Abstract 1 Introduction 2 Preliminaries 3 Constructing the Laminar Family References Appendix A Corollaries from Prior Work

Thin Trees for near Minimum Cuts

Nathan Klein ORCID Boston University, MA, USA    Neil Olver ORCID London School of Economics and Political Science, UK    Zi Song Yeoh ORCID Boston University, MA, USA
Abstract

The strong thin tree conjecture states that every k-edge-connected graph G contains an O(1/k)-thin spanning tree, meaning a spanning tree which contains at most an O(1/k) fraction of the edges across each cut in G. This conjecture is still open despite significant effort; the best current result by Anari and Oveis Gharan shows the existence of an O(polyloglogn/k)-thin tree.

In this work, we demonstrate that the conjecture is true if one only requires thinness for the set of η-near minimum cuts of the graph for η=1/40, in other words, for the set of cuts with fewer than (1+1/40)k edges. Our approach constructs such a tree in polynomial time. To show this, we utilize the structure of near minimum cuts, and in particular the polygon representation of Benczúr and Goemans, to reduce to the previously solved problem of finding a spanning tree that is O(1/k)-thin for all sets in a laminar family.

Keywords and phrases:
Graph Theory, Thin Trees, Polygon Representation, Near Minimum Cuts
Category:
Track A: Algorithms, Complexity and Games
Copyright and License:
[Uncaptioned image] © Nathan Klein, Neil Olver, and Zi Song Yeoh; licensed under Creative Commons License CC-BY 4.0
2012 ACM Subject Classification:
Theory of computation Algorithm design techniques
; Mathematics of computing Graph theory
Acknowledgements:
Initial discussions on this topic took place during the Trimester Program on Combinatorial Optimization at the Hausdorff Research Institute for Mathematics; we are grateful to the institute for its support.
Funding:
Nathan Klein and Zi Song Yeoh were supported by the NSF CAREER grant CCF-2442250. Neil Olver was supported by the NWO Vidi grant 016.Vidi.189.087.
Related Version:
ArXiv Version: https://arxiv.org/abs/2605.12669
Editors:
Sayan Bhattacharya, Danupon Nanongkai, Michael Benedikt, and Gabriele Puppis

1 Introduction

Given a graph G=(V,E), a spanning tree T of G is called α-thin if for every cut SV we have the inequality |Tδ(S)|α|δ(S)|. In other words, for each cut of G, T must have at most an α-fraction of the edges.

Goddyn [10] conjectured that there exists a function f with limkf(k)=0 so that every k-edge-connected graph G has an f(k)-thin spanning tree. Put another way, the conjecture is that for every α(0,1], there exists some k so that every k-edge-connected graph has an α-thin tree. This is known as the thin tree conjecture. Anari et al. [2] conjectured that one can pick f(k)=C/k for some constant k, which is the best one could hope for up to constant factors, as the minimum cut has k edges and any spanning tree must have at least one edge across the cut. This stronger version is known as the strong thin tree conjecture, and was shown by [2] to imply an O(1) integrality gap for the standard LP relaxation for the asymmetric traveling salesperson problem (ATSP). An O(1) integrality gap and approximation factor for ATSP was obtained by Svensson, Tarnawski, and Végh in a breakthrough result [23] (and further improved by [25]), but this would give an alternative approach to the problem. The thin tree conjecture would also imply an alternative proof of the weak 3-flow conjecture [12], which was resolved by Thomassen [24].

The thin tree conjecture is still open; the best known result is O(polyloglogn/k) thinness [1]. One challenge is the exponential number of cuts involved. So a natural direction to explore is to ask for thinness only for some cuts: given a collection of cuts 𝒞, say that a spanning tree T is α-thin with respect to 𝒞 if |Tδ(S)|α|δ(S)| for all S𝒞. Even here, results are limited. If 𝒞 is a laminar family111Recall a family is laminar if for all S,T we have ST{,S,T}., Klein and Olver [15] showed that an O(1)-thin spanning tree with respect to 𝒞 exists.

We give a positive result for another natural structured family. Given a k-edge-connected graph, a cut SV is an η-near minimum cut222Our use of a strict inequality here follows the convention of Benczúr [5]. if |δ(S)|<(1+η)k. We show that the strong thin tree conjecture is true for the set of near minimum cuts. In particular, we prove the following theorem:

Theorem 1.

Let G=(V,E) be a k-edge-connected graph. Then for η=1/40, there is a spanning tree T with |Tδ(S)|88 for all cuts S with |δ(S)|<(1+η)k (and therefore, |Tδ(S)|88k|δ(S)|). Furthermore, we can find such a tree in polynomial time.

We remark that the collection of near minimum cuts can be as large as Ω(|V|2) (on the cycle graph); by contrast, a laminar family has size at most 2|V|1.

Another motivation for considering the family of near minimum cuts comes from a conjecture due to Pritchard [21] concerning removable spanning trees. Pritchard’s conjecture states that there exists a function g(k) with limkg(k)/k0, such that every k-edge-connected graph G=(V,E) contains a spanning tree T so that the graph obtained after deleting T is kg(k) edge connected. The strong version of Pritchard’s conjecture states that one can choose g(k) to be an absolute constant. The best current bound for this conjecture is that there is a tree we can delete so that the resulting graph is k21 edge connected, and is an immediate consequence of the Nash-Williams theorem [17].

