Switching the Shannon Switching Game

Transcription

1 Switching the Shannon Switching Game A Senior Project sbmitted to The Diision of Science, Mathematics, and Compting of Bard College by Kimberly Wood Annandale-on-Hdson, New York May, 2012

2 Abstract The Shannon switching game is a combinatorial game for two players, which we refer to as the cop and the robber. In this project, we explore a few ariations of the original rles that make the game more interesting. One of these ariations is a game inoling mltiple cops and one robber. We present a formal recrsie definition of this game which we se to proe seeral basic theoretical reslts. Next, we consider this game on complete graphs and complete bipartite graphs. On each family of graphs, we inestigate the winning conditions for the players depending on who goes first. We describe these conditions as fnctions and proe seeral asymptotic reslts. Finally, after looking at the ariation with mltiple cops and one robber, we also stdy the game with mltiple cops and mltiple robbers and compare the two ariations.

3 Contents Abstract 1 Dedication 4 Acknowledgments 5 1 What is the Shannon Switching Game? Introdction Rles of the Game Types of Games N-Cop Game Introdction Deletion-Marking s. Deletion-Contraction Game Formal Definitions and Propositions Families of Graphs Complete Graphs Complete Bipartite Graphs N-Cop, N-Robber Game 49 Bibliography 56

4 List of Figres Example of a Game Example of a graph that is a Positie Game Positie Game Example of a graph that is a Negatie Game Example of a graph that is a Netral Game Netral Game where cop plays first Netral game where robber plays first Cops Complete Graph K K 3,3 with and on the same side and opposite side K 4,4 is a positie game Illstration of Theorem Illstration of Theorem

5 Dedication I wold like to dedicate this project to Bard College as withot this instittion I wold not hae been gien the opportnity to write a senior project.

6 Acknowledgments I wold like to thank my senior project adiser Jim Belk for his gidance throghot my project. He has been tremendosly helpfl and I am ery gratefl. Jim has been the best adiser a stdent cold ask for and has made this year-long jorney edcational and worth while. I wold also like to acknowledge Maria Belk and Sam Hsiao for agreeing to be on my senior project board. I wold like to thank Sam for his sggestions in or midway board meeting to inclde basic definitions and propositions abot the game. This trly contribted to my nderstanding of the project and sered to be ery important. All three professors hae taght me in arios classes and I wold like to thank them for their dedication and contribtion to my academic life. I hae the tmost respect for them.

7 1 What is the Shannon Switching Game? 1.1 Introdction Combinatorial game theory is an important branch of mathematics which combines the stdy of combinatorics, game theory and graph theory. A combinatorial game is a game with two players each haing separate trns and making different moes to achiee a winning positie. The players hae alternate trns with perfect information sch that the moes of each player are known by both players. Mathematicians stdy the pre strategies for each player inoled in a combinatorial game. They also stdy how these strategies affect the otcome of the game. They aim to explore sitations sch as: gien the two players play perfectly, who wins the game and how does the sitation define the reslt. Combinatorial games are so interesting as they proide the opportnity to explore seeral cases and qestions within the respectie game being stdied. One of the most notable books on combinatorial game theory is Winning Ways for yor Mathematical Plays [1]. The book incldes mathematical strategies and proofs for seeral combinatorial games therefore sering as a resorce for scholars to explore these games in greater depth.

8 1. WHAT IS THE SHANNON SWITCHING GAME? 7 The Shannon switching game is a combinatorial game inented by Clade Shannon and its soltion fond in 1964 by Alfred Lehman [3]. This game has been preiosly stdied by many mathematicians for its intrinsic interest and relationship to graph theory and matroid theory. As the game is played on a graph, we can stdy the positions of the two players while inestigating the game on different graphs. The game has also been generalized to a game on matroids and this was initially done by Alfred Lehman in his breakthrogh with the game [5]. Matroid theory and graph theory are also connected as the winning condition for one of the players inoles the presence of disjoint spanning trees within a graph which Lehman also proed sing matroids in his soltion [5]. The Shannon switching game is also closely related to the game of Hex sch that instead of sing edges, ertices are sed in each moe instead [2]. In 1976 [2], it was proen that this ersion of the Shannon switching game related to Hex is PSPACE-complete and the same was conclded in 1981 [2] for Hex as well. Therefore there is a ast pool of knowledge and mastery from mathematicians of this game. The game inoles two players that we will call the cop and the robber and is played on a graph. The Shannon switching game has been classified in a set of games called Maker-Breaker games [4]; which are games where the goal of one player, the robber, is to create a specific sbgraph and the goal of the other player, the cop, is to preent this from happening. The robber wants to mark a path while the cop wants to destroy it. In this project, we will be inestigating ariations of the original rles which make the game more exploratory. In Section 1.2 we will proide a description of the game and its rles as well as the types of games that exist. We will also show examples for the types of games and among these is a winning strategy for the robber mentioned earlier inoling disjoint spanning trees [6]. Thogh the game is traditionally played with only two players, we will stdy a game with mltiple cops and one robber. We will refer to this game as the n-cop game. Section 2.1

9 1. WHAT IS THE SHANNON SWITCHING GAME? 8 of Chapter 2 begins with an intitie definition of this ariation. We will proide a formal definition sing recrsion of the n-cop game which we will se to proe seeral fndamental properties. The game has been formally described with the method that the robber claims edges by marking them on his trn and the cop deletes edges on his trn. We will define this game as the Deletion-Marking Game. Howeer, we will also define a method inoling the robber contracting edges and the cop deleting edges. We will refer to this game as the Deletion-Contraction game. We will inclde in this chapter an illstration showing that these two methods reslt in the same game. As the game can be played on any graph, we will restrict or attention to specific families of graphs. As the n-cop game is a new ariation to the Shannon switching game, before it is possible to proe any generalization, we mst first focs on one class of graphs. This is becase the task of discoering winning strategies for the n-cop game on an arbitrary graph wold be ery difficlt if no test stdies were done. Therefore in Chapter 3 we will explore the n-cop game on complete graphs in Section 3.1. Or aim will be to research the winning conditions for the players and the relationship between different starting positions. We will describe these conditions sing fnctions and we will proe different reslts. After exploring the game on complete graphs, we will frther inestigate the n-cop game played on complete bipartite graphs in Section 3.2. Or goal here is similar as in the case of complete graphs. Howeer, or reslts will differ as complete bipartite graphs are also niqe. At the end of each section in Chapter 3, we will conclde it with obserations and qestions related to the reslts obtained and conjectres that we beliee to be tre. Chapter 4 inestigates a game played on complete bipartite graphs where there are mltiple cops and mltiple robbers of the same nmber. We will define this game as the n- cop, n-robber game and aim to find reslts for this as well. We will also compare the reslts fond in this chapter to those fond when playing the n-cop game. How adantageos is it to hae more robbers?

