Minimum-cost matching in a random graph with random costs

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1 Minimum-cost matching in a random grah with random costs Alan Frieze Tony Johansson Deartment of Mathematical Sciences Carnegie Mellon University Pittsburgh PA 523 U.S.A. November 8, 205 Abstract Let G n, be the standard Erdős-Rényi-Gilbert random grah and let G n,n, be the random biartite grah on n + n vertices, where each e [n] 2 aears as an edge indeendently with robability. For a grah G V, E, suose that each edge e E is given an indeendent uniform exonential rate one cost. Let CG denote the random variable equal to the length of the minimum cost erfect matching, assuming that G contains at least one. We show that w.h.. if d n log n 2 then w.h.. E [CG n,n, ] + o π2 6. This generalises the well-known result for the case G K n,n. We also show that w.h.. E [CG n, ] + o π2 2 along with concentration results for both tyes of random grah. Introduction There are many results concerning the otimal value of combinatorial otimization roblems with random costs. Sometimes the costs are associated with n oints generated uniformly at random in the unit square [0, ] 2. In which case the most celebrated result is due to Beardwood, Halton and Hammersley [3] who showed that the minimum length of a tour through the oints a.s. grew as βn /2 for some still unknown β. For more on this and related toics see Steele [9]. The otimisation roblem in [3] is defined by the distances between the oints. So, it is defined by a random matrix where the entries are highly correlated. There have been many examles considered where the matrix of costs contains indeendent entries. Aside from the Travelling Saleserson Problem, the most studied roblems in combinatorial otimization are erhas, the shortest ath roblem; the minimum sanning tree roblem and the matching roblem. As a first examle, consider the shortest ath roblem in the comlete grah K n where the edge lengths are indeendent exonential random variables with rate. We denote the exonential random variable with rate λ by Eλ. Thus PrEλ x e λx for x R. Janson [9] roved among other things that if X i,j denotes the shortest distance between vertices i, j in this model then E [X,2 ] Hn n where H n n i i. Research suorted in art by NSF Grant DMS alan@random.math.cmu.edu Research suorted in art by NSF Grant DMS tjohanss@andrew.cmu.edu

2 As far as the sanning tree roblem is concerned, the first relevant result is due to Frieze [6]. He showed that if the edges of the comlete grah are given indeendent uniform [0, ] edge weights, then the random minimum length of a sanning tree L n satisfies E [L n ] ζ3 i as i 3 n. Further results on this question can be found in Steele [8], Janson [8], Beveridge, Frieze and McDiarmid [4], Frieze, Ruszinko and Thoma [7] and Cooer, Frieze, Ince, Janson and Sencer [5]. In the case of matchings, the nicest results concern the the minimum cost of a matching in a randomly edge-weighted coy of the comlete biartite grah K n,n. If C n denotes the random minimum cost of a erfect matching when edges are given indeendent exonential E random variables then the story begins with Walku [20] who roved that E [C n ] 3. Later Kar [0] roved that E [C n ] 2. Aldous [], [2] roved that lim n E [C n ] ζ2 k. Parisi k 2 [3] conjectured that in fact E [C n ] n k. This was roved indeendently by Linusson and k 2 Wästlund [] and by Nair, Prabhakar and Sharma [2]. A short elegant roof was given by Wästlund [6], [7]. In the aer [4] on the minimum sanning tree roblem, the comlete grah was relaced by a d-regular grah G. Under some mild exansion assumtions, it was shown that if d then ζ3 can be relaced asymtotically by nζ3/d. Now consider a d-regular biartite grah G on 2N vertices. Here d dn as N. Each edge e is assigned a cost we, each indeendently chosen according to the exonential distribution E. Denote the total cost of the minimum-cost erfect matching by CG. We conjecture the following under some ossibly mild restrictions: Conjecture. Suose d dn as N. For any d-regular biartite G, Here the o term goes to zero as N. E [CG] + o N d π 2 6. In this aer we rove the conjecture for random grahs and random biartite grahs. Let G n,n, be the random biartite grah on n+n vertices, where each e [n] 2 aears as an edge indeendently with robability. Suose that each edge e is given an indeendent uniform exonential rate one cost. Theorem. If d n ωlog n 2 where ω then w.h.. E [CG n,n, ] π2 6. Here the statement is that for almost all grahs G G n,n, we have E [CG] π2 6. We will in fact show that in this case CG will be highly concentrated around π2 6. Here A n B n iff A n + ob n as n and the event E n occurs with high robability w.h.. if PrE n o as n. In the case of G n, we rove Theorem 2. If d n ωlog n 2 where ω then w.h.. E [CG n, ] π2 2. Alying results of Talagrand [4] we can rove the following concentration result. 2

