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2020 Прикладная теория графов №50
UDC 519.17 DOI 10.17223/20710410/50/7
THE CHROMATICITY OF THE JOIN OF TREE AND NULL GRAPH
L. X. Hung
Hanoi University for Natural Resources and Environment, Hanoi, Vietnam E-mail: [email protected]
The chromaticity of the graph G, which is join of the tree Tp and the null graph Oq, is studied. We prove that G is chromatically unique if and only if 1 ^ p ^ 3, 1 ^ q ^ 2; a graph H and Tp + Op-\ are %-equivalent if and only if H = Tp + Op-1, where Tp is a tree of order p; H and Tp + Op are %-equivalent if and only if H e {Tp + Op, Tp'+1 + Op-1}, where Tp is a tree of order p, Tp'+1 is a tree of order p + 1. We also prove that if p < q, then x'(G) = ch'(G) = A(G); if A(G) = |V(G)| - 1, then X'(G) = ch'(G) = A(G) if and only if G = K3.
Keywords: chromatic number, chromatically equivalent, chromatically unique graph, chromatic index, list-chromatic index.
1. Introduction
All graphs considered in the paper are finite undirected graphs without loops or multiple edges. If G is a graph, then V(G), E(G) (or V and E in short) and G denote its vertex set, edge set and its complementary graph, respectively. The set of all neighbours of a subset S С V(G) is denoted by NG(S) (or N(S) in short). If S = {v}, then N(S) is denoted by N(v). For a vertex v e V(G), the degree of v is denoted by degG(v) (or deg(v)), it equals |NG(v)|. The subgraph of G induced by W С V(G) is denoted by G[W]. Let R be a subset of edges in G, |R| = r; denote by G — R the graph obtained by deleting all edges in R from G.
The null graphs and complete graphs of order n are denoted by On and Kn, respectively. The K3 is called a triangle. Let t1(G), t2(G), and t3(G) be the numbers of triangles, of induced subgraphs C4, and of complete subgraphs K4 in G, respectively. Unless otherwise indicated, our graph-theoretic terminology follows [1].
An acyclic graph, one not containing any cycles, is called forest. A connected forest is called a tree, a tree of order n is denoted by Tn.
A graph G = (V, E) is called r-partite graph if V admits a partition into r classes V = V1 U V2 U ... U Vr such that the subgraphs of G induced by Vi, i = 1,..., r, are empty. If r = 2, then G is called bipartite graph, if r = 3, then G is called tripartite graph. An r-partite graph in which every two vertices from different partition classes are adjacent is called complete r-partite graph and is denoted by K\v1\,\v2\,..,,\vr|. The complete r-partite graph K\Vl|,|V2\,...,\Vr\ with |V1| = |V2| = ... = |VT| = s is denoted by KJ.
Let G1 = (V1,E1), G2 = (V2 ,E2) be two graphs such that V1 П V2 = 0. Their union G = G 1 U G2 has, as expected, V(G) = V1 U V2 and E(G) = E1 U E2. Their join is denoted G1 + G2 and consists of G1 U G2 and all edges joining V1 with V2.
Let G 1 = (V1,E1), G2 = (V2, E2) be two graphs. We call G1 and G2 isomorphic, and write G1 = G2, if there exists a bijection f : V1 ^ V2 with uv e E1 if and only if f (u)f (v) e E2 for all u,v e V1.
Let G = (V, E) be a graph and A is a positive integer.
A X-coloring of G is a bijection f : V(G) ^ {1, 2,..., A} such that f (u) = f (v) for any adjacent vertices u,v € V(G). The smallest positive integer A such that G has a A-coloring is called the chromatic number of G and is denoted by x(G). We say that a graph G is n-chromatic if n = x(G).
Let V(G) = {vi, v2,..., vn}, two A-colorings f and g are considered different if and only if f (vk) = g(vk) for some k € {1, 2,..., n}. Let P(G, A) (or simply P(G) if there is no danger of confusion) denote the number of distinct A-colorings of G. It is well-known that for any graph G, P(G, A) is a polynomial in A, called the chromatic polynomial of G. The notion of chromatic polynomials was first introduced by Birkhoff [2] in 1912 as a quantitative approach to tackle the four-color problem. Two graphs G and H are called chromatically equivalent (or, in short, x-equivalent), and we write G ~ H, if P(G, A) = P(H, A). A graph G is called chromatically unique (x-unique) if G' = G (i.e., G' is isomorphic to G) for any graph G' such that G' ~ G. For examples, all cycles are x-unique [3]. The notion of x-unique graphs was first introduced and studied by Chao and Whitehead [4] in 1978. The readers can see the surveys [3, 5, 6] for more information on x-unique graphs.
