound for the prmal problem. We here use a concept of complementary slackness to help us llustrate the connecton between the two solutons of the prmal and dualty problem: Defnton 1. The complementary slackness condton s as follows: P rmal : x > 0 a y = c Dual : y > 0 a x = b Then the followng theorem wll show ths connecton: Theorem 1. If (x, y) satsfes complementary slackness, then x and y are optmal solutons for prmal and dual problems, respectvely.
Proof: have From the forms of the prmal and dualty, f complementary slackness s satsfed, we wll c x = ( a y ) x b y = ( a x ) y It s easy to see that the RHS of the two equatons are equal. So we have c x = b y mplyng that x and y are both optmal by dualty. So f we obtan an LP soluton that satfes the complementary slackness, then the soluton s optmal. For nteger LPs n general, however, t s unlkely that the optmal soluton s ntegral. We apply the complementary slackness approach to approxmaton algorthms by defnng relaxed slackness as follows: Defnton 2. P rmal : x > 0 c α a y c Dual : y > 0 b a x βb And accordngly, we wll ft 1 nto the approxmate stuaton wth the descrpton below: Theorem 2. If (x, y) satsfes relaxed complementary slackness, then x and y are αβ-optmal for both prmal and dual problem. Proof: Based on the defnton of relaxed complementary slackness, we wll have c x α ( a y ) x (1) b y 1 ( a x ) y (2) β Then we can get cost of x cost of y = c x b αβ y whch ndcates that the dualty yelds a αβ-approxmaton soluton. Based on the property of the relaxed complementary slackness, the prmal-dual schema for LP soluton can be done n an teratve process: start wth a dual feasble soluton, then we try to mprove ths dual problem, untl the mproved soluton satsfes the relaxed complementary slack condtons. We wll now present an elegant prmal-dual algorthm of Goemans-Wllamson for the Stener Forest problem. 2
2 Stener Forest Problem 1. Gven: graph G = (V, E), edge cost functon C : E Q, collecton of sets {S 1, S 2,..., S k } V. Goal: Determne subgraph H of G wth least cost such that, S s connected n H. 3 Relatonshp to MST and Stener Tree We note that Mnmum spannng tree (MST) s a specal case of Stener Forest n whch k = 1 and S 1 = V (we want to connect all the nodes usng a mn-cost subset of the edges). Mnmum Stener tree s a specal case n k = 1 and S 1 s an arbtrary subset of V. Snce Stener Tree s NP-hard, Stener forest s also NP-Hard. There s a smple 2-approxmaton for Stener Tree. Create a new graph consstng of all the nodes of the set S that we want to connect, plus an edge connectng each par of nodes wth length = the shortest path length from the orgnal graph. Solve MST on ths new graph and use ths MST to create a Stener tree on the orgnal graph (by removng unnecessary edges). To verfy that ths s a 2-approxmaton: consder the mnmum Stener tree on the orgnal graph, and consder a tour of all nodes n S that vsts each edge of the Stener tree at most twce. If we vst the nodes n the same order n our new graph, we wll get a spannng tree that s at least as expensve as the MST, wth cost at most the cost of the tour on the orgnal graph, whch s at most twce the cost of the mn Stener tree. The best known (poly-tme) approxmaton factor s 1.55 (see [2]). Snce we have good approxmatons to Stener Tree, t s temptng to solve Stener Forest by mergng together Stener trees for each S. However, the followng example shows that ths could lead to an approxmaton rato as hgh as k, the number of sets n the gven nstance. 4 Lnear Program for Stener Forest There are two lnear program approaches to solve ths problem. One uses flows, the second uses cuts. To use flows, we would defne a varable representng the flow across each edge. There would be a constrant for each par (u, v) S (for all ) sayng that the flow between u and v must be at 3
least 1. The capacty of the flow across each edge would be another varable, representng whether or not that edge s ncluded n the soluton. Ths LP has a polynomal number of constrants and a polynomal number of varables. However, t s not clear how to use a prmal-dual approach wth the flow-based LPs. Therefore, we wll concentrate on the cut-based LP. We defne varables x e {0, 1}, for all edges e. x e = 1 f e s part of the forest, 0 otherwse. Notaton: c e = the cost of edge e. δ(s) = the set of all edges that are cut by cut (S, S). f(s) s a functon defned on the set of all cuts of G. f(s) = 1 f there exsts a par (u, v) such that (u, v) S for some, u S, v S. mnmze e E x e c e S, x e f(s) e δ(s) e, x e 0 Note that ths LP has an exponental number of constrants, and therefore cannot mmedately be solved as-s n polynomal tme. For the prmal-dual approach we do not need to solve the LP, so we defer the solvablty of the LP. When we consder the more general Stener Network problem, we wll n fact show that the LP s solvable n polynomal tme. Clam 1. An nteger soluton to the above LP s exactly Stener Forest. Proof. Gven Stener Forest H, construct soluton X for the LP as follows. { 1 e H x e = 0 e / H Consder any cut S such that f(s) = 1. Then (u, v) such that u S, v S, (u, v) S for some. u and v must have some path between them n H, snce H s a vald Stener forest. So there must be some edge e crossng cut S, x e = 1, satsfyng the constrant for S. So X s a vald soluton to the LP. Clearly, the cost of H = the value