Electronic Journal of Graph Theory and Applications 12 .
, 1Ae16 Properly even harmonious labeling of a union of stars Zachary M.
Henderson University of Minnesota - Duluth hende873@d.
Abstract A function f is defined as an even harmonious labeling on a graph G with q edges if f : V (G) Ie .
, 1, .
, 2.
is an injection and the induced function f O : E(G) Ie .
, 2, .
, 2.
Oe .
} defined by f O .
= f .
is bijective.
A properly even harmonious labeling is an even harmonious labeling in which the codomain of f is .
, 1, .
, 2q Oe .
, and a strongly harmonious labeling is an even harmonious labeling that also satisfies the additional condition that for any two adjacent vertices with labels u and v, 0 < u v O 2q.
In .
Gallian and Schoenhard proved that Sn1 O Sn2 O A A A O Snt is strongly even harmonious for n1 Ou n2 Ou A A A Ou nt and t < n21 2.
In this paper, we begin with the related question AuWhen is the graph of k n-star components.
G = kSn , properly even harmonious?Ay We conclude that kSn is properly even harmonious if and only if k is even or k is odd, k > 1, and n Ou 2.
We also conclude that Sn1 O Sn2 O A A A O Snk is properly even harmonious when k Ou 2, ni Ou 2 for all i and give some additional results on combinations of star and banana graphs.
Keywords: properly even harmonious, star graph, banana tree, graph labeling Mathematics Subject Classification : 05C78 DOI: 10.
5614/ejgta.
Introduction A harmonious graph labeling was first defined by Graham and Sloane .
A function f is a harmonious labeling of a graph G with q edges if f : V (G) Ie .
, 1, .
, q Oe .
is an injection and Received: 15 June 2021.
Revised: 28 May 2023.
Accepted: 24 October 2023.
Properly even harmonious labeling of a union of stars Zachary M.
Henderson the induced function f O : E(G) Ie .
, 1, .
, q Oe .
defined by f O .
= f .
mod q is bijective, and when G is a tree, exactly one vertex label may be used on two vertices.
An even harmonious labeling is a variation of harmonious labeling first defined by Sarasija and Binthiya .
such that a graph with p vertices and q edges is even harmonious if it is possible to label the vertices with distinct integers from the set .
, 1, .
, 2.
in such a way that the set of induced edge labels is a bijection with the set of even integers .
In these labelings, the label of a given edge xy is the sum of the labels of the end vertices x and y, f .
and f .
In our paper, we determine when some unions of stars and banana trees are properly even harmonious.
Preliminaries We start by presenting a formal definition of an even harmonious labeling.
Definition 2.
A function f is an even harmonious labeling of a graph G with q edges if f :
V (G) Ie .
, 1, .
, 2.
is an injection and the induced function f O : E(G) Ie .
, 2, .
, 2.
Oe .
} defined by f O .
= f .
mod 2q is bijective.
Now, we define two variations of an even harmonious labeling with added restrictions.
Definition 2.
A function f is a properly even harmonious labeling of a graph G with q edges if f :
V (G) Ie .
, 1, .
, 2q Oe .
is an injection and the induced function f O : E(G) Ie .
, 2, .
, 2.
Oe .
} defined by f O .
= f .
mod 2q is bijective.
Notice the only difference between Definition 2.
1 from Definition 2.
2 is that for f to be a properly even harmonious labeling, it cannot use the vertex label 2q.
Definition 2.
A strongly even harmonious labeling is an even harmonious labeling that also satisfies the additional condition that for any two adjacent vertices with labels u and v, we have 0 < u v O 2q.
The following result by Gallian and Schoenhard was the main motivation for our original research.
We asked the question AuIf we relax the condition of strongly even harmonious to properly even harmonious, when does the graph kSn have a properly even harmonious labeling?Ay Theorem 2.
(Gallian and Schoenhard .
) Sn1 O Sn2 O A A A O Snk is strongly even harmonious for n1 Ou n2 Ou A A A Ou nk and k < n21 2.
