Biclique Sandwich: bounding the size of a maximum biclique

I. Basic Concepts

1. Bipartitie graph

A bipartite graph, is an undirected graph G = (V, E) in which V can be partitioned into two sets V1 and V2 such that (u, v) ∈ E implies either u ∈ V1 and v ∈ V2 or u ∈ V2 and v ∈ V1. That is, the set of vertices is decomposed into two disjoint sets such that no two vertices within the same set are adjacent.

Example1:

2. Biclque

Given a bipartite graph B = (V1 ∪ V2, E), a biclique C = U1 ∪ U2 is a subset of the vertex set, such that U1 ⊆ V1, U2 ⊆ V2, and for every u ∈ U1, v ∈ U2, the edge (u, v) ∈ E. In other words, a biclique is a complete bipartite subgraph of B.

Example2:

II. Problem Definition

1. Maximum Edge Biclique Problem(MBP)

Given a bipartite graph B = (V1 ∪ V2, E), find the biclique with the maximum number of edges.

2. Hardness of the problem and related problems

A biclique C = U1 ∪ U2 clearly has |U1| + |U2| vertices and |U1| × |U2| edges. The computational complexity of deciding whether or not a bipartite graph contains a biclique of a certain size depends on what objective function is applied on the size of a biclique. The biclique problem with objective function max (|U1| + |U2|) is called Maximum Vertex Biclique problem. This problem can be solved in polynomial time because maximum vertex biclique is equivalent to the maximum independent set in bipartite graphs which, in turn, can be solved via the minimum cut algorithm. The biclique problem with objective function max |U1|, provided that |U1| = |U2|, is called Balanced Complete Bipartite Subgraph problem or Balanced Biclique problem. It is listed as GT24 among the NP-complete problems in Garey and Johnson's book. The biclique problem with objective function max (|U1| × |U2|) is called Maximum Edge Biclique problem which was proven to be NP-complete by induction from 3SAT[1].

3. Bounding the size of the maximum edge biclique in bipartite graphs

Idea: Given that the MAX-CLIQUE is upper bounded by the Lovasz theta function, if a reduction from biclique to clique exists, we can bound the size of maximum edge biclique by applying Lovasz theta function on the transformed graph.

III. Reduction from Biclique to Clique

1. Reduction from biclique to clique

Let Γ = (V1 ∪ V2, E) be a bipartite graph. Define

LQ(Γ) := (E, {((u, v),(w, s)) | Γ(u, v, w, s) is a biclique});

in words, the vertices of LQ(Γ) are edges of Γ, two of them are adjacent if they intersect or lie in a common C₄.

Note: A cycle on n vertices is denoted C_n

Example3:

2. Lemma

Let Ω be a maximal clique in LQ(Γ). Then the subgraph

Γ_Ω := (∪_{(u,v) ∈ Ω} {u,v}, Ω ) ⊆ Γ is a biclique.

In another words, a maximal clique in LQ(Γ) corresponds to a biclique in Γ = (V1 ∪ V2, E) and vice versa.

PROOF:

Let x_i be a vertex that is in a maximal clique in graph LQ(Γ). Since the edge in LQ(Γ) corresponds to two vertices in Γ(V1) and Γ(V2), therefore all the vertices that form a maximal clique in LQ(Γ) correspond to a set of edges of Γ. Let (x₁, y), (z, x₂) ∈ Ω. Since Ω is a maximal clique, they are adjacent as vertices of LQ(Γ). By the definition of LQ(Γ), {x₁, y, z, x₂} form a C₄ in Γ. Therefore the subgraph Γ (∪_{(u,v) ∈ Ω} {u, v}) is a biclique. By the maximality of Ω, the corresponding edges form a maximal edge biclique in Γ.

IV. Bounding the size of Maximum Edge Biclique

1. Lovasz theta function

Maximum Clique problem: Find a complete subgraph of G with maximum number of vertices.

Minimum Graph Coloring problem: Assign colors to vertices such that every pair of adjacent vertices has different colors, using the smallest possible number of colors.

Maximum Clique Number of graph G -- ω(G): the number of vertices in the maximum clique of G.

Chromatic Number of graph G -- χ(G): the least number of colors for which the vertices of the graph G can be colored so that the two endpoints of any edge have different colors.

Since the vertices in ω(G) must be colored differently in every legal coloration, we have ω(G) ≤ χ(G).

