{"title":"五常饱和游戏","authors":"Zhen He, Mei Lu","doi":"10.1007/s10255-024-1125-4","DOIUrl":null,"url":null,"abstract":"<p>Let <i>F, G</i> and <i>H</i> be three graphs with <i>G</i> ⊆ <i>H</i>. We call <i>G</i> an <i>F</i>-saturated graph relative to <i>H</i>, if there is no copy of <i>F</i> in <i>G</i> but there is a copy of <i>F</i> in <i>G</i> + <i>e</i> for any <i>e</i> ∈ <i>E</i>(<i>H</i>) <i>E</i>(<i>G</i>). The <i>F</i>-saturation game on host graph <i>H</i> consists of two players, named Max and Min, who alternately add edges of <i>H</i> to <i>G</i> such that each chosen edge avoids creating a copy of <i>F</i> in <i>G</i>, and the players continue to choose edges until <i>G</i> becomes <i>F</i>-saturated relative to <i>H</i>. Max wishes to maximize the length of the game, while Min wishes to minimize the process. Let sat<sub><i>g</i></sub>(<i>F, H</i>) (resp. sat′<sub><i>g</i></sub>(<i>F, H</i>)) denote the number of edges chosen when Max (resp. when Min) starts the game and both players play optimally. In this article, we show that sat<sub><i>g</i></sub>(<i>P</i><sub>5</sub>, <i>K</i><sub><i>n</i></sub>) = sat′<sub><i>g</i></sub>(<i>P</i><sub>5</sub>, <i>K</i><sub><i>n</i></sub>) = <i>n</i> + 2 for <i>n</i> ≥ 15, and sat<sub><i>g</i></sub>(<i>P</i><sub>5</sub>, <i>K</i><sub><i>m,n</i></sub>), sat′<sub><i>g</i></sub>(<i>P</i><sub>5</sub>, <i>K</i><sub><i>m,n</i></sub>) lie in <span>\\(\\left\\{ {m + n - \\left\\lfloor {{{m - 2} \\over 4}} \\right\\rfloor ,\\,m + n - \\left\\lceil {{{m - 3} \\over 4}} \\right\\rceil } \\right\\}\\)</span> if <span>\\(n \\ge {5 \\over 2}m\\)</span> and <i>m</i> ≥ 4, respectively.</p>","PeriodicalId":0,"journal":{"name":"","volume":null,"pages":null},"PeriodicalIF":0.0,"publicationDate":"2024-06-01","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":"{\"title\":\"The P5-saturation Game\",\"authors\":\"Zhen He, Mei Lu\",\"doi\":\"10.1007/s10255-024-1125-4\",\"DOIUrl\":null,\"url\":null,\"abstract\":\"<p>Let <i>F, G</i> and <i>H</i> be three graphs with <i>G</i> ⊆ <i>H</i>. We call <i>G</i> an <i>F</i>-saturated graph relative to <i>H</i>, if there is no copy of <i>F</i> in <i>G</i> but there is a copy of <i>F</i> in <i>G</i> + <i>e</i> for any <i>e</i> ∈ <i>E</i>(<i>H</i>) <i>E</i>(<i>G</i>). The <i>F</i>-saturation game on host graph <i>H</i> consists of two players, named Max and Min, who alternately add edges of <i>H</i> to <i>G</i> such that each chosen edge avoids creating a copy of <i>F</i> in <i>G</i>, and the players continue to choose edges until <i>G</i> becomes <i>F</i>-saturated relative to <i>H</i>. Max wishes to maximize the length of the game, while Min wishes to minimize the process. Let sat<sub><i>g</i></sub>(<i>F, H</i>) (resp. sat′<sub><i>g</i></sub>(<i>F, H</i>)) denote the number of edges chosen when Max (resp. when Min) starts the game and both players play optimally. In this article, we show that