Dual Charging for Half-Integral TSP
Abstract
In this extended abstract, we show that the max entropy algorithm is a randomized 1.49776 approximation for half-integral TSP, improving upon the previous known bound of 1.49993 from Karlin et al. This also improves upon the best-known approximation for half-integral TSP due to Gupta et al. Our improvement results from using the dual, instead of the primal, to analyze the expected cost of the matching. We believe this method of analysis could lead to a simpler proof that max entropy is a better-than-3/2 approximation in the general case.
Keywords and phrases:
Approximation Algorithms, Graph Algorithms, Randomized Rounding, Linear ProgrammingCategory:
APPROXCopyright and License:
2012 ACM Subject Classification:
Theory of computation Routing and network design problemsEditors:
Alina Ene and Eshan ChattopadhyaySeries and Publisher:
Leibniz International Proceedings in Informatics, Schloss Dagstuhl – Leibniz-Zentrum für Informatik
1 Introduction
In the metric traveling salesperson problem (TSP), we are given a weighted graph and aim to find the shortest closed walk that visits every vertex. Metric TSP is NP-Hard to approximate better than [25]. In the 1970s, Christofides and Serdyukov [5, 33] famously gave a approximation for the problem. This was not improved until recently, when Karlin, Klein and Oveis Gharan showed a approximation for [23, 24] in 2021. Gurvits, Klein, and Leake later showed that one can set [16].
These recent improvements originate in work from 2010 and 2011 by Asadpour et al. on asymmetric TSP [1] and Oveis Gharan, Saberi and Singh on graphic TSP [29]. These two papers introduced the so-called max entropy algorithm for TSP. In this algorithm, one first solves the subtour elimination [6] (or Held-Karp [19]) linear program for TSP to obtain a fractional point . Then, the max entropy distribution over spanning trees is computed whose marginals match (up to error , which can be made exponentially small in ). Finally, a tree is sampled from and the minimum cost matching is added to the odd vertices of . We know this algorithm is at worst a approximation in general, and we also know there are instances where it performs no better than 1.375 [21]. A fascinating open question is to determine what the worst case approximation ratio of the algorithm is.
Prior to the first improvement in the general case, Karlin, Klein and Oveis Gharan showed a randomized 1.49993 approximation for a special case of TSP. In particular, they showed that if for an instance there exists an optimal solution to the subtour LP with for all , then the max entropy algorithm outputs a solution of expected cost at most [22]. These “half-integral” points are of special interest due to a conjecture of Schalekamp, Williamson, and van Zuylen [31] that the integrality gap of the subtour LP is obtained by half integral points. Notably, the lower bound of 1.375 for max entropy in [21] is a half integral instance, as is the classical envelope graph which demonstrates an integrality gap of at least for the subtour LP (see e.g. [35]).
In 2021, Gupta, Lee, Li, Mucha, Newman, and Sarkar claimed to improve the bound for half integral points to 1.4983 [12] using a mix of the max entropy algorithm and a combinatorial one proposed by Haddadan and Newman [18]. After finding a small issue in the proof in [12], we had a discussion with the authors, and while the details are being checked, a fix appears to give a bound of 1.4990.
In this work, we show that the max entropy algorithm (with no adaptations) is a 1.49776 approximation for half-integral TSP. This improves over the state of the art for half integral TSP with a significantly simpler algorithm, as well as gives a large relative improvement for the analysis of max entropy. In particular, we show:
Theorem 1.
Given an optimal solution to the subtour LP with for all , the max entropy algorithm produces a solution of cost at most in expectation.
Therefore our result also bounds the integrality gap of the subtour LP in the half integral case by 1.49776 (and, should the conjecture of [31] hold, the general case as well). As discussed in more detail in Section 1.2, the primary reason for the improvement is a new dual-based analysis style. Due to the large relative improvement over the previous analysis of max entropy in the half integral case (in terms of the distance from 1.5, this is an improvement of a factor of about 30) and the fact we no longer need to analyze certain pathological cases, we believe these techniques may lead to a significant simplification of the analysis in the general case.
