A graph with vertex set and edge set is harmonious if there exists a harmonious labeling of ; which is an injective function provided that whenever are distinct with endpoints and , respectively, then . Using basic group theory, we prove in a different manner an already established result that a disjoint union of an odd cycle and a path is harmonious provided their lengths satisfy certain conditions. We apply the same basic idea to establish that, under the same conditions, a disjoint union of an odd cycle with a certain starlike tree is harmonious (where a starlike tree consists of a central vertex that is adjacent to an endpoint of a certain number of fixed length paths). Finally, we extend the latter result to include specifying that the central vertex in the tree be adjacent to different vertices in each of the -many -paths.
Keywords: Integer-magic spectrum, Tree diameter, Magic labeling
1. Introduction
Our notation is fairly standard, following [1]. Let be a simple graph with and . An
injection is said to be a harmonious labeling of
provided that whenever with respective
endpoints and , then . If there exists a harmonious labeling of
, then the graph is said to be harmonious. The
elements of the range of are the
vertex labels. For an edge with endpoints and , then is the edge label of.
In the case of trees, as there would not be enough labels of the
vertices for there to be such an injection, it is permitted for there to
be precisely two vertices having the same label; however, the edge
labels must still be all distinct.
Much is known about harmonious labelings of particular graphs; for a
comprehensive chronicling of known results, please refer to Gallian’s
work [2]. Although it is not
known whether all trees are harmonious, many classes of trees are known
to be harmonious. For example, all trees with at most 31 vertices are
known to be harmonious (see [3]). It is not difficult to show that all path
graphs and stars are harmonious. Caterpillars are known to be harmonious
(see [4]). As for cycle
graphs, odd cycles are known to be harmonious; even cycles are known to
not be harmonious (see [4]).
An injection is said to be even harmonious labeling of if the induced function defined by (mod ) is
bijective. Gallian and Schoenhard, [5], define an injective function
as properly even harmonious labeling of if the induced function defined by (mod ) is
bijective. An injective function is said to be -sequential labeling if for each
edge , the edge label induced by
is in and all the edge labels
are distinct. If the graph is a
tree then the domain of is . An injective function
is said to be even 2a-sequential if for each
edge , the edge label induced by
is in and all the
edge labels are distinct. If the graph is a tree then the domain of is .
In this paper, we first show that if the cycle is odd and the number
of vertices in the path is one more than an even multiple of the number
of vertices in the cycle is indeed harmonious; we achieve this using a
subgroup and its cosets. We are re-proving a known result using a
different method involving cosets of a subgroup. Then, we apply the
coset method used in that proof to the another family of graphs for
which it is unknown whether they are harmonious. A starlike
tree consists of a central vertex called a root and several
path graphs attached to it (at a leaf of the path). More formally, a
tree is starlike if it has exactly one vertex of degree greater
than or equal to 2. We show that the disjoint union of an odd cycle on
vertices with a starlike tree of
-many -paths (where is even) is harmonious. We also show
how a simple re-labeling of the root can lead to a harmonious labeling
of a disjoint union of an odd cycle with the root attached to -many -paths at selected vertices in the
paths.
2. Labeling and proofs
It has been determined that the is harmonious when
is odd (see [6]). Gallian
and Stewart (see [7]) proved the following:
Theorem 1. [7] is properly even harmonious if is an even -sequential graph when and (Mod 4).
There is an extension of Theorem 1 at
[8] that
states the following:
Theorem 2. is properly even harmonious if is an even -sequential graph with edges, , (mod
4) or and is even.
As odd cycles are even -sequential graphs, the graph is harmonious when
(mod 4).
Although already implied by Theorem 2 above, we
provide the following theorem and its proof (which is different than the
one provided by Gallian and Stewart) for completeness as well as the
reuse of part of the proof in the following theorems.
Let , where
is odd and is even. Since , we label the vertices of
with the elements of the additive
cyclic group ,
with one repeated vertex label permitted (allowing for the path). For
detailed information on Group Theory, see [9]. Our main goal is to prove the following
theorem.
Theorem 3. Let be odd and let be
even. Then is
harmonious.
Our proof will consist of a number of cases, which compare different
edge labels based on the labeling function. We shall label the vertices
of with the elements of the
subgroup of of order , i.e., . We do this by starting
from a vertex in the cycle and consecutively label the vertices starting
with the smallest element of the subgroup and finishing with the
greatest such element. We now show the induced edge labels are
distinct.
Case 1. With the labeling of the vertices of the
-cycle described above, the
resulting edge labels are distinct.