10 1. WHAT IS THE SHANNON SWITCHING GAME? Rles of the Game The Shannon switching game has two players, the cop and the robber, and is played on a graph. Gien an ndirected graph G = (V, E) with two distingished ertices and, the objectie of the robber is to mark a path from ertex to ertex. The objectie of the cop is to preent the robber from escaping by deleting edges to disconnect the robber s paths, prohibiting the robber from getting from to. The two players alternate trns and either player may begin the game. The cop can delete any edge as long as the robber has not preiosly claimed it. On the first trn, sppose that the cop plays first. He may delete any edge of his choosing. The robber plays next, marking his edge with red. The game ends once the robber has marked a complete path from to and therefore escapes or the cop sccessflly blocks all paths between and. w w w w x x x x (a) Graph (b) Cop s first trn w (c) Robber s first trn w (d) Cop s second trn w x (e) Robber s second trn x (f) Cop s third trn x (g) Robber s third trn Figre Example of a Game Figre is an example of a game where the cop has played first. It shows the seqence of trns. Let s now go throgh these plays from the figre to frther explain how the game works.

11 1. WHAT IS THE SHANNON SWITCHING GAME? 10 (a) The original graph we are playing on. (b) The cop has the first trn and chooses to delete the edge between ertices and. This moe is necessary for the cop to make in order to preent the robber from winning. (c) The robber now plays second and chooses to mark the edge between ertices and w. (d) The cop now contines on his trn and chooses to delete the edge between ertices w and. The cop is forced to delete this edge as otherwise the robber can win on his next trn. (e) The robber now proceeds and chooses to mark the edge between ertices x and. (f) The cop is again forced to delete the edge between ertices and x to preent the robber from winning. (g) The robber now makes the final trn of the game and marks the edge between ertices w and x. Now that the robber has sccessflly marked a path from ertex to ertex, he has won the game. 1.3 Types of Games On certain graphs, the robber will always be able to make an escape regardless of who plays first. Shown in Figre is a graph sch as this where the robber will always win een when the cop has the first trn. This is called a positie game [6]. Figre is an example of a positie game. There are two dashed edges marked in red shown in the graph. These two edges are between and so therefore there are two direct paths between and. If the cop goes first, he may only delete one of these paths therefore leaing the second path for the robber to mark on his trn. This is an example

12 1. WHAT IS THE SHANNON SWITCHING GAME? 11 Figre Example of a graph that is a Positie Game of a positie game as it shows that the robber can win een when the cop has the first trn. Definition A spanning tree in a graph G is a a selection of edges in G where there are no cycles and the edges are all connected. We say two spanning trees are disjoint when they share no common edges. Theorem Disjoint Spanning Tree Theorem A game is positie if and only if there is a sbgraph containing ertices and that has two disjoint spanning trees [3]. The Disjoint Spanning Tree Theorem is basically a strategy for the robber to se to garantee his escape when playing on a positie game. Let s now proe this theorem and therefore show the robber s strategy. We will show this by sing the following figre as or graph where the ble and orange edges each represent the two disjoint spanning trees. Figre Positie Game

13 1. WHAT IS THE SHANNON SWITCHING GAME? 12 Example The following illstrates the Spanning Tree Theorem We will show sing these two spanning trees and the following strategy that this is an example of a positie game and that the strategy works for positie games. If the robber can win playing second, he can also win playing first, (later proen in Proposition 2.3.4) so it is only necessary to show that the robber can win when the cop goes first. Therefore, the cop will always make the first moe in this strategy. Cop s First Trn: From Figre 1.3.2, the cop shold obiosly delete the direct path between and, edge b. b Robber s First Trn: The robber may now choose any edge a, except the edge that the cop has preiosly deleted: a Now the two spanning trees hae one edge in common, a. We hae made this edge green to make it clear that this edge is now part of both spanning trees. Cop s Second Trn: Now the cop may choose any edge b aailable to delete: b

14 1. WHAT IS THE SHANNON SWITCHING GAME? 13 Robber s Second Trn: Now the robber mst choose an edge a. The edge he chooses mst complete both spanning trees, which are now altered by the deletion of the edge b, so that a is now part of both spanning trees. We hae again changed its color to green to show this: a This strategy contines throgh the end of the game, garanteeing that there is always a path between and. Below are the rest of the steps, shown throgh the graphs: b a b a

15 1. WHAT IS THE SHANNON SWITCHING GAME? 14 Ths we hae shown the algorithm that garantees the robber a sccessfl escape in a positie game. On other graphs, the cop will always be able to preent the robber from escaping regardless of who plays first, shown in Figre A graph sch as this where the cop wins regardless of who plays first is called a negatie game [6]. Figre Example of a graph that is a Negatie Game Figre is an example of a negatie game. Notice that there are two edges in the graph highlighted in ble. If the cop deletes any of these edges the graph is disconnected and therefore the robber cannot escape. The cop s objectie is to disconnect the graph, which wold destroy all paths between and. Therefore if the robber goes first, the cop will disconnect the graph on his second trn. If the cop plays first, he will immediately disconnect the graph ths making this a negatie game. On other graphs, the winner of the game is determined by who plays first. A graph sch as this where either the cop or the robber wins contingently is called a netral game [6]. Figre Example of a graph that is a Netral Game

