Now, we turn to the proof of the main result.
Proof. We are sufficient to handle the situation
since cases of
are trivial. Let
be a
-function such that
contains the maximum number of components isomorphic to
. Suppose to the contrary that
. Then,
since
[
11], that is,
Formula (
1) indicates that every 2RiDF of
has weight at least
. We will complete our proof by constructing a 2RiDF of
of weight at most
or a 2RiDF of
G of weight less than
.
If
, then
, a contradiction; if
, then
and by Lemma 1
, also a contradiction. Therefore, by Lemma 4,
Then, by Lemma 5 (3) we have
for
. In addition, because, by definition,
is a clique,
, it follows that for every 2RiDF
of
,
Therefore, by Lemma 2 we can extend every
-function to a 2RiDF of
with weight at most
, i.e.,
by Formula (
1).
Claim 1.Denote by ℓ the number of vertices in , which have degree in . Then, where . If not, either ℓ is at least 2 or both ℓ and are equal to 1. Suppose that and take two vertices , such that they are adjacent to all vertices of in . Let be: for . Clearly, is a 2RiDF of and by Lemma 2 we can extend to a 2RiDF of , which has weight at most , a contradiction. Now, suppose that . Then, , which indicates that contains a component s.t. . Since , there is a vertex v, say , which is adjacent to every vertex of in . By Lemma 6 has a 2RiDF s.t. for some and . Observe that in v is adjacent to all vertices of ; by the rule of Lemma 2 we can extend to a 2RiDF g of under which there is at most one vertex in (and ) not assigned value 0. Thus, , a contradiction. This completes the proof of Claim 1.
Now, we WLOG assume
. Then,
by Formula (
2).
Claim 2. does not contain any isolated vertex v s.t. . Otherwise, define as: for , and . By Claim 1, in , has not more than one vertex adjacent to every vertex in ; say if such a vertex exists. We further let for (or for if exists). Since in G every vertex in (except for ) is adjacent to v and also , f is a 2RiDF of G of weight at most , a contradiction. This completes the proof of Claim 2.
We proceed by distinguishing two cases: and .
Case 1.. In this case, by Claim 1 each vertex of owns a neighbor belonging to in G where = [1,2]; by Theorem 2, has one component H isomorphic to one of (), (), () and , and other components of are isomorphic to or . Let be a vertex with . Clearly, . Let and be two vertices such that every vertex in has degree in H not exceeding . By the structure of H, for , we have that and if has a neighbor ) in H, then . Moreover, by Lemma 5 (1), , which implies that each vertex of is adjacent to or in G.
Claim 3.. Otherwise, let
. Then,
and
. Suppose that
(the case of
can be similarly discussed). Let
and clearly
. By Lemma 5 (2), a vertex
is adjacent to
in
. We WLOG assume that
. According to Lemma 6,
admits a 2RiDF
satisfying
and
for some
. Further, let
and
. So
is a 2RiDF of
, and by Lemma 2 and Formula (
3) we can extend
to a 2RiDF of
with weight at most
(since
), a contradiction. We therefore assume that
. By Lemma 5 (2) we have
. WLOG, suppose that in
,
has two neighbors belonging to
, say
and
. By Lemma 5 (1),
is not adjacent to both
and
, and
is not adjacent to both
and
in
, where
and
. Thus, it follows that
and
, or
and
, which contradicts to Lemma 5 (1) again. This completes the proof of Claim 3.
By Claim 3, we see that contains no component isomorphic to and contains at most one component.
Claim 4. contains a component. If not, we have = H.
Claim 4.1..
Otherwise, for , is adjacent to every vertex of in G, and by Lemma 5 (2) and . Set , ; then, . Let f be: , and for any x in . So, we get a 2RiDF f of G, which has weight 4, a contradiction. So, Claim 4.1 holds.
Claim 4.2..
