# Locating Chromatic Number of Powers of Paths and Cycles

^{*}

## Abstract

**:**

## 1. Introduction

## 2. Locating Chromatic Number of Powers of Paths

**Example**

**1.**

**Theorem**

**1.**

**Proof.**

## 3. Locating Chromatic Number of Powers of Cycles

**Lemma**

**1**

**.**Let c be a locating-coloring in a connected graph G. If u and v are distinct vertices of G such that $d(u,w)=d(v,w)$ for all $w\in v(G)\setminus \{u,v\}$, then $c(u)\ne c(v)$. In particular, if u and v are non-adjacent vertices of G such that $N(u)=N(v)$, then $c(u)\ne c(v)$.

**Theorem**

**2.**

**Proof.**

**Theorem**

**3.**

**Proof.**

**Lemma**

**2.**

**Proof.**

**Lemma**

**3.**

- (i)
- If $n\ge 7$, then ${\chi}_{L}({C}_{n}^{2})=5$.
- (ii)
- If $n\ge 9$, then ${\chi}_{L}({C}_{n}^{3})=6$ when $n\in \{4q,4q+1,4q+2\}$, and $\phantom{\rule{4.pt}{0ex}}6\le {\chi}_{L}({C}_{n}^{3})\le 7$ when $n=4q$.

**Proof.**

- (i)
- In view of Theorem 3 and Lemma 2, it is enough to show that ${\chi}_{L}({C}_{3q+2}^{2})\le 5$. Assume that $n=3q+2$, then $\pi =\{{R}_{1}=\{{x}_{1}\},{R}_{2}=\{{x}_{5},{x}_{8},\dots ,{x}_{3q+2}\},{R}_{3}=\{{x}_{3}\},{R}_{4}=\{{x}_{4},{x}_{7},\dots ,{x}_{3q+1}\},$${R}_{5}=\{{x}_{2},{x}_{6},{x}_{9},\dots ,{x}_{3q}\}\}$ is a partition of $V({C}_{n}^{2})$. Now, define $c:V({C}_{n}^{2})\u27f6\{1,2,3,4,5\}$ by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$. Then, it is easy to show that ${c}_{\pi}(u)\ne {c}_{\pi}(v)$ for any $u,v\in V({C}_{n}^{2})$.
- (ii)
- $\phantom{\rule{4.pt}{0ex}}\mathrm{Let}\phantom{\rule{4.pt}{0ex}}{\pi}_{1}=\{{R}_{1}=\{{x}_{1}\},{R}_{2}=\{{x}_{6},{x}_{10},\dots ,{x}_{4q+2}\},{R}_{3}=\{{x}_{3},{x}_{7},\dots ,{x}_{4q-1}\},{R}_{4}=\{{x}_{4}\},$${R}_{5}=\{{x}_{5},{x}_{9},\dots ,{x}_{4q+1}\},{R}_{6}=\{{x}_{2},{x}_{8},{x}_{12}\dots ,{x}_{4q}\}\},$ and ${\pi}_{2}={\pi}_{1}\cup {R}_{7}$, ${R}_{7}=\{{x}_{4q+3}\}$. Then, ${\pi}_{k}$ is a partition of $V({C}_{4q+1+k}^{3})$ for $k=1,2$. Now, let $c:V({C}_{4q+1+k}^{3})\u27f6\{1,2,\dots ,5+k\}$ defined by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$. Clearly, for any $u,v\in V({C}_{4q+1+k}^{3})$, ${c}_{{\pi}_{k}}(u)\ne {c}_{{\pi}_{k}}(v)$ for $k=1,2$.

**Lemma**

**4.**

- (i)
- Let $m=2t\ge 4$ and $n=q(m+1{}^{\xaf})+t,q\ge 2$. Then, $m+3\le {\chi}_{L}({C}_{n}^{m})\le m+t+1$.
- (ii)
- Let $m=2t-1\ge 5$ and $n=q(m+1)+(t-1),q\ge 2$. Then, $m+3\le {\chi}_{L}({C}_{n}^{m})\le m+t$.

