Incidence Labeling Games on Incidence-Symmetric Graphs

(2,∞)-INCIDENCE LABELING GAMES ON INCIDENCE-SYMMETRIC GRAPHS

MS. NATTHAWAN SRIPHONG

A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE (MATHEMATICS) DEPARTMENT OF MATHEMATICS AND STATISTICS FACULTY OF SCIENCE AND TECHNOLOGY THAMMASAT UNIVERSITY ACADEMIC YEAR 2018 COPYRIGHT OF THAMMASAT UNIVERSITY

Ref. code: 25616009031029DVT (2,∞)-INCIDENCE LABELING GAMES ON INCIDENCE-SYMMETRIC GRAPHS

MS. NATTHAWAN SRIPHONG

Ref. code: 25616009031029DVT

(1)

Thesis Title (2,∞)-INCIDENCE LABELING GAMES ON INCIDENCE-SYMMETRIC GRAPHS Author Ms. Natthawan Sriphong Degree Master of Science (Mathematics) Department/Faculty/University Mathematics and Statistics Faculty of Science and Technology Thammasat University Thesis Advisor Nantapath Trakultraipruk, Ph.D. Academic Year 2018

ABSTRACT

A(p,q)−incidence labeling game is a game of coloring on incidences of a graph G. There are two players, Alice and Bob (Alice begins). They alternately label an uncolored incidence of G with a color taken from a set of p colors, and each color can be used at most q times. Adjacent incidences must be labeled by diﬀerent colors. We determine a winning strategy of a (2,∞)−incidence labeling game on an incidence-symmetric graph.

Keywords : incidence labeling game, coloring game, incidence-symmetric graph

Ref. code: 25616009031029DVT (2)

ACKNOWLEDGEMENTS

First of all, I would like to specially thank my wonderful thesis advisor, Dr.Nantapath Trakultraipruk, for his direction, gentleness in giving advice, and support throughout my research. This helped me to complete this thesis.

I am also deeply thankful to my thesis committees, Associate Professor Dr.Wichai Witayakiattilerd, Assistant Professor Dr.Khajee Jantarakhajorn, and Assistant Professor Dr.Chanon Promsakon, who gave valuable comments.

I also recognize the ﬁnancial support of the Science Achievement Scholarship of Thailand. In addition, I would like to thank all the staﬀ members from the Department of Mathematics and Statistics, Faculty of Science and Technology, Thammasat University, and all of my friends for their friendship.

Finally, I would like to thank my parents for their love and support. This achievement would not have been possible without them.

Natthawan Sriphong

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LIST OF FIGURES

Figures Page

1.1 The graph G1 ...... 1 1.2 Pairs of adjacent incidences ...... 2 1.3 Pairs of nonadjacent incidences ...... 2

1.4 Graphs G2 and G3 ...... 2

1.5 A path P4 ...... 2

1.6 A cycle C5 ...... 3

1.7 A complete graph K6 ...... 3

1.8 A bipartite graph G4 ...... 3

1.9 The ladder graph L4 ...... 4

1.10 The crown graph C4,4 ...... 4

1.11 The friendship graph F4 ...... 5

3.1 The incidence-symmetric graph G5 ...... 7

3.2 The non incidence-symmetric graph G6 ...... 8

3.3 The path Pn ...... 9

3.4 The cycle Cn ...... 11

3.5 The complete graph Kn ...... 17

3.6 The ladder graph Ln ...... 22

3.7 The crown graph Cn,n ...... 26

3.8 The friendship graph Fn ...... 29

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TABLE OF CONTENTS

Page

ABSTRACT (1)

ACKNOWLEDGEMENTS (2)

LIST OF FIGURES (3)

CHAPTER 1 INTRODUCTION 1 1.1 Basic Knowledge in Graph Theory ...... 1 1.2 Basic Knowledge in Game Theory ...... 5

CHAPTER 2 LITERATURE REVIEW 6

CHAPTER 3 (2,∞)-INCIDENCE LABELING GAMES 7

CHAPTER 4 CONCLUSION 36

REFERENCES 37

BIOGRAPHY 38

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CHAPTER 1

INTRODUCTION

In this chapter, we give some basic knowledge in graph theory and in game theory, which are used in this thesis. For notations and terminology in graph theory, we in general follow [4] and [7].

1.1 Basic Knowledge in Graph Theory

A graph G is a triple, which consists of a vertex set V (G), an edge set E(G), and a relation associating with each edge and two vertices (not necessarily diﬀerent) called the endpoints. If an edge e has endpoints u and v, we write e = uv (the order does not count). Then u and v are said to be adjacent, and u and e are

said to be incident. For example, let G1 be the graph shown in Figure 1.1. Then

V (G1) = {v1,v2,v3,v4,v5} and E(G1) = {v1v2,v1v5,v2v3,v2v4,v3v4,v4v5}.

v1 v3

v2 G1 :

v5 v4

Figure 1.1: The graph G1

Let G be a graph. An incidence of G is deﬁned as a pair (v,e), where v ∈ V (G) is incident to e ∈ E(G). Let I(G) denote the set of all incidences in G. Two distinct incidences (v,e) and (u,f) are adjacent in G if one of the following conditions holds: 1. v = u and e 6= f; 2. e = f and v 6= u; and 3. e = vu, f = uw, and v 6= w. For example, some pairs of adjacent incidences are shown in Figure 1.2, and

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some pairs of nonadjacent incidences are shown in Figure 1.3. Note that a white dot denotes a vertex, and a black dot on an edge denotes an incidence.

Figure 1.2: Pairs of adjacent incidences

Figure 1.3: Pairs of nonadjacent incidences

A loop is an edge with the same endpoints. Multiple edges are edges, whose pair of endpoints are the same. A simple graph is a graph containing no loops or

multiple edges. For example, in Figure 1.4, the graph G2 is a simple graph, but

G3 is not a simple graph since it contains multiple edges.

G2 : G3 :

Figure 1.4: Graphs G2 and G3

A path is an alternating sequence of vertices and edges such that all vertices (except perhaps the ﬁrst and the last vertices) are distinct. A path with n vertices

is denoted by Pn. For example, a path P4 is shown in Figure 1.5.

P4 :

Figure 1.5: A path P4

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A cycle is a path such that the ﬁrst and the last vertices are the same.

A cycle with n vertices is denoted by Cn. For example, a cycle C5 is shown in Figure 1.6.

C5 :

Figure 1.6: A cycle C5

A complete graph is a simple graph whose vertices are pairwise adjacent.

A complete graph with n vertices is denoted by Kn. For example, a complete graph K6 is shown in Figure 1.7.

K6 :

Figure 1.7: A complete graph K6

A graph G is bipartite if we can partition V (G) into two subsets U and W , such that each edge of G joins a vertex in U and a vertex in W . For example,

a bipartite graph G4 is shown in Figure 1.8.

G4 :

Figure 1.8: A bipartite graph G4

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The ladder graph with 2n vertices is denoted by Ln. The vertex set of Ln

is V (Ln) = U ∪ V , where U = {u1,u2,...,un} and V = {v1,v2,...,vn}. The edge

set of Ln is E(Ln) = C ∪ D ∪ E, where C = {ci = uiui+1 : i ∈ {1,2,...,n − 1}},

D = {di = vivi+1 : i ∈ {1,2,...,n − 1}}, and E = {ei = uivi : i ∈ {1,2,...,n}}.

For example, the ladder graph L4 is shown in Figure 1.9.

e1 u1 v1

c1 d1 e2 u2 v2

L4 : c2 d2 e3 u3 v3

c3 d3 e4 u4 v4

Figure 1.9: The ladder graph L4

The crown graph with 2n vertices, denoted by Cn,n, is a bipartite graph with partite sets U = {u1,u2,...,un} and V = {v1,v2,...,vn}. The edge set is E(Cn,n) =

{ei,j = uivj : i,j ∈ {1,2,...,n} and i 6= j}. For example, the crown graph C4,4 is shown in Figure 1.10.

u1 v1

u2 v2 C4,4 : u3 v3

u4 v4

Figure 1.10: The crown graph C4,4

The friendship graph with 2n+1 vertices, denoted by Fn, is the graph with

vertex set V (Fn) = {v0,v1,...,v2n} and edge set E(Fn) = E ∪ F , where E = {ej =

v0vj : j ∈ {1,2,...,2n}} and F = {fk = vkvk+1 : k ∈ {1,3,...,2n−1}}. For example,

the friendship graph F4 is shown in Figure 1.11.

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v1 f1 v2

v8 e1 e2 v3

e8 e3 F4 : f7 e e f3 7 v0 4 v7 e6 e5 v4

v6 f5 v5

Figure 1.11: The friendship graph F4

1.2 Basic Knowledge in Game Theory

For notations and terminology in game theory, we in general follow [3] and [6].

Game theory is the logical analysis of situations of conflict and cooperation. A player is an agent who makes decisions in a game. A strategy is one of the possible actions of a player. A game is a situation which has at least two players, and each player has some possible strategies. A game is finite if these are a finite number of players and moves that a players can make.

A game has perfect information if all players are aware of all aspects of the game until the game is over. Chess and tic-tac-toe are examples of perfect information games. Card games such as bridge and poker where each player’s cards are hidden are examples of imperfect information games.

Theorem 1.1. [8] In any ﬁnite two-person game of perfect information, if the game cannot end in a draw, then one of the players must have a winning strategy.

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CHAPTER 2

LITERATURE REVIEW

A graph coloring game was introduced by Bodlaender [2]. There are two players, Alice and Bob (Alice begins). They alternately label an uncolored vertex of G, and two adjacent vertices must be ﬁlled by distinct colors. In one variant, Alice wins the game if the vertices of G are all colored; otherwise, Bob wins. In a second variant, the player who is unable to move, loses the game. The author determined the complexity of the problems for both variants in which Alice has a winning strategy.

