Multi-defender Security Games with Schedules

Abstract

Stackelberg Security Games are often used to model strategic interactions in high-stakes security settings. The majority of existing models focus on single-defender settings where a single entity assumes command of all security assets. However, many realistic scenarios feature multiple heterogeneous defenders with their own interests and priorities embedded in a more complex system. Furthermore, defenders rarely choose targets to protect. Instead, they have a multitude of defensive resources or schedules at its disposal, each with different protective capabilities. In this paper, we study security games featuring multiple defenders and schedules simultaneously. We show that unlike prior work on multi-defender security games, the introduction of schedules can cause non-existence of equilibrium even under rather restricted environments. We prove that under the mild restriction that any subset of a schedule is also a schedule, non-existence of equilibrium is not only avoided, but can be computed in polynomial time in games with two defenders. Under additional assumptions, our algorithm can be extended to games with more than two defenders and its computation scaled up in special classes of games with compactly represented schedules such as those used in patrolling applications. Experimental results suggest that our methods scale gracefully with game size, making our algorithms amongst the few that can tackle multiple heterogeneous defenders.

Z. Song—Independent Researcher.

Equal contribution between authors Song and Ling.

A Appendix

1.1 A.1 Proof of Lemma 1

Proof

Since $v^1\in V^1$ and $\widetilde{v}^1(j) \le v^1(j)$ for all $j \in [T]$, we have $\widetilde{v}^1\in V^1$ by Assumption 2. We now show that $(\widetilde{v}^1, v^2, t)$ is AIC, 1-IC, and 2-IC.

• 1. $(\widetilde{v}^1, v^2, t)$ is AIC. For all $j\in [T]$ and $j\ne t$, we have

  $$\begin{aligned} \widetilde{v}^1(j) + v^2(j) \underbrace{= v^1(j) + v^2(j)}_{\text {definition of }\widetilde{v}^1} \ge \underbrace{v^1(t) + v^2(t)}_{(v^1, v^2, t) \text { is AIC}} \ge \underbrace{\widetilde{v}^1(t) + v^2(t)}_{\text {by definition of }\widetilde{v}^1}. \end{aligned}$$

• 2. $(\widetilde{v}^1, v^2, t)$ is 1-IC. If not, there exists $\widehat{v}^1\in \mathcal {V}^1, j\succ _1 t$ where $(\widehat{v}^1, v^2, j)$ is 1-WAIC, implying $(v^1, v^2, t)$ is not 1-IC and not an NSE.

• 3. $(\widetilde{v}^1, v^2, t)$ is 2-IC. Suppose otherwise. Then there exists $\widehat{v}^2\in \mathcal {V}^2, j \succ _2 t$ where $(\widetilde{v}^1, \widehat{v}^2, j)$ is 2-WAIC, implying that $\widetilde{v}^1(k) +\widehat{v}^2(k) > \widetilde{v}^1(j) +\widehat{v}^2(j)$ for $k \prec _2 j$, and $\widetilde{v}^1(k) +\widehat{v}^2(k) \ge \widetilde{v}^1(j) +\widehat{v}^2(j)$ for $k \succ _2 j$. Then for any $k \succ _2 j (\succ _2 t)$, $\widetilde{v}^1(k) +\widehat{v}^2(k) \ge \widetilde{v}^1(j) +\widehat{v}^2(j)$ indicates that $v^1(k) +\widehat{v}^2(k) \ge v^1(j) +\widehat{v}^2(j)$ by definition of $\widetilde{v}^1$ that $\widetilde{v}^1(k) = v^1(k), \widetilde{v}^1(j) = v^1(j)$. Besides, for any $k\prec _2 j$,

  $$\begin{aligned} v^1(k) + \widehat{v}^2(k) \underbrace{\ge \widetilde{v}^1(k) + \widehat{v}^2(k)}_{\text {by definition of }\widetilde{v}^1} > \underbrace{\widetilde{v}^1(j) + \widehat{v}^2(j)}_{(\widetilde{v}^1, \widehat{v}^2, j)\text { is 2-WAIC}} = \underbrace{v^1(j) + \widehat{v}^2(j)}_{\text {by definition of }\widetilde{v}^1}. \end{aligned}$$

  Since $(v^1, \widehat{v}^2, j)$ is 2-WAIC, $(v^1, v^2, t)$ is not 2-IC and not an NSE.