Pritchard’s conjecture is implied by the thin tree conjecture (and its strong form is implied by the strong thin tree conjecture). Indeed, if T is f(k)-thin, then the connectivity of the graph obtained by removing T is at least

minSV|δ(S)||Tδ(S)|minSV|δ(S)|(1f(k))k(1f(k));

so we can set g(k)=kf(k). Observe though that Pritchard’s conjecture asks much less for large cuts than the thin tree conjecture. Consider the strong forms. For both conjectures, T must have only O(1) edges across every minimum cut, and across cuts close to minimum. But for Pritchard’s conjecture, T may have more than ηk edges on cuts with at least (1+η)k edges, i.e., thinness 11/(η+1) suffices on these cuts.

Our result shows that we can find a tree that is thin enough on the “most constrained” cuts, and so we view it as progress towards Pritchard removable spanning tree conjecture. A tree with thinness η/(η+1) for cuts of size (1+η)k, for all η above some constant, would miss at least k edges from every cut S with |δ(S)|(1+η)k, more than what Pritchard’s conjecture needs for these “less constrained” cuts. Notice that this requires a thinness of only 11k, for example, for cuts of size k2. The existence of such a spanning tree is a very interesting open problem; a resolution of this, together with this work, would likely lead to a proof of Pritchard’s conjecture.

1.1 Our Approach

To prove Theorem 1, we will reduce this problem to the strong thin tree conjecture for laminar families of cuts, which was shown to be true by Klein and Olver [15]. Their result is phrased in the language of finding spanning trees satisfying prescribed cut bounds on a given laminar family. Here we state a version of their theorem for our setting.

Theorem 2 (Implied by [15]).

Let G=(V,E) be a k-edge-connected graph and be a laminar family of cuts. Then, there is a spanning tree T with |Tδ(S)|66k|δ(S)| for all S.

If in addition |δ(S)|43k for all S, this can be improved to |Tδ(S)|11 for all S. Moreover, this spanning tree can be found in polynomial time.

This is a slightly different form of the theorem stated in [15], and the quantitative improvement for our setting where all sets in the laminar family are near-minimum cuts requires unpacking their proof slightly. We discuss these details in Appendix A.

Given the existence of thin trees for laminar families, it is very natural to try and leverage this. A plausible first attempt is to take an arbitrary maximal laminar family of near minimum cuts and find a tree that is thin with respect to using Theorem 2. In the case of minimum cuts, this works: taking a maximal laminar family of min cuts and finding a tree which is thin for implies that |Tδ(S)|O(1) for all minimum cuts S. This can be shown in a straightforward manner using the cactus representation of the minimum cuts [8]. It is perhaps unsurprising that this works given that minimum cuts can be uncrossed; if S and T are crossing minimum cuts, then so are ST and ST.

So, one might hope that this simple approach extends to the set of η-near minimum cuts for some η>0. Unfortunately, this is not the case. For an extreme example, consider the following graph with an even number of vertices n displayed in Figure 1. Begin with a Hamiltonian cycle H1=v1,,vn in clockwise order with k edges between adjacent vertices. Then, add a single copy of a second Hamiltonian cycle H2=v1,v2,vn,v3,vn1,v4,vn2,,vn/2+1,v1. In other words, to construct H2, begin with v1 and then alternately add the first new vertex clockwise of v1 and the first new vertex counterclockwise of v1, terminating at vn/2+1, at which point we return to v1.

Figure 1: A bad example for taking a maximal laminar family of near minimum cuts. In solid red is a tree which is thin with respect to the family described in this section ({{v1,v2},{v1,v2,v8},{v1,v2,v3,v8},{v1,v2,v3,v7,v8},}) but not thin with respect to the blue cut with interval {v2,,v5}.

Taking kn (although this is not strictly necessary for the example to work), it is the case that every interval of H1 is a near minimum cut if we consider the set of η-near minimum cuts for any constant η>0. Thus, we can choose to be any set of intervals. Define

={{v1,v2},{vn,v1,v2},{vn,v1,v2,v3},{vn1,vn,v1,v2,v3},}.

In other words, we choose n2 sets to be in , with the ith set consisting of the first i+1 vertices of H2 (starting with v1). But now, taking the edges of H2 (without the last one) is a thin tree for . But it is far from thin for the cut {v2,,vn/2}; in fact, it has every edge of the tree.

We remark that the set of η-near minimum cuts cannot be uncrossed; if A and B are crossing η-near minimum cuts, AB and AB are both necessarily 2η-near minimum cuts, but this enlargement of η can compound over taking combinations of many cuts. This is the key reason that near minimum cuts can give rise to much richer structures than minimum cuts.

Despite this, we are able to show that a very carefully constructed laminar family suffices. The construction is quite delicate. We make heavy use of the polygon representation of 1/5-near-minimum cuts [5, 6, 7], which we will discuss in detail in the next section.

We remark that we have prioritized a more structured, modular analysis over obtaining the best possible constants. We anticipate that a more refined analysis (possibly with some adjustments to the algorithm) can show the result for some larger η. The polygon representation applies to the set of η-near minimum cuts for any η1/5. Thus, a natural question is whether our techniques can be made to work up to this limit; this seems to require some additional ideas.

Another natural question is whether the use of the polygon representation can be avoided in order to obtain the result for η>1/5. A result of this type would be of significant interest, as it could shed light on how to handle much more general cut families.

