Mixed graph edge coloring
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June 2009, Volume309(Issue12) Page 4027To 4036
Consider the relevant (2, 3)regular bipartite graph ${G}_{M}$ constructed in the proof of Theorem 4.1 . Observe that all its degree 2 vertices are in the same partition, and the number of these vertices is divisible by three. Now, let ${G}_{M}^{\ast}=({V}^{\ast},{U}^{\ast},{E}^{\ast})$ be the graph resulting from connecting all the triples of degree 2 vertices in ${G}_{M}$ with the gadget depicted in Fig. 6 (triples are chosen in an arbitrary manner). By the construction, ${G}_{M}^{\ast}$ is cubic and bipartite, with $l\left({G}_{M}^{\ast}\right)=3$ and ${E}^{\ast}=0\u0338$ . And, taking into account the proof of Theorem 4.1 and the fact that ${G}_{M}$ is a subgraph of ${G}_{M}^{\ast}$ , all we need is to prove that 5colorability of ${G}_{M}$ implies 5colorability of ${G}_{M}^{\ast}$ .
Consider a vertex $v$ of degree 2 in ${G}_{M}$ . Observe that in any 5coloring of ${G}_{M}$ , none of the arcs incident to $v$ is assigned color 1. This follows from the fact that $in\left(v\right)=2$ , and that the outer degree of $v$ is 1 in ${G}_{M}$ . Next, consider again the gadget in Fig. 6 . It is 4colorable in a manner that the arcs incident to vertices $x,y$ , and $z$ are assigned color 1 (one of such colorings is depicted in the figure). Consequently, bearing in mind the aforementioned observation for any 5coloring of edges incident to a degree 2 vertex in ${G}_{M}$ , one can easily extend edge 5coloring of ${G}_{M}$ onto arcs of ${G}_{M}^{\ast}$ : all arcs of ${G}_{M}^{\ast}$ that are present in ${G}_{M}$ are just colored as in ${G}_{M}$ , and all the other arcs are colored in the manner depicted in Fig. 6 . And thus, edge 5colorability of ${G}_{M}$ implies edge 5colorability of ${G}_{M}^{\ast}$ .□
We use a reduction from the precoloring extension problem on edges ( $PrExtEd$ ), which has been shown to be $\mathcal{N}\mathcal{P}$ complete in cubic planar bipartite graphs with 3 colors (see [12] ). Consider an undirected cubic planar bipartite graph ${G}^{\prime}=({V}^{\prime},{E}^{\prime})$ and suppose that some of its edges are precolored with colors 1, 2 and 3. We will transform ${G}^{\prime}$ into ${G}_{M}$ by making the following replacements:

(i)
if $[{v}_{i},{v}_{j}]\in {E}^{\prime}$ is precolored with color 2, we replace it by the gadget shown in Fig. 7 ; this gadget is called the 2gadget;

(ii)
if $[{v}_{p},{v}_{q}]\in {E}^{\prime}$ is precolored with color 1, we replace it by the gadget shown in Fig. 8 , where we attach to the vertices $\{b,h\}$ , $\{d,e\}$ and $\{g,f\}$ a 2gadget (i.e., these pairs of vertices correspond to the vertices $\{{v}_{i},{v}_{j}\}$ in the 2gadget); this gadget is called a 1gadget;

(iii)
if $[{v}_{r},{v}_{s}]\in {E}^{\prime}$ is precolored with color 3, we replace it by the gadget shown in Fig. 9 , where we attach, as before, a 2gadget to the vertices $\{b,h\}$ , $\{d,e\}$ and $\{g,f\}$ ; this gadget is called a 3gadget.
The resulting mixed graph ${G}_{M}=(V,U,E)$ is clearly cubic planar bipartite and $l\left({G}_{M}\right)=2$ . Furthermore, all paths are vertex disjoint.
In any mixed edge 3 coloring of ${G}_{M}$ , the edges $[{v}_{i},a]$ and $[{v}_{j},u]$ in a 2 gadget must be assigned color 2 .
First notice that in any mixed edge 3coloring of ${G}_{M}$ , arcs $(b,d)$ and $(k,n)$ can only get colors 2 or 3, and arcs $(e,f)$ and $(o,p)$ can only get colors 1 or 2. This implies that the cycle $([d,e],[e,n],[n,o],[o,d])$ cannot use color 2, and then the cycle must be colored using twice color 1 and twice color 3. Thus arcs $(b,d),(k,n),(e,f)$ and $(o,p)$ must all get color 2. Consequently, $(a,b)$ will be assigned color 1, and arc $(p,u)$ will be assigned color 3. So edges $[p,q],[u,t]$ and $[s,r]$ get color 1, edges $[s,t]$ and $[r,q]$ color 2, and edges $[a,s]$ , $[q,t]$ and $[b,r]$ color 3. This means that both edges $[{v}_{i},a],[{v}_{j},u]$ are forced to be assigned color 2. Similarly, $(f,g),[h,m],[l,k]$ and $[i,j]$ are forced to be assigned color 3, $(j,k),[l,i],[f,m]$ and $[g,h]$ are forced to be assigned color 1, and finally, edges $[l,m],[h,i]$ and $[j,g]$ are forced to be assigned color 2.
In any mixed edge 3 coloring of ${G}_{M}$ , the edges $[{v}_{p},b],[{v}_{q},h]$ (resp. $[{v}_{s},b],[{v}_{r},h]$ ) in a 1gadget (resp. in a 3 gadget) must necessarily get color 1 (resp. color 3 ).
It follows from the fact that all the edges linking the vertices $b$ and $h$ , $d$ and $e$ , as well as $g$ and $f$ to the 2gadgets have color 2. Notice that once these colors are fixed, the remaining yet uncolored edges in a 1gadget or a 3gadget are 3colorable in the unique way.
So suppose now that $MGEC({G}_{M},3)$ has a positive answer. Then, as we have just explained above, edges $[{v}_{p},b],[{v}_{q},h]$ have color 1, edges $[{v}_{i},a],[{v}_{j},u]$ have color 2, and edges $[{v}_{s},b],[{v}_{r},h]$ have color 3. Thus by replacing each 1gadget by the original edge $[{v}_{p},{v}_{q}]$ and coloring it with color 1, each 2gadget by the original edge $[{v}_{i},{v}_{j}]$ and coloring it with color 2, and each 3gadget by the original edge $[{v}_{r},{v}_{s}]$ and coloring it with color 3, we get a positive answer for $PrExtEd\left({G}^{\prime}\right)$ .
Conversely, if $PrExtEd\left({G}^{\prime}\right)$ has a positive answer, then by replacing the precolored edges by the relevant gadgets and by coloring edges $[{v}_{p},b],[{v}_{q},h]$ with color 1, edges $[{v}_{i},a],[{v}_{j},u]$ with color 2 and edges $[{v}_{s},b],[{v}_{r},h]$ with color 3 will give us a positive answer for $MGEC({G}_{M},3)$ , since the remaining uncolored edges can be colored in the unique way as explained above. □