Given a graph G = (V, E), V = {1, 2, ..., n}. Let Δ be the set of all symmetric n × n matrices for which (A)_ij = 1 if i = j or if i and j are adjacent in G. Let the elements of A corresponding to non-adjacent position vary. For any symmetrix matrix A, let Λ(A) be its largest eigenvalue. (Since A is symmetric, its eigenvalues are real.) Define

θ(G) = min{Λ(A): A ∈ Δ}[2].

THEOREM: For very graph G, ω(G) ≤ θ(G) ≤ χ(G).

PROOF:

Without lose of generality, let vertex 1, 2, ..., w of G form a maximum complete subgraph of G. Then the upper left w × w submatrix A' of any A ∈ Δ consists of 1's, and so

Λ(A) ≥ Λ(A') = w.

Hence

θ(G) = min{Λ(A): A ∈ Δ} ≥ ω(G).

To prove the other inequality, let k = χ(G). It suffices to consider the case of "k-partite Turan graphs". i.e. the case where V(G) has a partition V(G) = V1 ∪ ... ∪ Vk such that |V1| = |V2| = ... = |Vk| = m (say) and two vertices are adjacent if and only if they belong to different classes Vi. Let B denote the adjacentcy matrix of the complement graph of G, G, i.e. let (B)_ij = 1 if and only if i and j are adjacent vertices in G. Then clearly A = J + tB belongs to Δ for all real t (where J is the mk × mk matrix of all 1's). Since G is a regular graph, B has constant row and column sums and hence J and B commute. Since J has eigenvalues km, 0, ..., 0 and B has eigenvalues m - 1 (k times) and -1 ((km - k) times), the matrix A has eigenvalues km + t(m + 1), t(m - 1) ((k - 1) times) and -t ((km - k) times). For the choice t = -k, largest of these is k. So for this t, Λ(A) = k and so θ(G) ≤ Λ(A) = k.

2. Alternative formulation of Lovasz theta function

The Lovasz theta function has the sandwich property, and can be computed with arbitrary precision in polynomial time via a semidefinite program. There are many alternative definitions to the Lovasz theta function. Here is a definition of the Lovasz theta function as a maximization problem in terms of Maximum Independent Set.

Assign a vector z_i in Rⁿ to each vertex v_i in G, where n is the number of vertices in G.

maximize (∑ z_i) • (∑ z_i)

subject to

∑ (z_i • z_i) = 1

z_i • z_j = 0     (i, j) ∈ E(G)

3. Bounding the size of Maximum Edge Biclique

An upper bound for the size of the Maximum Edge Biclique in a bipartite graph is obtained based on the following two facts:

a. A biclique in a bipartite graph Γ can be reduced to a maximal clique in a general graph LQ(Γ).

b. The size of the Maximum Clique in graph G (Maximum Independent Set in the complement graph of G) is upper bounded by Lovasz theta function.

Therefore the size of the Maximum Edge Biclique in a bipartite graph can be upper bounded by Lovasz theta function too.

The following is the formulation of Lovasz theta function in terms of Maximum Edge Biclique.

maximize (∑ x_uv) • (∑ x_uv)

subject to

∑ (x_uv • x_uv) = 1 ,     ∀ x_uv ∈ V(LQ(Γ))

x_uv • x_ws = 0,     (x_uv, x_ws) ¬∈ E(LQ(Γ))

where x_uv is a vector in Rⁿ, n is the number of vertices in LQ(Γ) (the number of edges in Γ).

The following additional contraint can be added to the above Lovasz theta function, since Γ is a bipartite graph:

x_uv • x_ws = x_us • x_wv,     (x_uv, x_ws), (x_us, x_wv) ∈ E(LQ(Γ))

V. Concluding Remark

It is well known that Maximum Clique is hard to approximate within n^1-ε. The hardness results for Maximum Edge Biclique are still unknown. In this tutorial, we showed that biclique can be reduced to clique. Are these two problems of the same hardness? That is, can clique be reduced to biclique? It's still an open question.

VI. Reference

1. R. Peeters. The maximum edge biclique problem is NP-complete. Tech. Rep. 789, Tilburg University: Faculty of Economics and Business administration, 2000.

2. L. Lovasz. An algorithm theory of numbers, graphs and convexity. Society for Industrial and Applied Mathematics(SIAM), Philadelphia, Pa., 1986.

3. D. V. Pasechnik. Bipartite sandwiches: bounding the size of a maximum biclique. arXiv:math.CO/9907109 v1, 1999