Discrete Mathematics 338 (2015) 674–687
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Discrete Mathematics
journal homepage: www.elsevier.com/locate/disc
The chromatic equivalence class of K1,n,n+2 ✩
Boon Leong Ng ∗ , Fengming Dong ∗
Mathematics and Mathematics Education, National Institute of Education, Nanyang Technological University, 1 Nanyang Walk,
Singapore 637616, Singapore
article
abstract
info
Article history:
Received 29 April 2014
Received in revised form 18 December 2014
Accepted 19 December 2014
Available online 13 January 2015
Keywords:
Complete tripartite graphs
Chromatic polynomials
Chromatic equivalence
The chromatic equivalence class of a graph G is the set of graphs that have the same
chromatic polynomial as G. We find the chromatic equivalence class of the complete
tripartite graphs K1,n,n+2 for all n ≥ 2. This partially answers a question raised in Chia
and Ho (2009), which asks for the chromatic equivalence class of the graph K1,m,n where
2 ≤ m ≤ n.
© 2014 Elsevier B.V. All rights reserved.
1. Definitions and background
We consider only finite, undirected and simple graphs. Let G be a graph and P (G, λ) be its chromatic polynomial. The
chromatic equivalence class of G, denoted by C (G), is the set of graphs that have the same chromatic polynomial as G. If
C (G) = {G}, then G is said to be chromatically unique. Also, let Tk denote the set of trees with k vertices.
The chromatic polynomial P (G, λ) of a finite undirected simple graph G is given by
P (G, λ) =
|
V (G)|
si (G)(λ)i
i=1
where G is the complement of G, si (G) is the number of spanning subgraphs of G with exactly i components, each of which
is a complete graph, and (λ)i = λ(λ − 1) · · · (λ − i + 1) (see Theorem 1.4.1 of [6]).
Define the join of two vertex-disjoint graphs G and H, written as G + H, as follows:
V (G + H ) = V (G) ∪ V (H ),
E (G + H ) = E (G) ∪ E (H ) ∪ {(u, v)|u ∈ V (G) and v ∈ V (H )}.
Then we have (see Theorem 1.5.1 of [6])
P (G + H , λ) =
|
V (G)| |V
(H )|

i =1
si (G)sj (H )(λ)i+j .
j=1
✩ This paper was partially supported by NIE AcRf (RI 2/12 DFM) of Singapore.
∗
Corresponding authors.
E-mail addresses: [email protected] (B.L. Ng), [email protected] (F.M. Dong).
http://dx.doi.org/10.1016/j.disc.2014.12.013
0012-365X/© 2014 Elsevier B.V. All rights reserved.
B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
675
We therefore have:
Lemma 1. For graphs G, H1 and H2 , the graphs G + H1 and G + H2 are chromatically equivalent if and only if H1 and H2 are
chromatically equivalent.
Let Km1 ,m2 ,...,mk be the complete k-partite graph with partite sets X1 , . . . , Xk where |Xi | = mi for 1 ≤ i ≤ k. Chao and
Novacky, Jr. [1] have shown that Km1 ,m2 ,...,mk is chromatically unique if |mi − mj | ≤ 1 for all i, j ∈ {1, 2, . . . , k}.
For k = 2, Teo and Koh [10,11] and, independently, Dong [5], have shown that Km1 ,m2 is chromatically unique if
m1 , m2 ≥ 2. If at least one of m1 , m2 is equal to 1, then Km1 ,m2 is a tree and hence we have:
Lemma 2. The chromatic equivalence class of K1,n is Tn+1 .
The case where k = 3 has been studied by many authors, beginning with Chia, Goh and Koh [2]. The following complete
tripartite graphs have been shown to be chromatically unique:
• Km1 ,m2 ,m3 where |mi − mj | ≤ 3, and mi ≥ 2 where i ∈ {1, 2, 3} (see [3]),
• Km,n,n where 2 ≤ m ≤ n (see [3]),
• Kn−m,n,n+k where m ≥ 2, k ≥ 0 and n ≥
m2 +k2 +mk+m−k+4
3
• Kn−m,n,n+k where m ≥ 0, 2 ≤ k ≤ 4 and n ≥
• Kn−3,n,n+1 where n ≥ 5 (see [4]).
(m+k)2
4
(see [9]),
+ 1 (see [8]), and
It is still an open problem whether all complete tripartite graphs Km1 ,m2 ,m3 where mi ≥ 2 for all i ∈ {1, 2, 3} are chromatically
unique (see Problem 4.3.1 of [6]).
On the other hand, the complete tripartite graphs K1,n,n+k , n ≥ 1, k ≥ 0, have received far less attention. It is clear that
the graph K1,n,n+k is isomorphic to both K n + K1,n+k and K n+k + K1,n . Applying Lemmas 1 and 2, we find that the set of graphs
J (n, n + k) = {T + K n , S + K n+k |T ∈ Tn+k+1 , S ∈ Tn+1 } is a subset of C (K1,n,n+k ).
Whitehead [12] has shown that when n = 1, then C (K1,n,n+k ) = C (K1,1,1+k ) is the set of two-trees with k + 3 vertices.
Chia and Ho [3] have shown that J (n, n + k) is indeed the chromatic equivalence class of K1,n,n+k for n ≥ 2, k = 0 as
well as the special cases (n, k) = (2, 2) and (3, 1). They then pose the question of finding the chromatic equivalence class
of K1,n,n+k where n ≥ 2, k ≥ 1. In a subsequent paper [4], they show that J (n, n + k) is the chromatic equivalence class of
K1,n,n+k for n ≥ 2, k = 1.
In this paper, we will present a proof for the cases where k ∈ {1, 2}, n ≥ 2. We will repeatedly make use of the fact that
the following parameters of a graph G can be deduced from P (G, λ) (see [6,7]):
• The number of vertices and the number of edges of G.
• The number of C3 -subgraphs of G. For the rest of this paper, C3 -subgraphs will be referred to as triangles.
• The difference between the number of K4 -subgraphs of G and twice the number of induced C4 -subgraphs of G. If G is a
tripartite graph, it has no K4 -subgraphs and hence we can deduce the number of induced C4 -subgraphs of G from P (G, λ).
• For every integer χ ≥ χ (G), the number of ways of partitioning V (G) into exactly χ independent sets. Equivalently, the
number of ways of proper colourings of G that use exactly χ colours.
If G and H are chromatically equivalent graphs, then the values of these parameters for G and H must be identical.
2. Preliminary lemmas
We follow the definitions and notations in Section 2 of [3]. Let G be a complete tripartite graph Km1 ,m2 ,m3 . The number of
vertices p is m1 + m2 + m3 and the number of edges q is m1 m2 + m2 m3 + m3 m1 . We try to obtain some necessary conditions
that graphs in C (G) must satisfy.
Suppose Y is obtained by deleting e edges from the complete tripartite graph Kr1 ,r2 ,r3 , where r1 + r2 + r3 = p and
e=

i<j
ri rj −

mi mj .
i <j
In other words, Y is the union of three complete subgraphs Kr1 , Kr2 , and Kr3 with e edges joining these subgraphs. Thus Y
and G have the same order and the same size.
Definition 3. Let ai to be the number of edges joining the subgraphs Krj and Krk in Y , where {i, j, k} = {1, 2, 3}.
Then
3
i=1
ai = e. The following is a result on G and Y from [3]:
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B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
Lemma 4. Let G be the complete tripartite graph Km1 ,m2 ,m3 and Y be obtained by deleting e edges from the complete tripartite
graph Kr1 ,r2 ,r3 , where r1 + r2 + r3 = p and |E (G)| = |E (Y )|. Then for each j ∈ {1, 2, 3},
sp−2 (G) − sp−2 (Y ) ≥
3