Additionally, we reference the following theorems to complete some remaining cases in our Notice that P3 = S2 .
P2 = S1 , and Sn is a tree, so Theorems 2.
2, 2.
3, and 2.
4 apply to unions of stars.
Theorem 2.
(Gallian and Stewart .
) Pm O Sn1 O Sn2 O A A A O Snk is properly even harmonious when m > 2 and at least one ni is greater than 1.
Theorem 2.
(Gallian and Schoenhard .
) kP2 is properly even harmonious if and only if k is Theorem 2.
(Gallian and Schoenhard .
) A tree cannot have a properly even harmonious labeling.
Properly even harmonious labeling of a union of stars Zachary M.
Henderson Results We begin by showing that Sn1 O Sn2 O A A A O Snk is properly even harmonious when ni Ou 2.
Lemma 3.
1 reduces this situation to two additional cases.
Then, after slightly altering the labeling method, weAore able to produce the desired result (Theorem 3.
Lemma 3.
Sn1 O Sn2 O A A A O Snk is properly even harmonious when ni Ou 2 for all i, k Ou 2, and n1 , n2 Ou 3.
Proof.
Consider G = Sn1 O Sn2 O A A A O Snk , the union of k ni -stars where ni Ou 2.
Then, there are q = ki=1 ni edges in G and the labeling is done modulo 2q.
Because all edge labels are even, all vertex labels in a given star must have the same even/odd parity.
Throughout the construction, we will use a to determine which parity the vertex labels of the stars have.
We define a for the remainder of the construction as follows:
A if k even, let a = k2 .
A if k odd, let a = kOe1 Without loss of generality, we can assume n1 Ou n2 Ou A A A Ou nk .
Now, label G in the following For 1 O i O k, let vi,0 be the center vertex of Sni , vi,j be the leaf vertices of Sni .
O j O ni ), and ei,j = vi,0 vi,j .
When no confusion may arise, we will identify a vertex/edge with its associated vertex/edge label.
instead of saying Auvertex vi,j with label f .
i,j ) = cAy, we will say vi,j = c.
Then, label the vertices and edges of Sn1 , .
Sna as follows .
ee Figures 1 and .
ll computations are assumed to be modulo 2.
Recursively label all edges with e1,1 = 0, ei,1 = eiOe1,ni 2, and ei,j = ei,jOe1 2 When 1 O i O a, recursively label the center vertices with v1,0 = q Oe 1 and vi,0 = viOe1,0 Oe 2.
Then label the leaf vertices .
= .
recursively with v1,1 = q 1, vi,j = vi,jOe1 2, and vi,1 = viOe1,niOe1 4.
Particularly, the closed form for these labels are vi,0 = q Oe 1 Oe 2.
Oe .
= q 1 Oe 2i vi,j = q 2( iOe1 nEe ) 2i 2j Oe 3 for j = 0 Ee=1 iOe1 ei,j = 2( nEe ) 2j Oe 2 Ee=1 Now, the labels of Sna 1 , .
Snk will depend on ea 1,1 , the first edge in Sna 1 .
Let q A = be the total number of edges in the first a stars and label Sna 1 , .
Snk as follows:
Ee=1 nEe Case 1.
if q Oe 1 and q A have the same even/odd parity:
When a 1 O i O k, recursively label the center vertices with va 1,0 = q A Oe 1 and vi,0 = viOe1,0 Oe 2.
Then label the leaf vertices recursively with va 1,1 = q A 1, vi,j = vi,jOe1 2, and vi,1 = viOe1,niOe1 4.
Properly even harmonious labeling of a union of stars Zachary M.
Henderson Particularly, the closed form for these labels are vi,0 = q A Oe 1 Oe 2.
Oe a Oe .
= q A 1 Oe 2.
Oe .
vi,j = 2( iOe1 nEe ) q A 1 2.
Oe a Oe .
Oe .
for j = 0 Ee=a 1 ei,j = 2q A 2( iOe1 nEe ) 2.
Oe .