**Proof.**

- (i)
- Assume that $m=2t$ and $n=q(m+1)+t,q\ge 2$. Notice that the length of the path ${x}_{q(m+1)-t}-{x}_{q(m+1)-(t-1)}-\cdots -{x}_{q(m+1)+t}-{x}_{1}-\dots -{x}_{t}$ is $m+t$ and the length of the path ${x}_{q(m+1)}-{x}_{q(m+1)+1}-\dots -{x}_{q(m+1)+t}-{x}_{1}-\dots -{x}_{t+1}$ is $m+1$, while the length of the path ${x}_{q(m+1)-(t-i+1)}-{x}_{q(m+1)-(t-i)}-\dots -{x}_{q(m+1)+t}-{x}_{1}-\dots -{x}_{t+i}$ is $m+t+1$ for $2\le i\le t$ Thus, $d({x}_{t},{x}_{q(m+1)-t})=d({x}_{t+1},{x}_{q(m+1)})=d({x}_{t+i},{x}_{q(m+1)-(t-i+1)})=2$. Now, let ${R}_{1}=\{{x}_{1}\},{R}_{2}=\{{x}_{2}\},\dots ,{R}_{t-1}=\{{x}_{t-1}\},$ ${R}_{t}=\{{x}_{t},{x}_{(t+1)+(m+1)},$${x}_{(t+1)+2(m+1)},\dots ,{x}_{q(m+1)-t}\},$${R}_{t+1}=\{{x}_{t+1},{x}_{2(m+1)},$${x}_{3(m+1)},\dots ,{x}_{q(m+1)}\},$ ${R}_{t+i}=\{{x}_{t+i},{x}_{(t+i)+(m+1)}$ $,{x}_{(t+i)+2(m+1)},$ $\dots ,{x}_{q(m+1)-(t-i+1)}\}$, $2\le i\le t$, ${R}_{m+1}=\{{x}_{m+1}\},$ ${R}_{m+i}=\{{x}_{m+i},{x}_{(m+i)+(m+1)}$ $,{x}_{(m+i)+2(m+1)},\dots ,{x}_{q(m+1)+(i-1)}\}$, $2\le i\le t+1$. Then, $\pi =\{{R}_{i}:i=1,2,\dots ,m+t+1\}$ is a partition of $V({C}_{n}^{m})$ and $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+t+1\}$ defined by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$. By using Theorem 3, we obtain $m+3\le {\chi}_{L}({C}_{n}^{m})\le m+t+1$.
- (ii)
- Assume that $m=2t-1$ and $n=q(m+1)+(t-1),q\ge 2$. Then, $d({x}_{t-1},{x}_{q(m+1)-t})=d({x}_{t},{x}_{q(m+1)-(t-1)})=d({x}_{t+1},{x}_{q(m+1)})=d({x}_{t+i},{x}_{q(m+1)-(t-i)})=2$ for all $2\le i\le t-1$. Set ${R}_{1}=\{{x}_{1}\},{R}_{2}=\{{x}_{2}\},\dots ,{R}_{t-2}=\{{x}_{t-2}\},{R}_{t-1}=\{{x}_{t-1},{x}_{t+(m+1)},{x}_{t+2(m+1)}$$,\dots ,{x}_{q(m+1)-t}\},$${R}_{t}=\{{x}_{t},{x}_{(t+1)+(m+1)},{x}_{(t+1)+2(m+1)},\dots ,{x}_{q(m+1)-(t-1)}\},$${R}_{t+1}=\{{x}_{t+1},{x}_{2(m+1)},{x}_{3(m+1)},$$\dots ,{x}_{q(m+1)}\},$ ${R}_{t+i}=\{{x}_{t+i},{x}_{(t+i)+(m+1)},$${x}_{(t+i)+2(m+1)},$$\dots ,{x}_{q(m+1)-(t-i)}\},$$2\le i\le t-1$, ${R}_{m+1}=\{{x}_{m+1}\}$, ${R}_{m+i}=\{{x}_{m+i},$${x}_{(m+i)+(m+1)},$ ${x}_{(m+i)+2(m+1)},$$\dots ,{x}_{q(m+1)+(i-1)}\},2\le i\le t$. Then, $\pi =\{{R}_{i}:i=1,\dots ,m+t\}$ is a partition of $V({C}_{n}^{m})$ and $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+t\}$ defined by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.