(ri − mj ) −
i=1
3

ai (ri − mj ).
i=1
Note that the left-hand side is a constant independent of our choice of j.
For the rest of this paper, we assume that Y ∈ C (G) where G = Km1 ,m2 ,m3 , and Y is a subgraph of Kr1 ,r2 ,r3 where
r1 ≤ r2 ≤ r3 . It is clear that si (Y ) = si (G) for all χ (G) ≤ i ≤ p, in particular, sp−2 (Y ) = sp−2 (G).
Lemma 5. Let G and Y be as described above. Then (r3 − m1 )(r3 − m2 )(r3 − m3 ) ≤ 0.
Proof. Substituting j = 3 in Lemma 4, we thus have
0 ≥
3

(ri − m3 ) −
i=1
≥
3

3

ai (ri − m3 )
i=1
(ri − m3 ) −
i=1
3

ai (r3 − m3 )
i=1
= (r3 − m3 )[(r1 − m3 )(r2 − m3 ) − e]




mi mj
ri rj +
= (r3 − m3 ) (r1 − m3 )(r2 − m3 ) −
i<j
i<j

= (r3 − m3 ) m1 m2 − (r1 + r2 )(r3 + m3 ) + m3
3


mi
i=1

= (r3 − m3 ) m1 m2 − (r1 + r2 )(r3 + m3 ) + m3
3


ri
i=1
= (r3 − m3 )[r3 (m3 − r1 − r2 ) + m1 m2 ]
= (r3 − m3 )[r3 (r3 − m1 − m2 ) + m1 m2 ]
= (r3 − m3 )(r3 − m2 )(r3 − m1 ). Corollary 6. Let G and Y be as described above. If m1 ≤ m2 ≤ m3 , then m2 ≤ r3 ≤ m3 .
Lemma 7. For any positive integers k, n where n ≥ 2, let G be K1,n,n+k and Y ∈ C (G). If Y is a subgraph of Kk+1,n,n and a1 = 0
(where ai is defined in Definition 3) then Y ∈ {T + K n |T ∈ Tn+k+1 }.
Proof. Let X1 , X2 and X3 represent the three partite sets of vertices of Y , with |X1 | = k + 1 and |X2 | = |X3 | = n. If a2 = 0
then Y = T + X3 = T + K n and, using Lemma 1, we get T ∈ Tn+k+1 . Similarly, if a3 = 0 then Y ∈ {T + K n |T ∈ Tn+k+1 }. We
will show that no graph exists satisfying the given conditions with a2 , a3 ≥ 1.
Suppose a2 , a3 ≥ 1. Let S1 ⊂ X1 (respectively, T1 ⊂ X1 ) be the set of vertices incident to deleted edges incident to vertices
in X2 (respectively, X3 ). Note that since a2 , a3 ≥ 1, we have S1 , T1 ̸= ∅.
Since G has n2 + kn + 2n + k edges and Kk+1,n,n has n2 + 2kn + 2n edges, Y is obtained from Kk+1,n,n by deleting k(n − 1)
edges, hence a2 + a3 = k(n − 1). Since G has n2 + kn triangles and Kk+1,n,n has kn2 + n2 triangles, Y is obtained from Kk+1,n,n
by deleting kn(n − 1) triangles. Since a1 = 0, the deleted edges are all incident to X1 . Each edge incident to X1 belongs to
exactly n triangles, hence no two deleted edges belong to the same triangle. Hence S1 ∩ T1 = ∅.
If G and Y are chromatically equivalent
  then they must have the same number of induced C4 -subgraphs. The number
of induced C4 -subgraphs in G is
n
2
n +k
2
. We now try to find an upper bound on the number of induced C4 -subgraphs in
Y , and finally show that the number of induced C4 -subgraphs in Y is strictly less than
a2 , a3 ≥ 1.
There are five types of induced C4 -subgraphs in Y :
(a)
(b)
(c)
(d)
(e)
Two vertices in X2 and two vertices in X3 ;
Two vertices in X1 and two vertices in X2 ;
Two vertices in X1 and two vertices in X3 ;
Two vertices in X3 , one vertex in X1 and one vertex in X2 ;
Two vertices in X2 , one vertex in X1 and one vertex in X3 .
 n   n +k 
2
2
under the condition that
B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
Note that the number of induced C4 -subgraphs of type (a) is
n n
2
2
677
and the number of induced C4 -subgraphs of types
(d) and (e) are a3 2 and a2 2 respectively; the sum of the number of types (d) and (e) is k(n − 1) 2 .
Now we consider the number of induced C4 -subgraphs of types (b) and (c).
Assume that X1 is partitioned into S = {u1 , u2 , . . . , us } and T = {v1 , v2 , . . . , vk+1−s } such that each ui ∈ S is adjacent to
all vertices in X3 and each vj ∈ T is adjacent to all vertices in X2 . Note that S1 ⊆ S and T1 ⊆ T .
Let di be the degree of ui in the subgraph Y [X1 ∪ X2 ] of Y induced by X1 ∪ X2 , and d′j be the degree of vj in the subgraph
Y [X1 ∪ X3 ] of Y induced by X1 ∪ X3 .
Then the number of induced C4 -subgraphs of type (b) is actually the number of C4 -subgraphs in Y [X1 ∪ X2 ], which is
bounded from above by
n

k
d1

2
n
+ (k − 1)

d2

2

+ · · · + (k + 1 − s)
ds


+
2
This is because, in the subgraph Y [X1 ∪ X2 ], there are at most k
2 ≤ i ≤ s, there are at most (k − i + 1)
Finally, there are exactly

 
k+1−s
n
2
2
 
di
2
n
k+1−s
 
n
2
 
d1
2
2
.
(1)
induced C4 -subgraphs containing u1 , and for each i where
induced C4 -subgraphs containing ui but not containing uj for all 1 ≤ j < i.
induced C4 -subgraphs not containing ui where 1 ≤ i ≤ s. Note that the expression in
(1) remains as an upper bound for the number of induced C4 -subgraphs of type (b) for any permutation of d1 , d2 , . . . , dk .
Similarly, the number of induced C4 -subgraphs of type (c) is bounded from above by

k
d′1

2
+ (k − 1)