= 2( iOe1 Ee=a 1 nEe ) 2j Oe 2 Ee=1 Claim.
This labeling is properly even harmonious when k is even.
Proof.
Considering each parity of vertex labels separately:
Since all edges are of the form ei,j = vi,0 vi,j , show f O .
i,j ) = f .
i,0 ) f .
i,j ).
For Sn1 .
Sna , 1 O i O a and 1 O j O ni .
Then, f O .
i,j ) = f .
i,0 ) f .
i,j ) mod 2q since:
f O .
i,j ) = f .
i,0 ) f .
i,j ) iOe1 nEe ) 2j Oe 2 = q 2( nEe ) 2i 2j Oe 3 q 1 Oe 2i Ee=1 iOe1 Ee=1 iOe1 iOe1 nEe ) 2j Oe 2 = 2q 2( nEe ) 2j Oe 2 Ee=1 Ee=1 which are equivalent modulo 2q.
For Sna , .
Snk , a 1 O i O k and 1 O j O ni .
Then, f O .
i,j ) = f .
i,0 ) f .
i,j ) mod 2q since:
f O .
i,j ) = f .
i,0 ) f .
i,j ) A 2q 2( iOe1 iOe1 nEe ) 2j Oe 2 = 2( Ee=a 1 nEe ) q A 1 2.
Oe a Oe .
Oe .
Ee=a 1 q Oe 1 Oe 2.
Oe a Oe .
A 2q 2( iOe1 nEe ) 2j Oe 2 = 2q 2( Ee=a 1 iOe1 nEe ) 2j Oe 2 Ee=a 1 which are equal .
nd hence equivalent modulo 2.
Additionally, f O : E(G) Ie .
, 2, .
, 2.
Oe .
} is a bijection since all desired edge labels are used exactly once.
Now, in order for f to be injective, we need to make sure no vertex label is duplicated.
Since we labeled the centers in decreasing order and the leaves in increasing order, we need to compare the smallest center label with the largest leaf label of Properly even harmonious labeling of a union of stars Zachary M.
Henderson each parity of labels.
Specifically, since our computations are modulo 2q and the center labels are smaller than the leaf labels .
ithout the modulu.
, we need to show va,na Oe 2q < va,0 and vk,nk Oe 2q < vk,0 .
For Sn1 , .
Sna , va,na Oe 2q < va,0 q 2( aOe1 nEe ) 2a 2na Oe 3 Oe 2q < q 1 Oe 2a Ee=1 aOe1 nEe ) 4a 2na Oe 4 < 2q Ee=1 aOe1 .
Ee ) 2a na Oe 2 < Ee=1 2a Oe 2 < nEe Ee=1 nEe Ee=a 1 and since nEe Ou 2 for all Ee and if k even, a = k2 , so the sum has k Oe .
1 = a terms.
if k odd, a = kOe1 , so the sum has k Oe .
1 = a 1 terms.
we get that nEe 2a Oe 2 < 2a O Ee=a Hence, no vertex label used in Sn1 , .
Sna is duplicated.
For Sna 1 , .
Snk , vk,nk Oe 2q < vk,0 kOe1 nEe ) q A 1 2.
Oe a Oe .
k Oe .
Oe 2q < q A 1 Oe 2.
Oe .
Ee=a 1 kOe1 nEe ) 4k Oe 4a 2nk Oe 4 < 2q Ee=a 1 kOe1 .
Ee ) 2k Oe 2a nk Oe 2 < Ee=a 1 nEe Ee=1 2k Oe 2a nk Oe 2 < ( nEe ) nk Ee=1 2k Oe 2a Oe 2 O nEe Ee=1 and since n1 , n2 Ou 3, nEe Ou 2 for all Ee, and Properly even harmonious labeling of a union of stars Zachary M.
