Matroid-Based TSP Rounding for Half-Integral Solutions

We show how to round any half-integral solution to the subtour-elimination relaxation for the TSP, while losing a less-than-1.5 factor. Such a rounding algorithm was recently given by Karlin, Klein, and Oveis Gharan based on sampling from max-entropy distributions. We build on an approach of Haddadan and Newman to show how sampling from the matroid intersection polytope, and a new use of max-entropy sampling, can give better guarantees.


Introduction
The (symmetric) traveling salesman problem asks: given an graph G = (V, E) with edge-lengths c e ≥ 0, find the shortest tour that visits all vertices at least once.The Christofides-Serdyukov algorithm [Chr76,Ser78] gives a 3 /2-approximation to this APX-hard problem; this was recently improved to a ( 3 /2−ε)-approximation by the breakthrough work of Karlin, Klein, and Oveis Gharan, where ε > 0 [KKO20b].A related question is: what is the integrality gap of the subtour-elimination polytope relaxation for the TSP ?Wolsey had adapted the Christofides-Serdyukov analysis to show an upper bound of 3 /2 [Wol80] (also [SW90]), and there exists a lower bound of 4 /3.Building on their above-mentioned work, Karlin, Klein, and Oveis Gharan gave an integrality gap of 1.5 − ε for another small constant ε > 0 [KKO21], thereby making the first progress towards the conjectured optimal value of 4 /3 in nearly half a century.
Both these recent results are based on a randomized version of the Christofides-Serdyukov algorithm proposed by Oveis Gharan, Saberi, and Singh [OSS11].This algorithm first samples a spanning tree (plus perhaps one edge) from the max-entropy distribution with marginals matching the LP solution, and adds an O-join on the odd-degree vertices O in it, thereby getting an Eulerian spanning subgraph.Since the first step has expected cost equal to that of the LP solution, these works then bound the cost of this O-join by strictly less than half the optimal value, or the LP value.The proof uses a cactus-like decomposition of the min-cuts of the graph with respect to the values x e , like in [OSS11].
Given the 3 /2 barrier has been broken, we can ask: what other techniques can be effective here?How can we make further progress?These questions are interesting even for cases where the LP has additional structure.The half-integral cases are particularly interesting due to the Schalekamp, Williamson, and van Zuylen conjecture, which says that the integrality gap is achieved on instances where the LP has optimal half-integral solutions [SWvZ14].The team of Karlin, Klein, and Oveis Gharan first used their max-entropy approach to get an integrality gap of 1.4998 for half-integral LP solutions [KKO20a], before they moved on to the general case in [KKO20b] and obtained an integrality gap of 1.5 − ε; the latter improvement is considerably smaller than in the half-integral case.It is natural to ask: can we do better for half-integral instances?
In this paper, we answer this question affirmatively.We show how to get tours of expected cost at most 1.4983 times the linear program value using an algorithm based on matroid intersection.Moreover, some of these ideas can be used to strengthen the max-entropy sampling approach in the half-integral case.The matroid intersection approach and the strengthened max-entropy approach each yield improvements over the bound in [KKO20a].Combining the techniques gives our final quantitative improvement: Theorem 1.1.Let x be a half-integral solution to the subtour elimination polytope with cost c(x).There is a randomized algorithm that rounds x to an integral solution whose cost is at most (1.5 − ε) • c(x), where ε = 0.001695.
We view our work as showing a proof-of-concept of the efficacy of combinatorial techniques (matroid intersection, and flow-based charging arguments) in getting an improvement for the half-integral case.We hope that these techniques, ideally combined with max-entropy sampling techniques, can give further progress on this central problem.
Our Techniques The algorithm is again in the Christofides-Serdyukov framework.It is easiest to explain for the case where the graph (a) has an even number of vertices, and (b) has no (nontrivial) proper min-cuts with respect to the LP solution values x e -specifically, the only sets for which x(∂S) = 2 correspond to the singleton cuts.Here, our goal is that each edge is "even" with some probability: i.e., both of its endpoints have even degree with probability p > 0. In this case we use an idea due to Haddadan and Newman [HN19]: we shift and get a { 1 /3, 1}-valued solution y to the subtour elimination polytope K T SP .Specifically, we find a random perfect matching M in the support of x, and set y e = 1 for e ∈ M , and 1 /3 otherwise, thereby ensuring E[y] = x.To pick a random tree from this shifted distribution y, we do one of the following: 1. We pick a random "independent" set M of matching edges (so that no edge in E is incident to two edges of M ).For each e ∈ M , we place partition matroid constraints enforcing that exactly one edge is picked at each endpoint-which, along with e itself, gives degree 2 and thereby makes the edge even as desired.Finding spanning trees subject to another matroid constraint can be implemented using matroid intersection.
2. Or, instead we sample a random spanning tree from the max-entropy distribution, with marginals being the shifted value y.(In contrast, [KKO20a] sample trees from x itself; our shifting allows us to get stronger notions of evenness than they do: e.g., we can show that every edge is "even-at-last" with constant probability, as opposed to having at least one even-at-last edge in each tight cut with some probabiltiy.)(Our algorithm randomizes between the two samplers to achieve the best guarantees.)For the Ojoin step, it suffices to give fractional values z e to edges so that for every odd cut in T , the z-mass leaving the cut is at least 1.In the special case we consider, each edge only participates in two mincuts-those corresponding to its two endpoints.So set z e = xe /3 if e is even, and xe /2 if not; the only cuts with z(∂S) < 1 are minimum cuts, and these cuts will not show up as O-join constraints, due to evenness.For this setting, if an edge is even with probability p, we get a ( 3 /2 − p /6)-approximation!It remains to get rid of the two simplifying assumptions.To sample trees when |V | is odd (an open question from [HN19]), we add a new vertex to fix the parity, and perform local surgery on the solution to get a new TSP solution and reduce to the even case.The challenge here is to show that the losses incurred are small, and hence each edge is still even with constant probability.
Finally, what if there are proper tight sets S, i.e., where x(∂S) = 2? We use the cactus decomposition of a graph (also used in [OSS11,KKO20a]) to sample spanning trees from pieces of G with no proper min-cuts, and stitch these trees together.These pieces are formed by contracting sets of vertices in G, and have a hierachical structure.Moreover, each such piece is either of the form above (a graph with no proper min-cuts) for which we have already seen samplers, or else it is a double-edged cycle (which is easily sampled from).Since each edge may now lie in many min-cuts, we no longer just want an edge to have both endpoints be even.Instead, we use an idea from [KKO20a] that uses the hierarchical structure on the pieces considered above.Every edge of the graph is "settled" at exactly one of these pieces, and we ask for both of its endpoints to have even degree in the piece at which it is settled.The z e value of such an edge may be lowered in the O-join without affecting constraints corresponding to cuts in the piece at which it is settled.
Since cuts at other levels of the hierarchy may now be deficient because of the lower values of z e , we may need to increase the z f values for other "lower" edges f to satisfy these deficient cuts.This last part requires a charging argument, showing that each edge e has z e that is strictly smaller than xe /2 in expectation.For our samplers, the naïve approach of distributing charge uniformly as in [KKO20a] does not work, so we instead formulate this charging as a flow problem.

