2.1.1. Path Graphs
Let
n be a positive integer and consider the path graph
which is defined as follows:
A well known fact is that , for and .The following lemmas characterize -sets for these graphs, they are properly counted and the action of the automorphism group on is analyzed.
Lemma 1. For , is a minimum dominating set if and only if , for , and , for .
Proof. Suppose that there is
with
or that there are
such that
, then
thus,
D is not a dominating set.
Conversely, if
, for
and no pair of vertices share neighbors, we have
which implies,
; that is,
D is a
-set. □
Observe that Lemma 1 implies that
is the unique
-set for
.
Figure 1 shows this set for
.
Lemma 2. For , is a minimum dominating set if and only if one of the following conditions hold:
- i.
, for , and for all but one vertex;
- ii.
there exists a unique pair of vertices such that , , for , and , for .
Proof. Suppose that D is a dominating set and let t denote the number of unordered pairs of distinct vertices such that ; that is, . We proceed by cases.
If
, then
which contradicts
D being dominating.
If
D contains just one vertex of degree 1, then
so
, that is,
D satisfies (
i).
Finally, suppose that all vertices in
D have degree 2, then
Note that . If , then necessarily , but this is not possible. Supposing contradicts D being a dominating set. Thus, and D satisfies condition (ii).
Conversely, if
, for
, and every vertex but one has degree 2, then
and
D is a dominating set. Now suppose that there is exactly one pair of vertices
such that
and
, for
, and
, for
, as before, we obtain
and, therefore,
D is dominating. □
Remark 1. By Lemma 2 we may deduce that there are exactly two minimum dominating sets satisfying the first condition and satisfying the second one In fact, if is a γ-set and we considerthere is a one-to-one correspondence between these γ-sets and triplets such that , and . Indeed, we may observe that the correspondence is given by Figure 2a shows the
-set
and
Figure 2b the
-set
, respectively, for
.
Lemma 3. For , is a minimum dominating set if and only if one of the following conditions hold:
- i.
, for , and ;
- ii.
there are exactly two distinct vertices sharing a neighbor, , for , and every vertex in D has degree 2, but one;
- iii.
for all and there are exactly two distinct vertices such that and , for ; or
- iv.
for all and there exist such that , and , for .
Proof. Suppose that D is a dominating set and, as in the proof of Lemma 2, let t denote the number of unordered pairs of distinct vertices which share neighbors.
Suppose that D contains vertices of degree 1.
If
, since
D is dominating and
necessarily
and condition (
i) is satisfied.
If
D contains just one vertex of degree one, then
implies
, which is condition (
ii).
Suppose for all .
If then and, since , necessarily and the two vertices in this intersection form an edge, moreover, no other vertices share neighbors, thus, D satisfies (iii).
If , we get , obtaining and D satisfies (iv).
Conversely, if
D satisfies (
i), then
If
D satisfies condition (
ii), we get
Analogously the other cases. Therefore, in any case, D is a dominating set. □
It is important to note that condition (iv) is true for . Considering the triplets of nonnegative integers such that , as above, this lemma implies the following facts.
There is just one
-set satisfying condition (
i), namely,
which corresponds to
.
Proposition 1. There is a one-to-one correspondence between minimum dominating sets that satisfies condition (ii) and triplets such that and or and . Thus, the number of γ-sets of this kind is .
Proof. If D satisfies (ii), it contains a vertex of degree 1, say , since there are exactly two vertices sharing a neighbor they must be of the form and ; moreover, preceding vertices to have disjoint closed neighborhoods, so that the first vertices should be and, by the same argument, the vertices after must be . Thus, the triplet corresponding to D is . Analogously, if , the triplet corresponding to D is .
Conversely, given a triplet
such that
and
or
, we may construct easily the corresponding set as follows
Therefore, there are -sets satisfying condition (ii). □
Proposition 2. There is a one-to-one correspondence between minimum dominating sets that satisfy condition (iii) and triplets of natural numbers such that and . Thus, the number of γ-sets corresponding to these triplets is .
Proof. Let D be a -set satisfying condition (iii), thus every vertex has degree 2. It contains two vertices forming an edge, say , and , for . Since every vertex has degree 2 and preceding vertices to have disjoint closed neighborhoods, the first vertices should be , which implies and the same argument shows that the following vertices are . Thus, the triplet corresponding to D is .
Conversely, given a triplet
such that
and
we may construct the
-set
which clearly satisfies condition (
iii).
Therefore, there are -sets satisfying this condition. □
Proposition 3. There is a one-to-one correspondence between minimum dominating sets that satisfy condition (iv) and triplets of natural numbers such that . Moreover, the number of γ-sets satisfying this condition is .
Proof. Let D be a -set satisfying condition (iv); that is, every vertex has degree 2, it contains four vertices (not necessarily different) sharing a neighbor, say , and , for i or . Applying an analogous argument as in the previous cases, we may infer that the first vertices are , the vertices after must be and the vertices following are . Thus, the triplet corresponding to D is .
Conversely, given a triplet
such that
and
, we get the following
-set
which fulfills condition (
iv).
Observe that there are triplets having , with and so on. When , there is just one triplet, namely, . Thus, there are -sets satisfying condition (iv). □
Figure 3a shows the
-set which satisfies condition (
i),
Figure 3b shows the one that satisfies condition (
ii) corresponding to
,
Figure 3c the one satisfying condition (
iii) corresponding to
and
Figure 3d the set of the fourth kind which corresponds to
, respectively, for
.