1.1 Other Related Work
There has been exciting recent progress on two important variants of TSP, graphic TSP, in which the input graph is unweighted, and path TSP in which the goal is to find the shortest - walk visiting every vertex. For graphic TSP, Oveis Gharan, Saberi, and Singh [29] first demonstrated that max entropy was a approximation for a small constant . Using different methods, Mömke and Svensson [27] then obtained a 1.461 approximation. This was further improved by Mucha [28] to and then to 1.4 by Sebö and Vygen [32]. For path TSP, Zenklusen showed a approximation [36] using a dynamic programming approach. Traub, Vygen, and Zenklusen then showed that given an approximation for TSP there is an approximation for Path TSP for any .
1.2 Overview
As discussed, in the max entropy algorithm we first solve the subtour LP to obtain a solution . The subtour LP is as follows, where for is the set of edges with exactly one endpoint in and for , .
| (1) | |||||
Then, using , we find a maximum entropy distribution over spanning trees111Technically, over spanning trees plus an edge. subject to the constraint that for all , up to a small multiplicative error (see Section 2.3 for more discussion on this sampling procedure). Finally, we sample a tree from and add a minimum cost matching on the odd vertices of the tree . is an Eulerian graph and thus contains an Eulerian walk, which is a solution to the metric TSP problem.222Using the fact that the costs form a metric, one can then shortcut the Eulerian tour to a Hamiltonian cycle of no greater cost if desired. The main goal, then, is to bound the expected cost of the matching over the randomness of the sampled tree . To do so, we find a function that given a tree constructs a vector so that , and then bound the expected cost of . In particular, will be a feasible point in the -Join polytope (where is the set of odd vertices in ); see Section 2 for further details.
This is the approach taken by all previous papers analyzing max entropy [29, 22, 23] or variants of this algorithm [12], and we do not deviate from this here. However, here we construct in a new way that streamlines the analysis and allows for a sharper guarantee. While previous works bounded by bounding the contribution to of each edge individually, we instead bound the contribution of each dual variable to . By complementary slackness, the dual variables correspond to tight cuts with , which contain multiple edges with . By bounding the cost of groups of edges in terms of , the argument becomes much more flexible. Previously, analyses had to deal with pathological cases where single edges had a high and create workarounds. By looking at groups of edges, issues due to these pathological edges are averaged out. Previously it was necessary to bound for all (as this would correspond to an expected cost of less than for the matching), in our construction there may be edges with .
There is one other big advantage of this approach, which is as follows.333Understanding this advantage is a bit technical, so we recommend readers unfamiliar with work on the max entropy algorithm come back to this point after having read the body of the paper. In the analysis of max entropy, typically one begins with for all edges. Then, depending on the parity of certain cuts in the tree, some edges have decreased and other edges have increased. In the per-edge argument, it was important to bound the expected increase of each edge carefully. However, when considering a cut, these increases and decreases often cancel out, simplifying things quite a bit. (Despite this, our analysis is not simpler than [22]. This is because we take care to improve the analysis in several places which require additional casework.)
One other aspect of our improvement is to incrementally improve some of the important probabilistic bounds from [22] using polynomial capacity (see e.g. [16] for usage of this tool in TSP) or more precise arguments. However the impact of this is relatively minor compared to that of the move to a dual-based analysis.
2 Preliminaries
2.1 Notation
Sets and Cuts.
Given a set , let be the set of edges with exactly one endpoint in and be the set of edges with both endpoints in . Given and , let . A set is tight if . A set is proper if . Two sets cross if are all non-empty.
Support Graph.
We will use to construct a 4-regular and 4-edge-connected graph which we will call . will have the same vertex set as our input. Then for every edge with , we add the corresponding edge to the support graph. For every edge with , we will add two parallel copies of to .
Minimum Cuts.
Tight sets are therefore minimum cuts of the support graph , as they have 4 edges. It is helpful to note that since is Eulerian, every cut which is not a minimum cut has at least 6 edges. For an overview of the structure of minimum cuts, we recommend an unfamiliar reader to look at the cactus representation of Dinits, Karzanov, and Lomonozov [7], or a succinct explanation of it by Frank and Fleiner [10]. This structure is very important to our analysis.
Trees.
Given a tree and a set of edges , we let denote the number of edges of contained in . We will use to denote the number of edges in contained in .
Parity.
For a sampled tree , we say a set is even if is even and odd otherwise.