1.2 A.2 Proof of Lemma 2

Proof

$v^1\in V^1$ and $\widetilde{v}^1$ has no more coverage than $v^1$, so $\widetilde{v}^1\in V^1$ by the SSAS assumption. By the definition of NSE, it is sufficient to show that $(\widetilde{v}^1, v^2, t)$ is AIC, 1-IC, and 2-IC to prove the Lemma.

• 1. $(\widetilde{v}^1, v^2, t)$ is AIC. $t\in B(\widetilde{v}^1, v^2)$ because $\widetilde{v}^1(t) + v^2(t) = 0$.

• 2. $(\widetilde{v}^1, v^2, t)$ is 1-IC. If not, then there exists $\widehat{v}^1\in V^1, j\succ _1 t$ such that $(\widehat{v}^1, v^2, j)$ is 1-WAIC. Therefore, $(v^1, v^2, t)$ is also not 1-IC, contradicting the assumption that $(v^1, v^2, t)$ is an NSE.

• 3. $(\widetilde{v}^1, v^2, t)$ is 2-IC. We prove this by contradiction. Suppose $(\widetilde{v}^1, v^2, t)$ is not 2-IC, then there exists $\widehat{v}^2\in V^2, j \succ _2 t$ where $(\widetilde{v}^1, \widehat{v}^2, j)$ is 2-WAIC. By definition of 2-WAIC, we have that $\widetilde{v}^1(k') +\widehat{v}^2(k') > \widetilde{v}^1(j) +\widehat{v}^2(j)$ for any target $k' \prec _2 j$, and $\widetilde{v}^1(k') +\widehat{v}^2(k') \ge \widetilde{v}^1(j) +\widehat{v}^2(j)$ for any target $k' \succ _2 j$. Consider $u^2\in V^2$ such that $u^2(k') = \widehat{v}^2(k')$ for any target $k' \preceq _2 t$ and $u^2(k') = 0$ for any target $k' \succ _2 t$. Notice that there always exist a unique target $k \in [T]$ such that $(v^1, u^2, k)$ is 2-WAIC. We claim that if k is the target that $(v^1, u^2, k)$ is 2-WAIC, then $k \succ _2 t$. We prove this by excluding other targets $e \preceq _2 t$. Notice that there is a target $m \succ _2 t$ that $v^1(m) = \min _{{k'}\succ _2 t }v^1({k'})$. Consider a target $e \preceq _2 t$, then

  $$\begin{aligned} v^1(e) + u^2(e) &\underbrace{\ge \widetilde{v}^1(e) + u^2(e)}_{\text {by definition of }\widetilde{v}^1} \underbrace{=\widetilde{v}^1(e) + \widehat{v}^2(e)}_{\text {by definition of }u^2} \underbrace{> \min _{{k'}\succ _2 t}\{\widetilde{v}^1({k'}) + \widehat{v}^2({k'})\}}_{(\widetilde{v}^1, \widehat{v}^2, j)\text { is 2-WAIC}} \\ & \underbrace{\ge \min _{{k'}\succ _2 t}\widetilde{v}^1({k'})}_{\widehat{v}^2({k'})\ge 0} \underbrace{ = v^1(m)}_{\text {by definition of }m} \underbrace{ = v^1(m) + u^2(m)}_{u^2(m) = 0}. \end{aligned}$$

  Thus, e has more coverage than m, so $(v^1, u^2, e)$ is not 2-WAIC. Then $k \preceq _2 t$ does not hold. There exists a target $k \succ _2 t$ such that $(v^1, u^2, k)$ is 2-WAIC, so $(v^1, v^2, t)$ is not 2-IC, which contradicts that $(v^1, v^2, t)$ is an NSE. $\blacksquare$

1.3 A.3 Proof of Theorem 1

To better understand the structure of $\mathcal {H}$, we introduce partial sets $\mathcal {H}^1_t\subset V^1, \mathcal {H}^2_t\subset V^2$ for $\mathcal {H}_t$ and show how they compose the set $\mathcal {H}_t$.

Definition 8

$\mathcal {H}_t^1$ is the set of $v^1\in V^1$ such that there exists $h^1\ge 0$, $v^1(j) = h^1$ for any target $j \in \mathcal {T}_t^{\succ _2}$, $v^1(j) = 0$ for any target $j \in \mathcal {T}_t^{\preceq _2}$,and there does not exist $j\succ _2 t$ and $\widehat{v}^2\in V^2$, $(v^1, \widehat{v}^2, j)$ is 2-WAIC. Similar for $\mathcal {H}_t^2$.

