半徑為3以下的中點圖

國 立 交 通 大 學

應用數學系

碩 士 論 文

半徑為 3 以下的中點圖

Median graphs of radius at most three

研 究 生：洪湧昇

指導教授：翁志文 教授

半徑為 3 以下的中點圖

Median graphs with radius at most 3

研 究 生：洪湧昇

Student：Yung-Sheng Hung

指導教授：翁志文

Advisor：Chih-Wen Weng

國 立 交 通 大 學

應 用 數 學 系

碩 士 論 文

A Thesis

Submitted to Department of Applied Mathematics

National Chiao Tung University

In partial Fulfillment of the Requirements

for the Degree of

Master

In

Applied Mathematics

June 2012

半徑為 3 以下的中點圖

學生：洪湧昇

指導教授：翁志文

國立交通大學應用數學系（研究所）碩士班

摘

要

中點圖已經被研究數十年，許多中點圖的理論及性質已經被發表，

而其大部分的理論都和其具”凸”性質的某些子圖有關。我們試著用不同

的角度去研究半徑不超過 3 的圖，過程中不用到”凸”性質並且試著找

出這些圖是中點圖的充分且必要條件。

ii

Median graphs with radius at most 3

Student：Yung-Sheng Hung

Advisor：Dr. Chih-Wen Weng

Department（Institute）of Applied Mathematics

National Chiao Tung University

ABSTRACT

Median graphs have been studied for decades. Many important theorems

and properties of median graphs have been found out and almost all

of those theorems are relative to convex. We try to study median graphs

in a different way. We consider median graphs with radius at most 3

and try to find out their necessary and sufficient conditions without using

convex property.

誌謝

這篇文章的完成，首先要感謝的就是我的指導教授───翁志文教授。從大

學時期就在代數這門課中，給了我不少啟發。也因為老師的鼓勵和支持，為我推

薦了應用數學所的碩士班，我才有這個機會來完成這篇文章。也因為對於代數與

圖論這兩個數學領域的熱愛，成為了翁老師的學生。在與老師討論研究的過程中，

老師對於問題的敏銳的洞察力以及題目發展的延伸性，都是非常值得學習的地方。

尤其往往我所寫出來複雜的論述，經過老師的修改之後，都能變成較於簡化易懂

的形式，這是最令我佩服的一點。但由於我的懶散，以至於將論文拖到了第三年

才完成，這點對於當初極力提拔我的老師，感到相當慚愧。但老師仍然盡心盡力

地協助我完成這篇論文，甚至老師也答應了我在第三年下學期出國交換的計畫，

幫我寫了推薦信，讓我在求學生涯獲得更豐富的經歷。實在相當地感謝翁老師，

所有為我付出的一切。

另外也相當感謝系主任───陳秋媛教授。在秋媛老師任職系主任期間，不

遺餘力地為系上勞心勞力，只為了多為系上爭取一些福利。而老師在身為系主任

繁忙之餘，也時常關心著我及其他學生的近況，並且給予許多建議與協助，令我

深深敬佩與感動。

當然，這段時間，有許多的好朋友在身邊支持鼓勵，分享心情，這也是這趟

旅程中不可或缺的一部份。首先就是同研究室的幾位戰友。段俊旭學長，我們在

彼此心情低潮時常常互相鼓勵，給予對方一些回應，支持對方繼續步上軌道。楊

嘉勝，范揚仁同學，以及蔡詩妤同學，從大學時期就認識的好朋友，也常常在課

業上互相討論，分享彼此的想法。並在對未來方向徬徨時，互相扶持，在論文上

也一起努力著。周彥伶同學，常常與大家分享有趣的事物，也常以點心來慰勞大

家疲憊的身心。而要特別感謝的，是同住長達五年半的室友───趙致平同學。

一樣同為大學時期就認識的朋友，在這段期間，互相鼓勵，分享日常生活，已在

這過程中建立起革命情感。感謝他常常在我遇到困難時，給予支持。並且多次協

助我克服了電腦技術與硬體上的問題，如果少了這個朋友，完成論文的期間肯定

乏味許多。

最後，要感謝所有我愛的家人與親人。感謝爸爸媽媽在我求學的路上，尊重

我的興趣與志願，讓我選擇自己想要的路來走，並且對我的幼稚與任性多所包容。

感謝哥哥，常在我意想不到時，給予我關心與無私的支持，謝謝你們。還有其他

的親人們，你們的愛，是我生活的動力，謝謝你們，我愛你們。

洪湧昇 民國一百零一年七月 於桃園南崁

Contents

Contents iv

1 Introduction 1

2 Preliminaries 2

3 Median graphs 3

4 Median graphs with radius 2 5

Median graphs of radius at most three

Chih-wen Weng

Associate Professor

Yung-Sheng Hung

Graduate student

Median graphs have been studied for decades. Many important theo-rems and properties of median graphs have been found out and almost all of those theorems are relative to convex. We try to study median graphs in a different way. We consider median graphs with radius at most 3 and try to find out their necessary and sufficient conditions without using convex property.