2.2 Polyhedral Background
The subtour LP is given in Equation 1. We will also crucially use the dual linear program in our analysis. By the parsimonious property [11], the subtour bound does not change after dropping the equality constraints. Thus, the dual can be seen to have the following formulation:
| (2) | |||||
We will also make significant use of the following characterization of the cost of the matching. It is well known (see [8], for example) that given a metric, the following LP bounds upper bounds the cost of an integral perfect matching on a set of vertices :
| (3) | |||||
This is known as the -Join polyhedron, and we call a feasible point in this polyhedron an -Join solution, where is the set of vertices in the sampled tree with odd degree. In this context, note that the set of cuts which have constraints in (3) are exactly those for which is odd.
2.3 Max Entropy Distribution
A distribution over spanning trees is -uniform if and for every spanning tree of the graph,
Given , we will let denote the resulting -uniform distribution. Given a point in the spanning tree polytope, [1] show that one can efficiently find a -uniform tree with marginals arbitrarily close to :
Theorem 2 ([1]).
Let be a point in the spanning tree polytope of a graph . Then, for any , there is an algorithm running in time polynomial in the size of the graph and that outputs a vector so that the -uniform distribution has the property:
Since can be chosen to be exponentially small in , following previous work on half integral TSP, we will assume for brevity that we may set . This error can be handled using the stability of max entropy distributions [34] (one can see this applied in [23]).
-uniform spanning tree distributions have maximum entropy over all distributions with the same marginal vector. Therefore, when one can set , they are indeed the distributions of maximal entropy respecting the constraints.
2.4 Algorithm and Critical Sets
We will study the version of the max entropy algorithm used by [22]. Here, we first fix an edge where have two parallel edges between them. Such an edge always exists in any extreme point solution (see [3]), or as noted in [22] one can create such an edge by splitting a vertex in two and putting an edge of value 1 and cost 0 between its two endpoints. After fixing , we iteratively select a minimal proper tight set that is not crossed (and ), compute the max entropy distribution inside , sample , and contract it. When no such set exists, the graph must be a cycle , at which point we sample a random cycle. In particular, for every that share two edges we sample one of them independently and uniformly at random. The sampled tree at the end is the union of all trees sampled during the procedure. See Algorithm 1 for a complete description and [22] for further details.
Following [22], we call every tight set contracted by the algorithm a critical set. In addition, we call vertices critical sets. For a critical set , we call a critical cut. The collection of critical sets is a laminar family , where recall a family is laminar if for all , .
There are two types of sets . If at the moment before contraction, there are no proper minimum cuts inside , then call a degree cut. Otherwise, call a cycle cut, in which case at the moment before contraction is a path with two edges between every pair of adjacent vertices. See Figure 1(b) for examples of each type of cut in , and see Section 3.1 for more details.
Definition 3 ().
For every critical cut , we let denote the support graph after contracting every critical cut lower than in the hierarchy as well as into vertices. We use to denote the graph , where is the vertex representing after contraction.
We also note here that by definition, edges which are in and for critical cuts are independent. This fact will be used crucially throughout the proof.
Remark.
We defer additional preliminaries concerning strongly Rayleigh distributions and polynomial capacity to after the proof overview.
3 Overview of our Method
In this section, we introduce the key tools used in our analysis and provide a high-level overview of our techniques. In Section 3.1, we define the necessary notation and describe the structural properties of the hierarchy of critical cuts. Understanding this hierarchy is crucial to our analysis. In Section 3.2, we describe the construction of an -Join solution with a small expected cost. Furthermore, we provide an explanation of our proof and describe our use of the dual formulation (Equation 2) in analyzing the cost of the -Join solution.
3.1 Hierarchy of Critical Cuts
The algorithm constructs a natural hierarchy of min-cuts. That is, the critical cuts form a laminar family of min cuts, and therefore can be arranged in a hierarchical structure. At the bottom of this hierarchy are the singleton vertex cuts.
Let be a critical cut. Based on the structure of the contracted graph , we classify into one of two types. Note that, we assumed is the support graph after contracting and all critical cuts lower than in the hierarchy.
-
1.
Degree cut: If contains no proper min-cuts, we call a degree cut.
-
2.