Observe that ; it is enough by showing that G admits a 2RiDF f s.t. . When , let f be: for , for , and for . By Lemma 5 (1), in , contains no vertex adjacent to and also . Therefore, f is a 2RiDF of G of weight . Now, suppose that . By Lemma 5 (1), contains at most one vertex adjacent to both and in ; say u if such a vertex exists. Let f be: (or if u exists), for (or ) and for . Notice that by Claim 1 every vertex of is adjacent to in G, and by the structure of H and the selection of and , every vertex of is adjacent to in G; f is a 2RiDF of G of weight at most . This completes the proof of Claim 4.2.
By Claim 4.2, we have . Let in the following.
Claim 4.3.In , for every vertex in has not more than one neighbor in .
If not, let
be adjacent to two vertices of
in
, say
. By Lemma 5 (1)
or
is not adjacent to
v in
, say
. If
, define
as:
for every
,
,
. If
, then
and let
be:
; further, let
when
, or let
and
when
. According to Lemma 2, in either case the
defined above can be extended to a 2RiDF
g of
under which
and
or
. Therefore, by Formula (
3)
, a contradiction. With a similar discussion, there is also a contradiction if we assume
contains a vertex that has two neighbors in
in
. This completes the proof of Claim 4.3.
Now, we consider . Suppose that and let . According to Claim 4.1, we WLOG assume that . This indicates that by Lemma 5 (1). If has a neighbor in , say , then according to Lemma 5 (1), , , and (resp. ) is not adjacent to (resp. ) in (otherwise or has two neighbors in or in , respectively. This contradicts to Claim 4.3). Let f be: and for . Observe that (resp. ) is not adjacent to (resp. ) in and by Lemma 5 (1) contains no vertex adjacent to both and for some . Hence, f is a 2RiDF of of weight 4 and we are able to extend f to a 2RiDF of G with weight at most according to Lemma 2, a contradiction. Therefore, we may assume that . In this case, when , let f be: , . By Lemma 5 (1) has not more than one vertex adjacent to both and in and has not more than one vertex adjacent to in ; for we further let . Then, f is a 2RiDF of of weight 3 and according to Lemma 2 we can extend f to a 2RiDF of G of weight at most , a contradiction. We therefore suppose that has a neighbor in in , say . With the same argument as , we can show that as well.
Then, if and , the function f: and for , is a 2RiDF of with weight 4, and according to Lemma 2, we are able to extend f to a 2RiDF of G with weight at most , a contradiction. Therefore, we suppose that by the symmetry. By Lemma 5 (1), it has that , and or , say by the symmetry. Let f be: and for . Since in G every vertex in has a neighbor in and also , f is a 2RiDF of of weight 4 and according to Lemma 2 we can extend f to a 2RiDF of G of weight at most , and a contradiction.
A similar line of thought leads to a contradiction if we assume that , and so Claim 4 holds.
By Claim 4, we see that contains one component isomorphic to . Let s be the vertex of the component. We first show that . If not, in we assume that s has two neighbors in , say . By Lemma 5 (1) for , (resp. ) can not be adjacent to and (resp. and ) simultaneously in . This implies that either , , or and , which violates Lemma 5 (1) as well. Thus, by Claim 2 and the vertex adjacent to s in belongs to . Let f be: for , , for . Observe that by Claim 1 all vertices in are adjacent to in G. Hence, every vertex in is adjacent to s and also in G. Therefore, f is a 2RiDF of G with weight (since ), a contradiction.
The foregoing discussion shows that there exists a contradiction if we assume that . In what remains, we handle the case when .
Case 2.. Then by Lemma 1 every component of
is isomorphic to
or
. Recall that
for
. Take two vertices
in
s.t.
if
contains a
component and
are isolated vertices in
otherwise. By Lemma 5 (1), we have
We deal with two subcases in terms of the adjacency property of u and v.
Case 2.1.. Then in , contains no vertex adjacent to .