16 1. WHAT IS THE SHANNON SWITCHING GAME? 15 Figre illstrates an example for a netral game where the robber is trying to mark a path between and. Let s now show why this is a netral game by playing it with the cop starting and then playing it with the robber starting. Figre Netral Game where cop plays first Figre shows the graph with a highlighted ble edge. If the cop plays first, he shold delete this edge and therefore disconnect the graph. Ths the cop wins playing first. Figre now shows the seqence of trns made by each player with the robber starting. The edges hae been labeled,..., z and the robber is trying to get from to. Here we are asserting that this is a netral game. Other cases to consider for the cop s moes in each trn are not explicitly shown. Let s now go throgh these trns from the figre to frther explain. (a) The robber on his first trn marks the edge between ertices and z. This is a necessary moe as otherwise the cop can disconnect the graph on his following trn. (b) The cop then chooses to delete the edge between ertices and y. There are other moes to consider howeer they do not increase his chances of winning. (c) The robber now plays second and chooses to mark the edge between ertices w and x.

17 1. WHAT IS THE SHANNON SWITCHING GAME? 16 w x y w x y w x y w x y z z z z (a) Robber s first trn (b) Cop s first trn (c) Robber s second trn (d) Cop s second trn w x y w x y w x y z z z (e) Robber s third trn (f) Cop s third trn (g) Robber s final trn Figre Netral game where robber plays first (d) The cop now contines on his trn and chooses to delete the edge between ertices w and. (e) The robber now proceeds and chooses to mark the edge between ertices x and. (f) The cop is now forced to delete two edges bt only has one moe so he deletes edge between ertices x and z. (g) The robber now makes the final trn of the game and marks the edge between ertices w and z. Now that the robber has sccessflly marked a path from ertex to ertex, the robber has won the game. Formal definitions of the types of game follow in the next chapter.

18 2 N-Cop Game 2.1 Introdction The Shannon switching game is traditionally played with only two distingished players: one robber and one cop. In this chapter, we introdce a ariation of the game inoling mltiples cops and one robber. We refer to the mltiple cops as n cops. Haing n cops intitiely means that one of the players has n moes per trn. Therefore when the n cops play, it is eqialent to one cop haing n moes and therefore deleting n edges on each trn. Therefore in the n-cop game, the two players alternate trns, howeer on the cop s trn, he has n cops and therefore n moes. The n cops in this sense therefore moe collectiely together to delete a total of n edges. The robber proceeds as before haing only one moe per trn. Therefore the ariation is consistent with the framework of combinatorial games of haing two players. With this ariation, a positie game is now a game where the robber still wins regardless of whether the n cops play first. In a negatie game, the n cops are able to win regardless of if the robber starts. Netral games remain the same sch that the winner is determined by whether the n cops start or the robber starts. Let s show an example of a game with

19 2. N-COP GAME 18 two cops and one robber. We will play the game on the complete graph K 5 with the 2 cops playing first. For the prposes of explaining the game, the ertices hae been labelled throgh y where the robber is trying to get from ertex to ertex. y w x y y y w x w x w x y y y w x w x w x y w x Figre Cops Figre shows the game played on K 5. On the cops first moe they delete the edge between ertices and as otherwise the robber wins. They also delete the edge between

20 2. N-COP GAME 19 ertices and y. Next, the robber marks with red the edge between ertices and w. The trns by the players contine ntil the final moe shown in the last graph where the cops hae deleted the edge between ertices and x. Here the cops did not een hae to se their second moe in order to disconnect the graph and therefore win the game. 2.2 Deletion-Marking s. Deletion-Contraction Game The Shannon switching game is normally defined with the robber s moes as marking edges and the cop s moes as deleting edges. When playing the game howeer, deleting and marking edges for each player is eqialent to deleting and contracting edges also. Therefore, when referring to the play of each player, either game can be sed. The deletionmarking game is a game where the robber s moes are marking edges and the cop s moes are deleting edges. The deletion-contraction game is a game where the robber s moes inole contracting edges and the cop s moes inole deleting edges. We will show in this section why the two ways are the same. Let s look at the complete graph K 4. We will se this graph to illstrate that the two games reslt in the same otcome. The first figre shows the complete graph K 4. The following figres show graph 1 and graph 2 where the deletion-marking game is played on graph 1 (g1) while simltaneosly the deletion-contraction game is played on graph 2 (g2). w Complete Graph K 4 : We will show the two ways by playing on the complete graph K 4 x

21 2. N-COP GAME 20 w w Cop s first moe: On g1 and g2 the cop has chosen to delete the edge between ertices and as otherwise the robber will win. x x x x x g1 g1 g1 g1 g1 w w w w x g2 g2 g2 g2 g2 w w Robber s first moe: On g1, the robber has chosen to mark with red the edges between ertices and x. On g2, the robber has chosen to contract the same edge between ertices and x reslting in the contracted graph. Cop s second moe: On g1, the cop is forced to delete the edge between ertices x and otherwise the robber will win. On g2, the cop deletes the edge between ertices and as otherwise the robber will win too. Robber s second moe: On g1 the robber has chosen to mark with red the edge between ertices w and. On g2 the robber has chosen to contact the same edge reslting in the contracted graph. Cop s third moe: On g1 in order to preent the robber from creating a path, the cop needs to delete two edges. The edge between ertices x and w and the edge between ertices and w howeer he only has one moe and so chooses to delete the edge between ertices x and w. The same goes for g2 where two edges between ertices and need to be deleted howeer the cop can only delete one.