1 Introduction

A median of vertices u, v, and w is a vertex lies on the shortest paths between any two of them. A graph is called median graph if any triple of vertices has a unique median. Trees and n-cubes Qn are well-known median graphs.

There are many important theorems and properties of median graphs today. Those theorems have been found out by some great mathematicians. In [1], Nebeský has proved a lot of basic properties and theorems of median graphs which let us have a basic understanding of median graphs. Mulder found out the structure of median graphs, which is median graphs could be obtained from a one-vertex graph by a so-called convex expansion procedure in [2]. Also, Mulder discovered the relations between n-cubes Qn and median graphs in [3].

There are still many important theorems of median graphs. However, almost all of those theorems are relative to convex property. In order to avoid the complicated condition as convex, we try to study median graphs in a different way. We consider median graphs with radius 1, 2, and 3 and try to find out their necessary and sufficient conditions without using convex property.

In section 2, we introduce some definitions and notations as preliminary knowledge. They are needed in the rest of this paper. In section 3, the definition and some basic properties of median graphs are introduced. Also, we prove some properties which will help us to prove the necessary and sufficient conditions of median graphs with radius 2.

We start to prove those necessary and sufficient conditions of median graphs with radius 1 and 2 in section 4. In order to prove the part of median graphs with radius 2, we have used the method which mentioned by W.Imrich, S. Klavžar, H.M.Mulder in [5]. In section 5, we give two conditions and prove that they are sufficient conditions of median graphs with radius 3. We only prove a part of the necessary part but we believe that they are also the necessary conditions of median graphs with radius 3.

2 Preliminaries

At the beginning, we recall some definitions and notations needed in the rest of this paper. Given G a simple connected graph, V (G) and E(G) are vertex set and edge set of G, respectively.

Let u, v∈ V (G). If uv ∈ E(G), we say u is incident to v, denoted by u ∼ v.

By a path from u to v of length t, we mean a sequence of vertices u0 = u, u1, . . . , ut= v such that ui are distinct with possible u0= ut and uiui+1 ∈ E(G)

for 0≤ i ≤ t − 1, where u and v are called the start vertex and the end vertex

of the path respectively. The distance function d(u, v) means the length of a shortest length among paths from u to v. A cycle with length n, denoted by

Cn, is a path with same start vertex and end vertex and has length n, n ≥ 3.

We call a Cn odd cycle if n = 2k + 1, and even cycle if n = 2k + 2 for k∈ N.

The interval between u and v is the set

I(u, v) ={w ∈ V | d(u, w) + d(w, v) = d(u, v)},

i.e. those vertices on the shortest paths from u to v. The set of neighbors of

u is denoted as N (u) and defined as N (u) ={x ∈ V (G) | d(u, x) = 1}. The

number d(u) =|N(u)| is called the degree of u and we call those vertices with

degree 1 in G the leaves.

Give two graphs G and H. We say H is a subgraph of G if V (H)⊆ V (G)

and E(H)⊆ E(G). Given X ⊆ V (G), we call < X > an induced subgraph of G

if u, v∈ X and uv ∈ E(< X >) if and only if uv ∈ E(G). A subgraph H is an isometric subgraph if dG(u, v) = dH(u, v), for all u, v∈ V (H), where dG(u, v) is

the distance if u and v in G and dH(u, v) is the distance of u and v in H. A

subgraph H is convex if for all u, v∈ V (H), I(u, v) ⊆ V (H).

Give x ∈ V (G). The eccentricity of x is denoted by e(x) and defined as e(x) = max{d(x, v) | v ∈ V (G)}. The radius of G is denoted by r(G) and

defined as r(G) = min{e(x) | x ∈ V (G)}. A vertex c ∈ V (G) is a central vertex 2

of G if e(c) = r(G). By periphery of G, is a set consists of all vertices in G which has distance r(G) from some central vertex c∈ V (G).

Given two graphs G and H. The Cartesian product of G and H, denoted by

GH, is the graph with vertex set V (G) × V (H) and (a, x)(b, y) ∈ E(GH)

whenever ab∈ E(G) and x = y, or a = b and xy ∈ E(H).

