Minimum cost edge blocker clique problem

Abstract

Given a graph with weights on its vertices and blocking costs on its edges, and a user-defined threshold $\tau >0$, the minimum cost edge blocker clique problem (EBCP) is introduced as the problem of blocking a minimum cost subset of edges so that each clique’s weight is bounded above by $\tau$. Clusters composed of important actors with quick communications can be effectively modeled as large-weight cliques in real-world settings such as social, communication, and biological systems. Here, we prove that EBCP is NP-hard even when $\tau$ is a fixed parameter, and propose a combinatorial lower bound for its optimal objective. A class of inequalities that are valid for the set of feasible solution to EBCP is identified, and sufficient conditions for these inequalities to induce facets are presented. Using this class of inequalities, EBCP is formulated as a linear 0–1 program including potentially exponential number of constraints. We develop the first problem-specific branch-and-cut algorithm to solve EBCP, which utilizes the aforementioned constraints in a lazy manner. We also developed the first combinatorial branch-and-bound solution approach for this problem, which aims to handle large graph instances. Finally, computational results of solving EBCP on a collection of random graphs and power-law real-world networks by using our proposed exact algorithms are also provided.

Appendix: Proofs of Lemmas 1–3

Lemma 1

Given a nonnegative integer a, $\gamma _a$ is the smallest nonnegative integer such that $a\le \frac{1}{2}\gamma _a(\gamma _a+1)$.

Proof

For a nonnegative integer a and $b\ge 0$, $a\le \frac{1}{2}b(b+1)$ if and only if $b\ge \frac{1}{2}(-1+\sqrt{1+8a})$. Therefore, $\lceil \frac{1}{2}(-1+\sqrt{1+8a})\rceil$ is the smallest nonnegative integer such that $a\le \frac{1}{2}\lceil \frac{1}{2}(-1+\sqrt{1+8a})\rceil (\lceil \frac{1}{2}(-1+\sqrt{1+8a})\rceil +1)$. $\square$

Lemma 2

Given a nonempty set $D\subseteq V$ and a positive integer a,

• 1. $\gamma _{a}\le a$.

• 2. $0\le \gamma _{a}-\gamma _{a-1}\le 1$.

• 3. if $a\le \frac{1}{2}|D|(|D|-1)$, then $\varPhi (D,a)-\varPhi (D,a-1)=w_{\mathbf{d }[\gamma _{a}+1]}$.

• 4. if $1\le a\le |D|-1$, then $\varPhi (D,a)\ge \sum _{k=1}^aw_{\mathbf{d }[k+1]}$.

• 5. if $\emptyset \ne S\subseteq {\mathcal {E}}(D)$, then $\sum _{(i,j)\in S}\min \{w_i,w_j\}\le \varPhi (D,|S|)$.

Proof

• 1. Notice that $\gamma _{a}$ is well-defined, because $a>0$. We know $a(a-1)\ge 0$. So, $1+8a\le (2a+1)^2$. Therefore,

  $$\begin{aligned} \frac{1}{2}(-1+\sqrt{1+8a})\le a. \end{aligned}$$

  Since a is an integer, then

  $$\begin{aligned} \lceil \frac{1}{2}(-1+\sqrt{1+8a})\rceil =\gamma _{a}\le a. \end{aligned}$$

• 2. Note that since $0\le a-1<a$, then $\gamma _{a}$ and $\gamma _{a-1}$ are well-defined. By Lemma 1, $a\le \frac{1}{2}\gamma _{a}(\gamma _{a}+1)$. So,

  $$\begin{aligned} a-1<\frac{1}{2}\gamma _{a}(\gamma _{a}+1), \end{aligned}$$

  and again by Lemma 1, $\gamma _{a-1}\le \gamma _{a}$. On the other hand, again by Lemma 1,

  $$\begin{aligned} a-1\le \frac{1}{2}\gamma _{a-1}(\gamma _{a-1}+1). \end{aligned}$$

  Since $\gamma _{a-1}\ge 0$, then

  $$\begin{aligned} a\le \frac{1}{2}\gamma _{a-1}(\gamma _{a-1}+1)+1\le \frac{1}{2}\gamma _{a-1}(\gamma _{a-1}+1)+1+\gamma _{a-1}. \end{aligned}$$

  Therefore,

  $$\begin{aligned} a\le \frac{1}{2}(\gamma _{a-1}+1)((\gamma _{a-1}+1)+1), \end{aligned}$$

  and by Lemma 1, $\gamma _{a}\le \gamma _{a-1}+1$.

