5.1. Case and
Theorem 1, the construction given in [5
] and the general lower bound (2
) imply the next proposition.
For , and , we have .
Follows directly from definitions and Theorems 1 and 2. As , we can write . Recall that , , and . □
In the sequel, we will define several S3RDFs based on the S3RDF for
. Therefore, we now give explicit definition of a 3RDF of weight
. It is obtained by lifting the S3RDF for
]. Let us define the generic function
, defined on integers as follows.
The values of
on the inner cycles are determined by the rule that
must be a 3RDF. It is easy to check that we must have
We know [5
restricted to indices
gives a 3RDF for
which, recalling the general lower bound, implies
(mod6). See Table 1
, Table 2
and Table 3
, recalled from [5
]. The assignment is extended by lifting, using Theorem 1, to
5.2. Lower Bounds for
First, we prove a property of any 3RD function f of with . Recall that, according to Lemma 1, implies that f must be a singleton 3RDF.
Assume and let f be a 3RD function of minimal weight . Then, exactly one-half of the vertices on the outer cycle are colored. WLOG, assume that these are vertices with even indices. Then, the following holds: (1) , for all odd i, (2) , , and (3) , , are pairwise different. Consequently, .
In other words, the lemma says that the vertices on the outer cycle are colored following the pattern , where are the three colors.
Recall that according to Lemma 1, any minimal 3RDF must be a singleton 3RDF, and hence that . Furthermore, since its weight is , exactly half of the vertices are unweighted. It is not possible to have two adjacent unweighted vertices, because the graph is three-regular, and so there would be at least one color missing in the neighborhood of some unweighted vertex. Hence, we may assume that on the outer cycle, exactly one-half of the vertices—WLOG, those with odd indices—are unweighted. Another simple, but useful observation is that vertices with two consecutive even indices must be colored differently, because if , would not have all three colors in the neighborhood.
We wish to prove (1), (2) and (3).
Assume that f is a singleton 3RDF with and that (3) does not hold, for example, that , , . Then, it follows that for the third neighbor of , we have and, similarly, . We also know that , since already has one neighbor, , colored by R. Similarly, , and . Thus, we know that , or . As and , we know that , or . Hence, . Consequently, for the third neighbor of , we have .
The same reasoning leads us to the conclusion that . Then, however, vertex has two neighbors colored by G, and f is not a singleton 3RDF.
In the argument above, we started with local pattern . The case , i.e., when two consecutive colors on the outer cycle are identical, clearly does not extend to a S3RDF assignment. This proves statement (3), that , , are pairwise different. As, by the same argument, , , and are pairwise different, we conclude that . Similarly, and . By induction, using, obviously, for odd i. Hence (1) and (2) also hold and the proof is complete. □
Assume . Let f be a 3RD function of minimal weight, . Then, exactly one half of the vertices on any inner cycle are colored and the coloring follows the pattern . Consequently, , and k must be odd.
Recall that the pattern on the outer cycle is given by Lemma 3. If k is even, there are inner cycles that have no colored neighbors, and this implies that any S3RDF must assign colors to all vertices of the inner cycle. Hence, k must be odd. To complete the proof, just observe that the coloring of the outer cycle exactly determines the colors of inner cycles. More precisely, if , then its neigbors are colored by different colors, and the color of is defined uniquely because . □
For later reference, we explicitly write the following proposition.
If then .
Follows directly from Lemma 4. □
Another case in which the general lower bound cannot be attained is given by the next Lemma.
Let , . If , then .
Let , , and assume that . As , we know that . Consider the outer cycle and an inner cycle, say . According to Lemma 3, exactly one half of the vertices on the outer cycle are colored, and it follows the pattern given above. Analogous reasoning as in the proof of Lemma 3 implies that one-half of the vertices on are colored and again, the coloring follows the same pattern, maybe in the other direction. WLOG, we can assume that . Now, distinguish two cases.
First, assume i is odd. As the neighbors of on the inner cycle are colored by B and G, and since the length of cycle is c, for the opposite vertex of , we have because . Furthermore, the pattern on the outer cycle gives . Hence, vertex has two neighbors colored by R, and so f is not a 3RDF. Contradiction.
Second, assume i is even. In this case, . Then, the two neighbors of , vertices and are not colored by R. Now, consider . The pattern on the outer cycle implies that its two neighbors, and , are colored by B and G. Recall that the third neighbor, , is also not colored by R. Hence, there is no neighbor of with color R, and therefore f is not a 3RDF. Again, this is a contradiction.
In both cases, the reasoning leads to contradiction, and we conclude that there is no S3RDF of weight . □
5.3. Case , General c
Now, let us assume that .
