Research Article | Open Access

# On Graphs of the Cone Decompositions for the Min-Cut and Max-Cut Problems

**Academic Editor:**Frank Werner

#### Abstract

We consider maximum and minimum cut problems with nonnegative weights of edges. We define the graphs of the cone decompositions and find a linear clique number for the min-cut problem and a superpolynomial clique number for the max-cut problem. These values characterize the time complexity in a broad class of algorithms based on linear comparisons.

#### 1. Introduction

We consider the well-known maximum and minimum cut problems.

*Instance 1. *Given an undirected graph and a weight function , it is required to find such a subset of the vertex set (cut) that the sum of the weights of the edges from with one endpoint in and another in is as small as possible (minimum cut or min-cut) or as large as possible (maximum cut, max-cut).

While our discussion of the cut problem [1] in this paper is focused on integral edge weights, we remark that results and their proofs remain valid with real weights.

It is known that the min-cut problem with nonnegative edges is polynomially solvable: Dinic-Edmonds-Karp algorithm based on the maximum flow problem has the running time of [2], Hao-Orlin modification has the running time of [3], and Stoer-Wagner algorithm that does not use the flow techniques has the complexity [4].

Nevertheless, the min-cut and max-cut problems with arbitrary edges and the max-cut problem with nonnegative edges are NP-hard with no known algorithms faster than an exhaustive search [1].

We will estimate the min-cut and max-cut complexity with the polyhedral approach and study the clique number of the graph of the cone decomposition for the cut problems with nonnegative edges. This value is known as a measure of complexity in a wide class of algorithms based on linear comparisons. The presented results were announced in [5]. Similar characteristics of the shortest path and 3-dimensional matching problems are considered in [6, 7].

#### 2. Cut Polytope and Cone Decomposition

With every subset (every cut in the complete graph on vertices) we associate a characteristic vector according to the following rule:Therefore, the coordinates of the characteristic vector (also known as the cut vector) indicate whether the corresponding edges are in the cut or not. The convex hull of all cut vectors is known as the cut polytope [8]:

Max-cut and min-cut problems are reduced to the linear programming on the polytope with objective vector containing the weights of the edges.

We introduce a dual construction. Let be a set of points in . Let . Denote Since is the set of solutions of a finite system of homogeneous linear inequalities, it is a convex polyhedral cone. Given that the set of all cones is called the cone decomposition of the space by the set . Cone decomposition is similar to Voronoi diagram, exactly coinciding with it if the Euclidean norm of all points in is equal.

We consider the graph of the cone decomposition with the cones being the vertices, and two cones and are adjacent if and only if they have a common facet:

Denote by the clique number, the number of vertices in a maximum clique, of the graph of the cone decomposition. It is known [6, 9] that the complexity of the direct type algorithms, based on linear comparisons, of finding the minimum (or maximum, if we change the sign of the inequality in the definition of the cone) of a linear objective function on the set , or, which is the same, finding the cone , which the vector belongs to, cannot be less than the value of .

Indeed, if some algorithm at each step performs a single linear comparison (verification of linear inequality or ), from a geometric point of view that means drawing a hyperplane and discarding a wrong half-space. But if cones are pairwise adjacent, then for any hyperplane there exist points of at least cones in one of the half-spaces; hence, such direct type algorithm (algorithm with direct type linear decision tree [6, 9]) can separate and discard at most one wrong cone at a time in the worst case (Figure 1). Thus, is a lower bound on the height of the decision tree and on the complexity of combinatorial optimization problems in the wide class of algorithms, including sorting algorithms, greedy algorithm, dynamic programming, and branch and bound.

We introduce four different cone decompositions: for the maximum cut and the minimum cut problems, as well as for the problems with nonnegative edges. Denote by the set of vertices of the cut polytope . Let . Define

It is known [10] that the graph of the polytope is complete, so the graphs of cone decompositions and are complete as well, because the adjacency of vertices means the adjacency of the corresponding cones, and, therefore, their clique numbers are exponential. However, if we consider the problem with nonnegative edges, then the situation is fundamentally different.

Note that these constructions are deeply connected with the cut polyhedron [11, 12]. In particular, we reestablish the results on the min-cut in terms of cones.