22 2. N-COP GAME 21 x g1 w = g2 Robber s third moe: On g2 the robber has chosen to mark with red the last remaining edge between ertices and w creating a direct path between ertices and therefore winning the game. On g2 the robber chose to contract the last remaining edge reslting in =. Therefore the robber has also won go g Formal Definitions and Propositions In this section we proide a formal recrsie definition of the n-cop game and proe seeral basic theoretical propositions. In order for s to proe or reslts obtained in chapters to come, we mst first hae a strctred definition to reference. We se a recrsie definition as we are defining the game for a fixed nmber of cop size n. Therefore it is sefl as it can be applied repeatedly to all terms of this seqence of n for all n N. Earlier in or introdction we intitiely discssed how the game works. Here we will define the game sing ordered triples. Formal definitions are ery sefl here becase in order for s to proe these propositions, we first need to hae definitions that can proide clarity and strctre. This formality will also help s to nderstand other theorems in Chapter 3. The types of games hae been originally described as positie, negatie and netral games. In this section howeer, we will formally define positie and non-negatie games. Intitiely the types of games are as follows: A non-negatie game is a game where the robber can win when the robber starts. A positie game is a game where the robber can win when the cops start. A negatie game is a game where the cop can win when the robber starts. A non-positie game is a game where the cop can win when the cop starts.

23 2. N-COP GAME 22 A netral game is a game that is both non-negatie and non-positie. NOTE: In the following definitions and proofs, we will consider a fixed nmber of n cops and one robber for all games mentioned. Therefore it can be assmed that we are referring to n cops and one robber. Definition A game is an ordered triple (G,, ) where G is a graph with edge set E(G) and ertex set V (G) sch that, V (G). By conention the game (G,, ) is eqal to the game (G,, ). In or definition that follows, the notation (G {e 1, e 2...e n }) means that a distinct set of n edges are deleted from G. The notation G/e means that the edge e is contracted in G. Definition Let (G,, ) be a game. If = then the game is always positie and non-negatie. If then we define positie and non-negatie games recrsiely as follows: (G,, ) is a positie game if G has at least n edges and (G {e 1, e 2...e n },, ) is a non-negatie game for all distinct edges e 1, e 2... e n E(G). (G,, ) is a non-negatie game if there exists an edge e E(G) sch that the graph (G/e,, ) is a positie game. Note that if G has zero edges and then the game is neither positie nor nonnegatie. Also, if G has fewer than n edges and then the game cannot be positie. Proposition Let (G,, ) be a game. Let H be a sbgraph of G containing ertices and. 1. If (H,, ) is a non-negatie game, then (G,, ) is a non-negatie game. 2. If (H,, ) is a positie game, then (G,, ) is a positie game.

24 2. N-COP GAME 23 Proof. We will proe this by indction on the nmber of edges. Sppose it holds tre for G with fewer than m edges. We need to show it holds tre for m edges. If = then it follows that G is both positie and non-negatie by definition. Therefore we can assme for a total of m edges. For the first statement in the proposition, sppose (H,, ) is a non-negatie game. Then by definition there exists an edge e E(H) sch that (H/e,, ) is a positie game. Howeer we know that the graph G/e has m 1 edges and H/e is a sbgraph of G/e. So therefore by or indction hypothesis (G/e,, ) is a positie game. Therefore by definition since (G/e,, ) is a positie game then (G,, ) is a non-negatie game. For the second statement in the proposition, sppose (H,, ) is a positie game. Then by definition (H {e 1, e 2,..., e n },, ) is non-negatie for all e 1, e 2,..., e n E(H). Now let e 1, e 2,..., e n E(G). Therefore (G {e 1, e 2,..., e n }) has m n edges. Sppose first that e 1, e 2,..., e n E(H). Since we know (H {e 1, e 2,..., e n }) is a sbgraph of (G {e 1, e 2,..., e n }) then by or indction hypothesis (G {e 1, e 2,..., e n },, ) is a nonnegatie game. Therefore by definition since (G {e 1, e 2,..., e n },, ) is a non-negatie game then (G,, ) is a positie game. Sppose now that {e 1, e 2,..., e n } are not all in E(H). This means that the edges chosen by the cop to delete are not all in the sbgraph H. Withot loss of generality, we may assme that e 1,..., e p E(H) and e p+1,..., e n / E(H) for some p sch that 0 < p n. Let d 1,..., d n p E(H) be different from e 1,..., e p E(H). These d 1... d n p edges represent dmmy moes made by the cop. Since (H,, ) is positie then (H {e 1,... e p, d 1,... d n p },, ) is a non-negatie game. NOTE: We know that (H,, ) is positie therefore by definition, so H has at least n edges. Since (H {e 1,... e p, d 1,..., d n p } (G {e 1..., e n } then this implies that (G {e 1,..., e n },, ) is also non-negatie and therefore (G,, ) is a positie game.

25 2. N-COP GAME 24 Proposition Let (G,, ) be a game with n > 1 cops and one robber. If (G,, ) is a positie game, then (G,, ) is a non-negatie game. This proposition is saying that if the robber can win playing second, then the robber can win playing first. Proof. Sppose (G,, ) is a positie game. If = then by definition, (G,, ) is a non-negatie game. Sppose. Assming (G,, ) is positie then we know it has at least n edges. Let d 1, d 2,..., d n be dmmy moes sch that d 1, d 2... d n E(G). We also know that (G {d 1, d 2,..., d n }) is a sbgraph of G. Since (G,, ) is positie then this implies (G {d 1, d 2,..., d n },, ) is non-negatie. Therefore by the first statement of Proposition 2.3.3, (G,, ) is a non-negatie game. Proposition Let (G,, ) be a game with n > 1 cops and one robber. 1. If (G,, ) is a positie game with n + 1 cops, then (G,, ) is a positie game with n cops. 2. If (G,, ) is a non-negatie game with n + 1 cops, then (G,, ) is a non-negatie game with n cops. P roof. We will proe this by indction on the nmber of edges. Sppose it holds tre for G with fewer than m edges. We need to show it holds tre for m edges. If = then it follows that G is both positie and non-negatie by definition. Therefore we can assme for a total of m edges. For the first statement of the proposition, sppose (G,, ) is positie with n + 1 cops then by definition (G {e 1, e 2,..., e n+1 },, ) is a non-negatie game for all e 1, e 2,..., e n+1 E(G). Let e 1, e 2,..., e n E(G). Since (G,, ) is positie with n + 1 cops then we know it has at least n + 1 edges. Sppose there exists d 1 sch that d 1 is a dmmy moe for d 1 E(G). Let the robber choose d 1 and e 1, e 2,..., e n. Therefore