• 3. Note that since $1\le a\le \frac{1}{2}|D|(|D|-1)$, then $\varPhi (D,a)$ and $\varPhi (D,a-1)$ are well-defined. By Lemma 2.2, either $\gamma _{a}=\gamma _{a-1}$, or $\gamma _{a}=\gamma _{a-1}+1$. If $\gamma _{a}=\gamma _{a-1}$, then by definition, $\varPhi (D,a)=\varPhi (D,a-1)+w_{\mathbf{d }[\gamma _{a}+1]}.$ Alternatively, suppose $\gamma _{a}=\gamma _{a-1}+1$. Then by Lemma 1,

  $$\begin{aligned} a-1\le \frac{1}{2}\gamma _{a-1}(\gamma _{a-1}+1)<a. \end{aligned}$$

  Because $\frac{1}{2}\gamma _{a-1}(\gamma _{a-1}+1)$ is an integer, then

  $$\begin{aligned} a-1=\frac{1}{2}\gamma _{a-1}(\gamma _{a-1}+1). \end{aligned}$$

  (10)

  Additionally, since $\gamma _{a}=\gamma _{a-1}+1$, then

  $$\begin{aligned} \begin{aligned} \varPhi (D,a)=&\left( \sum _{k=0}^{\gamma _{a-1}}kw_{\mathbf{d }[k+1]}\right) +\gamma _{a}w_{\mathbf{d }[\gamma _{a}+1]}\\&-(\frac{1}{2}\gamma _{a-1}(\gamma _{a-1}+1)-(a-1))w_{\mathbf{d }[\gamma _{a}+1]}\\&-(\gamma _{a}-1)w_{\mathbf{d }[\gamma _{a}+1]}. \end{aligned} \end{aligned}$$

  (11)

  Using Eqs. (10) and (11), we have $\varPhi (D,a)=\varPhi (D,a-1)+w_{\mathbf{d }[\gamma _{a}+1]}.$

• 4. Notice that since $0<a\le |D|-1\le \frac{1}{2}|D|(|D|-1)$, then both sides of the inequality are well-defined. We use induction on a to prove this result. If $a=1$, then it can be easily verified that the inequality holds as equality. Now, assume that the inequality holds for any integer l such that $1\le l<|D|-1$. We need to prove that the inequality is also valid for integer $l+1$. By Lemma 2.3, we have

  $$\begin{aligned} \varPhi (D,l+1)=\varPhi (D,l)+w_{\mathbf{d }[\gamma _{l+1}+1]}. \end{aligned}$$

  So, it is enough to show that

  $$\begin{aligned} \varPhi (D,l)+w_{\mathbf{d }[\gamma _{l+1}+1]}\ge w_{\mathbf{d }[l+2]}+\sum _{k=1}^lw_{\mathbf{d }[k+1]}. \end{aligned}$$

  By Lemma 2.1, we have $\gamma _{l+1}\le l+1$. Therefore, $w_{\mathbf{d }[\gamma _{l+1}+1]}\ge w_{\mathbf{d }[l+2]}$. On the other hand, by induction assumption, we have $\varPhi (D,l)\ge \sum _{k=1}^lw_{\mathbf{d }[k+1]}.$ So, the inequality is valid for integer $l+1$.

• 5. Note that since $1\le |S|\le \frac{1}{2}|D|(|D|-1)$, then $\varPhi (D,|S|)$ is well-defined. To show this, we use induction on |S|. Let $|S|=1$ (e.g., $S=\{(i_0,j_0)\}$). Suppose $\min \{w_{i_0},w_{j_0}\}>\varPhi (D,1)=w_{\mathbf{d }[2]}$. This means $w_{i_0}>w_{\mathbf{d }[2]}$ and $w_{j_0}>w_{\mathbf{d }[2]}$, which is a contradiction as both $i_0$ and $j_0$ belong to D. Now, we assume if $|S|=k$, where $1\le k\le \frac{1}{2}|D|(|D|-1)-1$, then

  $$\begin{aligned} \sum _{(i,j)\in S}\min \{w_i,w_j\}\le \varPhi (D,|S|), \end{aligned}$$

  and show the correctness of our claim for the case of $|S|=k+1$.

