3.1. Graovac–Ghorbani Index of Paths and Cycles
In 2014, Rostami and Sohrabi-Haghighat found trees that minimize the index.
Theorem 1 ([
18]).
The path is the vertex tree with the minimum Graovac–Ghorbani index. The Graovac–Ghorbani index of a path
is given by the following formula:
which can be written as
From part
of Lemma 1, we can observe that the gg-values of the edges in
decrease as we move from pendant edges to the central one (ones). For an even
n, the smallest gg-value is obtained for a single central edge and is equal to
, while for
n odd, we have two central edges with the smallest gg-value
.
In a cycle graph
, all edges have the same gg-value. For
n even, this is
, while for
n odd, we have
. Therefore,
In [
10], Dimitrov et al. investigated the
index of bipartite graphs. As an auxiliary result, they established that
for all even
, while for
, it holds
. Here, we examine the case where
n is odd. For this purpose, we need several auxiliary results.
Lemma 3. For and it holds Proof. Let us prove that
, that is
By expanding and simplifying both sides of (
4), we obtain
, which holds for any
. Therefore, the inequality (
4) and consequently (by taking square roots) (
3) holds. □
Proof. Numerical calculations show that for
n odd,
inequality holds. Let
. Then,
,
and inequality (
5) can be written as
It is easy to see that
is a decreasing function of
i. Therefore, all summands in (
6) are decreasing and the last one is equal to
. Next, we notice that
. Similarly,
. Bearing in mind that
, we have
Let
,
and
,
. Then,
,
and inequalities (
7)–(9) imply
Notice that (
10) is larger than
if and only if
If we analyze the inequality (
11) for each of the 12 possible pairs
, we come to the conclusion that it holds for
. The results are summarized in
Table 1 and the proof is complete. □
Now, we are ready to prove the main result.
Theorem 2. For it holds , while for and for we have .
Proof. As we mentioned above, for
n even,
inequalities were proven in [
10]. For
,
. Inequality
can be checked directly for
. Let
,
n odd. From Lemmas 3 and 4, it follows that
□
Graovac–Ghorbani indices of
and
for some
n are presented in
Table 2.
3.2. Unicyclic Graphs
As we mentioned in the introduction, unicyclic graphs maximizing the
index were found in [
16]. To the best of our knowledge, the problem of minimizing the
index for unicyclic graphs has not been solved in general. By studying the
index of bipartite graphs. Dimitrov et al. [
10] characterized unicyclic graphs with an even number of vertices and even girth in a non-explicit way that minimized the
index. By
we denote a unicyclic
vertex graph consisting of a cycle
with a pendant vertex, and by
we denote a graph with an odd number of vertices
n comprised of two even cycles
and
that have three common vertices and two common edges.
Theorem 3 ([
10]).
Among all bipartite graphs on vertices, the minimum Graovac–Ghorbani index is attained by the cycle for even n, by for odd , and by for odd . For , the graph that minimizes the Graovac–-Ghorbani index is the path on n vertices. Furthermore, these are the unique graphs with these properties. If we restrict ourselves to bipartite unicyclic graphs with an even number n of vertices, , then a direct consequence of Theorem 3 states that for such n, the cycle is a unicyclic graph with even girth and minimal index.
Pendant edge-moving transformation of a connected graph G. Let
and let
G be a connected graph with an induced path (induced subgraph that is a path)
, in which only one internal vertex has a degree of at least 3. Let
a be the number of vertices of
on one side of
w, and
b the number of vertices on the other side, see
Figure 1. By moving a pendant vertex from the
side of a path to its
side, we perform a so-called pendant edge-moving transformation of
G.
In [
18], Rostami and Sohrabi-Haghighat proved the following lemma for trees. We generalize it to connected graphs.
Lemma 5. Let G be a connected vertex graph that allows the pendant edge-moving transformation, and let be the resulting graph. Then Proof. Let
and let
be an induced path of
G with a single internal vertex
w, such that
. Then,
w is a cut-vertex in both
G and
. Let
. Then,
and the pendant edge-moving transformation preserves the gg-values of the edges in
H. We have
If we take
, then
and from Lemma 1
we obtain
□
For
,
, we denote by
an
vertex unicyclic graph consisting of a cycle
,
and paths
,
, such that
is an end vertex of
,
. The vertices
are positioned clockwise on
, see
Figure 2. Consequently,
.
