Scaling-Up Stackelberg Security Games Applications Using Approximations

Abstract

Stackelberg Security Games (SSGs) have been adopted widely for modeling adversarial interactions, wherein scalability of equilibrium computation is an important research problem. While prior research has made progress with regards to scalability, many real world problems cannot be solved satisfactorily yet as per current requirements; these include the deployed federal air marshals (FAMS) application and the threat screening (TSG) problem at airports. We initiate a principled study of approximations in zero-sum SSGs. Our contribution includes the following: (1) a unified model of SSGs called adversarial randomized allocation (ARA) games, (2) hardness of approximation for zero-sum ARA, as well as for the FAMS and TSG sub-problems, (3) an approximation framework for zero-sum ARA with instantiations for FAMS and TSG using intelligent heuristics, and (4) experiments demonstrating the significant 1000x improvement in runtime with an acceptable loss.

Appendix

Implementability: Viewing SSGs as ARAs provides an easy way of determining implementability using results from randomized allocation [7]. First, we define bi-hierarchical assignment constraints as those that can be partitioned into two sets $H_1, H_2$ such that two constraints $S, S'$ in the same partition ($H_1$ or $H_2$) it is the case that either $S \subseteq S'$ or $S' \subseteq S$ or $S \cap S' = \phi$. Further, as defined in [7], canonical assignment constraints are those that impose constraints on all rows and columns of the matrix. We obtain the following result

Proposition 1

All marginal strategies are implementable, or more formally $conv(P) = MgS$, if the assignment constraints are bi-hierarchical. Given canonical assignment constraints, if all marginal strategies are implementable then the assignment constraints are bi-hierarchical.

As Fig. 1 reveals, both FAMS and TSG have non-implementable marginals due to overlapping constraints. The proof of the proposition is straightforward applications of Theorems 1 and 2 in Budish et al. [7].

Fig. 6.

RAND modified heuristic comparison

Modified Heuristic is Bad: The modified RAND approach is compared to RAND in Fig. 6. It can be seen that the loss increases a lot with almost 35% loss over RAND for 110 flights.

Proof of Theorem 1: First we define some problems related to the DB problem.

• DBR is the problem $\max _{\mathbf {x} \in P} \mathbf {d} \cdot \mathbf {x}$ where $\mathbf {d}$ is a vector of positive constants. DBR is a combinatorial problem.

• The continuous version of DBR is DBR-C: $\max _{\mathbf {x} \in conv(P)} \mathbf {d} \cdot \mathbf {x}$.

• The unweighted version of the DBR is DBR-U: $\max _{\mathbf {x} \in P} \mathbf {1} \cdot \mathbf {x}$.

Proof

For the first part, given a NP hard DBR-U instance (for the decision version of DBR-U), we construct an ARA instance such that the feasibility problem for that ARA instance solves the hard DBR-U decision problem. Thus, as the feasibility is NP Hard, there exists no approximation. First, since the ARA problem is so general there exists DBR-U problems that are NP Hard. For example, the DBR-U problem for FAMS has been shown to be NP Hard [23]. Given the hard DBR-U problem, form an ARA problem with by adding the constraint $\mathbf {1} \cdot \mathbf {x} = k$. Also, let there be only one target t in the problem, so that the objective becomes $U(\mathbf {x}, t)$ instead of z and all constraints in the optimization are just the marginal space constraints and $\mathbf {1} \cdot \mathbf {x} = k$. Now, the existence of any solution of the optimization gives a feasible point $\mathbf {x} = \sum _m a_m \mathbf {P_m}$, where $\mathbf {P_m} \in P$ is integral. Also, it must be that $\mathbf {1} \cdot \mathbf {P_j} \ge \mathbf {1} \cdot \mathbf {x} = k$ for some j. Then, $\mathbf {P_j}$ is a solution to the decision version of the DBR-U problem, i.e., does there exist a solution of the DBR-U optimization problem with value $\ge k$? Thus, since finding the existence of any solution for ARA is NP Hard, thus, no approximation exists in poly time.