Theorem 3

$\mathcal {H}_t = \mathcal {H}^1_t \times \mathcal {H}^2_t \times \{t\}$.

Proof

For any $(v^1, v^2, t)\in \mathcal {H}_t$, $v^1$ and $v^2$ are t-standard coverage. Since $\mathcal {H}_t$ only contains NSEs, $(v^1, v^2, t)$ is 1-IC and 2-IC. Thus, $v^1\in \mathcal {H}^1_t, v^2\in \mathcal {H}^2_t$. We now show that for any coverage $v^1\in \mathcal {H}_t^1$ and $v^2\in \mathcal {H}_t^2$, $(v^1, v^2, t)\in \mathcal {H}_t$. First, $(v^1, v^2, t)$ is an NSE. $(v^1, v^2, t)$ is AIC because $v^1(t) = v^2(t) = 0$. Also, $(v^1, v^2, t)$ is 1-IC and 2-IC by definition of $\mathcal {H}^1_t$ and $\mathcal {H}^2_t$. Second, $v^1$ and $v^2$ are t-standard coverage. Thus, $(v^1, v^2, t) \in \mathcal {H}_t$. $\blacksquare$

Theorem 3 decomposes the space of $\mathcal {H}_t$. Next we consider how to compute $\mathcal {H}^1_t$ and $\mathcal {H}^2_t$, which provides us an NSE in $\mathcal {H}_t$. We first consider the reduction of the set containing deviation strategies.

Lemma 4

For a target t and a coverage $v^1 \in \mathcal {H}_t^1$, if there is a $\widehat{v}^2\in V^2$ and $j \succ _2 t$ such that $(v^1, \widehat{v}^2, j)$ is 2-WAIC, we construct $u^2\in V^2$ such that $u^2(k) = 0$ for $k \in \mathcal {T}_t^{\succ _2}$ and $u^2(k) = \min _{k' \preceq _2 t} \widehat{v}^2(k')$ for $k \in \mathcal {T}_t^{\preceq _2}$, then there exists a target $m\succ _2 t$ such that $(v^1, u^2, m)$ is 2-WAIC.

Proof

In a 2-WAIC strategy profile $(v^1, \widehat{v}^2, j)$ and $j\succ _2 t$, $v^1(k) + \widehat{v}^2(k) > v^1(j) + \widehat{v}^2(j)$ for any target $k\preceq _2 t$. Since $v^1\in \mathcal {H}_t^1$, $v^1(k) = 0$ for any target $k\preceq _2 t$. So, $\min _{k\preceq _2 t}\widehat{v}^2(k) > v^1(j) + \widehat{v}^2(j)$. We have $u^2(k) > v^1(j) + \widehat{v}^2(j)$ by definition of u, so any target $k\in \mathcal {T}_t^{\preceq _2}$ is not the attacked target. There always exists a target m such that $(v^1, u^2, m)$ is 2-WAIC, and here $m \succ _2 t$. $\blacksquare$

Lemma 4 reduces the deviation $\widehat{v}^2$ to $u^2$ with a canonical structure that $u^2(k)$ are equal for targets $k \in \mathcal {T}_t^{\preceq _2}$, but $u^2(k) = 0$ elsewhere. The computation of $u^2$ is related to the oracle $\textsc {MaximinCov}$. For convenience, we use $M^i(\mathcal {T})$ instead of $\textsc {MaximinCov}(\mathcal {T}, V^i)$ when the set of coverage is default to be $V^i$. Formally, $M^i(\mathcal {T}) = \max _{v^i \in V^i}\min _{j \in \mathcal {T}} v^i(j)$ for $\mathcal {T} \ne \emptyset$, and $M^i(\mathcal {T}) = + \infty$ for $\mathcal {T} = \emptyset$. With this notation, we give a sufficient and necessary condition for $\mathcal {H}_t^i \ne \emptyset$.

Theorem 4

$\mathcal {H}_t^1 \ne \emptyset$ if and only if $M^1(\mathcal {T}_t^{\succ _2}) \ge M^2(\mathcal {T}_t^{\preceq _2})$. Similarly, $\mathcal {H}_t^2 \ne \emptyset$ if and only if $M^2(\mathcal {T}_t^{\succ _1}) \ge M^1(\mathcal {T}_t^{\preceq _1})$.

Proof

We prove the first claim for $\mathcal {H}_t^1\ne \emptyset$, and it is same for $\mathcal {H}_t^2\ne \emptyset$.

