Efficient enumeration of the optimal solutions to the correlation clustering problem

Abstract

According to the structural balance theory, a signed graph is considered structurally balanced when it can be partitioned into a number of modules such that positive and negative edges are respectively located inside and between the modules. In practice, real-world networks are rarely structurally balanced, though. In this case, one may want to measure the magnitude of their imbalance, and to identify the set of edges causing this imbalance. The correlation clustering (CC) problem precisely consists in looking for the signed graph partition having the least imbalance. Recently, it has been shown that the space of the optimal solutions of the CC problem can be constituted of numerous and diverse optimal solutions. Yet, this space is difficult to explore, as the CC problem is NP-hard, and exact approaches do not scale well even when looking for a single optimal solution. To alleviate this issue, in this work we propose an efficient enumeration method allowing to retrieve the complete space of optimal solutions of the CC problem. It combines an exhaustive enumeration strategy with neighborhoods of varying sizes, to achieve computational effectiveness. Results obtained for middle-sized networks confirm the usefulness of our method.

Appendices

2-Partition and 2-chorded cycle inequalities

For the sake of completeness, we detail below the 2-partition and 2-chorded cycle inequalities:

• Let $S, T \subseteq V$ be two nonempty disjoint subsets of V. Then, the 2-partition inequality, illustrated in Fig. 11a, is defined as

  $$\begin{aligned} \sum \limits _{u \in S} \sum \limits _{v \in T} x_{uv} - \sum _{\begin{array}{c} (u,v) \in S\\ u \ne v \end{array}} x_{uv} - \sum _{\begin{array}{c} (u,v) \in T\\ u \ne v \end{array}} x_{uv} \le min\{|S|,|T|\}. \end{aligned}$$

• Let C $\subseteq$ E be a cycle of length at least 5, and $\overline{C} = \{v_i v_{i+2} | i=1,..,|C|-2\} \cup \{v_1 v_{|C|-1}, v_2 v_{|C|}\}$ be a 2-chorded cycle of C. Then, the 2-chorded cycle inequality, illustrated in Fig. 11b, is defined as

  $$\begin{aligned} \sum \limits _{(u,v) \in C} x_{uv} - \sum \limits _{(u,v) \in \overline{C}} x_{uv} \le \lfloor \frac{|C|}{2} \rfloor . \end{aligned}$$

Fig. 11

Illustrations of the 2-partition (left) and 2-chorded cycle (right) inequalities

Edit distance between two membership vectors

Before calculating the edit distance between two membership vectors, we need to select one of them as the reference vector, in order to adapt the module assignments of the other vector based accordingly. Hence, the edit distance is calculated between the reference vector and this newly changed one, that we call relative vector.

The task of adapting the relative vector w.r.t the reference one can be expressed as an assignment problem, also known as maximum weighted bipartite matching problem, as already done in the literature [36]. Let $\pi ^s$ and $\pi ^t$ be two membership vectors of length n, associated with the partitions $P^s$ and $P^t$, which contain $\ell ^s$ and $\ell ^t$ modules, respectively. Also, since the edit distance is symmetric, without loss of generality, let $\ell ^s \le \ell ^t$. Moreover, let CM be the $\ell ^s\ \times \ \ell ^t$ confusion matrix of $\pi ^s$ and $\pi ^t$. The term $CM_{ij}$, with $1\le i \le \ell ^s$ and $1\le j \le \ell ^t$, represents the number of vertices in the intersection of modules $M^s_{i}$ and $M^t_{j}$, i.e. $|M^s_{i} \cap M^t_{j}|$. Then, we look for a bijection $f: \{ 1, 2, \ldots , \ell ^s \} \rightarrow \{ 1, 2, \ldots , \ell ^t \}$ that maximizes the number of vertices common to pairs of modules from both membership vectors, i.e.

$$\begin{aligned} \max \sum _{i=1}^{\ell ^s} CM_{i,f(i)}. \end{aligned}$$

Since this problem can be modelled as an assignment or a maximum weighted bipartite matching problem, it can be solved in various ways. One of them is through the well-known Hungarian algorithm, whose complexity is $O(n^3)$ [34]. Nevertheless, the best polynomial time algorithm is currently based on the network simplex method, and it runs in $O(|V||E| + |V|^2 log(|V|))$ time using a Fibonacci heap data structure [25]. One final remark is about the case of $\ell ^s < \ell ^t$, in which there are $|\ell ^t - \ell ^s|$ unassigned module labels in $\pi ^t$. In this case, one can arbitrarily renumber these labels, starting from $\ell ^{s}+1$.