3 Theorem 3. Let ε > 0 be fixed, then Pr CG n,n, π2 6 ε n K, for any constant K > 0 and n large enough. Pr CG n, π2 2 ε n K 2 Proof of Theorem We find that the roofs in [6], [7] can be adated to our current situation. Suose that the vertices of G G n,n, are denoted A {a i, i [n]} and B {b j, j [n]}. Let Cn, r denote the cost of the minimum cost matching We will rove that w.h.. for r, 2,..., n m where M r {a i, φ r a i : i, 2,..., r} of A r {a, a 2,..., a r } into B. Using this we argue that w.h.. E [Cn, r Cn, r ] m n ω /2 log n r i0. rn i. E [CG] E [Cn, n] E [Cn, n Cn, n m + ] + + o We will then show that n m r r i0 Theorem follows from these two statements. rn i k n m r r i0 rn i. k 2 π E [Cn, n Cn, n m + ] o w.h Outline of the roof We first argue Lemma that B r φa r is a uniformly random set. This enables us to show Lemma 2 that w.h.. vertices v A r have aroximately n r neighbors in B \ B r. Then comes the beautiful idea of adding a vertex b n+ and joining it to every vertex in A by an edge of cost Eλ. The heart of the roof is in Lemma 3 that relates E [Cn, r Cn, r ] in a recise way to the robability that b n+ is covered by Mr, the minimum cost matching of A r into B B {b n+ }. The roof now focuses on estimating this robability P n, r. If r is not too close to n then this robability can be estimated see 6 by careful conditioning and the use of roerties of the exonential distribution. From thereon, it is a matter of analysing the consequences of the estimate for E [Cn, r Cn, r ] in 7. The final art of the roof involves showing that E [Cn, n Cn m + ] is insignificant. This essentially boils down to showing that w.h.. no edge in the minimum cost matching has cost more than Olog n/n. 3

4 2.2 Proof details Let B r {φ r a i : i, 2,..., r}. Lemma. B r is a random r-subset of B. Proof. Let L denote the n n matrix of edge costs, where Li, j W a i, b j and Li, j if edge a i, b j does not exist in G. For a ermutation π of B let L π be defined by L π i, j Li, πj. Let X, Y be two distinct r-subsets of B and let π be any ermutation of B that takes X into Y. Then we have PrB r L X PrB r L π πx PrB r L π Y PrB r L Y, where the last equality follows from the fact that L and L π have the same distribution. We use the above lemma and the Chernoff bounds to bound degrees. For reference we use the following: Let Bn, denote the binomial random variable with arameters n,. Then for 0 ε and α > 0, PrBn, εn e ε2 n/2. 4 PrBn, + εn e ε2n/3. 5 e αn PrBn, αn. α For v A let d r v {w B \ B r : v, w EG}. Then we have the following lemma: Lemma 2. d r v n r ω /5 n r w.h.. for v A, 0 r n m. Proof. This follows from Lemma i.e. B \ B r is a random set and the Chernoff bounds 4, 5 with ε ω /5 viz. Pr v : d r v n r ω /5 n r 2ne ω 2/5 n r/3 2n ω/0 /3. We can now use the ideas of [6], [7]. We add a secial vertex b n+ to B, with edges to all n vertices of A. Each edge adjacent to b n+ is assigned an Eλ cost indeendently, λ > 0. We now consider M r to be a minimum cost matching of A r into B B {b n+ }. We denote this matching by M r and we let B r denote the corresonding set of vertices of B that are covered by M r. Define P n, r as the normalized robability that v n+ articiates in M r, i.e. Its imortance lies in the following lemma: P n, r lim λ 0 λ Pr {b n+ Br }. 4

5 Lemma 3. E [Cn, r Cn, r ] P n, r. r Proof. Let X Cn, r and let Y Cn, r. Fix i [r] and let w be the cost of the edge a i, b n+, and let I denote the indicator variable for the event that the cost of the cheaest A r - assignment that contains this edge is smaller than the cost of the cheaest A r -assignment that does not use b n+. In other words, I is the indicator variable for the event {Y + w < X}. If a i, b n+ Mr then w < X Y. Conversely, if w < X Y and no other edge from b n+ has cost smaller than X Y, then a i, b n+ Mr, and when λ 0, the robability that there are two distinct edges from b n+ of cost smaller than X Y is of order Oλ 2. Since w is Eλ distributed, as λ 0 we have, E [X Y ] d dλ E [I] lim λ0 λ 0 λ Pr {w < X Y } P n, r. r The factor /r comes from each i [r] being equally likely to be incident to the matching edge containing b n+, if it exists. We now roceed to estimate P n, r. Lemma 4. Suose r < n m. Then Prb n+ B r b n+ / B r λ n r + + ε r + λ 6 where ε r ω /5. Proof. Assume that b n+ / Br. M r is obtained from Mr by finding an augmenting ath P a r,..., a σ, b τ from a r to B \ Br of minimum additional cost. Let α W σ, τ. We condition on i σ, ii the lengths of all edges other than a σ, b j, b j B \ Br and iii min { W σ, j : b j B \ Br } α. With this conditioning Mr Mr will be fixed and so will P a r,..., a σ. We can now use the following fact: Let X, X 2,..., X M be indeendent exonential random variables of rates α, α 2,..., α M. Then the robability that X i is the smallest of them is α i /α + α α M. Furthermore, the robability stays the same if we condition on the value of min {X, X 2,..., X M }. Thus Prb n+ B r b n+ / B r λ d r a σ + λ. Corollary. where ε r ω /5. P n, r n + n ε r 7 n r + 5