• 1. When $M^1(\mathcal {T}_t^{\succ _2}) \ge M^2(\mathcal {T}_t^{\preceq _2})$, we consider a coverage $v^1\in V^1$ of defender 1 such that $v^1(k) = M^1(\mathcal {T}_t^{\succ _2})$ for target $k \succ _2 t$, and $v^1(k) = 0$ elsewhere. For any coverage $\widehat{v}^2 \in V^2$, there is a target $m\in \mathcal {T}_t^{\preceq _2}$ such that $v^1(m) + \widehat{v}^2(m) \le M^2(\mathcal {T}_t^{\preceq _2}) \le v^1(k)$ for any target $k\in \mathcal {T}_t^{\succ _2}$. Any target $k\in \mathcal {T}_t^{\succ _2}$ has no less coverage than target m given coverage $(v^1, \widehat{v}^2)$, so there does not exist target $j\succ _2 t$ such that $(v^1, \widehat{v}^2, j)$ is 2-WAIC. Thus, $v^1\in \mathcal {H}_t^1$ by definition.

• 2. When $M^1(\mathcal {T}_t^{\succ _2}) < M^2(\mathcal {T}_t^{\preceq _2})$, we want to show that $\mathcal {H}_t^1 = \emptyset$. $M^1(\mathcal {T}_t^{\succ _2}) \ne +\infty$, so for any $v^1\in V^1$, there exists target $j\succ _2 t$ such that $v^1(j) \le M^1(\mathcal {T}_t^{\succ _2})$. Then consider a coverage $\widehat{v}^2\in V^2$ of defender 2 such that $\widehat{v}^2(k) = M^2(\mathcal {T}_t^{\preceq _2})$ for targets $k\in \mathcal {T}_t^{\preceq _2}$ and $\widehat{v}^2(k) = 0$ for other targets. In the coverage profile $(v^1, \widehat{v}^2)$, target $k\in \mathcal {T}_t^{\preceq _2}$ has strictly more coverage than target j. Notice that there always exists a target $j'$ such that $(v^1, \widehat{v}^2, j')$ is 2-WAIC. Thus, for any $v^1\in V^1$, there exists $\widehat{v}^2\in V^2$, such that there exists $j'\succ _2 t$ and $(v^1, \widehat{v}^2, j')$ is 2-WAIC. Therefore, $\mathcal {H}_t^1 = \emptyset$. $\blacksquare$

Corollary 1

For any defender $i\in [2]$, there exists $t\in [T]$ such that $\mathcal {H}_t^i \ne \emptyset$.

Proof

Let $i'\ne i$ be another defender. There exists a target t such that for any target $j \in [T]$, $t \succeq _{i'} j$. Then $M^{i}(\mathcal {T}_t^{\succ _{i'}}) = M^{i}(\emptyset ) = +\infty$. However, $M^{i'}(\mathcal {T}_k^{\preceq {i'}}) = M^{i'}([T])$ is finite. Therefore, $\mathcal {H}_t^i \ne \emptyset$. $\blacksquare$

Lemma 5

For any set of targets $\mathcal {T}' \subset \mathcal {T}$, $M_i(\mathcal {T}')\ge M_i(\mathcal {T})$ holds.

Proof

Let $v^{i*}$ be the coverage when $M_i(\mathcal {T})$ achieves the maximum. Let $v^i = v^{i *}$, then we have $\min _{t_j \in \mathcal {T}'} v^i(j) \ge M_{i}(\mathcal {T}).$ Therefore $M_i(\mathcal {T}') \ge M_i(\mathcal {T})$. $\blacksquare$

Lemma 6

For two defenders $i, i'$ and targets $t \succ _{i'} j$, if $\mathcal {H}_j^i \ne \emptyset$, then $\mathcal {H}_t^i \ne \emptyset$.

Proof

Since $t \succ _{i'} j$, we have $\mathcal {T}_t^{\succ _{i'}} \subset \mathcal {T}_j^{\succ _{i'}}$ and $\mathcal {T}_j^{\preceq _{i'}} \subset \mathcal {T}_t^{\preceq _{i'}}$. By $\mathcal {H}_j^i \ne \emptyset$, we have $M^i(\mathcal {T}_j^{\succ _{i'}}) \ge M^{i'}(\mathcal {T}_j^{\preceq _{i'}})$. Then using Lemma 5, we get $M^i(\mathcal {T}_t^{\succ _{i'}}) \ge M^i(\mathcal {T}_j^{\succ _{i'}}) \ge M^{i'}(\mathcal {T}_j^{\preceq _{i'}}) \ge M^{i'}(\mathcal {T}_t^{\preceq _{i'}})$. By Theorem 4, $\mathcal {H}_k^i \ne \emptyset$. $\blacksquare$

Theorem 5

$\mathcal {H}\ne \emptyset$.

