On independent cliques and linear complementarity problems

Abstract

In recent work (Pandit and Kulkarni [Discrete Applied Mathematics, 244 (2018), pp. 155–169]), the independence number of a graph was characterized as the maximum of the $\ell _1$ norm of solutions of a Linear Complementarity Problem (LCP) defined suitably using parameters of the graph. Solutions of this LCP have another relation, namely, that they corresponded to Nash equilibria of a public goods game. Thus the maximum $\ell _1$ norm has an important economic interpretation as the maximum total investment over all equilibria of this game. Motivated by this, we consider a perturbation of this LCP that corresponds to a public goods game with imperfect substitutability. We identify the combinatorial structures on the graph that correspond to the maximum $\ell _1$ norm of solutions (equivalently, investment maximizing equilibria) of the new LCP. We show that these solutions correspond to “independent cliques" – collections of cliques such that no two vertices from two distinct cliques are adjacent. When the perturbation becomes null, these cliques collapse to singletons and we recover the earlier relation to maximum independent sets. Independent cliques have been studied before as a generalization of independent sets. Our work establishes an intricate connection between independent cliques, LCPs and public goods games.

Appendices

Proofs from Section 3

1.1 Proof of Lemma 1

• (a) From (3), ${{\mathcal {C}}}_i(0) = 0$ which implies (5) is violated. Thus, $0 \not \in$ SOL$_\delta (G$).

• (b) From (3), ${{\mathcal {C}}}_i(x) = x_i + \delta \sum _{j \in N_G(i)}x_j$. Since $\delta \ge 0$ and $x_i \ge 0 \ \forall i \in V(G)$ from (4), ${{\mathcal {C}}}_i(x) \ge x_i$. Thus, ${{\mathcal {C}}}(x) \ge x \ \forall x \in SOL _\delta (G$).

• (c) We know that $x \ge 0$. Let $x_i > 0$ for some i. Then, from (6), ${{\mathcal {C}}}_i(x) = 1$. From part (b) of Lemma 1, $x_i \le {{\mathcal {C}}}_i(x) = 1$. Thus, $x_i \in [0,1] \ \forall i \in V(G)$.

• (d) First, consider $x^1 \in SOL _\delta (G_1)$ and $x^2 \in SOL _\delta (G_2)$, we prove that $x = (x^1,x^2) \in SOL _\delta (G)$. Since $x^1,x^2 \ge 0$ we have $x \ge 0$. Next, note that since $G_1$ and $G_2$ are disjoint, $N_{G_1}(i) = N_G(i) \ \forall i \in V(G_1)$ and $N_{G_2}(i) = N_G(i) \ \forall i \in V(G_2)$. Thus, for $i \in V(G_1)$, ${{\mathcal {C}}}_i(x) = x_i + \delta \sum _{j \in N_G(i)}x_j = x^1_i + \delta \sum _{j \in N_{G_1}(i)}x^1_j = {{\mathcal {C}}}_i(x^1)$. Similarly, for $i \in V(G_2)$, ${{\mathcal {C}}}_i(x) = {{\mathcal {C}}}_i(x^2)$. Thus we get ${{\mathcal {C}}}_i(x) \ge 1 \ \forall i \in V(G)$. Also, $x^1_i({{\mathcal {C}}}_i(x^1) - 1) = 0 \ \forall i \in V(G_1)$ and $x^2_i({{\mathcal {C}}}_i(x^2) - 1) = 0 \ \forall i \in V(G_2)$ imply $x_i({{\mathcal {C}}}_i(x) - 1) = 0 \ \forall i \in V(G)$. Thus, $x \in SOL _\delta (G)$. Next, given $x \in SOL _\delta (G)$, we show that $x_{G_1} \in SOL _\delta (G_1)$ and $x_{G_2} \in SOL _\delta (G_2)$, where $x_{G_1} \in {\mathbb {R}}^{|V(G_1)|}$, $x_{G_2} \in {\mathbb {R}}^{|V(G_2)|}$ and $x=(x_{G_1},x_{G_2})$. Clearly $x_{G_1} \ge 0$. Using $N_{G_1}(i) = N_G(i) \ \forall i \in V(G_1)$, we get ${{\mathcal {C}}}_i(x) = {{\mathcal {C}}}_i(x_{G_1})$ and thus $x_{G_1}$ satisfies both (5) and (6). Hence, $x_{G_1} \in SOL _\delta (G_1)$. Similarly, $x_{G_2} \in SOL _\delta (G_2)$.

• (e) From Lemma 1 (c), $0 \le x_i \le 1 \ \forall i \in V(G)$. Thus $\mathbf{1}_{\sigma (x)} \ge x$. Thus, for all $i \in V(G)$, ${{\mathcal {C}}}_i(\mathbf{1}_{\sigma (x)}) \ge {{\mathcal {C}}}_i(x) \ge 1$. Let $i \not \in \sigma (x)$. For this i we have, ${{\mathcal {C}}}_i(\mathbf{1}_{\sigma (x)}) = 0 + \delta \sum _{j \in N_G(i)}x_j = \delta |N_G(i) \cap \sigma (x)|$. Thus, $|N_G(i) \cap \sigma (x)| \ge \frac{1}{\delta }$. Since, $|N_G(i) \cap \sigma (x)|$ is a positive integer, $|N_G(i) \cap \sigma (x)| \ge \lceil \frac{1}{\delta } \rceil$. Thus, $\sigma (x)$ is a $\lceil \frac{1}{\delta } \rceil$-dominating set of G.

• (f) For $i \in V(G_{\sigma (x)})$, ${\tilde{x}}_i = x_i > 0$. Thus, $\sigma {({\tilde{x}})} = V(G_{\sigma (x)})$. Note that for a vertex i in $G_{\sigma (x)}$, the discounted sum of the closed neighborhood ${\tilde{{{\mathcal {C}}}}}({\tilde{x}}) = {\tilde{x}}_i + \delta \sum _{j \in N_{G_{\sigma (x)}}(i)}{\tilde{x}}_j = x_i + \delta \sum _{j \in N_G(i)}x_j = {{\mathcal {C}}}_i(x)$ since $x_i = 0$ for $i \not \in \sigma (x)$. Hence, ${\tilde{x}} = x_{\sigma (x)} \in SOL _\delta (G_{\sigma (x)})$.