1.2 Related Work

The thin tree conjecture is known to hold for planar and bounded genus graphs [19]. As already mentioned, the best known result for general graphs is O(polyloglogn/k) thinness [1]; this result is not constructive. The best known constructive result gives thinness of O(lognloglognk) [2].333We remark that since this is a randomized algorithm and it is not known how to test the thinness of a tree in polynomial time, this is not fully constructive.

The study of thinness for particular cut sets began with Fürer and Raghavachari [9], who constructed thin trees with respect to the singleton cuts. This predates the thin tree conjecture, but questions concerning the existence of spanning trees satisfying bounds across given cuts, in this case singleton cuts, are well-studied in their own right. Goemans [11] and Singh and Lau [22] studied minimum cost bounded degree spanning trees. Bansal et al. [4] studied the problem of designing a spanning tree which crosses each cut in a laminar family at most a specified number of times, and showed how to obtain a tree with additive O(logn) violation whenever the fractional relaxation is feasible. Olver and Zenklusen [18] showed one can achieve thinness for any chain of cuts S1S2SkV, and Linhares and Swamy [16] resolved the minimum cost version of this question up to a small approximation factor in cost.

Near minimum cuts have been studied in several contexts, for example, Karger and Stein [13] gave bounds on how many such cuts can exist, and Benczúr [5, 6] and Benczúr and Goemans [7] studied their structure and defined the polygon representation of near minimum cuts, which we will rely upon heavily in this work. Karlin, Klein, and Oveis Gharan [14] further studied the polygon representation to improve the integrality gap of the subtour polytope and proved some properties we will make use of in this work. Bansal, Cheriyan, Grout, and Ibrahimpur [3] studied the problem of finding a minimum cover of all α-near minimum cuts in a k-edge-connected graph (i.e. a minimum cardinality set of edges so that each near minimum cut includes an edge from the set), and give a 16-approximation for any α.

2 Preliminaries

2.1 Near Minimum Cuts

Given a (multi-)graph G=(V,E) and SV, we will use δ(S) to denote the set of edges with exactly one endpoint in S. Recall a graph is k-edge-connected if |δ(S)|k for all SV.

Definition 3 (η-Near Minimum Cut).

Given a multi-graph G=(V,E) with minimum cut size k, a cut (S,S¯) with SV is an η-near minimum cut if |δ(S)|<(1+η)k, where η>0. We will use η-NMC (or sometimes NMC, when η is clear from context) as a shorthand.

We will often identify a cut (S,S¯) with (an arbitrary) one of its shores S and S¯, as in the following definition.

Definition 4 (Crossing Cuts).

We say two cuts S and T cross if ST,ST,TS, and V(ST) are all non-empty sets.

The following is standard and we simply cite a place it is proved.

Lemma 5 ([20]).

For G=(V,E,x), let AV be an ηA-NMC and BV an ηB-NMC, with A and B crossing. Then AB,AB,AB and BA are all (ηA+ηB)-NMCs.

2.2 Polygon Representation

Definition 6 (Connected Components).

Construct the cross graph as follows. Create a vertex corresponding to each η-near minimum cut (represented as an arbitrary shore), and connect the vertices corresponding to two near minimum cuts S,T if S and T cross. We define a connected component of near minimum cuts as a collection of near minimum cuts that correspond to a connected component in the cross graph.

Definition 7 (Atoms and Containment).

Given a connected component 𝒞 of near minimum cuts, let {ai}i0 be the coarsest partition of the vertices of the original graph so that for every S𝒞, we have aiS or aiS= for every i. We use 𝒜(𝒞) to represent {ai}i0, and call this the set of atoms.

Benczúr and Goemans [7] showed that any connected component of η-near minimum cuts admits a so-called polygon representation so long as η1/5. Here we define a polygon P representing a connected component 𝒞 of η-near minimum cuts (for η1/5) of a k-edge-connected graph.

Figure 2: Consider the graph on the left, with minimum cut 7. On the right is the polygon representation of the connected component of all proper cuts with at most 8 edges. This component consists of all proper near minimum cuts of the graph excluding the cut {7,8}, which is in its own connected component of size 1. As defined above, 15 and 16 are inside atoms, the others are outside atoms. Note {7,8} is a single atom.
  1. 1.

    A polygon representation is a convex regular polygon with a collection of diagonals connecting vertices of the polygon. All polygon edges and diagonals are drawn using straight lines in the plane. The diagonals partition the polygon into cells.

  2. 2.

    Each atom a𝒜(𝒞) is mapped to a cell of the polygon. If one of these cells is bounded by some portion of the polygon boundary it is non-empty and we call its atom an outside atom. We call the atoms of all other non-empty cells inside atoms. Note that some cells may not contain any atom. WLOG label the outside atoms a0,,am1 in counterclockwise order, and label the inside atoms arbitrarily. We also label points of the polygon p0,,pm1 such that outside atom ai is on the side (pi,pi+1) and a0 is on the side (pm1,p0).

  3. 3.

    No cell has more than one incident outer polygon edge.

  4. 4.

    Each diagonal (often called a representing diagonal) defines a cut such that each side of the cut is given by the union of the atoms on each side. Furthermore, the collection of cuts given by these diagonals is exactly 𝒞.

For set of vertices S and a polygon P for connected component 𝒞, let OP(S) denote the set of outside atoms contained in S. By the definition of polygons, S,S𝒞 cross if and only if OP(S) and OP(S) cross. When the polygon is clear from context, we will sometimes simply use O(S).

2.3 Inside Atoms

Inside atoms only exist if certain structures called k-cycles appear in the connected component 𝒞 being represented by P.

