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5URAL MATHEMATICAL JOURNAL, Vol. 8, No. 2, 2022, pp. 94-114
DOI: 10.15826/umj.2022.2.008
ON LOCAL IRREGULARITY OF THE VERTEX COLORING OF THE CORONA PRODUCT OF A TREE GRAPH
Arika Indah Kristiana1^, M. Hidayat1, Robiatul Adawiyah1, D. Dafik1, Susi Setiawani1, Ridho Alfarisi2
department of Mathematics Education, University of Jember, Jalan Kalimantan 37, 68126, Jember, Jawa Timur, Indonesia
2Department of Elementary School Education, University of Jember, Jalan Kalimantan 37, 68126, Jember, Jawa Timur, Indonesia
t arika.fkip@unej.ac.id
Abstract: Let G = (V, E) be a graph with a vertex set V and an edge set E. The graph G is said to be with a local irregular vertex coloring if there is a function f called a local irregularity vertex coloring with the properties: (i) l : (V(G)) ^ {1, 2, ...,k} as a vertex irregular k-labeling and w : V(G) ^ N, for every uv € E(G), w(u) = w(v) where w(u) = '(i) and (ii) opt(l) = min{max{li : li is a vertex irregular labeling}}. The
chromatic number of the local irregularity vertex coloring of G denoted by Xlia(G), is the minimum cardinality of the largest label over all such local irregularity vertex colorings. In this paper, we study a local irregular vertex coloring of PmQ G when G is a family of tree graphs, centipede Cn, double star graph (S2,n), Weed graph (S3,n), and E graph (E3>n).
Keywords: Local irregularity, Corona product, Tree graph family.
1. Introduction
Let G(V, E) be a connected and simple graph with a vertex set V and an edge set E. In this paper, we combine two concepts, namely the local antimagic vertex coloring and the distance irregular labelling, with a local irregularity of vertex coloring. This concept firstly was introduced by Kristiana [2, 3], et. al. The latest research was conducted by Azzahra [4], who examined the local irregularity vertex coloring of a grid graph family. In this paper we study the local irregularity of vertex coloring of corona product graph of a tree graph family.
Definition 1. Suppose l : V(G) — {1,2, ...,k} and w : V(G) ^ N, where
w(u) = Y1 l(v)'
v€N (u)
then l(v) is called the vertex irregular k-labeling and w(u) is called the local irregularity of vertex coloring if
(i) opt(l) = min{max{li} : li vertex irregular labeling};
(ii) for every uv € E(G),w(u) = w(v).
Definition 2. The chromatic number of local irregular graph G denoted by Xus(G), is the minimum of cardinality of the local irregularity of vertex coloring.
In this paper, we will use the following lemma which gives a lower bound on the chromatic number of local irregular vertex coloring:
Lemma 1 [2]. Let G be a simple and connected graph, then xus(G) > x(G).
Proposition 1 [2]. Let G be a graph each two adjacent vertices of which have a different vertex degree then opt(l) = 1.
Proposition 2 [2]. Let G be a graph each two adjacent vertices have the same vertex degree then opt(l) > 2.
Definition 3 [1]. Let G and H be two connected graphs. Let o be a vertex of H. The corona product of the combination of two graphs G and H is defined as the graph obtained by taking a duplicate of graph G and |V(G)| a duplicate of graph H, namely Hf, i = 1,2,3,...|V(G)| then connects each vertex i in G to each vertex in Hi. The corona product of the graphs G and H is denoted by G H.
2. Result and discussion
In this paper, we analyze the new result of the chromatic number of local irregular vertex coloring of corona product by family of tree graph (PmQ) G) where G is centipede graph (Cn), double star graph (S2,n), and Weed graph (S3,n).
Theorem 1. Let G = PmQ Cpn, be a corona product of a path graph of order m and a centipede graph of order n for n,m > 2, then
5, for m = 3 and n = 2, 3,
6, for m = 2 and n = 2, 3 or for m = 3 and n > 4,
7, for m = 2 and n>4 or for m > 4 and n = 2, 3,
8, for m > 4 and n > 4,
1, for m =3 and n = 3,
1, 2, for m =2 and n = 2 or
for m =3 and n = 2 or
for m >3 and n > 4.
Xlis (PmQ Cpn) = with opt(l) defined as
opt(Z)(PmO CPn)
Proof. Vertex set is
V(Pm (¿) CPn) = {xi; 1 < i < m} U {xij; 1 < i < m, 1 < j < n} U {yij; 1 < i < m, 1 < j < n} and the edge set is
E(PmQ) CPn) = {xiXi+i; 1 < i < m - 1} U {xij Xj+i; 1 < i < m, 1 < j < n - 1} U{xijyj; 1 < i < m, 1 < j < n} U {xiXj; 1 < i < m, 1 < j < n} U {xiyj; 1 < i < m, 1 < j < n},
the order and size respectively are 2mn + m and 4mn — 1.
Case 1: m = p, m > 2, p > 2, n > 3.
First step to prove this theorem is to find the lower bound of V(Pm O CPn). Based on Lemma 1, we have Xlis(PmQ Cpn) > X(pmQ Cpn) = 3.
Assume Xiis(PmQ Cp.n) = 4, let Xiis(PmQ Cp.n) = 4, if l(^) = l(x3) = 1, l(x>) = 2, l(xij) = l(yij) = 1 then w(x1) = w(x2), then there are 2 adjacent vertices that have the same color, it contradicts the definition of vertex coloring. If
l(xij) = 1, l(yij) = 1, 1 < i < 3, j = 1, l(yij) =2, 1 < i < 3, j = 2, l(xi) = 1 — w(xi) = w(xi+1), w(xi1) = (xi2),
then xlis(PmQ) Cpn) > 5. Based on this, we have the lower bound Xus(PmQ) Cpn) > 5.
After that, we will find the upper bound of Xus(PmQ) Cpn). Furthermore, the upper bound for the chromatic number of local irregular (Pm O Cpn), we define l : V(Pm O Cpn) — {1,2} with the vertex irregular 2-labelling as follows:
l(xi) = 1, l(xij) = 1,
J 1, for 1 < i < 3 and j = 1, j) [2, for 1 < i < 3 and j = 2.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
6, for i = 1, 3,
w(xi) =
i 8, for i = 2,
3, for 1 < i < 3 and j = 1,
w(x'') = \
4, for 1 < i < 3 and j = 2, w(yij) = 2, for 1 < i < 3 and j = 1,2.
The upper bound is true: Xus(PmQ) Cpn) < 5, and we have 5 < Xlis(PmQ) Cpn) < 5, so Xiis(PmQ Cpn) = 5 for m = 3 and n = 2.
Case 2: m = n = 3.
Based on Proposition 1, opt(l) = 1. So the lower bound of (PmQ Cpn) is
Xlis (PmQ Cpn) > 5.