Notation and Preliminaries
Given a multigraph G = (V, E), and a set S ⊆ V , let ∂S denote the cut consisting of the edges connecting S to V \ S; S and S := V \ S are called shores of the cut.(For a singleton set {v}, we write ∂v instead of ∂{v}.) Two sets S and S are crossing if S ∩ S , S \ S , S \ S, and V \ (S ∪ S ) are all non-empty.
Define the subtour elimination polytope Let x be half-integral and feasible for (LP-TSP).W.l.o.g.we can focus on solutions with x e = 1 /2 for each e ∈ E, doubling edges if necessary.The support graph G is then a 4-regular 4-edge-connected (henceforth 4EC) multigraph.
The spanning tree polytope where E(S) is the set of edges with both endpoints in S.This is the graphic matroid polytope, and the convex hull of the spanning trees.
Definition 2.1 (r-Trees).Given a multigraph G and "root" vertex r ∈ V (G), an r-tree T is a connected subgraph with n edges: the vertex r has degree exactly 2, and the subgraph restricted to the other vertices V (G) \ {r} is a spanning tree on them.This is a matroid, though we do not use this fact.
The (integral) perfect matching polytope | is defined as follows: Lemma 2.3.Consider a sub-partition P = {P 1 , P 2 , . . ., P t } of the edge set of G. Let x be a fractional solution to (LP-spT) that satisfies x(P i ) ≤ 1 for all i ∈ [t].Then we can efficiently sample from a probability distribution D over spanning trees which contain at most one edge from each of the parts P i , such that Pr T ←D [e ∈ T ] = x e .
Proof.This follows from the integrality of the matroid intersection polytope.

The Max-Entropy Distribution over Spanning Trees
Definition 2.4 (Strong Rayleigh Distributions, [BBL09]).Let µ be a probability distribution on spanning trees.The generating polynomial for µ is p(z) = T ∈T P(e ∈ T ) e∈T z e .We say µ is strongly Rayleigh (SR) if p is a real stable polynomial (i.e., p(z) = 0 if z lies in the upper half plane).
A distribution µ over spanning trees is called λ-uniform or weighted uniform if there exist nonnegative weights λ : E → R such that P(T ) ∝ e∈T λ(e).Borcea et al. [BBL09] showed that λ-uniform spanning tree distributions are SR.

Theorem 2.5 ([OSS11]
).There exists a λ-uniform distribution µ over spanning trees such that given z in the spanning tree polytope of G = (V, E), The max-entropy distribution, which is the distribution µ on spanning trees maximizing the entropy of µ subject to preserving the marginals z e (that is, P µ (e ∈ T ) = z e ), is one such distribution.
Asadpour et al [AGM + 17] showed that we can find weights λ : E → R + which approximately respect the marginals given by a vector z in the spanning tree polytope in polynomial time.
Theorem 2.6 ([AGM + 17]).Given z in the spanning tree polytope of G = (V, E) and some ε > 0, values λe for e ∈ E can be found such that, for all e ∈ E , the λ-uniform distribution µ satisfies The running time is polynomial in |V |, log1 /mine ze and log 1 /ε.
2. For any set of edges S and e ∈ S,

Samplers
In this section, we describe the MaxEnt and MatInt samplers for graphs that contain no proper min-cuts.We give bounds on certain correlations between edges that will be used in §5 to prove that every edge is "even" with constant probability.The samplers for the case where the graph has an odd number of vertices are more technical and are deferred to Appendix B.
Suppose the graph H = (V, E) is 4-regular and 4-edge-connected (4EC), contains at least four vertices, and has no proper min-cuts. 1This means all proper cuts have six or more edges.We are given a dedicated external vertex r ∈ V (H); the vertices I := V \ {r} that are not external are called internal.(In future sections, this vertex r will be given by a cut hierarchy.)Call the edges in ∂r external edges; all other edges are internal.An internal vertex is called a boundary vertex if it is adjacent to r.An edge is said to be special if both of its endpoints are non-boundary vertices.
We show two ways to sample a spanning tree on H[I], the graph induced on the internal vertices, being faithful to the marginals x e , i.e., P T (e ∈ T ) = x e for all e ∈ E(H) \ ∂r.Moreover, we want that for each internal edge, both its endpoints have even degree in T with constant probability.This property will allow us to lower the cost of the O-join in §6.While both samplers will satisfy this property, each will do better in certain cases.The MatInt sampler targets special edges; it allows us to randomly "hand-pick" edges of this form and enforce that both of its endpoints have degree 2 in the tree.The MaxEnt sampler, on the other hand, relies on maximizing the randomness of the spanning tree sampled (subject to being faithful to the marginals); negative correlation properties allow us to obtain the evenness property, and in particular, better probabilities than MatInt for non-special edges, and a worse probability for the special edges. Our 3. Sample a spanning tree faithful to the marginals y, using one of two samplers: (a) MaxEnt Sampler: Sample from the max-entropy distribution on spanning trees with marginals y.
(b) MatInt Sampler: i. Color the edges of M using 7 colors such that no edge of H is adjacent to two edges of M having the same color; e.g., by greedily 7-coloring the 6-regular graph H/M .Let M be one of these color classes picked uniformly at random.Hence, P(e ∈ M ) = 1 /28, and P(∂v ∩ M = ∅) = 1 /7.ii.For each edge e = uv ∈ M , let L uv and R uv be the sets of edges incident at u and v other than e.Note that |L uv | = |R uv | = 3. Place partition matroid constraints y(L uv ) ≤ 1 and y(R uv ) ≤ 1 on each of these sets.Finally, restrict the partition constraints to the internal edges of H; this means some of these constraints are no longer tight for the solution y.
(c) Given the sub-matching M ⊆ M , and the partition matroid M on the internal edges defined using M , use Lemma 2.3 to sample a tree on H[V \ {r}] (i.e., on the internal vertices and edges of H) with marginals y e , subject to this partition matroid M.
Conditioned on the matching M , we have P(e ∈ T | M ) = y e ; now using (1), we have P(e ∈ T ) = x e for all e ∈ (E \ ∂r) .