d′2

2

+ ··· + s
d′k−s+1

2
+
 s  n
2
2
.
Now we consider an upper bound for
  k
 ′
s
−s+1

di
d
(k − i + 1)
+
(k − j + 1) j .
A=
2
i=1
(2)
2
j =1
By the rearrangement inequality, A is maximised when d1 ≥ d2 ≥ · · · ≥ ds and d′1 ≥ d′2 ≥ · · · ≥ d′k+1−s . We can assume
that this condition is satisfied.
Now
d1 + d2 + · · · + ds + d′1 + · · · + d′k+1−s = (k + 1)n − k(n − 1) = n + k.
We also know that di , d′j ≥ 1 for all 1 ≤ i ≤ s, 1 ≤ j ≤ k − s + 1, because Y is uniquely 3-colourable. Thus 1 ≤ di ≤ n and
1 ≤ d′j ≤ n, and if d1 = n then di = 1 for all i ≥ 2 and d′j = 1 for all j. Similarly, if d′1 = n then di = 1 for all i and d′j = 1 for
all j ≥ 2. Therefore A is maximised if d1 = n or d′1 = n, because for all a ≥ b ≥ 1, we have
a
2
 
b
+
2
<

a+1


+
2
b−1

2
.
Hence
A≤k
n
2
,
with equality only if d1 = n or d′1 = n.
Combining the above results on the number of induced C4 -subgraphs in Y , we find that the number of induced
C4 -subgraphs in Y is at most
B=
n n
2
2
+ k(n − 1)
n
2
+k
n
2

+
k+1−s
2
 
n
2
+
 s  n
2
2
′
with equality only if d1 = n or d1 = n.
Now we show that
 nthe
 number of induced C4 -subgraphs in Y is in fact strictly less than this. Suppose the contrary, then
we must have A = k 2 and so either d1 = n or d′1 = n. But we can show that the number of induced C4 -subgraphs in Y is
still strictly less than B even when d1 = n or d′1 = n.
Suppose that d1 = n. Then s ≥ 1 and di = 1 for all i ≥ 2 and d′j = 1 for all j. If s = 1, then a3 = 0, which is a contradiction,
so s ≥ 2. We now exchange the labels of u1 and us (i.e., exchange the values of d1 and ds ) to get another upper bound for the
number of induced C4 -subgraphs in Y . If d1 and ds are exchanged, by Eq. (2), we have
A ≤ (k + 1 − s)
n
2
<k
n
2
and thus the number of induced C4 -subgraphs in Y is strictly less than B. We reach the same conclusion if d′1 = n.
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B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
As G and Y have the same number of induced C4 -subgraphs, we have

n+k
2
 
n
2
<
n n
2
2
+ k(n − 1)
n
2
+k
n
2

+
k+1−s
2
 
n
2
+
 s  n
2
2
.
Now s = 0 implies that a3 = 0 and s = k + 1 implies that a2 = 0, therefore we have 1 ≤ s ≤ k. Then the right-hand side of
the above inequality is largest when s = 1 or s = k. Therefore,

n+k
2
 
n
2
<
n n
2
2
+ k(n − 1)
n
2
+k
n
2
+
  
k
n
2
2
.
However, the two sides of the above inequality are algebraically identical, a contradiction. Hence either a2 = 0 or a3 = 0
and thus Y ∈ {T + K n |T ∈ Tn+k }. Lemma 8. For any integers k and n where n ≥ 2 and 1 ≤ k ≤ n − 2, let G be K1,n,n+k and Y ∈ C (G). If Y is a subgraph of
Kr1 ,r2 ,r3 and r3 = n, then Y ∈ {T + K n |T ∈ Tn+k+1 }.
Proof. Substituting r3 = n = m2 into Lemma 4 with j = 2 gives us
0 ≥ −a1 (r1 − m2 ) − a2 (r2 − m2 )
≥ −(a1 + a2 )(r2 − m2 )
since r1 ≤ r2 . Since a1 and a2 are nonnegative, either a1 = a2 = 0 or r2 ≥ m2 = n. The former case implies that Y = T + K n
and, using Lemma 1, we get T ∈ Tn+k+1 , and the latter case implies that r2 = r3 = n so r1 = k + 1. Hence Y is a subgraph
of Kk+1,n,n .
Since G has n2 + kn + 2n + k edges and Kk+1,n,n has n2 + 2kn + 2n edges, Y is obtained from Kk+1,n,n by deleting k(n − 1)
edges. Since G has n2 + kn triangles and Kk+1,n,n has kn2 + n2 triangles, Y is obtained from Kk+1,n,n by deleting kn(n − 1)
triangles.
3
3
The number of deleted triangles is bounded from above by i=1 ai ri . However, we have i=1 ai = k(n − 1). Hence we
have
3