Lam et al. [5] introduced an edge coloring game. This game is similar to a graph coloring game, but where edges are colored instead of vertices. They gave the upper bounds of the smallest number of colors such that Alice has a winning strategy for trees and wheels.

Andres [1] introduced an incidence coloring game. This game is similar to a graph coloring game except that Alice and Bob label incidences. The author gave the upper bound of the minimum number of needed colors which can give Alice a winning strategy for k-degenerate graphs.

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CHAPTER 3

(2,∞)-INCIDENCE LABELING GAMES

We introduce a (p,q)-incidence labeling game on a graph G, where p and q are positive integers. There are two players, Alice and Bob, and Alice is the ﬁrst player. They alternately label an uncolored incidence of G by a color received from a set of p colors, and each color can be used at most q times. Adjacent incidences must be labeled by diﬀerent colors. The last player who can label is the winner. When p = 2 and each color can be used unlimited times, we call this game a (2,∞)-incidence labeling game.

Since all (2,∞)−incidence labeling games have perfect information and they cannot end in a draw, by Theorem 1.1 one of the players must have a winning strategy.

A graph G is incidence-symmetric if there is a bijection θ : I(G) → I(G) such that for all (v,e),(u,f) ∈ I(G), 1. θ(v,e) 6= (v,e); 2. θ(θ(v,e)) = (v,e); and 3. (v,e) and (u,f) are adjacent if and only if θ(v,e) and θ(u,f) are adjacent. We call this bijection, an incidence-symmetric bijection. For example, the incidence-symmetric graph G5 with the bijection θ(vi,ej) = (vn−i+1,en−j) for all

(vi,ej) ∈ I(G5) is shown in Figure 3.1.

e1 e2 e3 G5 : v1 v2 v3 v4

Figure 3.1: The incidence-symmetric graph G5

Let G6 be the graph shown in Figure 3.2. Since these are three (odd number)

pairwise non-adjacent incidences (v1,e1),(v3,e2), and (v4,e3), we can not ﬁnd any bijection satisfying the 3 properties, so G6 is not an incidence-symmetric graph.

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e3 G6 : v1 e1 v2 e2 v3

Figure 3.2: The non incidence-symmetric graph G6

We determine (2,∞)−incidence labeling games on incidence-symmetric graphs such as paths, cycles, complete graphs, ladder graphs, crown graphs and friendship graphs. Note that if a graph has no edges, there are no incidences to label, so we always assume that all graphs in this paper have at least one edge. To ﬁnd a winning strategy of these games, we use the following theorem.

Theorem 3.1. Let L be a (2,∞)−incidence labeling game on an incidence-symmetric graph with n vertices, where n ≥ 2 is an integer. Then Bob has a winning strategy on L.

Proof. Let θ be an incidence-symmetric bijection of G. Then Bob can win the game by the following iteration.

While Alice can label some incidence, say she labels an incidence (v,e) by some color, Bob then labels the incidence θ(v,e) by the other color.

Assume that Alice can label the incidence (v,e) by some color. Without loss of generality, we may assume that she labels (v,e) by color 1. This means that all neighbors of (v,e), which were already labeled, received the color 2. Since θ is incidence-symmetric, all neighbors of θ(v,e), which were already labeled, received the color 1. Suppose for a contradiction that θ(v,e) is already labeled. By the iteration, the incidence θ(θ(v,e)) = (v,e) must be already labeled as well. This contradicts the fact that Alice can label (v,e). Hence Bob can label the incidence θ(v,e) by the color 2. This completes the proof.

We ﬁrst consider a (2,∞)−incidence labeling game on a path. We show that a path is incidence-symmetric, so we get the following theorem.

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Theorem 3.2. Let L be a (2,∞)−incidence labeling game on a path with n ver-

tices, where n > 2 is an integer. Then Bob has a winning strategy on L.

Proof. By Theorem 3.1, it suﬃces to show that all paths are incidence-symmetric.

Let Pn be a path. Without loss of generality, we may assume that

Pn = v1e1v2e2 ···vn−1en−1vn, where vi is a vertex for all i, and ej is an edge for all j. This graph is shown in Figure 3.3.

e1 e2 en−1 Pn : v1 v2 v3 ··· vn−1 vn

Figure 3.3: The path Pn

Let θ : I(Pn) → I(Pn) deﬁned by θ(vi,ej) = (vn−i+1,en−j) for all (vi,ej) ∈

I(Pn).

First, we show that θ(vi,ej) ∈ I(Pn) for each (vi,ej) ∈ I(Pn). Let (vi,ej) ∈

I(Pn). Then 1 ≤ i ≤ n and 1 ≤ j ≤ n − 1. Thus n − i ≥ 0, so n − i + 1 ≥ 0. Also n − i + 1 ≤ n since i ≥ 1. Thus 1 ≤ n − i + 1 ≤ n. Similarly, we have

1 ≤ n − j ≤ n − 1. Furthermore, since (vi,ej) ∈ I(Pn), we have i = j or i = j + 1.

Then n − i + 1 = (n − j) + 1 or n − i + 1 = n − j. Thus θ(vi,ej) ∈ I(Pn).

Second, we show that θ is a function. Assume (vi,ej),(vk,el) ∈ I(Pn)

such that (vi,ej) = (vk,el). Then i = k and j = l, so θ(vi,ej) = (vn−i+1,en−j) =

(vn−k+1,en−l) = θ(vk,el).

Third, we show that θ is injective. Assume that (vi,ej),(vk,el) ∈ I(Pn)

such that θ(vi,ej) = θ(vk,el). Then we have (vn−i+1,en−j) = (vn−k+1,en−l), so

n−i+1 = n−k +1 and n−j = n−l. Hence i = k and j = l. Thus (vi,ej) = (vk,el).

We next show that θ is surjective. Let (vi,ej) ∈ I(Pn). Previously, we show

that θ(vi,ej) = (vn−i+1,en−j) ∈ I(Pn). Consider θ(vn−i+1,en−j) = (vn−(n−i+1)+1,en−(n−j))

= (vi,ej).

We then show that θ(vi,ej) 6= (vi,ej) for all (vi,ej) ∈ I(Pn). Suppose for a

contradiction that θ(vi,ej) = (vi,ej) for some (vi,ej) ∈ I(Pn). Then (vn−i+1,en−j)

= (vi,ej), so n − i + 1 = i and n − j = j. This implies that n = 2i − 1 and n = 2j.

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This contradicts the parity property.

We next show that, θ(θ(vi,ej)) = (vi,ej) for all (vi,ej) ∈ I(Pn). Let (vi,ej) ∈

I(Pn). Then θ(θ(vi,ej)) = θ(vn−i+1,en−j) = (vn−(n−i+1)+1,en−(n−j)) = (vi,ej).

Finally, we show that (vi,ej) and (vk,el) are adjacent if and only if θ(vi,ej) and θ(vk,el) are adjacent for all (vi,ej),(vk,el) ∈ I(Pn).

Necessity. Suppose (vi,ej),(vk,el) ∈ I(Pn) such that they are adjacent. We consider the following three cases. Case 1: k = i and l = j + 1 or k = i and j = l + 1. Without loss of generality, we may assume that k = i, and l = j +1.

Then θ(vi,ej) = (vn−i+1,en−j), and θ(vk,el) = (vn−k+1,en−l) = (vn−i+1,en−j−1).

Thus θ(vi,ej) and θ(vk,el) are adjacent. Case 2: k = i + 1 and l = j or i = k + 1 and l = j. Without loss of generality, we may assume that k = i+1 and l = j.

Then θ(vi,ej) = (vn−i+1,en−j), and θ(vk,el) = (vn−k+1,en−l) = (vn−(i+1)+1,en−(j+1))

= (vn−i,en−j−1). Thus θ(vi,ej) and θ(vk,el) are adjacent. Case 3. k = i + 1 and l = j + 1 or i = k + 1 and j = l + 1. Without loss of generality, we may assume that k = i+1 and l = j +

1. Then θ(vi,ej) = (vn−i+1,en−j), and θ(vk,el) = (vn−k+1,en−l) = (vn−(i+1)+1,en−(j+1))

= (vn−i,en−j−1). Thus θ(vi,ej) and θ(vk,el) are adjacent.

Suﬃciency. Suppose (vi,ej),(vk,el) ∈ I(Pn) such that θ(vi,ej) and

θ(vk,el) are adjacent. By necessity, we have θ(θ(vi,ej)) = (vi,ej) and θ(θ(vk,el)) =

(vk,el) are adjacent. Therefore, θ is an incidence-symmetric bijection. This completes the proof.

For (2,∞)−incidence labeling games on cycles, complete graphs, ladder graphs, crown graphs, and friend graphs, we follow the steps in Theorem 3.2. Then we get the following theorems.

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Theorem 3.3. Let L be a (2,∞)−incidence labeling game on a cycle with n vertices, where n > 3 is an integer. Then Bob has a winning strategy on L.

Proof. To prove this theorem, we show that all cycles are incidence-symmetric. Let

Cn be a cycle. Without loss of generality, we may assume that Cn = v1e1v2e2 ···vnenv1, where vi is a vertex for all i, and ej is an edge for all j. This graph is shown in Figure 3.4.

v1 e1 v2

en e2

Cn : vn v3 en−1 e3

vn−1 v4 ···

Figure 3.4: The cycle Cn

Deﬁne θ : I(Cn) → I(Cn) by

  (vn−i+1,en−j) if j 6= n; θ(vi,ej) =  (vn−i+1,en) otherwise.

