VII. MINOR REVISION REQUIRED
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 Theorem3we 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
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.
International Journal of Mathematics and Mathematical Sciences 3 Case 1 (π odd). According to Theorem4forπ 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 Theorem7 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)
4 International Journal of Mathematics and Mathematical Sciences
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.
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
0, for3π‘βcomponent, π even, 2 β€ π β€ π β 22
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
0, for3π‘βcomponent, π odd, 3 β€ π β€ π β 12
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.
International Journal of Mathematics and Mathematical Sciences 5
Data Availability
The data used to support the findings of this study are available from the corresponding author upon request.
3
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.
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