ai ri ≥ kn(n − 1) = n


3

i=1
ai
i=1
(k + 1)a1 + na2 + na3 ≥ na1 + na2 + na3
(k + 1)a1 ≥ na1 .
Since n > k + 1, we must have a1 = 0. By Lemma 7, Y ∈ {T + K n |T ∈ Tn+k+1 }.
Lemma 9. For any positive integers k, n where n ≥ 2, let G be K1,n,n+k and Y ∈ C (G). If Y is a subgraph of Kr1 ,r2 ,r3 and r3 = n + k,
then Y ∈ {S + K n+k |S ∈ Tn+1 }.
Proof. Substituting r3 = n + k = m3 into Lemma 4 with j = 3 gives us
0 ≥ −a1 (r1 − m3 ) − a2 (r2 − m3 )
≥ −(a1 + a2 )(r2 − m3 )
since r1 ≤ r2 . Since a1 and a2 are nonnegative, either a1 = a2 = 0 or r2 ≥ m3 = n + k. The former case implies that
Y = S + K n+k and, using Lemma 1, we get S ∈ Tn+1 , and the latter case implies that r1 ≤ 0, a contradiction. Lemma 10. For any positive integers k, n where n ≥ 2, let G be K1,n,n+k and Y ∈ C (G). If Y is a subgraph of Kr1 ,r2 ,r3 and
r3 = n + u, where 1 ≤ u ≤ k − 1 then
(a1 − u(k − u))(r3 − r1 ) + (a2 − u(k − u))(r3 − r2 ) ≤ u(k − u)(k − 2u).
Proof. As r3 = n + u and Y ∈ C (G), Y is obtained from Kr1 ,r2 ,r3 by deleting r1 r2 + uk + u − u2 − n − k edges. Also,
r1 r2 (n + u) − n2 − kn triangles will be deleted from Kr1 ,r2 ,r3 after deleting these edges. We also have r1 + r2 = n + k + 1 − u.
Substituting this into Lemma 4 with j = 1 gives us
0 ≥ (r1 − 1)(r2 − 1)(n + u − 1) − a1 (r1 − 1) − a2 (r2 − 1) − a3 (n + u − 1).
(3)
Also, Lemma 4 with j = 3 gives us
0 ≥ −(r1 − (n + k))(r2 − (n + k))(k − u) − a1 (r1 − (n + k)) − a2 (r2 − (n + k)) + a3 (k − u)
= −(r2 + u − 1)(r1 + u − 1)(k − u) + a1 (n + k − r1 ) + a2 (n + k − r2 ) + a3 (k − u).
(4)
B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
679
Adding Eq. (4) multiplied by n + u − 1 to Eq. (3) multiplied by k − u gives
0 ≥ (n + u − 1)[−(k − u)(r1 + r2 + (u − 1))(u − 1) + a1 (n + k − r1 ) + a2 (n + k − r2 )]
+ (k − u)[−(r1 + r2 − 1)(n + u − 1) − a1 (r1 − 1) − a2 (r2 − 1)]
= (n + u − 1)[−(k − u)(n + k)(u − 1) + a1 (n + k − r1 ) + a2 (n + k − r2 )]
+ (k − u)[−(n + k − u)(n + u − 1) − a1 (r1 − 1) − a2 (r2 − 1)]
= (k − u)(n + u − 1)[−(n + k)(u − 1) − (n + k − u)]
+ a1 [(n + u − 1)(n + k − r1 ) − (k − u)(r1 − 1)] + a2 [(n + u − 1)(n + k − r2 ) − (k − u)(r2 − 1)]
= (k − u)(n + u − 1)(−u)(n + k − 1) + (n + k − 1)[a1 (n + u − r1 ) + a2 (n + u − r2 )]
= (n + k − 1)[a1 (n + u − r1 ) + a2 (n + u − r2 ) − u(k − u)(n + u − 1)]
= (n + k − 1)[(a1 − u(k − u))(n + u − r1 ) + (a2 − u(k − u))(n + u − r2 )
+ u(k − u)((2n + 2u − r1 − r2 ) − (n + u − 1))]
= (n + k − 1)[(a1 − u(k − u))(n + u − r1 ) + (a2 − u(k − u))(n + u − r2 ) + u(k − u)(2u − k)].
Since n + k − 1 > 0, we get
(a1 − u(k − u))(r3 − r1 ) + (a2 − u(k − u))(r3 − r2 ) ≤ u(k − u)(k − 2u). 3. The chromatic equivalence classes of K1,n,n+1 and K1,n,n+2
We first determine C (K1,n,n+1 ). This result was first obtained by Chia and Ho [4].
Theorem 11. For any positive integer n ≥ 2, C (K1,n,n+1 ) = J (n, n + 1).
Proof. Let Y ∈ C (K1,n,n+1 ). The case where n = 2 is proven in [3]. Suppose n ≥ 3. From Corollary 6, we find that r3 = n or
r3 = n + 1. If r3 = n, by Lemma 8, Y ∈ {T + K n |T ∈ Tn+2 }. If r3 = n + 1, by Lemma 9, Y ∈ {S + K n+1 |S ∈ Tn+1 }. To determine C (K1,n,n+2 ), we first take care of a special case.
Lemma 12. For any positive integer n ≥ 3, let G be K1,n,n+2 and Y ∈ C (G). If Y is a subgraph of K2,n,n+1 then Y ∈ {T + K n |T ∈
Tn+3 }.
Proof. Y and G must have the same number of edges and triangles. Now G has q = n2 + 4n + 2 edges and K2,n,n+1 has
n2 + 5n + 2 edges, so the number of edges deleted from K2,n,n+1 is n. Also, G has n2 + 2n triangles and K2,n,n+1 has 2n2 + 2n
triangles. Therefore, by deleting the n edges, we must also remove n2 triangles from K2,n,n+1 . Let X1 , X2 and X3 represent the
three partite sets of vertices of K2,n,n+1 , with X1 = {v1 , v2 }, |X2 | = n and |X3 | = n + 1.
The number of deleted triangles is bounded from above by
3