Cycle cut: Otherwise, must form a cycle with two parallel edges between each pair of consecutive vertices. In this case, we call a cycle cut. We call the parallel edges in that share their endpoints companions. Note that a pair of companions has the property that exactly one is in the tree and the event of which edge is chosen is independent of all other events. The remaining two pairs of parallel edges sharing endpoints in are called cycle partners.
Fact 4.
Every min-cut in is either a critical cut or an interval of a cycle cut.
For an edge , let denote the minimal critical cut such that . We will distinguish edges into two types depending the structure of .
Definition 5 (Top Edge and Bottom Edge).
We call an edge a top edge if is a degree cut, and a bottom edge if is a cycle cut.
We define a similar notation for the critical cuts.
Definition 6 (Top Cut and Bottom Cut).
A min cut is said to be a top cut if its parent in the hierarchy is a degree cut. Otherwise, is called a bottom cut, i.e., if its parent in the hierarchy is a cycle cut. See Section 3.1 for an example.
Moreover, for an edge , we define the Last Cuts of as the two maximal min cuts such that . More precisely, let be a bottom edge where has child cuts with two edges between and for . If is between and , then, the last cuts of are and .
Note that the last cuts of a top edge are critical cuts, meanwhile, last cuts of a bottom edge are not necessarily a critical cut.
Definition 7 (Going Higher).
We say an edge goes higher in if the lowest critical cut such that satisfies . Additionally, when is a critical cut, by we denote the edges in that go higher in . Similarly, denotes the edges in that don’t go higher.
3.2 Constructing the -Join Solution
Given a tree sampled from the max-entropy distribution, we will describe a randomized process to construct a feasible solution for the -Join formulation (Equation 3) where is the odd degree vertices of .
Before describing the construction, we will restate the definition of even at last edges from [22].
Definition 8 (Definition 4.3 in [22]).
For an edge we say is even at last if the two last cuts of are even. If is a bottom edge, this is equivalent to defining to be even at last when all the min cuts containing on the cycle defined by the graph consisting of with contracted are even. If is a top edge, then it is even at last if its last cuts are even simultaneously.
Let be the optimal solution of the subtour LP (Equation 1). We will initialize so that satisfies the -Join constraints. Now, when an edge is even at last, we will decrease by , where is a parameter we will set later. Since an even at last edge can still cover lower level min-cuts in the hierarchy that are odd in , we will increase the value of accordingly to make a feasible -Join solution.
We utilize the fact that when an edge is even at last, lower level cuts such that are (in most cases) still even with probability . For an edge , let denote the probability of being even at last. Unfortunately, some edges can have (see [23]). This is an issue for arguments that bound the contribution of each edge individually. Therefore, deviating from prior work, we will instead look at the expected number of even at last edges in for a min cut . This value is denoted by . We show lower bounds for this value for every min cut , which in turn gives that decreases meaningfully for each min cut in the -Join solution we construct. Moreover, when we increase edges to cover the violated -Join constraint of a cut, we increase them according to (roughly) their value. This in turn means that edges that increase in the third step of our construction, should also decrease meaningfully in the second step. A more accurate and complete description of this process is provided in Section 5.
Our main goal is to show that the -Join solution of a tree sampled from the max entropy algorithm has cost at most in expectation. This immediately proves Theorem 1 as the -Join polyhedron (Equation 3) has an integrality gap of 1. To do so, we will show the -Join solution has expected cost at most . Instead of bounding the contribution of each edge, we will bound the expectation on each minimum cut as follows:
Lemma 9.
There exists a randomized -Join solution for the random tree sampled from the max entropy distribution such that for each min cut we have,
To analyze the cost of our solution, we utilize the dual formulation of the subtour LP (Equation 2). Now, we will use Lemma 9 to prove Theorem 1.
Theorem 1. [Restated, see original statement.]
Given an optimal solution to the subtour LP with for all , the max entropy algorithm produces a solution of cost at most in expectation.
Proof.
By complementary slackness we know for an edge in the support of , , therefore, the cost of the -Join solution can be written as,
Now, by Lemma 9,
where the final equality follows from strong duality.
To prove Lemma 9, we analyze top cuts and bottom cuts separately. For each cut , we show that either the value of decreases for every edge in , or, if there exists an edge for which does not decrease significantly, then the remaining edges in have a larger decrease.