Definition 8 ([7, Definition 3]).

A family of sets C1,,CkV, for some k3, forms a k-cycle if

  • Ci crosses both Ci1 and Ci+1 (we treat Ck+1 as C1 and C0 as Ck);

  • CiCj= for ji1,i or i+1; and

  • 1ikCiV.

  • If k=3, we have the additional condition (CiCi+1)Ci1 for i{1,2,3}.

Lemma 9 ([7, Lemma 22]).

Any k-cycle formed by cuts in a connected component 𝒞 of η-near min cuts satisfies k1/η. (Note if η=0 then there are no k-cycles.)

Lemma 10 ([7, Definition 4]).

An atom a𝒜(𝒞) is an inside atom if and only if there is a k-cycle C1,,Ck𝒞 such that aCi for all 1ik.

In [14], the following lemmas were shown. The following lemma shows that cells of P that can be described with few diagonals and do not intersect any side of P are empty.

Lemma 11 ([14, Lemma 4.30]).

Let η1/5 and let P be the polygon representation for a connected component |𝒞|>1 of η-NMCs of a graph. Suppose H is the intersection of half-planes H0,,H1 corresponding to diagonals D0,,D1 of P that has a positive area. If H does not contain any side of P (equivalently, it does not contain any outside atom) and <1/(2η) then H does not have any inside atoms.

It is clear that in any polygon representing a connected component of cuts 𝒞, for any S,T𝒞, S=T if and only if O(S)=O(T) and similarly S,T cross if and only if O(S),O(T) cross. The following two lemmas extend this notion to all 1/5-NMC, even in polygons which were constructed for smaller values of η.

Lemma 12 ([14, Lemma 4.23]).

Let P be the polygon representation for a connected component |𝒞|>1 of η-NMCs of a graph G with atom set 𝒜(𝒞). If A,B𝒜(𝒞) are two 1/5-NMCs with OP(A)=OP(B) and there is an atom r𝒜(𝒞) such that rA,B, then A=B.

Lemma 13 ([14, Lemma 4.24]).

Let P be the polygon representation for a connected component |𝒞|>1 of η-NMCs of a graph G with atom set 𝒜(𝒞). If A,B𝒜(𝒞) are two 1/5-NMCs with OP(A)=OP(B). Then, A and B cross if and only if O(A) and O(B) cross.

3 Constructing the Laminar Family

Say that a collection of cuts covers an η-NMC S if δ(S)Bδ(B). For η=1/40, we will show how to construct a laminar family of 4η-NMCs such that every η-NMC S is covered by a set (S) of constant size. This means that any spanning tree T and η-NMC S,

|δ(S)T|B(S)|δ(B)T|11|B|O(1),

where the second inequality follows from Theorem 2 (which we may apply as 4η<1/5). Hence, the existence of such implies Theorem 1.

We will prove a version of this coverage statement for each connected component of η-NMCs, with the help of the polygon representation. In Section 3.4 we will see that the laminar families for each connected component can be easily combined to construct the overall laminar family.

The coverage theorem for a single nontrivial connected component is the following.

Theorem 14.

Let η=1/40. Given a connected component 𝒞 of η-NMC with |𝒞|>1, there exists a laminar family such that every η-NMC in 𝒞 is covered by a collection of at most 8 cuts in . This collection can be computed in polynomial time given 𝒞, and for each atom a𝒜(𝒞) and every cut S, we have aS or aS=.

In Section 3.1, we describe our algorithm for constructing the laminar family. In Section 3.2, we show a weaker coverage statement that involves only outside atoms. Then in Section 3.3 we use properties of the polygon representation described in Section 2 to show that coverage holds fully. We find it slightly surprising that for the bulk of our analysis, we can eschew the complexity of the full polygon representation and focus only on the outside atoms, which have a simpler structure.

For the remainder, fix η=1/40, a connected component 𝒞 of η-NMCs with |𝒞|>1, and its polygon representation P. We let 𝒜 denote the atoms of P. Let the vertices of P be p0,p1,,pn1, and the outside atoms of P be a0,a1,,an1, in counterclockwise order such that ai is the only outside atom in one side of the cut with representing diagonal (pi,pi+1), where indices are taken modulo n. We will refer to a0 as the root atom; an will be another name for a0.

Definition 15 (Intervals).

Let l,r be such that 1l<rn. We define l,r to be the set of outside atoms {al,al+1,,ar1}; we will refer to such a set as an interval.

Note that l,r includes al but not ar; it can be thought of as the set of outside atoms between pl and pr on the polygon.

By Lemma 12, no two distinct 1/5-NMCs have the same set of outside atoms. Thus the following is well-defined.

Definition 16 (Shadows and near minimum intervals).

Call an interval l,r a shadow if there exists a 1/5-NMC S with O(S)=l,r, and in this case, define cut(l,r)=S. We refer to cut(l,r) as the canonical cut associated with the interval l,r.

If in addition cut(l,r) is an η-NMC, we call l,r a near minimum interval, or NMI for short.

The following is an immediate consequence of Lemma 13.

Corollary 17.

If cut(l,r) and cut(l,r) are crossing canonical cuts, then l,r and l,r are crossing intervals.

3.1 Algorithm

In this subsection, we describe a recursive top-down procedure to construct , the desired laminar family. Before describing the algorithm formally, we start by giving some intuition for the algorithm and an overview of its general steps. For this intuition, assume that there is a single polygon and there are no inside atoms.