ai r i ≥ n2 = n
i =1
3

3
i=1
ai ri . However, we have
3
i=1
ai = n. Hence we have

ai
i=1
2a1 + na2 + (n + 1)a3 ≥ na1 + na2 + na3
a3 ≥ (n − 2)a1 .
(5)
If a1 ≥ 2 then a3 ≤ n − 2. Then we have
n − 2 ≥ a3 ≥ (n − 2)a1 ≥ 2n − 4,
which is true only when n ≤ 2. Since we are considering n ≥ 3, we must have a1 ≤ 1.
We count the number of ways of partitioning V (G) into four nonempty independent sets. This can be done only by
partitioning either the partite set with n vertices or the partite set with n + 2 vertices (but not both) into two nonempty
sets, and hence there are a total of 2n−1 − 1 + 2n+1 − 1 = 5 × 2n−1 − 2 ways. We will show that if the number of ways of
partitioning Y into four nonempty independent sets is equal to 5 × 2n−1 − 2, then Y ∈ {T + K n |T ∈ Tn+3 }.
The case when a1 = 1 Returning to Eq. (5), we find that a3 ≥ n − 2 and so a2 ≤ 1.
3
If a2 = 1 then a3 = n − 2. Also i=1 ai ri = n2 which is the number of deleted triangles. Hence, no pair of deleted edges
belongs to the same triangle. Without loss of generality, the a3 deleted edges between X1 and X2 are incident to v1 ∈ X1 ,
the deleted edge between X2 and X3 is incident to w1 ∈ X2 and u1 ∈ X3 , and the deleted edge between X1 and X3 is incident
to v2 ∈ X1 and u2 ∈ X3 . Note that (i) there is one vertex in X2 that is still adjacent to every vertex in X1 ∪ X3 , and (ii) there
are r3 − 2 = n − 1 vertices in X3 that are still adjacent to every vertex in X1 ∪ X2 . We then count the number of ways of
partitioning V (Y ) into four nonempty independent sets (‘‘parts’’). We have two cases.
680
B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
1. Suppose none of the vertices in X1 ∪ X2 are in the same part as any vertex in X3 . Then we can either partition X3 into two
nonempty sets, and there are 2n − 1 ways of doing this, or we can partition X1 ∪ X2 into three nonempty sets. If v1 and
v2 are in the same part then we can only partition X2 into two nonempty sets, and there are 2n−1 − 1 ways of doing this.
If v1 and v2 are in different parts then we can select some (possibly none) of the a3 vertices in X2 which are not adjacent
to v1 to be in the same set as v1 . There are 2a3 = 2n−2 ways of doing this. Hence there are 3 × 2n−1 + 2n−2 − 2 ways
altogether.
2. Suppose at least one of the vertices in X1 ∪ X2 are in the same part as at least one vertex in X3 . Then the partition either
has {v2 , u2 } as one of the parts, or {w1 , u1 } as one of the parts. Both parts cannot exist in the same partition as it would
then require partitioning V (Y ) into five independent sets. If {v2 , u2 } is a part, then we can select some (possibly none) of
the a3 vertices in X2 which are not adjacent to v1 to be in the same part as v1 . There are 2a3 = 2n−2 ways of doing this. If
{w1 , u1 } is a part, then there is only 1 way of partitioning V (Y ) \ {w1 , u1 } into three nonempty independent sets. Hence
there are 2n−2 + 1 ways altogether.
Hence the total number of ways of partitioning V (Y ) into four nonempty independent sets is 4 × 2n−1 − 1. However, this is
less than 5 × 2n−1 − 2 for n > 2.
If a2 = 0 then a3 = n − 1. Hence Y [X1 ∪ X2 ] has n + 2 vertices and n + 1 edges. Since G is uniquely 3-colourable, so
3
2
is Y . Hence Y [X1 ∪ X2 ] is a tree. Also,
i=1 ai ri = n + 1 which is one more than the number of deleted triangles. Hence
one pair of deleted edges belongs to the same triangle. Hence some vertex w ∈ X2 is incident to a deleted edge between
X1 and X2 , and the deleted edge between X2 and X3 . Let the deleted edge between X2 and X3 be incident to u ∈ X3 . Since X2
has n vertices, there exists a vertex in X2 which is not incident to any deleted edge. We then count the number of ways of
partitioning V (Y ) into four nonempty independent sets (‘‘parts’’). We have two cases.
1. Suppose none of the vertices in X1 ∪ X2 are in the same part as any vertex in X3 . Then we can either partition X3 into two
nonempty sets, and there are 2n − 1 ways of doing this, or we can partition X1 ∪ X2 into three nonempty sets. Y [X1 ∪ X2 ] is
a tree with n + 2 vertices, so there are 3 × 2n+1 proper colourings with at most three colours. We divide by 3! to obtain the
number of ways of partitioning X1 ∪ X2 into at most three nonempty independent sets. There is also exactly one way of
partitioning X1 ∪ X2 into two nonempty independent sets. Hence the number of ways of partitioning X1 ∪ X2 into exactly
three nonempty independent sets is
3 × 2n+1
3!
− 1 = 2n − 1.
(6)
Hence there are 4 × 2n−1 − 2 ways altogether.
2. Suppose {w, u} is one of the parts. Then there is only 1 way of partitioning V (Y )\{w, u} into three nonempty independent
sets.
Hence the total number of ways of partitioning V (Y ) into four nonempty independent sets is 4 × 2n−1 − 1. However, this is
less than 5 × 2n−1 − 2 for n > 2.
Neither of these subcases leads to any graph that is chromatically equivalent to G. Since a1 ̸= 1, it can only be the case
that a1 = 0.
The case when a1 = 0 In this case, every edge deleted from K2,n,n+1 is incident with X1 . Since G is uniquely 3-colourable, so is
Y . Therefore every vertex in X2 and X3 is incident to at most one deleted edge. For i ∈ {1, 2}, j ∈ {2, 3}, let bij be the number
of deleted edges between vi and Xj .
Since Y is uniquely 3-colourable, the subgraph Y [X1 ∪ X2 ] of Y induced by X1 ∪ X2 is connected and so every vertex in X2
is incident with at most one deleted edge. Similarly, every vertex in X3 is also incident with at most one deleted edge.
We then count the number of ways of partitioning V (Y ) into four nonempty independent sets (‘‘parts’’). We have two
cases.
1. If v1 and v2 belong to the same part, then, since every vertex in X2 ∪ X3 is adjacent to one of them, they cannot belong
to the same part as {v1 , v2 }. Hence we must partition X2 ∪ X3 into three nonempty independent sets. But Y [X2 ∪ X3 ] is
a complete bipartite graph. Hence we can either partition X2 into two nonempty sets or partition X3 into two nonempty
sets. There are 2n−1 − 1 ways to do the former and 2n − 1 ways to do the latter, for a total of 3 × 2n−1 − 2 ways.
2. If v1 and v2 belong to different parts then some (possibly none) of the b12 vertices in X2 or some (possibly none) of
the b13 vertices in X3 , but not both together, can be in the same part as v1 . Hence there are 2b12 + 2b13 − 1 ways to
select the part containing v1 . Similarly, there are 2b22 + 2b23 − 1 ways to select the part containing v2 . Hence there are
(2b12 + 2b13 − 1)(2b22 + 2b23 − 1) ways altogether.
The total number of ways is thus
3 × 2n−1 + (2b12 + 2b13 − 1)(2b22 + 2b23 − 1) − 2.
Hence
5 × 2n−1 − 2 = 3 × 2n−1 + (2b12 + 2b13 − 1)(2b22 + 2b23 − 1) − 2
2n = (2b12 + 2b13 − 1)(2b22 + 2b23 − 1).
B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
681
Fig. 1. The two cases for n = 3. Dashed edges are deleted.
Therefore (2b12 + 2b13 − 1) and (2b22 + 2b23 − 1) both have to be powers of 2, so at least one of b12 and b13 is 0, and at least
one of b22 and b23 is 0. Hence b12 b13 = b22 b23 = 0.
The number of deleted triangles is given by
(b12 + b22 )(n + 1) + (b13 + b23 )n − b12 b13 − b22 b23 = (b12 + b22 )(n + 1) + (b13 + b23 )n.
2 3
Equating this to n2 and simplifying using
i=1
j=2 bij = e = n gives us b12 + b22 = 0. Hence, b12 = b22 = 0. Since
all the deleted edges are between X1 and X3 , we have Y = J + X2 = J + K n for some bipartite graph J. But since Y and G
are chromatically equivalent and G = K1,n+2 + K n , by Lemma 1, J must be chromatically equivalent to K1,n+2 , and hence
J ∈ Tn+3 . Now we can determine C (K1,n,n+2 ).
Theorem 13. For any positive integer n ≥ 2,
C (K1,n,n+2 ) = J (n, n + 2) = {T + K n , S + K n+2 |T ∈ Tn+3 , S ∈ Tn+1 }.
Proof. Let G be K1,n,n+2 and Y ∈ C (G). The case where n = 2 is proven in [3]. Now assume that n ≥ 3. As Y ∈ C (G), Y is a
spanning subgraph of Kr1 ,r2 ,r3 for some integers r1 , r2 , r3 with 1 ≤ r1 ≤ r2 ≤ r3 and r1 + r2 + r3 = 2n + 3.
From Corollary 6, we find that n ≤ r3 ≤ n + 2.
If r3 = n + 2, by Lemma 9, Y ∈ {S + K n+2 |S ∈ Tn+1 }.
If r3 = n then, by Lemma 8, Y ∈ {T + K n |T ∈ Tn+3 } for n ≥ 4. If r3 = 3, then Y is a subgraph of K3,3,3 . By Lemma 7, if
a1 = 0 then Y ∈ {T + K 3 |T ∈ T6 }. Suppose a1 ≥ 1. By symmetry, we only need to consider the cases where a1 , a2 , a3 ≥ 1.
Y is obtained from K3,3,3 by deleting 4 edges and 12 triangles. Since each edge belongs to 3 triangles, no two deleted edges
can belong to the same triangle. This leaves only the two graphs in Fig. 1 to consider. Counting the number of induced C4
subgraphs in each of them shows that neither one is chromatically equivalent to K1,3,5 . The one on top has 25 induced C4
subgraphs and the one on the bottom has 24, whereas K1,3,5 has 30. Hence for n = 3, Y ∈ {T + K 3 |T ∈ T6 }.
Now it remains to consider the case where r3 = n + 1. If r1 = 1 then r2 = n + 1 and so Y is a subgraph of K1,n+1,n+1 .
Then Y is obtained from K1,n+1,n+1 by deleting exactly one edge. Since exactly one triangle of K1,n+1,n+1 also vanishes, this
implies that a1 = 1 and a2 = a3 = 0. However, this means that the number of induced C4 -subgraphs in Y is
which is not equal to

 
n +2
n
2
2
for n ≥ 2, a contradiction.