1.2 Proof of Lemma 2

Let x be an integer solution of LCP$_\delta (G)$. Then, by Lemma 1 (c), x is a binary vector, and hence $x = \mathbf{1}_S$ for a subset S of vertices in G. If S is not an independent set, then there exist $i,j \in S$ such that $j \in N_G(i)$, i.e. $x_i = x_j = 1$, whereby ${{\mathcal {C}}}_i(x) = x_i + \delta \sum _{j \in N_G(i)}x_j \ge 1+\delta > 1$. Thus, ${{\mathcal {C}}}_i(x) - 1 > 0$ and $x_i > 0$ imply (6) is violated. This gives a contradiction. Hence, S must be an independent set. From Lemma 1 (e) we have that S must also be a $\lceil \frac{1}{\delta } \rceil$-dominating set, as required.

For the converse, let S be a $\lceil \frac{1}{\delta } \rceil$-dominating independent set of G. Then, for $i \in S$, ${{\mathcal {C}}}_i(x) = 1 + \delta \sum _{j \in N_G(i)}x_j = 1$ since $x_j = 0 \ \forall j \in N_G(i)$. Also, for $i \not \in S$, ${{\mathcal {C}}}_i(x) = 0 + \delta \sum _{j \in N_G(i)}x_j \ge \delta \lceil \frac{1}{\delta } \rceil \ge \delta \times \frac{1}{\delta } = 1$. Thus, we have ${{\mathcal {C}}}_i(x) \ge 1$ and $x_i({{\mathcal {C}}}_i(x) - 1) = 0$ for all $i \in V(G)$. Hence, if x is a characteristic vector of a $\lceil \frac{1}{\delta } \rceil$-dominating indepdendent set of G, then it is an integer solution of LCP$_\delta (G)$.

1.3 Proof of Lemma 3

Let $U = \sigma (x) \backslash S$. From (6), we know that

$$\begin{aligned} {{\mathcal {C}}}_i(x) = x_i + \delta \sum _{j \in V(G)}a_{ij}x_j = 1 \ \forall i \in \sigma (x). \end{aligned}$$

(13)

Summing over $i \in \sigma (x)$, we get

$$\begin{aligned} \sum _{i \in V(G)}x_i + \delta \sum _{j \in \sigma (x)}\sum _{j \in V(G)}a_{ij}x_j = |\sigma (x)| = |S| + |U|. \end{aligned}$$

On rearranging,

$$\begin{aligned} |S| - \mathbf{e}^\top x&= \delta \sum _{i \in \sigma (x)}\sum _{j \in V(G)}a_{ij}x_j - |U| \nonumber \\&= \delta \sum _{i \in S}\sum _{j \in U}a_{ij}x_j + \delta \sum _{i \in U}\sum _{j \in V(G)}a_{ij}x_j - |U| \end{aligned}$$

(14)

$$\begin{aligned}&= \delta \sum _{j \in U}\sum _{i \in S}a_{ij}x_j + \sum _{i \in U}(\delta \sum _{j \in V(G)}a_{ij}x_j - 1) \end{aligned}$$

(15)

$$\begin{aligned}&= \delta \sum _{j \in U}N_S(j)x_j - \sum _{i \in U}x_i \nonumber \\&\ge \sum _{j \in U}x_j - \sum _{i \in U}x_i = 0 \end{aligned}$$

(16)

The equality in (14) follows from the fact that $a_{ij} = 0 \ \forall i,j \in S$ and $x_j = 0 \ \forall j \not \in \sigma (x)$, (15) follows from (13). The inequality (16) follows since S is a $\lceil \frac{1}{\delta } \rceil$-dominating independent set of G.

1.4 Proof of Lemma 4

We find the Karush Kuhn Tucker (KKT) conditions of the following maximization problem:

$$\begin{aligned} \max \ \phi (x;\delta ,G) \ s.t. \ x_i \ge 0. \end{aligned}$$

We get x is a stationary point iff $\exists \ \mu \ge 0$ such that $\mathbf{e}- (I + \delta A)x = -\mu$ and $x^\top \mu = 0$. These conditions are equivalent to the LCP$_\delta (G)$ conditions in equations (4),(5) and (6).

Proofs from Section 4

1.1 Proof of Lemma 5

• (a) It suffices to show that $S' \subset V'$ with $S'$ and $V'$ as defined in Line 8 and Line 9. By definition, we have $S' \subset V$. To show that $S' \subset V'$, we need to show that in the vertices we removed from V to form $V'$, there is no member of $S'$. Suppose we have executed Line 11 of the $k^{\mathrm{th}}$ iteration of the algorithm for some $k \le |S|$. The set of vertices removed from V at this stage to form $V'$ is $\bigcup _{i = 1}^{k} ({{\bar{C}}}_{i} \cup Q_{i})$ where $Q_{i} = N_{{{\bar{G}}}}({{\bar{C}}}_{i})$ , and we need to show that $\Big (\bigcup _{i = 1}^{k} ({{\bar{C}}}_{i} \cup Q_{i})\Big ) \cap S' = \emptyset$. Now, for each i, $({{\bar{C}}}_{i} \cup Q_{i})$ contains i which is a member of S but not $S'$. The set $({{\bar{C}}}_{i} \cup Q_{i})$ also has neighbors of i which can not be in $S'$ since $S' \subset S$ and S is an independent set. Lastly, $({{\bar{C}}}_{i} \cup Q_{i})$ has neighbors of $C_{i}$ (defined on Line 4 of Algorithm 1) that are also not in $S'$ since $C_{i}$ is such that $\forall c \in C_{i}, \ N_G(c) \cap S = \{i\}$ and hence $N_G(c) \cap S' = \emptyset$. Since this is true for all $i \in \{1,2,\cdots , k\}$, $\bigcup _{i = 1}^{k} ({{\bar{C}}}_{i} \cup Q_{i}) \cap S' = \emptyset$, and hence $S' \subset V'$.