If the set of near minimum intervals is already laminar, we are done. At the other extreme, if all intervals l,r are near minimum, then we can set ={cut(1,2),cut(1,3),,cut(1,n)} because for all l,r, cut(l,r) is covered by cut(1,r) and cut(1,l). In fact,
={cut(1,2),cut(1,3),,cut(1,n)} would always work if 1,r is the shadow of a 4η-NMC for all r. Unfortunately, this is not always the case.

Instead, suppose we greedily add all cuts of the form cut(1,r) that are 4η-NMCs to . Let S=cut(lS,rS) be an η-NMC. Let

i=min{x:lSxrS,1,x is a shadow of a 4η-NMC}
j=max{x:lSxrS,1,x is a shadow of a 4η-NMC}.

Suppose i and j exists. Then, cut(1,i) and cut(1,j) covers all edges of δ(S) except the edges between 1,lS and lS,i and the edges between j,rS and rS,n. If we could add cut(lS,i) and cut(j,rS) (assuming they exist and are 4η-NMCs), then we can cover all edges of δ(S). We can almost do that: since cut(1,i) is a 4η-NMC and cut(lS,rS) is an η-NMC, by Lemma 5, cut(lS,i) exists and is a 5η-NMC. In the actual algorithm, instead of greedily taking all prefix 4η-NMCs, we will only take a subset of 2η-NMCs we define as special cuts (see Definition 18). We argue that in that case, cut(lS,i) and cut(j,rS) are also special.

It is still unclear whether we can add both cut(lS,i) and cut(j,rS) to . A natural way to proceed with the algorithm is to recurse on all subintervals xk,xk+1 where x1<x2<<xm are indices such that 1,xk is special. Suppose xa=i and xb=j. When we recurse on xa1,xa, we need to add special intervals from the right to add cut(lS,i), but when we recurse on xb,xb+1, we need to add special intervals from the left to add cut(j,rS). How does the algorithm know which direction to add intervals? A wishful attempt is to alternate between adding intervals from the left and right, and surprisingly this works (modulo some technical additions). It turns out that for any S=cut(lS,rS), we can pick O(1) sets added to in 3 recursion layers that covers S.

We define special cuts and intervals that are alluded in the outline.

Definition 18 (Special cuts and intervals).

We say that a canonical cut S=cut(l,r) is special if either

  1. (i)

    S is an η-NMC, or

  2. (ii)

    there exist two crossing canonical cuts A and B, both of which are η-NMCs, such that S is the intersection, union or set difference of A and B.

We say that an interval l,r is special if it is a shadow and cut(l,r) is a special cut.

By Lemma 5, any special cut is a 2η-NMC; however, not every 2η-NMC is necessarily special. Similarly, the set difference of two crossing special cuts is a 4η-NMC, since 4η<1/5. Note that by Corollary 17, a special interval is the intersection, union or set difference of crossing NMIs.

Case 2. The red intervals show the NMIs used to prove that AB and BA are special.

Case 3. The red intervals show the NMIs used to prove that AB and BA are special.

Case 4. x1 and x2 represent possible positions of x relative to rB. The red intervals show the NMIs used to prove that AB and AB are special if x<rB. The blue intervals show the NMIs used to prove that AB and BA are special if xrB.

Case 5. x1 and x2 represent possible positions of x relative to lB. The red intervals show the NMIs used to prove that AB and BA are special. The blue intervals show the NMIs used to prove that AB and AB are special.

Figure 3: Figures depicting the proof of Lemma 19.

Here is a useful lemma about special intervals that will be used repeatedly in our proofs.

Lemma 19.

Let A=lA,rA be a special interval and B=lB,rB be an NMI crossing A. Then, both of the following hold:

  • At least one of AB and AB is special.

  • At least one of AB and BA is special.

Proof.

Without loss of generality, assume that lA<lB<rA<rB. There are several cases depending on how A is special; we refer to Figure 3 for an illustration of all cases other than the first.

  1. 1.

    A is an NMI.

    Then, AB, AB, AB, and BA are all special by definition.

  2. 2.

    A=x,rAlA,y for some NMIs x,rA and lA,y with x<lA<rA<y.

    Then, AB=lB,rA=x,rAB and BA=rA,rB=Bx,rA are special.

  3. 3.

    A=y,rAx,lA for some NMIs y,rA and x,lA with x<y<lA<rA.

    Then, AB=lB,rA=y,rAB and BA=rA,rB=By,rA are special.

  4. 4.

    A=lA,xrA,y for some NMIs lA,x and rA,y with lA<rA<x<y.

    If x<rB, then AB=lA,lB=lA,xB and AB=lA,rB=lA,xB are special. Otherwise, xrB, so y>rB. Then, AB=lB,rA=BrA,y and BA=rA,rB=rA,yB are special.

  5. 5.

    A=lA,xy,rA for some NMIs lA,x and y,rA with lA<y<x<rA.

    If xlB, then y<lB, so AB=lB,rA=y,rAB and BA=rA,rB=By,rA are special. Otherwise, x>lB, so AB=lA,lB=lA,xB and AB=lA,rB=lA,xB are special.

