Overlapping Biclustering

TL;DR

The paper studies Bicluster Editing with Vertex Splitting on bipartite graphs, offering two main variants and showing APX-hardness for both. It advances the field by proving quadratic-size kernels (O(k^2) vertices) for both variants and correcting a flawed prior algorithm, delivering an O(k^{11k}+n+m)-time solution that combines kernelization with exhaustive search. These results resolve open questions on kernel size and algorithm correctness, and have practical implications for overlapping clustering tasks in recommender systems and biology. The work lays a foundation for further improvements in kernel bounds and exploration of non-bipartite inputs.

Abstract

We study the problem of transforming bipartite graphs into bicluster graphs. Abu-Khzam, Isenmann, and Merchad [IWOCA '25] introduced two variants of this problem. In both problems, the goal is to transform a bipartite graph into a bicluster graph with at most $k$ operations, where the allowed operations are inserting an edge, deleting an edge, and splitting a vertex. Splitting a vertex $v$ refers to replacing $v$ by two new vertices whose combined neighborhood equals the neighborhood of $v$. The latter models overlapping clusters, that is, vertices belonging to multiple clusters, and is motivated by several real-world applications. The versions differ in that one variant allows splitting any vertex, while the second variant only allows vertex splits on one side of the bipartition. Regarding computational complexity, they showed APX-hardness for both variants and a polynomial kernel (with $O(k^5)$ vertices) for the one-sided variant. They asked as open problems whether the polynomial kernel can be improved and whether it can also be extended for the other variant. We answer both questions in the affirmative and give kernels with $O(k^2)$ vertices for both variants. We also show that both problems can be solved in $O(k^{11k} + n + m)$ time, where $n$ and $m$ denote the number of vertices and edges in the input graph, respectively.

Figures (4)

• Figure 1: An example instance on the left where two operations are needed when only edge insertions and edge removals are allowed. On the right side a solution resulting from a single vertex split is shown.
• Figure 2: An illustration of two (inclusive) vertex splits (in a non-bipartite graph). In the first split, the vertex $v$ is replaced by $v_1$ and $v_2$, with the vertices $a_1,a_2,b_1$ and $b_2$ being adjacent to exactly one of the two vertices and $c$ being adjacent to both. In the second, $c$ is split into $c_1$ and $c_2$.
• Figure 3: Examples showing that previous kernelization techniques for Bicluster Editing do not extend to our setting.
• Figure 4: An example of a graph in which the algorithm by Abu-Khzam et al. abu-khzam2025biclusterediting can fail to find an optimal solution. Each node represents a critical independent set and the number above it shows how many vertices are contained in it. The nodes on the left side represent the vertices in $V_1$ and the nodes on the right represent $V_2$. The only optimal solution (with $k=4$) splits the vertices in $A_2$, $A_3$, and $A_4$ once each and deletes the edge between the vertex in $A_3$ and in $B_3$. Consider the case where the first $P_4$ found by the algorithm contains vertices from $A_1,B_1,A_2,$ and $B_2$. The only guess to consider is to mark the vertex in $A_2$. The second $P_4$ found contain vertices from $A_5,B_5,A_4$, and $B_4$ and the vertex in $A_4$ is marked. The third $P_4$ contains vertices from $A_6,B_6,A_3$, and $B_2$. Only marking the vertex in $A_3$ has to be considered. However, now no more $P_4$ with an unmarked vertex $c$ can be found and hence the algorithm computes a solution in which only vertices are split. The optimal way to do this splits the vertex in $A_2$ once and each vertex in $A_3 \cup A_4$ twice. Thus, the cost is $5 > 4$ showing that the algorithm by Abu-Khzam et al. abu-khzam2025biclusterediting is flawed.

Theorems & Definitions (8)

• Lemma 1
• proof
• Lemma 2
• proof
• Theorem 1
• proof
• Theorem 2
• proof