A tree is a simple connected graph which has no cycle. A star is a graph with a unique center c, and E(G) = {vc | v ∈ V (G) \ {c}}. Obviously, a star is also a tree. A cube of size n, denoted by Qn, is defined inductively as

Qn = Qn₋₁Qn₋₁, n≥ 2, where Q0 is a vertex, Q1is an edge.

For an edge e = uv in a graph G, the subdivision of e is obtained by replacing the edge e by a new vertex adjacent to both u and v. For convenience, we denote the new vertex by e and the new edges by ue and ev.

A graph G is a bipartite graph if there are two set A and B such that A̸= ∅, B ̸= ∅, V (G) = A ∪ B and A ∩ B = ∅. Also, uv is not an edge if u, v ∈ A or u, v∈ B. It is well-known that G is a bipartite graph if and only if G contains

no odd-cycle.

3 Median graphs

Let G be a simple connected graph, and V (G), E(G) are vertex set and edge set of G, respectively. For u, v, w∈ V (G) we use the abbreviation

I(u, v, w) = I(u, v)∩ I(u, w) ∩ I(v, w),

and for m∈ V (G), we call m a median of u, v and w if m ∈ I(u, v, w), i.e. m lies on the paths between each two of these three vertices. A connected graph

G is a median graph if there is exactly a median for all u, v, w ∈ V (G), i.e. |I(u, v, w)| = 1. Trees and n-cubes Qn are well-known median graphs.

Lemma 3.1. Suppose u, v, w∈ V (G). Then v ∈ I(u, w) if and only if I(u, v, w) = {v}.

Proof. (⇒) If a ∈ I(u, v, w) then

(d(u, a) + d(a, v)) + (d(v, a) + d(a, w)) = d(u, v) + d(v, w) = d(u, w), so 2d(a, v) = 0. This implies a = v to have the lemma.

(⇐) Clearly {v} = I(u, v, w) ⊆ I(u, w).

The following lemma is also proved in paper [1].

Lemma 3.2. Give a simple connected graph G, u, v, w∈ V (G) and vw ∈ E(G). If{u, v, w} has a median m then either m = v or m = w, not both.

Proof. Since I(u, v, w)⊆ I(v, w) = {v, w}, the lemma follows from Lemma 3.1.

Proposition 3.3. A median graph is bipartite.

Proof. We suppose that G is a median graph but not bipartite, i.e. G contains an

odd cycle, C2k+1for some k∈ N. If it is not isometric, it must contain a smaller

isometric odd cycle C2ℓ+1 for ℓ ∈ N with l < k. Hence we just suppose it is

isometric. Then we pick vertices v, w∈ V (C2k+1) where vw is an edge. Because C2k+1 is an odd cycle, there is a vertex u ∈ V (C2k+1) such that d(u, v) = d(u, w) = k. By Lemma 3.2, the median of{u, v, w} must be v or w, without loss

of generality, say v. By definition of median, we have d(w, v) + d(v, u) = d(w, u) which is a contradiction to d(w, u) = d(v, u). Therefore, we proved that if G is a median graph then G is also a bipartite graph.

From Proposition 3.3, if a median graph contains a cycle C4 or a complete

multiple graph K2,3, then these two subgraphs are indeed the induced

sub-graphs.

Lemma 3.4. A median graph does not contain K2,3.

Proof. Suppose the graph does contain K2,3 which has bipartition {u, v, w} ∪ {s, t}. Then I(u, v, w) = {s, t}, a contradiction to the median graph definition.

Since a median graph is well-known a bipartite graph, the graphs mentioned in this paper are supposed to be connected bipartite graphs.

Definition 3.5. G1,G2 are two graphs, x∈ V (G1), y∈ V (G2), we defined the

operation coalescenceu as u(G1, G2, x, y) which is a function combine G1 and G2to a new graph G1uxyG2by deleting y and adds edges between x and N (y).

From the above definition V (G1uxyG2) = V (G1)∪ V (G2)\ {y}. Another

way to view the new edges set of G1uxyG2is as E(G1uxyG2) = E(G1)∪E(G2),

where we replace those edges xz by yz when z∈ N(y).

Lemma 3.6. G1uxyG2 is a median graph if G1, G2 are median graphs. Proof. To complete this lemma, we need to show that every three vertices u, v, w

in V (G1uxyG2) has a unique median. Obviously, it holds when u, v, w∈ V (G1)

or u, v, w∈ (V (G2)\ {y}) ∪ {x} since < V (G1) >, < (V (G2)\ {y}) ∪ {x} > are

convex subgraphs in G1uxyG2.

