The Minimum Perfect Matching in Pseudo-dimension 0

It is known that for $K_{n,n}$ equipped with i.i.d. $exp(1)$ edge costs, the minimum total cost of a perfect matching converges to $\pi^2/6$ in probability. Similar convergence has been established for all edge cost distributions of pseudo-dimension $q \geq 1$, such as Wei(1,q) costs. In this paper we extend those results all $q>0$, confirming the M\'ezard-Parisi conjecture in the last remaining applicable case.


Introduction
There has been substantial interest over the past few decades in the minimum matching problem. Given a graph G, and a positive cost associated to each edge of G, we want to find a perfect matching of minimal total cost M (G). Of special interest is minimum matching on the complete bipartite graph K n,n on n + n vertices with random edge costs given by independent exp(1)variables, sometimes referred to as the random assignment problem. For this graph model, the lower bound 1 on the cost M (K n,n ) of minimum matching is trivial, and the upper bound 3 was established by Walkup [3] by finding perfect matchings using only cheap edges. Mézard & Parisi [2] conjectured that M (K n,n ) converge in probability to π 2 /6, based on non-rigorous replica symmetry calculations. This was later proven by Aldous [1].
A natural question is whether these results extend to other edge cost distributions. A random graph with i.i.d. edge costs given by the distribution function F is said to be of pseudo-dimension q if F (x) ∼ x q for small x. The exponential distribution is of pseudo-dimension 1, and more generally a Weibull distribution with shape parameter q is of pseudo-dimension q (regardless of mean). Mézard and Parisi [2] considered these distributions for real positive q, but most focus since then has been on the special case q = 1.
The motivation for the term pseudo-dimension is this: A geometric graph model is given by embedding the vertices as n random points in a hypercube [0, 1] q , and setting the edge costs to be the corresponding Euclidean distances. The mean field approximation, i.e. the graph model given by independently rerandomizing each edge, of a geometric graph model of dimension q is a graph model of pseudo-dimension q.
Let K q n,n denote K n,n equipped with independent Weibull(1, q) edge costs. The cost of minimum matching on it can be shown to be of order n 1−1/q , by a minor modification of [3]. This suggests studying the quantity n −1+1/q M (K q n,n ). Does it converge in probability to a constant for q = 1? This question was answered in the affirmative for q > 1 in 2011 by Wästlund [4], but it remained open for 0 < q < 1. Our main result is the following theorem, from which the desired result follows as a corollary: Theorem 1: Let T q be an infinite Galton-Watson tree where each vertex gives rise to offspring by a Poisson random measure on R + of intensity qt q−1 at t, and for any weighted graph G let G λ be the subgraph with all edges of cost more than λ removed. If G n is a graph model of pseudodimension q > 0, such that G n,λ has the local limit T q λ for any λ > 0, the following then holds: limprob n→∞ n −1 M (G n ) exists and is non-random.
In order to apply this theorem to a given graph model, we might need to multiply all its edge costs by a suitable number, possible depending on n. If we multiply the edge costs of K q n,n by Γ(1 + 1/q)n 1/q , we get a graph (with edge cost distribution Weibull(n, q)) whose local limit is precisely T q λ , leading to the corollary.
Examples of other graphs that, if equipped with edge costs of pseudodimension q, satisfy the conditions of theorem 1 include complete m-partite graphs for any m ≥ 1, Erdős-Rényi graphs, and quasi-random graphs.

Proof strategy
Following the work by Wästlund [4], we will not deal with matchings directly, but instead study the game Exploration. This zero-sum game is played in the following way: On an edge-weighted graph G, Alice and Bob takes turns picking the next edge of a self-avoiding walk starting from a designated vertex φ. When it is a player's turn (Alice's, say), and the current vertex is u, she can take one of two actions: (i) Pick any neighbour v of u that has not already been visited, and pay Bob the cost of the (u, v). Bob then continues the game from v.
(ii) Quit the game, and pay Bob a penalty of λ/2, for some fixed positive parameter λ.