4 Probabilistic and Structural Lemmas
In this section, we will provide some key probabilistic and structural lemmas on the max entropy algorithm. These lemmas will provide strong probabilistic bounds as well as crucial observations about the structure of critical cuts that are essential in proving Lemma 9.
Before doing so, in Section 4.1 and Section 4.2 we introduce some key additional preliminaries that were omitted in Section 2.
4.1 Strongly Rayleigh Distributions
Given a distribution over ground set , its generating polynomial is defined
where . is strongly Rayleigh (SR) [2] if is real stable. A polynomial is real stable if for all with for all . In other words, is strongly Rayleigh if it has no zeros in the upper half of the complex plane. -uniform spanning tree distributions are strongly Rayleigh (see e.g. [2, 29]).
Negative Association.
SR distributions are negatively associated [2, 9]. In particular, given any increasing functions that depend on disjoint coordinates:
An easy consequence is the following:
Fact 10 (Fact 3.16 in [22]).
For any -uniform spanning tree distribution , for any , any , and any we have:
-
1.
If then and .
-
2.
If , then and .
There is also a useful consequence of negative association when applied to a homogeneous distribution. (Recall that a polynomial is homogeneous when all terms have the same degree. A distribution is homogeneous when all outcomes have the same number of elements.)
Fact 11 (Fact 3.17 in [22]).
For any -uniform spanning tree distribution , for any set of edges and any , we have:
and similarly,
When we apply one of these two facts, we will often simply say we are using negative association.
Closure Properties.
A second consequence of real stability is that given an SR distribution , the following distributions are SR as well:
-
Projection. , the projection of to the coordinates in some .
-
Conditioning on a binary element to be 0 or 1. If , then and are SR.
Hoeffding’s Theorem.
For any subset , the law of for is distributed as the sum of independent Bernoulli random variables (not necessarily all with the same success probability). This is a consequence of the fact that the coefficients of any real rooted polynomial with positive coefficients can be described by a sum of Bernoullis [30, 2]. This makes the law of particularly easy to analyze for any , especially when one applies the following theorem of Hoeffding:
Theorem 12 ([20, Corollary 2.1]).
Let and for some integer . Let be independent Bernoulli random variables with success probabilities , where that minimizes (or maximizes)
over all such distributions. Then, for some . In particular, if only of ’s are nonzero and of ’s are 1, then the remaining are .
A very useful corollary is the following.
Lemma 13 (Lemma 3.23 of [22]).
Let with and . Then .
The proof is omitted in [22] as it follows straightforwardly from Theorem 12. One can see Lemma 3.6 in [26] for a proof.
Finally, we will need the following lemma.
Lemma 14 (Lemma 3.21 of [22]).
Let with . Furthermore, assume that Then,
4.2 Polynomial Capacity
The capacity at of a real stable polynomial with positive coefficients is defined as:
A classical result of Gurvits [14] relates the capacity of a polynomial to the coefficient of for -variate homogeneous polynomials of degree as follows (where is the vector consisting of 1s):
Theorem 15 ([15]).
Let be a homogeneous real stable polynomial of degree with non-negative coefficients. Then, where ,
Note that is exactly the coefficient of . There are various similar statements in the literature, and we will use the following, first stated as Theorem 5.1 of [15] and restated as follows in [4]:
Theorem 16 ([15, 4]).
Let be a homogeneous real stable polynomial of degree with positive coefficients. Let such that . For , let be the degree of in the polynomial
and the degree of in . Then, where is the coefficient of the term ,
Furthermore, the capacity of a real stable polynomial can be bounded using its gradient. In particular, we can apply the following theorem of Gurvits and Leake [17] (also see [16]) generalizing [13]. (We do not need the generalization here, but we state the stronger form regardless.)
Theorem 17 ([17]).
Let be a real stable polynomial in variables with non-negative coefficients, and fix any . If and , then
We will use the following corollary in this work, which follows easily from the above.
Corollary 18.
Let be the generating polynomial of a strongly Rayleigh distribution over ground set . If for all , or equivalently , then if is the maximum degree of ,
Proof.
Let be the maximum degree of . Let be the homogenization of . Then, . Set . By Theorem 17, . Applying Theorem 16 with , noting for , we obtain:
where we adjust the indices to match the degree of each variable . But , so the corollary follows.