Our algorithm for constructing a laminar family of near minimum cuts is described in Algorithm 1. The procedure is recursive; our family is generated by running MakeLam(1,n,0,0) on the outside atoms of our polygon P. Each call either proceeds “from the left” or “from the right”; crucially, this direction alternates as one goes down the recursion tree. Suppose we are proceeding from the left on some interval L,R (everything is symmetric). The algorithm finds a maximal chain L,x1L,x2L,xm of special sets within the interval. The algorithm recurses into the rings xi,xi+1 of this chain, including L,x1 and xm,R. Except for possibly xm,R, all these intervals are special or set differences of special intervals, and thus their corresponding cuts cut(xi,xi+1) are 4η-minimum cuts. These are all added to . The special cuts cut(L,xi) are also added.

One technical addition is the 𝑢𝑛𝑡𝑜𝑢𝑐ℎ𝑒𝑑 counter. It may be that the algorithm is applied to some interval L,R, say from the left, but there are no special cuts of the form L,x. The algorithm will then recurse on the same interval, but now from the right. If, however, there are also no special intervals of the form x,R, we shrink the interval to make sure the algorithm continues.

Algorithm 1 Procedure MakeLam(L,R,0pt,𝑢𝑛𝑡𝑜𝑢𝑐ℎ𝑒𝑑).

Let be the family obtained from MakeLam(1,n,0,0).

Lemma 20.

is a laminar family of 4η-near minimum cuts.

Proof.

All the cuts added to are either special or the setwise difference of two crossing special cuts, and so are 4η-near minimum cuts by Lemma 5.

To show that is a laminar family, firstly note that all the cuts added by Algorithm 1 are of the form cut(l,r) with 1l<rn. Hence, by Corollary 17, is laminar if and only if the corresponding intervals O():={l,r:cut(l,r)} form a laminar family.

We prove by induction on RL that MakeLam(L,R,,) returns a family of such that O() is a laminar family of intervals contained in L,R, which suffices by the above observation.

The base case RL=0 is obvious. For the inductive step, consider a call of the form MakeLam(L,R,0pt,𝑢𝑛𝑡𝑜𝑢𝑐ℎ𝑒𝑑). Either the algorithm calls MakeLam with a subinterval of L,R at a higher depth, or line 13 or line 20 was triggered. If line 13 was triggered (the other case is analogous), then the cuts added to before recursing are cut(L,x1), cut(L,x2),,cut(L,xm), cut(xm,R) and possibly cut(L,R), with corresponding intervals L,x1, L,x2,,L,xm and L,R included in O(). Furthermore, the algorithm then recurses into each subinterval of the form xi,xi+1, as well as L,x1 and xm,R. By the induction hypothesis, each of these adds a laminar family contained within some xi1,xi to O(), and so O() is laminar.

3.2 Analysis for Outside Atoms

Here, we will prove a weaker coverage statement than is claimed by Theorem 14, that only refers to outside atoms.

Definition 21.

Given a pair (u,v) of atoms, we say that a cut C covers (u,v) if u and v lie on different sides of C, and we say that a collection of cuts covers (u,v) if at least one cut in covers (u,v).

Lemma 22.

Let S be an η-near minimum cut. Then there exists a collection of cuts of size at most 8 such that every pair of outside atoms covered by S is covered by . Furthermore, all of the sets in are special except for at most one, which is the set difference of two crossing special cuts.

Note that if “every pair of outside atoms” was instead “every pair of atoms”, this would yield Theorem 14.

Proof.

Since S is an η-near minimum cut, S=cut(lS,rS) for some 1lS<rSn. We say that an interval L,R is active at depth d if MakeLam(L,R,d,) was called at some point. Let

MakeLam(l1,r1,0pt,𝑢𝑛𝑡𝑜𝑢𝑐ℎ𝑒𝑑)

be the call of MakeLam with the highest depth such that lS,rSl1,r1.

Firstly, we show that 𝑢𝑛𝑡𝑜𝑢𝑐ℎ𝑒𝑑<2. If this is not the case, then by maximality of 0pt we must have that l1=lS, and l1,x is not special for any x such that l1<xr1. However, lS,rS=l1,rS is an NMI, a contradiction.

Without loss of generality, assume that 0pt is even. Since 𝑢𝑛𝑡𝑜𝑢𝑐ℎ𝑒𝑑<2, there exists some x such that lS<x<rS and l1,x is special (or else lS,rS is contained in some active interval with depth 0pt+1). Let

i=min{x:lSxrS,l1,x is special}
j=max{x:lSxrS,l1,x is special}.

Since 0pt is even and l1,i, l1,j are special, cut(l1,i) and cut(l1,j) are added to by line 13 of Algorithm 1. These two cuts cover all pairs of outside atoms covered by lS,rS, except for pairs involving either

  1. (i)

    an outside atom in l1,lS and an outside atom in lS,i, or

  2. (ii)

    an outside atom in j,rS and an outside atom not in l1,rS.

Figure 4 shows the relative positions of the defined points.

Figure 4: A figure showing the relative locations of the points defined in the first part of the proof of Lemma 22. The cut we are analyzing is lS,rS. The cuts cut(l1,i) and cut(l1,j) are in and are included in .

Firstly, we prove that the pairs of outside atoms such that one is in l1,lS and the other is in lS,i are covered by a cut in .

Claim 23.

There is a special cut in that covers every pair of outside atoms such that one is in l1,lS and the other is in lS,i.

Proof.

If lS=i, no such pairs exist and we are done. Otherwise, l1,lS is not special. We show that cut(lS,i) is active at depth 0pt+1 and is added to ; this will be our desired cut. Note that there will be some active interval of depth 0pt+1 ending at i containing lS. Since 0pt+1 is odd, if lS,i is special, then line 20 of Algorithm 1 will add cut(lS,i) to . Thus, it is sufficient to show that lS,i is special.