• (b) We show this by induction on the iteration number. At the start of the first iteration of the for loop (Line 3 to Line 12 in Algorithm 1), ${\bar{S}}$ is a maximum independent set of ${{\bar{G}}}$. Assume that ${\bar{S}}$ is a maximum independent set of ${{\bar{G}}}$ after $k-1$ iterations (i.e., at the start of iteration k). To show that the claim is true at the start of the $(k+1)^{\mathrm{st}}$ iteration, it suffices to show that $S'$ is a maximum independent set of ${{\bar{G}}}_{V'}$ at the end of $k^{\mathrm{th}}$ iteration. Note that $|S'| = |{\bar{S}}| - 1$, since only one element (k) is removed from ${\bar{S}}$ to form $S'$. Suppose $S'$ is not a maximum independent set in ${{\bar{G}}}_{V'}$. Then there exists an independent set $S''$ of ${{\bar{G}}}_{V'}$ such that $|S''| > |S'| = |{\bar{S}}| - 1$ and hence $|S''| \ge |{\bar{S}}|$. Now note that $S' \subset V'$ and since $V'={{\bar{V}}}\backslash ({\bar{N}}_{{{\bar{G}}}}({{\bar{C}}}_{k}))$ there is no edge between any vertex in $V'$ and $\{k\}$. But then ${{\widehat{S}}}= S'' \cup \{k\}$ is an independent set of ${{\bar{G}}}$ with $|{{\widehat{S}}}| = |S''| + 1 > |{\bar{S}}|$, which is a contradiction. Thus, $S'$ is a maximum independent set in ${{\bar{G}}}_{V'}$.

1.2 Proof of Lemma 6

First we prove that ${{\bar{C}}}_{i}$ forms a clique for all $i \in \{1,2,\ldots , |S|\}$. At the $i^{\mathrm{th}}$ iteration of Algorithm 1 let $C_{i}$ be as defined in Line 4 and let ${\bar{S}}$ be a maximum independent set of ${{\bar{G}}}$ with ${{\bar{V}}}= V({{\bar{G}}})$. To show that ${{\bar{C}}}_i$ forms a clique, it suffices to show that $C_{i}$ forms a clique since all vertices of $C_{i}$ are neighbors of i. We prove this by contradiction. Suppose $C_{i}$ is not a clique. Then choose an independent set $S_{i}$ of $C_{i}$ such that it has at least two vertices. Consider ${{\widehat{S}}}= (S \backslash \{i\}) \cup S_{i}$. We claim that ${{\widehat{S}}}$ is an independent set. To prove this, it suffices to show that $\forall p \in S \backslash \{i\}, \forall q \in S_{i}, \ (p,q)$ is not an edge in G. Now, $S_{i} \subset L$ whereby each vertex in $S_{i}$ has only one neighbor in S. Moreover, $S_{i} \subset N_{G}(i)$, whereby for all vertices in $S_{i}$, their only neighbor in S is i. Consequently, ${{\widehat{S}}}$ is an independent set. Moreover, $|{{\widehat{S}}}| > |S|$, which is a contradiction since S must be a maximum independent set of G by definition. Hence, $C_{i}$ and therefore ${{\bar{C}}}_{i}$ must be a clique. Clearly, this holds for $\forall i \in \{1,2,\cdots , |S|\}$.

Now, consider two cliques ${{\bar{C}}}_{i}$ and ${{\bar{C}}}_{j}$ for $i \ne j$. Without loss of generality, we assume $i < j$. Then, ${{\bar{C}}}_{j} \subset V \backslash {\bar{N}}_G({{\bar{C}}}_{i})$. Thus, ${{\bar{C}}}_{j}$ can have no neighbors of ${{\bar{C}}}_{i}$ as its elements. Hence, ${{\bar{C}}}_{i}$ and ${{\bar{C}}}_{j}$ are independent.

1.3 Proof of Two Clique Lemma (Lemma 7)

We will show that for the x given the statement of Lemma 7, $C_i(x) \ge 1$ for the interval $\delta \in [\gamma (G),1)$. It will then follow that for the given x with $x_i = 0$, the LCP conditions (4),(5) and (6) are satisfied. Since i is connected to all vertices of $C_1$ and at least one vertex of $C_2$,

$$\begin{aligned} C_i(x)&\ge \delta \Bigg (\frac{n}{1+(n-1)\delta } + \frac{1}{1+(m-1)\delta }\Bigg ) \end{aligned}$$

Now, observe that

$$\begin{aligned}&\delta \Bigg (\frac{n}{1+(n-1)\delta } + \frac{1}{1+(m-1)\delta }\Bigg ) \ge 1 \\&\quad \iff (m + n - 2)\delta ^2 + (3-m)\delta -1 \ge 0. \end{aligned}$$

This in turn is true in some range $\delta \in (\gamma (G),1)$ if and only if the positive root, denoted $\gamma (m,n)$, of the above quadratic in $\delta$ is less than 1. We know,

$$\begin{aligned} \gamma (m,n) = \frac{m-3 + \sqrt{(3-m)^2 + 4(m + n - 2)}}{2(m+n-2)} \end{aligned}$$

Now, it is easy to check that

$$\begin{aligned} \gamma (m,n) \le 1 \iff m+n \ge 2, \end{aligned}$$

which is true. We have $\gamma (m,n)$ as a decreasing function in n and increasing in m. Thus, the largest value of $\gamma (m,n)$ is attained at $n = 1$ and $m = \omega (G)$. Thus we get the result is true for $\delta \in [\gamma (G),1)$ where

$$\begin{aligned}\gamma (G) :=\gamma (\omega (G),1)= \frac{\omega (G)-3 + \sqrt{(\omega (G)-3)^2 + 4(\omega (G)-1)}}{2(\omega (G)-1)}.\end{aligned}$$

This is as required, thereby completing the proof.

1.4 Proof of Theorem 2

Recall from Algorithm 1 that $S = \{1,2,\ldots ,{|S|}\}$ is a maximum indepdendent set of G, $L = \{l \in V| \ l \mathrm{\ has \ only \ one \ neighbor \ in} \ S\}$. Observe that $C_{i} = N_{{{\bar{G}}}}(i) \cap L$, and ${{\bar{C}}}_{i} = C_{i} \cup \{i\}$ get defined with respect to ${{\bar{G}}}$ at the start of iteration i in Line 4. Let $Q_{i} = N_{{{\bar{G}}}}({{\bar{C}}}_{i})$. We will show that after |S| iterations, all the vertices satisfy the LCP$_\delta (G)$ conditions (4),(5) and (6) for $\delta \in [\gamma (G),1)$.

We first make a few preliminary observations.

