Mixed graph edge coloring

Source Information
June 2009, Volume309(Issue12) Page4027To4036
Appendix A proof of Theorem 4.2

Proof

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 = ( V , U , E ) 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 is cubic and bipartite, with l ( G M ) = 3 and E = . And, taking into account the proof of Theorem 4.1 and the fact that G M is a subgraph of G M , all we need is to prove that 5-colorability of G M implies 5-colorability of G M .

Consider a vertex v of degree 2 in G M . Observe that in any 5-coloring of G M , none of the arcs incident to v is assigned color 1. This follows from the fact that i n ( v ) = 2 , and that the outer degree of v is 1 in G M . Next, consider again the gadget in Fig. 6 . It is 4-colorable 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 5-coloring of edges incident to a degree 2 vertex in G M , one can easily extend edge 5-coloring of G M onto arcs of G M : all arcs of G M 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 5-colorability of G M implies edge 5-colorability of G M .□

Appendix B proof of Theorem 5.1

Proof

We use a reduction from the precoloring extension problem on edges ( P r E x t E d ), which has been shown to be N P -complete in cubic planar bipartite graphs with 3 colors (see  [12] ). Consider an undirected cubic planar bipartite graph G = ( V , E ) and suppose that some of its edges are precolored with colors 1, 2 and 3. We will transform G into G M by making the following replacements:

  1. (i)

    if [ v i , v j ] E is precolored with color 2, we replace it by the gadget shown in Fig. 7 ; this gadget is called the 2-gadget;

  2. (ii)

    if [ v p , v q ] E 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 2-gadget (i.e., these pairs of vertices correspond to the vertices { v i , v j } in the 2-gadget); this gadget is called a 1-gadget;

  3. (iii)

    if [ v r , v s ] E is precolored with color 3, we replace it by the gadget shown in Fig. 9 , where we attach, as before, a 2-gadget to the vertices { b , h } , { d , e } and { g , f } ; this gadget is called a 3-gadget.

The resulting mixed graph G M = ( V , U , E ) is clearly cubic planar bipartite and l ( G M ) = 2 . Furthermore, all paths are vertex disjoint.

Claim 2

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 3-coloring 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.

Claim 3

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 1-gadget (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 2-gadgets have color 2. Notice that once these colors are fixed, the remaining yet uncolored edges in a 1-gadget or a 3-gadget are 3-colorable in the unique way.

So suppose now that M G E C ( 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 1-gadget by the original edge [ v p , v q ] and coloring it with color 1, each 2-gadget by the original edge [ v i , v j ] and coloring it with color 2, and each 3-gadget by the original edge [ v r , v s ] and coloring it with color 3, we get a positive answer for P r E x t E d ( G ) .

Conversely, if P r E x t E d ( G ) 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 M G E C ( G M , 3 ) , since the remaining uncolored edges can be colored in the unique way as explained above. □

Report a problem
doi: 10.1016/j.disc.2008.11.033