Theorem 4. Let G be a unicyclic graph with a cycle , , , and let be an vertex tree in G containing , . Then Proof. We repeatedly apply a pendant-edge moving transformation to G; i.e., to each , , we perform a sequence of pendant-edge moving transformations until we obtain a path . These transformations preserve the unicyclic property of G, while Lemma 5 implies a reduction in the index. □
Due to Theorem 4, unicyclic graphs with minimal index belong to the class of graphs . Due to a different behavior, vertex unicyclic graphs of girth 3 are considered separately.
The calculations show that among all unicyclic graphs with vertices, the graph has the smallest index.
Theorem 5. Let and let G be an vertex unicyclic graph of girth 3. Then Proof. The cases and can be tested directly. Let . From Theorem 4, it follows for every unicyclic graph G of girth 3 with trees , that . Therefore, we only focus on the graphs . Without loss of generality, we assume . We consider three cases:
Case 1: . Then,
and
. According to Lemma 2 and Theorem 2, we have
Case 2: and
. Then,
. Notice that the gg-value of the edge
is equal to
, and for
, the gg-value of the edge
is
. By applying Theorem 2 and parts
and
of Lemma 1, we obtain
Case 3: . Since
and
, we have
and
. From Lemma 1
, we have
,
and
. This and the part
of Lemma 1 imply
The last inequality holds since
,
. Application of Theorem 2 results in
and the proof is complete. □
Next, we find the smallest gg-values of the edges of a cycle , in any unicyclic vertex graph G.
Lemma 6. Let , and let G be a unicyclic graph with cycle , . Then we have for each edge The equality is given if , i.e., the edge e belongs to , or if , i.e., the edge e belongs to a cycle in . The graphs and are unique unicyclic graphs containing the maximum number of cycle edges with the smallest gg-values. Proof. Case 1: n and s are even. For each edge , we have , and the largest value of the product is obtained for . Therefore, , which is a gg-value of an arbitrary edge of .
Case 2: n is even and s is odd. Then, and at least one vertex of G does not lie on a cycle . For , we have and , since there is at least one vertex that is equidistant from u and v. (For , such a vertex is unique and belongs to ). If t is odd, then , so . Note that is a decreasing function of odd and reaches its minimum value for . (If , then and , which implies .) Therefore, . If t is even, then and , so . Function is a decreasing function of even . It follows that . However, for every (even) . Therefore, .
Let w be the vertex on a cycle for which (such a vertex exists since at least one vertex of G is not on a cycle). Then, there is a single edge whose end vertices are equidistant from w. Since a tree attached to w exists, we conclude that the gg-value of f is greater than and there exists at least one pendant edge in G having gg-value . Therefore, G contains at least 2 edges with a non-minimal gg-value. We conclude that the maximum number of cycle edges with the smallest gg-value is and they belong to .
Case 3: n and s are odd. Then, and for a cycle edge , we have , . Similarly to in Case 2, if t is odd, then and , since . It follows that . If t is even, then , and . Cycle is the unique graph in which all edges have the smallest gg-value.
Case 4: n is odd and s is even. Then, and for any cycle edge in G, we have and . It follows that . In , all cycle edges have the smallest gg-value. □
In the following, we compare gg-values of edges in an arbitrary vertex tree with the smallest gg-values of cycle edges in an vertex unicyclic graph G.
Lemma 7. Let , and . Thenand for , it holds Proof. We have
and from Lemma 1
, by taking
we obtain
and the inequality (
16) is proven.
To prove (17), notice that
is equivalent to
which is a quadratic inequality of variable
i. Its solutions are integers
i from the interval
. Therefore, (17) holds. Similarly, we note that
is equivalent to
which gives integer solutions
i from
and we have proven (18). □
We are ready to characterize unicyclic graphs with even girth that minimize the Graovac–Ghorbani index.
Theorem 6. For , let G be an vertex unicyclic graph of even girth. Then Proof. Let us consider the case where
n is even. The inequality (12) from Lemma 6 implies that the gg-value of each edge of a cycle in
G is greater than or equal to the gg-value of
, which is equal to
. Moreover, inequality (
16) from Lemma 7 implies that the gg-value of each edge of a tree in
G (if any) is greater than or equal to the gg-value of
.
For
n odd, the inequality (15) from Lemma 6 implies that the gg-value of each edge of a cycle in
G is greater than or equal to the gg-value of a cycle edge in
, which is equal to
. The inequality (
16) from Lemma 7 implies that the gg-value of each edge of a tree in
G (which exists) is greater than or equal to the gg-value of a cycle edge
. Since
contains a single pendant edge, we obtain
and the inequality (
19) is proven. □
Lemma 8. For odd , it holds .