• (a) The set of vertices removed in Algorithm 1 $R = \bigcup _{i = 1}^{|S|} ({{\bar{C}}}_{i} \cup Q_{i})$ equals V by Remark 2. Thus V can be divided into two disjoint sets, $V=R_C \cup R_Q$ where $R_C {:}{=}\bigcup _{i = 1}^{|S|} {{\bar{C}}}_{i}$ and $R_Q {:}{=}\bigcup _{i = 1}^{|S|} Q_{i}$. Moreover, $Q_i$’s are disjoint because if vertex j is chosen after vertex i in the for loop, $Q_{j} \subseteq V \backslash (Q_{i} \cup {{\bar{C}}}_{i})$. Similarly ${{\bar{C}}}_i$’s are disjoint.

• (b) We have that $R_Q \cap S = \emptyset$. This can be seen as follows. For some $i \in S$, let $q \in Q_{i}$. Now, $Q_{i} \subset N_G(i) \cup N_G(C_{i})$. If $q \in N_G(i)$, it cannot be an element of S since S is an independent set. Suppose $q \in N_G(c)$ for some $c \in C_{i}$. Now recall that $N_G(c) \cap S = \{i\}$ by definition of $C_{i}$, whereby $q \not \in S$ unless $q = i$, which is not true by definition of $Q_i$. Hence, $R_Q \cap S = \emptyset$.

• (c) Algorithm 1 returns $x_{j} = 0 \ \forall j \in R_Q$. To prove this note that each $x_j$ for $j \in R_Q$ is initialized as 0 since $R_Q \cap S=\emptyset$. Furthermore, in each iteration, we assign non-zero values only to vertices in $R_C$.

• (d) Let $c \in {{\bar{C}}}_i$ for $1\le i\le |S|$. Algorithm 1 returns $x_j = 0 \ \forall j \in N_{G}(c) \backslash N_{{{\bar{C}}}_{i}}(c)$. To see this note that $N_{G}(c)$ $\backslash N_{{{\bar{C}}}_{i}}(c)\subset N_{G}({{\bar{C}}}_{i})$ and since all cliques are independent, we have $N_{G}({{\bar{C}}}_{i}) \subset R_Q$. The claim then follows from part (c).

• (e) Consider $q \in Q_{i}$ for some $1\le i \le |S|$. For the vertex q, define its set of “protective vertices” as $P_{q} {:}{=}(N_{G}(q) \cap S) \backslash \{i\}$. Then $P_{q}$ is always non-empty. The definition of $P_q$ is valid because $Q_j$’s are disjoint. To show $P_q\ne \emptyset$ consider two cases: if $q \in N_G(i)$ then since $q\notin {{\bar{C}}}_{i}$, q must have at least two neighbors in S, one of which is i. On the other hand, if $q \in Q_{i} \backslash N_G(i)$, it needs to have at least one neighbor in S for S to be a maximum independent set and since q is not a neighbor of i, there must be a neighbor of q in $S\backslash \{i\}$.

Now we claim that, with the returned value of x conditions (4),(5) and (6) are satisfied for all the vertices in V. First we show this for vertices in $R_C$. For any vertex $c \in {{\bar{C}}}_{i}$, $1\le i\le |S|$, for the x returned by Algorithm 1, we have

$$\begin{aligned} {{\mathcal {C}}}_c(x)&= x_{c} + \delta \sum _{j \in N_{G}(c)}x_j \end{aligned}$$

(17)

$$\begin{aligned}&= x_{c} + \delta \sum _{j \in N_{{{\bar{C}}}_{i}}(c)}x_j + \delta \sum _{j \in N_{G}(c) \backslash N_{{{\bar{C}}}_{i}}(c)}x_j \end{aligned}$$

(18)

$$\begin{aligned}&= \frac{1}{1+(|{{\bar{C}}}_{i}|-1)\delta } + \frac{\delta (|{{\bar{C}}}_{i}|-1)}{1+(|{{\bar{C}}}_{i}|-1)\delta } + 0 = 1 \end{aligned}$$

(19)

(19) is justified because of (d) above. Thus, for any vertex $c \in {{\bar{C}}}_{i}$, ${{\mathcal {C}}}_c(x) = 1$ and the LCP conditions (4),(5) and(6) are satisfied. It follows that $\mathrm{LCP}_\delta (G)$ conditions hold for all vertices in $R_C$.

We now show that $\mathrm{LCP}_\delta (G)$ conditions hold for all vertices in $R_Q$. Let $q \in Q_i$ for $1\le i \le |S|$ and let $P_{q} = \{p_1,p_2,\cdots ,p_r\}$ be its set of protective vertices. $P_q \ne \emptyset$ by (e) above. We now have the following cases:

Case 1: ${{\bar{C}}}_{i} \subset N_{G}(q)$.

Recall that i is not in the set of protective vertices. Since ${{\bar{C}}}_{i} \subset N_{G}(q)$ we have that q is fully connected to ${{\bar{C}}}_i$ and adjacent to at least one protective vertex, say $p_s \in P_q$. Algorithm 1 assigns $x_{p_s} = \frac{1}{1+(|{{\bar{C}}}_{p_s}|-1)\delta }$ and $x_t=\frac{1}{1+(|{{\bar{C}}}_i|-1)\delta }$ for all $t \in {{\bar{C}}}_i.$ Thus using the Two Clique Lemma (Lemma 7) we get that for $\delta \in [\gamma (G),1)$, x satisfies the LCP conditions (4),(5) and (6) for vertex q.

Case 2: ${{\bar{C}}}_{p_s} \subset N_{G}(q)$ for some $s \in \{1,2,\ldots ,r\}$.

Since ${{\bar{C}}}_{p_s} \subset N_{G}(q)$ we have that q is fully connected to ${{\bar{C}}}_{p_s}$ and adjacent to some vertex $j \in {{\bar{C}}}_i$. Algorithm 1 assigns $x_t = \frac{1}{1+(|{{\bar{C}}}_{p_s}|-1)\delta } \ \forall t \in {{\bar{C}}}_{p_s}$ and $x_{j} = \frac{1}{1+(|{{\bar{C}}}_{i}|-1)\delta }$. Thus, using the Two Clique Lemma (Lemma 7) for $\delta \in [\gamma (G),1)$, x satisfies the LCP conditions (4),(5) and (6) for q.

Case 3: ${{\bar{C}}}_{p_s} \backslash N_{G}(q) \ne \emptyset \ \forall s \in \{1,2,\cdots ,r\}$ and ${{\bar{C}}}_{i} \backslash N_{G}(q) \ne \emptyset$.

