3.1. Block Decomposition of and General Lemmas
A leaf block in a graph G is a maximal 2-connected subgraph of G containing exactly one cut vertex of G. Let be a leaf block of . We use to denote the cut vertex of connecting and the remaining part in . We call e an interior edge in if e is not incident with . We call v an interior vertex in if . For any , let denote the tree generated by the vertices and edges deleted from G starting from vertex x so that is obtained.
For , let and and be the order of . For , let denote the edges incident with x in , where . Denote and . Especially, let if either or .
Lemma 1. Let be an edge in . Then at least one of the following conditions does not hold.
- (1)
;
- (2)
for .
Proof. By contradiction, assume that we have one edge satisfying conditions and . Let , where . Thus for any as G is minimized. We choose an arbitrary . If , then we take , and so by , contrary to the assumption of G. Next, we assume and let . Now we have and . Let be a partial strong edge coloring of with colors. Since , any two edges in A satisfying and are colored distinct in coloring . Especially, when , either or . By , we would choose one available color for edge e. Thus, has a partial strong edge coloring with colors which implies that , a contradiction. This proves the lemma. □
Lemma 2. Let x be an interior vertex in . Then is a star graph.
Proof. Arguing by contradiction, assume that is not a star graph. Pick an edge such that is the outermost edge of with . Delete the edge e and let . By , we obtain and , a contradiction to Lemma 1. □
Lemma 3. Let v be an interior vertex in with for . Then .
Proof. By contradiction, assume which implies . By Lemma 2, we have is a star graph and choose . Delete the edge e and let . By , we obtain and , which is a contradiction to Lemma 1. □
Lemma 4. Let v be an interior vertex contained in with , where . Then .
Proof. By contradiction, assume which implies . By Lemma 2, we have as a star graph and pick an edge . Let . By the minimized counterexample property, for any .
Choose an arbitrary edge set . When , choose , and then . Thus, for any , a contradiction. Otherwise , let and we have . Thus, we obtain and by , which is a contradiction to Lemma 1. □
Lemma 5. For an interior edge contained in with , we have , where and .
Proof. By contradiction, assume that there exists an interior edge such that . Let . By the minimized counterexample property, for any set .
Pick an arbitrary edge set . When , take , and then by for any . If , let and we have . Thus we obtain by Lemma 4 and It is a contradiction to by Lemma 1. □
Corollary 1. Every leaf block is not isomorphic to for .
Proof. Suppose and choose an interior edge . Let . We have by the definition of , which is a contradiction to Lemma 5. □
Lemma 6. For any interior vertex x contained in , .
Proof. By contradiction, suppose that there exists an interior vertex x with . By Lemma 3, we have and the vertex x has exactly two neighbours . We characterize the graph obtained from G by deleting x and adding and as follows: adding a 1-vertex and an edge for ; adding an edge if does not exist in .
Consider the core of , represented by . As , by the minimality of counterexample, for arbitrary . To show for arbitrary , let be the set gained from by replacing with , if exists in A for . Let be a desired coloring of with colors.
If
, then it implies
. If
, let
and then
If either
or
, then
and
Let
be a new coloring of
A defined as follows.
Thus
is a desired coloring of
with
for any
, a contradiction to the assumption of
G.
If
then
by definition. Note that
and
. If
, then
and
If either
or
, then
and
. In each case,
is obtained. Let
be a coloring of
defined as below.
Thus is a desired coloring of and for any , which is a contradiction. □
Note that all lemmas above in
Section 3.1 are valid for the minimal counterexample
G whenever
. Next we apply these lemmas and characterize graph structures to prove Theorem 7 for each
below in
Section 3.2,
Section 3.3 and
Section 3.4, respectively.
3.3. Proof of Strong Edge Coloring of -Minor Free Graphs
For a -minor free graph G, we need to prove for any edge set . First, we give the following useful lemma for a -minor free graph
Lemma 8. Let be an interior edge contained in with . Let . For any as a neighbour of where , if either or holds, then
Proof. Without loss of generality, assume that Since and , we have either or and . By the arbitrariness of , . □
By Proposition 3, we get or is a -minor free graph or . We analyze the leaf block case by case.
Case 1. .
The class of graphs
can be characterized in
(see
Figure 3). For
, we obtain
for
by Lemma 6. Choose an interior edge
and let
By simple computation, we have
, which is a contradiction to Lemma 5.