n +1
2
2
− n2 ,
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B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
Fig. 2. The case for n = 4, r1 = r2 = 3. Dashed edges are deleted.
If r1 = 2, then r2 = n and so Y is a subgraph of K2,n,n+1 . By Lemma 12, Y ∈ {T + K n |T ∈ Tn + 3}. Therefore, in the rest of
the proof, we can assume that r1 ≥ 3 and that n ≥ 4. Consequently, r2 ≤ n − 1 < r3 .
By Lemma 10, where k = 2 and u = 1,
(a1 − 1)(r3 − r1 ) + (a2 − 1)(r3 − r2 ) ≤ 0.
(7)
Now r3 − r1 and r3 − r2 are both strictly positive. Thus we have the following subcases.
1. a2 = a1 = 1.
2. a2 = 0.
3. a1 = 0.
For each of the three subcases, we try to find an upper bound on the number of ways of partitioning V (Y ) into four nonempty
independent sets or on the number of induced C4 -subgraphs. Now the number of ways of partitioning
 V(G) into four
nonempty independent sets, which is 5 × 2n−1 − 2, and the number of induced C4 -subgraphs in G is
n +2
2
n
2
. We will
show that one of these parameters for Y is strictly less than that for G. Hence this case does not lead to any graph that is
chromatically equivalent to G.
The subcase when a2 = a1 = 1 Then a3 = r1 r2 − n − 2. Let t be the number of pairs of deleted edges which belong to
the same triangle and t ′ be the number of triangles with all three edges deleted. Then the number of deleted triangles is
a3 (n + 1) + a1 r1 + a2 r2 − t + t ′ . Hence
r1 r2 (n + 1) − n2 − 2n = a3 (n + 1) + a1 r1 + a2 r2 − t + t ′
r1 r2 (n + 1) − n2 − 2n = r1 r2 (n + 1) − n2 − 3n − 2 + r1 + r2 − t + t ′
n + 2 = r1 + r2 − t + t ′ = n + 2 − t + t ′ .
Hence t = t ′ . However, each triangle has 3 pairs of edges, and each pair of edges uniquely defines a triangle, so t ≥ 3t ′ .
Hence t = t ′ = 0.
We find an upper bound on the number of ways to partition V (Y ) into four nonempty independent sets (‘‘parts’’). We
have two cases.
1. Suppose none of the vertices in X1 ∪ X2 are in the same part as any vertex in X3 . Then we can either partition X3 into
two nonempty sets, and there are 2n − 1 ways of doing this, or we can partition X1 ∪ X2 into three nonempty sets. Now
Y [X1 ∪ X2 ] is a connected graph with n + 2 vertices, so the number of ways of partitioning X1 ∪ X2 into three nonempty
independent sets is bounded from above by the number of ways of partitioning a tree with n + 2 vertices, so, using the
derivation of Eq. (6), we find that there are at most 2n − 1 ways of partitioning X1 ∪ X2 . Hence there are at most 4 × 2n−1 − 2
ways altogether.
2. Suppose at least one of the vertices in X1 ∪ X2 are in the same part as at least one vertex in X3 . Suppose the deleted edge
between X1 and X3 is incident to v ∈ X1 and u1 ∈ X3 , and the deleted edge between X2 and X3 is incident to w ∈ X2 and
u2 ∈ X3 . Note that t = 0 so u1 ̸= u2 . The partition either has {v, u1 } as one of the parts, or {w, u2 } as one of the parts. Both
parts cannot exist in the same partition as it would then require partitioning V (Y ) into five independent sets. If {v, u1 }
is a part, then we can partition (X1 ∪ X2 ) \ {v} into two nonempty independent sets. Since the degree of v in Y [X1 ∪ X2 ]
is r2 , deleting v would separate Y [X1 ∪ X2 ] into at most r2 connected components, each of which is bipartite. Hence the
number of possible such partitionings is at most 2r2 −1 . Similarly, if {w, u2 } is a part, then the number of possible such
partitionings is at most 2r1 −1 . Hence there are at most 2r2 −1 + 2r1 −1 ways altogether.
Hence the total number of ways of partitioning V (Y ) into four nonempty independent sets is at most 4 × 2n−1 + 2r2 −1 +
2r1 −1 − 2 ≤ 4 × 2n−1 + 2n−2 + 2r1 −1 − 2. However, this is less than 5 × 2n−1 − 2 for n > 4, and for n = 4 and r1 < r2 . This
leaves the case where n = 4 and r1 = r2 = 3. This graph is shown in Fig. 2, and has only 81 induced C4 -subgraphs, whilst
B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
683
K1,4,6 has 90 induced C4 -subgraphs, hence they are not chromatically equivalent. Therefore this subcase does not lead to any
graph that is chromatically equivalent to G.
The subcase when a2 = 0 Then a1 + a3 = r1 r2 − n. Let t be the number of pairs of deleted edges which belong to the same
triangle. Then
t = a1 r1 + a3 r3 − (r1 r2 (n + 1) − n2 − 2n)
= a1 r1 + a3 (n + 1) − ((a1 + a3 )(n + 1) − n)
= a1 (r1 − n − 1) + n
= n − a1 (r2 − 1).
Since t ≥ 0, r1 + r2 = n + 2 and r1 ≤ r2 , we have r2 ≥ (n + 2)/2. Therefore a1 ≤ 2. Now if a1 = 0 then t = n, which is
clearly not possible.
If a1 = 1 then t = r1 − 1 so there is a vertex w ∈ X2 which is incident to a deleted edge between X2 and X3 , and to r1 − 1
deleted edges between X1 and X2 . Hence there is exactly one vertex v ∈ X1 which is adjacent to w . Since Y [X1 ∪ X2 ] is a
connected graph with n + 2 vertices and n + 1 edges, it is a tree, and w is a leaf. We have the following ways to partition
V (Y ) into four nonempty independent sets (‘‘parts’’):
1. Suppose none of the vertices in X1 ∪ X2 are in the same part as any vertex in X3 . Then we can either partition X3 into
two nonempty sets, and there are 2n − 1 ways of doing this, or we can partition X1 ∪ X2 into three nonempty sets. Now
Y [X1 ∪ X2 ] is a tree, so using the derivation of Eq. (6), we find that there are 2n − 1 ways. Hence there are 4 × 2n−1 − 2
ways altogether.
2. Suppose the deleted edge between X2 and X3 is incident to w ∈ X2 and u ∈ X3 , and the partition has {w, u} as one of the
parts. Then, since Y [(X1 ∪ X2 ) \ {w}] is still a tree, there is only 1 way of partitioning V (Y ).
Hence the total number of ways of partitioning V (Y ) into four nonempty independent sets is 4 × 2n−1 − 1. However, this is
less than 5 × 2n−1 − 2 for n > 2.
If a1 = 2 then t = r1 − r2 ≤ 0. Hence it must be the case that r1 = r2 = (n + 2)/2 and t = 0. Now the two deleted edges
between X2 and X3 cannot be incident to two distinct vertices in X2 . If that were the case, then, since t = 0, all the deleted
edges between X1 and X2 would have to be incident to the remaining r2 − 2 vertices in X2 . But