We will show that this case is not possible. Define $D_{s} = {{\bar{C}}}_{p_s} \backslash N_{G}(q) \ \forall s \in \{1,2,\cdots ,$ $r\}$. Note that all $D_{s}, s=1,\ldots ,r$ are independent cliques since $D_{s} \subset {{\bar{C}}}_{p_s}$ and each ${{\bar{C}}}_{p_s}$ are independent cliques from Lemma 6.

Now, choose a maximal independent set $D = \{d_1,d_2,\cdots ,d_r\}$ of $G_{\cup _{s = 1}^{r}D_{s}}$. Each vertex in D is from a different clique $D_{s}$ and there is one vertex from each clique; thus, $|D| = r$. Also, let $l \in {{\bar{C}}}_{i} \backslash N_{G}(q)$. We define

$$\begin{aligned} S'' {:}{=}{\widetilde{S}}\cup D \cup \{q,l\}, \quad {\widetilde{S}}:= S \backslash (P_{q}\cup \{i\}), \end{aligned}$$

and show that $S''$ is an independent set with $|S''| > |S|$. To see that $S''$ is an independent set, we check the independence of each of the pairs of sets in the union above. ${\widetilde{S}}$ and D are indepdendent because $D \subset L \cap N_G(P_{q})$ and hence $N_G(D) \cap S \subset P_{q}$ whereby $N_G(D)\cap {\widetilde{S}}= \emptyset$. ${\widetilde{S}}$ and q are indepdendent because $N_G(q) \cap S \subset P_{q}\cup \{i\}$. Finally, $N_G(l) \cap S = \{i\}$ by definition whereby ${\widetilde{S}}$ and l are independent. Also, $D \cap N_{G}(q) = \emptyset$ by definition and $D \cap N_{G}(l) \subset = \emptyset$ since ${{\bar{C}}}_t, t=1,\ldots ,|S|$ are independent. Finally, $N_G(q) \cap \{l\} = \emptyset$ by definition. Thus, $S''$ is an independent set. Moreover, $|S''| > |S|$, which gives a contradiction since S was assumed to be a maximum independent set. Hence, Case 3 is not possible.

Thus, since q was chosen to be an arbitrary vertex in $R_Q$, the LCP conditions (4),(5) and (6) are satisfied for all vertices in $R_Q$. Thus, we have exhausted all cases and constructively shown the existence of an independent clique solution in any graph.

1.5 Proof of 1

Consider $x^{\mathcal {K}}(\cdot ):[\delta ,1)\rightarrow [0,1]^{|V|}$ such that,

$$\begin{aligned} x^{\mathcal {K}}_i(\delta ') = {\left\{ \begin{array}{ll} \frac{1}{1 + (|C_j|-1)\delta '}, \ &{} \ \mathrm{if} \ i \in C_j, \ C_j \in {\mathcal {K}}\\ 0, \ &{} \ \mathrm{otherwise}, \end{array}\right. } \end{aligned}$$

(20)

for $\delta ' \in [\delta ,1)$. Then, we claim that for any given $\delta ' \in [\delta ,1)$, $x^{\mathcal {K}}(\delta ')\in \mathrm{SOL}_{\delta '}(G)$.

We know that for $\delta ' = \delta$, $x^{\mathcal {K}}(\delta )$ is same as the one defined in (9) and $x^{\mathcal {K}}(\delta ) \in \mathrm{SOL}_{\delta }(G)$. Thus, for $c \in C_i$ for some $C_i \in {\mathcal {K}}$, ${{\mathcal {C}}}_c(x^{\mathcal {K}}(\delta )) = 1$ following the analysis in (17), (18) and (19). Since, the analysis in (17), (18) and (19) is independent of the value of $\delta$, ${{\mathcal {C}}}_c(x^{\mathcal {K}}(\delta ')) = 1$ for $c \in C_i$ for some $C_i \in {\mathcal {K}}$ for any $\delta ' \in [\delta ,1)$.

For any vertex $j \in V \backslash {\mathcal {K}}$, let $|N_G(k) \cap C_i| = n_i$ $\forall i$ such that $C_i \in {\mathcal {K}}$. Then,

$$\begin{aligned} {{\mathcal {C}}}_j(x^{\mathcal {K}}(\delta ')) = \delta '\sum _{C_i \in {\mathcal {K}}}^{} \frac{n_i}{1 + (|C_i|-1)\delta '}. \end{aligned}$$

We know that ${{\mathcal {C}}}_j(x^{\mathcal {K}}(\delta )) \ge 1$ since $x^{\mathcal {K}}(\delta ) \in \mathrm{SOL}_{\delta }(G)$. Since, ${{\mathcal {C}}}_j(x^{\mathcal {K}}(\delta '))$ is an increasing function in $\delta '$, ${{\mathcal {C}}}_j(x^{\mathcal {K}}(\delta ')) \ge 1, \ \forall \delta ' > \delta$. Thus, $x^{\mathcal {K}}(\delta ') \in \mathrm{SOL}_{\delta '}(G)$ for $\delta ' \in [\delta ,1)$.

1.6 Proof of Proposition 1

We show that for any two sets of independent cliques ${\mathcal {K}}_1 = \{C^1_1,\dots ,C^1_{|{\mathcal {K}}_1|}\}$ and ${\mathcal {K}}_2 = \{C^2_1,\dots ,C^2_{|{\mathcal {K}}_2|}\}$ such that $|{\mathcal {K}}_1| = \alpha (G)$ and $|{\mathcal {K}}_1| > |{\mathcal {K}}_2|$, we have $||x^{{\mathcal {K}}_1}(\delta )||_1 \ge ||x^{{\mathcal {K}}_2}(\delta )||_1$ for all $\delta \in [\eta (G),1)$. Note that such a ${\mathcal {K}}_1$ exists, since $\eta (G) \ge \gamma (G)$, by using Remark 4. Also, $||x^{{\mathcal {K}}_t}(\delta )||_1 = \sum _{i = 1}^{|{\mathcal {K}}_t|}\frac{|C^t_i|}{1 + (|C^t_i|-1)\delta }$ for $t = 1,2$. Each term in the sum is increasing with $|C^t_i|$. Thus, the least value of $||x^{{\mathcal {K}}_1}(\delta )||_1$ is when $|C^1_i| = 1, \forall C^1_i \in {\mathcal {K}}_1$, which is $\alpha (G)$. Similarly, the maximum value of $||x^{{\mathcal {K}}_2}(\delta )||_1$ is when $|C^2_i| = \omega (G), \forall C^2_i \in {\mathcal {K}}_2$ and $|{\mathcal {K}}_2| = \alpha (G) - 1$, which is $\frac{\omega (G)(\alpha (G)-1)}{1 + (\omega (G)-1)\delta }$. It is easy to see that

$$\begin{aligned} \delta \ge \kappa (G) {:}{=}\frac{\alpha (G)(\omega (G)-1) - \omega (G)}{\alpha (G)(\omega (G)-1)} \iff \alpha (G) \ge \frac{\omega (G)(\alpha (G)-1)}{1 + (\omega (G)-1)\delta }. \end{aligned}$$