Therefore, we may assume, without loss of generality, u, v ∈ V (G1), w ∈

(V (G2)\ {y}) ∪ {x}. Since < V (G1) >, < (V (G2)\ {y}) ∪ {x} > are convex,

any path from w to u or v must pass x. By this fact, we get I(w, u) = I(w, x)∪

I(x, u) and I(w, v) = I(w, x)∪ I(x, v). Since < V (G1) > is convex, we have I(u, v)∩ I(x, w) = {x} or ∅. Therefore,

I(u, v, w) = I(u, v)∩ I(u, w) ∩ I(v, w)

= I(u, v)∩ (I(u, x) ∪ I(x, w)) ∩ (I(v, x) ∪ I(x, w))

= (I(u, v)∩ I(u, x) ∩ I(v, x)) ∪ (I(u, v) ∩ I(x, w)) = I(u, v, x),

where the last equality is by Lemma 3.1. Since u, v, x∈ V (G1) and < V (G1) >

is a median graph, we have

|I(u, v, w)| = |I(u, v, x)| = 1.

Corollary 3.7. Gu T is a median graph, where G is a median graph and T is

a tree.

Proof. Since tree is also a median graph, it is immediately proved by Lemma 3.6.

Lemma 3.8. Give two graphs G, H and two vertices x∈ V (G) and y ∈ V (H). If H is not a median graph, then GuxyH is not a median graph.

Proof. Suppose H is not a median graph because|I(u, v, w)| ̸= 1, where u, v, w ∈ V (H). Since < (V (H)\ {y}) ∪ {x} > is a convex subgraph in G uxyH, i.e.

I(a, b) ⊂ (V (H) \ {y}) ∪ {x}, for all a, b ∈ (V (H) \ {y}) ∪ {x}. The convex

property keeps the result|I(u, v, w)| ̸= 1 in graph G uxyH. Therefore, we have

proved that GuxyH is not a median graph.

Corollary 3.9. G1uxyG2 is a median graph if and only if both G1, G2 are median graphs.

Proof. By Lemma 3.6 and Lemma 3.8.

Lemma 3.10. Let G be a median graph. Let G′ = G\ {v ∈ V (G) | d(v) = 1},

i.e. G′ is the graph obtained from G by deleting all the leaves in G. Then G′ is also a median graph.

Proof. We can see that G is the result of repeating the operation u between

G′ and those edges incident to leaves in G. Therefore, by the fact of edges are median graphs, this lemma are proved by Corollary 3.9.

4 Median graphs with radius 2

Throughout the remaining of the thesis fix a simple connected graph G = (V (G), E(G)) with at least three vertices and a center c∈ V (G). Note that the degree d(c) of c is at least 2. We shall define some notions needed for the rest of this paper. Let

Li={x | x ∈ V (G), d(x, c) = i}

and ℓ(p) = i if p∈ Li. For p∈ Li set

p+ = {u | p ∈ I(u, c)},

p− = {u | u ∈ I(p, c)}. 5

Proposition 4.1. If G is a bipartite graph with radius 1. Then G is a median

graph.

Proof. Since G is bipartite. Bipartite graph with radius 1 is a star which is a

median graph.

Before mentioning median graphs with radius 2, we have to see some concepts from [5]. Let G = (V, E) be a graph with|V | = n, |E| = m. The graph ˆG is

obtained from G by subdividing all edges of G and adding a new vertex c joined to all the original vertices of G. So we have ˆV = V ∪ E ∪ {c} and

ˆ

E ={cv | v ∈ V } ∪ {ue | e ∈ E, u ∈ V and u is incident with e in G}.

Furthermore, the paper proves the following result:

Lemma 4.2. A graph G is triangle-free if and only if its associated graph ˆG is a median graph.  Theorem 4.3. Let G be a bipartite graph with radius 2. Then G is a median

graph if and only if the following (i)-(ii) hold. (i) G does not contain the induced subgraph K2,3.

(ii) G does not contain the induced subgraph C6⊆ L1∪ L2.

Proof. The necessity (i) follows from Lemma 3.4. For (ii), if G does contain

induced subgraph C6⊆ L1∪ L2then the three vertices in C6∩ L2has a median m in L1, and then m together with any two of the three vertices in C6∩ L1 has c as a median and another median in L2, a contradiction.