Hence opt(l) = 1 and the labelling provides the vertex-weight as follows:
7, for i = 1, 3,
w(xi) = <
i 8, for i = 2,
| 3, for 1 < i < 3 and j = 1,3 (mod 4), w(yij) = < " "
4, for 1 < i < 3 and j = 2,
w(xij) = 2, for 1 < i < 3 and 1 < j < 3.
The upper bound is true: Xiis(PmQ Cpn) < 5. We have 5 < Xiis(PmQ Cpn) < 5, so Xiis(PmQCpn) = 5 for m = 3 and n = 3.
Case 3: m = n = 2.
First step here is to find the lower bound of V (PmQ) CPn). Based on Lemma 1, we have Xiis (PmQ Cpn) > X(PmQ Cpn) = 3.
Assume Xiis(PmQ) Cpn) = 5, if l(xi) = 1, 1(x2) = 2, l(xj) = l(yj) = 1, then w(xn) = w(xi2) and there are 2 adjacent vertices, that have the same color, it contradicts the definition of vertex coloring. If
l(xi) = 1, l(x2) = 2, l(xij ) = 1, l(yii) = 1, i = 1,2, l(yi2) = 2, i = 1,2,
then w(xi) = w(x2), w(xii) = w(xi2). Based on that we have the lower bound Xlis (Pm O Cpn) > 6.
After that, we will find the upper bound of Xus(PmQ) Cpn).
Furthermore, we define l : V (PmQ) Cpn) ^ {1,2} with the vertex irregular 2-labelling as follows:
1, for i = 1, 1, for i = 1, 2 and j = 1,
l(xi) = \2, for i = 2, l(xj) = 1, l(yj)^2, for i = 1, 2 and j = 2.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
6, for i = 2,
w(xi) = <
17, for i = 1,
{3, for i = 1 and j = 1
4, for i = 1 and j = 1, or for i = 2 and j = 1,
5, for i = 2 and j = 2,
2, for i = 1 and j = 1, 2,
w(yij) = <
3 [3, for i = 2 and j = 1,2.
We have the following upper bound Xus(PmQ) Cpn) < 6. We have 6 < Xus(PmQ) Cpn) < 6, so Xiis(PmQ Cpn) = 6 for m = 2 and n = 2.
Case 4: m = 2 and n = 3.
First step here is to find the lower bound of V (PmQ) CPn). Based on Lemma 1, we have Xlis(PmO Cpn) > X(PmO Cpn) = 3. Assume Xiis (PmQ Cpn) = 5, if
l(xi)= l(xij ) = 1, l(yij ) = 1, l(y2j ) = 1, j = 1,2, l(yi3)=2,
then w(x22) = w(x23), so there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = 1, l(x2) = 2, l(xij) = 1, l(yij) = 1,
then w(xi) = w(x2), w(xi,i) = w(xi)2). Based on that we have the lower bound Xus(Pm O Cpn) > 6. After that, we will find the upper bound of Xus(PmQ) Cpn).
Furthermore, we define l : V(PmQ Cpn) ^ {1,2} with the vertex irregular 2-labelling as follows:
1 , for i = 1 ,
l(xi) = 1 o i • o l(xij) = 1, w(yij) = 1
2, for i = 2,
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
7, for i = 2,
w(xi) = < V ^ [8, for i = 1,
{3, for i = 1 and j = 1,3,
4, for i = 1 and j = 2, or for i = 2 and j = 1, 3,
5, for i = 2 and j = 2,
2, for i = 1 and 1 < j < 3,
w(yij) = <
3, for i = 2 and 1 < j < 3.
The upper bound is true: Xiis(PmQ Cpn) < 6. So we have Xiis(PmQ Cpn) = 6 for m = 2 and n = 3.
Case 5: m = 3 and n > 4.
First step to prove this theorem in this case is to find the lower bound of V(Pm O CPn). Based on Lemma 1, we have Xiis(PmQ Cpn) > X(pmQ Cpn) = 3.
Assume Xiis(pmQ Cpn) = 5, if l(x1) = l(x3) = 1, l(x2) = 2, l(xij) = l(yij) = 1, then w(x1) = w(x2) so there are 2 adjacent vertices with the have same color, it contradicts the definition of vertex coloring. If
l(xi) = l(xij) = 1, l(yij) = 1, 1 < i < 3, j = 1,n, j = 0 (mod 2), l(yij) =2, 1 < i < 3, j = 1, 3 (mod 4), j = 1,n,
with the w(xi) = w(xi+1), w(xij) — w(xij+1). Therefore we have the lower bound
Xiis (PmQ Cpn) > 6.
After that, we will find the upper bound for Xus(PmQ Cpn).
Furthermore, we define l : V(PmQ Cpn) — {1,2} with the vertex irregular 2-labelling as follows:
l( ) 11, for 1 < i < 3 and j = 1,n or for 1 < i < 3 and j = 0 (mod2), %J 1 2, for 1 < i < 3 and j = 1,3 (mod 4), j = 1,n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
w(xi) =
2n + n/2, for i = 1,3 and n = 0 (mod2),
2n + |_n/2j , for i = 1,3 and n = 1,3 (mod 4),
3n + 1 — n/2, for i = 2 and n = 0 (mod 2), 3n + 1 — [n/2] , for i = 2 and n = 1,3 (mod 4),
{3, for 1 < i < 3 and j = 1, n,
4, for 1 < i < 3 and j = 0 (mod 2),
5, for 1 < i < 3 and j = 1,3 (mod 4), j = 1, n,
w(yij) = 2.
The upper bound is true: Xus(PmQ Cpn) < 6. So Xus(PmQ Cpn) = 6 for m = 3 and n > 4. Case 6: m = 0 (mod 2), m > 4 and n = 2.
First step here is to find the lower bound of V (PmQ) CPn). Based on Lemma 1, we have Xlis (Pm O Cpn) > X(Pm O Cpn) = 3.
Assume Xlis (PmO Cpn) = 5, let Xiis(PmQ Cpn) = 5, if
l(xi) = 1, i = 1, 3 (mod4), i = 2 (mod4), l(xi) = 2, i = 0(mod4), l(xj) = l(yj) = 1
then w(xij) = w(xij+i), then there are 2 adjacent vertices that have same color, this contradicts the definition of vertex coloring. If
l(xi) = 1, i = 1, 3 (mod 4), i = 2 (mod 4), l(xi) = 2, i = 0(mod4), l(xij ) = 1l(yij ) = 1, j =2, l(yj) = 2, j = 1,
then w(xj) = w(xij+i); w(xi+i) = w(xi+2). So we have the lower bound Xus(PmQ) Cpn) > 7. After that, we will find the upper bound of Xus(PmQ) Cpn).