Thus, $|x^{\mathcal {K}}_1(\delta )| \ge |x^{\mathcal {K}}_2(\delta )|$ for any ${\mathcal {K}}_2$ with $|{\mathcal {K}}_2| < \alpha (G)$. Hence, there must exist a solution of maxICS$_\delta (G)$ with its support as a union of $\alpha (G)$ independent cliques.

1.7 Proof of Lemma 9

We divide this proof into three cases depending on the neighborhood of the vertex i which is in $G'$ but not in G. Let ${\mathcal {K}} = \{C_1,C_2,\dots , C_{|{\mathcal {K}}|}\}$. The first case is that i is fully connected to some clique $C_j \in {\mathcal {K}}$ and has at least one more edge connecting at least one other clique $C_k \in {\mathcal {K}}$ with $k \ne j$. In this case, we explicitly construct a solution of maxICS$_\delta (G')$ with $\ell _1$ norm at least as much as $x^{\mathcal {K}}(\delta )$. The second case is when i is not fully connected to any clique $C_i \in {\mathcal {K}}$. In this case, we show that the size of maximum indepdendent set of $G'$ is bigger than that of G and hence we can construct an ICS of $G'$ using Algorithm 1 which has $\ell _1$ norm greater than that of $x^{\mathcal {K}}(\delta )$. The last case is when $N_{G'}(i)\cap \sigma (x^{\mathcal {K}}(\delta ))$ is equal to a clique $C_j \in {\mathcal {K}}$. In this case, we divide into two subcases and construct separate solutions with $\ell _1$ norm greater than that of $x^{\mathcal {K}}(\delta )$.

Case 1: i is fully connected to some clique $C_j \in {\mathcal {K}}$ and has at least one more edge connecting at least one other clique $C_k \in {\mathcal {K}}$ with $k \ne j$.

Let $y^{\mathcal {K}}(\delta ) \in {\mathbb {R}}^{|V(G')|}$ be defined as

$$\begin{aligned} y^{\mathcal {K}}_t(\delta ) = {\left\{ \begin{array}{ll} x^{\mathcal {K}}_{\max ,t}(\delta ) &{} \text {if } t \in V(G) \\ 0 &{} \text {if } t = i. \end{array}\right. } \end{aligned}$$

The LCP$_\delta (G')$ conditions (4),(5) and (6) are satisfied $y^{{{\mathcal {K}}}}(\delta )$ for all $j \in V(G)$ since $y^{\mathcal {K}}_i(\delta ) = 0$ and they were already satisfied under $x^{\mathcal {K}}(\delta )$ in G. For i, by Lemma 8, since $\delta \ge \eta (G') \ge \gamma (G')$ and since i is fully connected to $C_j$ and has at least one more edge connecting $C_k$, the LCP$_\delta (G')$ conditions (4),(5) and (6) are also satisfied for i by the Two Clique Lemma. Thus, in this case, $||x^{\mathcal {K}}(\delta )||_1 = ||y^{\mathcal {K}}(\delta )||_1 \le ||x^{{\mathcal {K}}'}(\delta )||_1$.

Case 2: $C_j \not \subset N_{G'}(i)\cap \sigma (x^{\mathcal {K}}(\delta )) \ \forall C_j \in {\mathcal {K}}$, i.e. i is not fully connected to any clique in ${\mathcal {K}}$.

In this case, we will construct a maximum independent set of the graph $G'$ which has size strictly bigger than the maximum independent set of G thereby arriving at a contradiction. Let S be a maximum independent set of G formed by choosing one node from each independent clique in ${{{\mathcal {K}}}}$. In $G'$ we construct an independent set with size $|S|+1$ in the following way. For all the cliques which have no vertex as a neighbor of i, choose any vertex to be in the independent set. For the cliques which intersect the neighborhood of i, choose a vertex in the clique which is not a neighbor of i in the independent set. This gives us an independent set of size at least |S|. Now, by construction i has no neighbors in the independent set found, hence including i still preserves the independence. Thus we have an independent set, say ${\widetilde{S}}$, of size $|S|+1$, and the maximum independent set size can increase at most by 1. Hence, $\alpha (G')=|S|+1$. Next, we construct an ICS of $G'$ using a maximum independent set and show that its $\ell _1$ norm is greater than that of $x^{\mathcal {K}}(\delta )$.

Using Algorithm 1 we can find a set of independent cliques $\widetilde{{\mathcal {K}}}$ in $G'$ and a corresponding ICS $x^{\widetilde{{\mathcal {K}}}}(\delta )$ of $\mathrm{LCP}_\delta (G')$ such that $|\widetilde{{\mathcal {K}}}| = \alpha (G') =|S|+1 > \alpha (G)$. We now find conditions on $\delta$ such that $||x^{\mathcal {K}}(\delta )||_1 \le ||x^{\widetilde{{\mathcal {K}}}}(\delta )||_1$. We consider the least possible value of $||x^{\widetilde{{\mathcal {K}}}}(\delta )||_1$ and the maximum possible value of $||x^{\mathcal {K}}(\delta )||_1$. Note that we have, $||x^{\widetilde{{\mathcal {K}}}}(\delta )||_1 = \sum _{C \in \widetilde{{\mathcal {K}}}}\frac{|C|}{1 + (|C|-1)\delta }$. Each term in the sum is increasing with |C|. Thus, the least value of $||x^{\widetilde{{\mathcal {K}}}}(\delta )||_1$ is when $|C| = 1, \forall C \in \widetilde{{\mathcal {K}}}$. In this case $\Vert x^{{\widetilde{{{{\mathcal {K}}}}}}}(\delta )\Vert _1=\alpha (G')$. Similarly, the maximum value of $||x^{{\mathcal {K}}}(\delta )||_1$ is when $|C_i| = \omega (G), \forall i=1,\ldots ,|{\mathcal {K}}|$, which gives $\Vert x^{{{\mathcal {K}}}}(\delta )\Vert _1=\frac{\omega (G)(\alpha (G')-1)}{1 + (\omega (G)-1)\delta }$. It is easy to see that

$$\begin{aligned} \delta \ge 1 - \frac{\omega (G)}{\alpha (G')(\omega (G)-1)} \implies \Vert x^{{\widetilde{{{{\mathcal {K}}}}}}}(\delta )\Vert _1\ge \alpha (G') \ge \frac{\omega (G)(\alpha (G')-1)}{1 + (\omega (G)-1)\delta } \ge \Vert x^{{{\mathcal {K}}}}(\delta )\Vert _1. \end{aligned}$$