Let and c be odd. Then
Observe that the Petersen graph
is obtained by one, three or five applications of Construction 1 starting from
. The 3RDF of
is given by
. In other words, we only alter the function
on the last row. More precisely, exactly the vertices in
are given the color that is provided by its neighbor in
in 3RDF of
(see Figure 2
). We also observe that, by definition, f
is a singleton 3RDF. There are
such vertices. Note that all other vertices are already dominated by
We continue with the case c even (and (model 6)). Now , restricted to does not properly dominate vertices for even j in the set [0,k − 1] ∪ [c(k − 1),ck − 1] (in the first and in the last row). Furthermore, the vertex uck−1 does not have all three colors in the neighborhood. We know that , , and .
Proposition 7. Let and c even, . Then
Define for , for . Furthermore, for the first and the last row, set for , and for .
Observe that is the only vertex that is left not properly dominated. By coloring with any (!) color, we obtain a singleton 3RDF of weight . □
Analogous reasoning applies to the case , and the results, which are analogues to Propositions 6 and 7 are stated in the next proposition.
Proposition 8. Let , . If c is odd, then
If c is even, , then
(sketch) The proof is analogous to the proofs of Proposition 6 and Proposition 7, using Construction 1. Start with the 3RDF for
, based on Table 2
. Apply Construction 1 and then define the 3RDF of
in a similar way to the proof of Proposition 6 for c
odd and in the proof of Proposition 7 for c
. We omit the details. □
5.5. Case k Even
First, we consider the lower bound for .
Let be an even number, and , . Then, .
An inner cycle together with the neighbors gives rise to subgraphs
, induced on vertices
, for some i
. The subgraph
, induced on
, is on Figure 5
Consider Figure 5
, and observe that there are exactly c
paths on the outer cycle between vertices of
. All these paths have length
. One-half of them, together with other vertices of
disjointed cycles of length
. Denote the union of cycles (as emphasized on Figure 5
(the index zero is chosen because
Clearly, the intersection of subgraphs and is a union of paths. One of them, which we denote as P, is on vertices . The union of and consists of connected components. Consider one of them, for example the component including P. According to Lemma 2, any S3RDF f has weight , and if , at most one of the vertices and is assigned a color. For the connected component K, including P, we have in this case. (This is because we need at least two colors for each of the “handles” of K, i.e., the paths on vertices and .) Otherwise, if both and are assigned a color, then , and . Therefore, for connected components, we need at least colors.
The paths on the outer cycle that do not meet the union of and have vertices each. According to Lemma 2, or colors are needed, depending on whether f assigns colors to the neighboring vertices on the outer cycle or not. It can be shown that if then both paths next to K are assigned at least colors. This implies that a component and one of the paths together are assigned at least colors.
Finally, we also have
inner cycles of length c
and need to color at least half of the vertices on each of them. In total, we need at least
, as claimed. □
Let be an even number, and , . Then, .
Recall the constructions that provide S3RDF of weight when .
First, we give a construction that uses S3RDF for odd k and provides a S3RDF for , of weight . Choose any pair of adjacent columns (two inner cycles and the corresponding vertices on the outer cycle. Repeat these two columns and merge them. The merging is the following operation: any two vertices on the outer cycle are merged into one vertex, which inherits the colors of the original vertices. Observe that exactly one of the original vertices has had a color, so the merged assignment is still a S3RDF. Delete one of the inner cycles. Clearly, the weight of the new assignment is . This proves the statement of lemma for cases . and .
The second construction uses S3RDF for odd k and provides a S3RDF for , of weight . Choose any pair of adjacent columns (two inner cycles and merge them (as above). Using analogous reasoning, as shown above, the weight of the new assignment is . This proves the statement of lemma for cases and . □
The case is considered as a special case.
If , then .
The idea is the following. Color one of the inner cycles (use the pattern 0-R-0-B-0-G). This coloring forces the colors of one-fourth of the vertices on the outer cycle. However, one-fourth of the vertices on the outer cycle already have one colored neighbor. Color the other half of the vertices on the outer cycle so that the second fourth is properly colored. Complete the coloring by assigning colors to one-half of the vertices on the second inner cycle. We omit the details. See example
of Figure 6
Combination of Lemmas 6–8 gives exact values of in some cases.
Let k be an even number, and , . Then, .
Note that if we delete two columns, as in the proof above, we obtain for the case , as already shown by Proposition 9.
Let k be an even number, and c odd. Then .
Sketch. Start with a S3RDF for with and . Delete one, three, or five lines (as in the proof of Proposition 6). Recall the merging operation from the proof of Lemma 10. □
Before stating the next proposition, recall that the case k even and is not of interest, because is not a three regular simple graph.
Let k be an even number, and c even, , . Then, .
The proof is analogous to the proof of Proposition 12. However, we have to delete an even number of rows (two or four) which adds k (instead of ) to the weight of the final S3RDF. The details are left to the reader. □