Proof. Observe that each edge label has the form , where . So, consider
the scenario, for , that
. This implies that for some
integer . Since are both odd, then is even. Thus, , where , which
is contrary to .
For the path, the idea is to eventually construct a path on vertices from smaller paths each of
which has vertices and each of
whose vertices are labeled using elements of a coset of . Here is the first piece of the
labeling function: A coset of
will correspond to a path of
-many vertices; working from one
end of the path consecutively to the other, we label the vertices in
order using the elements of the coset. In particular, consider the
cosets
of ; these correspond to paths
each with vertices. For future
reference, we shall denote the corresponding paths by . Based upon
this correspondence, sometimes we will refer to the edges within a
particular -path as “internal
coset” edges and edges that connect different -paths as “coset-to-coset” edges
We shall now compare an edge label of the -cycle to an edge label within an -path.
Case 2. For an edge with both ends in the cycle
and an edge with its ends in
(), the two
edges have different labels.
Proof. Consider an edge whose endpoints are in the , . Now observe that ; these elements are the vertex labels of . Then the edge label would be , for some . Thus,
since , then and since
is even (hence is odd), then term in the expression above cannot
equal . Consequently, the
edge-label cannot be an element of the subgroup generated by . And since, by closure of , all the edge labels of are elements of , then the edge label under
consideration is distinct from the edge labels of .
Next, we compare edge labels of a pair of edges within a single one
of these -paths; i.e., we compare
internal coset edges.
Case 3. For every , any pair of edges whose endpoints are
labeled with elements of have
distinct labels.
Proof. From the previous case, a pair of internal coset
edges have labels and for some ,
and . Without loss
of generality, assume .
By contradiction, assume . Consequently, we have , for some . Hence, , or . Therefore, . Since is odd, then is even. Thus, dividing by 2 gives
, where
. So, . And since , then , a
contradiction.
We next compare edge labels from different -paths.
Case 4. For an edge in and an edge in , where , their edge labels are not
equivalent.
Proof. In this case, the edge labels would be of the form
and , where
and both edge labels are taken . We may assume, without loss of generality, that
. To the contrary, we
assume these labels are equivalent. Thus, we have . Since are odd, then is even; hence, , where . It follows that is a multiple of . But this would imply that since
, then
, a contradiction.
Next, we start joining the paths by edges in the following manner:
For , create an
edge between the vertex with the label in the path and the vertex in the path with the label . Moreover, we create a
new vertex with label 0 (note
that 0 was already used once in the cycle labeling and is the only
re-used label) and create an edge between vertex and the vertex in with the label . Note that this induces the edge
label . Next, create an edge
between and the vertex in with label . Note that this induces
the edge label of .
Observe that even apart from this labeling, that joining the edges in
this manner will produce a path with vertices, as required. We still must prove the following
regarding this labeling: (1) The labels of edges joining vertices from
the cycle have labels are different from the labels of the
coset-to-coset edges, (2) the labels of a pair of distinct
coset-to-coset edges are different, (3) the label of a coset-to-coset
edge is different than the label of an internal coset edge, (4) the edge
label of an edge joining the vertex with label 0 to either of one of its
neighbors is different from (a) the label of an internal coset edge ,
and (b) the label of a coset-to-coset edge.
We begin with (1) above.
Case 5. The label of a coset-to-coset edge
cannot be an element of (and thus
cannot have the label of an edge label of the cycle).
Proof. Observe, by the construction above, that a
coset-to-coset edge has the label ,
where .
Thus, , all of which are taken ; thus, as the resulting label
could only be an element of the subgroup provided that . This is impossible since
is odd and is even.
Now to (2).
Case 6. A pair of distinct coset-to-coset edges
have different labels.
Proof. We note the form of the coset-to-coset edge label
above. In particular, consider edges with labels and
as well as , where . Without loss of generality, assume . By contradiction, assuming their
edge labels are equivalent modulo , it follows that for some . This contradicts
the fact that .
We now compare the label of a coset-to-coset edge to the label of an
internal coset edge.
Case 7. The label of a coset-to-coset edge is
not equivalent to the label of an internal coset edge.
Proof. Suppose, on the contrary, that , where and .
Thus, , for some integer . Therefore, . Now, ; since , then . Since is odd,
then is
even and positive, from which it follows that is positive and a positive
multiple of . As are odd, then is even, which implies that , where . This
contradicts the fact that .
Finally, we compare the label of the edges incident to to the labels of the cycle edges, the
internal coset edges, and the coset-to-coset edges. First, as mentioned
earlier, the labels of the edges incident to are and . It is clear that neither of these are edge labels
in the cycle, as the cycle only admits labels from the subgroup . We show that neither of these are the
labels of internal coset edges.