Henderson if k even, a = k2 , 2k Oe 2a Oe 2 O nEe Ee=1 k Oe 2 < 2a 1 O kOe2<k 1O nEe Ee=1 nEe Ee=1 where we have a lower bound of 2a 1 since if a = 1, n2 wouldnAot be in the if k odd, a = kOe1 2k Oe 2a Oe 2 < nEe Ee=1 k Oe 1 < 2a 1 O nEe Ee=1 kOe1<k O nEe Ee=1 where we again have a lower bound of 2a 1 since if a = 1, n2 wouldnAot be in the sum.
Hence, no vertex label used in Sna 1 , .
Snk is duplicated.
Furthermore, since the vertex labels used in Sn1 , .
Sna and Sna 1 , .
Snk are of different parities, there are no duplicate vertex labels.
Hence f is injective, f O is bijective, and f O .
i,j ) = f .
i,0 ) f .
i,j ), so the given labeling is properly even harmonious.
Figure 1: A properly even harmonious labeling for S5 O S4 O S3 O S2 .
q Oe 1 = 13 and q A = 9 have the same parity Properly even harmonious labeling of a union of stars Zachary M.
Henderson Case 2.
if q Oe 1 and q A have different even/odd parities:
When a 1 O i O k, recursively label the center vertices with vi,a 1 = q A Oe 2 and vi,0 = viOe1,0 Oe 2.
Then label the leaf vertices recursively with va 1,1 = q A 2, vi,j = vi,jOe1 2, and vi,1 = viOe1,niOe1 4.
Particularly, the closed form for these labels are:
vi,0 = q A Oe 2 Oe 2.
Oe a Oe .
= q A Oe 2.
Oe .
iOe1 vi,j = 2( nEe ) q A 2 2.
Oe a Oe .
Oe .
for j = 0 Ee=a 1 ei,j = 2q 2( iOe1 nEe ) 2.
Oe .
= 2( Ee=a 1 iOe1 nEe ) 2j Oe 2 Ee=1 Claim.
This labeling is properly even harmonious.
Proof.
The proof for Sn1 , .
Sna is the same as the previous proof since the cases didnAot affect this labeling.
So, consider Sna 1 , .
Snk .
For a 1 O i O k, f O .
i,j ) = f .
i,0 ) f .
i,j ) since f O .
i,j ) = f .
i,0 ) f .
i,j ) A 2q 2( iOe1 iOe1 nEe ) 2j Oe 2 = 2( Ee=a 1 nEe ) q A 2 2.
Oe a Oe .
Oe .
Ee=a 1 q Oe 2.
Oe .
A 2q 2( iOe1 nEe ) 2j Oe 2 = 2q 2( Ee=a 1 iOe1 nEe ) 2j Oe 2 Ee=a 1 which are equal .
nd hence equivalent mod 2.
Additionally, f O : E(G) Ie .
, 2, .
, 2.
Oe .
} is a bijection since all desired edge labels are used exactly once.
Properly even harmonious labeling of a union of stars Zachary M.
Henderson .
Now, in the same way as the previous proof, we must check that vk,nk Oe 2q < vk,0 .
vk,nk Oe 2q < vk,0 kOe1 nEe ) q A 2 2.
Oe a Oe .
k Oe .
Oe 2q O q A Oe 2.
Oe .
Ee=a 1 kOe1 nEe ) 4k Oe 4a 2nk Oe 2 O 2q Ee=a 1 kOe1 nEe ) 2k Oe 2a nk Oe 1 O Ee=a 1 nEe Ee=1 2k Oe 2a nk Oe 1 O ( 2k Oe 2a Oe 1 O nEe ) nk Ee=1 nEe Ee=1 Then, since n1 , n2 Ou 3 and nEe Ou 2, if k even, a = k2 , 2k Oe 2a Oe 1 < nEe Ee=1 k Oe 1 < 2a 1 O kOe1<k 1O nEe Ee=1 nEe Ee=1 where we have a lower bound of 2a 1 since if a = 1, n2 wouldnAot be in the if k odd, a = kOe1 if a Ou 2, we have 2k Oe 2a Oe 1 < nEe Ee=1 k < 2a 2 O nEe Ee=1 k <k 1O nEe Ee=1 and if a = 1, then k = 3 and we have 3 possibilities:
Properly even harmonious labeling of a union of stars Zachary M.