Claim 5.In , contains only vertices with degree at most . Suppose that contains a vertex w such that for every . If (or ), define a 2RiDF of as: (or and (). According to Lemma 2 we can extend to a 2RiDF of , under which contains at most two vertices not assigned 0. Thus, , a contradiction. We therefore assume that and . By Lemma 5 (2), there are at least three vertices in that are adjacent to u or v. We WLOG assume that contains a vertex s.t. . Construct a 2RiDF of as follows: , and . Then, by Lemma 2 can be extended to a 2RiDF g of , under which contains at most one vertex not assigned value 0. Therefore, , a contradiction. Similarly, we can also obtain a contradiction if we assume that contains a vertex adjacent to every vertex of . So, Claim 5 holds.
By Claim 5, for = [1,2], each vertex of is adjacent to a vertex of in G. If , then in G all vertices of are adjacent to . Let f be: for , for , and . Obviously, f is a 2RiDF of G s.t. = , a contradiction. We therefore assume that contains a vertex s s.t. and . Then, in , by Lemma 5 (1) no vertex in is adjacent to u and v simultaneously. Analogously, the function f: for , and for (and for , and for ) is a 2RiDF of G with weight (and ). This implies that and . Let = and = . Then, in , neither u nor v is a neighbor of and simultaneously; otherwise, we, by the symmetry, suppose that and . Let be: , and . Obviously, is a 2RiDF of with weight 2. According to Lemma 2, we can extend to a 2RiDF of with weight at most , a contradiction. In addition, in , , is not adjacent to u and v simultaneously according to Lemma 5 (1). Therefore, we may assume, by the symmetry, that and .
If no edge between and in exists, then by Lemmas 5 (2), and . Then, the function such that , and is a 2RiDF of with weight 2. According to Lemma 2, we can extend to a 2RiDF of with weight at most , a contradiction. We therefore assume that contains an edge connecting and , say by the symmetry.
If
, define
as:
. Then,
is a 2RiDF of
with weight 2. By Lemma 2 and Formula
3, we are able to extend
to a 2RiDF of
of weight at most
, a contradiction. Consequently, we have
. Then, the function
such that
is a 2RiDF of
with weight 3, and by Lemma 2 and Formula
3 we can extend
to a 2RiDF of
with weight at most
. This contradicts the assumption.
Case 2.2.. Then, by the selection of and , contains only isolated vertices and G does not admit a -function for which the induced subgraph of by vertices with value 0 contains components.
For every
, let
for
. Let
be:
for
,
, and
. Apparently,
is a 2RiDF of
with weight 2. According to Lemma 2, we can extend
to a 2RiDF of
G with weight at most
. To ensure
, we have
Claim 6. and the two vertices in are adjacent in . Define a 2RiDF of as: . Suppose that . Since and are cliques in and every vertex in is adjacent to u or v in , by Lemma 2 we are able to extend to a 2RiDF g of under which at most one vertex in , is not assigned value 0 (here if contains a vertex, say w, then let ). Clearly, , a contradiction. Moreover, if contains two nonadjacent vertices in , say , then and are not in the same set for some . Therefore, we can extend to a 2RiDF g of via letting when x is in and . However, , a contradiction. This completes the proof of Claim 6.
By Claim 6, contains two adjacent vertices in , say . If there exists a s.t. (or ), then set as: , (or ), (or ). Since in every vertex in has a neighbor in and every vertex in is a neighbor of , where for some , we can extend to a 2RiDF g of according to Lemma 2. Under g, every vertex in is assigned value 0 and at most one vertex in is not assigned value 0. Therefore, , a contradiction. This demonstrates that in no vertex in is adjacent to . Furthermore, if there is a , then by Claim 6 we have and , which implies that . Set as: and for . Then, is a 2RiDF of with weight 3, and we can extend to a 2RiDF of with weight at most according to Lemma 2, a contradiction. So far, we have shown that , that is, .
Now, we define a 2RiDF of as follows: , and . According to Lemma 2, we can extend to a 2RiDF f of G with weight at most . To ensure , f must be a -function (since ). However, is isomorphic to . This contradicts the selection of . Eventually, the proof of Theorem 3 is finished. □