Triangle minimization in large networks

Abstract

The number of triangles is a fundamental metric for analyzing the structure and function of a network. In this paper, for the first time, we investigate the triangle minimization problem in a network under edge (node) attack, where the attacker aims to minimize the number of triangles in the network by removing $k$ edges (nodes). We show that the triangle minimization problem under edge (node) attack is a submodular function maximization problem, which can be solved efficiently. Specifically, we propose a degree-based edge (node) removal algorithm and a near-optimal greedy edge (node) removal algorithm for approximately solving the triangle minimization problem under edge (node) attack. In addition, we introduce two pruning strategies and an approximate marginal gain evaluation technique to further speed up the greedy edge (node) removal algorithm. We conduct extensive experiments over 12 real-world datasets to evaluate the proposed algorithms, and the results demonstrate the effectiveness, efficiency and scalability of our algorithms.

Appendix

Proof of Theorem 2.1

First, it is known that the set cover with frequency constraint (SCFC) problem is NP-hard [19, 48]. Given a ground set $\mathcal {U}$, a collection of $n$ subsets $\mathcal {S}=\{S_1, S_2, \cdots , S_n\}$ where $\bigcup _i S_i=\mathcal {U}$, and a frequency parameter $t$ ($t < n$), the SCFC problem is to find the minimum number of subsets in $\mathcal {S}$ that covers all elements in $\mathcal {U}$. Here, the frequency parameter $t$ denotes that every element in $\mathcal {U}$ is included in $t$ subsets in $\mathcal {S}$. Let us consider a special case of the SCFC problem, which has an additional constraint that the intersection of any three subsets in $\mathcal {S}$ has at most one element (i.e., for any $i, j, k$ and $i \ne j \ne k$, $|S_i \cap S_j \cap S_k| \le 1$). For convenience, we refer to this problem as the intersection-bounded SCFC (IBSCFC) problem. Below, we show that the IBSCFC problem is also NP-hard. Suppose to the contrary that there is a polynomial algorithm $\mathcal {A}$ to solve the IBSCFC problem. For any $|S_i \cap S_j \cap S_k| > 1$ in the SCFC problem, we can discard the “redundant-common elements” in the subsets $S_i, S_j, S_k$ so that $| \tilde{S}_i \cap \tilde{S}_j \cap \tilde{S}_k|=1$ where $\tilde{S}_i$ denotes the subset $S_i$ after discarding the redundant-common elements (i.e., for $S_i, S_j, S_k$, only one common element is left). Then, the SCFC problem becomes the IBSCFC problem, and we invoke algorithm $\mathcal {A}$ to solve it. It is important to note that the optimal solution (the selected subsets ID) obtained by algorithm $\mathcal {A}$ is the optimal solution for the SCFC problem. The reason is as follows. For any $S_i, S_j, S_k$ with $|S_i \cap S_j \cap S_k| > 1$, the redundant-common elements are only in these three subsets (by our constraint, each element is included in three subsets), thus they do not affect the optimal solution. Moreover, the optimal solution obtained by algorithm $\mathcal {A}$ must contain at least one subset from $S_i, S_j, S_k$, because these three subsets have one common element left which must be covered by a subset in the optimal solution. By the above process, there is a polynomial algorithm for the SCFC problem, which is a contradiction.

Second, we consider the maximum coverage version of the IBSCFC problem, called IBMCFC, where the goal is to find $k$ subsets in $\mathcal {S}$ to maximize the cardinality of their union. It is easy to show that this problem is also NP-hard. Because if not, there is a polynomial algorithm $\mathcal {B}$ to solve the IBMCFC problem. Since $\bigcup _i S_i=\mathcal {U}$, we can invoke $\mathcal {B}$ at most $n$ times to get an optimal solution of the IBSCFC problem (enumerating $k$ from $1$ to $n$). That is to say, there is a polynomial algorithm for the IBSCFC problem, which is a contradiction.

Third, to prove the theorem, we show a reduction from the IBMCFC problem. Specifically, for each subset $S_i$, we create an edge $e_i$ with $2|S_i|$ stubs, which are used to combine the end nodes of different edges. Each end node of an edge is associated with $|S_i|$ stubs, and these stubs are labeled by the element ID in $S_i$. Then, for any three subsets $S_i$, $S_j$, and $S_k$ ($i \ne j \ne k$) with $|S_i \cap S_j \cap S_k| =1$, we combine the end nodes of their corresponding edges with the same stub labels so that they can form a triangle. As an example, let $\mathcal {U}=\{u_1, u_2\}$, $S_1=\{u_1\}$, $S_2=\{u_1\}$, $S_3=\{u_1, u_2\}$, $S_4=\{u_2\}$, $S_5=\{u_2\}$. Clearly, each element in $\mathcal {U}$ is in three subsets and any three subsets have at most one common element. Then, for each subset, we create an edge with stubs as shown in the left part of Fig. 5. Then, we can construct a graph as shown in the right part of Fig. 5. By this construction, each triangle is represented by an element in $\mathcal {U}$, and each edge $e_i$ in the resulting graph is represented by a subset $S_i$ in $\mathcal {S}$. As a result, the optimal solution of the triangle minimization problem (by edge removal) in the resulting graph is the optimal solution of the IBMCFC problem. Since IBMCFC is NP-hard, the triangle minimization problem by edge removal is also NP-hard. This completes the proof. $\square$

Fig. 5

Illustration of the graph construction

Proof of Theorem 2.2

Similar to the proof of Theorem 2.1, we can show a reduction from the IBMCFC problem. Following the notations used in the proof of Theorem 2.1, we create a graph $G$ for the instance of triangle minimization problem by node removal as follows. Specifically, for each $S_i$ in $\mathcal {S}$, we create a node $v_i$. For each pair $S_i$ and $S_j$ ($i \ne j$), we create an edge $(v_i, v_j)$ if and only if $S_i \bigcap S_j \ne \emptyset$. By this construction, each node is represented by a subset, and each triangle is represented by an element. One can easily check that the optimal solution of the triangle minimization problem by node removal is the optimal solution of the IBMCFC problem. Thus, the theorem is established.