The Price of Defense

Abstract

We consider a game on a graph $G=\langle V, E\rangle$ with two confronting classes of randomized players: $\nu$ attackers, who choose vertices and seek to minimize the probability of getting caught, and a single defender, who chooses edges and seeks to maximize the expected number of attackers it catches. In a Nash equilibrium, no player has an incentive to unilaterally deviate from her randomized strategy. The Price of Defense is the worst-case ratio, over all Nash equilibria, of $\nu$ over the expected utility of the defender at a Nash equilibrium.

We orchestrate a strong interplay of arguments from Game Theory and Graph Theory to obtain both general and specific results in the considered setting:

(1) Via a reduction to a Two-Players, Constant-Sum game, we observe that an arbitrary Nash equilibrium is computable in polynomial time. Further, we prove a general lower bound of $\frac{\textstyle |V|}{\textstyle 2}$ on the Price of Defense. We derive a characterization of graphs with a Nash equilibrium attaining this lower bound, which reveals a promising connection to Fractional Graph Theory; thereby, it implies an efficient recognition algorithm for such Defense-Optimal graphs.

(2) We study some specific classes of Nash equilibria, both for their computational complexity and for their incurred Price of Defense. The classes are defined by imposing structure on the players’ randomized strategies: either graph-theoretic structure on the supports, or symmetry and uniformity structure on the probabilities. We develop novel graph-theoretic techniques to derive trade-offs between computational complexity and the Price of Defense for these classes. Some of the techniques touch upon classical milestones of Graph Theory; for example, we derive the first game-theoretic characterization of König-Egerváry graphs as graphs admitting a Matching Nash equilibrium.

Appendix: Proof of Lemma  11.5

Set:

• $V^{\prime }:=\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$,

• $E^{\prime }: =\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ and $r: =d_{G(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v)$ for an arbitrary vertex $v\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$.

By Lemma 11.3, $V^{\prime }$, $E^{\prime }$ and r satisfy Conditions (1) and (2) in the characterization of Defender-Uniform Graphs (Theorem 11.1). We first prove:

Lemma A.1

$\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})=E$.

Proof

Assume, by way of contradiction, that $\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}) \subset E$. Then, there is an edge $(v_{i},v_{i+1}) \not \in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$, for some i, $1\le i \le n$. Since a Nash equilibrium is a Covering profile (Proposition 7.1), Condition (1) in the definition of a Covering profile implies that $\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ is an Edge Cover. Since n is odd, this implies that $\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ contains two adjacent edges. Assume, without loss of generality, that both edges $(v_n,v_1),(v_1,v_2)\in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$. Thus, $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v_1)=2$. On the other hand, since $\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ is an Edge Cover and $(v_{i},v_{i+1}) \not \in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$, it follows that $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v_i)= 1$. We continue to prove:

Claim A.2

For any vertex $v\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$, $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v)= 1$.

Proof

Since $\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ is an Edge Cover, $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v)\ge 1$. Assume, by way of contradiction, that $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v)>1$. It follows that $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v)> d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v_i)$. Conditions (1/a) and (1/b) in the characterization of Defender-Uniform Graphs (Theorem 11.1) together imply that $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v_i)\ge d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v)$. A contradiction. $\square$

${\textit{Proof that}}\; v_1\not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$ By Claim A.2, for any vertex $v\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$, $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v)$ $=1$. Since $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v_1) =2$, it follows that $v_1\not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$.

${\textit{Proof that}}\;v_n, v_2\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$ Since a Nash equilibrium is a Covering profile (Proposition 7.1), Condition (2) in the definition of a Covering profile implies that $\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$ is a Vertex Cover of the graph $\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))$. Since $v_1 \not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$ while both $(v_n,v_1),$ $(v_1,v_2)\in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$, this implies that both $v_n, v_2\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$.

We continue to prove:

Claim A.3

Consider any vertex $v_i$, $i\ge 2$ such that $v_i\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$, $( v_{i-1},v_i )\in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ and $v_{i-1}\not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$. Then, the following conditions hold: (1) $(v_i, v_{i+1})\not \in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$, (2) $v_{i+1}\not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$, (3) $(v_{i+1}, v_{i+2})\in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$, and (4) $v_{i+2} \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$.

An illustration of Claim A.3 is shown in Fig. 8.