To prove sufficiency, first note that the no K2,3 assumption and radius 2 of G assumption imply that each vertex in L2 has degree at most 2, and there

is no induced subgraph C4 in L1∪ L2. We delete those leaves in G which are {v | v ∈ V (G), d(v) = 1} and this will not impact the median property by

Lemma 3.10. Thus, we can assume d(y)≥ 2 for all v ∈ V (G). Now we try to

make G to a new graph G′ by doing below steps. We delete the vertex c which is the center of G. We let V (G′) ={u | u ∈ L1} and u, v are incident if they

have a common neighbor in L2. Since there is no C4 in L1∪ L2, there are no

multiple edges in G′. Also, since there is no C6 in L1∪ L2, there is no triangle

in G′. Thus, by Lemma 4.2, G = ˆG′ is a median graph. Now we have proved this whole theorem.

5 Median graph with radius 3

To study those median graphs of higher radius, we need to introduce some more definitions and notations. In [4], it mentioned the following definitions. For

uv∈ E(G), we call uv an up-edge of u if d(u, c) < d(v, c), that is, ℓ(u) < ℓ(v).

Otherwise, we call uv a down-edge of u. Notice that G is a bipartite graph so that there is no edge uv such that d(u, c) = d(v, c). Therefore, each edge

uv is either a edge or a down-edge to u. Let down-degree d(u) (resp. up-degree d(u)) denote the number of down-edges (resp. up-edges) of u, that is,

the number of those neighbors of u in Ll(u)₋₁. By [4], we have the proposition

below

Proposition 5.1. Let G be a median graph and let v∈ Li with d(v) = k. Then

i≥ k and v and its down-edges are contained in a cube of dimension k which meets the levels Li, Li−1, ..., Li−k. 

This proposition give us some clues to develop median graphs with radius 3.

Lemma 5.2. Let G be a bipartite graph of radius 3. Suppose the following (a),

(b) hold.

(a) (forbidden condition) G does not contain the induced subgraph K3,2. (b) (enforcing condition) Every induced subgraph C6is contained in an induced

cube of dimension 3. Then the following (i)-(v) hold.

(i) If x ∈ L3, then d(x) = d(x) = k ≤ 3. Also, x and its down-edges are contained in a cube of dimension k which contains an element in L3−k. (ii) |p+_{∩ q}+_{∩ L}

i+1| ≤ 1, p, q ∈ Li, i = 1, 2;

(iii) |p−_{∩ q}+_{∩ L}

i| ≤ 2, p ∈ Li+1, q∈ Li₋₁, i = 1, 2;

(iv) If there is a induced subgraph C6in L1∪ L2, then there is a vertex a such that{a} ∪ {c} ∪ C6 is a cube, where a∈ L3 and there is no C6 in L2∪ L3 and

(v) If there is a induced subgraph C6u− x − v − c − w − z − u, where u ∈ L3, x, z ∈ L2, w, v ∈ L1. Then there are two vertices a and b such that {a} ∪ {b} ∪ C6 is a cube, where a∈ L2, b∈ L1.

Proof. (i) This is clear if d(x) = 1. Suppose d(x)≥ 2. Pick distinct a1, a2 ∈ N (x). If there exists e∈ a−₁∩a−₂∩L1, then the subgraph induced on{x, a1, e, a2}

is C4. This finishes the proof when d(x) = 2. Suppose a−1 ∩ a−2 ∩ L1 =∅. Then

we find b1, b2∈ L1 such that the subgraph induced on{x, a1, b1, c, b2, a2} is C6.

By the enforcing condition we find b3∈ L1and a3∈ L2 such that the subgraph

induced on{x, a1, b1, c, b2, a2, s, a3, b3} is a cube of dimension 3. This finishes the

proof when d(x) = 3. Suppose that there is a vertex in a4∈ N(x) − {a1, a2, a3}.

Note that a4 is not adjacent to bi for 1 ≤ i ≤ 3 , otherwise there is a K3,2.

Choose b4∈ P1such that the subgraph induced on{x, a2, b3, c, b4, a4} is C6. Use

enforcing condition again we find b5 ∈ L1 and a5∈ L2 such that the subgraph

induced on{x, a2, b3, c, b4, a4, b5, a5} is a cube of dimension 3. Note that a5̸= ai

for 1 ≤ i ≤ 4 and x ∼ a5 and a5b3 ∈ E(G). Then the induced subgraph on

{x, b3, a1, a2, a5} is a K3,2, a contradiction to the forbidden condition. Hence d(x)≤ 3.