Thus, since $\delta \in [\eta (G'),1)$, we have

$$\begin{aligned} ||x^{\mathcal {K}}(\delta )||_1 \le ||x^{\widetilde{{\mathcal {K}}}}(\delta )||_1 \le ||x^{{\mathcal {K}}'}(\delta )||_1. \end{aligned}$$

Case 3: $N_{G'}(i)\cap \sigma (x^{\mathcal {K}}(\delta )) = C_j$ for some $C_j \in {\mathcal {K}}$, i.e., i is fully connected to exactly one clique $C_j \in {\mathcal {K}}$ and has no edges to any other cliques $C_t \in {\mathcal {K}}$ for $t \ne j$.

Define a set of indepdendent cliques of $G'$, $\widehat{{\mathcal {K}}}$, by replacing $C_j$ in ${\mathcal {K}}$ by $C_j \cup \{i\}$, i.e., $\widehat{{\mathcal {K}}} = ({\mathcal {K}} \backslash C_j) \cup \{C_j \cup \{i\}\}$. Let $M = N_{G'}(C_j \cup \{i\})$, which is the neighborhood of the clique $C_j \cup \{i\}$. Note that any $m \in M$ has neighbors in at least one more clique other than $C_j$. For if not, then $N_{G}(m) \cap \sigma (x^{\mathcal {K}}(\delta )) \subseteq C_j$ and ${{\mathcal {C}}}_m(x^{\mathcal {K}}(\delta )) \le \frac{|C_j|\delta }{1 + (|C_j|-1)\delta } < 1$, which is a contradiction since $x^{\mathcal {K}}(\delta )$ is a solution to LCP$_\delta (G)$.

Define $F = \{m \in M | C_l \subset N_{G'}(m) \cap \sigma (x^{\widehat{{\mathcal {K}}}}(\delta )), \ \mathrm{for \ some \ } C_l \in \widehat{{\mathcal {K}}}\}$, where $x^{\widehat{{\mathcal {K}}}}(\delta )$ is defined as in (9) for the collection of cliques ${\widehat{{{{\mathcal {K}}}}}}$. F is the set of neighbors of $C_j \cup \{i\}$ such that they are fully connected to at least one clique $C_l \in \widehat{{\mathcal {K}}}$. We now have two cases depending on whether $F= M$.

Case 3a: $F = M$.

In this case, we show that $x^{\widehat{{\mathcal {K}}}}(\delta )$ is a solution of LCP$_\delta (G')$. To claim this, we divide the vertex set $V'$ as

$$\begin{aligned} V' = (V' \backslash (F \cup C_j \cup \{i\})) \cup (F) \cup (C_j \cup \{i\}). \end{aligned}$$

Consider $v \in V' \backslash (F \cup (C_j \cup \{i\}))$. It follows from (9) that for such a v, we have $x^{\widehat{{\mathcal {K}}}}_i(\delta ) = x^{{\mathcal {K}}}_{i}(\delta )$ for all $i \in {\bar{N}}_{G}(v)$. Since $x^{{\mathcal {K}}}(\delta )$ is a solution to $\mathrm{LCP}_{\delta }(G)$, the conditions (4), (5) and (6) for $\mathrm{LCP}_\delta (G')$ are satisfied for all $v \in V' \backslash (F \cup (C_j \cup \{i\}))$ by $x^{{\widehat{{{{\mathcal {K}}}}}}}(\delta )$. Now consider $v \in F$. We note that if v is such that $C_l$ is any clique in $\widehat{{\mathcal {K}}}$ other than $C_j \cup \{i\}$, then v is fully connected to $C_l$ and has at least one neighbor in $C_j \cup \{i\}$. Hence, using Lemma 8, the $\mathrm{LCP}_\delta (G')$ conditions are satisfied with $x^{\widehat{{\mathcal {K}}}}(\delta )$ for v. Next, if $C_l = C_j \cup \{i\}$, since $v\in F=M$, we conclude that v has neighbors in at least one clique other than $C_j$ in $\widehat{{\mathcal {K}}}$. Again, using Lemma 8, LCP conditions are satisfied for v by $x^{{\widehat{{{{\mathcal {K}}}}}}}(\delta )$. Finally for any node in $C_j \cup \{i\}$, the LCP$_\delta (G')$ conditions are satisfied by definition of $x^{\widehat{{\mathcal {K}}}}(\delta )$. Thus, for any vertex in $V'$, the LCP$_\delta (G')$ conditions are satisfied with $x^{\widehat{{\mathcal {K}}}}(\delta )$.

Note that, $||x^{\widehat{{\mathcal {K}}}}(\delta )||_1 - ||x^{\mathcal {K}}(\delta )||_1 = \frac{n+1}{1+n\delta } - \frac{n}{1+(n-1)\delta } \ge 0 \ \forall \delta < 1$. Thus, $||x^{\mathcal {K}}(\delta )||_1 \le ||x^{\widehat{{\mathcal {K}}}}(\delta )||_1 \le ||x^{{\mathcal {K}}'}(\delta )||_1$.

Case 3b: $F \ne M$.

