Saturation Games for Odd Cycles

Abstract

Given a family of graphs F , we consider the F -saturation game. In this game, two players alternate adding edges to an initially empty graph on n vertices, with the only constraint being that neither player can add an edge that creates a subgraph that lies in F . The game ends when no more edges can be added to the graph. One of the players wishes to end the game as quickly as possible, while the other wishes to prolong the game. We let satg (F ; n) denote the number of edges that are in the final graph when both players play optimally.The {C 3 }-saturation game was the first saturation game to be considered, but as of now the order of magnitude of satg({C 3 }, n) remains unknown. We consider a variation of this game. Let C 2k+1 := {C 3 , C 5 , . . . , C 2k+1 }. We prove that satg (C 2k+1 ; n) ≥ ( 1 4 − ǫ k )n 2 + o(n 2 ) for all k ≥ 2 and that satg (C 2k+1 ; n) ≤ ( 1 4 − ǫ ′ k )n 2 + o(n 2 ) for all k ≥ 4, with ǫ k < 1 4 and ǫ ′ k > 0 constants tending to 0 as k → ∞. In addition to this we prove satg({C 2k+1 }; n) ≤ 4 27 n 2 + o(n 2 ) for all k ≥ 2, and satg (C∞ \ C 3 ; n) ≤ 2n − 2, where C∞ denotes the set of all odd cycles.

Linear Bounds for Cycle-free Saturation Games

“…Many of the results in this paper and in [12] focused on families of odd cycles. This is because in theory the game saturation number of a family of odd cycles could be anywhere between linear and quadratic.…”

“…It is well known that ex(n, C 2k ) = o(n 2 ) for all k, so by (1) a necessary condition to have sat g (n, C) = Θ(n 2 ) is that C consists only of odd cycles. The last author [12] showed that a sufficient condition for a set of odd cycles C to have quadratic game-saturation number is to have C 3 , C 5 ∈ C, in which case we have sat g (n, C) ≥ 6 25 n 2 + o(n 2 ). Carraher, Kinnersley, Reiniger, and West [2] showed that sat g (n, C o ) = n 2 /4 where C o is the set of all odd cycles [2], though the last author [12] showed that in general a set of odd cycles containing C 3 and C 5 need not have game saturation number asymptotic to 1 4 n 2 .…”

“…The last author [12] showed that a sufficient condition for a set of odd cycles C to have quadratic game-saturation number is to have C 3 , C 5 ∈ C, in which case we have sat g (n, C) ≥ 6 25 n 2 + o(n 2 ). Carraher, Kinnersley, Reiniger, and West [2] showed that sat g (n, C o ) = n 2 /4 where C o is the set of all odd cycles [2], though the last author [12] showed that in general a set of odd cycles containing C 3 and C 5 need not have game saturation number asymptotic to 1 4 n 2 . Similarly one could ask for necessary and sufficient conditions for a family of cycles C to have linear game saturation number, and for this problem much less is known.…”

Linear Bounds for Cycle-Free Saturation Games

“…It is well known that ex(n, C 2k ) = o(n 2 ) for all k, so by (1) a necessary condition to have sat g (n, C) = Θ(n 2 ) is that C consists only of odd cycles. The last author [13] showed that a sufficient condition for a set of odd cycles C to have quadratic game-saturation number is to have C 3 , C 5 ∈ C, in which case we have sat g (n, C) " 6 25 n 2 + o(n 2 ). Carraher, Kinnersley, Reiniger, and West [2] showed that sat g (n, C o ) = ⌊n 2 /4⌋ where C o is the set of all odd cycles [2], though the last author [13] showed that in general a set of odd cycles C containing C 3 and C 5 need not have sat g (n, C) = (1 + o(1)) 1 4 n 2 .…”

“…The last author [13] showed that a sufficient condition for a set of odd cycles C to have quadratic game-saturation number is to have C 3 , C 5 ∈ C, in which case we have sat g (n, C) " 6 25 n 2 + o(n 2 ). Carraher, Kinnersley, Reiniger, and West [2] showed that sat g (n, C o ) = ⌊n 2 /4⌋ where C o is the set of all odd cycles [2], though the last author [13] showed that in general a set of odd cycles C containing C 3 and C 5 need not have sat g (n, C) = (1 + o(1)) 1 4 n 2 . It is straightforward to show that sat(n, C) = Ω(n) for any set of cycles C, so it is natural to ask for necessary and sufficient conditions for sat g (n, C) to be linear.…”