(ii) Assume that there exist two distinct s, t ∈ p+_{∩ q}+_{∩ P}

c(i+1) for i = 1

or 2. In the case i = 1 we find K2,3 on the set{c, p, q, s, t}. For the case i = 2

if there exists a vertex u ∈ Pc1 adjacent to p and q, we still find K2,3 on the

set {u, p, q, s, t}. Suppose that there exists no vertex in Pc1 adjacent to p and

q. Then we find d, e∈ L1 such that the subgraph induced on{s, p, d, c, e, q, s}

is C6. By the enforcing condition we find b ∈ L1 and a ∈ L2 such that the

subgraph induced on{s, p, d, c, e, q, s, a, b} is a cube of dimension 3. Then the subgraph induced on{s, t, b, p, q} is K2,3, a contradiction.

(iii) This is clear from the forbidden condition assumption. (iv)-(v) It is clear from the enforcing condition.

t t t t t t t t t t t t x b1 b2 c a1 a2 a3 a4 a5 b3 b4 b5

Figure 1. A diagram to illustrate the above proof. The following is our conjecture.

Conjecture 5.3. Let G be a bipartite graph with radius 3. Then G is a median graph if and only if the following conditions (a), (b) hold.

(a) (forbidden condition) G does not contain the induced subgraph K3,2.

(b) (enforcing condition) Every induced subgraph C6 is contained in an

in-duced cube of dimension 3.

In fact the necessary condition of the above Conjecture holds for any median graphs.

Theorem 5.4. Let G be a median graph. Then the forbidden condition and the

enforcing condition hold in G.

Proof. We have proved in Lemma 3.4 for the forbidden of K23in a median graph. a1, a2, a3, a4, a5, a6, a1 be an induced C6 in G. Let m1∈ I(a1, a3, a5) and m2∈ I(a2, a4, a6). Clearly m1 ̸∈ {a1, a3, a5} by Lemma 3.1, and m1 ̸∈ {a2, a4, a6}

since the the subgraph C6 is induced. Similarly, m2 ̸∈ {a1, a2, a3, a4, a5, a6}.

Then the subgraph induced on{a1, a2, a3, a4, a5, a6, m1, m2} is an induced cube Q3.

To prove the sufficient condition of our Conjecture holds, we need more tools. Definition 5.5. A path u = u0, u1, . . . , ut = v is called a down-path from u

to v if there exist an integer 0 ≤ k ≤ t − 1 such that ℓ(u0) > ℓ(u1) > · · · > ℓ(uk) and ℓ(uk) < ℓ(uk+1) <· · · < ℓ(ut), and it is denoted by ˆPu,v.

Lemma 5.6. Let G be a bipartite graph such that for any two vertices u, v of

length two there exists a down-path for u to v. Then for any vertices u, v∈ V (G) there exists a down-path ˆPu,v.

Proof. We prove by induction on d(u, v). The case d(u, v)≤ 1 is clear since a

path of length at most 1 is a down-path. The case d(u, v) = 2 follows from our assumption. Suppose d(u, v) = t > 2. Pick a vertex u′t−1 ∈ I(u, v) ∩ N(v).

By induction there exists a down path u = u′0, u′1, . . . , u′t−1 from u to u′t−1.

Note that u′1∈ u− and d(u′1, v) = t− 1. By induction there exists a down-path u1= u′1, u2, . . . , ut= v from u′1 to v. Now the path u = u0, u1, u2, . . . , ut= v is

a down-path from u to v.

Lemma 5.7. Let G be a bipartite graph with radius 3 satisfying the forbidden

condition and the enforcing condition. Then for any two vertices u, v of length two there exists a down-path ˆPu,v.

Proof. This is clear from Lemma 5.2(i).

H.M. Mulder proved the following result in [6].

Lemma 5.8. Let G be a connected triangle free graph. If|I(u, v, w)| = 1 for any three vertices u, v, w such that d(u, v) = 2 then G is a median graph.  In fact, we only proved a part of the sufficient condition of the above Con-jecture as following theorem.

Theorem 5.9. Let G be a bipartite graph with radius 3 satisfying the forbidden

condition and the enforcing condition. Then |I(u, v, w)| ≤ 1 for all u, v, w ∈ V (G) with d(v, w) = 2.

Proof. To the contrary suppose|I(u, v, w)| > 1 for some u, v, w ∈ V (G) with d(v, w) = 2. Thus, there are two vertices a, a′ ∈ I(u, v, w). Note that d(u, v) ∈ {d(u, w), d(u, w) + 2, d(u, w) − 2}. By Lemma 3.1, we have a, a′ _{̸= u, v, w and}

d(u, v) = d(u, w) = d(u, a) + 1 = d(u, a′) + 1. If d(u, a) = 1 then the subgraph induced on {u, v, w, a, a′_{} is K2,3, a contradiction to the forbidden condition.}

Suppose 2 ≤ d(u, a) = d(u, v) − 1 ≤ 5 as G has diameter at most 6. Since d(v, w) = 2, we prove it in two situations.