Proof

Suppose $\mathcal {T}^1 = \{t\in [T] | \mathcal {H}_t^2 \ne \emptyset \}$ and $\mathcal {T}^2 = \{t\in [T] | \mathcal {H}_t^1 \ne \emptyset \}$. By Corollary 1, $\mathcal {T}^1, \mathcal {T}^2 \ne \emptyset$ and $\mathcal {T}^2 \ne \emptyset$. There is a unique ${j_1} \in \mathcal {T}^1$ (${j_2} \in \mathcal {T}^2$) such that for any $t \in \mathcal {T}^1$, $t \succeq _1 {j_1}$ (for any $t \in \mathcal {T}^2$, $t \succeq _2 {j_2}$). It is sufficient to show that $\mathcal {T}^1 \cap \mathcal {T}^2 \ne \emptyset$. If we show this, then there is a target $j\in \mathcal {T}^1\cap \mathcal {T}^2$. We have $\mathcal {H}_j^1\ne \emptyset , \mathcal {H}_j^2\ne \emptyset$, and thus $\mathcal {H}\ne \emptyset$. Next we prove $\mathcal {T}^1 \cap \mathcal {T}^2 \ne \emptyset$ by contradiction.

Suppose $\mathcal {T}^1 \cap \mathcal {T}^2 = \emptyset$. Since $\mathcal {T}^1, \mathcal {T}^2 \ne \emptyset$ and $\mathcal {T}^1 \cap \mathcal {T}^2 = \emptyset$, we have $\overline{\mathcal {T}^1} = [T]\setminus \mathcal {T}^1\ne \emptyset$, and $\overline{\mathcal {T}^2} = [T]\setminus \mathcal {T}^2\ne \emptyset$. There is a unique ${k_1}\in \overline{\mathcal {T}^1}$ (${k_2}\in \overline{\mathcal {T}^2}$) such that for any $t\in \overline{\mathcal {T}^1}$, ${k_1} \succeq _1 t$ (for any $t\in \overline{\mathcal {T}^2}$, ${k_2} \succeq _2 t$). By Lemma 6, for any target $t \succ _1 {j_1}$, $t \in \mathcal {T}^1$, and for any target $t \succ _2 {j_2}$, $t \in \mathcal {T}^2$. Therefore, ${j_1} \succ _1 {k_1}$ and ${j_2} \succ _2 {k_2}$. Then we have $\mathcal {T}_{k_1}^{\succ _1} = \mathcal {T}^1, \mathcal {T}_{k_1}^{\preceq _1} = \overline{\mathcal {T}^1}$, and $\mathcal {T}_{k_2}^{\succ _2} = \mathcal {T}^2, \mathcal {T}_{k_2}^{\preceq _2} = \overline{\mathcal {T}^2}$. By Theorem 4 and $\mathcal {H}^2_{k_1} = \emptyset , \mathcal {H}^1_{k_2} = \emptyset$, we have $M^2(\mathcal {T}^1) < M^1(\overline{\mathcal {T}^1})$ and $M^1(\mathcal {T}^2) < M^2(\overline{\mathcal {T}^2})$. By $\mathcal {T}^1\cap \mathcal {T}^2 = \emptyset$, $\mathcal {T}^2 \subset \overline{\mathcal {T}^1}$ and $\mathcal {T}^1 \subset \overline{\mathcal {T}^2}$. By Lemma 5, $M^1(\overline{\mathcal {T}^1})< M^1(\mathcal {T}^2)$ and $M^2(\overline{\mathcal {T}^2})< M^2(\mathcal {T}^1)$. Combining this with previous inequality, we get $M^2(\mathcal {T}^1)< M^1(\mathcal {T}^2)$ and $M^1(\mathcal {T}^2) < M^2(\mathcal {T}^1)$, which are contradictory.