By Lemma 19 with A=l1,i and B=lS,rS, either AB or AB is special. However, AB=l1,lS is not special from the definition of i, since lS<i. Hence, AB=lS,i is special, as claimed. Figure 5 shows the intervals used in the argument.

Figure 5: Figure depicting the proof that lS,i is special. The red interval denotes A and the blue interval denotes B.

Finally, we prove that the pairs of outside atoms such that one is in j,rS and the other is not in l1,rS are covered by at most 5 cuts in .

Claim 24.

There is a collection of at most 5 cuts in that covers every pair of outside atoms such that one is in j,rS and the other is not in l1,rS. Furthermore, all the cuts in are special except possibly one, which is the set difference of two crossing special cuts.

Proof.

If rS=j, no such pairs exist and we are done. Similar to the proof of Claim 23, we first show that j,rS is special. Note that unlike Claim 23, this does not immediately finish the proof since we cannot guarantee that cut(j,rS) is added to at depth 0pt+1 (rS lies in an active interval starting at j at depth 0pt+1 and line 20 of Algorithm 1 adds cuts from right to left). However, we will see that a similar analysis works to show that we can cover the desired pairs of atoms by adding at most 5 cuts from depths 0pt+1 and 0pt+2 in to the collection .

To show that j,rS is special, apply Lemma 19 to the crossing intervals A=l1,j and B=lS,rS. Then either AB or BA is special; but since j<rS, AB=l1,rS is not special by the definition of j. Hence, BA=j,rS is special. Figure 6 shows the intervals used in the argument.

Figure 6: Figure depicting the proof that j,rS is special. The red interval denotes A and the blue interval denotes B.

Let l2,r2 be the active interval at depth 0pt+1 such that l2rSr2. Note that l2=j and since rS>j, l2<rSr2r1. Let

l3 =max{x:l2xrS,x,r2 is special},
r3 =min{x:rSxr2,x,r2 is special}, and
z =min{x:l2xrS,x,r2 is special}.

If l3 does not exist, then there exists an active interval [l2,r3] at depth 0pt+2 such that l2rSr3. Since j,rS=l2,rS, the cut cut(j,rS) will be added to by line 13 of Algorithm 1 at depth 0pt+2 and we are done. Otherwise, l3,r3,z exist. Since l3,r2, r3,r2, and z,r2 are special, line 20 of Algorithm 1 adds the corresponding cuts to . We add these 3 cuts to . The only remaining pairs of outside atoms that are currently uncovered are pairs of outside atoms with:

  1. 1.

    one in j,z and the other not in l1,r2, and

  2. 2.

    one in l3,rS and the other in rS,r3.

Figure 7 shows the relative positions of the defined points, intervals corresponding to cuts added to , and pairs of intervals that are still uncovered.

Figure 7: A figure indicating the relative positions of the defined points and the cuts in we considered so far (denoted as red lines). The pairs of atoms between the pair of blue intervals and the pair of green intervals are currently unaccounted for.

At depth 0pt+2, the intervals j,z and l3,r3 are active. Note that j,z=j,rSz,r2 is the set difference of two crossing special intervals. Thus, line 8 of Algorithm 1 adds cut(j,z) to , which covers the pairs of atoms described in (i). We add cut(j,z) to .

It remains to show that the pairs of atoms described in (ii), where one is in l3,rS and the other is in rS,r3, are covered by some cut in . If l3=rS, we are done, so suppose l3<rS.

By Lemma 19 with A=l3,r2 and B=lS,rS, either AB or AB is special. Since l3<rS, AB=rS,r2 is not special. Hence, AB=l3,rS is special. This means that cut(l3,rS) will be added to by line 13 of Algorithm 1 during the call with active interval l3,rS, which covers the desired pairs of atoms. We add cut(l3,rS) to . Figure 8 shows the intervals used in the argument.

Figure 8: Figure depicting the proof that l3,rS is special. The red interval denotes A and the blue interval denotes B.

Taking to be the collection of cuts containing cut(l1,i), cut(l1,j), the cut from Claim 23 and the cuts in from Claim 24 completes the proof.

3.3 Extending the Analysis to Inside Atoms

Next, we show how to extend Lemma 22 to cover all pairs of atoms covered by an η-near minimum cut, not merely all pairs of outside atoms.

Lemma 25.

Let S be an η-near minimum cut with η=1/40 in a connected component of cuts 𝒞 with |𝒞|>1. Let be the laminar family constructed for 𝒞 by Algorithm 1. Then, there exists a collection of cuts of size at most 8 such that if (u,v) is a pair of atoms covered by S, then (u,v) is covered by .

Proof.

Let be the collection obtained from Lemma 22. We show that it covers all pairs of atoms covered by S even if some of them are inside atoms.

Note that cuts corresponding to special intervals can be represented by the intersection of at most 2 halfplanes bounded by diagonals of the polygon and cuts corresponding to set differences of two crossing special intervals can be represented by the intersection of at most 4 such halfplanes. Consider the set of diagonals 𝒟 of the polygon corresponding to either S or a halfplane bounding a cut in . Note that |𝒟|1+27+4=19, since by Lemma 22, contains at most 7 special cuts and one cut corresponding to set differences of two crossing special intervals. This set of diagonals divide the polygon into regions with at most 19 sides (since each diagonal in 𝒟 corresponds to at most one side). By Lemma 11, each such region with no outside atoms is empty. Hence, the only inside atoms are in regions containing at least one outside atom.