Case (1) ℓ(v) = ℓ(w) = i: By condition Lemma 5.2(ii), ℓ(a)̸= ℓ(a′_{). Without}

loss of generality, suppose ℓ(a) = i− 1 and ℓ(a′_{) = i + 1 as shown in Figure 2.}

Note that i = 1, 2. t t t t a′ a w v

Figure 2. A diagram to Case (1).

Since ℓ(a) = 1, we have d(x, a) ≤ 4 for all x ∈ V (G). Therefore, we only

have to consider that d(u, a) from 2 to 4.

• d(u, a) = d(u, a′) = 2: It will cause ℓ(u) = i + 1 or i− 1.

1. Suppose ℓ(u) = i + 1. Pick y∈ ˆPu,a∩ N(a) and y′ ∈ ˆPu,a′∩ N(a′).

As |a+_{∩ a}′ − _{∩ L}

i| ≤ 2, y ̸= y′ and by Lemma 5.7 there exists

x∈ y−∩ y′−∩ Li−1 as shown in Figure 3-1. Since d(a′)≥ 3, there is

a cube Q3 contains a′, y′, v, w by Lemma 5.2(i). Note that the cube

also contains x, otherwise |c+_{∩ y}′ −_{∩ L}

1| ≥ 3. Thus, v ∼ x or w∼ x. W.L.O.G., we suppose v ∼ x, which make a contradiction to

Lemma 5.2(ii) by|a+_{∩ x}+_{∩ L} 2| ≥ 2. t t t t t t t t u a′ y′ y v w x a

Figure 3-1. The case d(u, a) = d(u, a′) = 2 and ℓ(x) = i− 1. 2. If ℓ(u) = i− 1, then i = 2., pick y′ _{∈ ˆP}

u,a′ ∩ N(a′) as shown in

Figure 3-2. Since d(a′)≥ 3, there is a cube Q3contains a′, y′, v, w by

Lemma 5.2(i). Also, this cube contains u, otherwise|c+_{∩ y}′ −_∩L 1| ≥

3. W.L.O.G., let u∼ v, which is a contradiction. 10

t t t t tu t a′ y′ v w a

Figure 3-2. The case d(u, a) = d(u, a′) = 2 and ℓ(y′) = i. • d(u, a) = d(u, a′) = 3: If ℓ(u) ≤ 1 then d(u, a) ≤ 2. Thus, we have

ℓ(u) = 2, 3.

1. If ℓ(u) = 3, pick z∈ ˆPu,a′∩ N(a′) as shown in Figure 4-1. Note that

v, w are not in ˆPu,a′. In this case a = c and|a+∩ a′ −∩ L1| ≥ 3.

t t t t t t t u a′ z v w a

Figure 4-1. The case d(u, a) = d(u, a′) = 3 and ℓ(z) = 1. 2. If ℓ(u) = 2, pick z ∈ ˆPu,a′ ∩ N(u) as shown in 4-2. Since d(a′) ≥

3, there is a cube Q3 contains a′, v, w by Lemma 5.2(i). Also, Q3

contains z as above proof. W.L.O.G., let v ∼ z, then d(u, v) = 2

which is a contradiction. t t t t t t t z u a′ v w a

Figure 4-2. The case d(u, a) = d(u, a′) = 3 and ℓ(z) = i− 1. 11

• d(u, a) = d(u, a′) = 4 If ℓ(u) ≤ 2, then d(u, a) ≤ 3. Thus, ℓ(u) = 3. Consider ˆPu,a′ as shown in Figure 5. Since d(a′)≥ 3, there is a cube Q3

contains a′, v, w by Lemma 5.2(i). Also, Q3 contains z as above proof.

W.L.O.G., let v∼ z, then d(u, v) = 3 which is a contradiction.

t t t t t t t t z u a′ v w a

Figure 5. The case d(u, a) = d(u, a′) = 4 and ˆPu,a′.

Case (2) ℓ(v) = i + 1, ℓ(w) = i− 1 : In this case, ℓ(a) = ℓ(a′_{) = i as shown}

in Figure 6. Note that i = 1, 2. Since ℓ(w) ≤ 1, we have d(u, w) ≤ 4 and d(u, a) = d(u, a′)≤ 3. Then we only have to consider that d(u, a) from 2 to 3.

t t t t v w a a′

Figure 6. A diagram to Case (2).

• d(u, a) = d(u, a′) = 2: If ℓ(u)≤ 1, then d(u, w) ≤ 2. Thus, ℓ(u) = 2, 3.