Case 8. The labels of the edges incident to
are not equal to the edges
incident to the labels of the internal coset edges.
Proof. As earlier, the labels of the internal coset edges
have the form , for some and . By
contradiction, if we assume that , then we have
that for some integer . Thus, is a multiple of . As , we have . Thus, either
– which is contrary to being even
– or we have .
This latter case implies , a
contradiction.
Now, we show this internal coset edge label is different from the
label of the other edge incident to . So suppose, on the contrary, that
, where and . Hence, for some ; and so
must be a multiple of . Since
, then ; hence or . The former case is a
contradiction, as is even. The
latter case implies that ; so
we are working in the coset . In
this coset, the edge labels are of the form , where . This range of values for loops through the odd multiples of
through its first half of
values; thus the edge label will not equal (an even multiple of ) in this round. In the second and
final round of the loop values of , it will reach up to , and so, at this maximum, the
multiple of is , and thus the edge label in
this round reaches its maximum just short (and not equal to .
Finally, we compare the labels of edges incident to to those of the coset-to-coset edge
labels.
Case 9. The labels of the edges incident with
are different from the labels
of the edges of the form , where .
Proof. First, we consider the edge with label incident to . By contradiction, assume that . Thus, for some integer ; this implies that is a multiple of . Now, as , we have . Thus, ; hence is odd, a contradiction.
Next, we consider the edge with label incident to . To the contrary, we assume that
. Therefore, for some integer . Since , we have . It follows
that either or . The former contradicts the fact
that is even. Thus, . This is impossible as precludes this.
The following is an example the harmonious labeling of using the theorem
above.
Figure 1. Example of Odd Cycle Union Path
Now, we apply this subgroup and coset method to other disjoint unions
of odd cycles and trees. Again, let be even and be
odd. Then let be a
starlike tree consisting of -many
-paths; i.e., consists of a central vertex (root)
adjacent to a leaf of each of
paths with vertices. Let . As before, we label the
vertices of with the elements of
the additive group , with one repeated
vertex label permitted.
In light of the previous theorem, with a little adjustment, we have
the following theorem.
Theorem 4. Let be odd and let be even. Then is harmonious.
Again, we proceed by cases, comparing the labels of classes of edges.
As before, we label the vertices of with all the elements of the cyclic
subgroup
of order generated by . That is, we have . In light of case 1 of the proof of the
previous theorem, we omit the proof of the following:
Case 1.The edge labels of are distinct.
In a manner similar to using cosets to label smaller paths in the
proof of Theorem 1, we construct a starlike tree on vertices (along with its labeling
function) by labeling the -paths
(not including the root). Start with -many copies of , namely, . The
vertices of are labeled using
elements of the coset of
, where . For each such path,
working from one end of the path consecutively to the other, we label
the vertices in order with the elements of its corresponding coset (in
order of the elements of the coset). .
Again, since we have proved this earlier, we have that the labels on
the vertices in -cycle is
different than the label of an internal coset edge.
Case 2.For an edge with both endpoints in the
-cycle and an edge with its
endpoints in (), the two edges
have different labels.
Now, we again note that edge labels of a pair of edges within a
single -path are distinct. As with
other cases so far, we do not repeat the proof.
Case 3.For all , any pair of edges in whose endpoints are labeled with
elements of have distinct
labels.
As in theorem 1, the edge labels from different -paths are distinct.
Case 4.For an edge in and an edge in , where , their edge labels are
distinct.
Now, we label the root of the tree, call it , with vertex label (note that was already used once in the cycle
labeling and is the only re-used label). For odd (for even ), for i between 1 and , create an edge between and the vertex labeled (vertex labeled ). Obviously, the resulting
edge labels would then be (for
odd ) and (for even ).
It is obvious that the edges incident to have distinct labels. It is also
clear that the edges incident to are different than these edge labels
in the cycle, as the cycle only admits labels from the subgroup . It remains to show that the edges
incident to have different
labels than the internal edges of the paths.
Case 5.The labels of the edges incident to
are not equal to the labels of
the internal coset edges.
Proof. As earlier, the labels of the internal coset edges
have the form , for some and . Since we
the way we“hook” the root to each path depends on the parity of the path
label, call it , we look at the
proof of this case in terms of the parity of :
Case 5A: Let
be odd. For sake of contradiction assume that , where , , and
. Therefore,
for some positive integer .
Since , , , and are odd, is even. Hence .
We note that . Hence, . So, either , or . Since and have different parity, . If , then . Since , . Note that . Hence,
which contradicts our assumption that .