Correlation Properties of Samplers
Let T be a tree sampled using either the MatInt or the MaxEnt sampler.The following claims will be used to prove the evenness property in §5.Each Probability Statement MatInt MaxEnt P(exactly one of u, v has odd degree in T ) 1 /9 5 /18

Correlation Properties: |V (H)| even
We now prove the correlation properties for the even case: the numerical bounds for the even case are better than those claimed above (which will be dictated by the proofs of the odd case; see Appendix B).
Proof of Lemma 3.1, Even Case.To prove P(|T ∩ {f, g}| = 2) ≥ 1 /9, we need only knowledge of the marginals and not the specific sampler.If one of f, g lies in M (which happens w.p. 1 /2), then its y-value equals 1 and it belongs to T w.p. 1, and the other edge is chosen w.p. 1 /3, making the unconditional probability 1 /2 • 1 /3 = 1 /6 ≥ 1 /9.Similarly, conditioned on f lying in M and hence belonging to T , edge g is not chosen w.p.
The MaxEnt claim: It remains to show that P(T ∩ {f, g} = {f }) ≥ 12 /72.Conditioned on f ∈ M , we have g ∈ T w.p. 2 /3.Now condition on neither f nor g in M (happens w.p. 1 /2).By Theorem 2.8, By Theorem 2.9, . Putting all of this together, we get Proof of Lemma 3.2, Even Case.The MatInt claims: Each perfect matching M contains one of these four edges in ∂v.Say that edge is e.If e also belongs to M (w.p. 1 /7), then T contains exactly one of {f, g, h}.Since 1 /7 ≥ 2 /21, this gives us the first bound in Lemma 3.2.Moreover, the probability of this edge in T belonging to the other pair (in this case, {g, h}) is 2 /3.Hence The MaxEnt claims: For the first bound in Lemma 3.2, we have that one of the four edges must be in M and have y-value 1. W.l.o.g., call that edge e.The other three, f, g, h, will be 1 /3-valued edges.Since E[X f,g,h ] = 1, we may apply Theorem 2.9 and obtain a lower bound of For the second bound in Lemma 3.2, call S 1 = {e, f } and S 2 = {g, h}.W.l.o.g., we once again label e ∈ M .Then, P(f ∈ T ) = 2 /3.So for S 2 , we condition on f ∈ T .Then by Theorem 2.8, Hence, we use Theorem 2.9 again and obtain P( In total, we obtain Proof of Lemma 3.3a,Even Case.The MatInt claims: The event happens when e ∈ M , which happens w.p. 1 /28, which is at least 1 /36.
The MaxEnt claims: Consider the distribution over the edges in S 2 conditioned on a ∈ T ; this distribution is also SR.By Theorem 2.8, By Theorem 2.9, P( Using symmetry, we obtain Proof of Lemma 3.3b,Even Case.The MatInt claims: For part (b), suppose e ∈ M (w.p. 1 /4), then each of u, v have two other internal edges, each with y-value 1 /3.Let us say the good cases are when exactly one of these four is chosen; exactly one of u, v has degree 2 and the other has degree 1 in these cases.We cannot choose zero of these four edges, because of the connectivity of T , so all bad cases choose at least two of these four.Given the y-values of 1 /3 on all four edges, the expected number of these edges chosen are 4 /3, so the the probability of a bad case at most 1 /3.This means that with probability at least 1 /4 • (1 − 1 /3) = 1 /6, exactly one of u, v has odd degree in T .Since 1 /6 ≥ 1 /9, we have proven (b).
The MaxEnt claims: Observe that the edge e will always contribute either: 0 to both the degree of u in T and the degree of v in T OR 1 to both of these degrees.Let a, b be the two internal edges incident to u and c, d be the two edges incident to v. Taking the minimum of the bounds 2 /3, 4 /9, and 8 /27 in the three cases gives P(exactly one of u, v has odd degree in T ) ≥ 8 /27 ≥ 5 /18.

Sampling Algorithm for General Solutions; and Cut Hierarchy
Now that we can sample a spanning tree from a graph with no proper min-cuts, we introduce the algorithm to sample an r 0 -tree from a 4-regular, 4EC graph, perhaps with proper min-cuts.

The Algorithm
Assume that the graph G = (V, E) has a set of three special vertices {r 0 , u 0 , v 0 }, with each pair r 0 , u 0 and r 0 , v 0 having a pair of edges between them.This is without loss of generality (used in line let S be a minimal such set (and choose S such that r 0 ∈ S).

6:
if G is a double cycle then 7: Label S a cycle set.

8:
sample a random edge from each set of partner edges in G[S]; add these edges to T .

16:
sample a spanning tree on G[S] from µ and add its edges to T .

17:
let G = G/S 18: Due to r 0 , u 0 , v 0 , at this point G is a double cycle (Claim 4.1).Sample one edge between each pair of adjacent vertices in G.
18).Define a double cycle to be a cycle graph in which each edge is replaced by a pair of parallel edges, and call each such pair partner edges.
The sampling algorithm appears as Algorithm 1, and is very similar to that in [KKO20a]: they refer to the sets in line 4 as critical sets.Since G is a 4-regular, 4EC graph at every stage of the algorithm, if |S| = 2 or 3, then S must be a cycle set, whereas if |S| ≥ 4, then S may be a degree or cycle set.The following two claims are proved in Appendix A.1, and show that Algorithm 1 is well-defined.
Claim 4.1.The graph remaining at the end of the algorithm (line 18) is a double cycle.
Claim 4.2.In every iteration in Algorithm 1, G is either a double cycle or a graph with no proper min-cuts.
We will prove the following theorem in §6.This in turn gives Theorem 1.1.
Theorem 4.3.Let T be the r 0 -tree chosen from Algorithm 1, and O be the set of odd degree vertices in T .The expected cost of the minimum cost O-join for T is at most

The Cut Hierarchy
Recall our ultimate goal is to create a low cost, feasible solution in the O-join polytope, where O is the set of odd degree vertices in our sampled tree T .We start with the fractional solution z = x/2 and then reduce the z e value of some edges.In the process, we may violate some constraints corresponding to min-cuts.To fix these cuts, we need a complete description of the min-cuts of a graph.This is achieved by the implicit hierarchy of critical sets that Algorithm 1 induces.The hierarchy is given by a rooted tree T = (V T , E T ). 2 The vertex set V T corresponds to all critical sets found by the algorithm, along with a root node and leaf nodes representing the vertices in

