2 International Journal of Mathematics and Mathematical Sciences
al. [6] have succeeded in constructing tree on π vertices, π β₯ 5, with locating chromatic numbers varying from 3 to π, except for (π β 1). Then Behtoei and Omoomi [7]
have obtained the locating chromatic number of the Kneser graphs. Recently, Asmiati et al. [8] obtained the locating chromatic number of the generalized Petersen graphπ(π, 1) forπ β₯ 3. Baskoro and Asmiati [9] have characterized all trees with locating chromatic number 3. In [10] all trees of order π with locating chromatic number π β 1 were characterized, for any integers π and π‘, where π > π‘ + 3 and2 β€ π‘ < π/2. Asmiati et al. in [11] have succeeded in determining the locating chromatic number of homogeneous amalgamation of stars and their monotonicity properties and in [12] for firecracker graphs. Next, Wellyyanti et al. [13]
determined the locating chromatic number for complete π-ary trees.
The generalized Petersen graphπ(π, π), π β₯ 3 and 1 β€ π β€ β(π β 1)/2β, consists of an outer π-cycle π¦1, π¦2, . . . , π¦π, a set of π spokes π¦ππ₯π, 1 β€ π β€ π, and π edges π₯ππ₯π+π, 1 β€ π β€ π, with indices taken modulo π. The generalized Petersen graph was introduced by Watkins in [14]. Let us note that the generalized Petersen graphπ(π, 1) is a prism defined as Cartesian product of a cycleπΆπand a pathπ2.
Next theorems give the locating chromatic numbers for complete graphπΎπand generalized Petersen graphπ(π, 1).
Theorem 3 (see [6]). For π β₯ 2, the locating chromatic number of complete graphπΎπisπ.
Theorem 4 (see [8]). The locating chromatic number of generalized Petersen graphπ(π, 1) is 4 for odd π β₯ 3 or 5 for evenπ β₯ 4.
The barbell graph is constructed by connecting two arbitrary connected graphsπΊ and π» by a bridge. In this paper, firstly we discuss the locating chromatic number for barbell graphπ΅π,πforπ, π β₯ 3, where πΊ and π» are complete graphs onπ and π vertices, respectively. Secondly, we determine the locating chromatic number of barbell graphπ΅π(π,1)forπ β₯ 3, whereπΊ and π» are two isomorphic copies of the generalized Petersen graphπ(π, 1).
2. Results and Discussion
Next theorem proves the exact value of the locating chromatic number for barbell graphπ΅π,π.
Theorem 5. Let π΅π,π be a barbell graph forπ β₯ 3. Then the locating chromatic number ofπ΅π,πisππΏ(π΅π,π) = π + 1.
Proof. Letπ΅π,π,π β₯ 3, be the barbell graph with the vertex setπ(π΅π,π) = {π’π, Vπ : 1 β€ π β€ π} and the edge set πΈ(π΅π,π)
=βπβ1π=1{π’ππ’π+π : 1 β€ π β€ π β π} βͺ βπβ1π=1{VπVπ+π : 1 β€ π β€ π β π} βͺ {π’πVπ}.
First, we determine the lower bound of the locating chromatic number for barbell graph π΅π,π forπ β₯ 3. Since the barbell graphπ΅π,π contains two isomorphic copies of a complete graphπΎπ, then with respect to Theorem 3 we have ππΏ(π΅π,π) β₯ π. Next, suppose that π is a locating coloring
using π colors. It is easy to see that the barbell graph π΅π,π contains two vertices with the same color codes, which is a contradiction. Thus, we have thatππΏ(π΅π,π) β₯ π + 1.
To show that π + 1 is an upper bound for the locating chromatic number of barbell graphπ΅π,π it suffices to prove the existence of an optimal locating coloringπ : π(π΅π,π) σ³¨β
{1, 2, . . . , π + 1}. For π β₯ 3 we construct the function π in the following way:
π (π’π) = π, 1 β€ π β€ π
π (Vπ) = {{ {{ {{ {{ {
π, forπ = 1 π, for2 β€ π β€ π β 1 π + 1, otherwise.