of X. Now take nteger soluton X to the LP. The frst constrant of the LP combned wth the fact that the obectve s mnmzaton mples that no x e wll have a value greater than 1. Therefore, each (nteger) x e value s 0 or 1. Construct a Stener forest H by addng each edge that has x e = 1. Suppose H does not connect some S. Then, there exst (u, v) S such that there s no path n H connectng u and v. Let set S = the set of all nodes connected to u n H (and notce that v / S). Take cut (S, S). f(s) = 1 by defnton of S. Snce X s a vald soluton to the LP, there must be some edge e crossng S such that x e = 1. However, ths contradcts our defnton of S. So H must connect all S, and therefore be a vald Stener forest. Agan, clearly the cost of H = the value of X. Clam 2. The above LP has an ntegralty gap 2 1/n. 4
Proof. Consder a cycle of length n, all edges have cost 1, S 1 = V (MST on a cycle). The mnmum Stener forest has cost n 1. However, the LP can be solved by settng x e =.5 for all e, gvng a value n 2. 5 Dual to the LP maxmze S f(s)y S e, S:e δ(s) S, y S 0 y S c e 6 Geomans and Wllamson prmal-dual algorthm ([1]) Intuton: We wll grow a tree startng at each u S for each. We wll stop as soon as the prmal soluton s feasble. In each step, we ncrease the y S values for cuts that make the prmal nfeasble untl the dual constrant for some edge e becomes tght. We wll set x e = 1 for one such e, satsfyng at least one more prmal constrant, and remove any edges that mght create a cycle. When the prmal soluton s feasble, we wll go back and delete any unnecessary edges. However the above approach, stated as s, turns out to be smlar as solvng separate Stener tree problems at the same tme, whch we know wll not gve a good approxmaton rato. To fx ths, we wll carefully pck whch y S to grow at each step. In the algorthm below, an actve set s a mnmal set whose prmal constrant has not yet been satsfed. The prmal-dual algorthm follows: 1: F 0; 2: y S = 0, S; 3: ntally, actve sets = {{u} : u S } 4: whle actve set S for whch prmal constrant s not satsfed do 5: rase y S unformly for all actve sets S untl some edge e has S:e δ(s) y S = c e. (notce: ths may rase the y S values by 0 f an edge s already tght). 6: add e to F (f multple e became tght, pck only one of them) 7: remove all tght edges that haven t been added yet that would form a cycle n F 8: set F = F - {e : removng e does not volate any connectng constrant}. 9: return F. Observaton: actve sets are connected components n F. Lemma 1. F s prmal feasble and y s dual feasble. Proof. By desgn, F s a forest whch satsfes all constants. Let e be an edge removed from F at step 8. Then there are no pars on opposte sdes of e that want to be connected. So F s prmal feasble. 5
We now show that y s dual feasble. Consder edge e. If e never becomes tght, then the constrant for e must be obeyed. If e does become tght at some pont, then after ths pont, there s no actve set S such that e crosses S. To see ths, we note that for an actve set S (a) f e crosses S and we ncluded e n the soluton, then the cut S constrant s satsfed by e; (b) f e crosses S and we removed e from the soluton n step 7, then ncludng e would close a cycle; ether all nodes n the cycle are part of S, all nodes are not part of S, or some chosen edge n the cycle crosses S. Snce we ncrease y S only for actve sets, S:e δ(s) y S wll reman = c e. Lemma 2. cost(f ) 2 cost(y). That s, e F c e 2 S:f(S)=1 y S Proof. e F c e = = y S (snce we only pcked tght edges e) e F e δ(s) y S 1 S:f(S)=1 e F δ(s) If we could say that degree F (S) 2, we would be done (snce we can ust use the relaxed complementary slackness condtons. But ths sn t necessarly true. Instead, our goal s to prove that y S degree F (S) 2 y S S S We wll show ths usng nducton on the teratons of the algorthm. Intally, all y S values are 0, so t s true. At any teraton, we start wth F = the collecton of connected components. Some are actve sets, others are not. There are no other actve sets. (An actve set cannot be smaller than a connected component, because the cut constrant s satsfed for sets smaller than a connected component. An actve set cannot be larger than a connected component because our actve sets are the mnmal sets that don t satsfy ther constrants.) Suppose n an teraton, we ncrease y S for all actve S by. The rght hand sde of the above equaton ncreases by 2 tmes number of actve sets. To evaluate the left hand sde, we must consder the fnal forest F. Imagne F wth each current connected component collapsed nto a sngle node. In ths revsed F, none of the nactve connected components wll be leaves, because we would have removed edges connectng these components to the rest of the forest durng step 8. Therefore, the average degree of all actve connected components s the average degree of a subset of the nodes n a tree, ncludng all of the leaves. So the average degree s at most 2. So the left hand sde of the above equaton also ncreases by at most 2 tmes the number of actve sets, thus completng the proof of the lemma. 6
References [1] M.X. Goemans and D.P. Wllamson. A general approxmaton technque for constraned forest problems. SIAM Journal of Computng, 24(2): 296-317, 1995. [2] G. Robns and A. Zelkovsky. Tghter Bounds for Graph Stener Tree Approxmaton. SIAM Journal on Dscrete Mathematcs, 19:122-134, 2005. 7