Furthermore, we define l : V (PmQ) Cpn) — {1,2} with the vertex irregular 2-labelling as follows:
J 1, for i = 1, 3(mod4)orfor i = 2 (mod 4),
l(xi) =
I 2, for i = 0 (mod 4),
,, , ,, , | 1, for 1 < i < m and j = 1,
l(xij) = 1, l(yij) = <
2, for 1 < i < m and j = 2. Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
{6, for i = 1, m,
7, for i = 0 (mod 2), i = m,
8, for i = 1, 3 (mod 4), i = 1,
/
3, for i = 1,3 (mod 4) and j = 1 or for i = 2 (mod 4) and j = 1,
4, for i = 1, 3 (mod 4) and j = 2 or for i = 2 (mod 4) and j = 2 or for i = 0 (mod 4) and j = 1,
5, for i = 0 (mod 4) and j = 2,
I 2, for i = 1, 3 (mod 4) and j = 1, 2 or for i = 2 (mod 4) and j = 1, 2, w(yij) = s
I 3, for i = 0 (mod 4) and j = 1, 2.
We have the upper bound Xus(Pm O Cpn) < 7. So Xus(Pm O Cpn) = 7 for m > 4 and n = 2.
Case 7: m = 0 (mod 2), m > 4 and n = 3.
First step to prove this theorem is to find the lower bound of V(Pm Q CPn). Based on Lemma 1, we have Xlis (PmO Cpn) > X(pmQ Cpn) = 3. Assume Xiis(Pm O Cpn) = 5, in this case if
l(xi) = l(xij) = 1, l(yij) = 1, 1 < i < m, j = 3, l(yij) = 2, 1 < i < m, j = 1,2,
then w(xi) = w(xi+i), then there are 2 adjacent vertices that have the same color, this contradicts the definition of vertex coloring. If
l(xi) = 1 i = 1, 3 (mod4), i = 2 (mod4), l(xi) = 2, i = 0(mod2), l(yj) = l(xj) = 1,
w(xij) = <
w(xij) = <
then w(xi+1) = w(xi+2), w(xi1) = w(xi2), w(xi1) = w(yi2). Therefore we have the lower bound Xiis (Pm O Cpn) > 7.
After that, we will find the upper bound of Xus(PmQ) Cpn).
Furthermore, we define l : V (PmQ) Cpn) — {1,2} with the vertex irregular 2-labelling as follows:
,, | 1, for i = 1, 3 (mod 4) or for i = 2 (mod 4), l(xi) =
[2, for i = 0 (mod4),
l(xij) = 1, l(yij) = 1.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
{7, for i = 1, m,
8, for i = 0 (mod 2), i = m,
9, for i = 1, 3 (mod 4), i = 1,
/
3, for i = 1,3 (mod 4) and j = 1,3 or for i = 2 (mod 4) and j = 1,3,
4, for i = 1, 3 (mod 4) and j = 2 or for i = 2 (mod 4) and j = 2 or for i = 0 (mod 4) and j = 1,3,
5, for i = 0 (mod 4) and j = 2,
I 2, for i = 1,3 (mod 4) and 1 < j < 3 or for i = 2 (mod 4) and 1 < j < 3, w(yij) = <
I 3, for i = 0 (mod 4) and 1 < j < 3.
We have the upper bound Xus(PmQ) Cpn) < 7. So Xus(PmQ) Cpn) = 7 for m > 4 and n = 3.
Case 8: m = 2 and n > 4.
First step here is to find the lower bound of V (PmQ) CPn). Based on Lemma 1, we have Xiis (Pm O Cpn) > X(Pm O Cpn) = 3.
Assume Xiis (PmQ Cpn) < 7, let Xiis(PmQ Cpn) = 6, if
l(x1) = 1, l(x2) = 2, l(xij) = l(yij) = 1,
then w(xij+1) = w(xij+2), then there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(x1) = 1, l(x2) = 2, l(xij) = 1, l(yij) = 1, j = 0 (mod2), j = 1,n, l(yij) =2, j = 1,3 (mod4), j = 1, n — w(x1) = w(x2), w(xj+1) = w(xj+2),
then we have the lower bound Xus(PmQ) Cpn) > 7.
After that, we will find the upper bound of Xus(PmQ) Cpn).
Furthermore, we define l : V (PmQ) Cpn) — {1,2} with the vertex irregular 2-labelling as follows:
l(xi) = (1, for i = 1,
i 2, for i = 2, l(xij) = 1
j 11, for i = 1,2 and j = 1,n or for i = 1,2 and j = 0 (mod2), %J 1 2, for i = 1, 2 and j = 1, 3 (mod 4), j = 1, n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
w(xi)
3n + 1 — n/2, for i = 1 and n = 0 (mod 2),
3n + 1 — [n/2] , for i = 1 and n = 1,3 (mod 4),
2n + n/2, for i = 2 and n = 0 (mod 2),
2n + |_n/2j , for i = 2 and n = 1, 3 (mod 4),
/
w(xij)
3, for i = 1 and j = 1, n,
4, for i = 1 and j = 0 (mod 2), j = n or for i = 2 and j = 1, n,
5, for i = 1 and j = 1,3 (mod 4), j = 1, n or for i = 2 and j = 0 (mod 2), j = n,
6, for i = 2 and j = 1,3 (mod 4), j = 1, n,
w(Vij)
2, for i = 1 and 1 < j < n,
3, for i = 2 and 1 < j < n.
The upper bound Xlis(Pm O Cpn) < 7 is true. So Xlis(Pm O Cpn) = 7 for m = 2 and n > 4.
Case 9: m = 1, 3 (mod 4), m > 5 and n = 2.
First step to prove this theorem in this case is to find the lower bound of V(Pm Q CPn). Based on Lemma 1, we have Xlis(PmQ Cpn) > X(pmQ Cpn) = 3. Assume Xlis (PmQ Cpn) < 7, and let Xlis(pmQ Cp,n) = 6, if
l(xi) = 1, i = 1 (mod 4), i = 0 (mod 2), l(xi) = 2, i = 3(mod4), l(xj)= l(yj) = 1,
then w(xii) = w(xi2), w(xi+i) = w(xi+2), then there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = 1, i = 1 (mod4), i = 0 (mod2), l(xi) = 2, i = 3 (mod4), l(xij) = 1, l(yij) = 1, 1 < i < m, j = 1, l(yij) = 2, 1 < i < m, j = 2,
then w(xi+i) = w(xi+2), w(xij+i) = w(xij+2). Therefore we have the lower bound Xlis (PmQ Cpn) > 7.
After that, we will find the upper bound of Xus(PmQ) Cpn).
Furthermore, we define l : V (PmQ) Cpn) — {1,2} with the vertex irregular 2-labelling as follows:
l(Xi )
1, for i = 1 (mod 4) or for i = 0 (mod 2),
2, for i = 3 (mod 4),
1, for 1 < i < m and j = 1,
2, for 1 < i < m and j = 2.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
w(xij) = <
3, for i = 1 (mod 4) and j = 1 or for i = 0 (mod 2) and j = 1,
4, for i = 1 (mod 4) and j = 2 or for i = 0 (mod 2) and j = 2 or for i = 3 (mod 4) and j = 1,
5, for i = 3 (mod 4) and j = 2,
I 2, for i = 1 (mod 4) and j = 1,2 or for i = 0 (mod 2) and j = 1,2, w(yij) = s
I 3, for i = 3 (mod 4) and j = 1, 2.