An edge coloring of a graph G can be defined similarly. Namely, an edge A-coloring of a graph G is a mapping f : E(G) ^ {1, 2,... A} such that two adjacent edges have distinct images. The chromatic index of G, denoted by x'(G), is the smallest positive integer A such that G has an edge A-coloring. In 1964, Vizing [7] proved that x'(G) is equal to either A(G) or A(G) + 1, where A(G) is the maximum degree of G. A graph G is said to be Class one (resp., Class two) if x'(G) = A(G) (resp., A(G) + 1). For examples, all cycles Cn with n even are Class one; all cycles Cn with n odd are Class two. Let (L(e))eeE(G) be a family of sets. We call an edge coloring f of G with f (e) € L(e) for all e € E(G) a list edge coloring from the lists L(e). The least integer k such that G has an edge coloring from any family of lists of size k is the list-chromatic index of G and is denoted by ch'(G). The idea of list colorings of graphs is due independently to V. G. Vizing [8] and to P. Erdos, A. L. Rubin, and H. Taylor [9].
In [10] we have characterized chromatically uniqueness of the graph K£ + Ok, in [11] we have characterized chromatically uniqueness of the graph G = Kn + Kr, and in [6] we have determined chromatic index and characterized chromatically uniqueness split graphs.
In this paper, we study the chromaticity of G, which is join of the tree Tp and the null graph Oq. We prove that G is chromatically unique if and only if 1 ^ p ^ 3, 1 ^ q ^ 2; H and Tp + Op-1 are x-equivalent if and only if H = Tp + Op-1, where Tp is a tree of order p; H and Tp + Op are x-equivalent if and only if H € {Tp + Op, Tf'+1 + Op-1}, where Tp is a tree of order p, Tf'+1 is a tree of order p +1. We also prove that if p ^ q, then x'(G) = ch'(G) = A(G); if A(G) = |V(G)| - 1, then x'(G) = ch'(G) = A(G) if and only if G = K3.
2. Vertex colorings
For a graph G and a positive integer k, a partition {A1, A2,..., Ak} of V(G) is called a k-independent partition in G if each Aj is a non-empty independent set of G. Let a(G, k) denote the number of k-independent partitions in G. Hence, P(G, A) = a(G, k)(A)k,
where (A)k = A(A - 1)... (A - k + 1).
The polynomial a(G,x) = a(G, k)xk is called the a-polynomial of G.
The polynomial h(G,x) = a(G,k)xk is called the adjoint polynomial of G.
Let K+ be the vertex gluing of Kp and K2.
For convenience, denote a(G,x) by a(G), h(G,x) by h(G), and G = H by G = H. The following lemmas will be used to prove our main results.
Lemma 1 [3]. If G = Kn is the complete graph on n vertices, then x(G) = n and G is x-unique.
Lemma 2. If G = Krai ra2 .. )rar is the complete r-partite graph, then x(G) = r. Lemma 3 [12]. Let G and H be two x-equivalent graphs. Then
(i) |V(G)| = |V(H)|;
(ii) \E(G)| = \E(H)|;
(iii) x(G) = x(H);
(iv) G is connected if and only if H is connected;
(v) G is 2-connected if and only if H is 2-connected;
(vi) ti(G)= ti(H);
(vii) t2(G) - 2is(G)= t2(H) - 2ta(H);
(viii) a(G, k) = a(H, k) for each k = 1, 2,...
Lemma 4 [12].
(i) All trees of the same order are x-equivalent. Further, the graph G of order n is a tree if and only if P(G, A) = A(A - 1)n-1;
(ii) A tree Tn is x-unique if and only if 1 ^ n ^ 3;
(iii) If G = Tn is a tree of order n, then x(G) = 2. Lemma 5 [11]. The graph G = Km + Kn is x-unique. Lemma 6 [13]. Let G and H be two disjoint graphs. Then
(i) a(G + H,x) = a(G,x)a(H,x);
(ii) h(G U H,x) = h(G,x)h(H,x).
Lemma 7 [14]. Let G and H be two graphs. Then
(i) P(G, A) = P(H, A) if and only if a(G, x) = a(H, x);
(ii) P(G, a) = P(h, a) if and only if h(G, x) = h(H, x).