Properties of T :
1. Let G be a 4-regular, 4EC graph with associated hierarchy T .Let S ⊆ V G be a set labelling a node in T .If S is a degree node in T , then G S has at least five vertices and no proper min-cuts, and hence every proper cut in G S has at least 6 edges.If S is a cycle node in T , then G S is exactly a double cycle.
These follow from Claim 4.2 and the equivalence between G and G S .
2. Algorithm 1 can be restated as follows: For each non-leaf and non-root node S in T , sample a random path on G S \ v S if it is a double cycle or K 5 , and otherwise use the MaxEnt or MatInt samplers w.p. λ and 1 − λ, respectively, on G S \ v S .Sample a uniformly random cycle on the double cycle in line 18.
3. For a degree set S, the graph G S having no proper min-cuts implies that it has no parallel edges.In particular, no vertex has parallel edges to the external vertex in G S .Hence we get the following: ) is a compact representation of the min-cuts of a graph, and it can be constructed from the cut hierarchy; we defer the details to Appendix A.2.In turn we obtain the following complete characterization of the min-cuts of G in terms of local multigraphs.
Claim 4.6.Any min-cut in G is either (a) ∂S for some node S in T , or (b) ∂X where X is obtained as follows: for some cycle set S in T , X is the union of vertices corresponding to some contiguous segment of the cycle G S .
5 Analysis Part I: The Even-at-Last Property The proof of Theorem 4.3 proceeds in two parts: (1) In this section, we show that each edge e is "even-at-last" with constant probability.(This is an extension of the property that both of its endpoints have even degree.)(2) Then we construct the fractional O-join.As always, z = x/2 is a feasible join, but we show how to save a constant fraction of the LP value for an edge when it is even-at-last.This savings causes other cuts to be deficient, so other edges raise their z values in response.However, a charging argument shows that the z-value for an edge does decrease by a constant factor, in expectation.This argument appears in §6.
To address part (1), let us define a notion of evenness for every edge in G.In the case where G has no proper min-cuts, we called an edge even if both of its endpoints were even in T .Now, the general definition of evenness will depend on where an edge belongs in the hierarchy T .
Definition 5.1.We say an edge e ∈ E(G) is settled at S if S is the (unique) set such that e is an internal edge of G S ; call S the last set of e.If S is a degree or cycle set, we call e a degree edge or cycle edge, respectively.
Definition 5.2 (Even-at-Last).Let S be the last set of e, and T S be the restriction of T to G S .
1.A degree edge e is called even-at-last (EAL) if both of its endpoints have even degree in T S .
2. For a cycle edge e = uv, the graph G S \ {v S } is a chain of vertices v , . . ., u, v, . . ., v r , with consecutive vertices connected by two parallel edges.Let C := {v , . . ., u}, and C := {v, . . ., v r } be the partition of this chain.The cuts ∂C and ∂C are called the canonical cuts for e. Cycle edge e is called even-at-last (EAL) if both canonical cuts are crossed an even number of times by T S ; in other words, if there is exactly one edge in T S from each of the two pairs of external partner edges leaving v and v r .
Informally, a degree edge is EAL in the general case if it is even in the tree at the level at which it is settled.Also note that cycle edges settled at the same (cycle) set are either all EAL or none are EAL.
Definition 5.3 (Special and Half-Special Edges).Let e be settled at a degree set S. We say that e is special if both of its endpoints are non-boundary vertices in G S and half-special if exactly one of its endpoints is a boundary vertex in G S .
We now prove a key property used in §6 to reduce the z-values of edges in the fractional O-join.
Theorem 5. Proof.Let e be settled at S. Let T S be the spanning tree sampled on the internal vertices of G S (in the notation of Algorithm 1, the spanning tree sampled on G[S]).
First, assume that S is a degree set: 1.If none of the endpoints of e are boundary vertices in G S (i.e., e is special), then it is EAL exactly when both its endpoints have even degree in T S .By Lemma 3.3(a), this happens w.p. 1 /36 for the MatInt sampler and w.p. 128 /6561 for the MaxEnt sampler.
2. Now suppose that e is half-special, so that exactly one of the endpoints of e = uv (say u) is a boundary vertex in G S , with edge f incident to u leaving S. By Lemma 3.2, the other endpoint v is even in T S w.p. 2 /21 for the MatInt sampler and 8 /27 for the MaxEnt sampler.Moreover, the edge f is chosen at a higher level than S and is therefore independent of T S , and hence can make the degree of u even w.p. 1 /2.Thus e is EAL w.p. 1 /21 for the MatInt sampler and 4 /27 for the MaxEnt sampler.
3. Suppose both endpoints of e are boundary vertices of S, with edges f, g leaving S. Let q = be the probability that the degrees of vertices u, v in the tree T S chosen within S have the same parity, and q = = 1 − q = .Now, when S is contracted and we choose a r 0 -tree T on the graph G/S consistent with the marginals, let p = be the probability that either both or neither of f, g are chosen in T , and p = = 1 − p = .Hence Pr[e EAL] = q oo p 11 + q oe p 10 + q eo p 01 + q ee p 00 = 1 /2(p = q = + p = q = ), where q oo , q oe , q eo and q ee correspond to different parity combinations of u and v in T S and p 00 , p 01 , p 10 , p 11 correspond to whether f and g are chosen in T .The second inequality follows from Claim D.1 applied to the random r 0 -tree T .
(a) If f, g are settled at different levels, then they are independent.This gives p = = p = = 1 /2, and hence Pr[e EAL] = 1 /4 regardless of the sampler.
Next, let S be a cycle set.Let e = uv be an edge inside the cycle, and let {a, b}, {c, d} be the four edges crossing ∂S.Let v S be the vertex obtained by contracting down S (whose incident edges are then {a, b, c, d}).Now in order for e to be EAL, one each of {a, b} and {c, d} must belong to T ; call this event E. We again consider cases based on where these edges are settled.Let node P be the parent of node S, and let v S be the vertex in G P obtained from contracting S.
1.If all four edges are settled at P , and P is a degree set, then Lemma 3.2 says that for the MatInt sampler, we have Pr[E] ≥ 4 /63.In contrast, the MaxEnt sampler gives us Pr[E] ≥ 16 /81.
2. If all four edges are settled at P , and P is a cycle set, then no matter how these four edges are distributed, Pr[E] ≥ 1 2 .
3. If three of them {a, b, c} are settled at P , and the fourth (say d) at a higher level, then since P has a vertex v S with a single edge leaving it, P must be a degree set.Now exactly one of {a, b} is chosen in T w.p. at least 1 /9 in the MatInt sampler and 12 /72 in the MaxEnt sampler, by Lemma 3.1.And since d is independently picked at a different level, exactly one of {c, d} is chosen in T w.p. 1 /2, giving an overall probability of 1 /18 in the MatInt sampler and 1 /12 in the MaxEnt sampler.
4. Finally, if only two edges are settled at P , and two others go to higher levels, then P is a cycle set by Corollary 4.5.In this case, exactly one of the two edges that are settled in P is chosen.Now we want a specific one of the edges going to a higher level to be chosen into T (and the other to not be chosen), which in the worst case happens w.p. at least 1 /9, due to Lemma 3.1.

