Published: 21 August 2023

# Erratum. Dominator coloring of total graph of path and cycles

Minal Shukla1
Foram Chandarana2
1, 2Marwadi University, Rajkot, Gujarat, 360003, India
Corresponding Author:
Minal Shukla
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## The description of the correction

Authors have identified errors in the paper originally submitted and finally approved (after the acceptance) by the Authors.

On page 74, case 2 second line and third line highlighted part to be changed.

Case 2: $\mathbit{n}\ge 3$

By Proposition 2.2, $\chi \left[T\left({P}_{n}\right)\right]\ge 3$ as $T\left({P}_{n}\right)$ includes an odd cycle. Assign the proper coloring to the vertices as $f\left({v}_{i}\right)=1,3,2,1,3,2,\dots .,n$, $f\left({u}_{i}\right)=2,1,3,2,1,3,\dots .,n-1$. Thus, a minimum of three colors are required for proper coloring. Therefore $\chi \left[T\left({P}_{n}\right)\right]=3$.

On page 74, an error in the symbol of statement of Theorem 3.3.

Theorem 3.3

4
${\chi }_{d}\left[T\left({P}_{n}\right)\right]=\left\{\begin{array}{ll}\chi \left[T\left({P}_{n}\right)\right]+\gamma \left[T\left({P}_{n}\right)\right]-1,& n=2,3,4,6,\\ \chi \left[T\left({P}_{n}\right)\right]+\gamma \left[T\left({P}_{n}\right)\right],& n\ge 5,n\ne 6.\end{array}\right\$

On page 75, third paragraph second line of case when $n$ = 6, 4 is to be written as 3.

In this case the set $\left\{{v}_{1},{v}_{6},{u}_{4}\right\}$ or $\left\{{v}_{2},{v}_{5},{u}_{3}\right\}$ are only $\gamma$-sets of graph $T\left({P}_{6}\right)$. According to Lemma – 3.1, $\gamma \left[T\left({P}_{6}\right)\right]=3$ and by Lemma – 3.2, $\chi \left[T\left({P}_{6}\right)\right]=3$. Allocating several colors to the vertices of the $\gamma$-set that is equal to $\gamma \left[T\left({P}_{6}\right)\right]$ in order to determine its optimal coloring. Now we use $\chi \left[T\left({P}_{6}\right)\right]-1$ number of colors to color the remaining vertices.

On page 75, Case 1 last line in place of 6, it should be 5.

The coloring pattern can be defined as $f\left({v}_{1}\right)=f\left({v}_{4}\right)=f\left({u}_{2}\right)=f\left({u}_{5}\right)=3,f\left({v}_{3}\right)=1,f\left({v}_{6}\right)=f\left({u}_{1}\right)=f\left({u}_{4}\right)=4,f\left({v}_{2}\right)=1,f\left({u}_{4}\right)=2,f\left({v}_{5}\right)=2,f\left({u}_{3}\right)=5.$ Here every vertex dominates the vertices of at least one color class. As a result, the proper coloring creates a dominator coloring for the relevant graph. Therefore, ${\chi }_{d}\left[T\left({P}_{6}\right)\right]=5=\chi \left[T\left({P}_{6}\right)\right]+\gamma \left[T\left({P}_{6}\right)\right]-1$.

On page 75, Case 2 title is mentioned incorrect.

Case 2: $\mathbit{n}\ge 5$, $\mathbit{n}\ne 6$

On page 75, In the line just above the Fig. 1, in place of six colors there must be five colors.

A dominator coloring of $T\left({P}_{6}\right)$ using five colors is shown in Fig. 1.