We show that in this case we can construct an independent set in $G'$ with size strictly greater than $\alpha (G)$. Let $p \in M \backslash F$. We know that p has neighbors in at least one clique other than $C_j$, say $C_o$, and neither clique is fully connected to p since $p \not \in F$. Let S be a maximum independent set of G formed by choosing one node from each independent clique in ${{{\mathcal {K}}}}$. Suppose the nodes chosen from $C_o$ and $C_j$ are a and b, respectively. Now, consider ${\widetilde{S}} = (S \backslash \{a,b\}) \cup \{c,d,p\}$ where $c \in C_o \backslash N_{G'}(p)$ and $d \in (C_j \cup \{i\}) \backslash N_{G'}(p)$. It can be observed that ${\widetilde{S}}$ is an indepdendent set and $|{\widetilde{S}}| > |S|$. Thus we have found an independent set of $G'$ of size greater than that of the maximum independent set of G. Using the argument in the second paragraph of Case 2, we have a set of indepdendent cliques in $G'$, say $\widetilde{{\mathcal {K}}}$ generated from ${\widetilde{S}}$ by Algorithm 1, such that $||x^{\mathcal {K}}(\delta )||_1 \le ||x^{\widetilde{{\mathcal {K}}}}(\delta )||_1 \le ||x^{{\mathcal {K}}'}(\delta )||_1$.

Thus, having considered all possible cases for the neighborhood of the node i, we have shown that $||x^{\mathcal {K}}(\delta )||_1 \le ||x^{{\mathcal {K}}'}(\delta )||_1$.

Proof from Section 5

1.1 Proof of Theorem 3

Let G be a graph such that $|V(G)| = n$. Let $w = (w_1,w_2,\ldots ,w_n)$ denote the vector of given weights. We define

$$\begin{aligned} M_w(\delta ,G) = \max \{w^\top x \ | \ x \ \mathrm{solves } \ \mathrm{LCP}_\delta (G) \}. \end{aligned}$$

and prove that $M_w(\delta ,G) = \alpha _w(G)$ by showing inequalities in both directions.

Let S be a w-weighted maximum independent set. Then, it is a maximal independent set and $\mathbf{1}_S$ solves $\mathrm{LCP}_\delta (G)$ for $\delta \ge 1$ by Lemma 2. Thus, $M_w(\delta ,G) \ge w^\top \mathbf{1}_S = \alpha _w(G)$.

Next, we want to show that $M_w(\delta ,G) \le \alpha _w(G)$. We prove this by induction on the size of the graph. For the base case, consider the graph with 1 vertex and no edges. In this case, it is trivial to see that $M_w(\delta ,G) \le \alpha _w(G)$.

Induction Hypothesis: For any graph $G'$ with $|V(G')| < n$, for a given set of weights w and $\delta \ge 1$, we have $M_w(\delta ,G') \le \alpha _w(G')$.

Let G be a graph with vertex set $V(G) = \{1,2,\ldots ,n\}$ and let $x^*$ be the maximizer of $w^\top x$ where $x \in \mathrm{SOL}_\delta (G)$. Let S be a w-weighted maximum indepdendent set of G.

Case 1: $\sigma (x^*) = V(G)$

Here, we have $x^*_i > 0 \ \forall i \in V(G)$. Thus, the LCP conditions (4),(5) and (6) dictate

$$\begin{aligned} {{\mathcal {C}}}_i(x^*) = x^*_i + \delta \sum _{j=1}^{n}a_{ij}x^*_j = 1 \ \forall i \in V(G). \end{aligned}$$

Taking the w-weighted sum over $i \in S$, we get

$$\begin{aligned} \sum _{i \in S}w_ix^*_i = \sum _{i \in s}w_i - \delta \sum _{i \in S}w_i\sum _{j=1}^{n}a_{ij}x^*_j. \end{aligned}$$

(21)

Now,

$$\begin{aligned} M_w(\delta ,G)&= \sum _{i \in V(G)}w_ix^*_i \nonumber \\&= \sum _{i \in S}w_ix^*_i + \sum _{i \in V(G) \backslash S}w_ix^*_i \nonumber \\&= \sum _{i \in S}w_i - \delta \sum _{i \in S}w_i\sum _{j=1}^{n}a_{ij}x^*_j + \sum _{i \in V(G) \backslash S}w_ix^*_i \end{aligned}$$

(22)

$$\begin{aligned}&= \alpha _w(G) - \sum _{j \in V(G) \backslash S}x^*_j(\delta \sum _{i \in S}w_i a_{ij} - w_j), \end{aligned}$$

(23)

where Equation ((22)) follows by using Equation ((21)) and Equation ((23)) follows because $a_{ij} = 0$ when $i,j \in S$. We claim that $\sum _{i \in S}w_i a_{ij} - w_j \ge 0 \ \forall j \in V(G) \backslash S$. Suppose not, then we construct an independent set with higher weight than that of S. Let $l \in V(G) \backslash S$ be such that $w_l - \sum _{i \in S}w_i a_{il} > 0$. Consider ${\widetilde{S}} = (S \backslash N_G(l)) \cup {l}$. It is easy to see that ${\widetilde{S}}$ is an independent set and the difference between weights of ${\widetilde{S}}$ and S is $w_l - \sum _{i \in S}w_i a_{il} > 0$. This contradicts the fact that S is a w-weighted maximum independent set. Thus, $\sum _{i \in S}w_i a_{ij} - w_j \ge 0 \ \forall j \in V(G) \backslash S$. Since $\delta \ge 1$, this implies $\delta \sum _{i \in S}w_i a_{ij} - w_j \ge 0 \ \forall j \in V(G) \backslash S$ which gives $M_w(\delta ,G) \le \alpha _w(G)$.

Case 2: $\sigma (x^*) \subset V(G)$, a strict subset.

From Lemma 1, we know that since $x^* \in \mathrm{SOL}_\delta (G)$, we have $x^*_{\sigma (x^*)} \in$

$\mathrm{SOL}_\delta (G_{\sigma (x^*)})$. For brevity let $y^*:=x^*_{\sigma (x^*)}$. Now,

$$\begin{aligned} M_w(\delta ,G) = \sum _{i \in V(G)}w_i x^*_i = \sum _{i \in V(G_{\sigma (x^*)})}w_i y^*_i&\le M_w(\delta ,G_{\sigma (x^*)}) \\&\le \alpha _w(G_{\sigma (x^*)}) \le \alpha _w(G), \end{aligned}$$

where $M_w(\delta ,G_{\sigma (x^*)}) \le \alpha _w(G_{\sigma (x^*)})$ follows from the induction hypothesis since $|V(G_{\sigma (x^*)})| < n$. Hence, by the principle of mathematical induction, we have that for any graph G, with given weights w and $\delta \ge 1$, we have $M_w(\delta ,G) \le \alpha _w(G)$.