1. If ℓ(u) = 3, pick y ∈ ˆPu,a∩ N(a) and y′ ∈ ˆPu,a′∩ N(a′). Note that

y ̸= y′ and y  a′ _{and y}′ _{a, otherwise |a}+_{∩ a′ + ∩ L}

i| ≥ 2.

Now the subgraph induced on {u, y′_{, a}′_{, w, a, y} is a C}

6 as shown in

Figure 7-1. Thus, there is a vertex x ∈ L2 and x′ ∈ L1 such that {x, x′_{, u, y}′_{, a}′_{, w, a, y} is a Q}

3by Lemma 5.2(v). Observe that v = x,

otherwise |a+_{∩ a}′ +_{∩ L}

i| ≥ 2. Thus, we have u ∼ v which is a

contradiction. t t t t t t t u y y′ v a a′ w 12

Figure 7-1. The case d(u, a) = d(u, a′) = 2 and C6.

2. If ℓ(u) = 2, then i = 2. Pick y∈ ˆPu,a∩ N(a) and y′ ∈ ˆPu,a′ ∩ N(a′).

Note that y ̸= y′ _{and y  a}′ _{and y}′ _{a, otherwise |w}+_{∩ y}′ +_∩ L2| ≥ 2. Now the subgraph induced on {u, y′, a′, w, a, y} is a C6

as shown in Figure 7-2. Thus, there is a vertex x ∈ L3 such that {x, c, u, y′_{, a}′_{, w, a, y} is a Q3. Observe that v = x, otherwise it violate}

Lemma 5.2(ii) by|a+_{∩ a′ +∩ L}

3| ≥ 2. Thus, we have u ∼ v which

is a contradiction to d(u, v) = 3. t t t t t t t v u a a′ y y′ w

Figure 7-2. The diagram to illustrate above proof.

• d(u, a) = d(u, a′) = 3: If ℓ(u)≤ 2, then d(u, w) ≤ 3. Thus, ℓ(u) = 3. Pick

z∈ ˆPu,a∩ N(a) and z′∈ ˆPu,a′∩ N(a′). Note that z̸= z′ and z a′ and

z′ a, otherwise it will violate Lemma 5.2(ii) |w+_{∩ z}′ +_{∩ L} 2| ≥ 2.

Suppose there exists y ∈ ˆPu,a′ ∩ ˆPu,a such that y ∼ z and y ∼ z′, as

Figure 8-1. Now the subgraph induced on{u, z′_{, z, w, a, a}′_{} is a C6. Thus,}

there is a vertex x ∈ L3 such that {x, c, z, z′, y, a′, a, w} is a Q3. As |a+_{∩ a′ +∩ L}

3| ≤ 2, we have x = v. Therefore, d(u, v) = 2 which is a

contradiction to d(u, v) = 4. t t t t t t t t t v u y a a′ z z′ c w

Figure 8-1. The case d(u, a) = d(u, a′) = 3 and ℓ(y) = 2. If there does not exist y ∈ ˆPu,a′ ∩ ˆPu,a such that y ∼ z and y ∼ z′,

then there exists y, y′ ∈ ˆPu,a′∩ ˆPu,a such that y ∼ z and y′ ∼ z′. Now

the subgraph induced on {u, z′_{, z, c, y, y}′_{} is a C6as shown in Figure 8-2.}

Thus, there is a cube contains{u, z′_{, z, c, y, y}′_{} by the enforcing condition.}

Therefore, there exists a x∈ L2 such that x∈ ˆPu,a′∩ ˆPu,aand x∼ z and

x∼ z′. Now the situation is similar to above proof.

t t t t t t t t t t v u y _y′ a a′ z z′ c w

Figure 8-2. The case d(u, a) = d(u, a′) = 3 and C6.

Now we have proved this theorem.

References

[1] L. Nebeský, Median graphs, Comment. Math. Univ. Carolinae, 12 (1971), 317-325.

[2] H.M. Mulder, The structure of median graphs, Discrete Math., 24 (1978), 197-204.

[3] H.M. Mulder, n-Cubes and median graphs, J. Graph Theory, 4 (1980), 107-110.

[4] J. Hagauer, W. Imrich, S. Klavžar, Recognizing median graphs in sub-quadratic time, Theoretical Computer Science, 215 (1999), 123-126. [5] W. Imrich, S. Klavžar, H.M. Mulder, Median graphs and triangle-free

graphs, SIAM J. Discrete Math., 12 (1999), 111-118

[6] H.M. Mulder, The interval function of graph, Mathematical Centre Tracts 132 (1980).