a3 = r 1 r 2 − n − 2 =
n+2
2
2
−n−2=
n2 − 4
4

=
n−2

n+2
2

2
= r1 (r2 − 2)
meaning that all except two of the vertices in X2 are not adjacent to any vertices in X1 . This contradicts the fact that Y has
to be uniquely 3-colourable like G. Therefore the two deleted edges between X2 and X3 must be incident to the same vertex
w in X2 .
Claim. There is exactly one induced C4 subgraph in Y [X1 ∪ X2 ].
Now Y [X1 ∪ X2 ] is a connected bipartite graph where each of the partite sets X1 and X2 has (n + 2)/2 vertices. Since
there are no deleted edges which belong to the same triangle, the vertex w ∈ X2 is adjacent to every vertex in X1 . There are
[(n + 2)/2]2 − a3 = n + 2 edges in Y [X1 ∪ X2 ]. But Y [X1 ∪ X2 ] has n + 2 vertices, hence it contains exactly one cycle. Since
Y [X1 ∪ X2 ] is connected, each vertex in X2 is adjacent to at least one vertex in X1 . Therefore, by the pigeonhole principle,
there is a vertex w ′ ̸= w in X2 which is adjacent to two vertices in X1 , and the vertices in X2 \ {w, w ′ } are each adjacent to
one vertex in X1 . Hence the cycle in Y [X1 ∪ X2 ] is an induced C4 subgraph containing w and w ′ , proving the claim.
There are five types of induced C4 -subgraphs in Y .
(a) Two vertices in X2 and two vertices in X3 . There are

n/2
2

n +1
2

  
+ n n−2 1 /2 of these.
(b) Two vertices in X1 and two vertices in X2 . We have shown that there is exactly one.
(c) Two vertices in X1 and two vertices in X3 . There are

(n+2)/2
2

n +1
2

of these.

 
= (n2 − 4) n+2 1 /4 of these.




(n+2)/2
(e) Two vertices in X1 , one vertex in X2 and one vertex in X3 . There are a1
= 2 (n+22)/2 of these.
2
(d) Two vertices in X3 , one vertex in X1 and one vertex in X2 . There are a3

n +1
2

4
3
2
Adding
  these up, we find that Y has (n + 2n − 4n + 2n + 4)/4 induced C4 -subgraphs, which is strictly less than
n +2
2
to G.
n
2
= (n4 + 2n3 − n2 − 2n)/4 for n > 2, so this subcase does not lead to any graph that is chromatically equivalent
684
B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
The subcase when a1 = 0 We can assume a2 ≥ 1. Now then a2 + a3 = r1 r2 − n. Let t be the number of pairs of deleted edges
which belong to the same triangle. Then
t = a2 r2 + a3 r3 − (r1 r2 (n + 1) − n2 − 2n)
= a2 r2 + a3 (n + 1) − ((a2 + a3 )(n + 1) − n)
= a2 (r2 − n − 1) + n
= n − a2 (r1 − 1).
Now, since t ≥ 0, we have n ≥ a2 (r1 − 1). Label the vertices in X1 as v1 , . . . , vr1 . For each i between 1 and r1 , let xi
(respectively, yi ) be the number of deleted edges between X1 and X3 (respectively, X2 ) incident to vi . We have
r1

x i = a2 ,
(8)
yi = a3 = r1 r2 − n − a2 = r1 (n + 2 − r1 ) − n − a2 ,
(9)
i =1
r1

i =1
r1

xi yi = t = n − a2 (r1 − 1).
(10)
i =1
Without loss of generality, let x1 ≤ · · · ≤ xr1 . We first determine the structure of Y , then find bounds for the number of
ways to partition V (Y ) into four independent nonempty sets and the number of induced C4 -subgraphs in Y .
Claim 1. x1 = x2 = · · · = xr1 −1 = 0.
Subtracting Eq. (9) from Eq. (10) gives us
r1

(xi − 1)yi = (r1 − 2)(r1 − n − a2 ) < 0
i =1
and since the yi are nonnegative, x1 = 0. We proceed to prove Claim 1 by contradiction. Suppose, on the contrary, that there
exists some b, where 1 ≤ b ≤ r1 − 2, such that xi = 0 for 1 ≤ i ≤ b and xi ≥ 1 for b + 1 ≤ i ≤ r1 . From Eq. (8), xr1 must be
at least as large as the mean of the nonzero terms, which is a2 /(r1 − b). Substituting these into Eqs. (8) and (10) give us
1 + ··· + 1+



r1 −b−1 terms
a2
r1 − b
yb+1 + · · · + yr1 −1 +
≤
r1

x i = a2 ,
(11)
i=1
a2
r1 − b
yr1 ≤
r1

xi yi = n − a2 (r1 − 1).
(12)
i=1
Adding (11) multiplied by (r1 − 1) to (12) gives us
yb+1 + · · · + yr1 −1 +
a2
r1 − b
(yr1 + r1 − 1) + (r1 − 1)(r1 − b − 1) ≤
r1

xi (yi + r1 − 1) = n.
i=1
Rearranging terms gives us
yb+1 + · · · + yr1 −1 +
a2
r1 − b
yr1 ≤ n − (r1 − 1)(r1 − b − 1) −
a2 (r1 − 1)
r1 − b
.
Subtracting this from Eq. (9) gives us

y1 + · · · + yb +
1−
a2
r1 − b


yr1 ≥ nr − 2n + b − br + 1 + a2
b−1
r1 − b

.
Now since Y is uniquely 3-colourable, yi ≤ r2 − 1 = n + 1 − r1 for all 1 ≤ i ≤ r1 , so y1 + · · · + yb ≤ b(n + 1 − r1 ). Hence