The upper bound Xiis(PmQ) Cpn) < 7 is true. So Xiis(PmQ) Cpn) = 7 for m = 1,3 (mod 4), m > 5 and n = 2.
Case 10: m = 1, 3 (mod 4), m > 5 and n = 3.
First step here is to find the lower bound of V(PmQ CPn). Based on Lemma 1, we have Xiis (Pm O Cpn) > X(Pm O Cpn) = 3.
Assume Xiis (PmQ Cpn) < 7, let Xiis(PmQ Cpn) = 6, if
l(xi) = l(xj) = 1, l(yn) = 1, l(yj) = 2, j = 2,3,
then w(xi+1) = w(xi+2), then we have that there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = 1, i = 1 (mod 4), i = 0 (mod 2), l(xi) = 2, i = 3 (mod 4), l(xj) = 1; l(yij) = 1,
then w(xi+1) = w(xi+2), w(xij+1) = w(xij+2). Based on that we have the lower bound Xiis(PmQ Cp,n) > 7.
After that, we will find the upper bound of Xus(PmQ Cpn).
Furthermore, we define l : V (PmQ Cpn) — {1,2} with the vertex irregular 2-labelling as follows:
,, | 1, for i = 1 (mod 4) or for i = 0 (mod 2), l(xi) =
[2, for i = 3 (mod4),
l(xij) = 1, l(yij) = 1.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
{7, for i = 1, m,
8, for i = 1, 3 (mod 4), i = 1,m,
9, for i = 0 (mod 2),
/
3, for i = 1 (mod 4) and j = 1,3 or for i = 0 (mod 2) and j = 1,3,
4, for i = 1 (mod 4) and j = 2 or for i = 0 (mod 2) and j = 2 or for i = 3 (mod 4) and j = 1,3,
5, for i = 3 (mod 4) and j = 2,
. | 2, for i = 1 (mod 4) and 1 < j < 3 or for i = 0 (mod 2) and 1 < j < 3,
w(yij) = <
I 3, for i = 3 (mod 4) and 1 < j < 3.
The upper bound is true: Xiis(PmQ) Cpn) < 7. So Xiis(PmQ) Cpn) = 7 for m = 1,3 (mod 4), m > 5 and n = 3.
w(xij) = <
Case 11: m = 0 (mod 2) m > 4 and n > 4.
First step to prove this theorem is to find the lower bound of V(Pm Q CPn). Based on Lemma 1, we have Xlis(PmQ Cp,n) > X(pmQ Cp,n) = 3.
Assume Xlis (PmQ Cpn) < 8, let Xlis (PmQ Cpn) = 7, if
l(xi) = 1, i = 1, 3 (mod4), i = 2 (mod4), l(xi) = 2, i = 0(mod4), l(xj) = l(yj) = 1,
then w(xij+i) = w(xij+2), so there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = 1, i = 1, 3 (mod4), i = 2 (mod4), l(xi) = 2, i = 0(mod4), l(xj) = 1, l(yij) = 1, 1 < i < m, j = 1,n, j = 0 (mod 2), l(yj) = 2, 1 < i < m, j = 1,3 (mod 4), j = 1,n,
then w(xi+i) = w(xi+2), w(xij+i) = w(xij+2), w(xj) = w(yj). Based on that we have the lower bound Xlis (PmQ) Cpn) > 8.
After that, we will find the upper bound of Xlis(PmQ Cpn).
Furthermore, we define l : V(PmQ Cpn) — {1,2} with the vertex irregular 2-labelling as follows:
l(Xi) =
l(Vij) =
1, for i = 1,3 (mod 4) and 1 < j < n or for i = 2 (mod 4) and 1 < j < n,
2, for i = 0 (mod 4) and 1 < j < n,
l(xij) = 1
1, for 1 < i < m and j = 1, n or for 1 < i < m and j = 0 (mod 2),
2, for 1 < i < m and j = 1,3 (mod 4), j = 1, m.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
w(xi) = <
2n + n/2, 2n + | n/2j , 3n + 1 — n/2, 3n + 1 — [n/2] 3n + 2 — n/2, 3n + 1 — | n/2j
w(xij) = <
3,
4,
5,
6,
w(Vij) =
for i = 1,m and n = 0 (mod 2), for i = 1, m and n = 1, 3 (mod 4), for i = 1, 3 (mod 4), i = 1 and n = 0 (mod 2), for i = 0 (mod2), i = m and n = 1,3 (mod4), for i = 1, 3 (mod 4), i = 1 and n = 0 (mod 2), for i = 0 (mod2), i = m and n = 1,3 (mod4),
for i = 1,3 (mod 4) and j = 1, n or for i = 2 (mod 4) and j = 1, n, for i = 1, 3 (mod 4) and j = 0 (mod 2), j = n or
for i = 2 (mod 4) and j = 0 (mod 2), j = n or for i = 0 (mod 4) and j = 1, n, for i = 1, 3 (mod 4) and j = 1, 3 (mod 4), j = 1, n or for i = 2 (mod 4) and j = 1, 3 (mod 4), j = 1, n or for i = 0 (mod 4) and j = 0 (mod 2), j = n, for i = 0 (mod 4) and j = 1, 3 (mod 4), j = 1, n,
2, for i = 1,3 (mod 4) and 1 < j < n or for i = 2 (mod 4) and 1 < j < n,
3, for i = 0 (mod 4) and 1 < j < n.
The upper bound is true: Xlis(PmQ Cpn) < 8. So Xlis(PmQ Cpn) = 8 for m = 0 (mod 4), m > 4 and n > 4.
Case 12: m = 1, 3 (mod 4), m > 5 and n > 4.
First step to prove this theorem in this case is to find the lower bound of V(Pm O CPn). Based on Lemma 1, we have Xiis (PmQ Cp,n) > X(PmQ Cpn) = 3. Assume Xiis (PmQ Cpn) < 8, let Xiis(PmQ Cpn) = 7, if
l(xi) = 1, i = 1 (mod 4), i = 0 (mod 2), l(xi) = 2, i = 3 (mod 4), l(xj)= l(yij) = 1,
then w(xij+1) = w(xij+2), so there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = 1, i = 1 (mod 4), i = 0 (mod 2), l(xi) = 2, i = 3(mod4), l(xj) = 1, l(yj) = 1, 1 < i < m, j = 1,n, j = 0 (mod 2), l(yj) = 2, 1 < i < m, j = 1,3 (mod 4), j = 1,n — w(xi+1) = w(xi+2), w(xij+1) = w(xij+2), w(xij) = w(yij),
therefore we have the lower bound Xus(PmQ Cpn) > 8.
