In the multiway cut problem, we are given an undirected graph with non-negative edge weights and a collection of terminal nodes and the goal is to find a minimum weight subset of edges to delete so that the input terminals cannot reach each other. For convenience, we will use -way cut to denote this problem when we would like to highlight the dependence on and multiway cut to denote this problem when grows with the size of the input graph. The -way cut problem is the classic minimum -cut problem which is solvable in polynomial time. For , Dahlhaus, Johnson, Papadimitriou, Seymour and Yannakakis [DJP94] showed that the -way cut problem is APX-hard and gave a -approximation. Owing to its applications in partitioning and clustering, -way cut has been an intensely investigated problem in the algorithms literature. Several novel rounding techniques in the approximation literature were discovered to address the approximability of this problem.
The known approximability as well as inapproximability results are based on a linear programming relaxation, popularly known as the CKR relaxation in honor of the authors—Călinescu, Karloff and Rabani—who introduced it[CKR00]. The CKR relaxation takes a geometric perspective of the problem. For a graph with edge weights and terminals , the CKR relaxation is given by
where is the -dimensional simplex and is the extreme point of the simplex along the -th coordinate axis, i.e., if and only if .
Călinescu, Karloff and Rabani designed a rounding scheme for the relaxation which led to a -approximation thus improving on the -approximation by Dahlhaus et al. For -way cut, Cheung, Cunningham and Tang [CCT06] as well as Karger, Klein, Stein, Thorup and Young [KKS04] designed alternative rounding schemes that led to a -approximation factor and also exhibited matching integrality gap instances. We recall that the integrality gap of an instance to the LP is the ratio between the integral optimum value and the LP optimum value. Determining the exact integrality gap of the CKR relaxation for has been an intriguing open question. After the results by Karger et al. and Cunningham et al., a rich variety of rounding techniques were developed to improve the approximation factor of -way cut for [BNS13, SV14, BSW17]. The known approximation factor for multiway cut is due to Sharma and Vondrák [SV14].
On the hardness of approximation side, Manokaran, Naor, Raghavendra and Schwartz [MNRS08] showed that the hardness of approximation for -way cut is at least the integrality gap of the CKR relaxation assuming the Unique Games Conjecture (UGC). More precisely, if the integrality gap of the CKR relaxation for -way cut is , then it is UGC-hard to approximate -way cut within a factor of for every constant . As an immediate consequence of this result, we know that the -approximation factor for -way cut is tight. For -way cut, Freund and Karloff [FK00] constructed an instance showing an integrality gap of . This was the best known integrality gap until last year when Angelidakis, Makarychev and Manurangsi [AMM17] gave a remarkably simple construction showing an integrality gap of for -way cut. In particular, this gives an integrality gap of for multiway cut.
We note that the known upper and lower bounds on the approximation factor for multiway cut match only up to the first decimal digit and thus the approximability of this problem is far from resolved. Indeed Angelidakis, Makarychev and Manurangsi raise the question of whether the lower bound can be improved. In this work, we improve on the lower bound by constructing an instance with integrality gap 1.20016.
For every constant , there exists an instance of multiway cut such that the integrality gap of the CKR relaxation for that instance is at least .
The above result in conjunction with the result of Manokaran et al. immediately implies that multiway cut is UGC-hard to approximate within a factor of for every constant .
2 Background and Result
Before outlining our techniques, we briefly summarize the background literature that we build upon to construct our instance. We rely on two significant results from the literature. In the context of the -way cut problem, a cut is a function such that for all , where we use the notation . The use of labels as opposed to labels to describe a cut is a bit non-standard, but is useful for reasons that will become clear later on. The approximation ratio of a distribution over cuts is given by its maximum density:
Karger et al. [KKS04] define
and moreover showed that there exists that achieves the infimum. Hence, . With this definition of , Karger et al. [KKS04] showed that for every , there is an instance of multiway cut with terminals for which the integrality gap of the CKR relaxation is at least . Thus, Karger et al.’s result reduced the problem of constructing an integrality gap instance for multiway cut to proving a lower bound on .
Next, Angelidakis, Makarychev and Manurangsi [AMM17] reduced the problem of lower bounding further by showing that it is sufficient to restrict our attention to non-opposite cuts as opposed to all cuts. A cut is a non-opposite cut if for every . Let . For a distribution over cuts, let
Angelidakis, Makarychev and Manurangsi showed that for all . Thus, in order to lower bound , it suffices to lower bound . That is, it suffices to construct an instance that has large integrality gap against non-opposite cuts.