The payoff for Alice, once the game has finished, is the total amount Bob has payed to her minus the total amount she has payed to Bob. Each player's aim is to maximize their payoff. It is never optimal for Alice to choose an edge more expensive than λ, since even in the best case scenario where Bob quits immediately afterwards, Alice will have lost more than the penalty of λ/2 it would have cost her to quit instead. Similarly, Bob will never want to pick such an edge either. The game itself is entirely non-random on any fixed graph, and the game value of any game position is well defined if the graph is finite. For infinite graphs, however, the latter may not be true. Wästlund [4] established that for any q > 0 and G n satisfying the conditions of theorem 1, the limit of n −1 M (G n ) exists if the game value of Exploration played on T q λ is well defined for any λ. He proceeded to prove it was indeed well defined for q ≥ 1, but the proof did not work for 0 < q < 1. Consider the most optimistic valuations of the game Alice and Bob could possibly have (f A and f B , respectively). These valuations exist, and any other valuation lies between them, so it suffices to show that they are identical [4, p.1072].
The main idea of Wästlund's proof was to view f A as known and f B as unknown, and use f A to predict which moves Bob could not possibly consider to be optimal. It turns out that the difference |f A − f B | is non-increasing along the game path if each player plays optimally with respect to their valuation, and for any δ > 0 the moves by Bob will be within δ of being optimal w.r.t. f A eventually -after the vertex u, say, has been reached. Such moves are called δ-reasonable. For q ≥ 1, δ can be chosen sufficiently small so that the tree rooted in u consisting of all δ-reasonable moves have branching number bounded away from 1. This tree is guaranteed to contain the game path from u onwards, so the game will almost surely end after a finite number of moves, and f A and f B must agree.
We will employ a similar strategy to extend the result to q > 0, but the proof will be somewhat more involved. By keeping more careful track of the game history from the root φ to a vertex u, we will be able to rule out more move options from u. Instead of considering a single move as reasonable, we consider entire game paths, and call a game path (u, t)-reasonable if the sum of all of Bob's deviations from f A , starting from u, is at most t. A move is called (u, t)-reasonable if it belongs to some (u, t)-reasonable path from u. Let ∆ t (u) be the tree of (u, t)-reasonable moves. The actual game path is a (φ, λ)-reasonable game path, so it is contained in the tree ∆ λ (φ).
However, ∆ t does not have branching number uniformly bounded away from 1. We will instead bound the size of subtrees of ∆ t , using that the subtree of T q λ rooted in any vertex has the same distribution as the whole T q λ , and thus bound ∆ t inductively. Note that the game value of a vertex is determined by the game values of its children together with the cost of the edges to them, and if the game value of a vertex is very close to either λ/2 or −λ/2 it is likely that the game will end soon. We will therefor consider the expected size of the tree of (u, t)-reasonable moves, conditioned on the game value of the root being z, so that the bound is a function of both z and t. We will show that the inductive step involves applying a certain linear (in z) operator L B • L A to this function. The problem then amounts to finding functions ψ t (z) such that L B • L A (ψ t ) ≤ ψ t holds (with sufficient margin to deal with some technicalities). We will first bound the norm of this operator away from 1, and then use it to construct ψ t that satisfies the inequality.

Main result
Our main theorem 1 follows from proposition 3, which in turn uses lemmas 4 to 7. Proof of theorem 1. Assume q < 1, the theorem is already known to be true otherwise. For any λ > 0, by proposition 3 there exists a tree ∆ λ ⊆ T q λ which almost surely is finite and contains the actual game path. f A and f B satisfy identical recursive relationships, and if ∆ λ is finite they have the same boundary conditions. Hence f A (φ) = f B (φ) almost surely. The argument of Wästlund [4, p.1076-1087] can now be applied, giving that the limit of In what follows, we will use as short-hand for λ/2 −λ/2 if no other integration limits are given, and we will let |G| denote the number of edges of the graph G. We will assume 0 < q < 1 and λ > 0 are fixed, and often suppress dependence on them in our notation. Figure 1: Each point vi corresponds to a child of some fixed vertex (u, say). Its coordinates are fi, the game value of the vertex vi, and ℓi, the cost of the edge (u, vi). For a fixed i, ℓi and fi are independent, so we can give the ℓf -square the product measure µ given by the distributions of ℓ and fA(φ). We can then view the points as having arisen from a Poisson random measure with intensity measure µ, so that the number of points in any set S is a Poisson random variable with mean µA(S).