Analysis Part II: The O-Join and Charging
To prove Theorem 4.3 and thereby finish the proof, we construct an O-join for the random tree T , and bound its expected cost via a charging argument.The structure here is similar to [KKO20a]; however, we use a flow-based argument to perform the charging instead of the naive one, and also use our stronger notion of evenness (EAL).
Let O denote the (random set of) odd-degree vertices in T .The dominant of the O-join polytope Observation 6.2.If e is a special, half-special, other degree, or cycle edge, then z e is reduced with probability exactly p sp , p hs , p, or p, respectively.
This reduction scheme may make z infeasible for the O-join polytope.We now discuss how to maintain feasiblity.

Maintaining Feasibility of the Fractional O-join via Charging
Suppose f is EAL and that we reduce edge z f (per its coin flip B f ).Say f is settled at S. If S is a degree set, then the only min-cuts of G S are the degree cuts.So the only min-cuts that the edge f is part of in G S are the degree cuts of its endpoints, call them U, V , in G S (U and V are vertices in G S representing sets U and V in G).But since |∂U ∩ T | and |∂V ∩ T | are both even by definition of EAL, we need not worry that reducing z f causes z(∂U ) ≥ 1 and z(∂V ) ≥ 1 to be violated.Likewise, if S is a cycle set, then by definition of EAL all min-cuts S in G S containing e have |∂S ∩ T | even, so again we need not worry.
Since f is only an internal edge for its last set S, the only cuts S for which the constraint z(∂S ) ≥ 1, |∂S ∩ T | odd, may be violated as a result of reducing z f are cuts represented in lower levels of the hierarchy.Specifically, lef f be an external edge for some G X (meaning X is lower in the hierarchy T than S) and S be a min-cut of G X (either a degree cut or a canonical cut).By Claim 4.6, cuts of the form S are the only cuts that may be deficient as a result of reducing z f .Call the internal edges of ∂S lower edges.When z f is reduced and |∂S ∩ T | is odd, we must distribute an increase (charge) over the lower edges totalling the amount by which z f is reduced, so that z(∂S ) = 1.We say these lower edges receive a charge from f .How the charge is distributed will depend on whether the lower edges are degree edges or cycle edges (see §6.3 and §6.4 below).
We claim that this procedure maintains feasibility of the O-join solution z.Indeed, by Claim 4.6, no constraint corresponding to a min-cut in the O-join polytope is violated.Further, by capping the amount an edge can be reduced at 1 /4 − 1 /6 = 1 /12, we will ensure that none of the constraints are violated, since every other cut has at least 6 edges.
To prove Lemma 6.1, we will lower bound the expected decrease to z e , using different strategies for charging e when e is a cycle edge versus a degree edge.In particular, when e is a cycle edge, we distribute charge from an external edge evenly between e and its partner.When e is a degree edge, charge will be distributed according to a maximum-flow solution.

Charging of Cycle Edges
We now analyze the expected net decrease for each edge, starting with cycle edges.The expected decrease for a cycle edge starts off as exactly p•β by Observation 6.2.However, we need to control the amount of charge the edge receives from edges settled at higher levels.Note that in the calculations, we assume the worst-case scenario that such edges settled at higher levels are reduced w.p. exactly p.The special/half-special edges are reduced with lower probability, which is only better for us because the expected amount of charge will be lower in these cases.Note also that throughout the calculations we may assume that an edge settled at a higher level is a cycle edge.This will follow from the assumption that β ≥ 2τ, 2γ.
Consider a cycle edge e that is settled at set S. Let its partner on the cycle be e , and let the four external edges for S be {a, b} (leaving the vertex u) and {c, d} (leaving vertex v).When S is contracted, vertex v S has these four edges {a, b, c, d} incident to it.Let P be S's parent node in T .Moreover, let C and C be the canonical cuts, as in the figure below.We consider the following cases as in [KKO20a].(i) Two of the four edges incident to v S also leave P -and therefore P is a cycle set by Corollary 4.5.Of the two edges leaving P , one must be from {a, b} and one from {c, d}.To see this, say both c and d (leaving node u) both did not leave P .Consider the set of vertices S which c and d lead to.S ∪ u is a proper min cut which crosses S; however, this contradicts the fact that S is a critical set (namely S is not supposed to be crossed by another proper min cut).
Suppose {a, d} are the external edges for P , and {b, c} are internal (and hence partners in this cycle set).Now, consider the event that b is reduced; the important observation is that c is also reduced, because both are EAL at the same time and their coins are perfectly correlated.Not only that, the event that b is reduced implies that |δ(S) ∩ T | is even.This means that both cuts C, C are deficient by the same amount and at the same time, due to this event of reducing b, c, and raising e helps both of them.This means the net expected charge to e is at most (Note that there is no term for c here because of the discussion above; moreover, the β /2 term reflects that the charge of β is split between e, e .) Since a and b are settled at different levels, the parity of cut C is unbiased, even either conditioned on choices within P , or else conditioned on choices outside P .The same holds for the parity of C , so each of the probabilities above is at most 1 /2, making the net expected decrease (ii) Only one of the edges incident to v S , say edge a, is external for P .By Corollary 4.5, P is a degree set, and in the worst case, a K 5 degree set since γ ≥ τ .The charge is maximized when a is a cycle edge, in which case the expected charge to e is at most Since a and b are settled at different levels, the parity of the cut C is unbiased, either conditioning on the process inside P or else outside P , which makes the above expression p( β /4 + 5γ /4), and the net expected decrease to be (iii) All four of the edges are internal to P , which is a cycle set.The partners must now be one edge from each of the two pairs {a, b} and {c, d}-let's say {a, d} and {b, c} are partners (using an argument similar to that in (i)).The cut C is odd in S if and only if C is odd; indeed, C is odd means we choose both or neither of {a, b}, and since {c, d} are the partners of {a, b}, we choose both or neither of them.Hence, increasing e fixes both cuts at the same time, similar to the argument in (i), so we focus only on the increase due to cut C. Since a and b belong to different pairs and are chosen independently into the tree T , Pr[C odd] = 1 /2.This means both a and b can give a expected charge of p( β /2) • 1 /2 to e.The net expected decrease is , which is no worse than (3).
(iv) All four of the edges are internal to P , and P is a degree set.Note that G P cannot be a K 5 since the vertex v S has four internal edges incident to it.We simply bound the charge due to each edge by τ /2, and hence get net expected decrease of