As a central contribution, Angelidakis, Makarychev and Manurangsi constructed an instance showing that . Now, by setting , we see that is at least . Furthermore, they also showed that their lower bound on is almost tight, i.e., . The salient feature of this framework is that in order to improve the lower bound on , it suffices to improve for some .
The main technical challenge towards improving is that one has to deal with the -dimensional simplex . Indeed, all known gap instances including that of Angelidakis, Makarychev and Manurangsi are constructed using the -dimensional simplex. In the -dimensional simplex, the properties of non-opposite cuts are easy to visualize and their cut-values are convenient to characterize using simple geometric observations. However, the values of non-opposite cuts in the -dimensional simplex become difficult to characterize. Our main contribution is a simple argument based on properties of lower-dimensional simplices that overcomes this technical challenge. We construct a -dimensional instance that has gap larger than 1.2 against non-opposite cuts.
3 Outline of Ideas
Let be the graph with node set and edge set
, where the terminals are the four unit vectors. In order to lower bound, we will come up with weights on the edges of such that every non-opposite cut has cost at least and moreover the cumulative weight of all edges is . This suffices to lower bound by the following proposition.
Suppose that there exist weights on the edges of such that every non-opposite cut has cost at least and the cumulative weight of all edges is . Then, .
For an arbitrary distribution over non-opposite cuts, we have
where the last inequality follows from the hypothesis that every non-opposite cut has cost at least and the cumulative weight of all edges is . ∎
We obtain our weighted instance from four instances that have large gap against different types of cuts, and then compute the convex combination of these instances that gives the best gap against all non-opposite cuts.
All of our four instances are defined as edge-weights on the graph . We identify with the facet of defined by . Our first three instances are 2-dimensional instances, i.e. only edges induced by have positive weight. The fourth instance has uniform weight on .
We first explain the motivation behind Instances 1,2, and 4, since these are easy to explain. Let
Instance 1 is simply the instance of Angelidakis, Makarychev and Manurangsi [AMM17] on . It has gap against all non-opposite cuts, since non-opposite cuts in induce non-opposite cuts on . Additionally, we show in Lemma 4.5 that the gap is strictly larger than by a constant if the following two conditions hold:
there exist such that contains only one edge whose end-nodes have different labels (a cut with this property is called a non-fragmenting cut), and
has a lot of nodes with label 5.
Instance 2 has uniform weight on , and , and 0 on all other edges. Here, a cut in which each contains at least two edges whose end-nodes have different labels (a fragmenting cut) has large weight. Consequently, this instance has gap at least against such cuts.
Instance 4 has uniform weight on all edges in . A beautiful result due to Mirzakhani and Vondrák [MV15] implies that non-opposite cuts with no node of label 5 have large weight. Consequently, this instance has gap at least against such cuts. We extend their result in Lemma 4.4 to show that the weight remains large if has few nodes with label 5.
At first glance, the arguments above seem to imply that a convex combination of these three instances already gives a gap strictly larger than 1.2 for all non-opposite cuts. However, there exist two non-opposite cuts such that at least one of them has cost at most 1.2 in every convex combination of these three instances (see Section 7.1). One of these two cuts is a fragmenting cut that has almost zero cost in Instance 4 and the best possible cost, namely 1.2, in Instance 1. Instance is constructed specifically to boost the cost against this non-opposite cut. It has positive uniform weight on 3 equilateral triangles, incident to , and on the face . We call the edges of these triangles red edges. The side length of these triangles is a parameter, denoted by , that is optimized at the end of the proof. Essentially, we show that if a non-opposite cut has small cost both on Instance 1 and Instance 4 (i.e., weight on Instance 1 and weight on Instance 4), then it must contain red edges.
Our lower bound of is obtained by optimizing the coefficients of the convex combination and the parameter . By Proposition 3.1 and the results of Angelidakis, Makarychev and Manurangsi, we obtain that , i.e., the integrality gap of the CKR relaxation for -way cut is at least . We complement our lower bound of by also showing that the best possible gap that can be achieved using convex combinations of our four instances is (see Section 7.2).
4 A -dimensional gap instance against non-opposite cuts
We will focus on the graph with the node set being the discretized -dimensional simplex and the edge set . The four terminals will be the four extreme points of the simplex, namely for . In this context, a cut is a function such that for all . The cut-set corresponding to is defined as
For a set of nodes, we will also use to denote the set of edges with exactly one end node in . Given a weight function , the cost of a cut is . Our goal is to come up with weights on the edges so that the resulting -way cut instance has gap at least against non-opposite cuts.