Henderson If G = S3 O S3 O S2 , then q = 8 so q Oe 1 = 7 and q A = 3 have the same parity so G doesnAot belong to this case.
If G = S3 O S3 O S3 , this labeling method doesnAot work.
However, with a slight adjustment, we get that G is properly even harmonious (Figure .
Figure 2: A properly even harmonious labeling for 3S3 .
Otherwise, n1 Ou 4, so 2a 2 O Ee=1 nEe as above, so we get k <k 1O nEe Ee=1 Hence, no vertex label used in Sna 1 , .
Snk is duplicated.
Furthermore, since the vertex labels used in Sn1 , .
Sna and Sna 1 , .
Snk are of different parities, there are no duplicate vertex labels.
Hence f is injective, f O is bijective, and f O .
i,j ) = f .
i,0 ) f .
i,j ), so the given labeling is properly even harmonious.
Figure 3: A properly even harmonious labeling for S3 O S3 O S2 O S2 O S2 .
q Oe 1 = 11 and q A = 6 have different Lemma 3.
S3 O S2 O A A A O S2 is properly even harmonious when there are an odd number of S2 .
Proof.
Let k be the number of stars in G = S3 O S2 O A A A O S2 .
Since there are an odd number of S2 , k is even.
Let q = 2k 1 be the number of edges in G, so calculations are done modulo 2q = 4k 2.
Properly even harmonious labeling of a union of stars Zachary M.
Henderson Figure 4: A properly even harmonious labeling for S3 O S2 O A A A O S2 for an odd number of S2 Label the vertices of the first k2 copies of S2 as given in Figure 4:
Label the centers with 2k 2 Oe 2i for 1 O i O k2 .
Label the top vertices with 2k Oe 4 6i for 1 O i O k2 .
Label the bottom vertices with 2k Oe 2 6i for 1 O i O k2 .
Since the labels k 2, k 4, .
, kOe4, kOe6 are in ascending order and kOe6 < k 2, all even vertex labels are distinct modulo 2q.
Additionally, label the edges of these stars with .
, 2, .
, 2k Oe .
as in Figure 4.
Now, consider the remaining k2 Oe 1 2-stars and 3 star.
Label the vertices of the k2 Oe 1 2-stars as given in Figure 4:
Label the centers with k 1 Oe 2i for 1 O i O k2 Oe 1.
Label the top vertices with k Oe 5 6i for 1 O i O k2 Oe 1.
Label the bottom vertices with k Oe 3 6i for 1 O i O k.
2 Oe 1.
and the 3-star as given in Figure 4:
Label the center with 1.
Label the vertices with 4k Oe 5, 4k Oe 3, 4k Oe 1.
Since the labels 1, 3, .
, 4k Oe 3, 4k Oe 1 are in ascending order and 4k Oe 1 < 2q = 4k 2, all odd vertex labels are distinct modulo 2q.
Additionally, label the edges with the set .
k, 2k 2, .
, 4.
as in Figure 4.
Now, since all vertex labels are distinct, f : V (G) Ie .
, 1, .
, 2q Oe .
is injective.
Furthermore, f O : E(G) Ie .
, 2, .
, 2.
Oe .
} is bijective and, by construction, f O .
= f .
for edge uv between vertices u and v.
Hence, this labeling is properly even harmonious.
Lemma 3.
S3 O S2 O A A A O S2 is properly even harmonious when there are an even number of S2 .
Proof.
Let k be the number of stars in G = S3 O S2 O A A A O S2 .
Since there are an even number of S2 , k is odd and k = 2a 1 for some integer a.
Let q = 2k 1 = 4a 3 be the number of edges in G, so calculations are done modulo 2q = 4k 2 = 8a 6.
Properly even harmonious labeling of a union of stars Zachary M.
Henderson Figure 5: A properly even harmonious labeling for S3 O S2 O A A A O S2 for an even number of S2 Label the vertices of the first a 1 copies of S2 as given in Figure 5:
Label the centers with 4a 4 Oe 2i for 1 O i O a 1.