4.3 Structure of Critical Cuts
In this section, we recall some basic facts about the structure of critical cuts. For proofs, we refer the interested reader to Section 3 of [22]. These facts are used solely to ensure that our case analysis is exhaustive and covers all possible scenarios.
Fact 19 (Fact 3.10 [22]).
Suppose that is a critical set. If some (contracted) vertex has two edges to , then is a cycle cut.
Fact 20 (Fact 3.11 [22]).
Suppose that and are two distinct tight sets. Then .
Fact 21 (Fact 3.12 [22]).
Suppose that and are two critical sets such that . Then if , then is a cycle cut.
Fact 22 (Fact 3.13 [22]).
Suppose that are two critical cycle cuts. Then any two edges are cycle partners on at most one of these (cycle) cuts.
The following two corollaries are immediate.
Corollary 23.
Suppose is a degree cut. Then has at most one edge that goes higher.
Corollary 24.
Suppose is a cycle cut. Then has either exactly two edges or no edges that go higher.
We will also prove the following simple fact.
Fact 25.
Let be a min cut that is not the last cut for any edge in the support graph. is always even in the tree .
Proof.
Let be the parent of in the hierarchy of critical cuts. By Fact 4, must be a cycle cut with child cuts with two edges between and for . Since is not the last cut of any edge, it must be of the form for .
Since exactly one of the two companions between and are in , and is even.
4.4 Bounding the Expected Number of Even at Last Edges on Min Cuts
As previously mentioned, for an edge , there cannot be any meaningful lower bound on the probability of being even at last as there can be edges with . In contrast, we will show that for any top cut , there are strong lower bounds on . This intuitively shows that in expectation, can be decreased for every min cut.
The following lemma can be thought of as as the analog of Lemma 5.3 of [22] (which showed at least one edge in each cut has ) when adapted to our framework, and uses similar proof ideas.
Lemma 26.
For any top cut with no edge going higher, . Moreover, if . Then, .
Proof.
Let be the vertex corresponding to after contraction. For an edge , with (contracted) last cuts and , we have,
Where in the last line we used Lemma 13. Now we will bound .
Conditioning on is identical to conditioning on , as . Therefore, by negative association, we have . Moreover,
as the measure conditioned on first samples a tree from and then chooses an edge in according to its conditional probability. Depending on the degree of in being even or odd, we can make odd by either conditioning on being in the tree or out of the tree. By negative association, , therefore,
Which in turn gives,
As , summing over gives,
By Theorem 12, this term attains its minimum when the Bernoullis are at . This shows the first claim in the lemma, that .
For the second part, consider the inequality,
Since for every edge in we have , . By summing this inequality over all we get,
By Theorem 12, this term attains its minimum at which concludes the proof.
Now, we will prove a similar result for top cuts with an edge that goes higher. But first, we will prove the following simple lemma.
Lemma 27.
Let be a top edge with (contracted) last cuts and . Suppose has an edge going higher, and does not. Then is even at last with probability at least .
Proof.
By Lemma 13, is even at last with probability at least . Additionally, as is in a different level of the hierarchy than the edges in , by letting be in or out of respectively, we can fix the parity of with probability . This concludes the proof.
Lemma 28.
Let be a top cut with an edge that goes higher and let be any two of the edges in . Then, .
Proof.
Denote the two edges in by . If the other last cut of any of these two doesn’t go higher, by Lemma 27 which satisfies the lower bound. Now, assume the other last cut of each of and has an edge that goes higher. Call these other last cuts by and the edges going higher from them . Now, condition on . Since goes higher in , it’s independent from the edges that are inside . Therefore, by Lemma 14, with probability at least exactly one of the edges in are in . This makes the degree of even.
Similarly, condition on which happens with probability . By Lemma 14, with probability at least exactly 2 edges in are in which makes even. Thus,
where we have used the fact that is independent from the edges in .
Now, in both cases, depending on the parity of inside , we can choose (or ) to be inside or outside to make (or ) even at last. Since and , we have,
Where the last inequality follows from negative association.
For bottom cuts we can prove stronger bounds. In fact, bottom edges inside a cycle cut are all simultaneously even at last. This symmetry enables us to individually bound for each edge .