26 2. N-COP GAME 25 (G {e 1, e 2,..., e n+1 + d 1 } is a sbgraph of G. Therefore (G {e 1, e 2,..., e n+1 + d 1 },, ) is a non-negatie game. As the dmmy moe is a fake moe, this is eqialent to saying that (G {e 1, e 2,..., e n },, ) is non-negatie for n cops. Therefore (G,, ) is a positie game fore n cops. For the second statement of the proposition, sppose (G,, ) is non-negatie with n+1 cops then by definition there exists an edge e sch that the graph (G/e,, ) is a positie game with n + 1 cops. Therefore by or indction hypothesis, since (G/e,, ) is positie with n+1 cops then it follows that (G/e,, ) is also positie for n cops. Therefore (G,, ) is a non-negatie game with n cops. Proposition For a gien graph (G,, ), if there exists a direct edge between and then the game is non-negatie. Proof. If a game is non-negatie then this means the robber can win when the robber starts. Ths on the robber s first trn if there is a direct edge between and, the robber can mark this edge and therefore win the game.

27 3 Families of Graphs 3.1 Complete Graphs The Shannon switching game can be played on any graph G. There are many families of graphs. Ths in order for s to learn how the n-cop game is related to graphs, we mst start with applying it to a specific class of graphs. In this section we therefore explore the n-cop game played on complete graphs. The reason we need to specialize to complete graphs is de to the fact that we cannot yet make any conjectres regarding the n-cop game on an arbitrary graph. If we were to choose any arbitrary graph to stdy this game, it wold be ery challenging to proe anything abot it. Therefore in order to generalize this ariation to all graphs if possible, we mst first start ot small with test cases and proe what we can find abot the game on complete graphs before inestigating a different class of graphs. Stdying the n-cop game on different types of graphs will contribte towards nderstanding how the n-cop game relates to all graphs. Definition Let n N. A complete graph K n is a graph where there are n ertices and eery ertex is connected to eery other ertex in the graph.

28 3. FAMILIES OF GRAPHS 27 Figre Complete Graph K 5 NOTE: For all the games played on complete graphs in this section, we will assme throghot that. Therefore from Proposition 2.3.6, this indicates that eery complete graph is a non-negatie game. Since we assme then for all complete graphs, de to atomorphism as the graph has symmetry, the position of and does not affect the game. Definition Let φ: N N be the fnction defined by: φ(n) = min{m K m is positie with n cops and one robber} We will denote each edge of a graph as a two element set {a, b} sch that the edge is between ertex a and ertex b. Theorem φ(1) = 4. Proof. In order for s to illstrate this Theorem, we need to show that K 4 is a positie game and K 3 is a non-positie game. The following seqence of graphs illstrate that K 4 is a positie game. The ertices are labelled 1, 2, 3, 4. The robber is trying to get from 1 to 4. As we are trying to show that this is a positie game, we mst show that the robber wins when the cop starts.

29 3. FAMILIES OF GRAPHS Complete Graph K Cop s first trn: The cop is forced to delete {1, 4} to preent the robber from winning Robber s first trn: The robber shold mark {2, 4} Cop s second trn: The cop is forced to delete {1, 2} Robber s second trn: The robber shold mark {1, 3} Cop s third trn: The cop is forced to delete two edges: {2, 3} and {3, 4} bt only has one moe. Therefore the robber wins on his next trn. 4 3 We now need to show that K 3 is a non-positie game. This means that the cop can win if the cop starts. The following seqence of graphs illstrate this game. The goal of the robber is to get from ertex 1 to ertex 3.

30 3. FAMILIES OF GRAPHS 29 1 Complete Graph K Cop s first moe: The cop shold delete {1, 3} to preent the robber from winning Ths we hae shown that φ(1) = 4. Robber s first moe: The robber needs to mark two edges: {1, 2} and {2, 3} bt only has one moe. Therefore the cop will disconnect the graph on his next trn and win the game. Theorem K n+3 is a non-positie game for n cops and one robber for n 2. Proof. Let the ertices of K n+3 be labelled 1,..., n + 3. Sppose the robber is trying to get from 1 to n + 3. We mst show that the cops win when they start. Cops First Trn: The cops shold delete the edge {1, n + 3}. With the remaining n 1 moes, the cops shold delete {2, n + 3},...,{n, n + 3}. Robber s First Trn: The robber is forced to mark one of the two remaining edges connected to ertex n + 3, say {n + 1, n + 3}. De to symmetry it does not make a difference if the robber chose the other edge. Cops Second Trn: The cops are forced to delete the edge between ertex 1 and the ertex the robber preiosly sed so {1, {n + 1} With the n 1 moes remaining, the cops shold delete n 1 edges connected to ertex 1 so {1, 2},...,{1, n}. Robber s Second Trn: The robber is forced to mark the last remaining edge connected to ertex 1, say {1, n + 2}. Cops Third Trn: The cops are forced to delete the edge between the two marked edges done by the robber so {n + 1, n + 2}. The cops are also forced to delete the edge