Therefore, $\mathcal {T}^1\cap \mathcal {T}^2 \ne \emptyset$. There is a target $j\in \mathcal {T}^1\cap \mathcal {T}^2$. By definition of $\mathcal {T}^1, \mathcal {T}^2$, we have $\mathcal {H}_j^1, \mathcal {H}_j^2\ne \emptyset$. Thus, $\mathcal {H}_j\ne \emptyset$ by Theorem 3. $\mathcal {H}_j\subset \mathcal {H}$, so $\mathcal {H}\ne \emptyset$. $\blacksquare$

1.4 A.4 Proof of Lemma 3

Proof

By Theorem 3 in the appendix, $\mathcal {H}_t = \mathcal {H}^1_t \times \mathcal {H}^2_t \times \{t\}$. $\mathcal {H}_t \ne \emptyset$, so $\mathcal {H}^1_t, \mathcal {H}^2_t \ne \emptyset$. Since $j \succ _1 t, j \succ _2 t$, further by Lemma 6 in the appendix, we have $\mathcal {H}^1_j, \mathcal {H}^2_j \ne \emptyset$. Use Theorem 3 again, $\mathcal {H}_j = \mathcal {H}^1_j \times \mathcal {H}^2_j \times \{j\}$. Therefore, $\mathcal {H}_j \ne \emptyset$. $\blacksquare$

1.5 A.5 Proof of Theorem 2 where $F(t)\ne F(t')$

Proof

Here, $\mathop {\textrm{Argmin}}\limits _{t} F(t)$ may not be a singleton, and we handle the edge case of selecting a specific $k^*$ from $\mathop {\textrm{Argmin}}\limits _{t} F(t)$. Consider the set $\mathcal {F} = \mathop {\textrm{Argmin}}\limits _{t} F(t)$ and let $\underline{F} = \min _{t'\in [T]} F(t')$. For any $t, j\in \mathcal {F}, t\ne j$, we say $t \triangleleft j$ if for any defender i, $\mathcal {M}_{i, \pi _i(t)} = \underline{F} \implies t \succ _i j$. We show that $\triangleleft$ is transitive. Suppose $t\triangleleft j$ and $j\triangleleft j'$. Hence, $t \succ _i j$ for any defender i where $\mathcal {M}_{i, \pi _i(t)} = \underline{F}$. On one hand, $\mathcal {M}_{i, \pi _i(j)}\ge \mathcal {M}_{i, \pi _i(t)} \ge \underline{F}$ since $t\succ _i j$. On the other hand, $j\in \mathcal {F}$, so $\mathcal {M}_{i, \pi _i(j)} \le \underline{F}$. Therefore, $\mathcal {M}_{i, \pi _i(j)} = \underline{F}$. Using a similar argument on $j\triangleleft j'$ gives us $\mathcal {M}_{i, \pi _i(t)} = \mathcal {M}_{i, \pi _i(j')} = \underline{F}$ and $t\succ _i j\succ _i j'$. So, $t\triangleleft j'$ and $\triangleleft$ is transitive.

We claim there exists a target $k^*\in \mathcal {F}$ such that $\not \exists t\in \mathcal {F}$, $t \triangleleft k^*$. If not, then for any $k_q\in \mathcal {F}$, there exists $k_{q+1}\in \mathcal {F}$ where $k_{q+1} \triangleleft k_q$. Thus, we can construct an infinite sequence $\cdots \triangleleft k_2 \triangleleft k_1$. Since $\triangleleft$ is transitive, each $k_q$ must be distinct, which is not possible with a finite number of targets. Now we select a target $k^*$.

Once $k^*$ is selected, we set $v^i(k^*) = 0$ for all defenders i. For any $t\ne k^*$, we have $F(t) \ge F(k^*)$. For a fixed t, there are two cases: (i) if $F(t) > F(k^*)$, for any defender i such that $\mathcal {M}_{i, \pi _i(t)} = F(t)$, we have $k^* \succ _i t$, since $\mathcal {M}_{i, \pi _i(t)} > F(k^*) \ge \mathcal {M}_{i, \pi _i(k^*)}$. (ii) If $F(t) = F(k^*)$, there is at least one defender i satisfying $\mathcal {M}_{i, \pi _i(t)} = F(t)$ such that $k^* \succ _i t$, otherwise $t\triangleleft k^*$, contradicting the choice of $k^*$. Thus, for any $t\ne k^*$, there always exists a defender i where $\mathcal {M}_{i, \pi _i(t)} = F(t)$ and $k^* \succ _i t$. Our construction sets $v^i(t) = F(k^*)$ and $v^{i'}(t) = 0$ for $i'\ne i$. Showing $\textbf{v}$ is feasible and $(\textbf{v}, k^*)$ is an NSE is same with the main proof of Theorem 2.$\blacksquare$