Finally, the edit distance between two membership vectors is calculated by simply counting the number of cases where the module labels of the vertices in the reference and relative vectors are different.

Proofs

1.1 All proofs related to the MVMO property for 3-edit operations on complete unweighted signed graphs

We complete the proof of Lemma 2 by verifying below the conditions of all atomic 3-edit operations for unweighted complete signed networks (illustrated in Fig. 12), where and $(u,v), (u,z), (v,z) \in \widetilde{E}$. Recall that .

Fig. 12

All atomic 3-edit operations

a) :

We have $(\gamma ^{left}_{u} = a_{uv} + a_{uz}) > (\gamma ^{right}_{u} = -a_{uv} -a_{uz})$, $(\gamma ^{left}_{v} = a_{uv} + a_{vz}) > (\gamma ^{right}_{v} = -a_{uv} -a_{vz})$ and $(\gamma ^{left}_{z} = a_{uz} + a_{vz}) > (\gamma ^{right}_{z} = -a_{uz} -a_{vz})$. We see that $a_{uv}$, $a_{uz}$ and $a_{vz}$ cannot be negative.

b) :

We have $(\gamma ^{left}_{u} = a_{uv} + a_{uz}) > (\gamma ^{right}_{u} = 0)$, $(\gamma ^{left}_{v} = a_{uv} + a_{vz}) > (\gamma ^{right}_{v} = -a_{vz})$ and $(\gamma ^{left}_{z} = a_{uz} + a_{vz}) > (\gamma ^{right}_{z} = -a_{vz})$. We see that $a_{uv}$, $a_{uz}$ and $a_{vz}$ cannot be negative.

c) :

We have $(\gamma ^{left}_{u} = a_{uv} + a_{uz}) > (\gamma ^{right}_{u} = 0)$, $(\gamma ^{left}_{v} = a_{uv} + a_{vz}) > (\gamma ^{right}_{v} = 0)$ and $(\gamma ^{left}_{z} = a_{uz} + a_{vz}) > (\gamma ^{right}_{z} = 0)$. We see that $a_{uv}$, $a_{uz}$ and $a_{vz}$ cannot be negative.

d) :

We have $(\gamma ^{left}_{u} = a_{uv}) > (\gamma ^{right}_{u} = -a_{uv} -a_{uz})$, $(\gamma ^{left}_{v} = a_{uv}) > (\gamma ^{right}_{v} = -a_{uv} -a_{vz})$ and $(\gamma ^{left}_{z} = 0) > (\gamma ^{right}_{z} = -a_{uz} -a_{vz})$. We see that $a_{uv}$, $a_{uz}$ and $a_{vz}$ must be positive.

e) :

We have $(\gamma ^{left}_{u} = a_{uv} - a_{uz}) > (\gamma ^{right}_{u} = -a_{uv} + a_{uz})$, $(\gamma ^{left}_{v} = a_{uv} - a_{vz}) > (\gamma ^{right}_{v} = -a_{uv} + a_{vz})$ and $(\gamma ^{left}_{z} = -a_{uz} - a_{vz}) > (\gamma ^{right}_{z} = a_{uz} + a_{vz})$. We see that $a_{uv}$ (resp. $a_{uz}$ and $a_{vz}$) cannot be negative (resp. positive).

f) :

We have $(\gamma ^{left}_{u} = a_{uv} - a_{uz}) > (\gamma ^{right}_{u} = -a_{uv})$, $(\gamma ^{left}_{v} = a_{uv} - a_{vz}) > (\gamma ^{right}_{v} = -a_{uv})$ and $(\gamma ^{left}_{z} = 0) > (\gamma ^{right}_{z} = a_{uz} + a_{vz})$. We see that $a_{uv}$, $a_{uz}$ and $a_{vz}$ cannot be negative.

g) :

We have $(\gamma ^{left}_{u} = a_{uv}) > (\gamma ^{right}_{u} = a_{uz} -a_{uv})$, $(\gamma ^{left}_{v} = a_{uv}) > (\gamma ^{right}_{v} = a_{vz} - a_{uv})$ and $(\gamma ^{left}_{z} = -a_{uz} - a_{vz}) > (\gamma ^{right}_{z} = 0)$. We see that $a_{uv}$, $a_{uz}$ and $a_{vz}$ cannot be negative.

h) :

We have $(\gamma ^{left}_{u} = a_{uv}) > (\gamma ^{right}_{u} = -a_{uv})$, $(\gamma ^{left}_{v} = a_{uv}) > (\gamma ^{right}_{v} = - a_{uv})$ and $(\gamma ^{left}_{z} = -a_{uz} - a_{vz}) > (\gamma ^{right}_{z} = 0)$. We see that $a_{uv}$, $a_{uz}$ and $a_{vz}$ cannot be negative.