**Lemma**

**5.**

**Proof.**

- (1)
- For $n=q(m+1)+2$, let ${R}_{1}=\{{x}_{1}\},{R}_{2}=\{{x}_{2},{x}_{2+(m+1)},\dots ,{x}_{q(m+1)-(m-1)}\},$${R}_{3}=\{{x}_{3},{x}_{3+(m+1)},\dots ,{x}_{q(m+1)-(m-2)}\},\dots ,$ ${R}_{m}=\{{x}_{m},{x}_{m+(m+1)},\dots ,{x}_{q(m+1)-1}\},$${R}_{m+1}=\{{x}_{m+1}\},$${R}_{m+2}=\{{x}_{1+(m+1)},{x}_{1+2(m+1)},\dots ,{x}_{q(m+1)+1}\},$ ${R}_{m+3}=\{{x}_{2(m+1)},{x}_{3(m+1)},\dots ,{x}_{q(m+1)}\},$${R}_{m+4}=\{{x}_{q(m+1)+2}\}$. Then, $\pi =\{{R}_{i}:i=1,2\dots ,m+4\}$ is a partition of $V({C}_{n}^{m})$ and $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+t\}$ defined by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.
- (2)
- For $n=q(m+1)+i$, where $3\le i<t$ let ${R}_{1},{R}_{2},\dots ,{R}_{m+4}$ similar to the case $n=q(m+1)+2$. Set ${R}_{m+j}=\{{x}_{(m+1)q+(j-2)}\},5\le j\le i+2$. Then, $\pi =\{{R}_{i}:i=1,2\dots ,m+i+2\}$ is a partition of $V({C}_{n}^{m})$ and $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+i+2\}$ given by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.
- (3)
- By part (i) of Lemma 4, ${\chi}_{L}({C}_{n}^{m})\le m+t+1$ when $n=q(m+1)+t,q\ge 2$.
- (4)
- For $n=q(m+1)+(t+1)$, take ${R}_{1},{R}_{2},\dots ,{R}_{m+t+1}$ similar to the case $n=q(m+1)+(t-1)$ except ${R}_{t+1}$ and ${R}_{m+3}$. Let ${R}_{t+1}=\{{x}_{t+1+(m+1)},{x}_{t+1+2(m+1)},\dots ,$${x}_{t+1+q(m+1)}\}$, ${R}_{m+3}=\{{x}_{t+1},{x}_{2(m+1)},{x}_{3(m+1)},\dots ,{x}_{q(m+1)}\}$, and ${R}_{m+t+2}=\{{x}_{t+q(m+1)}\}$. This implies that $\pi =\{{R}_{i}:i=1,2\dots ,m+i+2\}$ is a partition of $V({C}_{n}^{m})$ and $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+t\}$ defined by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.
- (5)
- For $n=q(m+1)+(t+i)$ where $2\le i\le t$, take ${R}_{1},{R}_{2},\dots ,{R}_{m+t+2}$ similar to the case $n=q(m+1)+(t+1)$ except ${R}_{t+j}$ and ${R}_{m+j+2}$, where $2\le j\le i$. Set ${R}_{t+j}=\{{x}_{(t+j)+(m+1)},{x}_{(t+j)+2(m+1)},$$\dots ,{x}_{(t+j)+q(m+1)}\}$, and ${R}_{m+j+2}=\{{x}_{q(m+1)+j},{x}_{t+j}\}$. Note that $d({x}_{q(m+1)+j},{x}_{t+j})=2$, $d({x}_{t+j},{x}_{m+1})=1$ and $d({x}_{q(m+1)+j},{x}_{m+1})=2$. Then, it is easy to show that the function $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+t+2\}$ defined by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.