After that, we will find the upper bound of Xus(PmQ Cpn).
Furthermore, we define l : V(PmQ Cpn) — {1,2} with the vertex irregular 2-labelling as follows:
l(xi) =
1, for i = 1 (mod 4) or for i = 0 (mod 2),
l(yij) =
2, for i = 3 (mod 4), l(xij) = 1
1, for 1 < i < m and j = 1,n or for 1 < i < m and j = 0 (mod2),
2, for 1 < i < m and j = 1,3 (mod 4), j = 1, n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows: /
2n + n/2, for i = 1,m and n = 0 (mod2),
w(xi) =
3n + 2 - n/2, for i = 0
3n + 1 — n/2, for i = 1
2n + Ln/2J , for i = 1.
3n — |_n/2j , for i = 1
3n + 1 — |n/2j , for i = 0
w(xij) = <
5
6
w(yij) =
, for i = 1 (mod 4) and j , for i = 1 (mod 4) and j for i = 0 (mod 2) and for i = 3 (mod 4) and , for i = 1 (mod 4) and j for i = 0 (mod 2) and for i = 3 (mod 4) and , for i = 3 (mod 4) and j
for i = 1 (mod 4) and 1 < for i = 3 (mod 4) and 1 <
(mod 2) and n = 0 (mod 2), 3 (mod 4) and n = 0 (mod 2), m and n = 1, 3 (mod 4), 3 (mod 4), i = 1 and n = 1, 3 (mod 4), (mod 4) and n = 1, 3 (mod 4),
= 1,n or for i = 0 (mod 2) and j = 1,n, = 0 (mod 2), j = n or j = 0 (mod 2), j = n or j = 1,n,
= 1, 3 (mod 4), j = 1, n or j = 1, 3 (mod 4), j = 1,n or j = 0 (mod 2), j = n, = 1, 3 (mod 4), j = 1,n,
j < n or for i = 0 (mod 2) and 1 < j < n, j < n.
The upper bound is true: Xus(PmQ Cpn) < 8. So Xus(PmQ Cpn) = 8 for m > 5 and n > 4. □
Theorem 2. Let G = Pm Q S2,n for n,m > 2, then the chromatic number of local irregular G
for m > 4 and n > 2, with opt(l)(Pm O S2,n) = 1,2, for m > 2 and n > 2. Proof. Vertex set is
V(PnQ S2,n) = {xi; 1 < i < m} U {ai; 1 < i < m} U {bi; 1 < i < m} U{aij; 1 < i < m, 1 < j < n} U {bij; 1 < i < m, 1 < j < n}
and the edge set is
E(Pn Q S2,n) = {xixi+i, 1 < i < m - 1} U {aibi; 1 < i < m} U {xiai; 1 < i < m} U{xibi; 1 < i < m} U {xiaj; 1 < i < m, 1 < j < n} U {x^j; 1 < i < m, 1 < j < n} U{aiaij; 1 < i < m, 1 < j < n} U {bibij; 1 < i < m, 1 < j < n}.
The order and the size respectively are 2mn + 3m and 4mn + 4m — 1. This proof is divided into 4 cases as follows.
Case 1: m = 3 and n > 2.
First step to prove this theorem is to find the lower bound of V (PmQ S2,n). Based on Lemma 1, we have Xlis (PmQ S2,n) > X(pmQ S2.n) = 3.
Assume Xlis (PmQ S2,n) = 4, if l(ai) = l(bi) = 1, l(xi) = l(aj) = l(bj) = 1 then w(ai) = w(bi), then there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
therefore we have the lower bound Xlis(PmQ S2,n) > 5.
After that, we will find the upper bound of Xus(PmQ S2,n).
Furthermore, we define l : V(PmQ S2,n) — {1,2} with the vertex irregular 2-labelling as follows:
5, for m = 3 and n > 2,
6, for m = 2 and n > 2,
l(xi) = l(ai) = l(bi) = l(aij) = l(bij) = 1, 1 < j < n — 1, l(bin) = 2,
then
w(ai) = w(bi), w(xi) = w(x3) = w(x2),
l(xi) = 1, l(a) = 1, l(bi) = 1, l(aij) = 1,
. . I 1, for 1 < i < 3 and 1 < j < n — 1, l(bij) = < " " "
2, for 1 < i < 3 and j = n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
l(bij)
w(xi)
w(ai) = n + 2, for 1 < i < 3, w(bi) = n + 3, for 1 < i < 3, w(aij) = 2, w(bij) = 2.
The upper bound Xiis(PmQ) S2,n) < 5 is true. So Xiis(PmQ) S2,n) = 5 for m = 3 and n > 2.
Case 2: m = 2 and n > 2.
First step here is to find the lower bound of V (PmQ S2,n). Based on Lemma 1, we have Xiis(PmQ S2,n) > X(PmQ S2,n) =3. Assume Xiis (PmQ S2,n) = 5, if
l(xi)= l(ai)= l(bi) = l(aij) = l(b2j) = 1, l(b1j) = 1, 1 < j < n — 1, l(M = 2,
and then w(a2) = w(b2), and there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = l(ai) = l(bi) = l(aij) = 1, l(b1j) = 1, l(b1,n) = 2, l(b2J) =2, j = 1, n, l(b2j) = 1, 2 < j < n — 1,
then w(ai) = w(bi), w(x1) = w(x2). Based on that we have the lower bound Xiis(PmQ S2,n) > 6. After that, we will find the upper bound of Xiis(PmQ) S2,n).
Furthermore, we define l : V (PmQ S2,n) — {1,2} with the vertex irregular 2-labelling as follows:
l(xi) = 1, l(ai) = 1, l(bi) = 1, l(aij) = 1,
1, for i = 1 and 1 < j < n — 1 or for i = 2 and 2 < j < n — 1,
l(bij) = s
2, for i = 1 and j = n or for i = 2 and j = 1, n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
. . f 2n + 4, for i = 1, w(xi) =
2n + 5, for i = 2,
w(ai) = n + 2, for i = 1,2,
,, N In + 3, for i = 1, w(bi) =
\n + 4, for i = 2, w(aij) = 2, w(bij) = 2.
The upper bound is true: Xiis(PmQ) S2,n) < 6. So Xiis(PmQ) S2,n) = 6 for m = 2 and n > 2. Case 3: m = 0 (mod 4), m > 4 and n > 2.