Lemma 8. If p ^ 2, then x(Tp + Oq) = 3.
Proof. If p ^ 2, then the complete graph K3 is a subgraph of G = Tp + Oq. So x(G) ^ 3. Let V(G) = Vi U V2 is a partition of V(G) such that G[Vi] = Tp, G[V2] = Oq. The graph G[V1] is a tree, by (iii) of Lemma 4, G[V1] has a coloring f1 using two colors 1, 2. Set mapping
f : V(G) ^{1, 2, 3}
such that f (v) = f1(v) if v G V1, f (v) = 3 if v G V2. Then f is a 3-coloring of G, i.e., x(G) ^ 3. Thus, x(G) = 3. ■
Theorem 1. G = Tp + Oq is x-unique if and only if 1 ^ p ^ 3, 1 ^ q ^ 2. Proof. First we prove the necessity. Suppose that G = Tp + Oq is x-unique. Suppose the contrary, that p ^ 4. Set G1 = (K1 U 11, E1) with
K1 = {v1, v2, . . . , vp}, 11 = {«1,«2, . . . ,Uq}, E1 = {v1v2, v1v3,... ,v1vp} U {viUj : i = 1, 2,.. .p, j = 1, 2,.. .q}.
Set G2 = (K2 U 12,E2) with
K2 = {v1, v2, . . . , vp}, 12 = {«1,«2, . . . ,Uq}, E2 = {v1v2, v2v3,... , vp-1vp} U {viUj : i =1, 2,.. .p,j = 1, 2,... q}.
By (i) of Lemma 4, (i) of Lemma 6, and (i) of Lemma 7, it follows that
P (G\A) = P (G2,A) = P (G, A).
It is not difficult to see that
A(G1) = max{deg(u) : u G V(G1)} = deg(v1) = p + q - 1
and
A(G2) = max{deg(u) : u G V(G2)} = max{p, q + 2}.
If q ^ 2, then max{p, q + 2} <p + q - 1, it follows that A(G2) < A(G1). So G1 ^ G2 and
G is not x-unique, a contradiction.
If q =1, then A(G2) = A(G1) = p. It is not difficult to see that
|{u G V(G1) : degGi(u) = p}| = 2 and |{u G V(G2) : degG2(u) = p}| = 1.
It follows that G1 ^ G2 and G is not x-unique, a contradiction. Thus, 1 ^ p ^ 3.
Suppose that q ^ 3. For p = 3, we set G3 = (K3 U I3, E3) with
K3 = {V1 ,V2, V3}, I3 = {u1,u2, . . . ,uq}, E3 = {V1V2, V2V3} U {v^- : i = 1, 2, 3, j = 1, 2,... q},
and set G4 = (K4 U I4, E4) with
K4 = {V1 ,V2, V3}, I4 = {u1,u2, . . . ,uq}, E4 = {V2u1,u1u2,u2u3,. . . ,uq-1uq} U {^1^2,^2V3} U {v^, V3uj : j = 1, 2,..., q}.
It is not difficult to see that G = G3 = T3 + Oq and G4 = Tq+1 + O2. By (i) of Lemma 4,
(i) of Lemma 6, and (i) of Lemma 7, we have
a(G3,x) = a(T3 + Oq ,x) =
= a(T3,x)a(Oq ,x) =
= a(O1 + O2, x)a(Oq ,x) =
= a(O1, x)a(O2, x)a(Oq ,x) =
= a(O1,x)a(Oq ,x)a(O2,x) =
= a(O1 + Oq ,x)a(O2,x) =
= a(Tq+1,x)a(O2,x) =
= a(Tq+1 + O2,x) =
= a(G4, x).
It follows that P (G3, A) = P (G4,A). Otherwise,
A(G3) = max{deg(u) : u G V(G3)} = deg(v2) = q + 2, A(G4) = max{deg(u) : u G V(G2)} = deg(v1) = deg(v3) = q + 1.
So G3 ^ G4 and G is not x-unique, a contradiction.
For p = 2, we set G5 = (K5 U 15, E5) with
K5 = {V1,V2}, I5 = {u1,u2, . . . ,uq}, E5 = {V1V2} U {viuj : i = 1, 2, j = 1, 2,... q},
and set G6 = (K6 U I6, E6) with
K6 = {V1,V2}, I6 = {U1,U2, . . . }, E6 = {v2Ul,UlU2,U2U3, . . . ,Uq-lUq} U {^2} U {ViUj : j = 1, 2,... q}.