i) :

We have $(\gamma ^{left}_{u} = a_{uv}) > (\gamma ^{right}_{u} = a_{uz})$, $(\gamma ^{left}_{v} = a_{uv} - a_{vz}) > (\gamma ^{right}_{v} = a_{vz})$ and $(\gamma ^{left}_{z} = -a_{uz} - a_{vz}) > (\gamma ^{right}_{z} = + a_{vz})$. We see that $a_{uv}$ (resp. $a_{uz}$ and $a_{vz}$) cannot be negative (resp. positive).

j) :

We have $(\gamma ^{left}_{u} = a_{uv}) > (\gamma ^{right}_{u} = - a_{uz})$, $(\gamma ^{left}_{v} = a_{uv}) > (\gamma ^{right}_{v} = -a_{vz})$ and $(\gamma ^{left}_{z} = 0) > (\gamma ^{right}_{z} = -a_{uz} + a_{vz})$. We see that $a_{uv}$ and $a_{vz}$ (resp. $a_{uz}$) must be positive (resp. negative).

k) :

We have $(\gamma ^{left}_{u} = a_{uv}) > (\gamma ^{right}_{u} = 0)$, $(\gamma ^{left}_{v} = a_{uv} - a_{vz}) > (\gamma ^{right}_{v} = 0)$ and $(\gamma ^{left}_{z} = 0) > (\gamma ^{right}_{z} = a_{vz})$. We see that $a_{uv}$ and $a_{vz}$ (resp. $a_{uz}$) must be positive (resp. negative).

l) :

We have $(\gamma ^{left}_{u} = -a_{uv}) > (\gamma ^{right}_{u} = a_{uz})$, $(\gamma ^{left}_{v} = -a_{vz}) > (\gamma ^{right}_{v} = a_{uv})$ and $(\gamma ^{left}_{z} = -a_{uz}) > (\gamma ^{right}_{z} = a_{vz})$. We see that $a_{uv}$ and $a_{vz}$ (resp. $a_{uz}$) must be positive (resp. negative).

m) :

We have $(\gamma ^{left}_{u} = -a_{uv}) > (\gamma ^{right}_{u} = 0)$, $(\gamma ^{left}_{v} = -a_{vz}) > (\gamma ^{right}_{v} = a_{uv})$ and $(\gamma ^{left}_{z} = 0) > (\gamma ^{right}_{z} = a_{vz})$. We see that $a_{uv}$ and $a_{vz}$ (resp. $a_{uz}$) must be positive (resp. negative).

n) :

We have $(\gamma ^{left}_{u} = -a_{uv}) > (\gamma ^{right}_{u} = a_{uv} - a_{uz})$, $(\gamma ^{left}_{v} = -a_{uv}) > (\gamma ^{right}_{v} = a_{uv} + a_{vz})$ and $(\gamma ^{left}_{z} = -a_{vz}) > (\gamma ^{right}_{z} = -a_{uz})$. We see that $a_{uv}$ (resp. $a_{uz}$ and $a_{vz}$) cannot be negative (resp. positive).

o) :

We have $(\gamma ^{left}_{u} = -a_{uv}) > (\gamma ^{right}_{u} = 0)$, $(\gamma ^{left}_{v} = 0) > (\gamma ^{right}_{v} = a_{uv} - a_{vz})$ and $(\gamma ^{left}_{z} = 0) > (\gamma ^{right}_{z} = -a_{vz})$. We see that $a_{uv}$ (resp. $a_{uz}$ and $a_{vz}$) cannot be negative (resp. positive).

p) :

We have $(\gamma ^{left}_{u} = -a_{uv}) > (\gamma ^{right}_{u} = -a_{uz})$, $(\gamma ^{left}_{v} = 0) > (\gamma ^{right}_{v} = a_{uv} + a_{vz})$ and $(\gamma ^{left}_{z} = -a_{vz}) > (\gamma ^{right}_{z} = -a_{uz})$. We see that $a_{uv}$ (resp. $a_{uz}$ and $a_{vz}$) cannot be negative (resp. positive).

r) :

We have $(\gamma ^{left}_{u} = 0) > (\gamma ^{right}_{u} = -a_{uv} - a_{uz})$, $(\gamma ^{left}_{v} = 0) > (\gamma ^{right}_{v} = -a_{uv} - a_{vz})$ and $(\gamma ^{left}_{z} = 0) > (\gamma ^{right}_{z} = -a_{uz} - a_{vz})$. We see that $a_{uv}$, $a_{uz}$ and $a_{vz}$ must be positive.