Consider a pair of atoms (u,v) covered by S. Since each inside atom is in the same region as some outside atom, there exist outside atoms u,v such that u,u and v,v lie in the same region. However, a diagonal separates u,v if and only if it separates u,v, so a cut in separates u,v if and only if it separates u,v (since each cut can be defined by the intersection of some halfplanes defined by 𝒟). In particular, u,v are outside atoms separated by S. By Lemma 22, u and v are separated by some cut in . Hence, u and v are also separated by some cut in .

Figure 9: The blue lines are the diagonals in 𝒟 and the red line is the representing diagonal of S. In the argument of the proof of Lemma 25, we have u=a6 and v=a0.

Lemma 25 finishes the proof of Theorem 14.

3.4 Constructing the Full Laminar Family

Now that we have shown how to construct a laminar family P on the atoms of a polygon P for a connected component of cuts 𝒞 so that all cuts in 𝒞 are implied by P, we will describe how to construct the overall laminar family. We will begin by constructing a cross-free family .

In particular, initialize a cross-free family . Add to one side of each η-near minimum cut which is not crossed (i.e. every connected component 𝒞 with |𝒞|=1). Then, construct a polygon P for each connected component 𝒞 of near minimum cuts with |𝒞|>1. Finally, run our construction procedure on each polygon P (setting one of the outside atoms as root), and add the resulting laminar family of cuts P to .

Here we make use of the following from Benczúr:

Lemma 26 ([6, Lemma 4.1.7]).

Let 𝒞,𝒞 be two distinct connected components of crossing cuts for a family of cuts of G=(V,E). Then, there exists an atom a𝒜(𝒞) and a𝒜(𝒞) so that aa=V.

Lemma 27.

is a cross-free family.

Proof.

Every cut added to is the union of some set of atoms in 𝒜(𝒞) for some connected component 𝒞 (note if |𝒞|=1 then 𝒜(𝒞)=C for the unique C𝒞).

By construction, every S,T over the same set of atoms 𝒜(𝒞) do not cross. So, suppose S is the union of a set of atoms in 𝒜(𝒞) and T is the union of a set of atoms in 𝒜(𝒞) for 𝒞𝒞. By Lemma 26, there is an atom a𝒜(𝒞) and a𝒜(𝒞) so that aa=V. Recall that 𝒜(𝒞) and 𝒜(𝒞) both form partitions over the vertex set.

  • If aS and aT, then ST=V.

  • If aS and aT=, then TS.

  • If aS= and aT, then ST.

  • If aS= and aT=, then ST=.

In each possible case they do not cross, so is cross-free.

Finally we will let , our final laminar family, be the result of fixing an arbitrary node rV and for each S, adding whichever of S and VS does not contain r. Given this, we can prove our main theorem. See 1

Proof.

We first compute the set of η-near minimum cuts of G, and for each connected component 𝒞, we compute the polygon representation. This can be done in polynomial time [5]. For each connected component (with given polygon representation P) we apply Theorem 14 and produce a laminar family as described above using Lemma 27 so that every cut S𝒞 is covered by at most 8 cuts in .

Then, apply Theorem 2 to to find a spanning tree T which has at most 11 edges in every cut in and return T.

Fix an η-near minimum cut S. If it does not lie in a connected component of size at least 2, then |Tδ(S)|11 immediately since S. Otherwise, S lies in some polygon P. By Theorem 14, the edges of S are contained in the edges of at most 8 cuts from . So, |Tδ(S)|88.

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Appendix A Corollaries from Prior Work

Recall the theorem mentioned in the introduction: See 2 As mentioned, this is not exactly the statement from [15], and further the second statement for laminar families requires an additional argument. We provide these now.

Klein and Olver [15] show that given a feasible solution x0E to the LP relaxation which requires that x is in the spanning tree polytope and x(δ(S))bS for integral bounds bS, one can round x to a spanning tree that violates the bounds by at most a factor of 22. One can exhibit a point in the dominant of this polytope in a general k-edge-connected graph by setting xe=2k for all edges, thus giving bS=2k|δ(S)| and an overall violation of 222k|δ(S)|222k|δ(S)|+2266k|δ(S)|, where we used that |δ(S)|k.

Next we consider the case where |δ(S)|43k for all S. [15] call a point x -aligned if x(E(S))=|S|1 for all S, where E(S) is the set of edges with both endpoints in S. For -aligned points, they give a guarantee of 2x(δ(S))+3. Shortly, we will show that the vector x¯0E defined by x¯e=3k for all edges dominates a point x which is -aligned. Using that |δ(S)|43k, this allows us to obtain a guarantee of

23k|δ(S)|+311.

Now we show that x¯ dominates an -aligned point x. In particular, it suffices to show that x¯(E(S))|S|1 for all S (see [15] for further details). The standard formulation of the dominant of the spanning tree polytope requires that for any partition Π of the vertex set, we have PΠx¯(δ(P))2(|Π|1). Now, let P1,,P be any partition of a set S. Then

PΠx¯(δ(P))3|Π|x¯(δ(S)),

where we use that x¯(δ(P))3 for all partitions, as G is k-edge-connected. Since x¯(δ(S))=3k|δ(S)|43k3k, and |Π|2,

PΠx¯(δ(P))3|Π|42|Π|2

as desired.