We recall that denotes the boundary edges between terminals and , i.e.,
We will denote the boundary nodes between terminals and as , i.e.,
Let be a constant to be fixed later, such that is integral. For each , we define node sets and and edge set as follows:
We will refer to the nodes in as red111We use the term “red” as a convenient way for the reader to remember these nodes and edges. The exact color is irrelevant. nodes near terminal and the edges in as the red edges near terminal (see Figure 0(b)). Let denote the subgraph of induced by the nodes whose support is contained in . We emphasize that red edges and red nodes are present only in and that the total number of red edges is exactly .
4.1 Gap instance as a convex combination
Our gap instance is a convex combination of the following four instances.
Instance . Our first instance constitutes the -way cut instance constructed by Angelidakis, Makarychev and Manurangsi [AMM17] that has gap against non-opposite cuts. To ensure that the total weight of all the edges in their instance is exactly , we will scale their instance by . Let us denote the resulting instance as . In , we simply use the instance on and set the weights of the rest of the edges in to be zero.
Instance . In this instance, we set the weights of the edges in to be and the weights of the rest of the edges in to be zero.
Instance . In this instance, we set the weights of the red edges to be and the weights of the rest of the edges in to be zero.
Instance . In this instance, we set the weight of every edge in to be .
We note that the total weight of all edges in each of the above instances is . For multipliers to be chosen later that will satisfy , let the instance be the convex combination of the above four instances, i.e., . By the properties of the four instances, it immediately follows that the total weight of all edges in the instance is also .
4.2 Gap of the Convex Combination
The following theorem is the main result of this section.
For every and such that is integer, every non-opposite cut on has cost at least the minimum of the following two terms:
Before proving Theorem 4.1, we see its consequence.
There exist constants and with such that the cost of every non-opposite cut in the resulting convex combination is at least .
For every constant and every with , there exists a non-opposite cut whose cost in the resulting convex combination is at most .
In light of Corollary 4.2 and Theorem 4.3, if we believe that the integrality gap of the CKR relaxation is more than , then considering convex combination of alternative instances is a reasonable approach towards proving this.
The rest of the section is devoted to proving Theorem 4.1. We rely on two main ingredients in the proof. The first ingredient is a statement about non-opposite cuts in the -dimensional discretized simplex. We prove this in Section 5, where we also give a generalization to higher dimensional simplices, which might be of independent interest.
Let be a non-opposite cut on with nodes from labeled as , , or for some . Then, .
The constant that appears in the conclusion of Lemma 4.4 is the best possible for any fixed (if ). To see this, consider the non-opposite cut obtained by labeling to be for every , all nodes at distance at most from to be , and all remaining nodes to be . The number of nodes from labeled as , , or is . The number of edges in the cut is .
The second ingredient involves properties of the -way cut instance constructed by Angelidakis, Makarychev and Manurangsi [AMM17]. We need two properties that are summarized in Lemma 4.5 and Corollary 4.6. We prove these properties in Section 6. We define a cut to be a fragmenting cut if for every distinct ; otherwise it is a non-fragmenting cut. We recall that denotes the instance obtained from the 3-way cut instance of Angelidakis, Makarychev and Manurangsi by scaling it up by .
The first property is that non-opposite non-fragmenting cuts in that label a large number of nodes with label have cost much larger than .
Let be a non-opposite cut with nodes labeled as . If is a non-fragmenting cut and , then the cost of on is at least .
We show Lemma 4.5 by modifying to obtain a non-opposite cut while reducing its cost by . By the main result of [AMM17], the cost of every non-opposite cut on is at least . Therefore, it follows that the cost of on is at least . We emphasize that while it might be possible to improve the constant that appears in the conclusion of Lemma 4.5, it does not lead to much improvement on the overall integrality gap as illustrated by the results in Section 7.2.
The second property is that non-opposite cuts which do not remove any of the red edges, but label a large number of nodes in the red region with label have cost much larger than .
Let be a non-opposite cut and . For each , let
Then, the cost of on is at least .
In order to show Corollary 4.6, we first derive that the cost of the edges in the instance is at least using Lemma 4.5. Next, we modify to obtain a non-opposite cut such that . By the main result of [AMM17], the cost of every non-opposite cut on is at least . Therefore, it follows that the cost of on is at least .
We now have the ingredients to prove Theorem 4.1.