31 3. FAMILIES OF GRAPHS 30 between {n + 3} and the ertex the robber jst sed so {n + 2, n + 3}. With the remaining n 2 moes, the cops shold delete n 2 edges connected to the ertex the robber sed to connect to 1. So therefore {n + 2, 2},...,{n + 2, n 1}. Robber s Third Trn: The robber is forced to mark the last remaining edge connected to the ertex he connected to 1 so {n + 2, n}. Cops Forth Trn: The cops shold now delete the n 1 edges remaining that are connected to the ertex the robber preiosly sed. Therefore the cops are deleting n 1 edges connected to ertex n. The cops hae now disconnected the graph and therefore win. Ths K n+3 is non-positie. Corollary φ(n) > n + 3. Theorem K m is a positie game for m = 2n 2 + n + 1 with n cops and one robber. Proof. By Theorem we know that φ(1) = 4 so we can assme here that n 2. Let the ertices of K m be labelled 1,..., m. Withot loss of generality, sppose the robber is trying to get from ertex 1 to ertex 2. We mst show that the robber wins when the cops start. We claim the robber wins on his n + 2 trn. Cops First Trn: On eery trn, the cops will be deleting a total of n edges which ses a total of 2n ertices. On his first moe he is forced to delete edge {1, 2} to preent the robber from winning. The cops next moes will inole deleting edges sing most 2n 2 nsed ertices. Robber s First Trn: The robber shold now mark an edge between ertex 2 and an nsed ertex. Withot loos of generality, we can assme {2, 3}. Cops Second Trn: The cops are now forced to delete {1, 3}. With the cops s remaining n 1 moes, they will se at most 2n 2 nsed ertices.

32 3. FAMILIES OF GRAPHS 31 The plays contine in this fashion ntil the end of the cops n th trn. The seqence of plays contine for the robber where he is to mark an edge between his preiosly sed ertex to another nsed ertex. Therefore withot loss of generality, the robber is marking {k + 1, k + 2} on his k th trn. This contines ntil he has marked a path of n 1 edges say 2,..., n and has made n 1 trns. The cops also contine their seqence of plays and on the k th trn, are forced to delete {1, k + 1}. The cops then contine to delete edges sing at most 2n 2 nsed ertices. After the cops hae made their n th trn let s cont the nmber of ertices that hae been sed p this far. ertex 1 and 2 in the cops first trn 2n 2 ertices in the cops first trn after the first trn, the cops hae n 1 trns and deleted at most 2n 2 nsed ertices on each trn so sed a total of (n 1) (2n 2) ertices n 1 ertices from the robber This gies a total of 2 + (2n 2) + (n 1) (2n 2) + (n 1) = 2n 2 n + 1 sed ertices ths far. Therefore there remain (2n 2 + n + 1) (2n 2 n + 1) = 2n nsed ertices. Robber s nth Trn: The robber shold now mark an edge between his preiosly sed ertex from his n 1 trn to one of the 2n nsed ertices, leaing 2n 1 nsed ertices. Therefore the robber now has a path of (n 1) + 1 = n edges. Cops n+1 Trn: The cops are now forced to delete the edge between ertex 1 and the preiosly sed ertex by the robber to preent the robber from winning. With the cops s remaining n 1 moes, they will se at most 2n 2 nsed ertices. Robber s n+1 Trn: There now remain (2n1) (2n 2) = 1 nsed ertex. The robber shold mark the edge between ertex 1 and this nsed ertex.

33 3. FAMILIES OF GRAPHS 32 Cops n+2 Trn: The cops are now forced to delete n + 1 edges bt only hae n moes. Therefore the robber wins on his n + 2 trn. Ths K m is a positie game for m = 2n 2 + n + 1. Corollary φ(n) 2n 2 + n + 1. From Corollary and Corollary 3.1.7, we can obsere the following ineqalities: n + 4 φ(n) 2n 2 + n + 1. These ineqalities proide lower and pper bonds for φ(n). The following table shows nmerical ales for n + 4 and 2n 2 + n + 1. n n n 2 + n The ales indicate that φ(n) is somewhere between a wide range of ales as n gets larger. Let s look at the same table bt for larger ales of n. n n n 2 + n ,101 80, , ,401 As we can see the larger the ale of n, the wider the range for φ(n) becomes. We can also obsere that φ(n) lies in the range between a linear eqation and a qadratic eqation. I beliee φ(n) is a qadratic eqation for larger ales of n. De to the fact that n + 3 is the ale for which the game is non-positie and n + 4 is ery close to this ale it gies frther reason to beliee that φ(n) is closer to being qadratic especially for larger ales of n. It also seems as thogh for larger integer ales of n, as the seqence goes from n, the ale of n does not depend on any nmber of finite nmbers in the seqence. Conjectre φ(2) 9 and φ(3) 14.

34 3. FAMILIES OF GRAPHS 33 The following are some qestions that can be later on explored: 1. Is φ(n) linear or qadratic? For what ales of n does φ(n) become qadratic? 2. What is the relationship, if any, between φ(n) and the Disjoint Spanning Tree Theorem 1.3.2? Does φ(n) contain three or more disjoint spanning trees and how does this affect the robber s strategy? 3.2 Complete Bipartite Graphs After playing on complete graphs and trying to find ales for φ(n), it seemed time to consider a different family of graphs to inestigate. Complete bipartite graphs are similar to complete graphs so therefore I chose to inestigate the n-cop game on complete bipartite graphs. Definition Let n N. A complete bipartite graph K m,n is a bipartite graph with edges in two disjoint sets say A and B. All the ertices in A are connected to all the ertices in B howeer the ertices within the same set are not connected. Here we only consider complete bipartite graphs when m = n. Gien that the robber s goal is to get from ertex to ertex and, we may consider two cases: 1. and are in the same set so, A or, B (same side) 2. and are in different sets so A and B (opposite side) Definition Let α, β, γ : N N be the fnctions defined by: α(n) = min{m K m,m is positie with n cops and, on opposite side} β(n) = min{m K m,m is positie with n cops and, on same side} γ(n) = min{m K m,m is non-negatie with n cops and, on same side}

35 3. FAMILIES OF GRAPHS 34 Figre K 3,3 with and on the same side and opposite side NOTE: All graphs K m,m with, on opposite sides are non-negatie by Proposition as there will be a direct edge between the two ertices regardless of where, are positioned. Therefore when inestigating these games, the cops shold always start first as otherwise the robber wins. Theorem α(1) = 4. Proof. For s to proe that α(1) = 4 we mst show that K 4,4 is positie and K 3,3 is non-positie for, on opposite sides. The following seqence of graphs illstrates that K 4,4 is a positie game sing the deletion-contraction method. We mst show that the robber wins when the cop starts. The robber is trying to get from ertex 1 to ertex 4. So in this example ertices 1 = and 4 =. 1 1' Complete Bipartite Graph K 4,4 2 2' 3 3' 4 4' 1 1' 2 2' 3 3' Cop s first trn: The cop is forced to delete the edge {1,4 } as otherwise the robber can contract this edge and win the game. 4 4' 1/1' 2 2' 3 3' 4 4' Robber s first trn: The robber shold choose to contract an edge between ertices 1 and an nsed edge in B say {1, 1 }. Now ertex 1 becomes 1/1.