Label the top vertices with 4a Oe 2 6i for 1 O i O a 1.
Label the bottom vertices with 4a 6i for 1 O i O a 1.
Since the labels 2a 2, 2a 4, .
, 2aOe4 are in ascending order and 2aOe4 < 2a 2, all even vertex labels are distinct modulo 2q.
Additionally, label the edges of these stars with .
, 2, .
, 4a Oe .
as in Figure 5.
Now, consider the remaining a Oe 1 2-stars and 3-star.
Label the vertices of the a Oe 1 2-stars as given in Figure 5:
Label the centers with 2a 3 Oe 2i for 1 O i O a Oe 1.
Label the top vertices with 2a Oe 3 6i for 1 O i O a Oe 1.
Label the bottom vertices with 2a Oe 1 6i for 1 O i O a Oe 1.
and the 3-star as given in Figure 5:
Label the center with 3.
Label the vertices with 8a Oe 3, 8a Oe 1, 8a 1.
Since the labels 3, 5, .
, 8a 4 are in ascending order and 8a 4 < 2q = 8a 6, all odd vertex labels are distinct modulo 2q.
Additionally, label the edges with the set .
a 4, 4a 6, .
, 8a .
as in Figure 5.
Now, since all vertex labels are distinct, f : V (G) Ie .
, 1, .
, 2q Oe .
= .
, 1, .
, 8a .
is injective.
Furthermore, f O : E(G) Ie .
, 2, .
, 2.
Oe .
} = .
, 2, .
, 8a .
is bijective and, by construction, f O .
= f .
for edge uv between vertices u and v.
Hence, this labeling is properly even harmonious.
Lemma 3.
kS2 is properly even harmonious.
Proof.
This was proved by Gallian and Stewart in .
as Theorem 3.
4: Authe graph Pn OSt1 OA A AOStk is properly even harmonious for n > 2 and at least one ti is greater than 1.
Ay Here, we use the fact that P3 = S2 and let ti = 2 for all i to get the result.
Properly even harmonious labeling of a union of stars Zachary M.
Henderson Theorem 3.
Sn1 O Sn2 O A A A O Snk is properly even harmonious when ni Ou 2 for all i and k Ou 2.
Proof.
The result follows from Lemmas 3.
1, 3.
2, 3.
3, and 3.
Now, if we consider kSn = Sn O Sn O A A A O Sn , we can completely determine which graphs of this type are properly even harmonious.
Using Theorem 3.
1, we need only look at kS1 and the graph of a single star Sn .
Lemma 3.
Sn is not properly even harmonious.
Proof.
This was proved by Gallian and Schoenhard in .
as a case of Theorem 3.
1: AuA tree cannot have a properly even harmonious labeling.
Ay Since Sn is a tree, we get the result.
Lemma 3.
kS1 is properly even harmonious if and only if k is even.
Proof.
This was proved by Gallian and Schoenhard in .
as Theorem 4.
1: AunP2 is properly even harmonious if and only if n is even.
Ay Here, we use the fact that P2 = S1 to get the result.
Now, by collecting the results from above, we can prove the following theorem.
Theorem 3.
gives necessary and sufficient condition for the values of n and k that make G = kSn properly even harmonious, which is the solution to our original research question.
Theorem 3.
kSn is properly even harmonious if and only if:
k is even and n Ou 1, or k is odd, k > 1, and n Ou 2.
Proof.
This follows from Theorem 3.
Lemma 3.
5, and Lemma 3.
Now, we can make an interesting observation from Theorem 3.
In .
Barrientos and Youssef cite YoussefAos previous result in .
that if G is harmonious, then nG is also harmonious given that n > 0 is odd.
This same relationship is also applicable to the properly even harmoniousness of G = kSn as given in the following observation.
Observation 3.
If G = kSn is properly even harmonious, then k A G is also properly even harmonious for any k A > 0.
Proof.