6 Proof. Let νj n j + ε j, ε j ω /5. Then Prb n+ Br ν0 ν0 + λ ν ν + λ νr νr + λ + λ λ + ν0 νr ν0 + ν + + νr λ + Oλ 2 n + ε 0 + n + ε + + n r + + ε r and each error factor satisfies / + ε j ω /5. Letting λ 0 gives the lemma. Lemma 5. If r n m then where ε k ω /5. E [Cn, r Cn, r ] + o r Proof. This follows from Lemma 3 and Corollary. r i0 n i This confirms and we turn to 2. We use the following exression from Young [2]. n i i log n + γ + 2n + On 2, Let m ω /4 m. Observe first that m i0 n i n m ri+ r O log n n /4 + m in 3/4 m n i log o + n m in 3/4 o + 2 n m n log m! o + 2m ne n log o. m where γ is Euler s constant. n m ri+ r n i + 2n m + On 3/2 λ + Oλ 2 6

7 Then, n m r r i0 n m rn i i0 n i n m n i im n m n i im n m ri+ n m ri+ log n m n i log im n m jm+ n m xm+ j log x log r, r + o, n m i n m i n m n j n m n x + 2n m 2i + Oi 2 + o, + o, + o, 8 dx + o. We can relace the sum in 8 by an integral because the sequence of summands is unimodal and the terms are all o. Continuing, we have Observe next that n m m So, y 0 n m xm+ n m n m x log n x xm+ n m k xm+ n m m y x log y + m kn m k dy klog n If k log n then we write Now n m m y y + m kn m k dy n m m y y k y k n m xm+ dx x m n m x m k x kn m k dx dx dy. 9 y + m kn m k y k n m m y x m k x kn m k dx n m m y y k kn m k dy k 2. klog n y + m k n m m kn m k dy + y y + m k kn m k dy n m k m + k k 2 n m k k 2 + O 7 o. 0 k2 y k y + m k y + mkn m k dy. kω /4 log n.

8 If k then n m m y And if 2 k log n then It follows that 0 n m m y log n k y + m k y k y + mkn m k dy m logn m o. n m y + m k y k k y + mkn m k dy n m m y n m m l y n m m k l y0 Equation 2 now follows from 9, 0, and 2. Turning to 3 we rove the following lemma: k l y k l m l y + mkn m k dy k y k l m l l kn m k dy k k m l n m m k l l kk ln m k l km O O kk n kω /2 log n y + m k y k log n dy o + O y + mkn m k kω /2 o. 2 log n k2 Lemma 6. If r n m then 0 Cn, r + Cn, r O This will rove that m log n 0 E [Cn, n Cn m + ] O n and comlete the roof of 3 and hence Theorem. log n n. n O ω /2 o n 2.3 Proof of Lemma 6 Let we denote the weight of edge e in G. Let V r A r+ B and let G r be the subgrah of G induced by V r. For a vertex v V r order the neighbors u, u 2,..., of v in G r so that wv, u i wv, u i+. Define the k-neighborhood N k v {u, u 2,..., u k }. Let the k-neighborhood of a set be the union of the k-neighborhoods of its vertices. In articular, for S A r+, T B, N k S {b B : a S : y N k a}, N k T {a A r+ : b T : a N k b}. Given a function φ defining a matching M of A r into B, we define the following digrah: let Γr V r, X where X is an orientation of X {{a, b} G : a A r+, b N 40 a} {{a, b} G : b B, a N 40 b} {φa i, a i : i, 2,..., r}. 8

9 An edge e M is oriented from B to A and has weight w r e we. The remaining edges are oriented from A to B and have weight equal to their weight in G. The arcs of directed aths in Γ r are alternately forwards A B and backwards B A and so they corresond to alternating aths with resect to the matching M. It hels to know Lemma 7, next that given a A r+, b B we can find an alternating ath from a to b with Olog n edges. The ab-diameter will be the maximum over a A r+, b B of the length of a shortest ath from a to b. Lemma 7. W.h.., for every φ, the unweighted ab-diameter of Γ r is at most k 0 3 log 4 n. Proof. For S A r+, T B, let NS {b B : a S such that a, b X}, NT {a A r+ : b T such that a, b X}. We first rove an exansion roerty: that wh, for all S A r+ with S n/5, NS 4 S. Note that NS, NT involve edges oriented from A to B and so do not deend on φ. Pr S : S n/5, NS < 4 S o + n/5 s n/5 s o. n/5 ne s s 4s r + n 40 s 4s n 40 s ne 4s e s 35 n 35 4s 4s 40s n Exlanation: The o term accounts for the robability that each vertex has at least 40 neighbors in Γ r. Condition on this. Over all ossible ways of choosing s vertices and 4s targets, we take the robability that for each of the s vertices, all 40 out-edges fall among the 4s out of the n ossibilities. Similarly, w.h.., for all T B with T n/5, NT 4 T. Thus by the union bound, w.h.. both these events hold. In the remainder of this roof we assume that we are in this good case, in which all small sets S and T have large vertex exansion. Now, choose an arbitrary a A r+, and define S 0, S, S 2,... as the endoints of all alternating aths starting from a and of lengths 0, 2, 4,.... That is, S 0 {a} and S i φ NS i. Since we are in the good case, S i 4 S i rovided S i n/5, and so there exists a smallest index i S such that S is > n/5, and i S log 4 n/5 log 4 n. Arbitrarily discard vertices from S is to create a smaller set S i S with S i S n/5, so that S i S NS i S has cardinality S i S 4 S i S 4n/5. Similarly, for an arbitrary b B, define T 0, T,..., by T 0 {b} and T i φnt i. s s 9