First, we show that θ(vi,ej) ∈ I(Cn) for each (vi,ej) ∈ I(Cn). Let (vi,ej) ∈

I(Cn). Then 1 ≤ i ≤ n and 1 ≤ j ≤ n. Thus n − i ≥ 0, so n − i + 1 ≥ 1. Also, n − i + 1 ≤ n since i ≥ 1. Hence 1 ≤ n − i + 1 ≤ n. If j 6= n, then 1 ≤ n − j < n.

Otherwise, we have 1 ≤ j ≤ n. Furthermore, since (vi,ej) ∈ I(Cn), we consider the following two cases. Case 1: j 6= n. Then i = j or i = j +1. Hence n−i+1 = (n−j)+1 or n − i + 1 = n − j. Case 2: j = n. Then i = n or i = 1. Hence n − i + 1 = n − n + 1 = 1 or n − i + 1 = n − 1 + 1 = n.

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Thus θ(vi,ej) ∈ I(Cn). Second, it is obvious that θ is a function.

Third, we show that θ is injective. Assume that (vi,ej),(vk,el) ∈ I(Cn) such that θ(vi,ej) = θ(vk,el). We consider the following four cases. Case 1: j 6= n and l 6= n.

Then we have (vn−i+1,en−j) = θ(vi,ej) = θ(vk,el) = (vn−k+1,en−l), which implies that n − i + 1 = n − k + 1 and n − j = n − l. Hence i = k and j = l.

Thus (vi,ej) = (vk,el). Case 2: j 6= n and l = n.

Then we have (vn−i+1,en−j) = θ(vi,ej) = θ(vk,el) = (vn−k+1,en), which implies that n−i+1 = n−k +1 and n−j = n. Hence i = k and j = 0. This case is impossible. Case 3: j = n and l 6= n.

Then we have (vn−i+1,en) = θ(vi,ej) = θ(vk,el) = (vn−k+1,en−l), which implies that n−i+1 = n−k +1 and n = n−l. Hence i = k and l = 0. This case is impossible. Case 4: j = n and l = n.

Then we have (vn−i+1,en) = θ(vi,ej) = θ(vk,el) = (vn−k+1,en), which implies that n − i + 1 = n − k + 1 and j = l. Hence i = k and j = l. Thus

(vi,ej) = (vk,el).

We then show that θ is surjective. Let (vi,ej) ∈ I(Cn). We consider the following two cases. Case 1: j 6= n.

Then we have θ(vi,ej) = (vn−i+1,en−j) ∈ I(Cn). Consider θ(vn−i+1,en−j)

= (vn−(n−i+1)+1,en−(n−j)) = (vi,ej). Case 2: j = n.

Then we have θ(vi,ej) = (vn−i+1,en) ∈ I(Cn). Consider θ(vn−i+1,en)

= (vn−(n−i+1)+1,en) = (vi,ej). Thus θ is surjective.

We next suppose for a contradiction that θ(vi,ej) = (vi,ej) for some (vi,ej) ∈

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I(Cn). We consider the following two cases. Case 1: j 6= n.

Then (vn−i+1,en−j) = θ(vi,ej) = (vi,ej), so n−i+1 = i and n − j = j. This implies that n = 2i − 1 and n = 2j. This contradicts the parity property. Case 2: j = n.

Thus i = 1 or i = n. Then θ(vi,ej) = (vn−i+1,en). Thus i = n−i+1 and j = n. Then n = 2i−1 and j = n. Note that n ≥ 3. We consider the following two subcases. Subcase 2.1: i = 1. Then n = 2i − 1 = 2(1) − 1 = 1. This contradicts the fact that n ≥ 3. Subcase 2.2: i = n. Then n = 2n − 1, a contradiction.

We next show that, θ(θ(vi,ej)) = (vi,ej) for all (vi,ej) ∈ I(Cn). Let (vi,ej) ∈

I(Cn). We consider the following two cases. Case 1: j 6= n.

Then θ(θ(vi,ej)) = θ(vn−i+1,en−j) = (vn−(n−i+1)+1,en−(n−j)) = (vi,ej). Case 2: j = n.

Then θ(θ(vi,ej)) = θ(vn−i+1,en) = (vn−(n−i+1)+1,en) = (vi,ej).

Finally, we show that (vi,ej) and (vk,el) are adjacent if and only if θ(vi,ej) and θ(vk,el) are adjacent for all (vi,ej),(vk,el) ∈ I(Cn).

Necessity. Suppose (vi,ej),(vk,el) ∈ I(Cn) such that they are adjacent. We consider the following sixteen cases. Case 1: i = k = j = l = n.

There is only one incidence satisfying these conditions, so (vi,ej) =

(vn,en) = (vk,el). This contradicts the fact that (vi,ej) and (vk,el) are adjacent. Hence this case is impossible. Case 2: i = k = j = n and l 6= n.

Since (vi,ej) and (vk,el) are adjacent, we have (vi,ej) = (vn,en) and (vk,el) = (vn,en−1). Then θ(vi,ej) = (vn−i+1,en) = (vn−n+1,en) = (v1,en),

Ref. code: 25616009031029DVT 14 and θ(vk,el) = (vn−k+1,en−l) = (vn−n+1,en−(n−1)) = (v1,e1). Thus θ(vi,ej) and

θ(vk,el) are adjacent. Case 3: i = k = n, j 6= n, and l = n. The proof in this case is similar to Case 2. Case 4: i = k = n, j 6= n, and l 6= n.

There is only one incidence satisfying these conditions, so (vi,ej) =

(vn,en−1) = (vk,el). This contradicts the fact that (vi,ej) and (vk,el) are adjacent. Hence this case is impossible. Case 5: i = n, k 6= n, and j = l = n.

Since (vi,ej) and (vk,el) are adjacent, we have (vi,ej) = (vn,en) and (vk,el) = (v1,en). Then θ(vi,ej) = (vn−i+1,en) = (vn−n+1,en) = (v1,en), and

θ(vk,el) = (vn−k+1,en) = (vn−1+1,en) = (vn,en). Thus θ(vi,ej) and θ(vk,el) are adjacent. Case 6: i = n, k 6= n, j = n, and l 6= n.

Thus, we get (vi,ej) = (vn,en). Since (vi,ej) and (vk,el) are adjacent, we only have the following two subcases.

Subcase 6.1:(vk,el) = (v1,e1).

Then θ(vi,ej) = θ(vn,en) = (vn−n+1,en) = (v1,en), and θ(vk,el)

= θ(v1,e1) = (vn−1+1,en−1) = (vn,en−1). Thus θ(vi,ej) and θ(vk,el) are adjacent.

Subcase 6.2:(vk,el) = (vn−1,en−1).

Then θ(vi,ej) = θ(vn,en) = (vn−n+1,en) = (v1,en), and θ(vk,el)

= θ(vn−1,en−1) = (vn−(n−1)+1,en−(n−1)) = (v2,e1). Thus θ(vi,ej) and θ(vk,el) are adjacent. Case 7: i = n, k 6= n, j 6= n, and l = n.

Since (vi,ej) and (vk,el) are adjacent, we have (vi,ej) = (vn,en−1) and (vk,el) = (v1,en). Then θ(vi,ej) = θ(vn,en−1) = (vn−n+1,en−(n−1)) = (v1,e1), and θ(vk,el) = θ(v1,en) = (vn,en). Thus θ(vi,ej) and θ(vk,el) are adjacent. Case 8: i = n, k 6= n, j 6= n, and l 6= n.

Thus, we get (vi,ej) = (vn,en−1). Since (vi,ej) and (vk,el) are adjacent, we only have the following two subcases.

Ref. code: 25616009031029DVT 15

Subcase 8.1:(vk,el) = (vn−1,en−1).

Then θ(vi,ej) = θ(vn,en−1) = (vn−n+1,en−(n−1)) = (v1,e1), and

θ(vk,el) = θ(vn−1,en−1) = (vn−(n−1)+1,en−(n−1)) = (v2,e1). Thus θ(vi,ej) and

θ(vk,el) are adjacent.

Subcase 8.2:(vk,el) = (vn−1,en−2).

Then θ(vi,ej) = θ(vn,en−1) = (vn−n+1,en−(n−1)) = (v1,e1), and

θ(vk,el) = θ(vn−1,en−2) = (vn−(n−1)+1,en−(n−2)) = (v2,e2). Thus θ(vi,ej) and

θ(vk,el) are adjacent. Case 9: i 6= n and k = j = l = n. The proof in this case is similar to Case 5. Case 10: i 6= n, k = n, j = n, and l 6= n. The proof in this case is similar to Case 7. Case 11: i 6= n, k = n, j 6= n, and l = n. The proof in this case is similar to Case 6. Case 12: i 6= n, k = n, j 6= n, and l 6= n. The proof in this case is similar to Case 8. Case 13: i 6= n, k 6= n, and j = l = n.

There is only one incidence satisfying these conditions, so (vi,ej) =

(v1,en) = (vk,el). This contradicts the fact that (vi,ej) and (vk,el) are adjacent. Hence this case is impossible. Case 14: i 6= n, k 6= n, j = n and l 6= n.

Thus, we get (vi,ej) = (v1,en). Since (vi,ej) and (vk,el) are adjacent, we only have the following two subcases.

Subcase 14.1:(vk,el) = (v1,e1).

Then θ(vi,ej) = θ(v1,en) = (vn−1+1,en) = (vn,en), and θ(vk,el) =

θ(v1,e1) = (vn−1+1,en−1) = (vn,en−1). Thus θ(vi,ej) and θ(vk,el) are adjacent.

