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Deterministic Security Games in Software Connect USS Code 128 in Software Deterministic Security Games




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3. Deterministic Security Games generate, create code128 none on software projects GS1 Glossary Proof. The Code 128B for None existence of subgame-perfect equilibria in the revocation game can be obtained by applying the technique of backward induction, which is also called Zermelo s algorithm in dynamic programming. Backward induction works by eliminating sub-optimal actions, i.

e., yielding higher costs than the other actions in the same subtree and at the same stage of the game tree, beginning at the leaves of the extensive-form game tree. The obtained path (sequence of actions) in the game tree de nes the backward induction solution and any strategy pro le that realizes this solution is a subgameperfect equilibrium [67].

Here, the one-stage-deviation principle is used to prove that deviating from each of the strategies in the theorem under the corresponding conditions will not result in a gain. First, assume that c < v, i.e.

, voting is more expensive than enduring the attackinduced cost. If at any stage, player i deviates from the strategy A, then playing V or S would result in a cost of v or 1, respectively. In both cases, the cost is higher than c since v < 1.

If v < c < 1 and ni nr , i.e., voting is less expensive than the attack-induced cost and the number of remaining players is higher than the required number of voters, then playing S or V would result in a cost of 1 or v, respectively.

These costs are greater than 0 and the attacker will be revoked anyway since v < c. Hence, the player i cannot gain by deviating from the action A. Another case that makes action A the best response is when the attack-induced cost c is bigger than 1, the cost of self-sacri ce, and the number of remaining players is higher than 1 (ni 1), i.

e. the attacker will be revoked by another player anyway. The proof is similar to the previous cases.

Next, assume that v < c < 1, ni = nr 1, and player i that is supposed to play V according to strategy Ai above, deviates in a single stage. If it plays S or A, it loses 1 or c, respectively, both bigger than v. In both cases, the player cannot gain by deviating from V .

Finally, if c > 1, ni = 0, and the player deviates from S by playing A or V , then the player s cost will be c or c + v, respectively. Both costs are greater than 1 and deviation results in a loss for the player. Thus, a deviation from any action Ai under the corresponding conditions results in a loss for the deviating player i which completes the proof that the strategy Ai leads to a subgame-perfect equilibrium.

Theorem 3.7 essentially states that, since the attacker cost is xed, the only objective of the players is to remove the attacker, regardless of the actual game stage of this happening. Thus, the revocation decision is left to the last players, either by voting or by self-sacri ce, whichever induces less cost.

For example, a node plays S only if the attack-induced cost is higher than the cost of self-sacri ce and if it is the last player in the sequential game. The solution in Theorem 3.7 is not robust to changes in the system.

For example, if some of the last players move out of the system before their turn to play, then a revocation decision cannot be reached. To overcome this limitation, a variable cost version of the revocation game is de ned next..

Deterministic Security Games De ne a se Code 128 Code Set C for None quential game with variable costs where c j = j , where 1 j < n is the stage of the game and > 0 is the stage cost of attack. Let the attack cost at the nal stage of the revocation game grow in nitely, cn = , if the attacker is not revoked. Furthermore, assume that v < .

The subgame-perfect equilibrium strategies in the de ned variable cost sequential game are characterized in the following theorem. Theorem 3.8.

In the sequential revocation game with variable costs, for any given values of ni , nr , v < , and , the strategy of player i that results in a subgame-perfect equilibrium is: 1 A, if [(1 ni < min{nr 1, }) and (v + (nr 1) < 1)] 1 or [(1 ni < ) and (v + (nr 1) > 1)] Ai = V, if [(ni nr 1) and (v + (nr 1) < 1)] S, otherwise. Proof. The proof follows an argument similar to the one of Theorem 3.7.

If there are 1 not enough voters but at least another player can self-sacri ce, 1 ni < min{nr 1, } and self-sacri ce is more expensive than voting, v + (nr 1) < 1, then deviating from playing A has a cost of 1. This is due the fact that playing S is the only possible option in this case as the number of voters is insuf cient. Hence, the player cannot gain by a deviation from action A.

1 If 1 ni < and v + (nr 1) > 1, deviating from playing A will cause player i a cost of 1 if it plays S and a cost of v + (nr 1) if it plays V . Hence, the player again does not gain by deviating from A. If there are enough voters, ni nr 1, and voting is less expensive than self-sacri ce, v + (nr 1) < 1, then deviating from playing V by choosing A or S costs the player ni or 1, respectively.

In both cases, the cost will be greater than v + (nr 1) , assuming v is negligible. Hence, the player does not gain by one-stage deviation. The explicit condition for playing S is: > 1 or 1 )) and (v + (nr 1) < 1) or 1 (ni = 0) (ni )) ( < 1 < v + (nr 1) ) (ni < nr 1) and ((ni = 0) or (ni If the attack-induced cost is more expensive than self-sacri ce, > 1, and player i deviates from strategy S, it can play V or A.

If it plays V , it loses v + (nr 1) > 1 and if it plays A, it loses ni > 1. 1 If the current player is the last one, ni = 0, or ni (alternately ni 1, which means that the cost of abstaining is higher than the cost of self-sacri ce), and voting is cheaper than self-sacri ce but there are not enough voters, v + (nr 1) < 1 and 1 ni < nr 1, deviating from S by playing A would result in a cost of ni if ni or if ni = 0 as the attacker will not be revoked. Playing V would not lead to revocation (as 1 there are not enough voters) and hence result in a cost of v + ni if ni or if ni = 0.

Hence, deviation from S would not pay off. A similar argument can be made if ni = 0.
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