Charging Degree Edges: Max-Flow Formulation
So far we have shown that no cycle edge will receive too much charge.We did this by considering four different configurations for the external edges of S, and distributing charge evenly between a cycle edge and its partner.Before showing that no degree edge receives too much charge, we will define a charging scheme for degree edges that achieves better bounds than distributing charge uniformly.
Let G S be the local multigraph for some degree set S, so G S is 4-regular and has no proper min-cuts.The following lemma follows from the Max-Flow Min-Cut theorem: Lemma 6.3.Given a bipartite graph G = (B, F, E) with positive integers b u for each u ∈ B, and c ≥ 0, there exists x : where b(R) = u∈R b u .
Lemma 6.3 now gives the following result, which follows from straightforward casework.
Lemma 6.4.Let H = (B, F, E) be the bipartization of G S , as described above.If G S is not a K 5 , then the smallest c satisfying (8) is given by c = 1 /2.If G S is a K 5 , then the smallest c satisfying (8) is given by c = 2 /3.
The fact that we may achieve a value of c = 1 /2 instead of c = 2 /3 as long as G S is not a K 5 motivates why we utilize a different sampler for K 5 .In fact, c = 2 /3 on K 5 is achieved by distributing uniform charge over internal edges.

Charging of non-K 5 Degree Edges
We now analyze the expected net decrease for degree edges.By Observation 6.2, the expected decrease for a non-special degree edge starts off as at least p hs • τ or p hs • γ, since p hs ≤ p.We will analyze the charge on non-K 5 degree and K 5 degree edges separately, as a different sampler is used when G S = K 5 .We begin with the former.

Charging of non-special edges
1. Case 1: All external edges are non-K 5 degree edges.Let x : B × F → R + be any function satisfying the constraints in Lemma 6.3 for c = 1 /2 (by Lemma 6.4 such an x exists, and can be found efficiently using max-flow min-cut).The charging scheme is as follows: for each external edge e and boundary vertex u e , if the cut δ(u) is odd and the edge e is reduced, charge each internal edge f incident to u e by τ • x(e, f ).The constraint (6) ensures that the charge neutralizes the reduction of e by τ .For any internal edge f , let δ(f ) denote the external edges with which f shares an endpoint, i.e., the neighbors of f in the bipartization H = (B, F, E).We have

4.
The external edges are a combination of non-K 5 degree edges, K 5 degree edges, and cycle edges.An averaging argument as in the previous section shows that the three cases above are the only possible worst cases.
We have ε = 2δ = 0.001695, proving Theorem 4.3.to each u T for which T is a child of S. See Figure 7.All of the terminal vertices in U , except for u V G \r 0 , correspond to all of the leaf nodes in T , which in turn correspond to all of the vertices in V G \ {r 0 }.We identify the vertex u V G \r 0 in U with the vertex r 0 in V G .In other words, the mapping φ in Theorem A.3 can be defined as φ(v) = u v for all v ∈ V G \ r 0 , and φ(r 0 ) = u V G \r 0 .
Since the construction of the cactus in the proof of Theorem A.3 given in [FF09] and the construction of C above are equivalent, φ satisfies the conclusions of Theorem A.3.Since the min-cuts of a cactus are obtained by cutting any two edges belonging to the same cycle, Claim 4.6 follows.