Recall that
, for
, where 1 is the identity and
is the automorphism given (on subindices) by
and note that
Theorem 1. Let n be a positive integer, considering the action of on , we have:
- i.
is γ-transitive;
- ii.
for there are orbits;
- iii.
for and , there are orbits.
Proof. By the comment made after Lemma 1, we know that there is just one -set for , so that is trivially -transitive.
If
, by Lemma 2 we know that there are
-sets of two kinds: those which contain a vertex of degree 1, denoted by
and
, and those which are in correspondence with triplets
; moreover,
. Clearly,
and the image under
of the
-set corresponding to
is the one corresponding to
; that is,
Thus, there are orbits.
If , Lemma 3 implies that there are four kinds of -sets and . We proceed by cases.
In the first case there is just one -set which corresponds to and its image under is itself; thus, .
In the second case, -sets are in correspondence with triplets such that and or and . If D corresponds to , then corresponds to , and viceversa. A -set corresponding to is mapped to that corresponding to . This implies that the orbit of one of these sets contains exactly two of them. Thus, there are orbits.
In the third one, the correspondence is with triplets such that and . The image of the set corresponding to is the one that corresponds to . Note that the orbit is a singleton if and only if , which happens when k is even and , so there are orbits.
Finally, in the last case the correspondence is with triplets such that . We may observe that if D corresponds to , then corresponds to , and, as in the last case, the orbit is a singleton if and only if . In this sub-case the triplets are and, since , there are of these sets. Thus, there are orbits with two elements. □
The remaining cases
for
are as follows:
is trivially
-transitive and we may observe easily that
has four
-sets
and three orbits
2.1.2. Cycle Graphs
For an integer
, consider the cycle graph
, where
and
,
Figure 4,
Figure 5 and
Figure 6 show the
,
and
graphs, respectively. In this section, we use freely that the domination number of
is
k, for
and
.
The proof of the following lemma is completely analogous to that of Lemma 1 and it is omitted.
Lemma 4. For , is a minimum dominating set if and only if , for .
This lemma implies that there are exactly three
-sets for
, namely,
Lemma 5. For , is a minimum dominating set if and only if there is exactly one pair of vertices such that and , for and .
Proof. Suppose that D is a -set and, as in the previous section, let t denote the number of unordered pairs of distinct vertices which share neighbors. D dominating implies and since, in general, , we have . Thus, there is exactly one pair of vertices sharing a neighbor.
Now, if there are exactly two vertices sharing a neighbor, then
thus,
D is a
-set. □
By Lemma 5, we may observe that there is a one-to-one correspondence between
and
V given by
thus, there are
n sets of this kind:
where subindices are taken modulo
. In
Figure 5 is shown the set
for
.
Lemma 6. For and , is a minimum dominating set if and only if one of the following conditions hold:
- i.
there exists a unique pair of vertices such that and , for ; or
- ii.
there are exactly two pairs of vertices and in D sharing a neighbor and no other vertices share neighbors.
Proof. Let
be a
-set, as above,
t denotes the number of unordered pairs of vertices that shares a neighbor. Since
D is a
-set, we have
which implies
. We proceed by cases.
If , and these vertices form an edge, which is condition (i).
is not possible, as we noted in introduction.
If , and there are two pairs of distinct vertices and sharing a neighbor, satisfying condition (ii).
Conversely, if
is a set satisfying condition (
i), we get
which implies that
D is a
-set. Analogously if
D satisfies condition (
ii). □
There is a one-to-one correspondence between
-sets satisfying the first condition and
E. This correspondence is given by
Figure 6a shows the set
corresponding to
for
.
Proposition 4. For those γ-sets that satisfy (ii), we have two cases:
, obtaining nγ-sets of this kind; or
and are all different, getting γ-sets of this kind for k odd and for k even.
Proof. In the first case, there is a one-to-one correspondence between these
-sets and
V as follows
thus, there are
sets of this kind.
In the second one, we may obtain a one-to-one correspondence between these
-sets and pairs of vertices
such that
, for
, namely,
Note that if , there are exactly n pairs of vertices verifying this condition: and , for and taken modulo n. When , which happens if and only if and , there are just pairs of these vertices: and , for . Thus, there are -sets of this kind for k odd and for k even. □
Figure 6b shows the set
, corresponding to
, and in
Figure 6c is shown the set
, corresponding to the pair
, for
.
It is known that
, where
may be considered as a rotation by
and
the reflection across the symmetry line that fixes
of the regular
n-gon, thus,
Theorem 2. Let n be a positive integer, considering the action of on , we have:
- i.
is γ-transitive;
- ii.
is γ-transitive;
- iii.
for there are orbits.
Proof. By Lemma 4, we know that and are the three -sets for , and clearly and ; thus, is -transitive.
If , the correspondence between and V ensures that , for , so that is -transitive.
For , Lemma 6 implies that there are two kinds of -sets by cases.
In the first case, the correspondence between -sets and E implies that , for and ; thus, this orbit has n elements.
In the second case, there are two subcases. In the first one, -sets are in correspondence with the vertex set and clearly the orbit has n elements. In the second, the correspondence is between -sets and unordered pairs of vertices whose distance is , for . Let be the -set corresponding to the pair of vertices , thus and . Therefore, comprises all -sets of this kind, n if and if . □