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)}.
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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
E =ˆ {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
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
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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+∩ Li+1| ≤ 1, p, q ∈ Li, i = 1, 2;
{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+∩ Pc(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
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.
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Proof. We have proved in Lemma 3.4 for the forbidden of K23in a median graph.
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.
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′ − ∩ Li| ≤ 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′ −∩ L1| ≥ 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+∩ L2| ≥ 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′ ∈ ˆPu,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′ −∩L1| ≥ 3. W.L.O.G., let u∼ v, which is a contradiction.
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t
• d(u, a) = d(u, a′) = 4 If ℓ(u) ≤ 2, then d(u, a) ≤ 3. Thus, ℓ(u) = 3.
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.
Figure 7-1. The case d(u, a) = d(u, a′) = 2 and C6.
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
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).
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