Using the fact that k A G = k A .
Sn ) = .
A .
Sn .
Case 1.
if k is even, then k A k is even and .
A .
Sn is properly even harmonious.
Case 2.
if k is odd, k A is even, then k A k is even and .
A .
Sn is properly even harmonious.
Case 3.
if k is odd, k A is odd, then since kSn is properly even harmonious, k > 1 and n Ou 2.
Thus, k A k is odd and k A k Ou k > 1 and n Ou 2, so .
A .
Sn is also properly even harmonious.
Additionally, we give a properly even harmonious labeling for B2,k O B2,k where Bn,k is an .
, .
Oebanana tree.
Combining the labeling method from Theorem 3.
3 with the method from Theorem 3.
1, weAore able to produce a properly even harmonious labeling in Theorems 3.
4 and 3.
Properly even harmonious labeling of a union of stars Zachary M.
Henderson Theorem 3.
B2,k O B2,k is properly even harmonious.
Proof.
Since we have two disconnected components, we label the first banana tree.
B2,k , with even vertex labels and the second banana tree.
B2,k , with odd vertex labels as given in Figure 6.
Figure 6: A properly even harmonious labeling for G = B2,k O B2,k Let q = 4k 4 be the number of edges in G, so calculations are done modulo 2q = 8k 8.
Consider B2,k Label the vertices of a path of length 7 in B2,k with the sequence 4, 2q Oe 4, 6, 2q Oe 2, 8, 0, 10, the remaining leaf vertices of the second star with 12, 14, .
, 10 2.
Oe .
= 2k 6, and the remaining leaf vertices of the first star with 2k 12, 2k 14, .
, 4k 6.
Since the leaf vertices are labeled in increasing order, the largest leaf label is 4k 6 < 8k 4 = 2q Oe 4 .
he AusmallestAy label of B2,k if considered as Oe.
for all k > 1, so all even vertex labels are distinct modulo 2q.
Additionally, label the edges of B2,k with the set .
, 2, .
, 4k .
as given in Figure 6.
Now, consider B2,k .
Label the vertices of a path of length 7 in B2,k with the sequence 2k 5, 2k Oe 1, 2k 7, 2k 1, 2k 9, 2k 3, 2k 11, the additional leaf vertices of the second star with 2k 13, 2k 15, .
, 4k 7, and the additional leaf vertices of the first star with 4k 13, 4k 15, .
, 6k 7.
Since the leaf vertices are labeled in increasing order, the largest leaf label is 6k 7 < 8k 8 = 2q and the smallest label is 2k Oe 1 > 0, so all odd vertex labels are distinct modulo 2q.
Additionally, label the edges of B2,k with the set .
k 4, 4k 6, .
, 8k .
as given in Figure 6.
Now, since all vertex labels are distinct, f : V (G) Ie .
, 1, .
, 2q Oe .
is injective.
Furthermore, f O : E(G) Ie .
, 2, .
, 2.
Oe .
} is bijective and, by construction, f O .
= f .
for edge uv between vertices u and v.
Hence, this labeling is properly even harmonious.
Theorem 3.
B2,k1 O Sk2 is properly even harmonious if k2 > 2.
Proof.
Since we have two disconnected components, label B2,k1 with even labels and Sk2 with odd vertex labels as given in Figure 7.
Properly even harmonious labeling of a union of stars Zachary M.
Henderson Figure 7: A properly even harmonious labeling for G = B2,k1 O Sk2 Let q = 2k1 k2 2 be the number of edges in G, so calculations are done modulo 2q = 4k1 Label the vertices of a path of length 7 in B2,k1 with the sequence 3, 2qOe3, 5, 2qOe1, 7, 1, 9, the remaining leaf vertices of the second star with 11, 13, .
, 2k1 5, and the remaining leaf vertices of the first star with 2k1 11, 2k1 13, .
, 4k1 5.
Since the leaf vertices are labeled in increasing order, the largest leaf label is 4k1 5 < 2q Oe 3 .
he AusmallestAy label of B2,k1 if considered as -.
