In this article, a new algorithm, RF coloring algorithm, will be designed to evaluate chromatic index of loopness graph. RF coloring algorithm is introduced as follows:
Consider a graph G of order n and size m. List its vertices as v1, v2, v3, … , vn and its edges as e1, e2, e3, … , em.
$$ I(G)=\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& \dots \kern0.5em {a}_{1m}\\ {}{a}_{21}& {a}_{22}& \dots \kern0.5em {a}_{2m}\\ {}\begin{array}{c}\vdots \\ {}{a}_{n1}\end{array}& \begin{array}{c}\vdots \\ {}{a}_{n2}\end{array}& \begin{array}{cc}\begin{array}{c}\ddots \\ {}\dots \end{array}& \begin{array}{c}\vdots \\ {}{a}_{nm}\end{array}\end{array}\end{array}\right]. $$
$$ {a}_{11}^{\ast }=\kern0.75em \left\{\begin{array}{c}0\kern0.5em \mathrm{if}\kern0.75em {a}_{11}=0\\ {}\ \\ {}1\ \mathrm{if}\kern0.75em {a}_{11}=1\end{array}\right.. $$
After that put
$$ {a}_{12}^{\ast }=\left\{\begin{array}{ccc}0& \mathrm{if}& {a}_{12}=0\ \\ {}1& \mathrm{if}& {a}_{12}=1,{a}_{11}=0\ \\ {}2& \mathrm{if}& {a}_{11}={a}_{12}=1\end{array}\right.. $$
In the same way, for entry \( {a}_{1j}^{\ast } \), where 1 < j ≤ m, put
$$ {a}_{1j}^{\ast }=\kern0.75em \left\{\begin{array}{c}\kern4.25em 0\kern10.50em \mathrm{if}\kern0.75em {a}_{1j}=0\\ {}\ \\ {}\max \left\{{a}_{11}^{\ast },{a}_{12}^{\ast },\dots, {a}_{1\left(j-1\right)}^{\ast}\right\}+1\kern1em \mathrm{if}\kern0.75em {a}_{1j}=1\end{array}\ \right.. $$
-
(b)
Any column k has the entry \( {a}_{1k}^{\ast }=h \) put
$$ {a}_{ik}^{\ast }=\kern0.75em \left\{\begin{array}{c}h\kern1em \mathrm{if}\kern0.75em {a}_{ik}=1\\ {}\ \\ {}0\kern1em \mathrm{if}\kern0.5em {a}_{ik}=0\end{array}\ \right.. $$
where 1 < i ≤ n and 1 < k ≤ m.
-
(c)
Now start from the second row put
$$ \left\{\begin{array}{ccc}{a}_{2l}^{\ast }=0& \mathrm{if}& {a}_{2l}=0\\ {}\min \left(\left\{{a}_{1j}^{\ast }:{a}_{1j}^{\ast}\ne 0,\right\}\backslash \left\{{a}_{2j}^{\ast },{a}_{sj}^{\ast }:{a}_{2j}^{\ast}\ne 0,{a}_{sj}^{\ast}\ne 0\right\}\right)& \mathrm{if}& {a}_{2l}=1,{a}_{Sl}=1,\left\{{a}_{1j}^{\ast }:{a}_{1j}^{\ast}\ne 0\right\}\backslash \left\{{a}_{2j}^{\ast },{a}_{Sj}^{\ast }:{a}_{2j}^{\ast}\ne 0\ne {a}_{Sj}^{\ast}\right\}\ne \AE \\ {}\max \left(\left\{{a}_{1j}^{\ast },{a}_{2j}^{\ast },{a}_{sj}^{\ast }:{a}_{1j}^{\ast}\ne 0,{a}_{2j}^{\ast}\ne 0\ne {a}_{sj}^{\ast },\right\}\right)+1& \mathrm{if}& {a}_{2l}=1,{a}_{Sl}=1,\left\{{a}_{1j}^{\ast }:{a}_{1j}^{\ast}\ne 0\right\}\backslash \left\{{a}_{2j}^{\ast },{a}_{Sj}^{\ast }:{a}_{2j}^{\ast}\ne 0\ne {a}_{Sj}^{\ast}\right\}=\AE \end{array}\right. $$
where 1 ≤ l ≤ m, 1 ≤ j ≤ m and 1 ≤ s ≤ n.