#### 3. Graph of the Cone Decomposition for the Min-Cut Problem with Nonnegative Edges

Two sets and are called intersecting if

Lemma 1. *Cones and are adjacent if and only if cuts and are not intersecting.*

*Proof. *Assume that the cuts and are intersecting, but the cones and are adjacent. Adjacency of the cones means that there exists a nonnegative vector that belongs both to and , but does not belong to any other cone from the cone decomposition :It is well known that the cut function of a nonnegatively weighted undirected graph is submodular [12]: Since cuts and are intersecting, both cuts and are nonempty, and at least one of them is less than and . So we have a* contradiction*: cones and are not adjacent.

Now we consider two cuts and that are not intersecting. Without loss of generality we assume that . We assign the weights of edges in the following way: all edges that connect subsets and have the total sum equal to some positive integer , the sum of all edges connecting and is also equal to , edges between and have zero weight, and any other edge is equal to (Figure 2).

We have cuts and equal to , cut equals , and any other cut contains at least one edge of the weight . Thus, by inequality (8), cones and are adjacent.

Theorem 2. *Clique number of the graph of the cone decomposition for the min-cut problem with nonnegative weights of edges is linear by the number of vertices of the original graph and equal to *

*Proof. *A family is called laminar if no two sets are intersecting. It is known [11] that a laminar family of subsets of that does not contain , , and both a subset and its complement has at most subsets. Since, by Lemma 1, a laminar family of cuts corresponds to a clique in the graph of the cone decomposition, we obtain the required value.

As an example of pairwise adjacent cones we can consider the following cuts:(i)cuts of the form , where ;(ii) cuts of the form , where .

#### 4. Graph of the Cone Decomposition for the Max-Cut Problem with Nonnegative Edges

Lemma 3. *Cones and are adjacent if and only if one of the following conditions is true: *(i)*cuts and are intersecting;*(ii)*the difference between two cuts and has exactly one element: *

*Proof. *We consider two cuts and that are not intersecting and have more than one element in the difference. Without loss of generality, we assume that . We divide the set into two subsets and , since it has at least two vertices, and consider two additional cuts and (Figure 3).

Hence, by submodularity of the cut function, we have Thus, cones and can not be adjacent.

Now we assume that cuts and are intersecting and assign the weights in such way that all the edges, connecting sets and , have positive weights with the total sum equal to some integer ; similarly, the edges, connecting sets and , also have positive weights with the same total sum , while all the remaining edges are equal to zero (Figure 4).

Cuts and have the weight equal to . Since is the total weight of all edges in the graph, no other cuts can have weight exceeding . As for the cuts equal to , they have to split vertex sets and , as well as and . Only cuts and meet this requirement. Thus, cones and are adjacent.

The last configuration has cuts and not intersecting with exactly one element in the difference. Without loss of generality, we assume that and , where . We assign the weights of the edges in the following way: the weight of all edges connecting and is equal to , the same is for the edges connecting and , every edge between and has the weight , with the total number of these edges denoted by , and the weights of all other edges are assigned to be zero (Figure 5).

The total weight of all edges in the graph is equal to . The weight of both cuts and is . If some cut does not include at least one edge of the weight , then it is less than and : Thus, any maximum cut should split vertex sets and , while vertex can be included (cut ) or excluded (cut ). Hence, cones and are adjacent.

Theorem 4. *Clique number of the graph of the cone decomposition for the max-cut problem with nonnegative weights of edges is superpolynomial by the number of vertices of the original graph and equalsfor even andfor odd .*

*Proof. *The adjacency criterion of Lemma 3 is very similar to the Erdös-Ko-Rado theorem about the largest number of pairwise intersecting subsets. Therefore, we will use the construction from the Katona short proof [13].

First of all, we will consider only subsets of the size . Otherwise, we can replace with its complement that will have the required number of vertices.

Suppose we have some family of cuts with pairwise adjacent cones. We arrange the elements of into any cyclic order and consider the sets from that form intervals within this cyclic order. The question is how many intervals of the cyclic order may belong to . For example, consider the cyclic order and the interval . There are two intervals and inside and two intervals and containing , with cones adjacent to . From these four intervals, only two may be pairwise adjacent. Then, for any there are at most two adjacent intervals, starting at , with one element in the difference, and , and two adjacent intervals, ending at , and . Again from these four intervals only two may be pairwise adjacent. Thus, there are intervals in the cyclic order with the cones adjacent to .