B Samplers for Odd |V (H)|
The main change from the case where V (H) is even is that x = 1 /2 • 1 is not a point in the perfect matching polytope: in fact, H has no perfect matchings because it has an odd number of vertices.
To fix this, split the external vertex r into two external vertices r 1 , r 2 , each incident to some two of the edges in ∂r, and then add two parallel edges between r 1 , r 2 .Call this multigraph H. Again, all vertices and edges between vertices in V \ {r 1 , r 2 } are called internal, vertices in {r 1 , r 2 } and edges in ∂r 1 ∪ ∂r 2 are external, and internal vertices adjacent to external vertices are boundary vertices.
Fact B.1.The graph H is also 4-regular and 4EC; it has an even number of vertices.However, it contains a single proper min-cut separating the internal vertices I := V \ {r 1 , r 2 } from the two external vertices.Any perfect matching M of H contains either zero or two edges in ∂I.
The sampling procedure for trees on H[I] = H[I] is conceptually similar to the one above, but with some crucial changes because of the proper min-cut.
1. Since H is 4-regular and 4EC, and |V ( H)| is even, setting 1 /4 = xe /2 on each edge gives a solution K P M ( H) in the perfect matching polytope.Sample a perfect matching M such that for each edge e, we have Pr[e ∈ M ] = 1 /4.Again define the fractional solution This ensures E[y] = x, but y| I may no longer be in the spanning tree polytope K spT ( H[I]).Indeed, y(∂I) ∈ { 4 /3, 8 /3} depending on whether M has zero or two edges crossing the cut, so y(E(I)) ∈ {|I| − 2 /3, |I| − 4 /3} and not |I| − 1 as required.Hence we cannot directly sample a spanning tree on H[I], as we did above.We fix this in step 3 below.
Proof of Lemma 3.1, Odd Case.The MatInt claims: As in the even case, to prove P(|T ∩{f, g}| = 2) ≥ 1 /9, we need only knowledge of the marginals and not the specific sampler.If one of f, g lies in M (which happens w.p. 1 /2) y-values of these two edges remaining unchanged with probability at least 2 /3, due to Lemma B.4.Now, conditional on these y-values remaining unchanged, the other edge (not in M ) is chosen with probability 1 /3, making the unconditional probability 1 /2• 2 /3• 1 /3 = 1 /9, as desired.Similarly, conditioned on f lying in M and hence belonging to T , edge g is not chosen w.p. 1−y e = 2 /3, giving a probability at least 1 /4• 2 /3• 2 /3 = 1 /9, proving the second part of Lemma 3.1 for the MatInt sampler.
The MaxEnt claim: It remains to show that P(T ∩ {f, g} = {f }) ≥ 12 /72.In the case that f ∈ M , we once again have a probability of at least 2 /3 that f and g are untouched, due to Lemma B.4. Sincce P(g ∈ T ) = 2 /3 in this case, we have In the case that neither f nor g is in M , the probability f or g is chosen in a local decrease is at most 1 /3 and the probability f or g is chosen in a local increase is at most 1 /2.(For the first bound, the worst case is when each of the three endpoints of f, g is incident to an external edge.Then the probability that neither f nor g is chosen for a local decrease is at least 1 /2 For the second bound, the worst case is when each edge in M ∩ ∂I is incident to an endpoint of f , w.l.o.g.Then the probability that neither f nor g is chosen for a local increase is at least ) Hence, the edges remain untouched with probability at least 1 /2.In total, this gives where the bound P(T ∩ {f, g} = {f } | f, g ∈ M ) ≥ 1 /9 was computed in the even case (3.First suppose it is a local decrease, then with probability at least1 /3 a non-interesting edges is reduced, and hence the matroid constraint ensures we always succeed.With probability at most2 /3 an interesting edge is reduced, in which case we still have a 2 /3 chance of getting degree 2 at both u and v.So we have 1 /3 + 2 /3 • 2 /3 = 7 /9 of success, conditioned on the choice of the edge.
Else, suppose it is a local increase; the argument is similar.With probability at least 1 /3 a noninteresting edges is increased, and hence the matroid constraint ensures we always succeed.With probability at most 2 /3 an interesting edge is increased, in which case we change the matroid constraint to drop one of the other interesting edges.This dropped edge can be added in with its y-value 1 /3, and hence we have a 2 /3 probability of getting degree 2 at both u and v. Again, the probability of success is 1 /3 + 2 /3 • 2 /3 = 7 /9 of success.Removing the conditioning on e ∈ M , the net probability of success is 1 /28 • 7 /9 = 1 /36.
The MaxEnt claims: Let U = δ(u) \ {e} and V = δ(v) \ {e}.Condition on e ∈ M .Since u, v are both non-boundary vertices, the value of e will remain unchanged regardless of whether there is a local increase or decrease, so we will have y e = 1 always.Recall that y is changed at a single edge when the local increase or decrease is performed.We take cases on whether a local increase or local decrease occurs.
A local decrease occurs.If none of the edges in U ∪ V are decreased, then (conditioned on e ∈ M ) the analysis from the even case (3.2.1) gives Now consider the case that an edge in U ∪ V is decreased.Let S 1 = U and let S 2 = {a, b} denote the set of two nonzero edges in V \ {e} (see Figure 8).Applying Theorem 2.9 and the fact that which is the bound we sought.
Proof of Lemma 3.3b, Odd Case.The MatInt claims.Fix a matching M such that e ∈ M (w.p. 1 /4); each of u, v have two other internal edges, each with y-value 1 /3.Call these four interesting edges.Suppose M ∩ ∂I has two edges, then we perform a local increase step.Condition on the edge in M ∩ ∂I chosen as part of the local increase: in the worst case it is incident to the endpoint of two of the interesting edges.
• Then w.p. 2 /3 it increases one of the interesting edges from y f = 1 /3 to y f = 2 /3.The total y-value of interesting edges becomes 5 /3.By the same argument as for the even case, the tree T contains exactly one interesting edge with probability at least 1 /3.
• With the remaining probability 1 /3, the increase is not to an interesting edge, and hence we succeed with the original probability 2 /3.
On the other hand, if M ∩ ∂I is empty, then we perform a local decrease step.Condition on the boundary edge chosen in the first part of the local decrease.In the worst case it is one of the boundary edges incident to either u or v.One of the interesting edges is reduced w.p. 2 /3, from y f = 1 /3 to y f = 0.If that happens, the total y-value of interesting edges sums to 1.The connectivity of the spanning tree means that every tree we sample with these marginals will have exactly one interesting edge.
Hence, in either case, we have exactly one of these four internal edges and satisfy the condition of the claim with probability 4 /9, conditioned on having e ∈ M .The overall probability is therefore at least 1 /4 • 4 /9 = 1 /9.
The MaxEnt claims.Observe that the edge e will always contribute either: 0 to both the degree of u in T and the degree of v in T OR 1 to both the degree of u in T and the degree of v in T .Thus, to lower bound the probability that u and v have different parities in T , we will not need to consider whether or not e is in T .
Let a, b be the two internal edges in δ(u) \ {e} and c, d be the two internal edges in δ(v) \ {e}.(We know each set has exactly two edges because u, v are each boundary vertices.)Let f, g be the two external edges incident to u, v, respectively.We take cases based on (a) whether a local increase or decrease is performed, (b) whether or not any of a, b, c, d are changed by the local increase or decrease, and (c) whether or not e ∈ M .Note that for (b), we do not take subcases on whether or not e is changed, by the above paragraph.Also note that for (c), we consider whether e ∈ M in order to determine the marginals of {a, b, c, d} prior to a local increase or decrease.
Case I.N.1: local increase occurs, {a, b, c, d} not touched, e ∈ M .In this case, we obtain the same bound from Case 1 of the even case in the proof of Lemma 3.3b (see 3.2.1),that is, P(exactly one of u, v has odd degree in T ) ≥ P(X a,b,c,d = 1) ≥ 2 /3 .Lemma 6.3 in order to show there exists a function x : B × F → R + satisfying constraint (7) for c = max{ τ 2 , γ 2 , β 4 }.Then we can apply a charging scheme like the one described in Case 1 using the function x.The flow network is as follows: take the graph H = (B, F ) and add a source node s and a sink node t.Connect s to every vertex b in B, and label (s, b) with capacity τ, γ, or β 2 based on whether b corresponds to a non-K 5 degree, K 5 degree, or cycle edge, respectively.To find x satisfying the properties in the above paragraph, by Lemma 1 we need to show that for each T ⊆ B, where n τ , n γ , and n β are the numbers of vertices in T corresponding to non-K 5 degree, K 5 degree, and cycle edges, respectively.Note n τ + n γ + n β = |T |.
Lemma 6.4 showed that . Now taking a convex combination of the three inequalities above we obtain: which is precisely what we sought to show.This concludes Case 4.

Figure 1 :
Figure 1: The matching M consists of all highlighted edges (both brown and green), one possible choice of M has edges {a, c} highlighted in brown, and the constraints are placed on the edges adjacent to those in M (marked in gray).
21 > 4 /63, giving us the other bound in Lemma 3.2 for the MatInt sampler.