-
(d)
Any column l has the entry \( {a}_{2l}^{\ast }=f \) put
$$ {a}_{il}^{\ast }=\kern0.75em \left\{\begin{array}{c}f\kern1em \mathrm{if}\kern0.75em {a}_{il}=1\\ {}\ \\ {}0\kern1em \mathrm{if}\kern0.5em {a}_{il}=0\end{array}\right. $$
-
(e)
Again repeat steps (c) and (d) to complete the RF coloring matrix
$$ \mathrm{RCI}(G)=\left[\begin{array}{ccc}{a}_{11}^{\ast }& {a}_{12}^{\ast }& \dots \kern0.5em {a}_{1m}^{\ast}\\ {}{a}_{21}^{\ast }& {a}_{22}^{\ast }& \begin{array}{cc}\dots & {a}_{2m}^{\ast}\end{array}\\ {}\begin{array}{c}\vdots \\ {}{a}_{n1}^{\ast}\end{array}& \begin{array}{c}\vdots \\ {}{a}_{n2}^{\ast}\end{array}& \begin{array}{cc}\begin{array}{c}\ddots \\ {}\dots \end{array}& \begin{array}{c}\vdots \\ {}{a}_{nm}^{\ast}\end{array}\end{array}\end{array}\right] $$
In the following, we give examples solved by the new algorithm:
Example1. Let G be a graph shown below (Fig. 1).
The chromatic index of the graph G will be calculated by using the RF coloring algorithm as follows:
$$ I(G)=\left[\begin{array}{ccc}1& 1& 1\kern0.5em 0\\ {}1& 1& \begin{array}{cc}0& 1\end{array}\\ {}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}& \begin{array}{cc}\begin{array}{c}1\\ {}0\end{array}& \begin{array}{c}0\\ {}1\end{array}\end{array}\end{array}\right]\kern1.25em \overset{\begin{array}{c}\mathrm{Since}\ {a}_{11}=1\\ {}\mathrm{Put}\ {a}_{11}^{\ast }=1\kern0.5em \end{array}}{\to}\left[\begin{array}{ccc}1& & \begin{array}{cc}& \end{array}\\ {}& & \begin{array}{cc}& \end{array}\\ {}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}& \begin{array}{cc}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}\end{array}\end{array}\right]\kern1.5em \overset{\begin{array}{c}\mathrm{Since}\kern0.5em {a}_{11}=1\\ {}{a}_{12}=1\\ {}\mathrm{Put}\ {a}_{12}^{\ast }=2\end{array}}{\to}\kern1.5em \left[\begin{array}{ccc}1& 2& \begin{array}{cc}& \end{array}\\ {}& & \begin{array}{cc}& \end{array}\\ {}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}& \begin{array}{cc}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}\end{array}\end{array}\right] $$
$$ \overset{\begin{array}{c}\mathrm{Since}\ {a}_{11}={a}_{12}={a}_{13}=1\\ {}\mathrm{Put}\ {a}_{13}^{\ast }=3\end{array}}{\to}\left[\begin{array}{ccc}1& 2& 3\kern0.5em \\ {}& & \begin{array}{cc}& \end{array}\\ {}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}& \begin{array}{cc}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}\end{array}\end{array}\right]\overset{\begin{array}{c}\mathrm{Since}\ {a}_{14}=0\\ {}\mathrm{Put}\ {a}_{14}^{\ast }=0\end{array}}{\to}\left[\begin{array}{ccc}1& 2& \begin{array}{cc}3& 0\end{array}\\ {}& & \begin{array}{cc}& \end{array}\\ {}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}& \begin{array}{cc}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}\end{array}\end{array}\right]\overset{\begin{array}{c}\mathrm{Column}\ 1\\ {}\mathrm{Put}\ {a}_{21}^{\ast }=1,\\ {}\ {a}_{31}^{\ast }={a}_{41}^{\ast }=0\kern0.5em \end{array}}{\to}\left[\begin{array}{ccc}1& 2& \begin{array}{cc}3& 0\end{array}\\ {}1& & \begin{array}{cc}& \end{array}\\ {}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}\\ {}\end{array}& \begin{array}{cc}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}\end{array}\end{array}\right] $$
$$ \overset{\begin{array}{c}\mathrm{Column}\ 2\\ {}\mathrm{Put}\ {a}_{22}^{\ast }=2,\\ {}\ {a}_{32}^{\ast }={a}_{42}^{\ast }=0\kern0.5em \end{array}}{\to}\left[\begin{array}{ccc}1& 2& 3\kern0.5em 0\\ {}1& 2& \begin{array}{cc}& \end{array}\\ {}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}& \begin{array}{cc}\begin{array}{c}\\ {}\end{array}& \begin{array}{c}\\ {}\end{array}\end{array}\end{array}\right]\kern0.5em \overset{\begin{array}{c}\mathrm{Column}3\\ {}\mathrm{Put}\ {a}_{33}^{\ast }=3,\\ {}\ {a}_{23}^{\ast }={a}_{43}^{\ast }=0\kern0.5em \end{array}}{\to}\kern0.5em \left[\begin{array}{ccc}1& 2& \begin{array}{cc}3& 0\end{array}\\ {}1& 2& \begin{array}{cc}0& \end{array}\\ {}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}& \begin{array}{cc}\begin{array}{c}3\\ {}0\end{array}& \begin{array}{c}\\ {}\end{array}\end{array}\end{array}\right]\overset{\begin{array}{c}{2}^{\mathrm{nd}}\mathrm{row}\\ {}\mathrm{Since}\ {a}_{24}=1\\ {}\mathrm{then}\ {a}_{24}^{\ast }=\max \left\{1,2,3\right\}\backslash \left\{1,2\right\}=3\end{array}}{\to }\ \left[\begin{array}{ccc}1& 2& \begin{array}{cc}3& 0\end{array}\\ {}1& 2& \begin{array}{cc}0& 3\end{array}\\ {}\begin{array}{c}0\\ {}0\end{array}& \begin{array}{c}0\\ {}0\end{array}& \begin{array}{cc}\begin{array}{c}3\\ {}0\end{array}& \begin{array}{c}\\ {}\end{array}\end{array}\end{array}\right] $$
$$ \kern0.5em \overset{\begin{array}{c}\mathrm{Column}\ 4\\ {}\mathrm{Put}\ {a}_{44}^{\ast }=3,\\ {}\ {a}_{43}^{\ast }=0\kern0.5em \end{array}}{\to}\kern0.5em \left[\begin{array}{ccc}1& 2& 3\kern0.5em 0\\ {}1& 2& \begin{array}{cc}0& 3\end{array}\\ {}\begin{array}{c}0\\ {}1\end{array}& \begin{array}{c}0\\ {}2\end{array}& \begin{array}{cc}\begin{array}{c}3\\ {}0\end{array}& \begin{array}{c}0\\ {}3\end{array}\end{array}\end{array}\right] $$
Then, the RF coloring matrix is given by
$$ \mathrm{RCI}(G)=\left[\begin{array}{ccc}1& 2& 3\kern0.5em 0\\ {}1& 2& \begin{array}{cc}0& 3\end{array}\\ {}\begin{array}{c}0\\ {}1\end{array}& \begin{array}{c}0\\ {}2\end{array}& \begin{array}{cc}\begin{array}{c}3\\ {}0\end{array}& \begin{array}{c}0\\ {}3\end{array}\end{array}\end{array}\right]. $$
From the above matrix, we find the chromatic index χ ΄ (G) of the graph G is equal to 3.