1−
a2
r1 − b


yr1 ≥ nr1 − 2n + b − br1 + 1 + a2
= n(r1 − b − 2) + 1 + a2
≥1

b−1
r1 − b
b−1
r1 − b


− b(n + 1 − r1 )
B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
685
since r1 − b − 2 ≥ 0. Therefore 1 − a2 /(r1 − b) > 0. But this means
r1 − b > a2 = xb+1 + · · · + xr1 ≥ r1 − b,


r1 −b terms

a contradiction. Hence Claim 1 is true, therefore xr1 = a2 , so
Claim 2. a2 ≥
r1
i =1
xi yi = xr1 yr1 = a2 yr1 and thus yr1 = n/a2 − (r1 − 1).
√
n.
We find an upper bound on the number of ways to partition V (Y ) into four nonempty independent sets (‘‘parts’’). We
have two cases.
1. Suppose none of the vertices in X1 ∪ X2 are in the same part as any vertex in X3 . Then we can either partition X3 into
two nonempty sets, and there are 2n − 1 ways of doing this, or we can partition X1 ∪ X2 into three nonempty sets. Now
Y [X1 ∪ X2 ] is a connected graph with n + 2 vertices, so the number of ways of partitioning X1 ∪ X2 into three nonempty
independent sets is bounded from above by the number of ways of partitioning a tree with n + 2 vertices, so, using the
derivation of Eq. (6), we find that there are at most 2n − 1 ways of partitioning X1 ∪ X2 . Hence there are at most 4 × 2n−1 − 2
ways altogether.
2. Let uvr1 be a deleted edge in Y [X1 ∪X2 ] and suppose the partition has {vr1 , u} as a part. Then we can partition (X1 ∪X2 )\{vr1 }
into two nonempty independent sets. Since the degree of vr1 in Y [X1 ∪ X2 ] is r2 − yr1 , deleting vr1 would separate Y [X1 ∪ X2 ]
into at most r2 − yr1 connected components, each of which is bipartite. Hence the number of possible such partitionings
is at most (2a2 − 1) × 2r2 −yr1 −1 = 2n+a2 −n/a2 − 2n−n/a2 .
Adding these up, we find that the number of ways of partitioning V (Y ) into four nonempty independent sets is at most
4 × 2n−1√+ 2n+a2 −n/a2 − 2n−n/a2 − 2. This should be greater than or equal to 5 × 2n−1 − 2 for n > 2. Hence a2 − n/a2 ≥ 0,
so a2 ≥ n. This proves Claim 2.
Claim 3. Y has strictly fewer induced C4 subgraphs than G.
There are five types of induced C4 -subgraphs in Y .
(a) Two vertices in X2 and two vertices in X3 . There are

n+2−r1
2

n+1
2

of these.
(b) Two vertices in X1 and two vertices in X2 . We will find
for how
there are.
 an upper
 bound

 many 
+ (r1 − 1) n+12−a2 of these.
 

 
n+1
(d) Two vertices in X3 , one vertex in X1 and one vertex in X2 . There are a3
− yr1 a2 n − a22 of these.
2
 r2 −yr 
(c) Two vertices in X1 and two vertices in X3 . There are
r1 −1
2
n +1
2
(e) Two vertices in X2 , one vertex in X1 and one vertex in X3 . There are a2
1
2
of these.
Now we try to find an upper bound for the number of induced C4 -subgraphs of type (b).
We proceed along similar lines to the proof of Lemma 7. Let di be the degree of vi in the induced subgraph Y [X1 ∪ X2 ].
Then the number of induced C4 -subgraphs of type (b) is actually the number of C4 -subgraphs in Y [X1 ∪ X2 ], which is bounded
from above by
C = (r1 − 1)

d1
2

+ (r1 − 2)

d2


+ ··· + 2
2
dr1 −2
2


+
dr1 −1
2
This is because, in the subgraph Y [X1 ∪ X2 ], there are at most (r1 − 1)
each i where 2 ≤ i ≤ r1 , there are at most (r1 − i)
 
di
2

.
 
d1
2
induced C4 -subgraphs containing v1 , and for
induced C4 -subgraphs containing vi but not containing vj for all
1 ≤ j < i. Note that this expression remains as an upper bound for the number of induced C4 -subgraphs of type (b) for any
permutation of d1 , d2 , . . . , dr1 −1 .
By the rearrangement identity, C is maximised when d1 ≥ d2 ≥ · · · ≥ dr1 −1 . We can assume that this condition is
satisfied. Now
r 1 −1

di = r2 (r1 − 1) −
i=1
r1

yi + yr1 = r2 (r1 − 1) − a3 + n/a2 − (r1 − 1) = a2 + n/a2 − 1.
i =1
We also know di ≥ 1 for all 1 ≤ i ≤ r1 , because Y is uniquely 3-colourable. We can maximise C by letting d2 = · · · =
dr1 −1 = 1 and so d1 = a2 + n/a2 − 1 − (r1 − 2) = a2 + n/a2 − r1 + 1. Hence
C ≤ (r1 − 1)

a2 + n/a2 − r1 + 1
2

,
with equality only if d1 = a2 + n/a2 − r1 + 1.
686
B.L. Ng, F.M. Dong / Discrete Mathematics 338 (2015) 674–687
Comparing the number of induced C4 -subgraphs in Y and G, we have

n+2
 
n
2

≤
2
n + 2 − r1

n+1


a2 + n/a2 − r1 + 1


r1 − 1

n+1
+ (r1 − 1)
+
2
2
2








r2 − yr1
n+1
a2
n + 1 − a2
.
+ a2
+ a3
− yr1 a2 n −
+ (r1 − 1)
2

2

2
2
2
2
We make the following substitutions:
yr1 = n/a2 − (r1 − 1),
a3 = r1 r2 − n − a2 ,
r2 = n + 2 − r1 ,
so that the entire expression is in terms of only n, a2 and r1 . This simplifies to
n2 (a + r − a3 − 1) + na(1 − 2a − 2r ) + a2 (r − 1)(a + r )(a + r − 1) ≥ 0
(13)
where we have dropped the subscripts
√ on a2 and r1 .
Now, from Claim 2, we have a ≥ n. We also have yr1 ≥ 0. This gives us the chain of inequalities
√
n ≤ a ≤ n/(r − 1) ≤ n/2,
2≤r −1≤
which we will use to simplify the left-hand side of (13).
n2 (a + r − a3 − 1) + na(1 − 2a − 2r ) + a a(r − 1)(a + r )(a + r − 1)
  
≤n
≤ n (a + r − a − 1) + na((r − 1) +a + 2ar − 3a − r )
  
2
3
2
2
≤n
≤ n (2a + r − a − 1) + na(a + 2ar − 3a − r )
= n 
a3 (1 − n) + n 
a (2n + 2 a(r − 1) − 
r ) + n2 (r − 1)
a − 
  
  
√
√
3
≥3
≤ 2n
≥ n
≤n
≤ n
≤( 2n )
 n 3
√
√
n2
≤n
(1 − n) + (2n + 2n − n − 3) + n2 n
2
2
√
n4
n2 n
3n2
n5
3
+ 2n +
−
.
=− +
2
3
8
8
2
2
2
This expression is negative for n ≥ 5, so the inequality in (13) cannot hold for n ≥ 5. Therefore it suffices to check n =
√3
and n = 4, which we can do by verifying directly that the inequality in (13) does not hold for all values of r ≥ 3 and a ≥ n
where a(r − 1) ≤ n. This contradicts (13), and proves Claim 3.
Hence this subcase does not lead to any graph that is chromatically equivalent to G. 4. Conclusion
We have shown that C (K1,n,n+2 ) = J (n, n + 2) for the case where n ≥ 3. We hope that future work can determine
C (K1,n,n+k ) for k ≥ 3.
Acknowledgements
The authors wish to thank the referees for their careful reading of the submitted manuscript, and the detailed comments
and suggestions on improving this paper.
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