In this picture, v7 is the optimal move as it minimizes the quantity ℓ − f .
As a first step, we will need to modify slightly how we generate the tree T = T q λ , so that we condition on f A in each step, and label each vertex u with f A (u). Consider the distribution of Note that for a vertex u at even distance from the root, f A (u) has the same distribution as f A (φ), since the subtree rooted in u has the same distribution as the entire tree T and Alice is the player to move both from φ and u. If, on the other hand, u is at odd distance from the root, Bob is the player to move from u and f A (u) has the same distribution as f B (φ).
We start by using F A to generate the label of the root of a new treeT . For any vertex u ∈T (with label z, say) at even distance from the root, we generate its children v i , their edge costs ℓ i , and their labels f i concurrently, conditioned on f A (u) = z. Let the ℓf -square be the set [0, λ] × [−λ/2, λ/2] of all possible such pairs (ℓ i , f i ), see fig. 1. In other words, we condition on there being one point on the diagonal ℓ − f = z in the ℓf -square and no points below it. Letting µ A be the product measure on the ℓf -square given by the distribution of f A (φ) and the measure on [0, λ] of density qt q−1 at t, we generate the point on the diagonal ℓ−f = z by using its normalized marginal measure as probability measure. We then generate the points above the diagonal as a Poisson random measure with intensity measure µ A | {ℓ−f >z} .
For u at odd distance from the root, we generate its children in almost the same way. Starting from such a vertex, the roles of Alice and Bob are interchanged, so we construct the intensity This construction gives a treeT identical in distribution to T , but with vertex labels. We identifyT with T . Since we conditioned on exactly the information Alice needs to make her choice, and labelled the vertices accordingly, her move from any vertex is determined by the labels of its children and the cost of the edges to them. Bob's moves, however, may depend on the entirety of the tree, which is why it makes sense to see his strategy as unknown.
By [4, p.1076], the difference between f A and f B is non-increasing along the game path, and it decreases precisely when either player makes a move that is suboptimal with respect to the other player's valuation. We will only need to consider the times when Bob's f B -optimal move is not f A -optimal, and we will refer to them as mistakes. Let the cost of the move u → v to be If eq. (1) holds, and Bob moves from u to v (a mistake of cost δ, say), we can conclude that Possibly f B -optimal move Figure 2: A portion of the tree ∆t(φ) of reasonable game paths. Alice starts by moving from φ to u, then expects Bob to move to v0. Bob, however, may consider moving to some other vertex to be optimal. The vertices v1, 1 ≤ i ≤ 4, are those of his move options that would be mistakes costing at most t, and which we cannot rule out using the bound |fA(φ) − fB(φ)| ≤ t. Similarly for the subtree rooted in v0, where we have the bound |fA(v0) − fB(v0)| ≤ t, and can use it to rule out all move options but v 1 i , 0 ≤ i ≤ 3. But for the subtree rooted in v3, we know that the move (u → v3) was a mistake (costing δ, say) and we therefore get the slightly better bound |fA(v3) − fB(v3)| ≤ t − δ. Using this bound, we can rule out more move options than we would have been able to using only In this way we can get an upper bound on |f A (w)−f B (w)| for any vertex w that was reached from u by f A -optimal moves by Alice and f B -optimal moves by Bob: t minus the sum of the costs of Bob's mistakes. And to find such a bound for the root is easy: We can therefore use eq. (1) to rule out some of Bob's move options from u: those that are mistakes costing more than t. Ruling out move options in this way can be done without actually knowing the valuation f B at any vertex, as long as we have some bound on |f A −f B |. The better the bound, the more move options we can rule out.
Thus we can construct the subtree ∆ t (u) of game paths starting from u consisting of only optimal moves by Alice and (u, t)-reasonable moves by Bob. Figure 2 shows a portion of the tree ∆ t (φ). Our aim is to bound the size of ∆ t (u), and we will do this by finding a bound for the size of k-level truncations ∆ k t (u). Conditioned on f A (u), the distribution of ∆ k t (u) is the same for every u at even distance from the root, so we let Proposition 3: There exists a family of continuous functions ψ t s.t. R k t (z) < ψ t (z) for all z and even k, and satisfying sup z,t ψ t (z) < ∞. Hence ∆ λ is almost surely finite.