**Lemma**

**6.**

**Proof.**

- (1)
- If $n=q(m+1)+2$, define ${R}_{1}=\{{x}_{1}\},{R}_{2}=\{{x}_{2},{x}_{2+(m+1)},{x}_{2+2(m+1)},\dots ,{x}_{q(m+1)-(m-1)}\},$${R}_{3}=\{{x}_{3},{x}_{3+(m+1)},\dots ,{x}_{q(m+1)-(m-2)}\},\dots $, ${R}_{m}=\{{x}_{m},{x}_{m+(m+1)},\dots ,{x}_{q(m+1)-1}\},{R}_{m+1}=\{{x}_{m+1}\},$${R}_{m+2}=\{{x}_{1+(m+1)},{x}_{1+2(m+1)},\dots ,{x}_{1+q((m+1)}\},$${R}_{m+3}=\{{x}_{2(m+1)},{x}_{3(m+1)},\dots ,{x}_{q(m+1)}\},$${R}_{m+4}=\{{x}_{q(m+1)+2}\}$. Clearly, $\pi =\{{R}_{i}:i=1,2,\dots ,m+4\}$ is a partition of $V({C}_{n}^{m+1})$ and the function $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+4\}$ given by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.
- (2)
- If $n=q(m+1)+i$, where $3\le i\le t-2$, take ${R}_{1},{R}_{2},\dots ,{R}_{m+4}$ similar to the case $n=q(m+1)+2$. Set ${R}_{m+j}=\{{x}_{(j-2)+q(m+1)}\},5\le j\le i+2$. Then, $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+i+2\}$ defined by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.
- (3)
- From part (ii) of Lemma 4, we conclude that ${\chi}_{L}({C}_{n}^{m})\le m+t$ when $n=q(m+1)+(t-1),q\ge \phantom{\rule{3.33333pt}{0ex}}2$.
- (4)
- If $n=q(m+1)+t$, take ${R}_{1},{R}_{2},\dots ,{R}_{m+t}$ similar to the case $n=q(m+1)+(t-2)$ except ${R}_{t}$ and ${R}_{m+3}$. Set ${R}_{t}=\{{x}_{t+(m+1)},{x}_{t+2(m+1)},\dots ,{x}_{t+q(m+1)}\}$ and ${R}_{m+3}=\{{x}_{t},{x}_{2(m+1)},\dots ,{x}_{q(m+1)}\}$. Then, $\pi =\{{R}_{i}:i=1,2,\dots ,m+t\}\cup {R}_{m+t+1}$, where ${R}_{m+t+1}=\{{x}_{(t-1)+q(m+1)}\}$ is a partition of $V({C}_{n}^{m})$. Notice that $d(t,q(m+1))=2,d(t,m+1)=1$ and $d(q(m+1),m+1)=2$. Thus, the function $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+t+1\}$ given by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.
- (5)
- If $n=q(m+1)+(t+i)$, where $1\le i\le t-2$, define ${R}_{1},{R}_{2},\dots ,{R}_{m+t+1}$ similar to the case $n=q(m+1)+t$ except ${R}_{t+j}$ and ${R}_{m+3+j},$ where $1\le j\le i$. Set ${R}_{t+j}=\{{x}_{(t+j)+(m+1)},{x}_{(t+j)+2(m+1)},\dots ,{x}_{q(m+1)+(t+j)}\}$ and ${R}_{m+j+3}=\{{x}_{t+j},{x}_{(t+j)+q(m+1)}\}$. Then, it is easy to show that $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+t+1\}$ given by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.
- (6)
- If $n=q(m+1)+m$, take ${R}_{1},{R}_{2},\dots ,{R}_{m+t+1}$ similar to the case $n=q(m+1)+(m-1)$. Then, $\pi =\{{R}_{i}:i=1,2,\dots ,m+t+1\}\cup {R}_{m+t+2}$, where ${R}_{m+t+2}=\{{x}_{q(m+1)+m}\}$ is a partition of $V({C}_{n}^{m+1})$ and the function $c:V({C}_{n}^{m})\u27f6\{1,2,\dots ,m+t+2\}$ given by $c({x}_{i})=j$ for any ${x}_{i}\in {R}_{j}$ is a locating coloring of ${C}_{n}^{m}$.