36 3. FAMILIES OF GRAPHS 35 1/1' 2 2' Cop s second trn: Cop chooses to delete edge {4, 3 } 3 3' 4 4' 1/1'/2' 2 3 3' 4 4' 1/1'/2' 2 Robber s second trn: The robber shold choose to contract an edge between ertices 1 and an nsed edge in B say {1, 2 }. Now ertex 1/1 becomes 1/1 /2 Cop s third trn: The cop chooses to delete edge {3, 4 } 3 3' 4 4' 1/1'/2' 2 3 3' Robber s third trn: The robber shold now contract an edge between ertices 4 and any nsed ertex in A say {4, 4 }. Now ertex 4 becomes 4/4. 4/4' 1/1'/2' 2 3 3' 4/4' 1=4' 2 3 3' Cop s forth trn: At this point it is clear the robber will win on his next trn becase the cop needs to delete two edges bt only has one moe so chooses to delete edge {1, 4 }. Robber s forth trn: The robber now wins by contracting the edge between ertices 1 and 4 ths 1 = 4. We now need to show that K 3,3 is a non-positie game. We mst therefore show that the cop wins then the cop starts. Assme the robber is trying to get from 1 to ' Complete Bipartite Graph K 3,3 2 2' 3 3'

37 3. FAMILIES OF GRAPHS ' 2 2' Cop s first trn: The cop is forced to delete {1, 3 }. 3 3' 1/1' 2 2' 3 3' g1 1/1' 2 2' 3 3' 1 1' 2/2' 3 3' g2 1 1' 2/2' 3 3' Robber s first trn: Withot loss of generality we can assme the robber contracts either {1, 1 } or {2, 2 } (de to symmetry any other moe is eqialent to one of these two moes). Graph g1 shows the graph for the first case and g2 shows the graph for the second case. Cop s second trn: In either case the cop shold now delete an edge connected to 3 so {3, 3 } on both g1 and g2. g1 g2 1/1' 2/3' 2' 3 g1 1 2/2'/3' 3 g2 1' Robber s second trn: On both g1 and g2, the robber is forced to contract {2, 3 } as it is the only remaining edge connected to 3. 1/1' 2/3' 2' 1 2/2'/3' 1' Cop s third trn: In either case, the cop is now forced to delete {2, 1 } 3 3 g1 g2 1/1' 2/3' 3 g1 2' 1 2/2'/3' 3 g2 1' Robber s third trn: The robber only has one edge aailable to contract {2, 2 } on g1 and {2, 1 } on g2. Howeer on the cop s next trn he will disconnect the graph by deleting {1, 2 } on g1 and {1, 1 } on g2 and therefore the cop wins the game. Theorem β(1) = 4. Proof. For s to proe that β(1) = 4 we mst show that K 4,4 is positie and K 3,3 is non-positie for, on the same side. The following graphs show the graph K 4,4. By

38 3. FAMILIES OF GRAPHS 37 Theorem 1.3.2, to show that K 4,4 is positie, me mst show two disjoint spanning trees. Assme the robber is trying to get from 1 to ' ' 3' 4' 1 1' 1 1' 2 2' 3 3' 2 3 2' 3' 4 4 Figre K 4,4 is a positie game Now we mst show that K 3,3 is a non-positie game with, on the same side. We mst show that the cop wins when the cop starts. Let the set A = {1, 2, 3} and the set B = {1, 2, 3 }. The robber is trying to get from 1 to 3. Trn Cop Robber First {1, 1 } {2, 2 } Second {1, 2 } {1, 3 } Third {3, 3 } {2, 3 } Forth {3, 2 } {3, 1 } Trn Cop Robber First {1, 1 } {3, 3 } Second {3, 1} {1, 2 } Third {2, 3} {2, 2 } Forth {2, 3 } robber loses The tables aboe show the seqence of moes for two different games played on K 3,3. De to symmetry of the graph, withot loss of generality the robber on his first trn may choose to mark {3, 3 } or {2, 2 }. We will only be describing the moes for the first table. Howeer, the reader is inited to play the game following the seqence in table 2 on their own. On the first trn the cop shold delete {1, 1 }. The robber then marks {2, 2 }. The cop on his second trn shold delete {1, 2 } forcing the robber to mark {1, 3 } as this is the

39 3. FAMILIES OF GRAPHS 38 only remaining edge connected to 1. The cop on his third trn shold delete {3, 3 } forcing the robber to mark {3, 2} as otherwise the cop on his next trn can delete it disconnecting the graph. On the forth trn, the cop shold delete {3, 2 } forcing the robber to mark {3, 1 } as this is the only remaining edge connected to 3. On the cop s final trn, we shold delete {2, 1 } therefore winning the game. Theorem γ(1) = 3. Proof. Assme we are playing on K 3,3. We mst show that the robber wins when the robber starts. Let the set A = {1, 2, 3} and the set B = {1, 2, 3 }. The robber is trying to get from 1 to 3. Trn Robber Cop First {1, 1 } {1, 3} Second {3, 3 } {3, 1} Third {3, 2} {2, 1 } Forth {1, 2 } {2, 2 } {3, 2 } On the first trn the robber shold mark {1, 1 } forcing the cop to delete {1, 3}. The robber on his second trn shold then mark {3, 3 } forcing the cop to delete {3, 1}. The robber on his third trn shold mark {3, 2} forcing the cop to delete {2, 1 }. On the forth trn, the robber shold mark {1, 2 }. The cop needs to delete {2, 2 } and {3, 2 } bt only has one moe so therefore the robber wins on his next trn. Now we mst show that K 2,2 is a negatie game therefore the cop wins when the robber starts. Howeer, K 2,2 is obiosly a negatie game as there exists two edges that once deleted, disconnect the graph sch as in Example Theorem K m,m is a positie game when m = (n + 1) 2 for n cops with, on opposite sides.