Example2. Given a graph G, shown in Fig. 2.
Applying the RF coloring algorithm step by step to evaluate the colors of edges and chromatic index as shown in the sequence of matrices below:
Hence, the RF coloring matrix is given by
$$ \mathrm{RCI}(G)=\left[\begin{array}{c}1\\ {}1\\ {}\begin{array}{c}0\\ {}0\\ {}\begin{array}{c}0\\ {}0\end{array}\end{array}\end{array}\kern0.5em \begin{array}{c}0\\ {}3\\ {}\begin{array}{c}3\\ {}0\\ {}\begin{array}{c}0\\ {}0\end{array}\end{array}\end{array}\kern0.5em \begin{array}{ccc}\begin{array}{c}2\\ {}0\\ {}\begin{array}{c}2\\ {}0\\ {}\begin{array}{c}0\\ {}0\end{array}\end{array}\end{array}& \begin{array}{c}0\\ {}2\\ {}\begin{array}{c}0\\ {}2\\ {}\begin{array}{c}0\\ {}0\end{array}\end{array}\end{array}& \begin{array}{ccc}\begin{array}{c}3\\ {}0\\ {}\begin{array}{c}0\\ {}3\\ {}\begin{array}{c}0\\ {}0\end{array}\end{array}\end{array}& \begin{array}{c}4\\ {}0\\ {}\begin{array}{c}0\\ {}0\\ {}\begin{array}{c}0\\ {}4\end{array}\end{array}\end{array}& \begin{array}{ccc}\begin{array}{c}0\\ {}0\\ {}\begin{array}{c}1\\ {}0\\ {}\begin{array}{c}0\\ {}1\end{array}\end{array}\end{array}& \begin{array}{c}0\\ {}0\\ {}\begin{array}{c}5\\ {}0\\ {}\begin{array}{c}5\\ {}0\end{array}\end{array}\end{array}& \begin{array}{ccc}\begin{array}{c}0\\ {}4\\ {}\begin{array}{c}0\\ {}0\\ {}\begin{array}{c}4\\ {}0\end{array}\end{array}\end{array}& \begin{array}{c}0\\ {}0\\ {}\begin{array}{c}0\\ {}1\\ {}\begin{array}{c}1\\ {}0\end{array}\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}0\\ {}0\\ {}\begin{array}{c}0\\ {}5\\ {}\begin{array}{c}0\\ {}5\end{array}\end{array}\end{array}& \begin{array}{c}0\\ {}0\\ {}\begin{array}{c}0\\ {}0\\ {}\begin{array}{c}2\\ {}2\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\right]. $$
It is clear from the RF coloring matrix that the chromatic index of the graph G is 5, i.e., χ ΄ (G) = 5 and the color of e1, e7, e10 is 1, the color of e3, e4, e12 is 2, the color of e2, e5 is 3, the color of e6, e9 is 4, and the color of e8, e11 is 5.
In the following section, we reprove some theorems by using RF coloring algorithm
Theorem 1 Let Sn be a star graph of order n. Then, χ ΄ (Sn) = n − 1, where χ ΄ (Sn) is the chromatic index of Sn.
Proof Let Sn be a star graph of order n as shown in Fig. 3.