10 Again, we will find an index i T log 4 n whose modified set has cardinality T i T 4n/5. With both S i S and T i T larger than n/2, there must be some a S i S for which b φa T i T. This establishes the existence of an alternating walk and hence removing any cycles an alternating ath of length at most 2i S + i T 2 log 4 n from a to b in Γ r. We will need the following lemma, Lemma 8. Suose that k +k 2 + +k M a log N, and X, X 2,..., X M are indeendent random variables with Y i distributed as the k i th minimum of N indeendent exonential rate one random variables. If µ > then Pr X + + X M µa log N N a+log µ µ. N a log N Proof. Let Y k denote the kth smallest of Y, Y 2,..., Y N, where we assume that k Olog N. Then the density function f k x of Y k is N f k x k e x k e xn k+ k and hence the ith moment of Y k is given by [ ] N E Yk i kx i e x k e xn k+ dx 0 k N kx i+k e xn k+ dx 0 k N i + k! k k N k + i+k k 2 kk + i + k + O N N k + i. Thus, if 0 t < N k +, E [ e ty ] k k 2 + O N i0 t N k + i k + O i If Z X + X X M then if 0 t < N a log N, It follows that Pr Z E [ e tz] M E [ e tx ] i i µa log N N a log N k 2 N t a log N. N a log N t a log N { ex N a log N We ut t N a log N /µ to minimise the above exression, giving µa log N Pr Z µe µ a log N. N a log N t k. N k + tµa log N N a log N }. 0

11 Lemma 9. W.h.., for all φ, the weighted ab-diameter of log n Γ r is at most c n for some absolute contant c > 0. Proof. Let { k } k Z max wx i, y i wy i, x i+, i0 where the maximum is over sequences x 0, y 0, x,..., x k, y k where x i, y i is one of the 40 shortest arcs leaving x i for i 0,,..., k k 0 3 log 4 n, and y i, x i+ is a backwards matching edge. We comute an uer bound on the robability that Z is large. For any ζ > 0 we have Pr Z ζ log n n where i0 k 0 + o k+ on 4 + r + n k+ k n k0 y log n k qρ 0, ρ,..., ρ k ; ζ + y dy k! n y0 ρ 0 +ρ + +ρ k 40k+ qρ 0, ρ,..., ρ k ; η Pr X 0 + X + + X k η log n, n X 0, X,..., X k are indeendent and X j is distributed as the ρ j th minimum of r indeendent exonential random variables. When k 0 there is no term k! y log n n k. Exlanation: The on 4 term is for the robability that there is a vertex in V r that has fewer than on neighbors in V r. We have at most r + n k+ choices for the sequence k x 0, y 0, x,..., x k, y k. The term y log n k! n dy bounds the robability that the sum of k indeendent exonentials, wy 0, x + + wy k, x k, is in log n n [y, y + dy]. The density function for the sum of k indeendent exonentials is xk e x +o n k!. We integrate over y. is the robability that x i, y i is and edge of G and is the ρ i th shortest edge leaving x i, and these events are indeendent for 0 i k. The factor k is the robability that the B to A edges of the ath exist. The final summation bounds the robability that the associated edge lengths sum to at least ζ+y log n n. It follows from Lemma 8 with a 3, N + on, µ ζ + y/a that if ζ is sufficiently large then, for all y 0, qρ,..., ρ k ; ζ + y n ζ+y log n/2 log n n ζ+y/2. Since the number of choices for ρ 0, ρ,..., ρ k is at most 4k+40 k+ the number of ositive integral

12 solutions to a 0 + a a k+ 40k + we have Pr Z ζ log n k 0 on 4 + 2n 2 ζ/2 n k0 for ζ sufficiently large. k 0 on 4 + 2n 2 ζ/2 k0 log n k k! k 0 on n 2 ζ/2 log n on 4, 4k + 40 k + y k n y/2 dy y0 log n k 2 k 2 2 4k+40 z k e z dz k! log n z0 Lemma 9 shows that with robability on 4 in going from M r to M r+ we can find an augmenting ath of weight at most c log n n. This comletes the roof of Lemma 6 and Theorem. Note that to go from w.h.. to exectation we use the fact that w.h.. we Olog n for all e A B, Notice also that in the roof of Lemmas 7 and 9 we can certainly make the failure robability less than n K for any constant K > 0. k0 2 4k 3 Proof of Theorem 2 Just as the roof method for K n,n in [6], [7] can be modified to aly to G n,n,, the roof for K n in [5] can be modified to aly to G n,. 3. Outline of the roof This has many similarities with the roof of Theorem. The differences are subtle. The first is to let Mr be the minimum cost matching of size r after ading a secial vertex v n+. It is again imortant Lemma 0 to estimate the robability that v n+ Mr. The aroach is similar to that for Theorem, excet that we now need to rove searate lower and uer bounds for this robability P n, r. 3.2 Proof details Consider G G n,, and denote the vertex set by V {v, v 2,..., v n }. Add a secial vertex v n+ with Eλ-cost edges to all vertices of V, and let G be the extended grah on V V {v n+ }. Say that v,..., v n are ordinary. Let M r be the minimum cost r-matching one of size r in G, unique with robability one. Note the change in definition. Define P n, r as the normalized robability that v n+ articiates in M r, i.e. P n, r lim λ 0 λ Pr {v n+ articiates in Mr } 2