To prove proposition 3 we will need the following lemmas. The proofs of the technical lemmas 4 and 5 have been relegated to the appendix. Lemma 7 is central to this paper. Furthermore, is continuous in z, and for some constant α and all −λ/2 < z < λ/2, we have the bounds Equation (5) also holds for λ/2 < z ≤ 3λ/2, and analogous results hold for F ′ B . We will use eq. (5) in another form: Lemma 5: For z ∈ (−λ/2, λ/2), let ρ z A (t) be the density along the diagonal ℓ − f = z: and let the positive linear operator L A and function I A be defined by on (−λ/2, λ/2), and by their continuous extensions at ±λ/2. Let also ρ z B , L B and I B be defined similarly. If the move φ → u by Alice is f A -optimal and the move u → v by Bob is both f A -and f B -optimal, the following holds: Furthermore, I A satisfies these properties: (i) I A is continuous, (ii) I A (z) < 1 for z ∈ [−λ/2, λ/2), and (iii) I A (±λ/2) is well defined. Analogous statements hold for I B .
Proof of lemma 6. The lemma follows by letting t = λ in fig. 2.
Proof of lemma 7. L A is a substochastic operator 1 , and to be able to fully leverage this property we will factorize it into a stochastic operator that has almost all the structure of L A and a substochastic operator that is a diagonal map. Start by defining the kernel κ z A (t), as ρ z A normalized for (z, t) ∈ (−λ/2, λ/2) 2 : (11) Using this kernel, we write L A (h)(z) as I A (z)h(t)κ z A (t)dt. The factor I A (z) does not depend on t, so it can be factored out of the integral. We therefore see L A as the composition of the operators S A and D A , defined by The proof of the lemma will be divided into two cases, depending on whether I A (λ/2) = I B (λ/2) = 1 or not.
Assume without loss of generality that I A (λ/2) < 1. Then I A (z) < 1 for all z. By lemma 5, I A is a continuous function on a closed interval, so it attains its maximum θ, which must be less than 1. Thus we get that D A ≤ θ < 1. In order to show that this integral operator has norm less than 1, we will bound the integral of its kernel along the line s = z, where z ∈ (−λ/2, λ/2] is arbitrary but fixed. (We do not need to consider the case z = −λ/2, since I B (−λ/2) < 1.) Since I B (λ/2) = 1, we know that is the product of q(z + t) q−1 and −F ′ B (t), and both λ/2 −z q(z + t) q−1 dt and − λ/2 −z F ′ B (t)dt are finite. Each of these three integrands have a pole at either t = −z or t = −λ/2, and q(z + t) q−1 has no other pole. Hence the reason that the first integral λ/2 −z ρ z B (t)dt goes to infinity is that poles of q(z + t) q−1 and F ′ B get close to each other when z is close to λ/2, and those poles must be the one at t = −z for q(z + t) q−1 and t = −λ/2 for F ′ B , respectively.
While each of those two poles are integrable by themselves, their product is not. Therefore ρ z B (t) is integrable away from z = λ/2, and most of its mass must be concentrated near t = −z for z near λ/2. κ z B is just ρ z B normalized, so its mass must also be concentrated in this way. This implies that we can find 0 < η < λ/2 such that for all z > η, By (ii) of lemma 5, we can find δ > 0, such that when t ≤ η we have Then, for −λ/2 ≤ z ≤ η, we apply the bound to I B to get while for η < z ≤ λ/2 we apply it to I A This means that the weight along each line of As we have finished the proof in these two cases, we can conclude that L B • L A < 1.
Now that we have lemmas 4 to 7, we can proceed with the proof of proposition 3.