Proof of Theorem 4.1.
Let be a non-opposite cut. Let be the cut restricted to , i.e., for every with , let
We consider two cases.
Case 1: is a non-fragmenting cut.
Let the number of nodes in that are labeled by as (equivalently, labeled by as ) be for some . Since , Lemma 4.5 implies that the cost of on , and hence the cost of on , is at least . Moreover, the cost of on is at least since at least one edge in should be in for every pair of distinct
. To estimate the cost on, we observe that the number of nodes on labeled by as , , or is . By Lemma 4.4, we have that and thus, the cost of on is at least . Therefore, the cost of on the convex combination instance is at least
Case 2: is a fragmenting cut.
Then, the cost of on is at least as a fragmenting cut contains at least edges from each for distinct .
We will now compute the cost of on the other instances. Let , i.e., is the number of red triangles that are intersected by the cut . We will derive lower bounds on the cost of the cut in each of the three instances and based on the value of . For each , let
and let . Since , the sets and are disjoint for distinct . We note that since and .
In order to lower bound the cost of on , we will use Corollary 4.6. We recall that is the cut restricted to , so the cost of on is the same as the cost of on . Moreover, by Corollary 4.6, the cost of on is at least , because . Hence, the cost of on is at least .
The cost of on is at least by the following claim.
Let . If , then .
The subgraph is a cycle. If for some , then the path must also contain two consecutive nodes labeled differently by . ∎
Next we compute the cost of on . If , then the cost of on is at least . Suppose . For a red triangle with , we have at least nodes from that are labeled as , , or . Moreover, the nodes in and are disjoint for distinct . Hence, the number of nodes in that are labeled as , , or is at least , which is at least , since . Therefore, by Lemma 4.4, we have and thus, the cost of on is at least .
Thus, the cost of on the convex combination instance is at least for some , where
In particular, the cost of on the convex combination instance is at least , where
Now, Claim 4.8 completes the proof of the theorem. ∎
Let . If , then the claim is clear. We consider the three remaining cases.
Say . Then,
Say . Then,
where the last inequality is from the identity for all .
Say . Then,
where the last inequality is from the identity for all .
5 Size of non-opposite cuts in
In this section, we prove Lemma 4.4. In fact, we prove a general result for , that may be useful for obtaining improved bounds by considering higher dimensional simplices. Our result is an extension of a theorem of Mirzakhani and Vondrák [MV15] on Sperner-admissible labelings.
A labeling is Sperner-admissible if for every . We say that has an inadmissible label if . Let denote the hypergraph whose node set is and whose hyperedge set is
Each hyperedge has nodes, and if , then there exist distinct such that . We remark that has nodes and hyperedges. Geometrically, the hyperedges correspond to simplices that are translates of each other and share at most one node. Given a labeling , a hyperedge of is monochromatic if all of its nodes have the same label. Mirzakhani and Vondrák showed the following result that a Sperner-admissible labeling of does not have too many monochromatic hyperedges, which also implies that the number of non-monochromatic hyperedges is large.
Theorem 5.1 (Proposition 2.1 in [Mv15]).
Let be a Sperner-admissible labeling of . Then, the number of monochromatic hyperedges in is at most , and therefore the number of non-monochromatic hyperedges is at least .
Our main result of this section is an extension of the above result to the case when there are some inadmissible labels on a single face of . We show that a labeling in which all inadmissible labels are on a single face still has a large number of non-monochromatic hyperedges. We will denote the nodes with as .
Let be a labeling of such that all inadmissible labels are on and the number of nodes with inadmissible labels is for some . Then, the number of non-monochromatic hyperedges of is at least
Let , i.e. is the set of nodes in having an inadmissible label. Let us call a hyperedge of inadmissible if the label of one of its nodes is inadmissible.
There are at most inadmissible monochromatic hyperedges.
Let be the set of inadmissible monochromatic hyperedges. Each hyperedge has exactly nodes from and they all have the same label as is monochromatic. Thus, each contains nodes from . We define an injective map by letting to be the node with the largest st coordinate. Notice that if , then the other nodes of are (), and all but the last one are in . In particular, is positive.
Let be the image of . For and , let
Since and for every , the nodes of are on a line containing . It also follows that if . Let
We observe that if , then for each , the node is in and hence, with . In particular, this implies that . We now compute an upper bound on the size of , which gives an upper bound on the size of and hence also on the size of , as . For each node and for every , let be the node in with the largest th coordinate. Clearly if , because .
For given and