40 3. FAMILIES OF GRAPHS 39 Proof. By Theorem we know that α(1) = 4 so we can assme here that n 2. We mst show that the robber wins when the cops start. Let s assme the set A = {1,..., m} and the set B = {1,..., m }. The robber is trying to get from ertex 1 to ertex m. The game begins with n 2 + 2n + 1 nsed ertices in each ertex set A and B respectiely. We claim that the robber wins on his n + 3 trn. Cops First Trn: The cops are forced to delete {1, m } to preent the robber from winning. The cops can choose to delete any n 1 edges and will se at most n 1 nsed ertices from each set. Robber s First Trn: The robber shold choose to mark an edge between ertex 1 and any nsed ertex in B. Withot loss of generality we can assme {1, 1 }. Cops Second Trn: The cops can choose to delete any n edges and will se at most n nsed ertices from each set. Robber s Second Trn: The robber shold now choose to mark an edge between ertex 1 and any nsed ertex in B. Withot loss of generality we can assme {1, 2 }. Cops Third Trn: The cops can choose to delete any n edges and will se at most n nsed ertices from each set. The plays contine in this fashion ntil the end of the cops n + 1 trn. This seqence of plays contine for the robber where he is to mark an edge between ertex 1 and an nsed ertex in B. This contines ntil the robber has marked n edges from 1 to B. Therefore withot loss of generality we can assme his last marked edge in this seqence of plays is {1, n }. The cops also contine their seqence of plays ntil their n + 1 trn and will delete at most n nsed ertices. After the cops hae made their n + 1 trn let s cont the nmber of ertices that hae been sed p ths far in A and B respectiely: Set A is as follows:

41 3. FAMILIES OF GRAPHS 40 ertex 1 from the cop s first moe n 1 ertices from the cop s first moe after the first trn, the cops at this point hae n trns and deleted at most n nsed ertices on each trn so sed a total of n 2 ertices This gies a total of 1 + (n 1) + n 2 = n 2 + n sed ertices ths far in A. Set B is as follows: ertex m from the cop s first moe n 1 ertices from the cop s first moe n ertices from the robber after the first trn, the cops at this point hae n trns and deleted at most n nsed ertices on each trn so sed a total of n 2 ertices This gies a total of 1 + (n 1) + n + n 2 = n 2 + 2n sed ertices ths far in B. The game started with a total of n 2 + 2n + 1 ertices. Therefore after the cops n + 1 trn there remain (n 2 + 2n + 1) (n 2 + n) = n + 1 nsed ertices in A and (n 2 + 2n + 1) (n 2 + 2n) = 1 nsed ertex in B. Robber s n+1 Trn: The robber shold now mark the edge between 1 and the one remaining ertex in B. Let s refer to this ertex as nsed. Therefore the robber marks {1, nsed} Cops n+2 Trn: There now remain no nsed ertices in B ths with the cops n moes they will delete at most n nsed ertices in A. Robber s n+2 Trn: There is now at least (n + 1) n = 1 nsed ertex remaining in A. The robber shold mark the edge between m and this one remaining nsed ertex in

42 3. FAMILIES OF GRAPHS 41 A. Withot loss of generality we can assme this nsed ertex is m. Therefore the robber marks {m, m }. Now on the cops n + 3 trn, they are forced to delete n + 1 edges connected from m to B bt they only hae n moes. Therefore the robber wins on his n+3 trn. Figre is an illstration of how this proof works. The robber s n + 2 trn is shown as the dotted red lines from m to m. As yo can see, the cop needs to delete {m, [{1,..., n } + nsed]} bt only has n moes. Therefore the robber wins. n{ 2' n' 1 1' nsed m m' Figre Illstration of Theorem Corollary α(n) (n + 1) 2. Theorem K m,m is a positie game when m = (n + 1) 2 for n cops with, on the same side. Proof. By Theorem we know that β(1) = 4 so we can assme here that n 2. We mst show that the robber wins when the cops start. Let s assme A = {1,..., m} and B = {1,..., m }. The robber is trying to get from ertex 1 to ertex m. The game begins with n 2 + 2n + 1 nsed ertices in each ertex set A and B respectiely. The robber wins on his n + 4 trn.

43 3. FAMILIES OF GRAPHS 42 Cops First Trn: The cops can choose to delete any n edges and will se at most n nsed ertices from each set. Robber s First Trn: The robber shold choose to mark an edge between ertex 1 and any nsed ertex in B. Withot loss of generality we can assme {1, 1 }. Cops Second Trn: The cops are now forced to delete the edge {1, m} to preent the robber from winning. The cops can choose to delete any n 1 edges and will se at most n 1 nsed ertices from each set. Robber s Second Trn: The robber shold now choose to mark an edge between ertex 1 and any nsed ertex in B. Withot loss of generality we can assme {1, 2 }. Cops Third Trn: The cops are now forced to delete {2, m} to preent the robber from winning. The cops can choose to delete any n 1 edges and will se at most n 1 nsed ertices from each set. The plays contine in this fashion ntil the end of the cops n + 2 trn.this seqence of plays contine for the robber where he is to mark an edge between ertex 1 and an nsed ertex in B. This contines ntil the robber has marked n + 1 edges from 1 to B. Therefore withot loss of generality we can assme his last marked edge in this seqence of plays is {1, (n + 1) }. The cops also contine their seqence of plays ntil their n + 2 trn where they are forced to delete {1, (n + 1) } and at most n 1 nsed ertices. After the cops hae made their n + 2 trn let s cont the nmber of ertices that hae been sed p ths far in A and B respectiely: Set A is as follows: ertex 1 and m n ertices from the cops first trn