Proof of Proposition 3. We will use L B • L A < 1 to construct suitable ψ t . In a slight abuse of notation, let 1 denote both the natural number 1 and the function on [−λ/2, λ/2] with constant value 1. Let the functions ψ t (for any 0 ≤ t ≤ λ and some large constants K, m > 0 to be determined later) be defined by By lemma 7, L B • L A < 1, so the above series is absolutely convergent. Hence its norm is at most K exp(mt)/(1 − L B • L A ). K exp(mt) is a constant with respect to z, so if we apply the operator L B • L A to ψ t we can apply it termwise to the sum in eq. (16).
Crucially, L B • L A (ψ t ) is less than ψ t , and with a sizeable margin. We will do induction on even k to establish the main claim. Fix some −λ/2 ≤ z ≤ λ/2 and 0 ≤ t ≤ λ, and consider the first two moves of the game conditioned on f A (φ) = z. Let u be Alice's optimal move from the root (if such a move exists, the result holds trivially otherwise), and v 0 the vertex she expects Bob to move to after that. Assume there are n of Bob's move options from u that are mistakes costing at most t. Let v i , 1 ≤ i ≤ n, be the vertices those moves lead to, t i the cost of each mistake, ℓ i the cost of the edge (u, v i ), and let f i := f A (v i ). Note that t i , v i , f i and n are random variables, with distribution conditional on f A (φ) = z.
For the base case k = 2, the root has at most one child u in ∆ 2 t (φ). Its expected number of children, conditioned on any value of f A (u), is at most 1 + λ q , so we let K = 2(2 + λ q ). Then R 2 t ≤ 2 + λ q < ψ t , establishing the base case. Next, assume R k s < ψ s for some even k > 2 and all 0 ≤ s ≤ λ. We want to show that R k+2 t < ψ t as well, and to do that we will bound the expected size of ∆ k+2 t . The tree ∆ k+2 t can be written as an edge-disjoint union of copies of ∆ in the following way: We already have a bound on ∆ 2 t (φ), and we continue by bounding the conditional expected sizes of ∆ k t (v 0 ) and , by the definition of R k t and lemma 5, Finally, we bound the expected size of the union of the trees ∆ k t−ti (v i ). To do this we condition first on the random variables n, f i and t i and then on the event f A (φ) = z, so that the first conditional expectation is itself a random variable.
since the subtree rooted in Let σ A be the Poisson random measure generated by µ A . It is a sum of Dirac measures, each corresponding to a point in the ℓf -square. Among these points, (ℓ i , f i ), 1 ≤ i ≤ n, are exactly those that lie in the diagonal strip z < ℓ − f ≤ z + t. Note that for any bounded µ A -measurable function h, we have that E[ h dσ A ] = h dµ A (which can be seen by approximating h by simple functions). The expression 20 is then at most We have that ψ s ≤ K exp(ms)/(1 − L B • L A ) for all s, so if we use this bound and integrate first along diagonals ℓ − f = x for x fixed, and then integrate over x, we get where ε m → 0 as m → ∞, and does not depend on k, t or z. We now have a bound on the expected size of each term in the right hand side of eq. (18). The bounds from eqs. (19) and (22) give that We pick m large enough so that ε m < 1/2. The expression (23) is then less than ψ t (z), and the inductive step is complete. Hence R k t < ψ t for all even k and 0 ≤ t ≤ λ.

Appendix
In this appendix, all functions considered will be real-valued functions on [−λ/2, λ/2]. For f and g functions, we will henceforth use f ≤ g to mean that f (z) ≤ g(z) for all z.
Proof of lemma 4. We will consider some bounded non-increasing function 0 < G ≤ 1, derive the desired results for G, and then show that F A and F B satisfy the necessary conditions. This proof hinges on the relationships where V is the non-linear operator defined by Note that V (G)(−λ/2) = 1, while 0 < V (G)(z) < 1 for all other z. Bounded monotone functions have bounded variation, so we can integrate with respect to the measure dG, in the sense of a Riemann-Stieltjes integral. Let dG also have a point measure of mass −G(λ/2) at λ/2, as if G had a jump discontinuity there. We claim that, given this definition of dG, To verify eq. (24), start by integrating −z q(z +t) q−1 dG(t) = d dz ln(V (G)(z)), which implies eq. (24) and that V (G) is differentiable. Next, define (for some a > 0 to be determined later) the function g as We will show that if G is differentiable and satisfies −G ′ ≤ g, V (G) also satisfies this bound. We need to calculate (and then estimate) d dz V (G)(z). Using eq. (24), we find that Setting G = F A or F B gives eq. (3). Next, we will establish eqs. (4) and (5). The integrand on the right hand side of eq. (26) has a pole at t = −z, and the bound g has a pole at t = λ/2. For some positive parameter r < 16 −1/q , we will deal separately with the cases when the poles are within 2r of each other and when they are further apart. We will establish that the following inequality holds in both cases: from which it follows that − d dz V (G)(z) < g(z) for all z by letting a > max(8q, λ/r).