For the second part, we present a AP approximation preserving reduction (with problem mapping that does not depend on approximation ratio); such a reduction preserves membership in PTAS, APX, log-APX, Poly-APX (see [1]). Given any DBR problem, we construct the ARA problem with one target such that $T = \{1, \ldots , k\}\times \{1, \ldots , n\}$. Choose the weights $w_{i,j}$’s such that $w_{i,j} \propto d_{i,j}$ and $w_{i,j} \le 1/\max _{\mathbf {x} \in MgS} \sum _{i,j} x_{i,j}$. Observe that $\max _{\mathbf {x} \in MgS} \sum _{i,j} x_{i,j}$ is computable efficiently and $\max _{\mathbf {x} \in MgS} \sum _{i,j} x_{i,j} \ge \max _{\mathbf {x} \in conv(P)} \sum _{i,j} x_{i,j}$, thus, the ARA is well-defined. Thus, due to just one target, the ARA optimization is same as $\max _{\mathbf {x} \in conv(P)} \mathbf {w} \cdot \mathbf {x}$. Suppose we can solve this problem with r approximation with the solution mixed strategy being $\mathbf {x}^\epsilon = \sum _{i=1}^m a_i \mathbf {P_i}$ for some pure strategies $\mathbf {P_i}$. Now, since $w_{i,j} \propto d_{i,j}$ we also know that this solution also provides r approximation for DBR-C. Let the optimal solution for DBR-C be OPT; note that OPT is also the optimal solution for DBR. $\mathbf {x}^\epsilon$ provides a solution value $\mathbf {w} \cdot \mathbf {x}^\epsilon \ge OPT/r$. Further, as the objective is linear in $\mathbf {x}$ and $\mathbf {x}^\epsilon = \sum _{i=1}^m a_i \mathbf {P_i}$, it must be the case that there exists a $j \in \{1, \ldots , m\}$ such that $\mathbf {w} \cdot \mathbf {P_j} \ge \mathbf {w} \cdot \mathbf {x}^\epsilon \ge OPT/r$. Thus, since $\mathbf {P_j} \in P$, $\mathbf {P_j}$ provides r approximation for DBR. Since, m the number of the pure strategies in support of $\mathbf {x}^\epsilon$ is polynomial, $\mathbf {P_j}$ can be found in polynomial time by a linear search.

Proof of Theorem 2.

Proof

Given an independent set problem with V vertices, we construct a TSG with $\{1, \ldots , V + 1\}$ team types, where each team type in $1, \ldots , V$ corresponds to a vertex. The $V+1$ team is special in the sense that it does not correspond to any vertex and it is made up of just one resource with a very large resource capacity 2V. Construct just one passenger category with passengers $N = V+1$. Since, there is just one passenger category (and target) we will use $x_i$ as the matrix entries instead of $x_{i,j}$. Choose $U^t_s = V+1$ and $U^t_u = 0$ and efficiencies $E_i = 1$ for all teams, except $E_{V+1} = 0$. Then, the objective of the integer LP is $\sum _{i=1}^V x_i = \mathbf {1}_V \cdot \mathbf {x}$ where $\mathbf {1}_V$ is a vector with first V components as 1 and last component as 0.

Next, have resources for every edge $(i,k) \in E$ with resource capacity 1. This provides the inequality $\sum _{(i,k) \in E} x_i + x_j \le 1$. Also, we have $x_{V+1} \le 2V$. Inspection of every passengers provides the constraints $\sum _{i=1}^{V+1} x_i = V+1$. Treating $x_{V+1}$ as a slack, we can see that the constraint $x_{V+1} \le 2V$ and $\sum _{i=1}^{V+1} x_i = V+1$ are redundant. For the left over constraints $\sum _{(i,k) \in E} x_i + x_j \le 1$, we can easily check that any valid integral assignment (pure strategy) is an independent set. Moreover, the objective $\sum _{i=1}^V x_i$ tries to maximize the independent set. The optimal value of this optimization over conv(P) is an extreme point which is integral and equal to the maximum independent set OPT. Thus, suppose a solution $\mathbf {x}^\epsilon$ to the SSE problem with value $\ge OPT/r$. Further, as the objective is linear in $\mathbf {x}$ and $\mathbf {x}^\epsilon = \sum _{i=1}^m a_i \mathbf {P_i}$, it must be the case that there exists a $j \in \{1, \ldots , m\}$ such that $\mathbf {1}_V \cdot \mathbf {P_j} \ge \mathbf {1}_V \cdot \mathbf {x}^\epsilon \ge OPT/r$. Thus, since $\mathbf {P_j} \in P$, $\mathbf {P_j}$ provides r approximation for maximum independent set. Since, m the number of the pure strategies in support of $\mathbf {x}^\epsilon$ is polynomial, $\mathbf {P_j}$ can be found in poly time by a linear search.