Proof. A simple calculation shows that the inequality holds for
. Let
. The Lemma 3 implies
Therefore,
Since
, it is sufficient to prove the inequality
which is equivalent to
For
, we have
and for
, it holds
. Therefore, the inequality (
20) holds and this completes the proof. □
Corollary 1. Let and let G be an vertex unicyclic graph of girth , s is even. Then Proof. The result follows directly from Theorem 6 and Lemma 8. □
We continue our studies by examining unicyclic graphs G with odd girth s, where . We say that the edge of a tree in G (if any) is gg-small if its gg-value satisfies the inequality (17) (if n is even) or the inequality (18) (if n is odd).
Theorem 7. Let and let G be an vertex unicyclic graph of odd girth with zero gg-small edges. Then Proof. Case 1: n is odd. If , then, from the inequality (14) of Lemma 6, we have . Let . The assumption of zero gg-small edges in G means that the reversed inequality in (18) holds; i.e., . We conclude .
Case 2: n is even. Then,
G contains at least one vertex that is not on the cycle. Consequently, it contains at least one pendant edge and for at least one cycle edge
there are
vertices equidistant from
u and
v. We have
. Note that
p and
t have the same parity and
if
t is even, while
if
t is odd. We omit the details and refer to Case 2 of Lemma 6 to conclude that
. The above considerations in combination with the inequality (13) and the reversed inequality in (17) result in
□
Lemma 9. For even , it holds .
Proof. The proof follows directly from the inequalities
□
Corollary 2. Let and let G be an vertex unicyclic graph of odd girth with zero gg-small edges. Then Proof. The result follows directly from Theorem 7 and Lemma 9. □
Now, we focus on graphs with odd girth that contain gg-small edges. Note that pendant edge-moving transformations of such graphs also contain gg-small edges. Therefore, we consider with gg-small edges.
Lemma 10. There exist at most two paths and , , in , which contain gg-small edges.
Proof. If contains paths, then the maximum number of vertices that are not on a cycle is equal to . Suppose that there are at least three paths in that contain gg-small edges. Then, the number of vertices on these paths is at least , where, according to Lemma 7, if n is odd, and if n is even. In both cases, we show that . For n odd, can be written as , while for n even, can be written as . Both inequalities are valid for every , and we obtain a contradiction. □
Theorem 8. If contains two disjoint paths with gg-small edges, then Proof. For the sake of simplicity, we use the notation . Let us consider a graph . Then, contains the maximum number of vertices that do not lie on a cycle (it is ), and it consequently has the maximum number of gg-small edges. Let and S be the set of gg-small edges in and H, respectively. Then, .
Case 1:
n is odd. For
, at most one path in
contains gg-small edges, so we assume
. Let us calculate
. For simplicity, we assume that the paths in
are balanced, i.e., each contains
edges (moving a pendant vertex from a path
P to a path
Q in
decreases the number of gg-small edges of
P by one, while simultaneously increasing the number of gg-small edges in
Q by one). Then, from the inequality (18) of Lemma 7, the number of gg-small edges on each path in
is equal to the number of integers
i satisfying the condition
. If
is an integer, then the number of gg-small edges on both paths is
From Lemma 3, we know
Notice that
, for each
. By dividing this inequality by
, we obtain
By inserting (
22) into (
21), we obtain
The above inequality shows that sum of the four largest gg-values on paths in
H is greater than
gg-values of
.
If
, then the number of gg-small edges on both paths in
is
It is easy to see that
for each for
, that is
Combining (
24) with (
21), we obtain (
23). We conclude that for
n odd, we have
. From (
23) and from Lemmas 6 and 7, we obtain
Case 2:
n is even. For
, at most one path in
contains gg-small edges, so let
. Similarly to Case 1, we consider
with balanced paths. Then, one path contains
edges and the other one contains
edges. Note that
. Therefore, on one path, the number of gg-small edges is equal to the number of integers
i that satisfy
, while on the other path, this is the number of integers
i that satisfies
. We have
From Lemma 3, we have
Since for each
it holds
, combining this with (
25) gives
From (
26) and from Lemmas 6 and 7, we obtain
□
Corollary 3. If a pendant edge-moving transformation of an unicyclic graph G with odd girth yields with gg-small edges on two disjoint paths and , then .
Proof. Theorems 4, 8 and Lemma 9 give
□
Now, a single type of unicyclic graph with an odd girth remains to be investigated. This is a graph with gg-small edges whose pendant edge-moving transformation gives with gg-small edges on a single path. Numerical experiments indicate that many such graphs have an index larger than . However, at this moment, we are not able to provide a general proof of this conjecture, so we leave this for future research.