Fig. 8

An illustration of the sets $\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$ and $\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ in a Defender-Uniform Nash equilibrium for the odd cycle graph $\mathcal{C}_n$, where n is an odd integer. The edges in $\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ are shown with bold edges and the vertices in $\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$ are shown with surrounding circles

Proof

We separately prove each of the four claims.

(1): Since $v_i\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$, by Claim A.2, $d_{\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))}(v_i) =1$. Since $(v_i,v_{i-1})$ $\in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$, it follows that $(v_{i},v_{i+1})$ $\not \in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$.

(2): Assume, by way of contradiction, that $v_{i+1}\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$. Then, $\mathsf{A_{\varvec{\sigma }}}(v_{i+1})>0$, so that

$$\begin{aligned} \mathsf{A_{\varvec{\sigma }}}(v_{i},v_{i+1}) &> \mathsf{A_{\varvec{\sigma }}}(v_i) \\ &=\mathsf{A_{\varvec{\sigma }}}(v_{i-1},v_i)\quad({\text {since}}\;v_{i-1}\not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})). \end{aligned}$$

Since $(v_i,v_{i-1})\in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$, Condition (C.2) in the characterization of Nash equilibria (Proposition 3.2) implies that $\mathsf{A_{\varvec{\sigma }}}(v_{i},v_{i-1})\ge \mathsf{A_{\varvec{\sigma }}}(v_i,v_{i+1})$. A contradiction.

(3): Recall that $\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ is an Edge Cover of $\mathcal{C}_n$. Consider vertex $v_{i+1}$; since $(v_{i},v_{i+1}) \not \in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$, it follows that $(v_{i+1},v_{i+2}) \in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$.

(4): Recall that $\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$ is a Vertex Cover of the graph $\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))$. Since $(v_{i+1},v_{i+2}) \in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ while $v_{i+1} \not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$, it follows that $v_{i+2}\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$.

The claim follows. $\square$

Consider now vertex $v_2\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$. By Claim A.3, $(v_1,v_2) \in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$ but $v_1\not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$. Hence, Claim A.3 applies repeatedly (in a clockwise fashion) to yield that $\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}}) =\left\{ v_2,v_4,\right.$ $\left. \ldots , v_{n-1},v_1 \right\}$; thus, $v_1\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$. A contradiction. $\square$

We continue to prove:

Lemma A.4

$\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})=V$.

Proof

Since a Nash equilibrium is a Covering profile (Proposition 7.1), Condition (2) in the definition of a Covering profile implies that $\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$ is a Vertex Cover of the graph $\mathcal{C}_n(\mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d}))$. Since n is odd, this implies that $\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$ contains at least two adjacent vertices of $\mathcal{C}_n$. We prove:

Claim A.5

Assume that vertices $v_{i}, v_{i+1}\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$. Then, $v_{i+2}\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$.

Proof

Assume, by way of contradiction, that $v_{i+2}\not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})$. Then,

$$\begin{aligned} \mathsf{A_{\varvec{\sigma }}}(v_i,v_{i+1})& = \mathsf{A_{\varvec{\sigma }}}(v_i) + \mathsf{A_{\varvec{\sigma }}}(v_{i+1})\\> & {} \mathsf{A_{\varvec{\sigma }}}(v_{i+1}) \quad ({\text {since}}\;v_i\in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}}))\\& = \mathsf{A_{\varvec{\sigma }}}(v_{i+1},v_{i+2}) \quad ({\text {since}}\;v_{i+2}\not \in \mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})) \end{aligned}$$

Lemma A.1 implies that both edges $(v_{i},v_{i+1}), (v_{i+1},v_{i+2}) \in \mathsf{Supp_{\varvec{\sigma }}}(\mathsf{d})$. Hence, by Condition (C.2) in the characterization of Nash equilibria (Proposition 3.2), $\mathsf{A_{\varvec{\sigma }}}(v_{i},v_{i+1})= \mathsf{A_{\varvec{\sigma }}}(v_{i+1},v_{i+2})$. A contradiction. $\square$

By repeated application of Claim A.5, it follows that $\mathsf{Supp_{\varvec{\sigma }}}({{\mathcal {A}}})= V$. $\square$

By Lemmas A.1 and A.4, the claim follows. $\square$