By applying the RF coloring algorithm, we found the RF coloring matrix is given by
$$ \mathrm{RCI}\left({S}_n\right)=\left[\begin{array}{ccc}1& 2\kern0.5em 3& \kern0.5em \begin{array}{cc}\dots & n-1\end{array}\\ {}\begin{array}{c}1\\ {}0\end{array}& \begin{array}{c}\begin{array}{cc}0& 0\end{array}\\ {}\begin{array}{cc}2& 0\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}\dots \kern1em \\ {}\dots \kern1.25em \end{array}& \begin{array}{c}0\\ {}0\end{array}\end{array}\\ {}\begin{array}{c}\vdots \\ {}0\end{array}& \begin{array}{c}\begin{array}{cc}\vdots & \vdots \end{array}\\ {}\begin{array}{cc}0& 0\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}\kern0.5em \ddots \\ {}\dots \end{array}& \begin{array}{c}\kern0.75em \vdots \\ {}n-1\end{array}\end{array}\end{array}\right] $$
The greatest number in the RF coloring matrix is n − 1, then the chromatic index of Sn equals n − 1.
Theorem 2 Let Cn be a cycle graph of order n. Then \( \chi \prime \left({C}_n\right)=\left\{\begin{array}{c}2\kern1em if\kern0.75em n\ is\ even\\ {}\ \\ {}3\kern1em if\kern1em n\ is\ odd\end{array}\right. \), where χ ΄ (Cn) is the chromatic index of Cn.
Proof Let Cn be a cycle graph of order n. Applying the RF coloring algorithm, we will stop when the RF coloring matrix will become
$$ \mathrm{RCI}\left({C}_n\right)=\left[\begin{array}{c}\begin{array}{c}\begin{array}{cc}1& 0\end{array}\kern0.5em \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}0& 2\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{cc}\begin{array}{cc}1& 2\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}0& 0\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}0& 2\end{array}& \begin{array}{cc}1& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}0& 0\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}\vdots & \vdots \kern0.5em \end{array}& \begin{array}{cc}\vdots & \begin{array}{cc}\ddots & \begin{array}{cc}\vdots &\ \begin{array}{cc}\vdots & \vdots \end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}0& 0\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}l& \begin{array}{cc}0& 0\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}0& 0\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}l& \begin{array}{cc}m& 0\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{cc}\begin{array}{cc}0& 0\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}m& 2\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\right]. $$
Now we want to evaluate the values of l and m and we have two cases:
Case 1. When n is even then l is the entry in \( {a}_{\left(n-1\right)\left(n-2\right)}^{\ast } \), i.e., l = 2. Since \( {a}_{\left(n-1\right)\left(n-2\right)}^{\ast }=2 \) and \( {a}_{nm}^{\ast }=2 \) then m = 1; hence, the RF coloring matrix is
$$ \mathrm{RCI}\left({C}_n\right)=\left[\begin{array}{c}\begin{array}{c}\begin{array}{cc}1& 0\end{array}\kern0.5em \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}0& 2\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{cc}\begin{array}{cc}1& 2\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}0& 0\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}0& 2\end{array}& \begin{array}{cc}1& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}0& 0\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}\vdots & \vdots \kern0.5em \end{array}& \begin{array}{cc}\vdots & \begin{array}{cc}\ddots & \begin{array}{cc}\vdots &\ \begin{array}{cc}\vdots & \vdots \end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}0& 0\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}2& \begin{array}{cc}0& 0\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}0& 0\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}2& \begin{array}{cc}1& 0\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{cc}\begin{array}{cc}0& 0\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}1& 2\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\right] $$
and the chromatic index of Cn from the RF coloring matrix is 2, i.e., χ ΄ (Cn) = 2.