2 .
Case 2: e ∈ M , exactly two of a, b, c, d are in M .W.l.o.g., say a, c ∈ M , so a and c are in T .E[X b,d ] = 2 /3, which implies (by Theorem 2.9) that P(X b,d = 1) ≥ 4 /9.3. Case 3: e ∈ M , exactly one of a, b, c, d is in M .W.l.o.g., say a ∈ M , so we have E[X b,c,d ] = 1.We condition on e ∈ T , which happens w.p. 2 /3.Since we sample a spanning tree on the internal vertices, this means that at least one of c, d is in T .W.l.o.g., say c ∈ T .So in order to bound the probability that the internal parities of u, v are different, equivalently we would like to bound P(X b,d = 1 | e ∈ T ).Since 2 /3 ≤ E[X b,d | e ∈ T ] ≤ 2 /3 + 1 /3, Theorem 2.9 implies that P(X b,d = 1 | e ∈ T ) ≥ 4 /9.So removing the conditioning on e ∈ T gives a lower bound of 2 /3 • 4 /9 = 8 /27.

Figure 2 :
Figure 2: A portion of the cut hierarchy T and the local multigraph G S .

Definition 4 . 4 (
a critical set, we label the node in V T with S, where we view S ⊆ V G and not V G .The root node is labelled V G \ {r 0 } and each leaf nodes is labelled by the vertex in V G \ {r 0 } corresponding to it.Now we define the edge set E T .A node S is a child of S if S ⊂ S and S is the first superset of S contracted after S in the algorithm.In addition, the root node is a parent of all nodes corresponding to critical sets that are not strictly contained in any other critical set (i.e., the critical sets corresponding to the vertices in the graph G from line 17 when the while loop terminates).Finally, each leaf node is a child of the smallest critical set that contains it (or if no critical set contains it, is a child of the root node).Thus by construction, vertex sets labelling the children of a node are a partition of the vertex set labelling that node.A node in V T is a cycle or degree node if the corresponding critical set labelling it is a cycle or degree set, respectively.We say the root node is a cycle node (since the graph G in line 18 is a double cycle), and accordingly call V G \ {r 0 } a cycle set.(The leaf nodes are not labelled as degree or cycle nodes.)Local multigraph).Let S ⊆ V G be a set labelling a node in T .Define the local multigraph G S to be the following graph: take G and contract the subsets of V G labelling the children of S in T down to single vertices and contract S to a single vertex v S .Remove any selfloops.The vertex v S is called the external vertex; all other vertices are called internal vertices.An internal vertex is called a boundary vertex if it is adjacent to the external vertex.The edges in G S \ v S are called internal edges.Observe that G S is precisely the graph G in line 5 of Algorithm 1 when S is a critical set, and is a double cycle when S = V G \ {r 0 }.

Corollary 4 . 5 .
For a set S labeling a non-leaf node in T and any internal vertex v ∈ G S : if S is a cycle set then |∂v ∩ ∂S| ∈ {0, 2}, and if S is a degree set then |∂v ∩ ∂S| ∈ {0, 1}.Finally, we show how the hierarchy T allows us to characterize the min-cut structure of G.The cactus representation of min-cuts ([FF09]

Figure 6 :
Figure 5: Case (iii) For an external edge e, let u e denote the internal boundary vertex to which e is incident.We create a bipartite graph H = (B, F, E) from G S as follows.The vertices of B are labelled with the external edges of G S .The vertices of F are labelled with the internal edges of G S .So |B| = 4. Place an edge between e ∈ B and f ∈ F in the edge set E if e and f share u e as an endpoint.Set b e = 1 for all e ∈ B. We call H the bipartization of G S .
f ) • P(u e odd ∧ e reduced) ≤ τ • c • P(u e odd | e reduced) • P(e reduced) (9) Using the naive bound that P(u e odd | e reduced) ≤ 1, we obtain that the the expected decrease is at least p hs τ − τ 2 • p.

Figure 7 :
Figure 7: Construction of the cactus C from the hierarchy T in the proof of Claim 4.6.
2.7 ([BBL09]).Let µ be an SR distribution.Let F ⊂ E be a subset of edges.Then the projection of µ onto F , i.e., µ |F (A) = S:S∩F =A µ(S), is also SR.Moreover, for any edge e ∈ E, conditioning on e ∈ T or on e ∈ T preserves the SR property.Let µ be an SR distribution on spanning trees.1.Let S be a set of edges and X S = |S ∩ T |, where T ∼ µ.Then, X S ∼ |S| i=1 Y i , where the Y i are independent Bernoulli random variables with success probabilities p i and i samplers will depend on the parity of |V |: when |V | is even, the MatInt sampler is the one given by [HN19, Theorem 13], which we describe in §3.1.They left the case of odd |V | as an open problem, and in Appendix B we extend their procedure to the odd case.Define a new fractional solution y (that depends on M ): set y e = 1 for e ∈ M , and y e = 1 /3 otherwise.We have y ∈ K T SP (H) (and hence y| I ∈ K spT (H[I]) by Fact 2.2): indeed, each vertex has y(∂v) = 1 + 3 • 1 /3 = 2 because M is a perfect matching.Moreover, every proper cut U in H has at least six edges, so y(∂U Algorithm 1 Sampling Algorithm for a Half-Integral Solution 1: let G be the support graph of a half-integral solution x. 2: let T = ∅.3: while there exists a proper tight set of G that is not crossed by another proper tight set do 4: 4(The Even-at-Last Property).The table below gives lower bounds on the probability that special, half-special, and all other types of degree edges are EAL in each of the two samplers.
Proof of Lemma 3.2, Odd Case.The MatInt claims: Each perfect matching M contains one of these four edges in ∂v.Say that edge is e.The vertex v has no external edges, so regardless of whether we do a local increase or decrease, these four edges remain untouched w.p. 2 /3, due to Lemma B.4.If this happens, and if e also belongs to M (w.p. 1 /7), tree T contains exactly one of {f, g, h}, which gives us the bound 2 /21.Moreover, the probability of this edge in T belonging to the other pair (in this case, {g, h}) is 2 /3, giving the bound 4 /63.The MaxEnt claims: Since v is not a boundary vertex, none of e, f, g, h are picked for a local increase or decrease w.p. at least 2 /3, due to Lemma B.4. Conditioning on e, f, g, h remaining }| = |T ∩ {g, h}| = 1) ≥ 2 /3 • 8 /27 = 16 /81.Proof of Lemma 3.3a, Odd Case.The MatInt claims: In this proof, we will often use that G is a simple graph.Suppose e ∈ M , which happens w.p. 1 /28.Call the edges (∂u ∪ ∂v) \ {e} interesting edges.Condition on the matching M (which decides whether the change is a local increase or a decrease), and the edge f in ∂I whose incident internal edge is increased/decreased.At most two of the three internal edges incident to f can be interesting.Since u, v are non-boundary vertices, the value of e is never changed by the alteration.