, so all odd vertex labels are distinct modulo 2q.
This comes from the inequality 4k1 5 < 2q Oe 3 4k1 5 < 2.
k1 k2 .
Oe 3 2 < k2 since k2 > 2.
Additionally, label the edges of B2,k1 with .
, 2, .
as given in Figure 7.
Now, consider Sk2 .
Label the center vertex with 2k1 which is even and the leaf vertices with 2k1 2 2i for 1 O i O k2 as in Figure 7.
Since the leaves are labeled in increasing order and 2k1 2k2 2 < 2q, all of the even vertex labels are distinct.
Additionally, label the edges of Sk2 with .
k1 4, 4k1 6, .
, 4k1 2k2 .
as in Figure 7.
Now, since all vertex labels are distinct, f : V (G) Ie .
, 1, .
, 2q Oe .
is injective.
Furthermore, f O : E(G) Ie .
, 2, .
, 2.
Oe .
} is bijective and, by construction, f O .
= f .
for edge uv between vertices u and v.
Hence, this labeling is properly even harmonious.
Theorem 3.
B2,k1 O Sk2 O Sk3 is properly even harmonious if k2 > 1.
Proof.
Label B2,k1 with odd vertex labels and label Sk2 .
Sk3 with even vertex labels as given in Figure 8 Properly even harmonious labeling of a union of stars Zachary M.
Henderson Figure 8: A properly even harmonious labeling for G = B2,k1 O Sk2 O Sk3 Let q = 2k1 k2 k3 2 be the number of edges in G, so calculations are done modulo 2q = 4k1 2k2 2k3 4.
Label the vertices of a path of length 7 in B2,k with the sequence 3, 2q Oe 3, 5, 2q Oe 1, 7, 1, 9, the remaining leaf vertices of the second star with 11, 13, .
, 2k1 5, and the remaining leaf vertices of the first star with 2k1 11, 2k1 13, .
, 4k1 5.
Since the leaf vertices are labeled in increasing order, the largest leaf label is 4k1 5 < 2q Oe 3 .
he AusmallestAy label of B2,k1 if considered as -.
, so all odd vertex labels are distinct modulo 2q.
This comes from the inequality 4k1 5 < 2q Oe 3 4k1 5 < 2.
k1 k2 k3 .
Oe 3 2 < k2 k3 since k2 > 1.
Additionally, label the edges of B2,k1 with .
, 2, .
as given in Figure 8.
Now, consider Sk2 and Sk3 .
Label the center vertex of Sk2 with 2k1 and the center vertex of Sk3 with 2k1 Oe 2, which are both even.
Label the leaf vertices of Sk2 with 2k1 2 2i for 1 O i O k2 and the leaf vertices of Sk3 with 2k1 2k2 4 2i for 1 O i O k3 as in Figure Since the leaves are labeled in increasing order, the largest leaf is 2k1 2k2 2k3 4 < Properly even harmonious labeling of a union of stars Zachary M.
Henderson 2q, and the smallest center is 2k1 Oe 2 Ou 0, so all of the even labels are distinct.
Additionally, label the edges of Sk2 with .
k1 4, 4k1 6, .
, 4k1 2k2 .
and the edges of Sk3 with .
k1 2k2 4, 4k1 2k2 6, .
, 4k1 2k2 2k3 .
as in Figure 8.
Now, since all vertex labels are distinct, f : V (G) Ie .
, 1, .
, 2q Oe .
is injective.
Furthermore, f O : E(G) Ie .
, 2, .
, 2.
Oe .
} is bijective and, by construction, f O .
= f .
for edge uv between vertices u and v.
Hence, this labeling is properly even harmonious.
Acknowledgement The author would like to thank his advisor Dr.
Dalibor Froncek for his help and guidance on this project.
Dr.
Joseph Gallian and Dr.
Xuan Li for serving on his committee with their valuable comments and feedback, his professors and classmates who gave him a great experience at UMD, his friends, family, and wife who have supported him, and God for His grace throughout the entire References