Lemma 29.
Every bottom edge is even at last with probability at least .
Proof.
Consider a bottom edge , so that is a cycle cut with child cuts with two edges between each and for each . When a tree on is chosen, we will obtain exactly one edge among every pair of adjacent child cuts. The edges in are comprised of two sets of edges, and . So, is even at last exactly when .
Project to . The resulting distribution has generating polynomial . Symmetrize so that . By Corollary 18, where is the degree of and the degree of ,
as desired, since and so and .
We remark that this lemma is tight: there are instances where this probability is exactly , and the relevant edges have generating polynomial .
4.5 Structure of Degree Cuts and
In this section, we show that if a degree cut satisfies a certain structure, then must be the graph . We use this fact in Section 5 to show that if many edges in have conditions that prevent them from decreasing reasonably, the degree cut must have a simple structure. In particular, must be a . We note that Gupta et al. [12] also treated as a special case, and adapted the algorithm to treat these cuts differently. While we do not change the algorithm, we similarly analyze this case separately.
Lemma 30.
Let be a degree cut with (contracted) cuts each with an edge that goes higher. If, forms a triangle, then is .
Proof.
Consider the set . , so is a min-cut. However, , and since there are no proper min-cuts inside a degree-cut, it must be the case that . Therefore is a 4-regular graph with 5 vertices which concludes the proof.
The following results from the fact that setting is the -uniform distribution for when for all edges. (Note is unique for a fixed , so this is the only possible distribution.)
Fact 31.
The max entropy distribution on the graph is the uniform spanning tree distribution.
We will now complement Lemma 30 with the following.
Lemma 32.
Let be a degree cut of the support graph such that . Every edge in is even at last with probability at least .
We omit the simple proof in this extended abstract.
5 Analysis
Now, we describe the construction of the O-Join solution in full detail.
For an edge , recall is the probability of being even at last. For top edges define and for bottom edges for constants we will set momentarily. Note that for denotes . For each edge with , let be an independent Bernoulli with success probability , if , let . Moreover, for two bottom edges in the same cycle cut , let with probability one. Note that since these bottom edges are always simultaneously even at last this is well-defined.
Finally, let be a critical cut and . By we denote . In other words, represents the fraction of even at last edges in that is responsible for.
We will now construct in 3 different steps, done sequentially. The construction is done after sampling a tree and sampling the Bernoullis for every .
-
1.
Let for each .
-
2.
For any even at last edge with , reduce by .
-
3.
For each odd min cut , let . For each edge and its last cuts , increase by .
Throughout the analysis let . The third step of our construction along with Fact 25, ensure that satisfies the O-Join constraint for all odd min cuts. Moreover, since , the O-Join constraint for every non min cut is also satisfied, even if all edges in are simultaneously reduced, as such cuts have at least edges covering them.
Now, we will consider different cases for a min-cut and show that for every case is at least reduced by in expectation. For brevity, we do not consider cuts of the final cycle considered in Algorithm 1, as these cuts can easily be seen to obey the bounds described here.
Lemma 33.
Let be a top cut with an edge that goes higher, then, .
Proof.
Let be the three edges in that don’t go higher. If the probability of being even at last on none of these edges is more than , then,
where inequality () follows from Lemma 28. Otherwise, assume , then, by Lemma 28, . Therefore, in all cases .
A crucial consequence of Lemma 33 is that for any top cut and an edge , we have as by definition.
Lemma 34.
Let be a top cut with no edge that goes higher, then,
Proof.
Let be the edges in and denote the other last cuts of these edges respectively by . For an edge , if has an edge that goes higher, then by Lemma 27, . Let be the edge in that goes higher.
If is a bottom edge, then
since by switching with its companion, we can change the parity of while remains even at last.
To cover for the deficit caused by , increases by at most in expectation. This is because by Lemma 33, is at most responsible for half of .
Otherwise, if is a top edge, by Lemma 14 and our construction of the O-Join solution,
Consequently, increases by at most . Therefore, for any edge such that has an edge that goes higher, .
-
Case 1: Two or more of have an edge that goes higher. As ,
-
Case 3: None of has an edge that goes higher. Since , by Lemma 26,
Hence, for every case,
Lemma 35.
