Computing and Combinatorics: 19th International Conference, by Susanne Albers (auth.), Ding-Zhu Du, Guochuan Zhang (eds.)

By Susanne Albers (auth.), Ding-Zhu Du, Guochuan Zhang (eds.)

This ebook constitutes the refereed complaints of the nineteenth overseas convention on Computing and Combinatorics, COCOON 2013, held in Hangzhou, China, in June 2013. The fifty six revised complete papers provided have been conscientiously reviewed and chosen from a hundred and twenty submissions. there has been a co-organized workshop on discrete algorithms of which eight brief papers have been authorised and a workshop on computational social networks the place 12 papers out of 25 submissions have been accepted.

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Sample text

Motivated by the above reasons, in this paper we study the performance of Subgame Perfect Equilibria in sequential isolation games. More precisely, we focus on nearest-neighbor and total-distance sequential isolation games, and we show upper and lower bounds on the Sequential Price of Anarchy under the two social functions defined as the minimum utility per player (MIN) and the sum of the players’ utilities (SUM). , every Subgame Perfect Equilibrium is an optimal solution, both for nearest-neighbor and total-distance sequential isolation games, under the two considered social functions.

Proof. For k = 3, let I3 = (G, 3) be the instance of nearest-neighbor sequential isolation games yielded by the case in which G is the unweighted path v1 , v2 , . . , v5 . We show that there exists an SPE s such that a(s) = (v2 , v4 , v5 ) for which MIN(a(s)) = 1. Since the action profile oMIN (I3 ) = (v1 , v3 , v5 ) yields MIN(oMIN (I3 )) = 2, it then follows that SeqPoAMIN (I3 ) = 2. First of all, it is not difficult to see that, for any SPE s of I3 , it must be Ui (a(s )) ≤ 2 for any i ∈ [3]. Hence, since U1 (a(s)) = 2, it follows that player 1 is happy with her choice in s.

Player 2 is getting a utility of 1 in a(s). The only deviation that does not immediately give her a utility of 1 is to choose node v5 . In such a case, player 3 cannot get more that 1 and so we can correctly assume that she chooses node v4 which again gives a utility of 1 to player 2. So player 2 cannot improve her utility by deviating from s. Finally, player 3 is getting a utility of 1 in a(s). Since she cannot get more than 1 once that the first two players have chosen nodes v2 and v4 , it follows that player 3 cannot improve her utility by deviating from s and this shows that s is an SPE.

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