Proof of Theorem 5.

Proof

Consider the event of a target t having an infeasible assignment after the comb sampling. Call this event $E_t$. Let $C_{t,i}$ be the event that resource i covers this target t. Then, $P(E_t) = \sum _{i} P(E_t|C_{t,i})P(C_{t,i})$. From the guarantees of comb sampling we know that $P(C_{t,i}) = \sum _{j: (i,j) \in T} x^m_{i,j} \le 1$ and $P(x_{i,j} = 1) = x^m_{i,j}$. Also, by comb sampling if $x_{i,j} = 1$ then $x_{i,j'} = 0$ for any $j'\ne j$. Next, we know that $P(E_t|C_{t,i})$ is the probability that the any of the other $x_{i',j}$ is assigned a one, which is $1 -$ the probability that all other $x_{i',j}$ are assigned 0. Thus,

$$P(E_t|C_{t,i}) = 1 - \prod _{i' \ne i} (1- P(C_{t,i})) $$

Let $p_{t,i} = P(C_{t,i})$. Considering the fact that $\prod _i (1 - p_{t,i}) > 1 - \sum _i p_{t,i}$, we get

$$1 - \prod _{i' \ne i} (1- P(C_{t,i})) \le \sum _{(i',j): i'\ne i \wedge (i',j) \in T} x^m_{i',j} \le 1 - \sum _j x^m_{i,j}$$

where the last inequality is due to the fact that $\sum _{(i,j) \in T} x^m_{i,j} \le 1$.

Thus, $P(E_t) \le \sum _{i} (1 - p_{t,i})p_{t,i} \le \sum _{i} p_{t,i} - \sum _{i} (p_{t,i})^2$. Next, we know from standard sum of squares inequality that $\sum _{i} (p_i)^2 \ge (\sum _{i} p_i)^2/k$. Thus, we get $P(E_t) \le (\sum _{i} p_i) (1 - \sum _{i} p_i/k)$ The RHS is maximized when $\sum _{i} p_i = 1$, thus, $P(E_t) \le 1 - 1/k$. Also, then $P(\lnot E_t) \ge 1/k$

Now consider the coverage of target t: $x^m_t = \sum _{(i,j) \in T} x^m_{i,j}$. According to our algorithm the allocation for target t continues to remain 1 with probability $(1/2)^C$ if its allocation is already feasible after comb sampling (and we always obtain a pure strategy). This is because this target shares schedules with C other targets and thus in the worst case may be reduced with 1/2 probability for each of the C targets. We do a worst case analysis and assume that no resource is allocated to a target when the sampled allocation is infeasible for that target. Thus, let $y_t$ denote the random variable denoting that target t is covered. Thus, $E(y_t) = P(y_t = 1) = P(y_t = 1|E_t) P(E_t) + P(y_t = 1|\lnot E_t)P(\lnot E_t)$. Now, $P(y_t = 1|\lnot E_t)$ is same as $x^m_t/2^C$ and we assumed the worst case of $P(y_t = 1|E_t) = 0$. Thus, we have $E(y_t) \ge x^m_t/2^Ck$. As the utilities are linear in $y_t$, we have the utility for t as $U_t \ge U_t^m/2^Ck$, where $U_t^m$ is the utility under the marginal $\mathbf {x}^m$. Thus, if $t^*$ is the choice of adversary under the marginal $\mathbf {x}^m$ we know that $U_{t^*}^m$ is the lowest utility for the defender over all targets t. Hence, we can conclude that the utility with the approximation is at least $U_{t^*}^m/2^Ck$

Proof of Theorem 4.

Proof

The main assumption in the proof is that the steps after comb sampling changes the probability of detecting an adversary in passenger category j by at most 1/c. Also, by assumption of the theorem since Algorithm 1 does not fail ever, the change in utility for any passenger category j is at most a factor of 1/c. By similar reasoning as for FAMS, we conclude that this provides a c-approximation.