Subcase 14.2:(vk,el) = (v2,e1).

Then θ(vi,ej) = θ(v1,en) = (vn−1+1,en) = (vn,en), and θ(vk,el) =

θ(v2,e1) = (vn−2+1,en−1) = (vn−1,en−1). Thus θ(vi,ej) and θ(vk,el) are adjacent.

Ref. code: 25616009031029DVT 16

Case 15: i 6= n, k 6= n, j 6= n and l = n. The proof in this case is similar to Case 14. Case 16: i 6= n, k 6= n, j 6= n and l 6= n.

Since (vi,ej) and (vk,el) are adjacent, we consider the following three subcases. Subcase 16.1: k = i and l = j +1 or k = i and j = l +1. Without loss of generality, we assume that k = i, and l = j +1.

Then θ(vi,ej) = (vn−i+1,en−j), and θ(vk,el) = (vn−k+1,en−l) = (vn−i+1,en−j−1).

Thus θ(vi,ej) and θ(vk,el) are adjacent. Subcase 16.2: k = i+1 and l = j or i = k +1 and l = j. Without loss of generality, we assume that k = i+1 and l = j.

Then θ(vi,ej) = (vn−i+1,en−j), and θ(vk,el) = (vn−k+1,en−l) = (vn−(i+1)+1,en−(j+1)) =

(vn−i,en−j−1). Thus θ(vi,ej) and θ(vk,el) are adjacent. Subcase 16.3: k = i + 1 and l = j + 1 or i = k + 1 and j = l + 1. Without loss of generality, we assume that k = i+1 and l = j +

1. Then θ(vi,ej) = (vn−i+1,en−j), and θ(vk,el) = (vn−k+1,en−l) = (vn−(i+1)+1,en−(j+1))

= (vn−i,en−j−1). Thus θ(vi,ej) and θ(vk,el) are adjacent.

Suﬃciency. Suppose (vi,ej),(vk,el) ∈ I(Cn) such that θ(vi,ej) and

θ(vk,el) are adjacent. By necessity, we have θ(θ(vi,ej)) = (vi,ej) and θ(θ(vk,el)) =

(vk,el) are adjacent. Therefore, θ is an incidence-symmetric bijection. This completes the proof.

To prove Theorem 3.5, we apply the deﬁnition of adjacency of incidences to a complete graph. Then we easily get the following lemma.

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Lemma 3.4. Let Kn be the complete graph with vertex set V (Kn) = {v1,v2,··· ,vn}

and edge set E(Kn) = {ej,k = vjvk : j,k{1,2,...,n} and j < k}. Then two distinct incidences (vi,ej,k) and (vl,er,s) are adjacent if one of the following conditions holds:

1. vi = vl and ej,k 6= er,s;

2. ej,k = er,s and vi 6= vl; and

3. i ∈ {r,s} or l ∈ {j,k} and ej,k 6= er,s.

Theorem 3.5. Let L be a (2,∞)−incidence labeling game on a complete graph with n vertices, where n > 2 is an integer. Then Bob has a winning strategy on L.

Proof. To prove this theorem, we show that all complete graphs are incidence- symmetric. Let Kn be a complete graph. Without loss of generality, we may assume that its vertex set is V (Kn) = {v1,v2,··· ,vn} and its edge set is E(Kn) =

{ej,k = vjvk : j,k{1,2,...,n} and j < k}. This graph is shown in Figure 3.5.

v1 v2

Kn : vn v3

··· vn−1 v4

Figure 3.5: The complete graph Kn

Let θ : I(Kn) → I(Kn) deﬁned by θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1) for all

(vi,ej,k) ∈ I(Kn).

First, we show that θ(vi,ej,k) ∈ I(Kn) for each (vi,ej,k) ∈ I(Kn). Let

(vi,ej,k) ∈ I(Kn). Then 1 ≤ i ≤ n, 1 ≤ j ≤ n, and 1 ≤ k ≤ n. Thus n − i ≥ 1, so n − i + 1 ≥ 0. Also n − i + 1 ≤ n since i ≥ 1. Thus 1 ≤ n − i + 1 ≤ n. Similarly, we

have 1 ≤ n−k +1 ≤ n and 1 ≤ n−j +1 ≤ n. Furthermore, since (vi,ej,k) ∈ I(Kn), we consider the following two cases.

Ref. code: 25616009031029DVT 18

Case 1: i = j. Then n − i + 1 = n − j + 1. Case 2: i = k. Then n − i + 1 = n − k + 1.

Thus θ(vi,ej,k) ∈ I(Kn). Second, it is obvious that θ is a function.

Third, we show that θ is injective. Assume that (vi,ej,k),(vl,er,s) ∈ I(Kn) such that θ(vi,ej,k) = θ(vl,er,s). Then we have (vn−i+1,en−k+1,n−j+1) = θ(vi,ej,k) =

θ(vl,er,s) = (vn−l+1,en−s+1,n−r+1), which implies that n−i+1 = n−l +1, n−k + 1 = n − s + 1 and n − j + 1 = n − r + 1. Hence i = l, k = s, and j = r. Thus

(vi,ej,k) = (vl,er,s).

We then show that θ is surjective. Let (vi,ej,k) ∈ I(Kn). Then we have

θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1) ∈ I(Kn). Consider θ(vn−i+1,en−k+1,n−j+1) =

(vn−(n−i+1)+1,en−(n−j+1)+1,n−(n−k+1)+1) = (vi,ej,k).

We next suppose for a contradiction that θ(vi,ej,k) = (vi,ej,k) for some

(vi,ej,k) ∈ I(Kn). Then (vn−i+1,en−k+1,n−j+1) = θ(vi,ej,k) = (vi,ej,k). Hence n − i + 1 = i, n − k + 1 = j, and n − j + 1 = k. Thus n = 2i − 1, n = k + j − 1, which implies that 2i = j + k. We consider in two cases. Case 1: i = j. Then 2j = 2i = j + k, which implies that j = k. This contradicts the fact that j < k. Case 2: i = k. Then 2k = 2i = j + k, which implies that k = j. This contradicts the fact that j < k.

We next show that, θ(θ(vi,ej,k)) = (vi,ej,k) for all (vi,ej,k) ∈ I(Kn). Let

(vi,ej,k) ∈ I(Kn). Then θ(θ(vi,ej,k)) = θ(vn−i+1,en−k+1,n−j+1)

= (vn−(n−i+1)+1,en−(n−j+1)+1,n−(n−k+1)+1) = (vi,ej,k).

Finally, we show that (vi,ej,k) and (vl,er,s) are adjacent if and only if

θ(vi,ej,k) and θ(vl,er,s) are adjacent for all (vi,ej,k),(vl,er,s) ∈ I(Kn).

Necessity. Suppose (vi,ej,k),(vl,er,s) ∈ I(Kn) such that they are adja-

Ref. code: 25616009031029DVT 19

cent. We consider the following eight cases. Case 1: i = l, j = r, and k = s.

Then (vi,ej,k) = (vl,er,s). This contradicts the fact that (vi,ej,k) and (vl,er,s) are adjacent. Hence this case is impossible. Case 2: i = l, j = r, and k 6= s.

Then θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1), and θ(vl,er,s) =

(vn−l+1,en−s+1,n−r+1) = (vn−i+1,en−s+1,n−j+1). Thus θ(vi,ej,k) and θ(vl,er,s) are adjacent. Case 3: i = l, j 6= r, and k = s. The proof in this case is similar to Case 2. Case 4: i = l, j 6= r, and k 6= s.

Then θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1), and θ(vl,er,s) =

(vn−l+1,en−s+1,n−r+1) = (vn−i+1,en−s+1,n−r+1). Thus θ(vi,ej,k) and θ(vl,er,s) are adjacent. Case 5: i 6= l, j = r, and k = s.

Then θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1), and θ(vl,er,s) =

(vn−l+1,en−s+1,n−r+1) = (vn−l+1,en−k+1,n−j+1). Thus θ(vi,ej,k) and θ(vl,er,s) are adjacent. Case 6: i 6= l, j = r, and k 6= s.

Then θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1), and θ(vl,er,s) =

(vn−l+1,en−s+1,n−r+1) = (vn−l+1,en−s+1,n−j+1). By Lemma 3.4 (3), we have θ(vi,ej,k)

and θ(vl,er,s) are adjacent. Case 7: i 6= l, j 6= r, and k = s. The proof in this case is similar to Case 6. Case 8: i 6= l, j 6= r, and k 6= s.

Since (vi,ej,k),(vl,er,s) ∈ I(Kn), we have i ∈ {j,k} and l ∈ {r,s}. We also have i ∈ {r,s} or l ∈ {j,k} by Lemma 3.4 (3). Then we consider the following sixteen subcases. Subcase 8.1: i = j, l = r, and i = r. Then i = j = l = r, this contradicts the fact that i 6= l. Hence

Ref. code: 25616009031029DVT 20 this case is impossible. Subcase 8.2: i = j, l = r, and i = s.

Then i = j = s. Consider θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1)

= (vn−i+1,en−k+1,n−i+1), and θ(vl,er,s) = (vn−l+1,en−s+1,n−r+1) = (vn−l+1,en−i+1,n−l+1).

By Lemma 3.4 (3), we have θ(vi,ej,k) and θ(vl,er,s) are adjacent. Subcase 8.3: i = j, l = r, and l = j. Then i = j = l = r, this contradicts the fact that i 6= l. Hence this case is impossible. Subcase 8.4: i = j, l = r, and l = k.