13 Let Cn, r denote the cost of the cheaest r-assignment of G. To estimate Cn, r, we will again need to estimate P n, r, by the following lemma. Lemma 0. E [Cn, r Cn, r ] P n, r n Proof. Let X Cn, r and Y Cn, r. Fix i [n] and let w be the cost of the edge v i, v n+, and let I denote the indicator variable for the event that the cost of the cheaest r- assignment that contains this edge is smaller than the cost of the cheaest r-assignment that does not use v n+. The rest of the roof is identical to the roof of Lemma 3, excet that there are now n choices for i as oosed to r in the revious lemma. In this case, unlike the biartite case, we are unable to directly find an asymtotic exression for P n, r, as we did in Lemma 6 and Corollary. Here we will have to turn to bounding P n, r from below and above. 3.3 A lower bound for P n, r We will consider an algorithm that finds a set A s V which contains the set B s of vertices articiating in M r, s A s B s 2r. Call A s the set of exosed vertices. Initially let A s B s and r s 0. At stage s of the algorithm we condition on A s, B s and the existence and cost of all edges within A s. In articular, we condition on r and the minimum r-matching M r. Given a minimum matching M r, we decide how to build a roosed r +-matching by comaring the following numbers and icking the smallest. a z a equals the cost of the cheaest edge between a air of unexosed vertices. b z b min{c v : v A s \ B s }, where c v is the cost of the cheaest edge between v and a vertex τ v / A s. c z c min{c v + c u + δu, v : u, v B s } where δu, v denotes the cost of the cheaest alternating ath from u to v, with the cost of edges in M r taken as the negative of the actual value. Let z min min {z a, z b, z c }. If z min z a then we reveal the edge {v, w} and add it to Mr to form Mr+. Once v, w have been determined, they are added to A s and B s, and we move to the next stage of the algorithm, udating s s + 2, r r +. If z min z b then let v A s \ B s be the vertex with the cheaest c v. We reveal w τ v and add w to A s and to B s while adding v to B s. Now M r+ M r {v, w}. We move to the next stage of the algorithm, udating s s +, r r +. 3

14 If z min z c then reveal w τ u, w 2 τ v. If w w 2, we say that we have a collision. In this case, the vertex w is added to A s but not B s, and we move to the next stage with s s +. If there is no collision, we udate M r by the augmenting ath w, u,..., v, w 2 to form M r+. We add w, w 2 to A s and B s, and move on to the next stage with s s + 2 and r r +. It follows that A s \ B s consists of unmatched vertices that have been the subject of a collision. It will be helful to define A s for all s, so in the cases where two vertices are added to A s, we add them sequentially with a coin toss to decide the order. The ossibility of a collision is the reason that not all vertices of A s articiate in M r. However, the robability of a collision at v n+ is Oλ 2, and as λ 0 this is negligible. In other words, as λ 0, Pr {v n+ M r } Pr {v n+ B 2r } Pr {v n+ A 2r } Oλ 2 and we will bound Pr {v n+ A 2r } from below. Lemma. Conditioning on v n+ / A s, A s is a random s-subset of V. Proof. Trivial for s 0. Suose A s is a random s -subset of V. Define N s v {w / A s : v, w E}. In stage s, if we condition on d s v N s v, then under this conditioning N s v is a random d s v-subset of V \ A s. This is because the constructon of A s does not require the edges from A s to V \ A s to be exosed. So, if A s \ A s {w} where w is added due to being the cheaest unexosed neighbor of an exosed v, then w is a random element of N s v and hence a random element of V \ A s. If we are in case a, i.e. Mr+ is formed by adding an edge between two ordinary unexosed vertices v, w, then since we only condition on the size of the set {v, w : v, w / A s }, all airs v, w V \ A s are equally likely, and after a coin toss this can be seen as adding two random elements sequentially. We conclude that A s is a random s-subset of V. Recall that m n/ω /2 log n. Corollary 2. W.h.., for all 0 s n m and all v V, d s v n s ω /5 n s. Proof. This follows from the Chernoff bounds as in Lemma 2. We now bound the robability that A s \ A s {v n+ } from below. There are a few different ways this may haen. We now have to address some cost conditioning issues. Suose that we have just comleted an iteration. First consider the edges between vertices not in A s. For such an edge e, all we know is we ζ where ζ z min of the just comleted iteration. So the conditional cost of such an edge can be exressed as ζ +E or ζ +Eλ in the case where e is incident with v n+. The exonentials are indeendent. We only need to comare the exonential arts of each edge cost here to decide the robability that an edge incident with v n+ is chosen. We can now consider case a. Suose that an edge {u, v} between unexosed vertices is added to A s. By Corollary 2, there are at most n s+ 2 + ω /5 ordinary such edges. There are 4