Let be equal to . Since , at most intervals for a single cyclic order may belong to the family . We will count the number of pairs , where is a set in and is a cyclic order for which is an interval, in two ways. First, for each set we may generate by choosing one of permutations of and permutations of the remaining elements. The smallest value is for . And second, there are cyclic orders, each of which has at most intervals of . Thus, we have We consider two random cuts and of the size . Since they have the same size, and cannot be nested. Therefore, if they do not intersect, then , and cones and are adjacent. Thus, the set of all possible cuts of the size form a clique in the graph of the cone decomposition, and we have for odd .

Let be equal to . We again consider the cyclic order and estimate the maximum number of intervals with pairwise adjacent cones. Since any interval of the size may be adjacent to at most intervals, the maximum value will be for . However, there are only intervals of the size that form a cut, as all other intervals of the size will be complementary to them. So we have to include some smaller intervals as well. We assume that contains only intervals of the sizes and , and at most intervals for a single cyclic order may belong to the family .

We divide into two subsets and , containing only cuts of the corresponding size. Cuts of and cannot be nested, as they have the same size, and the difference with complementary cuts are not equal to one. Therefore, they are pairwise intersecting. Considering thatBy Erdös-Ko-Rado theorem we haveWe consider all the cuts of the sizes and that include vertex . If two such cuts and do not intersect, then, without loss of generality, and . There are exactly of such cuts, so we have Now suppose that there exists a family of cuts with pairwise adjacent cones that contains intervals of the size less than and is greater than . Then, for any cyclic order with at least one interval of the size less than there are at most intervals that may belong to the family . At the same time, for a smaller cut we can construct more cyclic orders containing as an interval. And so we get the* contradiction*: Thus, the family of and sized cuts with pairwise adjacent cones is the largest, and for even .

We can estimate these values using the central binomial coefficient:for even andfor odd .

#### 5. Conclusion

Min-cut and max-cut problems with arbitrary weights of edges and max-cut problem with nonnegative edges are NP-hard, while there are efficient polynomial algorithms for the min-cut problem with nonnegative edges. This situation corresponds to the geometric properties of the problem. Clique number of the graph of the cone decomposition is known as the lower bound on complexity in the class of algorithms that are based on linear comparisons. Graphs of the cone decompositions for the cut problems with arbitrary edges are complete, so their clique numbers are exponential. Clique number for the max-cut problem with nonnegative edges is superpolynomial and for the min-cut problem with nonnegative edges is linear.

Note that both obtained values are below the complexity of the best known algorithms for the cut problems with nonnegative edges. Hence, at least on this approach, there are no geometric obstacles to the construction of the more efficient algorithms.

Described results are based on the properties of the cut problem. For other combinatorial problems, it can be different.

Theorem 5. *If all points of the set belong to a sphere centered at the origin and for a pair of points cones and are adjacent, then nonnegative cones and are adjacent as well.**The same is true for cones and and nonnegative cones and .*

*Proof. *Consider a set of points such that Suppose that for a pair of points and of cones and are adjacent: Then there exists a vector () such that Now we have that Cones and are adjacent.

For cones proof can be obtained from the above by changing the sign in the corresponding inequality.

As a result, for the problem where Euclidean norm of the characteristic vectors is a constant, graphs of the cone decompositions with arbitrary objective vectors and nonnegative objective vectors will completely coincide, and, hence, the same problem on the minimum and maximum will have the same characteristics.

This is true, for example, for the traveling salesman problem, where all the possible solutions can be encoded as 0/1 vectors with exactly units, or the spanning tree problem, where all the possible solutions have units. Indeed, all four instances of the traveling salesman problem (minimum with arbitrary edges, minimum with nonnegative edges, maximum with arbitrary edges, and maximum with nonnegative edges) are known to be NP-hard, while all four instances of the spanning tree problem are polynomially solvable. At the same time, for the path problem in the graph with nonnegative edges the path can contain any number of edges, and possible solutions do not belong to a sphere centered at the origin, so, as well as for the cut problem, the shortest path problem is polynomially solvable, while the longest path problem is NP-hard [1].

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The research was partially supported by the Russian Foundation for Basic Research, Project 14-01-00333, and the President of Russian Federation Grant MK-5400.2015.1.

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Copyright © 2016 Vladimir Bondarenko and Andrei Nikolaev. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.