Online hypergraph matching: hiring teams of secretaries

Transcription

1 Olie hypergraph matchig: hirig teams of secretaries Rafael M. Frogillo Advisor: Robert Kleiberg May 29, 2008 Itroductio The goal of this paper is to fid a competitive algorithm for the followig problem. We are give a hypergraph G = (X, E), E =, with weighted edges ad maximum edge size k. We wish to maximize the weight of a matchig M E of G, give that edges are revealed i a radom order, ad that whe a edge e is revealed the algorithm must decide either e M or e / M. Note that this is a geeralizatio of the classic secretary problem; give a set {w i } of weights for the cadidates i the secretary problem, we create a hypergraph with oe vertex v ad edges, where the ith edge e i = {v} has cardiality ad weight w i. Clearly a matchig ca be oly oe edge e i, correspodig to choosig cadidate i. I fact, whe k =, the hypergraph matchig problem is just a collectio of idepedet secretary problems, oe for each vertex. 2 Algorithm The algorithm we preset is very simple. For some a [0, ] to be determied, the algorithm does othig while the first a edges are revealed. After that, a ew edge is selected if it is i the maximum matchig amog all edges see so far, ad does ot itersect ay edge already selected. Formally, this is the

2 procedure give as Algorithm, where maxmat(e ) is the maximum weight matchig of ay edge set E. Algorithm The algorithm. Ẽ M for t = to a do pick e from E \ Ẽ uiformly at radom Ẽ Ẽ {e} ed for for t = a + to do pick e from E \ Ẽ uiformly at radom Ẽ Ẽ {e} if e maxmat(ẽ) ad Γ(e) M = the M M {e} ed if ed for 3 Aalysis 3. Defiitios Let E t0 be a radom variable represetig the value of Ẽ i the algorithm whe t = t 0 ; i other words, E t is the first t edges revealed. Similarly, e t is the edge revealed at time t. We will deote the weight of a matchig M as w(m). Fially, usig the defiitio of maxmat from the previous sectio, defie M t = maxmat(e t ) as simply the max matchig the algorithm sees at time t. Note that M is the maximum matchig overall; we therefore are strivig to come withi a costat factor of w(m ). Some graph-theoretic shorthad will be useful; we deote the edges eighborig of v i G by Γ(v) = {e E v e}. Similarly, the eighbors of a edge Γ(e) = v e Γ(v) are all edges that share a edpoit with (i.e. itersect) e. We ow defie some more complicated radom variables that describe the behavior of the algorithm. Defiitio. We say t is a critical time for v if v is o the edge revealed at time t, ad that edge is i the matchig M t, which is the max matchig 2

3 of all edges see so far. Let crit(v, t) be the evet that t is critical for v; the formally, crit(v, t) := {e t M t, v e t }. Similarly, crit(e, t) := v e crit(v, t) is evet that t is critical for e. Fially, ocrit(e, T ) is the evet that o t T is critical for e. Defiitio 2. Let avail(e, t) be the evet that edge e is available to be selected by the algorithm at time t; by the defiitio of the algorithm, this meas that e has ot bee see before ad o edge adjacet to the edpoits of e has bee selected. Let sel(e, t) be the evet that the algorithm selects edge e at time t. Similarly, sel(e, T ) = sel(e, t) is the evet that e is picked i a set T of times. 3.2 Proof To uderstad the behavior of the algorithm, we first ask how ofte a ode v chages its assigemet from oe edge to aother. Note that for our reductio from the origial secretary problem, the ucoditioal versio of this boud is exactly the boud for how ofte a better cadidate appears. Lemma 3. Let e be give ad let v e. The for all s < t, Pr[crit(v, s) e M t e t = e] s, Proof. Let e(f, v) = maxmat(f ) Γ(v) be the (possibly empty) set cotaiig the edge adjacet to v i the max matchig of the edge set F. Note that the evet e M t e t = e ca be determied soley from E t ad e. Thus, for some edge set F let q(f, e) be the predicate which is true whe e / F ad e maxmat(f e), ad let Q(F, e) be the eve that q(f, e) is true ad e t = e. The by this costructio Q(E t, e) = e M t e t = e. Now let F s ad F be edge sets with F s = s ad F = t. Usig this otatio, we see Pr[crit(v, s) Q(F, e) E s = F s ] = Pr[e s e(f s, v) E t = F e t = e E s = F s ] { 0 if e(f s, v) = = otherwise s, t 3