By Corollary 1, we have . Take an interior edge and let . As for any , is obtained by Lemma 8. By Lemma 5, we have for any
Case 2. .
If , then we gain , and by Lemma 6. If , then we obtain , and by Lemma 6. In these two conditions, by Case 1, the desired result is achieved.
Case 3. is a 2-connected -minor free graph.
By Proposition 1 and Corollary 1, is a -minor free graph. By Lemma 7, contains at least one 2-vertex as interior vertex, contradicting Lemma 6.
3.4. Proof of Strong Edge Coloring of -Minor Free Graphs
For a -minor free graph G, we need to prove for any . By Proposition 3, we obtain or () or is a -minor free graph. We prove the statement based on the following three cases.
Case 1. .
For any , we have by Lemma 6. Let be an arbitrary interior edge contained in leaf block and let . By Lemma 5, does not hold. Thus there exist and satisfying By simple computation, we enumerate the following results.
- (1)
, when .
- (2)
and .
- (3)
for any
By symmetry and Lemma 6, we have either
or
for Fact A(2). If
and
, then we characterize all possible graphs
for
in
Figure 4. If
, then
and take
. Let
and we have
, which is a contraction to Lemma 5. If
, then choose a thick edge
marked in
Figure 4 such that
e is an interior edge contained in
. Let
and we have
by simple computation for some
, which is a contraction to Lemma 5.
Let
for
. If
and
, then
for some
. If either
where
or
(see
Figure 5) and
, then we can do the same analysis as
where
. If
and
, then we can first pick a cycle
. Then delete
C and let
, by the minimized counterexample property,
for arbitrary
. Choose an arbitrary
, then we consider the following two subcases.
Subcase 1.1 for .
Let . Then . We can get for any , a contradiction.
Subcase 1.2 and .
Let . Then . Take as a partial strong edge coloring of graph with colors. Choose different colors from colors to assign distinct edges in so that those colors are different from any , where . Therefore, we gain an desired coloring of with for any , a contradiction.
If
and
, then we have
for some
by Lemma 6. We characterize all possible graphs
for
in
Figure 6 corresponding to
by Fact A. If
and
, then we have all possible graphs
for
in the
Figure 7 when
.
For each of the graphs in
Figure 6 and
Figure 7, we know that
in each graph by Lemma 6. We would gain a new graph
by deleting the thick edge
. We have
by simple computation, which is a contradiction to Lemma 5. When
,
and
, we get the desired result as
since
contains
as a subgraph for some
.
Case 2. .
Recall that and . By construction of and , there is at least one 2-vertex as interior vertex in , which is a contradiction to Lemma 6.
Case 3. .
Recall with . For an arbitrary , there is at most one 2-vertex in by Lemma 6. We discuss it in the following four subcases.
Subcase 3.1 and .
The possible graphs are either
or
shown in the
Figure 8. By Lemma 6,
is the 2-vertex. We choose an edge
.
Subcase 3.2 and .
The possible graphs are either
or
shown in the
Figure 8. By Lemma 6,
is the 2-vertex. We choose an edge
.
Subcase 3.3 and .
The possible graphs are either
or
shown in the
Figure 8. We choose an edge either
or
which is not associated with
.
Subcase 3.4 and .
The possible graphs are either
or
shown in the
Figure 8. We choose
if it is an interior edge; otherwise, we choose
.
For each subcase, we pick the appropriate edge as discussed above. Let . By the minimized counterexample property, for any . By Lemma 3 and simple computation, we know and . Thus, by Lemma 1, for any , a contradiction.
Case 4. .
Recall that . For an arbitrary . By Lemma 6, we obtain and Applying Lemma 4, we have for arbitrary Choose an interior edge and let Thus we gain . By Lemma 5, is obtained for any , a contradiction.
Case 5. is a 2-connected -minor free graph.
By Proposition 3, if
, we obtain
for any
by Case 1 in
Section 3.3. If
, by Case 2 in
Section 3.3, we have
for any
. If
is a 2-connected
-minor free graph, by Case 3 in
Section 3.3, we have
for any
, and the proof is completed.
Based on the proof in
Section 3.2–
Section 3.4, we prove Theorem 7 when
, respectively and complete the proof of Theorem 7.