Case 5B: Let
be even. In this case the root will be “hooked” to the vertex labeled
. For sake of
contradiction assume that , where , , and .
Hence, for some integer . Observe . Hence,
equals or . Now, if , then . Hence
. Since and both sides are positive, we have . Thus, the equation implies
that is even, a
contradiction.
On the other hand, consider the case where . This is impossible since the left side is even, and the
right side is odd.
The following is an example of a harmonious labeling of using the theorem
above.
Figure 2. Example of Odd Cycle Union Star-like tree
Finally, we consider another disjoint union of an odd cycle and a
tree and its harmonious labeling that results from a simple re-labeling
of only the root in the labeling of the odd cycle with the starlike tree
described in the proof of the previous theorem.
Again, consider the labeling of the graph with odd and even described in the proof of Theorem
2. Now, we change the label of the root to , for some . Since , we can “unhook” from the ends of the paths described
above, and “re-hook” to a new
vertex of for each , again depending on the
parity of . If is odd, then we add an edge between
and the vertex in with label , creating the edge
label of . If is even, then we add an edge between
and the vertex in with label ; thus, we have
as the corresponding
new edge label. From the discussion before case 5 of the proof of
Theorem 4, it is clear that this will
yield the same edge labels between and the new vertices to which they
are adjacent as before. We note that no labels of the other edges have
changed. These graphs are obviously different than the (admittedly, more
symmetric) union of starlike graphs with paths. In particular, let be the resulting tree. In
particular, will have
vertex that is adjacent to
-many -paths at the vertex in each path that
is from one end of the
path; and is adjacent to -many -paths at the vertex in each of these
paths that is from one
end of the path, where .
To illustrate the argument and the labeling, we consider the case of
, , and . The cycle is labeled of
course with the elements of the subgroup . The root has label . The path on the
lower-left and lower-right would be, using the notation of the theorems,
and , respectively.
Thus, we have established the following result:
Figure 3. Example of Odd Cycle Union Tree
Theorem 5. Let be odd, be
even, and be the tree
described above. Then is harmonious.
Regarding the proofs, it should be noted that clearly the vertex
labels (with the exception of the recycled label of 0 occurring once in
the cycle and once again as the label of ) are distinct. Thus, the families of
graphs , , and , where is odd and is even, are indeed harmonious. The
proofs work nicely to a large extent because of the existence of the
subgroup whose order divides the order of the group; this occurred by
design, as we chose the order of the cycle (and so the cyclic subgroup)
to divide the order of the graph (and so the cyclic group). In
particular, since , we could use the elements of the
subgroup for the labels of the cycle vertices. We seek
answers to further questions that arise such as the following: Could the
proofs here be modified so to include paths and starlike trees with an
odd number of paths emanating from the root? Or, what about starlike
trees with an even number of paths, but with the paths having a
different specified length other than . These questions would surely require
further investigation. Curiously, we can still apply the same subgroup
labeling to the vertices of the odd cycle when is odd, but the labeling function
presented here yields repeated edge labels on the trees. We leave it to
the reader to verify this. Thus, for the future, we are left to ponder
the question as to whether this disjoint union would be harmonious when
is odd.
References:
West, D. B., 1996. Introduction to Graph Theory. Upper Saddle River, NJ: Prentice Hall.
Gallian, J. A., 2017. A dynamic survey of graph labeling. Electronic Journal of Combinatorics, 20, \# DS6.
Fang, W., 2011. New computational result on harmonious trees. arXiv:1106.3490v1 [cs.DM], 17 June 2011.
Graham, R. L. and Sloane, N. J. A., 1980. On additive bases and harmonious graphs. SIAM Journal on Algebraic and Discrete Methods, 1, pp.382-404.
Gallian, J. A. and Schoenhard, L. A., 2014. Even harmonious labeling. AKCE International Journal of Graphs and Combinatorics, 11(1), pp.27-49.
Renuka, J., Balaganesan, P. and Selvaraju, P., 2012. On Harmonious Labeling. International Journal of Advances in Mathematics and Mathematical Sciences, 1(2), pp.65-70.
Gallian, J. A. and Stewart, D., 2015. Properly even harmonious labelings of disjoint union with even sequential graphs. Journal of Graph Labeling, 1(1), pp.1-10.
Gallian, J.A. and Stewart, D., 2015. Properly even harmonious labelings of disjoint unions with even sequential graphs. AKCE J. Graphs Comin, pp.193-203.
Gallian, J., 2021. Contemporary Abstract Algebra. Chapman and Hall/CRC.