We apply the bound −G ′ (t) ≤ g(t), and use that the resulting integrand is symmetric around t = z/2 + λ/4: We will later be using the tighter bound in eq. (28), but for now eq. (29) suffices. Again using eq. (26), this gives a bound on d dz V (G)(z): which is less than the bound from eq. (27).
If g(t) is larger than g(−z), for −z ≤ t ≤ −z + r, it can be at most twice as large, since g is increasing fastest at λ/2 − 2r and g(λ/2 − r) ≤ 2g(λ/2 − 2r). Hence the first integral on the right hand side of eq. (30) is at most while second integral on the right hand side of eq. (30) is at most since q(z + t) q−1 is a decreasing function in t. Putting eqs. (31) and (32) together with eq. (26), this gives us that − d dz V (G)(z) is at most which is also less than the bound from eq. (27).
Let F 0 (z) = 0 for all z, and F k+1 = V (F k ). The operator V maps non-increasing functions to non-increasing functions, so all F k are nonincreasing. We know by [4, p.1078] that lim k F 2k and lim k F 2k+1 exist, and equals the nonincreasing functions F A and F B respectively. These functions are continuously differentiable, since V (F A ) = F B and vice versa. If −F ′ k ≤ g, then −F ′ k+1 ≤ g, and F ′ 0 ≤ g holds trivially. By induction, −F ′ k ≤ g for all k, and −F ′ A ≤ g ≥ −F ′ B follows.
The next step is to show that λ/2 −z ρ A (t)dt is continuous in z on (−λ/2, λ/2). It suffices to show that it is continuous on any closed subinterval I ⊂ (−λ/2, λ/2). Since F ′ A is bounded on I, there exists K I such that for any x, y ∈ I we have We will let ε := |x − y| → 0. Suppose (without loss of generality) that x < y and y + ε ∈ I. We estimate the difference Hence λ/2 −z ρ z A (t)dt is continuous in z, and so is F ′ B . To establish the bound for − λ/2 −z ρ z A (t)dt, we use that −F ′ A ≤ g. Then F A satisfies the conditions necessary for eq. (28) to hold for z near −λ/2 with G = F A . For other z, note that the integrand is at most −F ′ B (z), for which the weaker bound g suffices. In other words, for some constant b and any −λ/2 < z < λ/2, we have that while −z −λ/2 ρ z A (t)dt = 0. Finally, for z > λ/2, bound (z + t) q−1 ≤ (z − λ/2) q−1 gives that ρ z A (t)dt ≤ q(z − λ/2) q−1 . Setting α = max(a, b, q) gives the desired result.
Proof of lemma 5. Assume that φ → u is an f A -optimal move by Alice, and u → v is an f A -and f B -optimal move by Bob. We will begin by showing that E |∆ k t (v)| f A (u) = z depends linearly (in z) on E |∆ k t (v)| f A (v) = z for any fixed t. Suppose we had some integer valued graph function 2 g, that u → v was optimal with respect to both f A and f B , and that we wanted to find H(z) := E[g(v)|f A (u) = z] in terms of G(z) := E[g(v)|f B (v) = z]. How would we go about doing that?
If we let P z (t) := P(f A (v) ≤ t|f A (u) = z) and ℓ = f A (u) + f A (v) be the cost of the edge (u, v), we can write H(z) the following way: since the subtree rooted in v is independent of ℓ. Thus H(z) is given by some functional (dependent on z) applied to G, and hence H depends linearly on G.
Next we will show that L A and L B are indeed the right linear operators. Conditioning on