(1)
By using the coloringπ, we obtain the color codes of π(π΅π,π) as follows:
πΞ (π’π)
= {{ {{ {{ {{ {
0, for ππ‘β component, 1 β€ π β€ π
2, for (π + 1)π‘β component, 1 β€ π β€ π β 1 1, otherwise,
πΞ (Vπ) = {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
0, for ππ‘β component, 2 β€ π β€ π β 1 forππ‘β component, π = 1, and for (π + 1)π‘β component, π = π, 3, for 1π π‘ component, 1 β€ π β€ π β 1 2, for 1π π‘ component, π = π 1, otherwise.
(2)
Since all vertices inπ(π΅π,π) have distinct color codes, then the coloringπ is desired locating coloring. Thus, ππΏ(π΅π,π) = π + 1.
Corollary 6. For π, π β₯ 3, and π ΜΈ= π, the locating chromatic number of barbell graphπ΅π,πis
ππΏ(π΅π,π) = max {π, π} . (3)
Next theorem provides the exact value of the locating chromatic number for barbell graphπ΅π(π,1).
Theorem 7. Let π΅π(π,1)be a barbell graph forπ β₯ 3. Then the locating chromatic number ofπ΅π(π,1)is
International Journal of Mathematics and Mathematical Sciences 3
ππΏ(π΅π(π,1)) ={ {{
4, for odd π
5, for even π. (4)
Proof. Letπ΅π(π,1),π β₯ 3, be the barbell graph with the vertex setπ(π΅π(π,1)) = {π’π, π’π+π, π€π, π€π+π: 1 β€ π β€ π} and the edge set πΈ(π΅π(π,1)) = {π’ππ’π+1, π’π+ππ’π+π+1, π€ππ€π+1,π€π+ππ€π+π+1 : 1 β€ π β€ π β 1} βͺ {π’ππ’1, π’2ππ’π+1, π€ππ€1, π€2ππ€π+1} βͺ {π’ππ’π+π, π€ππ€π+π: 1 β€ π β€ π} βͺ {π’ππ€π}.
Let us distinguish two cases.
Case 1 (π odd). According to Theorem 4 for π odd we have ππΏ(π΅π(π,1)) β₯ 4. To show that 4 is an upper bound for the locating chromatic number of the barbell graphπ΅π(π,1) we describe an locating coloringπ using 4 colors as follows:
π (π’π) = {{ {{ {{ {{ {
1, for π = 1 3, for even π, π β₯ 2 4, for odd π, π β₯ 3.
π (π’π+π) = {{ {{ {{ {{ {
2, for π = 1 3, for odd π, π β₯ 3 4, for even π, π β₯ 2.
π (π€π) = {{ {{ {{ {{ {
1, for odd π, π β€ π β 2 2, for even π, π β€ π β 1 3, for π = π.
π (π€π+π) = {{ {{ {{ {{ {
1, for even π, π β€ π β 1 2, for odd π, π β€ π β 2 4, for π = π.
(5)
Forπ odd the color codes of π(π΅π(π,1)) are
πΞ (π’π)
= {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
π, for2ππ component, π β€ π + 1 2 π β 1, for1π π‘component, π β€π + 1
2 π β π + 1, for 1π π‘component, π >π + 1
2 π β π + 2, for 2ππ component, π > π + 1 0, for3π‘β component, π even, π β₯ 22
for4π‘β component, π odd, π β₯ 3 1, otherwise.
πΞ (π’π+π)
= {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
π, for1π π‘ component, π β€π + 1 2 π β 1, for2ππ component, π β€ π + 1
2 π β π + 1, for 2ππ component, π > π + 1
2 π β π + 2, for 1π π‘ component, π >π + 1 0, for4π‘β component, π even, π β₯ 22
for3π‘β component, π odd, π β₯ 3 1, otherwise.
πΞ (π€π)
= {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
π, for3π‘β component, π β€ π β 1 2 π + 1, for4π‘β component, π β€ π β 1
2 π β π, for3π‘β component, π β₯ π + 1
2 π β π + 1, for 4π‘β component, π β₯ π + 1 0, for2ππ component, π even, π β€ π β 12
for1π π‘ component, π odd, π β€ π β 2 1, otherwise.
πΞ (π€π+π)
= {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
π, for4π‘β component, π β€ π β 1 2 π + 1, for3π‘β component, π β€ π β 1
2 π β π, for4π‘β component, π β₯ π + 1
2 π β π + 1, for 3π‘β component, π β₯ π + 1 0, for1π π‘ component, π even, π β€ π β 12
for2ππ component, π odd, π β€ π β 2 1, otherwise.