  Assume $|S|=k+1$ and

  $$\begin{aligned} \sum _{(i,j)\in S}\min \{w_i,w_j\}> \varPhi (D,|S|). \end{aligned}$$

  (12)

  Consider an arbitrary edge $(i_1,j_1)\in S$. Using Inequality (12) and Lemma 2.3,

  $$\begin{aligned} \min \{w_{i_1},w_{j_1}\}+\sum _{(i,j)\in S\setminus \{(i_1,j_1)\}}\min \{w_i,w_j\}> \varPhi (D,|S|-1)+w_{\mathbf{d }[\gamma _{|S|}+1]}. \end{aligned}$$

  On the other hand, by induction assumption,

  $$\begin{aligned} \sum _{(i,j)\in S\setminus \{(i_1,j_1)\}}\min \{w_i,w_j\}\le \varPhi (D,|S|-1). \end{aligned}$$

  So, $\min \{w_{i_1},w_{j_1}\}>w_{\mathbf{d }[\gamma _{|S|}+1]}$. Therefore, for any $(i,j)\in S$, we have $w_{i}>w_{\mathbf{d }[\gamma _{|S|}+1]}$ and $w_{j}>w_{\mathbf{d }[\gamma _{|S|}+1]}$. Consequently, we have $w_{i}>w_{\mathbf{d }[\gamma _{|S|}+1]}$ for any $i\in D^\prime$, where $D^\prime =\cup _{(i,j)\in S}\{i,j\}.$ Since $D^\prime \subseteq D$ and $w_{i}>w_{\mathbf{d }[\gamma _{|S|}+1]}$ for any $i\in D^\prime$, then $|D^\prime |\le \gamma _{|S|}$.

  Notice that since $|S|\le \frac{1}{2}|D^\prime |(|D^\prime |-1)$, then

  $$\begin{aligned} |D^\prime |\ge \lceil \frac{1}{2}(1+\sqrt{1+8|S|})\rceil = \lceil \frac{1}{2}(-1+\sqrt{1+8|S|})+1\rceil =\gamma _{|S|}+1, \end{aligned}$$

  which is a contradiction.

$\square$

Lemma 3

Given a clique D in graph G and a set $S\subseteq E$, graph $G\langle E{\setminus }S\rangle$ contains a clique $D^\prime \subseteq D$ such that

$$\begin{aligned} {\mathcal {W}}(D^\prime )\ge {\mathcal {W}}(D)-\sum _{(i,j)\in S\cap {\mathcal {E}}(D)}\min \{w_i,w_j\}. \end{aligned}$$

Proof

Let $S^\prime =S\cap {\mathcal {E}}(D)$. The correctness of this lemma is proved by induction on $|S^\prime |$ as follows. If $|S^\prime |=0$, then $D^\prime =D$ is a clique in graph $G\langle E{\setminus }S\rangle$ for which the above inequality is valid. Now, we assume if $|S^\prime |=k\ge 0$, then graph $G\langle E{\setminus }S\rangle$ contains a clique ${\tilde{D}}\subseteq D$ such that

$$\begin{aligned} {\mathcal {W}}({\tilde{D}})\ge {\mathcal {W}}(D)-\sum _{(i,j)\in S^\prime }\min \{w_i,w_j\}, \end{aligned}$$

and show the correctness of this lemma for the case of $|S^\prime |=k+1$. Assume $|S^\prime |=k+1$. Consider an arbitrary edge $(i_1,j_1)\in S^\prime$ (note that $(i_1,j_1)$ exists), and let ${\tilde{D}}\subseteq D$ denote the clique in $G\langle E{\setminus }(S\setminus \{(i_1,j_1)\})\rangle$ such that

$$\begin{aligned} {\mathcal {W}}({\tilde{D}})\ge {\mathcal {W}}(D)-\sum _{(i,j)\in S^\prime \setminus \{(i_1,j_1)\}}\min \{w_i,w_j\}. \end{aligned}$$

(13)

If $i_1\not \in {\tilde{D}}$ or $j_1\not \in {\tilde{D}}$, then $D^\prime ={\tilde{D}}$ is a clique in $G\langle E{\setminus }S\rangle$ and by Inequality (13),

$$\begin{aligned} {\mathcal {W}}(D^\prime )> {\mathcal {W}}(D)-\sum _{(i,j)\in S^\prime }\min \{w_i,w_j\}. \end{aligned}$$

Alternatively, if $i_1\in {\tilde{D}}$ and $j_1\in {\tilde{D}}$, then ${\tilde{D}}\setminus \{i_1\}$ and ${\tilde{D}}\setminus \{j_1\}$ are both cliques in $G\langle E{\setminus }S\rangle$, which means $G\langle E{\setminus }S\rangle$ contains a clique $D^\prime \subset {\tilde{D}}$ such that

$$\begin{aligned} {\mathcal {W}}(D^\prime )={\mathcal {W}}({\tilde{D}})-\min \{w_{i_1},w_{j_1}\}. \end{aligned}$$

Again by Inequality (13), we have

$$\begin{aligned} {\mathcal {W}}(D^\prime )\ge {\mathcal {W}}(D)-\sum _{(i,j)\in S^\prime }\min \{w_i,w_j\}. \end{aligned}$$

This completes the proof of this lemma. $\square$