It is clear that P(G5, A) = P(G6, A) and
|{u G V(G5) : degG5(u) = q + 1}| = |{vi,v2}| = 2, |{u G V(G6) : degGa(u) = q + 1}| = |{vi}| = 1.
So G5 ^ G6 and G is not x-unique, a contradiction.
If p = 1, then G is a tree Tn with n = q +1 ^ 4. By (ii) of Lemma 4, G is not x-unique, a contradiction.
Now we prove the sufficiency. If p =1 and q =1, then G = K2, if p = 2 and q =1, then G = K3. By Lemma 1, G is x-unique.
If p =1 and q = 2, then G = T3. By (ii) of Lemma 4, G is x-unique. If p = 2 and q = 2 or p = 3 and q =1, then G = K^ + K2, if p = 3 and q = 2, then G = K| + K1. By Lemma 5, G is x-unique. ■
Theorem 2.
(i) H and Tp + Op-1 are x-equivalent if and only if H = Tp + Op-1, where Tp is a tree of order p;
(ii) H and Tp + Op are x-equivalent if and only if H G {Tp + Op, Tp+1 + Op-1}, where Tp is a tree of order p, Tp'+1 is a tree of order p +1.
Proof. If p = 2, then, by Theorem 1, Tp + Op-1 and Tp + Op are x-unique. It follows that the theorem is obviously true. Hence we may assume that p ^ 3.
Suppose that H and G = Tp + Oq are x-equivalent, where p — 1 ^ q ^ p. By (iii) of Lemma 3 and Lemma 8, x(H) = 3. So H is a tripartite graph. We may assume that D = Ka,b,c and R = {e1, e2,..., er} Ç E(D) such that H = D — R and a ^ b ^ c. It is clear that
r = |E (D) — |E (H )| = |E (D)| — |E (G)| = ab + ac + bc — pq — p + 1
and
a + b + c = |V(D)| = |V(H)| = |V (G)| = p + q.
Denote by t1(ei) the number of triangles containing the edge e^ in D for every i = 1, 2,..., r. It is not difficult to see that t1(ei) ^ c for every i = 1, 2,..., r. Then
t1(H) ^ t1(D) — rc,
and the equality holds only if t1(ei) = c for every i = 1, 2,..., r. By (vi) of Lemma 3, t1(G) = t1(H), it follows that
t1(D) — t1(G)= t1(D) — t1(H) ^ rc.
Since t1(D) = abc, t1 (G) = (p — 1)q, we have
f (c) = t1(D) — t1(G) — rc =
= abc — (p — 1)q — (ab + ac + bc — pq — p + 1)c =
= abc — (p — 1)q — [ab + (p + q — c)c — pq — p + 1]c =
= (c — 1)(c — p +1)(c — q) ^ 0.
By p ^ 3 and p — 1 ^ q ^ p, it follows that c ^ (p + q)/3 > 1. Let {V, V2, V3} be the 3-independent partition in H such that |Vi| = a, |V2| = b, |V3| = c. It is not difficult to see that if f (c) = 0, then t1(ei) = c for every i =1, 2,..., r, so edge e^ G R has one end vertex in Vi and another end vertex in V2. It follows that H = H[V U V2] + Oq
(i) If q = p — 1, then G = Tp + Op_. In this case, f (c) = (c — 1)(c — p + 1)2 ^ 0, f (c) = 0 if and only if c = p — 1. So t1(ei) = c = p — 1 for every i = 1, 2,..., r. By (i) of Lemma 6 and (i) of Lemma 7, we have
a(H,x) = a(H[Vi U V2] + Op_i,x) =
= a(H[Vi U V2],x)a(Op_i,x) = = a(G,x) = = a(Tp + Op_i,x) = = a(Tp,x)a(Op_i,x).
It follows that a(H[Vi U V2],x) = a(Tp,x). So P(H[Vi U V>],A) = P(Tp,A). By (i) of Lemma 4, H[Vi U V2] = Tp, where Tp is a tree of order p. Thus, H = Tp + Op-i.
It is not difficult to see that if H = Tp + Op-i, then P(H, A) = P(Tp + Op-i, A).
(ii) If q = p, then G = Tp + Op. So f (c) ^ 0 if and only if p — 1 ^ c ^ p, f (z) = 0 if and only if c = p — 1 or c = p. Now we consider separately two cases.