(6) Since all vertices inπ΅π(π,1)have distinct color codes, then the coloringπ with 4 colors is an optimal locating coloring and it proves thatππΏ(π΅π(π,1)) β€ 4.
Case 2 (π even). In view of the lower bound from Theorem 7 it suffices to prove the existence of a locating coloringπ : π(π΅π(π,1)) σ³¨β {1, 2, . . . , 5} such that all vertices in π΅π(π,1) have distinct color codes. Forπ even, π β₯ 4, we describe the locating coloring in the following way:
π (π’π) = {{ {{ {{ {{ {{ {{ {{ {
1, for π = 1
3, for even π, 2 β€ π β€ π β 2 4, for odd π, 3 β€ π β€ π β 1 5, for π = π.
4 International Journal of Mathematics and Mathematical Sciences
π (π’π+π) = {{ {{ {{ {{ {
2, for π = 1 3, for odd π, π β₯ 3 4, for even π, π β₯ 2.
π (π€π) = {{ {{ {{ {{ {{ {{ {{ {
1, for odd π, π β€ π β 3 2, for even π, π β€ π β 2 3, for π = π β 1 4, for π = π.
π (π€π+π) = {{ {{ {{ {{ {
1, for even π, π β€ π β 2 2, for odd π, π β€ π β 1 5, for π = π.
(7) In fact, our locating coloring of π΅π(π,1), π even, has been chosen in such a way that the color codes are
πΞ (π’π)
= {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
π, for2ππ and5π‘βcomponents, π β€ π 2 π β 1, for1π π‘component, π β€π
2 π β π, for5π‘βcomponent, π > π 2 π β π + 1, for 1π π‘component, π >π 2 π β π + 2, for 2ππ component, π >π
2
0, for3π‘βcomponent, π even, 2 β€ π β€ π β 2 for4π‘βcomponent, π odd, 3 β€ π β€ π β 1 2, for4π‘βcomponent, π = 1
for3π‘βcomponent, π = π 1, otherwise.
πΞ (π’π+π)
= {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
π, for1π π‘component, π β€π 2 π β 1, for2ππ component, π β€π
2 π + π, for5π‘βcomponent, π β€ π 2
π β π + 1, for 2ππ and5π‘βcomponents, π > π 2 π β π + 2, for 1π‘βcomponent,π >π
2
0, for3π‘βcomponent, π odd, 3 β€ π β€ π β 1 for4π‘βcomponent, π even, 2 β€ π β€ π 2, for3π‘βcomponent, π = 1
1, otherwise.
πΞ (π€π)
= {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
π, for4π‘βcomponent, π β€ π 2 π + 1, for5π‘βcomponent, π β€ π 2 for3π‘βcomponent, π β€ π
2β 1 π β π, for4π‘βcomponent, π > π
2 π β π + 1, for 5π‘βcomponent, π > π 2 π β π β 1, for 3π‘βcomponent, π
2β€ π β€ π β 1 0, for1π π‘component, π odd, π β€ π β 3
for2ππ component, π even, π β€ π β 2 2, for1π π‘component, π = π β 1
for2ππ component, π = π 1, otherwise.
πΞ (π€π+π)
= {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {{ {
π, for5π‘βcomponent, π β€ π 2 π + 1, for4π‘βcomponent, π β€ π 2 π + 2 for3π‘βcomponent, π β€ π
2β 1 π β π, for3π‘βcomponent, π
2β€ π β€ π β 1 for5π‘βcomponent, π > π
2 π β π + 1, for 4π‘βcomponent, π > π
0, for1π π‘component, π even, π β€ π β 22
for2ππ component, π odd, π β€ π β 1 2, for1π π‘and3π‘βcomponents, π = π 1, otherwise.
(8)
Since forπ even all vertices of π΅π(π,1)have distinct color codes then our locating coloring has the required properties and ππΏ(π΅π(π,1)) β€ 5. This concludes the proof.
Data Availability
The data used to support the findings of this study are available from the corresponding author upon request.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
The authors are thankful to DRPM Dikti for the Fundamental Grant 2018.
International Journal of Mathematics and Mathematical Sciences 5
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