sat<sub><i>g</i></sub>(<i>P</i><sub>5</sub>, <i>K</i><sub><i>n</i></sub>) = sat′<sub><i>g</i></sub>(<i>P</i><sub>5</sub>, <i>K</i><sub><i>n</i></sub>) = <i>n</i> + 2 for <i>n</i> ≥ 15, and sat<sub><i>g</i></sub>(<i>P</i><sub>5</sub>, <i>K</i><sub><i>m,n</i></sub>), sat′<sub><i>g</i></sub>(<i>P</i><sub>5</sub>, <i>K</i><sub><i>m,n</i></sub>) lie in <span>\\\\(\\\\left\\\\{ {m + n - \\\\left\\\\lfloor {{{m - 2} \\\\over 4}} \\\\right\\\\rfloor ,\\\\,m + n - \\\\left\\\\lceil {{{m - 3} \\\\over 4}} \\\\right\\\\rceil } \\\\right\\\\}\\\\)</span> if <span>\\\\(n \\\\ge {5 \\\\over 2}m\\\\)</span> and <i>m</i> ≥ 4, respectively.</p>\",\"PeriodicalId\":0,\"journal\":{\"name\":\"\",\"volume\":null,\"pages\":null},\"PeriodicalIF\":0.0,\"publicationDate\":\"2024-06-01\",\"publicationTypes\":\"Journal Article\",\"fieldsOfStudy\":null,\"isOpenAccess\":false,\"openAccessPdf\":\"\",\"citationCount\":\"0\",\"resultStr\":null,\"platform\":\"Semanticscholar\",\"paperid\":null,\"PeriodicalName\":\"\",\"FirstCategoryId\":\"100\",\"ListUrlMain\":\"https://doi.org/10.1007/s10255-024-1125-4\",\"RegionNum\":0,\"RegionCategory\":null,\"ArticlePicture\":[],\"TitleCN\":null,\"AbstractTextCN\":null,\"PMCID\":null,\"EPubDate\":\"\",\"PubModel\":\"\",\"JCR\":\"\",\"JCRName\":\"\",\"Score\":null,\"Total\":0}","platform":"Semanticscholar","paperid":null,"PeriodicalName":"","FirstCategoryId":"100","ListUrlMain":"https://doi.org/10.1007/s10255-024-1125-4","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
引用次数: 0
摘要
假设 F、G 和 H 是三个图,其中 G ⊆ H。如果 G 中没有 F 的副本,但对于任意 e∈E(H) E(G),G + e 中有 F 的副本,则我们称 G 为相对于 H 的 F 饱和图。主图 H 上的 F 饱和博弈由名为 Max 和 Min 的两个玩家组成,他们交替将 H 的边添加到 G 中,使得所选的每条边都能避免在 G 中创建 F 的副本。让 satg(F,H)(或 sat′g(F,H))表示当 Max(或 Min)开始博弈且双方都以最优方式下棋时所选择的边的数量。在本文中,我们将证明当n≥15时,satg(P5,Kn)= sat′g(P5,Kn)= n + 2,并且satg(P5,Km,n), sat′g(P5,Km,n)位于({m + n - \left\lfloor {{m - 2} \over 4}}\right\rfloor ,\m + n -\left\lceil {{m - 3}\over 4}}\right\rceil }如果n(ge {5 \over 2}m\)和m≥4,就分别是right\rfloor ,\,n -\left\lceil {{m -3}\over 4} \right\rceil }。
Let F, G and H be three graphs with G ⊆ H. We call G an F-saturated graph relative to H, if there is no copy of F in G but there is a copy of F in G + e for any e ∈ E(H) E(G). The F-saturation game on host graph H consists of two players, named Max and Min, who alternately add edges of H to G such that each chosen edge avoids creating a copy of F in G, and the players continue to choose edges until G becomes F-saturated relative to H. Max wishes to maximize the length of the game, while Min wishes to minimize the process. Let satg(F, H) (resp. sat′g(F, H)) denote the number of edges chosen when Max (resp. when Min) starts the game and both players play optimally. In this article, we show that satg(P5, Kn) = sat′g(P5, Kn) = n + 2 for n ≥ 15, and satg(P5, Km,n), sat′g(P5, Km,n) lie in \(\left\{ {m + n - \left\lfloor {{{m - 2} \over 4}} \right\rfloor ,\,m + n - \left\lceil {{{m - 3} \over 4}} \right\rceil } \right\}\) if \(n \ge {5 \over 2}m\) and m ≥ 4, respectively.