15 n s edges between v n+ and V \ A s, each at rate λ. As λ 0, the robability that one of the endoints of the edge chosen in case a is v n+ is therefore at least λn s λn s + n s+ 2 + ω /5 2λ n s ω /5 + Oλ 2 We toss a fair coin to decide which vertex in the edge {u, v} goes in A s. Hence the robability that A s \ A s {v n+ } in case a is at least λ n s ω /5 + Oλ 2. We may also have A s \ A s {v n+ } if case a occurs at stage s 2 and v n+ loses the coin toss, in which case the robability is at least λ n s + ω /5 + Oλ 2. Now consider case b. Here only one vertex is added to A s, the cheaest unexosed neighbor w of some v A s \ B s. The cost conditioning here is the same as for case a, i.e. that the cost of an edge is ζ + E or ζ + Eλ. By Corollary 2, this v has at most n s + + ω /5 ordinary unexosed neighbors, so the robability that w v n+ is at least λ n s + + ω /5 + λ λ n s + ω /5 + Oλ 2. Finally, consider case c. To handle the cost conditioning, we condition on the values c v for v B s. By well-known roerties of indeendent exonential variables, the minimum is located with robability roortional to the rates of the corresonding exonential variables. A collision at v n+ has robability Oλ 2, so assume we are in the case of two distinct unexosed vertices w, w 2. Suose that w is revealed first. Exactly as in b, the robability that w v n+ is at least λ n s + ω /5 + Oλ 2. If w v n+, the robability that w 2 v n+ i.e. A s+ \ A s {v n+ } is at least λ n s ω /5 + Oλ 2, so by considering the ossibility that v n+ is the second vertex added from A s 2, we again have robability at least λ n s + ω /5 + Oλ 2. We conclude that no matter which case occurs, the robability is at least So Write P n, r lim λ 0 λ 2r s λ n s + ω /5 + Oλ 2. Pr {A s \ A s {v n+ }} ω /5 Ln, r 2r s n s +. 2r s n s +. 5

16 3.4 An uer bound for P n, r We now alter the algorithm above in such a way that A 2r B 2r. We do not consider A s for odd s here. At a stage with s 2r, we condition on A s, and the aearance and cost of all edges within A s. In articular, we condition on r and the minimum r-matching Mr. set C s A s, where each v C s has been involved in a collision. This changes how we calculate a candidate for Mr+. We now take the minimum of A a z a equals the cost of the cheaest edge between unexosed vertices. b z b min{c u + c v + δu, v : u, v A s, {u, v} C s }, where c and δ are as defined in Section 3.3. c z c min{c u + c 2 v + δu, v : u, v C s, τ u τ v}, where τ is defined in Section 3.3 and c 2 v is the cost of the second cheaest edge between v and a vertex τ 2 v / A s. Let z min min {z a, z b, z c }. If z min z a then we reveal the edge {v, w} and add it to Mr to form Mr+. Once v, w have been determined, they are added to A s and we move to the next stage of the algorithm, udating s s + 2. If z min z b then reveal w τ u, w 2 τ v. If w w 2 then we add u, v to C s and go to the next stage of the algorithm without changing s. The robability that τ u τ v v n+ is Oλ 2, and we can safely ignore this as λ 0. If at some later stage w is added to A s and u say is still in A s then we remove u from C s. If w w 2 then we udate Mr by the augmenting ath w, u,..., v, w 2 to form Mr+. We add w, w 2 to A s, and move on to the next stage with s s + 2. If z min z c then we udate Mr by the augmenting ath w τ u, u,..., v, w 2 τ 2 v to form Mr+. We add w, w 2 to A s, and move on to the next stage with s s + 2. Eventually we will construct M r+ since case b with τ u τ v can haen at most s times before C s A s. The cost conditioning is the same as we that for comuting the lower bound in Section 3.3, excet for the need to deal with c 2 v, v C s. For this we condition on c 2 v and argue that the robability δ 2 v x is roortional to the exonential rate for the edge v, x. At this oint we know that δ v v n+, since we are assuming λ is so small that this ossibility can be ignored. So, in this case, we can only add v n+ as δ 2 v for some v C s. To analyze this algorithm we again need to show that A 2r is a uniformly random subset of V. Lemma 2. Conditioning on v n+ / A 2r, A 2r is a random 2r-subset of V. Proof. Let L denote the n n matrix of edge costs, where Li, j wv i, v j and Li, j if edge v i, v j does not exist in G. For a ermutation π of V let L π be defined by L π i, j Lπi, πj. 6

17 Let X, Y be two distinct 2r-subsets of V and let π be any ermutation of V that takes X into Y. Then we have PrA 2r L X PrA 2r L π πx PrA 2r L π Y PrA 2r L Y, where the last equality follows from the fact that L and L π have the same distribution. This shows that A 2r is a random 2r-subset of V. Let d 2r v {w / A 2r : v, w E}. Corollary 3. W.h.., for all 0 r n m/2 and all v V, d 2r v n 2r ω /5 n 2r. Proof. The roof is again via Chernoff bounds, see Lemma 2. We bound the robability that v n+ A 2r \ A 2r 2 from above. Suose we are at a stage where a collisionless candidate for M r has been found. In case a, as in the revious section the robability that v n+ is one of the two unexosed vertices is at most λn 2r + 2 λn 2r n 2r+2 2 ω /5 2λ n 2r + + ω /5 + Oλ 2 Now suose we are in case b with u, v / C 2r 2. If no collision occurs, the robability that one of τ u, τ v is v n+ is at most λ λ + d s u + λ 0 s λ λ + d s v 2λ n 2r + + ω /5 + Oλ 2 Finally, if we find Mr+ by alternating aths where one exosed vertex uses its second-cheaest edge to an unexosed vertex, the robability of that vertex being v n+ is even smaller at λ/n 2r +. So, r P n, r lim Pr {v n+ A 2s \ A 2s 2 } 2 + ω /5 r n 2s + Write Un, r r s 2 n 2s +. s 7