First step to prove this theorem in this case is to find the lower bound of V(PmQ S2,n). Based on Lemma 1, we have Xiis (PmQ S2,n) > X(pmQ S2.n) = 3. Assume Xiis(PmQ S2,n) = 6, if
l(xi) = l(ai) = l(bi) = l(aij) = 1, l(bij) = 1, i = 1,3 (mod 4), i = 2 (mod 4), l(bij) = 1, i = 0 (mod 4), j = 1, n, l(bj) = 2, i = 0 (mod 4), j = 1,n,
then w(ai) = w(bi), so there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = l(ai) = l(bi) = l(aj) = 1, l(bij) = 1, i = 1,3 (mod 4), i = m, 1 < j < n — 1, i = 0 (mod 2), i = m, 2 < j < n — 1, l(bj) = 2, i = 1, 3 (mod 4), i = m, j = n, i = 0 (mod 2), i = m, j = 1,n,
l(bij) = <
then w(ai) = w(bi), w(xi) = w(xi+i). Based on that we have the lower bound Xlis(Pm O S2,n) > 7. After that, we will find the upper bound of Xus(PmQ S2,n)
Furthermore, we define l : V (PmQ S2,n) — {1,2} with the vertex irregular 2-labelling as follows:
l(xi) = 1, l(ai) = 1, l(bi) = 1, l(aj) = 1, /
1, for i = 1,3 (mod 4) and 1 < j < n — 1 or for i = 0 (mod 2), i = m and 2 < j < n — 1 or for i = m, and 1 < j < n — 1,
2, for i = 1,3 (mod 4) and j = n or for i = 0 (mod 2), i = m and j = 1, n or for i = m, and j = n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
{2n + 4, for i = 1, m, 2n + 5, for i = 0 (mod 2), i = m, 2n + 6, for i = 1, 3 (mod 4), i = 1, w(ai) = n + 2, for 1 < i < m,
. In + 3, for i = 1,3 (mod4), i = m, w(bi) = <
In + 4, for i = 0 (mod 2), i = m,
w(aij) = 2, w(bij) = 2.
The upper bound is true: Xlis (PmQ S2,n) < 7. So Xlis(pmQ S2,n) = 7 for m = 0 (mod 2), m > 4 and n > 2.
Case 4: m = 1, 3 (mod 4), m > 5 and n > 2.
First step here is to find the lower bound of V(PmQ S2,n). Based on Lemma 1, we have
Xlis(PmQ S2,n) > X(PmQ S2,n) = 3.
Assume Xlis(PmQ S2,n) = 6, if
l(xi) = l(ai) = l(bi) = l(aij) = 1, l(bij) = 1, i = 1 (mod4), i = 0 (mod2), l(bij) = 1, i = 3 (mod4), j = n, l(bij) = 2, i = 3 (mod4), j = n,
then w(ai) = w(bi), and there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = l(ai) = l(bi) = 1, l(aij) = 1, l(bij) = 1, i = 1,3 (mod 4), 1 < j < n — 1, i = 0 (mod 2), 2<j<n — 1, l(bj) = 2, i = 1, 3(mod4), j = n, i = 0(mod2), j = 1,n,
then w(ai) = w(bi),w(xi) = w(xi+i). Based on that we have the lower bound Xlis(PmQ S2,n) > 7. After that, we will find the upper bound of Xlis(PmQ S2,n).
Furthermore, we define l : V(PmQ S2,n) — {1,2} with the vertex irregular 2-labelling as follows:
l(xi) = 1, l(ai) = 1, l(bi) = 1, l(aij) = 1,
J 1, for i = 1,3 (mod 4) and 1 < j < n — 1 or for i = 0 (mod 2), and 2 < j < n — 1,
l (bij ) — |
I 2, for i = 1,3 (mod 4) and j = n or for i = 0 (mod 2) and j = 1, n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
{2n + 4, for i = 1, m,
2n + 5, for i = 1, 3 (mod 4),
2n + 6, for i = 0 (mod 2),
w(ai) = n + 2, for 1 < i < m,
. In + 3, for i = 1, 3 (mod 4), w(bi) = <
In + 4, for i = 0 (mod 2),
w(aij) = 2, w(bij) = 2.
The upper bound is true: Xiis(PmQ S2,n) < 7. So Xiis(PmQ S2,n) = 7 for m = 1,3 (mod 4), m > 5 and n > 2. □
Theorem 3. Let G = PmQ S3,n for n,m > 2, then the chromatic number of local irregular G
is
{5, for m = 3 and n > 2,
6, for m = 2 and n > 2,
7, for m > 4 and n > 3,
with
{1, for m = 3 and n = 3,
1, 2, for m = 2 and n = 2 or for m = 3 and n = 2 or for m > 4 and n > 2.
Proof. The vertex set is
V(Pm (¿) S3,n) = {xi; 1 < i < m} U {ai; 1 < i < m} U {bi; 1 < i < m} U {ci; 1 < i < m} U{aij; 1 < i < m, 1 < j < n} U {bij; 1 < i < m, 1 < j < n} U {cj; 1 < i < m, 1 < j < n}
and the edge set is
V(PmQ S3,n) = {xiXi+1; 1 < i < m - 1} U {xiyi; 1 < i < m} U {xiai; 1 < i < m}
U{xibi; 1 < i < m} U {xici; 1 < i < m} U {yiai; 1 < i < m} U {yibi; 1 < i < m} U{yici; 1 < i < m} U {xiaj,j; 1 < i < m; 1 < j < n} U {xibj,j; 1 < i < m; 1 < j < n} U{xicij; 1 < i < m; 1 < j < n} U {aiaij; 1 < i < m; 1 < j < n} U{bibj; 1 < i < m; 1 < j < n} U {ciCj; 1 < i < m; 1 < j < n}.
The order and size respectively are 3mn + 5m and 6mn + 8n — 1. This proof can be divided into 8 following cases.
Case 1: m = 3 and n = 2.
First step to prove this theorem is to find the lower bound of V (PmQ S3,n). Based on Lemma 1, we have xiis (PmQ S3,n) > x(PmQ S3-n) = 3. Assume xiis (PmQ S3,n) = 4, if
l(ai) = l(bi) = l(ci) = l (yi) = 1(aij) = l(bij) = l(cAj) = 1,
then w(ai) = w(bi) = w(ci) = w(yi), and there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(ai) = l(bi) = l(ci) = 1, l(aj) = l(bij) = l(Cij) = 1, l(yi) = 2,
then (w(ai) = w(bi) = w(ci)) = w(yi), w(xi) = w(x2). Therefore we have the lower bound Xlis(PmQ S3,n) > 5.
After that, we will find the upper bound of Xlis(PmQ S3,n).
Furthermore, we define l : V(Pm Q S3,n) — {1,2} with vertex irregular 2-labelling as follows: l(xi) = 1, l(yi) = 2, l(ai) = 1, l(bi) = 1, l(Ci) = 1, l(aj) = 1, l(bj) = 1, l(cy) = 1. Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
12, for i = 1, 3,
w(xi) =
[13, for i = 2,
w(yi) = 4, w(ai) = 5, w(bi) = 5, w(ci) = 5, w(aj) = 2, w(bj) = 2, w(cj) = 2. The upper bound is true: Xlis(PmQ) S^,n) < 5. So Xlis(PmQ) S^n) = 5 for m = 3 and n = 2. Case 2: m = 3 and n = 3.