Case 1: c = p. We have
a(H,x) = a(H[Vi U V2] + Op,x) =
= a(H[Vi U V2],x)a(Op,x) =
= a(G, x) =
= a(Tp + Op ,x) =
= a(Tp,x)a(Op ,x).
It follows that a(H[Vi U V2],x) = a(Tp,x). So P(H[Vi U Vs],A) Lemma 4, H[Vi U V2] = Tp, where Tp is a tree of order p. Thus, H =
Case 2: c = p — 1. We have
a(H,x) = a(H[Vi U V2] + Op_i,x) =
= a(H[Vi U V2],x)a(Op_i,x) =
= a(G,x) =
= a(Tp + Op, x) =
= a(Tp,x)a(Op,x) =
= a(Oi + Op_i,x)a(Op,x) =
= a(Oi, x)a(Op_i ,x)a(Op,x) =
= a(Oi, x)a(Op, x)a(Op_i, x) =
= a(Oi + Op ,x)a(Op_i,x) =
= P(Tp, A). By (i) of Tp + Op.
= ^(Tp+i,x)a(Op_i,x).
It follows that a(H[Vi U V>],x) = a(Tp+i,x). So P(H[Vi U V2], A) = P(Tp+i, A). By (i) of Lemma 4, H[Vi U V2] = Tp+i, where Tp+i is a tree of order p + 1. Thus, H = Tp+i + Op_i. It is not difficult to see that if H G {Tp + Op, Tp+i + Op_i}, then P(H, A) = P(Tp + Op, A). ■
3. Edge colorings
We need the following lemmas 9-13 to prove our results.
Lemma 9 [15]. Every bipartite graph G satisfies x'(G) = A(G).
Lemma 10 [15]. ch'(G) ^ x'(G) for all graphs G.
Lemma 11 [15]. Every bipartite graph G satisfies ch'(G) = x'(G).
Lemma 12 [16]. If G is a graph of order 2n + 1 and A(G) = 2n, then G is Class one if and only if |E(G)| ^ n.
Lemma 13 [12]. If G = Tn is a tree of order n, then |E(G)| = n — 1.
Theorem 3. If p ^ q, then graph G = Tp + Op satisfies
(i) x'(G) = A(G);
(ii) ch'(G) = A(G).
Proof. Let V(G) = Vi U V2 is a partition of V(G) such that G[Vi] = Tp, G[V2] = Oq, Vi = {vi,V2,...,Vp}, V2 = {ui,M2,...,Mq}. Set Gi = G[Vi] and G2 = G — E(G[Vi]). It is not difficult to see that Gi and G2 are bipartite graphs, A(G2) = q = degG2 (v) for every vertex v G V and A(G) = A(Gi) + A(G2) = A(Gi) + q = deg(v) for some vertex v G Vi.
(i) By (iii) of Lemma 4, x(Gi) = 2, so Gi is a bipartite graph. By Lemma 9, Gi has an edge coloring fi using A(Gi) colors 1, 2,..., A(Gi). Again by Lemma 9, G2 has an edge coloring f2 using A(G2) = q colors A(Gi) + 1,..., A(Gi) + q. Since A(G) = deg(v) for some vertex v G Vi, it is clear that the mapping
f : E(G) ^ {1, 2,... , A(Gi), A(Gi) + 1,... , A(Gi) + q}
such that f (e) = fi(e) if e G E(Gi) and f (e) = f2(e) if e G E(G2) is an edge coloring of G. Since A(G) = A(Gi) + q, it follows that x'(G) = A(G).
(ii) By Lemma 10 and (i), we have ch'(G) ^ A(G) = A(Gi) + q. Now we prove that ch'(G) ^ A(G). Let L(e) be the lists of colors of e G E(G) such that |L(e)| = A(G).
Let Li(e) C L(e) such that |Li(e)| = A(Gi) for every e G E(Gi). Since Gi is a bipartite graph, by Lemma 9 and Lemma 11, there exists gi being a list edge coloring of Gi with the lists of colors Li(e) for every e G E(Gi).