Then l = r = k. Consider θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1) =

(vn−i+1,en−k+1,n−i+1), and θ(vl,er,s) = (vn−l+1,en−s+1,n−r+1)

= (vn−k+1,en−s+1,n−k+1). Thus θ(vi,ej,k) and θ(vl,er,s) are adjacent. Subcase 8.5: i = j, l = s, and i = r. Then i = j = r. This contradicts the fact that j 6= r. Hence this case is impossible. Subcase 8.6: i = j, l = s, and i = s. Then i = j = l = s, this contradicts the fact that i 6= l. Hence this case is impossible. Subcase 8.7: i = j, l = s, and l = j. Then i = j = l = s, this contradicts the fact that i 6= l. Hence this case is impossible. Subcase 8.8: i = j, l = s, and l = k. Then k = l = s. This contradicts the fact that k 6= s. Hence this case is impossible. Subcase 8.9: i = k, l = r, and i = r. Then i = k = l = r, this contradicts the fact that i 6= l. Hence this case is impossible. Subcase 8.10: i = k, l = r, and i = s. Then i = k = s. This contradicts the fact that k 6= s. Hence this case is impossible.

Ref. code: 25616009031029DVT 21

Subcase 8.11: i = k, l = r, and l = j. Then j = l = r. This contradicts the fact that j 6= r. Hence this case is impossible. Subcase 8.12: i = k, l = r, and l = k. Then i = k = l = r. This contradicts the fact that i 6= l. Hence this case is impossible. Subcase 8.13: i = k, l = s, and i = r.

Then i = k = r. Consider θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1)

= (vn−i+1,en−i+1,n−j+1), and θ(vl,er,s) = (vn−l+1,en−s+1,n−r+1)

= (vn−l+1,en−s+1,n−i+1). By Lemma 3.4 (3), we have θ(vi,ej,k) and θ(vl,er,s) are adjacent. Subcase 8.14: i = k, l = s, and i = s. Then i = k = l = s, this contradicts the fact that i 6= l. Hence this case is impossible. Subcase 8.15: i = k, l = s, and l = j.

Then j = l = s. Consider θ(vi,ej,k) = (vn−i+1,en−k+1,n−j+1),

and θ(vl,er,s) = (vn−l+1,en−s+1,n−r+1) = (vn−j+1,en−j+1,n−r+1). Thus θ(vi,ej,k)

and θ(vl,er,s) are adjacent. Subcase 8.16: i = k, l = s, and l = k. Then i = k = l = s, this contradicts the fact that i 6= l. Hence this case is impossible.

Suﬃciency. Suppose (vi,ej,k),(vl,er,s) ∈ I(Kn) such that θ(vi,ej,k) and

θ(vl,er,s) are adjacent. By necessity, we have θ(θ(vi,ej,k)) = (vi,ej,k) and θ(θ(vl,er,s))

= (vl,er,s) are adjacent. Therefore, θ is an incidence-symmetric bijection. This completes the proof.

Ref. code: 25616009031029DVT 22

Theorem 3.6. Let L be a (2,∞)−incidence labeling game on a ladder graph with

2n vertices, where n > 1 is an integer. Then Bob has a winning strategy on L.

Proof. To prove this theorem, we show that all ladder graphs are incidence-

symmetric. Let Ln be a ladder graph with 2n vertices. Without loss of generality, we may assume that its vertex set is V (Ln) = U ∪ V , where U = {u1,u2,··· ,un}

and V = {v1,v2,··· ,vn}, and its edge set is E(Ln) = C ∪ D ∪ E, where C =

{ci = uiui+1 : i ∈ {1,2,...,n − 1}}, D = {di = vivi+1 : i ∈ {1,2,...,n − 1}}, and

E = {ei = uivi : i ∈ {1,2,...,n}}. This graph is shown in Figure 3.6.

e1 u1 v1

c1 d1 e2 u2 v2

Ln : c2 d2 e3 u3 v3

. . . .

en−1 un−1 vn−1

cn−1 dn−1 en un vn

Figure 3.6: The ladder graph Ln

Deﬁne θ : I(Ln) → I(Ln) by   (vi,dj) if (w,f) = (ui,cj);    (ui,cj) if (w,f) = (vi,dj); θ(w,f) =  (vi,ei) if (w,f) = (ui,ei);    (ui,ei) if (w,f) = (vi,ei); for all (ui,cj),(vi,dj),(ui,ei),(vi,ei) ∈ I(Ln).

First, it is clear that θ is a function. 0 0 Next, we show that θ is injective. Assume that (w,f),(w ,f ) ∈ I(Ln) such

Ref. code: 25616009031029DVT 23

0 0 that θ(w,f) = θ(w ,f ). We consider the following four cases. 0 0 Case 1: θ(w,f) = θ(w ,f ) = (ui,cj) for some i ∈ {1,2,...,n} and j ∈ {1,2,...,n − 1}. 0 0 Then (w,f) = (vi,dj) = (w ,f ). 0 0 Case 2: θ(w,f) = θ(w ,f ) = (vi,dj) for some i ∈ {1,2,...,n} and j ∈ {1,2,...,n − 1}. 0 0 Then (w,f) = (ui,cj) = (w ,f ). 0 0 Case 3: θ(w,f) = θ(w ,f ) = (ui,ei) for some i ∈ {1,2,...,n}. 0 0 Then (w,f) = (vi,ei) = (w ,f ). 0 0 Case 4: θ(w,f) = θ(w ,f ) = (vi,ei) for some i ∈ {1,2,...,n}. 0 0 Then (w,f) = (ui,ei) = (w ,f ). Thus θ is injective.

We next show that θ is surjective. Let (w,f) ∈ I(Ln). If (w,f) = (ui,cj) for some i,j, then θ(vi,dj) = (ui,cj) = (w,f). For the other cases, we do in similar way. Thus θ is surjective.

Next, it is obvious that θ(w,f) 6= (w,f) for all (w,f) ∈ I(Ln).

We next show that, θ(θ(w,f)) = (w,f) for all (w,f) ∈ I(Ln). Let (w,f) ∈

I(Ln). If (w,f) = (ui,cj) for some i,j, then θ(θ(w,f)) = θ(θ(ui,cj)) = θ(vi,dj) =

(ui,cj) = (w,f). For the other cases, we do in similar way. 0 0 Finally, we show that (w,f) and (w ,f ) are adjacent if and only if θ(w,f) 0 0 0 0 and θ(w ,f ) are adjacent for all (w,f),(w ,f ) ∈ I(Ln). 0 0 Necessity. Suppose (w,f),(w ,f ) ∈ I(Ln) are adjacent. We consider the following sixteen cases. 0 0 Case 1:(w,f) = (ui,cj) and (w ,f ) = (uk,cl) for some i,k ∈ {1,2,...,n} and j,l ∈ {1,2,...,n − 1}. We consider the following three subcases. Subcase 1.1: k = i and l = j + 1 or k = i and j = l + 1. Without loss of generality, we assume that k = i, and l = j +1. 0 0 Then θ(w,f) = θ(ui,cj) = (vi,dj), and θ(w ,f ) = θ(uk,cl) = (vk,dl) = (vi,dj+1). 0 0 Thus θ(w,f) and θ(w ,f ) are adjacent.

Ref. code: 25616009031029DVT 24

Subcase 1.2: k = i + 1 and l = j or i = k + 1 and l = j. Without loss of generality, we assume that k = i+1 and l = j. 0 0 Then θ(w,f) = θ(ui,cj) = (vi,dj), and θ(w ,f ) = θ(uk,cl) = (vk,dl) = (vi+1,dj). 0 0 Thus θ(w,f) and θ(w ,f ) are adjacent. Subcase 1.3: k = i + 1 and l = j + 1 or i = k + 1 and j = l + 1. Without loss of generality, we assume that k = i + 1 and 0 0 l = j + 1. Then θ(w,f) = θ(ui,cj) = (vi,dj), and θ(w ,f ) = θ(uk,cl) = (vk,dl) = 0 0 (vi+1,di+1). Thus θ(w,f) and θ(w ,f ) are adjacent. 0 0 Case 2:(w,f) = (ui,cj) and (w ,f ) = (vk,dl) for some i,k ∈ {1,2,...,n} and j,l ∈ {1,2,...,n − 1}. 0 0 This contradicts the fact that (w,f) and (w ,f ) are adjacent. Hence this case is impossible. 0 0 Case 3:(w,f) = (ui,cj) and (w ,f ) = (uk,ek) for some i,k ∈ {1,2,...,n} and j ∈ {1,2,...,n − 1}. We consider the following three subcases. Subcase 3.1: k = i = j. 0 0 Then θ(w,f) = θ(ui,cj) = (vi,dj) = (vi,di), and θ(w ,f ) = 0 0 θ(uk,ek) = (vk,ek) = (vi,ei). Thus θ(w,f) and θ(w ,f ) are adjacent. Subcase 3.2: k = i + 1 = j + 1 0 0 Then θ(w,f) = θ(ui,cj) = (vi,dj) = (vi,di), and θ(w ,f ) = 0 0 θ(uk,ek) = (vk,ek) = (vi+1,ei+1). Thus θ(w,f) and θ(w ,f ) are adjacent. Subcase 3.3: k = i − 1 = j 0 0 Then θ(w,f) = θ(ui,cj) = (vi,dj) = (vi,di−1), and θ(w ,f ) = 0 0 θ(uk,ek) = (vk,ek) = (vi−1,ei−1). Thus θ(w,f) and θ(w ,f ) are adjacent. 0 0 Case 4:(w,f) = (ui,cj) and (w ,f ) = (vk,ek) for some i,k ∈ {1,2,...,n} and j ∈ {1,2,...,n − 1}.