4 sice eve uder the coditios o E t ad e t, all orderigs of the edges e,..., e s are equally likely. Hece, we have Pr[crit(v, s) e M t e t = e] = F : q(f,e) = F : F : F s F q(f,e) Pr[crit(v, s) Q(F, e)] F s F q(f,e) Pr[crit(v, s) Q(F, e) E s = F s ] Pr[E s = F s Q(F, e)] s Pr[E s = F s Q(F, e)] = Pr[E s = F s e t = e] = s s. F s: e/ F s Next, we fid a costat which bouds the probability that a edge is utouched i a time iterval T. Lemma 4. If T = [t +, t 2 ] is some time iterval ad t T, the ( ) Pr[avail(e, t) e M t e t = e] + k l t t2. Proof. Recall that k is the maximum size of a hyperedge. First, from the defiitio of avail, we have Pr[avail(e, t) e M t e t = e] By Lemma 3 we have = Pr[e / E t ocrit(e, [t +, t ]) e M t e t = e] = Pr[ocrit(e, [t +, t ]) e M t e t = e]. () Pr[ocrit(e, T ) e M t e t = e] v e v e t s=t + t s=t + Pr[crit(v, s) e M t e t = e] s k 4 t2 s=t + s (2)

5 Usig Riema sums to boud the harmoic series, we see that t 2 s=t + s l t 2 l t = l t t 2 (3) Combiig (), (2), ad (3), we obtai the desired boud. Usig the boud from Lemma 4, we ca boud how likely the algorithm is to select a give edge i a give time iterval. This is expressed i terms of the probability that the revealed edge is i the curret max matchig M t, which i tur is bouded i Lemma 6. Lemma 5. If T = [t +, t 2 ] is a time iterval, Pr[sel(e, T )] ( ) + k l t t2 Pr[e M t e t = e] Proof. For a edge e to be selected, it must have bee revealed at some time t, at which poit e was both available ad i the max matchig. Formally, Pr[sel(e, T )] = Pr[e M t e t = e avail(e, t)] = Pr[avail(e, t) e M t e t = e] Pr[e M t e t = e] Pr[e t = e] ( ) + k l t t2 Pr[e M t e t = e], by Lemma 4 ad the fact that Pr[e t = e] = / always, sice it depeds oly o the orderig of the edges. The result follows. Lemma 6. Pr[e M t e t = e] w e w(m ) e E Proof. Let ˆMt = M E t. Clearly ˆM t is a matchig, ad sice M t = maxmat(e t ), we must have w(m t ) w( ˆM t ). Thus, E[w(M t )] E[w( ˆM t )] = e M Pr[e E t ] w e = 5 e M t w e = t w(m ). (4)

6 Observe that E[w({e t } M t )] = e E Pr[e {e t } M t ] w e = e E Pr[e M t e t = e] Pr[e t = e] w e O the other had, = Pr[e M t e t = e] w e. (5) e E E[w({e t } M t )] t E[w(M t)] w(m ), (6) where the last iequality is by (4). The result follows from combiig (5) ad (6). We ca ow put all of the lemmas together to obtai the competitive costat for our algorithm. Theorem 7. Algorithm is a competitive algorithm for all k, with competitive costat c(k) = Ω(/k). Proof. Let T = [a +, ]. The E[weight gai i T ] = e E Pr[sel(e, T )] w e (7) α(t, k) Pr[e M t e t = e] w e (8) e E ( ) + k l a = Pr[e M t e t = e] w e (9) e E + k l a w(m ) (0) t=a+ = ( a)( + k l a) w(m ), () where (8) is by Lemma 5 ad (0) is by Lemma 6. To show our costat c(k) = max a [0,] {( a)( + k l a)} is Ω(/k), we first give a upper boud. Sice we must have a < (otherwise, we select o edges), we have + k l a <. (2) 6

7 Sice we eed c(k) > 0, ad certaily a 0, we must have + k l a > 0 = l a > k = a > e /k > k = a < k. (3) From (2) ad (3) we coclude c(k) < /k. To show a lower boud, we choose a particular value a k of a; set a k = e /2k. The usig the power series for e x we have a = ( 2k + ) 8k... 2 = 2k + O ( k 2 ). (4) Sice + k l a k = /2 = /2 by costructio, from (4) we have c(k) = 2 2k + O ( k 2 ) > 5k for sufficietly large k. Thus, c(k) is Ω(/k). (5) 7