Case 2. When n is odd then l is the entry in \( {a}_{\left(n-1\right)\left(n-2\right)}^{\ast } \), i.e., l = 1. So, m = 3 because \( {a}_{\left(n-1\right)\left(n-2\right)}^{\ast }=1 \) and \( {a}_{\mathrm{nm}}^{\ast }=2 \); hence, the RF coloring matrix in this case is given by
$$ \mathrm{RCI}\left({C}_n\right)=\left[\begin{array}{c}\begin{array}{c}\begin{array}{cc}1& 0\end{array}\kern0.5em \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}0& 2\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{cc}\begin{array}{cc}1& 2\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}0& 0\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}0& 2\end{array}& \begin{array}{cc}1& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}0& 0\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}\vdots & \vdots \kern0.5em \end{array}& \begin{array}{cc}\vdots & \begin{array}{cc}\ddots & \begin{array}{cc}\vdots &\ \begin{array}{cc}\vdots & \vdots \end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}0& 0\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}1& \begin{array}{cc}0& 0\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{c}\begin{array}{cc}\begin{array}{cc}0& 0\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}1& \begin{array}{cc}3& 0\end{array}\end{array}\end{array}\end{array}\end{array}\\ {}\begin{array}{cc}\begin{array}{cc}0& 0\end{array}& \begin{array}{cc}0& \begin{array}{cc}\dots & \begin{array}{cc}0& \begin{array}{cc}3& 2\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\end{array}\right], $$
and the chromatic index of Cn equals 3, i.e., χ ΄ (Cn) = 3.
One of the most famous applications on edge coloring of a graph is the timetable; let us give an example and solve it by our new algorithm as follows:
In a particular faculty of science, the Mathematics department has five teachers t1, t2, t3, t4, t5. The teaching assignments of the five teachers are given by the array:
|
I Year
|
II Year
|
III Year
|
IV Year
|
|
Y
1
|
Y
2
|
Y
3
|
Y
4
|
t
1
|
1
|
2
|
__
|
__
|
t
2
|
1
|
1
|
1
|
__
|
t
3
|
1
|
__
|
__
|
2
|
t
4
|
__
|
__
|
1
|
__
|
t
5
|
__
|
__
|
1
|
1
|
We want to find the minimum period timetable. To solve this problem, first we draw the graph corresponding to this problem, see Fig. 4
By applying the RF coloring algorithm, we find the RF coloring matrix is given by:
$$ \left[\begin{array}{ccc}\begin{array}{c}\begin{array}{c}1\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}1\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}0\end{array}\end{array}& \begin{array}{c}\begin{array}{c}2\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}2\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}0\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}\begin{array}{c}3\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}3\\ {}0\\ {}0\end{array}\end{array}& \begin{array}{c}\begin{array}{c}0\\ {}2\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}2\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}0\end{array}\end{array}\end{array}\end{array}\kern0.5em \begin{array}{ccc}\begin{array}{c}\begin{array}{c}0\\ {}3\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}3\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}0\end{array}\end{array}& \begin{array}{c}\begin{array}{c}0\\ {}1\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}1\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}0\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}\begin{array}{c}0\\ {}0\\ {}3\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}3\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}0\end{array}\end{array}& \begin{array}{c}\begin{array}{c}0\\ {}0\\ {}1\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}1\\ {}0\end{array}\end{array}\end{array}\end{array}\kern0.5em \begin{array}{ccc}\begin{array}{c}\begin{array}{c}0\\ {}0\\ {}2\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}2\end{array}\end{array}& \begin{array}{c}\begin{array}{c}0\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}1\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}1\\ {}0\\ {}0\end{array}\end{array}& \begin{array}{cc}\begin{array}{c}\begin{array}{c}0\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}2\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}2\\ {}0\\ {}0\end{array}\end{array}& \begin{array}{c}\begin{array}{c}0\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}3\\ {}0\\ {}0\end{array}\\ {}\begin{array}{c}0\\ {}0\\ {}3\end{array}\end{array}\end{array}\end{array}\right] $$
It is clear from the RF coloring matrix above that we have a 3-period timetable given by:
|
Period I
|
Period II
|
Period III
|
t
1
|
Y
1
|
Y
2
|
Y
2
|
t
2
|
Y
2
|
Y
3
|
Y
1
|
t
3
|
Y
4
|
Y
1
|
Y
4
|
t
4
|
__
|
__
|
Y
3
|
t
5
|
Y
3
|
Y
4
|
__
|