For every i = 1, 2, ...,p, the subgraph induced by the edges of Gi incident with v is denoted by Gi(vi). It is clear that |gi(Gi(vi))| ^ A(Gi). For every i = 1, 2,... ,p, j = = 1, 2,..., q, set L'(viWj) = L(viWj)\gi(Gi(v,)). It follows that |L/(viWi)| ^ A(G) — A(Gi) = = q. Let L2(viUj) C ¿'(v^Wj) such that |L2(viUj )| = A(G2) = q for every i = 1, 2, ...,p, j = 1, 2,..., q. By Lemma 11, there exists g2 being a list edge coloring of G2 with the lists of colors L2(viWj) for every i =1, 2,... ,p, j = 1, 2,..., q. Let g be the edge coloring of G such that g(e) = gi(e) if e G E(Gi) and g(e) = g2(e) if e G E(G2). Then g is a list edge coloring of G with the lists of colors L(e) for every e G E(G), i.e., ch'(G) ^ A(G). Thus, ch'(G) = A(G). ■
Theorem 4. Let G = Tp + Op be a graph with A(G) = p + q — 1. Then
x'(G) = ch' (G) = A(G)
if and only if G = .
Proof. Let V(G) = Vi U V2 is a partition of V(G) such that G[Vi] = Tp, G[V2] = Oq, Vi = {vi,v2,...,vp}, V2 = {ui,u2,...,uq}. Set Gi = G[Vi] and G2 = G — E(G[Vi]). It is not difficult to see that Gi and G2 are bipartite graphs and A(G) = p + q — 1 = deg(v) for some vertex v G Vi. It follows that A(Gi) = p — 1.
Suppose that x'(G) = ch'(G) = A(G). We have chi'(K3) = ch'(Ks) = 3. So G = K3.
Now suppose that G = K3. If p ^ q, then by Theorem 3, x'(G) = ch'(G) = A(G). So we may assume that p > q. If p =2, then q = 1, so G = K3, a contradiction. It follows that p ^ 3. Without loss of generality we may assume that A(Gi) = degGl(vi), so A(G) = = deg(vi). Since A(Gi) = p — 1, it is not difficult to see that E(Gi) = {viv2, viv3,..., vivp}. We consider separately two cases.
Case 1: p = q +1.
If q = 1, then p = 2, so G = K3, a contradiction. So we may assume that q ^ 2. By Lemma 13, it is not difficult to see that |E(G)| = q2 — q. Since q ^ 2, it follows that |E(G)| ^ q. By Lemma 12, G is Class one.
By Lemma 10, ch'(G) ^ x'(G) = A(G). Let L(e) be the lists of colors of e G E(G) such that |L(e)| = A(G). Set G3 = G — E(G[{v2, V3,..., Vp} U V2]) and G4 = G[{v2, V3,... ,Vp} U U V2]. It is clear that G3 and G4 are bipartite graphs with A(G3) = deg(vi) = A(G) and A(G4) = q. By Lemma 9 and Lemma 11, there exists g3 being a list edge coloring of G3 with the lists of colors L(e) for every e G E(G3). For every i = 2, 3,... ,p, j = 1, 2,..., q, set L'(vjUj) = L(viMj)\{g3(vivi),g3(viUj)}. It follows that |L'(vjUj)| ^ A(G) — 2 = p + q — 3 ^ q for every i = 2, 3,... ,p, j = 1, 2,..., q. Let L4(viUj) C L'(vjUj) such that |L4(viUj)| = q for every i = 2, 3,... ,p, j = 1, 2,..., q. By Lemma 11, there exists g4 being a list edge coloring of G4 with the lists of colors L2(viUj) for every i = 2, 3,... ,p, j = 1, 2,..., q. Let g be the edge coloring of G such that g(e) = g3(e) if e G E(G3) and g(e) = g4(e) if e G E(G4). Then g is a list edge coloring of G with the lists of colors L(e) for every e G E(G), i.e., ch'(G) ^ A(G). Thus, ch'(G) = A(G).
Case 2: p ^ q + 2.
It is clear that G[vi U V2] = Tq+i, where Tqpi is a tree of order q + 1. Therefore, G = = Tqpi + Op_i. Since q +1 < p — 1, by Theorem 3, x'(G) = ch'(G) = A(G). ■
Conclusion
The coloring problems are interesting topics in graph theory. Coloring graphs found application in many practical problems, for example, coding theory or security. Clearly, to estimate the chromatic as well as the chromatic uniqueness is very important. So far there have been many research results on this topic for different graph layers. However, the problem has not been generally solved, and further research is needed. This paper explores some of the coloring problems with graph G, which is join of the tree Tp and the null graph Oq, contributes to enriching the research results on the coloring problems.
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