Then i = j = k. Hence θ(w,f) = θ(ui,cj) = (vi,dj) = (vi,di), and 0 0 0 0 θ(w ,f ) = θ(vk,ek) = (uk,ek) = (ui,ei). Thus θ(w,f) and θ(w ,f ) are adjacent. 0 0 Case 5:(w,f) = (vi,dj) and (w ,f ) = (uk,cl) for some i,k ∈

Ref. code: 25616009031029DVT 25

{1,2,...,n} and j,l ∈ {1,2,...,n − 1}. The proof in this case is similar to Case 2. 0 0 Case 6:(w,f) = (vi,dj) and (w ,f ) = (vk,dl) for some i,k ∈ {1,2,...,n} and j,l ∈ {1,2,...,n − 1}. The proof in this case is similar to Case 1. 0 0 Case 7:(w,f) = (vi,dj) and (w ,f ) = (uk,ek) for some i,k ∈ {1,2,...,n} and j ∈ {1,2,...,n − 1}. The proof in this case is similar to Case 4. 0 0 Case 8:(w,f) = (vi,dj) and (w ,f ) = (vk,ek) for some i,k ∈ {1,2,...,n} and j ∈ {1,2,...,n − 1}. The proof in this case is similar to Case 3. 0 0 Case 9:(w,f) = (ui,ei) and (w ,f ) = (uk,cl) for some i,k ∈ {1,2,...,n} and l ∈ {1,2,...,n − 1}. The proof in this case is similar to Case 3. 0 0 Case 10:(w,f) = (ui,ei) and (w ,f ) = (vk,dl) for some i,k ∈ {1,2,...,n} and l ∈ {1,2,...,n − 1}. The proof in this case is similar to Case 4. 0 0 Case 11:(w,f) = (ui,ei) and (w ,f ) = (uk,ek) for some i,k ∈ {1,2,...,n}. 0 0 This contradicts the fact that(w,f) and (w ,f ) are adjacent. Hence this case is impossible. 0 0 Case 12:(w,f) = (ui,ei) and (w ,f ) = (vk,ek) for some for some i,k ∈ {1,2,...,n}. 0 0 Then i = k. Hence θ(w,f) = θ(ui,ei) = (vi,ei), and θ(w ,f ) = 0 0 θ(vk,ek) = (uk,ek) = (ui,ei). Thus θ(w,f) and θ(w ,f ) are adjacent. 0 0 Case 13:(w,f) = (vi,ei) and (w ,f ) = (uk,cl) for some i,k ∈ {1,2,...,n} and l ∈ {1,2,...,n − 1}. The proof in this case is similar to Case 4. 0 0 Case 14:(w,f) = (vi,ei) and (w ,f ) = (vk,dl) for some i,k ∈ {1,2,...,n} and l ∈ {1,2,...,n − 1}.

Ref. code: 25616009031029DVT 26

The proof in this case is similar to Case 3. 0 0 Case 15:(w,f) = (vi,ei) and (w ,f ) = (ui,ei) for some i ∈ {1,2,...,n}. The proof in this case is similar to Case 12. 0 0 Case 16:(w,f) = (vi,ei) and (w ,f ) = (vk,ek) for some i,k ∈ {1,2,...,n}. The proof in this case is similar to Case 11. 0 0 0 0 Suﬃciency. Suppose (w,f),(w ,f ) ∈ I(Ln) such that θ(w,f) and θ(w ,f ) 0 0 0 0 are adjacent. By necessity, we have θ(θ(w,f)) = (w,f) and θ(θ(w ,f )) = (w ,f ) are adjacent. Therefore, θ is an incidence-symmetric bijection. This completes the proof.

Theorem 3.7. Let L be a (2,∞)−incidence labeling game on a crown graph with

2n vertices, where n > 2 is an integer. Then Bob has a winning strategy on L.

Proof. To prove this theorem, we show that all crown graphs are incidence-symmetric.

Let Cn,n be a crown graph with 2n vertices. Without loss of generality, we

may assume that its vertex set is U ∪ V , where U = {u1,u2,··· ,un} and V =

{v1,v2,··· ,vn}, and its edge set is E(Cn,n) = {ei,j = uivj : i,j ∈ {1,2,...,n} and i 6= j}. This graph is shown in Figure 3.7.

u1 v1

u2 v2 . . Cn,n : . . un−1 vn−1

un vn

Figure 3.7: The crown graph Cn,n

Deﬁne θ : I(Cn,n) → I(Cn,n) by

  (vi,ej,i) if (w,e) = (ui,ei,j); θ(w,e) =  (ui,ei,j) if (w,e) = (vi,ej,i);

Ref. code: 25616009031029DVT 27 for all (w,e) ∈ I(Cn,n). First, it is clear that θ is a function. 0 0 Next, we show that θ is injective. Assume that (w,e),(w ,e ) ∈ I(Cn,n) such 0 0 that θ(w,e) = θ(w ,e ). We consider the following two cases. 0 0 Case 1: θ(w,e) = θ(w ,e ) = (ui,ei,j) for some i,j ∈ {1,2,...,n}. The proof in this case is similar to Case 1. 0 0 Case 2: θ(w,e) = θ(w ,e ) = (vi,ej,i) for some i,j ∈ {1,2,...,n}. 0 0 Then (w,e) = (ui,ei,j) = (w ,e ).

We next show that θ is surjective. Let (w,e) ∈ I(Cn,n). If (w,e) = (ui,ei,j) for some i,j ∈ {1,2,...,n}, then θ(vi,ej,i) = (ui,ei,j) = (w,e). Otherwise, we do in similar way.

Next, it is obvious that θ(w,e) 6= (w,e) for all (w,e) ∈ I(Cn,n).

We next show that, θ(θ(w,e)) = (w,e) for all (w,e) ∈ I(Cn,n). Let (w,e) ∈

I(Cn,n). If (w,e) = (ui,ei,j) for some i,j ∈ {1,2,...,n}, then θ(θ(w,e)) = θ(θ(ui,ei,j))

= θ(vi,ej,i) = (ui,ei,j) = (w,e). Otherwise, we do in similar way. 0 0 Finally, we show that (w,e) and (w ,e ) are adjacent if and only if θ(w,e) 0 0 0 0 and θ(w ,e ) are adjacent for all (w,e),(w ,e ) ∈ I(Cn,n). 0 0 Necessity. Suppose (w,e),(w ,e ) ∈ I(Cn,n) such that they are adjacent. We consider the following four cases. 0 0 Case 1:(w,e) = (ui,ei,j) and (w ,e ) = (uk,ek,l) for some i,j,k,l ∈ {1,2,...,n}. 0 0 This contradicts the fact that (w,e) and (w ,e ) are adjacent. Hence this case is impossible. 0 0 Case 2:(w,e) = (ui,ei,j) and (w ,e ) = (vk,el,k) for some i,j,k,l ∈ {1,2,...,n}. Then i 6= k. We consider the following four subcases. Subcase 2.1: i = l and j = k. 0 0 Then θ(w,e) = θ(ui,ei,j) = (vi,ej,i), and θ(w ,e ) = θ(vk,el,k) = 0 0 (uk,ek,l) = (uj,ej,i), so θ(w,e) and θ(w ,e ) are adjacent. Subcase 2.2: i = l and j 6= k.

Ref. code: 25616009031029DVT 28

0 0 Then θ(w,e) = θ(ui,ei,j) = (vi,ej,i), and θ(w ,e ) = θ(vk,el,k) = 0 0 (uk,ek,l) = (uk,ek,i), so θ(w,e) and θ(w ,e ) are adjacent. Subcase 2.3: i 6= l and j = k. 0 0 Then θ(w,e) = θ(ui,ei,j) = (vi,ej,i), and θ(w ,e ) = θ(vk,el,k) = 0 0 (uk,ek,l) = (uj,ej,l), so θ(w,e) and θ(w ,e ) are adjacent. Subcase 2.4: i 6= l and j 6= k. Then we have the following properties.

1. Since i 6= k, we have ui 6= vk.

2. Since i 6= l and j 6= k, we get ei,j = uivj 6= ulvk = el,k.

3. The edges ei,j and el,k do not share any endpoints. By the deﬁnition of adjacency of incidences, we can conclude that (ui,ei,j) and (vi,ej,i) are not adjacent, a contradiction. Hence this case is impossible. 0 0 Case 3:(w,e) = (vi,ej,i) and (w ,e ) = (uk,ek,l) for some i,j,k,l ∈ {1,2,...,n}. The proof in this case is similar to Case 2. 0 0 Case 4:(w,e) = (vi,ej,i) and (w ,e ) = (vk,el,k) for some i,j,k,l ∈ {1,2,...,n}. The proof in this case is similar to Case 1. 0 0 Suﬃciency. Suppose (w,e),(w ,e ) ∈ I(Cn,n) such that θ(w,e) and 0 0 0 0 θ(w ,e ) are adjacent. By necessity, we have θ(θ(w,e)) = (w,e) and θ(θ(w ,e )) = 0 0 (w ,e ) are adjacent. Therefore, θ is an incidence-symmetric bijection. This completes the proof.

Ref. code: 25616009031029DVT 29

Theorem 3.8. Let L be a (2,∞)−incidence labeling game on a friendship graph

with 2n + 1 vertices, where n > 1 is an integer. Then Bob has a winning strategy on L.