18 3.5 Calculating E [CG n, ] From Lemma 0 and 3.3 we have E [Cn, n m/2] by 2.2. n m/2 r + o + o + o P n r +, n m/2 r + n r + n m/2 r n m/2 r n m/2 r Ln r +, n m/2 r + n r + n m 2r+2 n r + s log n r + n r + s + n r + m + r + 2n r 2m + r + Om 2, 3 The correction terms are easily taken care of. First we have n m/2 n r + 2n r 2m + r + Om 2 r O n m/2 m n r + r n m O mn o. Now we want to relace the m + r term in the logarithm in the RHS of 3 by r. For this we let m n/ω /4 log n mω /4. Then n m/2 r n r + log m + r r m r m log m n r + log + m n m/2 + r r n r + + log log n m + m n m/2 n m m n/2 o. rm n r + log + m r + mm n m/2 rm n r + 8

19 So, Substituting y log/α we have n m/2 n r RHS3 o + n r + log r /2 0 α log o + α α r /2 0 α log α α y dα e y + dy j0 ye y + e 0 y dy j+ j 2 j k π2 2. k 2 dα. ye y e y j dy This roves a lower bound for E [Cn, n m/2]. It also shows that n m/2 r Ln r +, n m/2 r + + oπ2 n r For an uer bound, note that for r n m/2, r Un, r Ln, r n 2s + n 2s + 2 s r n 2s + n 2s + 2 s r O n 2r 2 9

20 So, E [Cn, n m/2] + o + o + o n m/2 r n m/2 r o + Un r +, n m/2 r + n r + Ln r +, n m/2 r + + O n r + n m/2 r n r + r n 2r 2 Ln r +, n m/2 r + 5 π2 + o. 2 6 To get from 5 to 6 we use 4. We show that for n even, E [Cn, n/2 Cn, n m/2] o/ to conclude that E [CG n, ] E [Cn, n/2 Cn, n m/2] + E [Cn, n m/2] π2 + o. 2 As above, this will follow from the following lemma. Lemma 3. Suose n is even. If n m/2 r n/2 then 0 E [Cn, r + Cn, r] O. log n n 3.6 Proof of Lemma 3 This section will relace Section 2.3. Let M {v, φv, v [n]}, φ 2 v v for all v [n] be an arbitrary erfect matching of G n,. We let M {u, v : v φu} consist of two oositely oriented coies of each edge of M. We then randomly orient the edges of G n, that are not in M and then add M to obtain the digrah G G n,. Because n ωlog n 2 we have that w.h.. the minimum in- or out-degree in G n, is at least ωlog n 2 /3. Let D be the event that all in- and out-degrees are at least this large. Let the M-alternating diameter of G be the maximum over airs of vertices u v of the minimum length of an odd length M-alternating ath w.r.t. M between u and v where i the edges are oriented along the ath in the direction u to v, ii the first and last edges are not in M. Given this orientation, we define Γ r to be the subdigrah of G consisting of the r cheaest non-m out-edges from each vertex together with M. Once we can show that the M-alternating diameter of Γ 20 is at most 3 log 3 n, the roof follows the roof of Lemma 9 more or less exactly. Lemma 4. W.h.., the alternating diameter of Γ 20 is at most k 0 3 log 3 n. Proof. We first consider the relatively simle case where n n /3 log n. Let N + u be the set of out-neighbors of u in G and let N v be the set of in-neighbors of v in G. If there is an edge of M from N + u to N v then this creates an alternating ath of length three. Otherwise, let N ++ u be the other endoints of the matching edges incident with N + u and define N v analogously. 20

21 Note that now we have N ++ u N v and given D, the conditional robability that there is no edge from N ++ u to N v in G is at most + o n/32 e log n3 /0 on 2. Thus in this case there will be an alternating ath of length five. Now assume that n < n /3 log n. In which case we can rove W.h.. S n 7/2 imlies that es 6 S. 7 This follows from Pr S violating 7 n 7/2 s3 n 7/2 s3 n 7/2 s3 o. n s ne s s 2 6s 6s s se 6s 2 e 7 s 5 log n n 3 s Imitating Lemma 7, we rove an exansion roerty for Γ 20 : Pr S : S n 2/3, N 20 S < 0 S o + n 2/3 s n 2/3 s n s ne s ne s 0s n 2/3 e 20 s n 9 s n s 0s s 20 n 20 0s s 20s n s s o. 8 Fix an arbitrary air of vertices a, b. Define S i, i 0,,... to be the set of vertices v such that there exists a directed M-alternating ath of length 2i in Γ 20 from a to v. We let S 0 {a} { and given S i we let S i N + S i \ {b} and S i {w b : {v, w} M : v S i }. Here N + S w / S : v S, v, w E } Γ is the set of out-neighbors of S. N S is similarly defined as the set of in-neighbors. It follows from 7 and 8 that w.h.. S i 3 S, so long as S i on 7/2. We therefore let S i+ be a subset of S i of size 3 S i. So w.h.. there exists an i a log 3 n such that S ia [n 3/24, 3n 3/24 ]. Reeat the rocedure with vertex b, letting T 0 {b}, T j+ N T j etc. By the same argument, there exists an j b log 3 n such that T jb is of size in [n 3/24, 3n 3/24 ]. Finally, the robability that there is no S ia T jb edge is at most n3/2 on 2. This comletes the roof of Lemma 4. The remainder of the roof of Lemma 3 is now exactly as in Section 2.3. This concludes the roof of Theorem 2. 2