Based on Proposition 1, we have opt(l) = 1. So the lower bound (PmQ S3,n) is Xlis(PmQ S3,n) > 5
Since opt(l) = 1, the labelling provides the vertex-weight as follows:
3n + 5, for i = 1, 3,
w(xi) =
3n + 6, for i = 2,
w(yi) = 4, w(ai) = n + 1 for 1 < i < 3, w(bi) = n + 1 for 1 < i < 3, w(ci) = n + 1 for 1 < i < 3, w(aij) = 2, w(bij) = 2, w(cij) = 2.
The upper bound is true: Xlis(PmQ) S^,n) < 5. So Xlis(PmQ) S^,n) = 5 for m = 3 and n > 2.
Case 3: m = 2 and n = 2.
First step to prove this theorem is to find the lower bound of V(PmQ S3,n). Based on Lemma 1, we have Xlis (PmQ S3,n) > X(PmQ S3,n) = 3. Assume Xlis (PmQ S3,n) = 5, if
l(ai) = l(bi) = l(ci) = l(aj) = l(bij) = l(cj) = 1, l(yi) = 1, l(y2) = 2,
then w(a2) = w(b2) = w(c2) = w(y2) and there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi ) = l(ai) = l(bi)= l(ci) = l(bij ) = 1, l(yi) = 2, l(cij ) = 1, l(c2,i) = 1, l(c2,2) = 2,
then w(xi) = w(x2), w(yi) = ((w(ai) = w(bi) = w(ci)). Based on that we have the lower bound Xlis (PmQ S3,n) > 6.
After that, we will find the upper bound of Xlis(Pm O S3,n).
Furthermore, we define l : V(Pm O S3,n) — {1,2} with vertex irregular 2-labelling as follows: l(xi) = 1, l(yi) = 2, l(ai) = 1, l(bi) = 1, l(c) = 1, l(aij) = 1, l(bj) = 1, | 1, for i = 1 and j = 1,2 or for i = 2 and j = 1,
l(cij) = s
2, for i = 2 and j = 2. Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
12, for i = 1,
w(xi) = <
[13, for i = 2,
w(yi) = 4, w(ai) = 5, w(bi) = 5,
5, for i = 1,
w(ci) = <
i 6, for i = 2,
w(aij) = 2, w(bij) = 2, w(cij) = 2.
The upper bound is true: Xlis (PmQ S2,n) < 6. So Xlis(PmQ) S2,n) = 6 for m = 2 and n = 2.
Case 4: m = 2 and n > 3.
First step here is to find the lower bound of V(PmQ S3,n). Based on Lemma 1, we have Xlis(PmO S3,n) > X(Pm O S3,n) =3. Assume Xlis (PmQ S3,n) = 5, if
l(ai) = l(bi) = l(ci) = l(yi) = 1, l(aij) = l(bij) = 1, l(cj) = 1, i = 1,2, 1 < j < n - 1
l(c,j) = 2, i = 1, 2, j = n,
then w(xi) = w(x2), then there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = l(aj) = l(bij) = 1, l(cj) = 1, i = 1, 1 < j < n, i = 2, 1 < j < n — 1,
l(cij) = 2, i = 2, j = n,
then w(xi) = w(x2),w(yi) = ((w(ai) = w(bi) = w(ci)). Therefore we have the lower bound Xlis (Pm CD S3,n) > 6.
After that, we will find the upper bound of Xus(PmQ) S3,n).
Furthermore, we define l : V (PmQ S3,n) — {1,2} with vertex irregular 2-labelling as follows: l(xi) = 1, l(yi) = 1, l(ai) = 1, l(bi) = 1, l(ci) = 1, l(aij) = 1, l(bj) = 1, ^ f 1, for i = 1 and 1 < j < n or for i = 2 and 1 < j < n — 1, ij 2, for i = 2 and j = n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
. | 3n + 5, for i = 1, w(xi) =
3n + 6, for i = 2, w(yi) = 4, w(ai) = n + 1, for i = 1,2, w(bi) = n + 1, for i = 1,2, , , In + 1, for i = 1,
w(ci) =
[n + 2, for i = 2, w(aij) = 2, w(bij) = 2, w(cij) = 2.
The upper bound is true: Xiis(PmQ) S3,n) < 6. So xiis (PmQ S3,n) = 6 for m = 2 and n > 3. Case 5: m = 0 (mod 2) m > 4 and n = 2.
First step to prove this theorem in this case is to find the lower bound of V(PmQ S3,n). Based on Lemma 1, we have xiis (PmQ S3,n) > x(pmQ S3,n) = 3. Assume xiis (PmQ S3,n) = 6, if
l(ai) = l(bi) = l(ci) = l(yi) = 1, l(aij) = l(bij) = 1, l(cj) = 1, i = 1,3 (mod 4), j = 1, 2, i = 0 (mod 4), j = 1, 2, l(cj) = 1, i = 2 (mod 4), j = 1, l(cij) = 2, i = 2 (mod 4), j = 2,
then w(yi) = w(ai). Then there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = 1, l(ai) = l(bi) = l(a) = 1, l(yi) = 2, l(aij) = l(bj) = 1, l(cj) = 1, i = 0 (mod 2), j = 1, i = m, i = 1,3 (mod 4), j = 1,2, l(cj) = 2, i = 0 (mod 2), i = m, j = 2,
then w(xi+1) = w(xi+2; w(yi) = w(ai). Therefore we have the lower bound xus(PmQ S3,n) > 7. After that, we will find the upper bound of xus(PmQ S3,n).
Furthermore, we define l : V(PmQ S3,n) ^ {1,2} with vertex irregular 2-labelling as follows:
l(xi) = 1, l(yi) = 2, l(a) = 1, l(bi) = 1, l(Ci) = 1, l(aij) = 1, l(bjj) = 1, 1, for i = 1,3 (mod 4) and j = 1,2 or for i = m and j = 1,2 or
The upper bound is true: xiis(PmQ S3,n) < 7. So xiis(PmQ S3,n) = 7 for m = 0 (mod 2); m > 4 and n = 2.
Case 6: m = 1, 3 (mod 4), m > 5 and n = 2.