Proof. By Theorem 3.1, it suﬃces to show that all friendship graphs are incidence-

symmetric. Let Fn be a friendship graph with 2n + 1 vertices. Without loss of

generality, we may assume that its vertex set is V (Fn) = {v0,v1,··· ,v2n} and its

edge set is E(Fn) = E ∪F , where E = {ej = v0vj : j ∈ {1,2,...,2n}} and F = {fk = vkvk+1 : k ∈ {1,3,...,2n − 1}}. This graph is shown in Figure 3.8.

v1 f1 v2

v2n e1 e2 v3

e2n e3 F : f2n−1 f n e2n−1 e 3 v0 4 v2n−1 ··· v4

Figure 3.8: The friendship graph Fn

Deﬁne θ : I(Fn) → I(Fn) by   (v2n−i+1,e2n−i+1) if (w,d) = (vi,ei);   θ(w,d) = (v0,e2n−j+1) if (w,d) = (v0,ej);    (v2n−i+1,f2n−k) if (w,d) = (vi,fk);

for all i,j ∈ {1,2,...,2n},k ∈ {1,3,...,2n − 1} and (w,d) ∈ I(Fn).

First, we show that θ(w,d) ∈ I(Fn) for each (w,d) ∈ I(Fn). Let (w,d) ∈

I(Fn). We consider the following three cases.

Case 1:(w,d) = (vi,ei) for some i ∈ {1,2,...,2n}. Then 2n − i ≥ 0, so 2n − i + 1 ≥ 1. Also, 2n − i + 1 ≤ 2n since i ≥ 1.

Thus 1 ≤ 2n − i + 1 ≤ 2n. Hence θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1) ∈ I(Fn).

Case 2:(w,d) = (v0,ej) for some j ∈ {1,2,...,2n}. Similar to Case 1, we have 1 ≤ 2n − j + 1 ≤ 2n. Hence θ(w,d) =

θ(v0,ej) = (v0,e2n−j+1) ∈ I(Fn).

Ref. code: 25616009031029DVT 30

Case 3:(w,d) = (vi,fk) for some i ∈ {1,2,...,2n} and k ∈ {1,3,...,2n − 1}. Similar to Case 1, we have 1 ≤ 2n−i+1 ≤ 2n. Since k ∈ {1,3,...,2n − 1},

we get 2n − k ∈ {1,3,...,2n − 1}. Furthermore, since (vi,fk) ∈ I(Fn), we have i = k or i = k + 1. Then 2n − i + 1 = (2n − k) + 1 or 2n − i + 1 = 2n − k. Hence

θ(w,d) = θ(vi,fk) = (v2n−i+1,f2n−k) ∈ I(Fn). 0 0 Second, we show that θ is a function. Assume (w,d),(w ,d ) ∈ I(Fn) such 0 0 that (w,d) = (w ,d ). We consider the following three cases. 0 0 Case 1:(w,d) = (w ,d ) = (vi,ei) for some i ∈ {1,2,...,2n}. 0 0 Then θ(w,d) = (v2n−i+1,e2n−i+1) = θ(w ,d ). 0 0 Case 2:(w,d) = (w ,d ) = (v0,ej) for some j ∈ {1,2,...,2n}. 0 0 Then θ(w,d) = (v0,e2n−j+1) = θ(w ,d ). 0 0 Case 3:(w,d) = (w ,d ) = (vi,fk) for some i ∈ {1,2,...,2n} and k ∈ {1,3,...,2n − 1}. 0 0 Then θ(w,d) = (v2n−i+1,f2n−k) = θ(w ,d ). Thus θ is a function. 0 0 Third, we show that θ is injective. Assume that (w,d),(w ,d ) ∈ I(Fn) such 0 0 that θ(w,d) = θ(w ,d ). We consider the following nine cases. 0 0 Case 1:(w,d) = (vi,ei) and (w ,d ) = (vl,el) for some i,l ∈ {1,2,...,2n}. 0 0 Then we have (v2n−i+1,e2n−i+1) = θ(vi,ei) = θ(w,d) = θ(w ,d ) = 0 0 θ(vl,el) = (v2n−l+1,e2n−l+1), which implies that i = l. Thus (w,d) = (w ,d ). 0 0 Case 2:(w,d) = (vi,ei) and (w ,d ) = (v0,em) for some i,m ∈ {1,2,...,2n}. 0 0 Consider θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1), and θ(w ,d ) = θ(v0,em) 0 0 = (v0,e2n−m+1). Then θ(w,d) 6= θ(w ,d ). Hence this case is impossible. 0 0 Case 3:(w,d) = (vi,ei) and (w ,d ) = (vl,fp) for some i,l ∈ {1,2,...,2n} and p ∈ {1,2,...,2n − 1}. 0 0 Consider θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1), and θ(w ,d ) = θ(vl,fp) 0 0 = (v2n−l+1,f2n−p). Then θ(w,d) 6= θ(w ,d ). Hence this case is impossible. 0 0 Case 4:(w,d) = (v0,ej) and (w ,d ) = (vl,el) for some j,l ∈ {1,2,...,2n}. The proof in this case is similar to Case 2.

Ref. code: 25616009031029DVT 31

0 0 Case 5:(w,d) = (v0,ej) and (w ,d ) = (v0,em) for some j,m ∈ {1,2,...,2n}. 0 0 Then we have (v0,e2n−j+1) = θ(v0,ej) = θ(w,d) = θ(w ,d ) = θ(v0,em) 0 0 = (v0,e2n−m+1), which implies that j = m. Thus (w,d) = (w ,d ). 0 0 Case 6:(w,d) = (v0,ej) and (w ,d ) = (vl,fp) for some j,l ∈ {1,2,...,2n} and p ∈ {1,2,...,2n − 1}. 0 0 Consider θ(w,d) = θ(v0,ej) = (v0,e2n−j+1), and θ(w ,d ) = θ(vl,fp) = 0 0 (v2n−l+1,f2n−p). Then θ(w,d) 6= θ(w ,d ). Hence this case is impossible. 0 0 Case 7:(w,d) = (vi,fk) and (w ,d ) = (vl,el) for some i,l ∈ {1,2,...,2n} and k ∈ {1,2,...,2n − 1}. The proof in this case is similar to Case 3. 0 0 Case 8:(w,d) = (vi,fk) and (w ,d ) = (v0,em) for some i,m ∈ {1,2,...,2n} and k ∈ {1,2,...,2n − 1}. The proof in this case is similar to Case 6. 0 0 Case 9:(w,d) = (vi,fk) and (w ,d ) = (vl,fp) for some i,l ∈ {1,2,...,2n} and k,p ∈ {1,2,...,2n − 1}. 0 0 Then we have (v2n−i+1,f2n−k) = θ(vi,fk) = θ(w,d) = θ(w ,d ) =

θ(vl,fp) = (v2n−l+1,f2n−p), which implies that i = l and k = p. Thus (w,d) = 0 0 (w ,d ). Thus θ is injective.

We next show that θ is surjective. Let (w,d) ∈ I(Fn). We consider the following three cases.

Case 1:(w,d) = (vi,ei) for some i ∈ {1,2,...,2n}.

Then we have θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1) ∈ I(Fn). Con-

sider θ(v2n−i+1,e2n−i+1) = (v2n−(2n−i+1)+1,e2n−(2n−i+1)+1) = (vi,ei) = (w,d).

Case 2:(w,d) = (v0,ej) for some j ∈ {1,2,...,2n}.

Then we have θ(w,d) = θ(v0,ej) = (v0,e2n−j+1) ∈ I(Fn). Consider

θ(v0,e2n−j+1) = (v0,e2n−(2n−j+1)+1) = (v0,ej) = (w,d).

Case 3:(w,d) = (vi,fk) for some i ∈ {1,2,...,2n} and k ∈ {1,3,...,2n − 1}.

Then we have θ(w,d) = θ(vi,fk) = (v2n−i+1,f2n−k) ∈ I(Fn). Con-

Ref. code: 25616009031029DVT 32 sider θ(v2n−i+1,f2n−k) = (v2n−(2n−i+1)+1,f2n−(2n−k)) = (vi,fk) = (w,d). We next suppose for a contradiction that θ(w,d) = (w,d) for some (w,d) ∈

I(Fn). We consider the following three cases.

Case 1:(w,d) = (vi,ei) for some i ∈ {1,2,...,2n}.

Then (vi,ei) = (w,d) = θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1). This implies that 2n = 2i − 1. This contradicts the parity property.

Case 2:(w,d) = (v0,ej) for some j ∈ {1,2,...,2n}.

Then (v0,ej) = (w,d) = θ(w,d)θ(v0,ej) = (v0,e2n−j+1). This implies that 2n = 2j − 1. This contradicts the parity property.

Case 3:(w,d) = (vi,fk) for some i ∈ {1,2,...,2n} and k ∈ {1,3,...,2n − 1}.

Then (vi,fk) = (w,d) = θ(w,d) = θ(vi,fk) = (v2n−i+1,f2n−k). This implies that 2n = 2i − 1 and n = k. This contradicts the parity property.

We next show that, θ(θ(w,d)) = (w,d) for all (w,d) ∈ I(Ln). Let (w,d) ∈

I(Ln). We consider the following three cases.

Case 1:(w,d) = (vi,ei) for some i ∈ {1,2,...,2n}.

Then θ(θ(w,d)) = θ(θ(vi,ei)) = θ(v2n−i+1,e2n−i+1)

= (v2n−(2n−i+1)+1,e2n−(2n−i+1)+1) = (vi,ei) = (w,d).

Case 2:(w,d) = (v0,ej) for some j ∈ {1,2,...,2n}.

Then θ(θ(w,d)) = θ(θ(v0,ej)) = θ(v0,e2n−j+1) = (v0,e2n−(2n−j+1)+1)

= (v0,ej) = (w,d).

Case 3:(w,d) = (vi,fk) for some i ∈ {1,2,...,2n} and k ∈ {1,3,...,2n − 1}.