22 4 Proof of Theorem 3 The roof of Lemma 9 allows us to claim that for any constant K > 0, with robability On K log n the maximum length of an edge in the minimum cost erfect matching of G is at most µ c 2 n for some constant c 2 c 2 K > 0. We can now roceed as in Talagrand s roof of concentration for the assignment roblem. We let ŵe min {we, µ} and let ĈG be the assignment cost using ŵ in lace of w. We observe that and so it is enough to rove concentration of ĈG. PrĈG CG On K 9 For this we use the following result of Talagrand [4]: consider a family F of N-tules α α i i N of non-negative real numbers. Let Z min α i X i α F where X, X 2,..., X N are an indeendent sequence of random variables taking values in [0, ]. Let σ max α F α 2. Then if M is the median of Z and u > 0, we have i N } Pr Z M u 4 ex { u2 4σ We aly 20 with N n 2 and X e ŵe/µ. For F we take the n! {0, } vectors corresonding to erfect matchings and scale them by µ. In this way, e α ex e will be the weight of a erfect matching. In this case we have σ 2 nµ 2. Alying 20 we obtain Pr ĈG M ε } { 4 ex { ε2 4 2 nµ 2 ex ε2 n c 2 log n 2 where M is the median of ĈG. Theorem 2 follows easily from 9 and 2. }, 2 5 Final remarks We have generalised the result of [2] to the random biartite grah G n,n, and the result of [5] to the random grah G n,. It would be of some interest to extend the result in some way to random regular grahs. In the absence of roving Conjecture we could maybe extend the results of [2], [5] to some secial class of secial grahs e.g. to the hyercube. References [] D. Aldous, Asymtotics in the random assignment roblem, Probability Theory and Related Fields [2] D. Aldous, The ζ2 limit in the random assignment roblem, Random Structures and Algorithms

23 [3] J. Beardwood, J. H. Halton and J. M. Hammersley, The shortest ath through many oints, Mathematical Proceedings of the Cambridge Philosohical Society [4] A. Beveridge, A.M. Frieze and C.McDiarmid, Random minimum length sanning trees in regular grahs, Combinatorica [5] C. Cooer, A.M. Frieze, N. Ince, S. Janson and J. Sencer, On the length of a random minimum sanning tree, see arxiv.org. [6] A.M. Frieze, On the value of a random minimum sanning tree roblem, Discrete Alied Mathematics [7] A.M. Frieze, M. Ruszinko and L. Thoma, A note on random minimum length sanning trees, Electronic Journal of Combinatorics R4. [8] S. Janson, The minimal sanning tree in a comlete grah and a functional limit theorem for trees in a random grah, Random Structures Algorithms [9] S. Janson, One, two and three times log n/n for aths in a comlete grah with random weights, Combinatorics, Probability and Comuting [0] R.M. Kar, An uer bound on the exected cost of an otimal assignment, Discrete Algorithms and Comlexity: Proceedings of the Jaan-US Joint Seminar D. Johnson et al., eds., Academic Press, New York, 987, -4. [] S. Linusson and J. Wästlund, A roof of Parisi s conjecture on the random assignment roblem, Probability Theory and Related Fields [2] C. Nair, B. Prabhakar and M. Sharma, Proofs of the Parisi and Coersmith-Sorkin random assignment conjectures, Random Structures and Algorithms [3] G. Parisi, A conjecture on Random Biartite Matching, Pysics e-print archive 998. [4] M. Talagrand, Concentration of measure and isoerimetric inequalities in roduct saces, Publications Mathmatiques de L IHÉS 995. [5] J. Wästlund, Random matching roblems on the comlete grah, Electronic Communications in Probability [6] J. Wästlund, A simle roof of the Parisi and Coersmith-Sorkin formulas for the random assignment roblem, Linköing Studies in Mathematics [7] J. Wästlund, An easy roof of the ζ2 limit in the random assignment roblem, Elec. Comm. in Probab [8] J. Michael Steele, On Frieze s ζ3 limit for lengths of minimal sanning trees, Discrete Alied Mathematics [9] J. Michael Steele, Probability Theory and Combinatorial Otimization, SIAM CBMS series, 996. [20] D.W. Walku, On the exected value of a random asignment roblem, SIAM Journal on Comuting [2] R.M. Young, Euler s constant, Mathematical Gazette