First step here is to find the lower bound of V(PmQ S3,n). Based on Lemma 1, we have
for i = 0 (mod 2), i = m and j = 1, 2, for i = 0 (mod 2), i = m and j = 2.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
1 6, for i = 0 (mod 2), i = m, w(aij) = 2, w(bij) = 2, w(cij) = 2,
5, for i = 1, 3 (mod 4),
Xlis(PrnQ S3,n) > X(PmQ S3,n) = 3. Assume Xlis (PmQ S3,n) = 6, if
l(ai) = l(bi) = l(ci) = l(yi) = 1, l(aij) = l(bij) = 1, l(cj) = 1, i = 1 (mod 4), j = 1,2, i = 0 (mod 2), j = 1,2, l(cj) = 1, i = 3 (mod4), j = 1, l(cij) =2, i = 3 (mod4), j = 2,
then w(yi) = w(ai), then there are 2 adjacent vertices that have same color, it contradicts to definition of vertex coloring. If
l(xi) = 1, l(aj) = l(bij) = 1, l(cj) = 1, i = 1,3 (mod4), j = 1, 2, i = 0 (mod2), j = 1, l(cAj) = 2, i = 0(mod2), j = 2, l(yi) = 2,
then
w(xi+i) = w(xi+2, w(yi) = w(ai), w(yi) = w(bi), w(yi) = w(c).
We have the lower bound Xlis(Pm O S3,n) > 7.
After that, we will find the upper bound of Xlis(Pm O S3,n).
Furthermore, we define l : V(Pm O S3,n) — {1,2} with vertex irregular 2-labelling as follows:
l(xi) = 1, l(yi) = 2, l(ai) = 1, l(bi) = 1, l(c) = 1, l(aij) = 1, l(bj) = 1,
{1, for i = 1,3 (mod 4) and j = 1,2 or
for i = 0 (mod 2), i = m and j = 1, 2, for i = 0 (mod 2), i = m and j = 2.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
{12, for i = 1, m
13, for i = 1,3 (mod4), i = 1,m,
14, for i = 0 (mod 2),
w(yi) = 4, w(ai) = 5, w(bi) = 5,
. . | 5, for i = 1, 3 (mod 4), i = 1, m, w(Ci) = <
I 6, for i = 0 (mod 2),
w(aij) = 2, w(bij) = 2, w(cij) = 2.
The upper bound is true: Xlis(PmQ) S^n) < 7. So Xlis(PmQ) S^,n) = 7 for m = 1, 3 (mod 4); m > 5 and n = 2.
Case 7: m = 0 (mod 2) m > 4 and n > 3.
First step to prove this theorem is to find the lower bound of V(PmQ S3,n). Based on Lemma 1, we have Xlis (PmQ S3,n) > X(PmQ S3,n) = 3. Assume Xlis (PmQ S3,n) = 6, it is true if
l(ai) = l(bi) = l(ci) = l(yi) = l(aij) = l(bij) = 1, l(cj) = 1, 1 < i < m, 1 < j < n — 1, l(cj) = 2, 1 < i < m, j = n,
then w(xi+i) = w(xi+2), then there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = 1, l(aij)= l(bij)= l(yi) = 1, l(cij) = 1, i = 1,3 (mod 4), 1 < j < n, i = 0 (mod 2), i = m, 1 < j < n — 1, i = m, 1 < j < n, l(c-ij) = 2, i = 0 (mod 2), i = m, i = m, j = n, w(xi+i) = w(xi+2),
we have the lower bound of Xus(PmQ) S3,n) > 7. After that, we will find the upper bound Xlis (PmQ S3 ,n).
l(cij) = <
Furthermore, we define l : V(Pm O S3,n) ^ {1,2} with vertex irregular 2-labelling as follows:
l(xi) = 1, l(yi) = 1, l(ai) = 1, l(bi) = 1, l(Ci) = 1, l(aij) = 1, l(bij) = 1, /
1, for i = 1,3 (mod 4) and 1 < j < n or for i = 0 (mod 2), i = m, and 1 < j < n — 1 or for i = m, and 1 < j < n,
2, for i = 0 (mod 2), i = m and j = n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
{3n + 5, for i = 1, m, 3n + 6, for i = 1, 3 (mod 4), i = 1, 3n + 7, for i = 0 (mod 2), i = m, w(yi) = 4, w(ai) = n + 2, for 1 < i < m, w(bi) = n + 2, for 1 < i < m,
. . In + 2, for i = m, or for i = 1,3 (mod4), w(ci) = <
In + 3, for i = 0 (mod2), i = m,
w(aij) = 2, w(bij) = 2, w(cij) = 2,
The upper bound xiis(PmQ S^,n) < 7. So xiis(PmQ S^n) = 7 for m = 0 (mod 2); m > 4 and n > 3.
Case 8: m = 1, 3 (mod 4), m > 5 and n > 3.
First step to prove the theorem in this case is to find the lower bound of V(PmQ S3,n). Based on Lemma 1, we have xiis (PmQ S3,n) > x(pmQ S3,n) = 3. Assume xiis (PmQ S3,n) = 6, if
l(xi) = l(yi) = l(aij) = l(bij) = l(cij) = 1, l(ai) = l(bi) = 1, l(ci) = 2,
then w(xi+1) = w(xi+2). Then there are 2 adjacent vertices that have same color, it contradicts the definition of vertex coloring. If
l(xi) = l(ai) = l(bi) = l(ci) = 1, l(aij)= l(bij)= l(yi) = 1, l(cj) = 1, i = 1,3 (mod 4), 1 < j < n, i = 0 (mod 2), 1 < j < n — 1, l(cj) = 2, i = 0 (mod 2), j = n,
then w(xi+1) = w(xi+2). Based on that we have the lower bound xiis(PmQ S3,n) > 7. After that, we will find the upper bound of xus(PmQ S3,n).
Furthermore, we define l : V(PmQ S3,n) ^ {1,2} with vertex irregular 2-labelling as follows: l(xi) = 1, l(yi) = 1, l(ai) = 1, l(bi) = 1, l(ci) = 1, l(aij) = 1, l(bij) = 1,
{1, for i = 1,3 (mod4), and 1 < j < n or
for i = 0 (mod 2), and 1 < j < n — 1, 2, for i = 0 (mod 2), and j = n.
Hence, opt(l) = 2 and the labelling provides the vertex-weight as follows:
{3n + 5, for i = 1, m,
3n + 6, for i = 1,3 (mod 4), i = 1,m,
3n + 7, for i = 0 (mod 2),
w(Vi ) = 4,
w(ai) = n + 2, for 1 < i < m, w(bi) = n + 2, for 1 < i < m,
. . In + 2, for i = 1, 3 (mod 4), w(ci ) = <
In + 3, for i = 0 (mod2),
w(aij ) = 2, w(bij ) = 2, w(cij ) = 2.
The upper bound is true: Xiis(PmQ S3,n) < 7. So Xiis(PmQ S3,n) = 7 for m = 1,3 (mod 4), m > 5 and n > 3. □
3. Conclusion
In this paper, we have studied the coloring of the vertices of the local irregular corona product by the graph of the family tree. We determined the exact value of the local irregular chromatic number of the corona product from the graph of the family tree, namely Xlis(Pm O Cpn), Xus(PmQ) S2,n) and Xlis(PmQ S3,n).
Acknowledgements
We gratefully acknowledge the support from University of Jember of year 2023.
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