Then θ(θ(w,d)) = θ(θ(vi,fk)) = θ(v2n−i+1,f2n−k)

= (v2n−(2n−i+1)+1,f2n−(2n−k)) = (vi,fk) = (w,d). 0 0 Finally, we show that (w,d) and (w ,d ) are adjacent if and only if θ(w,d) 0 0 0 0 and θ(w ,d ) are adjacent for all (w,d),(w ,d ) ∈ I(Fn). 0 0 Necessity. Suppose (w,d),(w ,d ) ∈ I(Fn) such that they are adjacent. We consider the following nine cases. 0 0 Case 1:(w,d) = (vi,ei) and (w ,d ) = (vl,el) for some i,l ∈ {1,2,...,2n}. 0 0 This contradicts the fact that (w,d) and (w ,d ) are adjacent. Hence this case is impossible.

Ref. code: 25616009031029DVT 33

0 0 Case 2:(w,d) = (vi,ei) and (w ,d ) = (v0,em) for some i,m ∈ {1,2,...,2n}. We consider the following ﬁve subcases. Subcase 2.1: i = m. 0 0 Then θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1), and θ(w ,d ) = 0 0 θ(v0,em) = (v0,e2n−m+1) = (v0,e2n−i+1). Thus θ(w,d) and θ(w ,d ) are adjacent. Subcase 2.2: i = 1 and m = 2n.

Consider θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1) = (v2n−1+1,e2n−1+1) 0 0 = (v2n,e2n), and θ(w ,d ) = θ(v0,em) = (v0,e2n−m+1) = (v0,e2n−2n+1) = (v0,e1). 0 0 Thus θ(w,d) and θ(w ,d ) are adjacent. Subcase 2.3: i − 1 = m and i ∈ {2,3,...,2n}. 0 0 Consider θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1), and θ(w ,d ) = 0 0 θ(v0,em) = (v0,e2n−m+1) = (v0,e2n−(i−1)+1) = (v0,e2n−i+2). Thus θ(w,d) and θ(w ,d ) are adjacent. Subcase 2.4: i = 2n and m = 1.

Consider θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1) = (v2n−2n+1,e2n−2n+1) 0 0 = (v1,e1), and θ(w ,d ) = θ(v0,em) = (v0,e2n−1+1) = (v0,e2n). Thus θ(w,d) and 0 0 θ(w ,d ) are adjacent. Subcase 2.5: i + 1 = m and i ∈ {1,2,...,2n − 1}. 0 0 Consider θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1), and θ(w ,d ) = 0 0 θ(v0,em) = (v0,e2n−m+1) = (v0,e2n−(i+1)+1) = (v0,e2n−i). Thus θ(w,d) and θ(w ,d ) are adjacent. 0 0 Case 3:(w,d) = (vi,ei) and (w ,d ) = (vl,fp) for some i,l ∈ {1,2,...,2n} and p ∈ {1,2,...,2n − 1}. We consider the following three subcases. Subcase 3.1: i = l = p. 0 0 Then θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1), and θ(w ,d ) = 0 0 θ(vl,fp) = (v2n−l+1,f2n−p) = (v2n−i+1,f2n−i), so θ(w,d) and θ(w ,d ) are adjacent. Subcase 3.2: i = l + 1 and p = l and l ∈ {1,3,...,2n − 1}. 0 0 Consider θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1), and θ(w ,d ) =

Ref. code: 25616009031029DVT 34

θ(vl,fp) = (v2n−l+1,f2n−p) = (v2n−l+1,f2n−l) = (v2n−(i−1)+1,f2n−(i−1)) = (v2n−i+2,f2n−i+1). 0 0 Thus θ(w,d) and θ(w ,d ) are adjacent. Subcase 3.3: i = l − 1 and p = i and l ∈ {2,4,...,2n}. 0 0 Consider θ(w,d) = θ(vi,ei) = (v2n−i+1,e2n−i+1), and θ(w ,d ) =

θ(vl,fp) = (v2n−l+1,f2n−p) = (v2n−(i+1)+1,f2n−i) = (v2n−i,f2n−i). Thus θ(w,d) 0 0 and θ(w ,d ) are adjacent. 0 0 Case 4:(w,d) = (v0,ej) and (w ,d ) = (vl,el) for some j,l ∈ {1,2,...,2n}. The proof in this case is similar to Case 2. 0 0 Case 5:(w,d) = (v0,ej) and (w ,d ) = (v0,em) for some j,m ∈ {1,2,...,2n}. 0 0 Since (w,d) and (w ,d ) are adjacent, we get j 6= m. Then θ(w,d) = 0 0 θ(v0,ej) = (v0,e2n−j+1) and θ(w ,d ) = θ(v0,em) = (v0,e2n−m+1). Since 2n−j +1 6= 0 0 2n − m + 1, we have θ(w,d) and θ(w ,d ) are adjacent. 0 0 Case 6:(w,d) = (v0,ej) and (w ,d ) = (vl,fp) for some j,l ∈ {1,2,...,2n} and p ∈ {1,2,...,2n − 1}. We consider the following two subcases. Subcase 6.1: j = l = p and j ∈ {1,3,...,2n − 1}. 0 0 Then θ(w,d) = θ(v0,ej) = (v0,e2n−j+1), and θ(w ,d ) 0 0 = θ(vl,fp) = (v2n−l+1,f2n−p) = (v2n−j+1,f2n−j). Thus θ(w,d) and θ(w ,d ) are adjacent. Subcase 6.2: j = l = p + 1 and j ∈ {2,4,...,2n}. 0 0 Consider θ(w,d) = θ(v0,ej) = (v0,e2n−j+1), and θ(w ,d ) =

θ(vl,fp) = (v2n−l+1,f2n−p) = (v2n−j+1,f2n−(j−1)) = (v2n−j+1,f2n−j+1). Thus θ(w,d) 0 0 and θ(w ,d ) are adjacent. 0 0 Case 7:(w,d) = (vi,fk) and (w ,d ) = (vl,el) for some i,l ∈ {1,2,...,2n} and k ∈ {1,2,...,2n − 1}. The proof in this case is similar to Case 3. 0 0 Case 8:(w,d) = (vi,fk) and (w ,d ) = (v0,em) for some i,m ∈ {1,2,...,2n} and k ∈ {1,2,...,2n − 1}. The proof in this case is similar to Case 6.

Ref. code: 25616009031029DVT 35

0 0 Case 9:(w,d) = (vi,fk) and (w ,d ) = (vl,fp) for some i,l ∈ {1,2,...,2n} and k,p ∈ {1,2,...,2n − 1}. 0 0 Since (w,d) and (w ,d ) are adjacent, we get l = i + 1 and p = k or i = l + 1 and p = k. Without loss of generality, we assume that l = i + 1 0 0 and p = k. Then θ(w,d) = θ(vi,fk) = (v2n−i+1,f2n−k), and θ(w ,d ) = θ(vl,fp) = 0 0 (v2n−l+1,f2n−p) = (v2n−(i+1)+1,f2n−k) = (v2n−i,f2n−k). Thus θ(w,d) and θ(w ,d ) are adjacent. 0 0 0 0 Suﬃciency. Suppose (w,d),(w ,d ) ∈ I(Fn) such that θ(w,d) and θ(w ,d ) 0 0 0 0 are adjacent. By necessity, we have θ(θ(w,d)) = (w,d) and θ(θ(w ,d )) = (w ,d ) are adjacent. Therefore, θ is an incidence-symmetric bijection. This completes the proof.

Ref. code: 25616009031029DVT 36

CHAPTER 4

CONCLUSION

In this thesis, we introduce incidence-symmetric graphs such as paths, cycles, complete graphs, ladder graphs, crown graphs, and friendship graphs. In fact, there are so many incidence-symmetric graphs. Furthermore, we introduce a (p,q)−incidence labeling game. When p = 2 and each color can be used unlimited times, we call this game a (2,∞)-incidence labeling game. Then we show that Bob has a winning strategy in (2,∞)-incidence labeling game of incidence-symmetric graphs.

Ref. code: 25616009031029DVT 37

REFERENCES

[1] Andres, S. D. (2009). The incidence game chromatic number. Discrete Applied Mathematics, 157(9), 1980-1987. doi:10.1016/j.dam.2007.10.021

[2] Bodlaender, H. L. (1991). On the complexity of some coloring games. Graph- Theoretic Concepts in Computer Science Lecture Notes in Computer Science, 30-40. doi:10.1007/3-540-53832-1 29

[3] Carse, J. P. (1987). Finite and inﬁnite games. New York: Ballantine Books.

[4] Chartrand, G., & Zhang, P. (2012). Introduction to graph theory. New York: Dover Publications.

[5] Lam, P. C. B., Shiu, W. C., & Xu, B. (1999). Edge game coloring of graphs. âĂİ Graph Theory Notes of New York, 37, 17-19.

[6] Straﬃn, P. D. (2010). Game theory and strategy. Washington: The Mathe- matical Association of America.

[7] West, D. B. (2001). Introduction to graph theory. Upper Saddle River: Pren- tice Hall.

[8] Zermelo, E (1913). Uber eine Anwendung der Mengenlehre auf die Theorie des Schachspiels. Proc. Fifth Congress Mathematicians, 501-504.

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BIOGRAPHY

Name Ms. Natthawan Sriphong Date of Birth March 13, 1995 Educational Attainment Academic Year 2016: Bachelor of Science (Mathematics), Thammasat University, Thailand Scholarships 2013-2018: Scholarship from the Science Achievement Scholarship of Thailand

Ref. code: 25616009031029DVT