**Theorem**

**4.**

**Theorem**

**5.**

**Theorem**

**6.**

**Proof.**

**Theorem**

**7.**

**Proof.**

- (1)
- If ${x}_{n}\in {R}_{m+1}$, then ${c}_{\pi}({x}_{n})={c}_{\pi}({x}_{m+1})$.
- (2)
- If ${x}_{n}\in {R}_{l},l\ge m+2$ and ${x}_{n-1}\in {R}_{k}$, where $k=m$ or $m+1$. Then, $(T\setminus \{{x}_{2m+1}\})\cap {R}_{i}\ne \varphi $ for all $i\ge m+3$ and $i\ne l$. However, $k=m$, which gives $N({x}_{k})=(\{{x}_{n},{x}_{1},{x}_{2},\dots ,{x}_{m+2}\}\setminus \{{x}_{m}\})\cup (T\setminus \{{x}_{2m+1}\})$ and $k=m+1$, which gives $N({x}_{k})=(\{{x}_{1},{x}_{2},\dots ,{x}_{m+2}\}\setminus \{{x}_{m+1}\})\cup T$, while $N({x}_{n-1})=\{{x}_{1},{x}_{2},\dots ,{x}_{m-1}\}\cup ({S}_{2}\setminus \{{x}_{n-1}\})$. Thus, $N({x}_{j})\cap {R}_{i}\ne \varphi $ whenever $j=k,$ or $n-1$ and $i\ne k$. Thus, ${c}_{\pi}({x}_{n-1})={c}_{\pi}({x}_{k})$.
- (3)
- If ${x}_{n},{x}_{n-1},\dots ,{x}_{n-r+1}\in {\cup}_{j=1}^{r}{R}_{{l}_{j}}$, where $2\le r\le t-1$, ${l}_{j}\ge m+2$ and ${x}_{n-r}\in {R}_{k},$ where $m-r+1\le k\le m+1$. Then, $(T\setminus \{{x}_{2m+1},{x}_{2m},\dots ,{x}_{2m-r+2}\})\cap {R}_{i}\ne \varphi $ for all $i\ge m+3$ and $i\notin {\cup}_{j=1}^{r}{l}_{j}$. Thus, $N({x}_{j})\cap {R}_{i}\ne \varphi $ whenever $j=k,$ or $n-r$ and $i\ne k$ Then, ${c}_{\pi}({x}_{n-r})={c}_{\pi}({x}_{k})$.

## Author Contributions

## Funding

## Acknowledgments

## Conflicts of Interest

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**MDPI and ACS Style**

Ghanem, M.; Al-Ezeh, H.; Dabbour, A.
Locating Chromatic Number of Powers of Paths and Cycles. *Symmetry* **2019**, *11*, 389.
https://doi.org/10.3390/sym11030389

**AMA Style**

Ghanem M, Al-Ezeh H, Dabbour A.
Locating Chromatic Number of Powers of Paths and Cycles. *Symmetry*. 2019; 11(3):389.
https://doi.org/10.3390/sym11030389

**Chicago/Turabian Style**

Ghanem, Manal, Hasan Al-Ezeh, and Ala’a Dabbour.
2019. "Locating Chromatic Number of Powers of Paths and Cycles" *Symmetry* 11, no. 3: 389.
https://doi.org/10.3390/sym11030389