Let be a bottom edge inside a cycle cut with child cuts with two edges between each and for each . Also, let be the parent of in the cut hierarchy. Then, we will bound the amount increases to cover the deficit on its last cuts caused by decreasing edges in in three cases:
-
1.
If is a degree cut with no edge going higher, increases by at most .
-
2.
If is a degree cut with an edge going higher, increases by at most .
-
3.
If is a cycle cut, increases by at most .
By Corollaries 23 and 24, the structure of falls into one of these three categories.
We omit the proof in this extended abstract.
Lemma 36.
Let be a bottom cut with no edge that goes higher, then,
Proof.
Now that we have dealt with min cuts having no edges going higher, we now turn to those with at least one such edge. The structure of these cuts is known from Corollaries 23 and 24.
Lemma 37.
Let be a top cut with an edge that goes higher. Let be the edges in , and let , , and be the other last cuts of , , and , respectively. If each of , , and have an edge that goes higher, then . More generally,
Proof.
Consider the min-cut , and let . Let be the other last cuts of . Suppose has an edge that goes higher. Since at most four (contracted) min-cuts can have such an edge inside a degree cut, must be either or .
In this case, the cuts , , and form a triangle. By Lemma 30, forms a , and Lemma 32 implies that (and every edge in ) is even at last with probability . Therefore, .
Now, suppose that the other last cut of every edge in does not have an edge that goes higher. In this case, by Lemma 27, these edges are even at last with probability at least . Thus concluding . Moreover, in both cases it is clear that .
We will utilize Lemma 37 to show decreases meaningfully when is a top cut with an edge going higher.
Lemma 38.
Let be a top cut with an edge that goes higher, then,
We omit the proof in this extended abstract, which has extended casework.
We have proved upper bounds on for all top cuts and all bottom cuts with no edge going higher. We next handle bottom cuts with edges going higher. By Corollary 24, such cuts have exactly two edges going higher.
Lemma 39.
Let be a bottom cut with two edges that go higher, then,
Proof.
Let be the parent of in the hierarchy, by denote the two edges in that go higher, denote the other edges in by . Additionally, let be the critical cut inside which go higher from and let be the edges in . We will now carefully analyze .
-
Case 1: is a bottom cut. By Corollary 24, at least two of the edge in are companions that don’t go higher in , WLOG, let be these edges. Since they are companions, are even at last simultaneously. Moreover, if is even at last, then, is even. This implies the parity of and should be the same and if is odd, should be odd too. Therefore, increasing simultaneously covers the deficit caused by reducing .
Another key observation is that in total increase exactly as much as has decreased. This implies that when (and simultaneously ) is even at last and is odd, remains the same and we don’t need to consider this scenario. By Lemma 35, at most increases by . Meanwhile, since , if can also decrease by on expectation. The same bound holds for if it is a bottom edge. Now, assume is a top edge and let be the edges going higher from the last cuts of . By Lemma 33, can at most contribute to half of the even at last edges in its last cuts. Therefore, in expectation, can at most increase by for each of its last cuts. Recall that one can swap , or , with their companion to fix the parity of the last cuts of while retaining the even at last property for , or . Since , the latter case is worse.
Finally, consider , if is a bottom edge, . Moreover, . Thus, each increase by in step (3) of our construction because of . (Remember that .) If was a top edge, then . Since , in the worst case is a bottom edge. Using and all the preceding observations gives us,
-
Case 2: is a top cut. Since is a top cut, by Corollary 23, at least three of are top edges in . First, suppose are those edges. Because of , (and similarly ) needs to increase by at most . For its other last cut, it increases by at most by Lemma 33. Finally, can increase in total by at most in expectation. is maximized when giving
Otherwise, WLOG, are top edges. Similar to the previous case, is maximized when . If is a bottom edge, similar to previous cases, it can decrease by , so:
However, if is a top edge it can decrease by at most . Meanwhile, the edge that goes higher from (if such an edge exists) is a top edge, therefore, to cover for the deficit on , needs to increase by at most (where we have used Lemma 14 to show ). This gives
as desired.
Now, we are ready to prove the main lemma of the paper.
Lemma 9. [Restated, see original statement.]
There exists a randomized -Join solution for the random tree sampled from the max entropy distribution such that for each min cut we have,
Proof.
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