ENUMERATION OF 1-REGULAR SEMIGRAPHS OF ORDER p WITH THREE MUTUALLY ADJACENT m-VERTICES

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International Journal of Advanced Research in Engineering and Technology (IJARET)
Volume 11, Issue 1, January 2020, pp. 235-256, Article ID: IJARET_11_01_027
Available online at https://iaeme.com/Home/issue/IJARET?Volume=11&Issue=1
ISSN Print: 0976-6480 and ISSN Online: 0976-6499
DOI: 10.34218/IJARET.11.1.2020.027
© IAEME Publication Scopus Indexed
ENUMERATION OF 1-REGULAR SEMIGRAPHS
OF ORDER p WITH THREE MUTUALLY
ADJACENT m-VERTICES
M.S. Chithra
Post Graduate and Research Department of Mathematics,
Thiagarajar College, Madurai, Tamil Nadu, India.
ABSTRACT
For a vertex 𝑣v in a semigraph 𝐺 = (𝑉, 𝑋), 𝑑𝑒𝑔 𝑣 G=(V,X), deg v is the number of
edges having v𝑣 as an end vertex. A semigraph G𝐺 is said to be k-regular if all its
vertices have degree k. In this paper, we classify the class of all 1-regular semigraphs
with exactly three 𝑚m-vertices into four categories and we enumerate the non-
isomorphic semigraphs in one of these categories.
2010 Mathematics Subject Classification: 05C30
Key words: Semigraph, regular semigraph, degree, m-vertex.
Cite this Article: M.S. Chithra, Enumeration of 1-Regular Semigraphs of Order p with
Three Mutually Adjacent m-Vertices, International Journal of Advanced Research in
Engineering and Technology, 11(1), 2020, pp. 228-234.
https://iaeme.com/Home/issue/IJARET?Volume=11&Issue=1
1. INTRODUCTION
Sampathkumar [14,15,16] introduced a new generalization of graphs called
Semigraphs, in 2000. Semigraphs look like graphs when drawn on a plane, where every
concept/result in graph can be easily generalized yielding a rich variety of corresponding
results. In fact, the beauty of semigraphs lies in the variety of definitions/concepts, all of which
coincide for graphs.
In last two decades, lot of research has been done in this area of Semigraph. Domination in
semigraphs, Matrix representation of semigraphs, Complete semigraphs, Edge complete
semigraphs, e-Adjacency matrix and e-Laplacian matrix of semigraphs, Adjacency matrix of
semigraphs, Energy of semigraphs are studied in [1,2,3,4,7,11,12,17].
Graph Theory as an applied science, its concepts and results have a wide range of
applications in Network Theory, Modelling of Chemical Phenomena, Electrical Circuits,
Ecological System, Logistics and so on. Semigraph Theory, as an extension of graph theory, is
applied in various fields such as DNA splicing system, Neural Networks, Extractive
Summarization, Wireless Sensor Networks [5,6,13,18] and so on.

Enumeration of 1-Regular Semigraphs of Order p with Three Mutually Adjacent m-Vertices
The main objective of this paper is to study the structure and to enumerate 1-regular
semigraphs with three m-vertices.
We first classify 1-regular semigraphs with three m-vertices into four types A, B, C, D,
according to their structure.
Next, Type A is classified into two subclasses 𝒜1 and 𝒜2; further 𝒜1 is categorized into 3
categories and 𝒜2 into 7 categories. In particular, to categorize the semigraphs in 𝒜2, we apply
a new technique for labelling the edges and the semigraphs in 𝒜2.
Finally, we study the isomorphism and enumerate the semigraphs of order p, in each of
these categories.
2. PRELIMINARIES
Definition 2.1. A semigraph G is a pair (V,X), where V is a nonempty set, whose elements are
called vertices of G, and X is a set of n-tuples, called edges of G, of distinct vertices, for various
n ≥ 2, satisfying the following conditions:
(i) Any two edges have at most one vertex in common.
(ii) Two edges (u1,u2,…,un) (𝑢1, 𝑢2, … , 𝑢𝑛)and (v1,v2,…,vm) (𝑣1, 𝑣2, … , 𝑣𝑚)are considered to
be equal if and only if m = n and either ui = vi 𝑢𝑖 = 𝑣𝑖for 1 ≤ i ≤ n, or ui = vn-i+1 𝑢𝑖 = 𝑣𝑛−𝑖+1
for 1 ≤ i ≤ n .
((Thus the edge (u1,u2,…,un) is the same as the edge (un,un-1,…,u1)).
Definition 2.2. If E = (v1,v2,…,vn) is an edge of G𝐺, then v1 and vn are the end vertices of E
𝐸 and vi’s are the middle vertices (or m-vertices) of E𝐸, for 2 ≤ i ≤ n-1. A vertex is said to be
a m−vertex, if it is a middle vertex of at least one edge. A vertex is said to be an end vertex, if
it is not a middle vertex of any edge.
In semigraphs, the end vertices and middle vertices are represented by thick dots and small
circles respectively.
Definition 2.3. Two vertices are adjacent if both of them belong to an edge, and two edges are
adjacent if they have a common vertex.
Definition 2.4. Two semigraphs G1=(V1,X1) and G2=(V2,X2)𝐺2 = (𝑉2, 𝑋2) are isomorphic if
there exists a bijection f : V1 → V2 such that E = (v1,v2,…,vn) is an edge in G1𝐺1 iff
(f(v1),f(v2),…,f(vn)) is an edge in G2.𝐺2.
Definition 2.5. A n-semiedge is an edge containing n vertices.
Definition 2.6. For a vertex v𝑣 in a semigraph 𝐺 = (𝑣, 𝑥) G = (V,X), various types of degrees
are defined as follows:
Degree : 𝑑𝑒𝑔 𝑣 deg v is the number of edges having v as an end vertex.
Edge Degree : dege v is the number of edges containing v.𝑣.
Adjacent Degree : 𝑑𝑒𝑔 𝑣 dega v is the number of vertices adjacent to v.𝑣.
Consecutive Adjacent Degree : 𝑑𝑒𝑔 𝑣 degca v is the number of vertices which are
consecutively adjacent to v.
Definition 2.7. A semigraph is said to be k-regular if all its vertices have degree k.
3. KNOWN RESULTS
The following results [8] deal with 1-regular semigraphs with one or two m-vertices.
The number of non-isomorphic 1-regular semigraphs of order 𝑝 p with
Result 1.
exactly one m-vertex is
.
1
2
−
p

M.S. Chithra
Result 2. The number of non-isomorphic 1-regular semigraphs of order p with exactly two m-
vertices is
(𝑝−4)(5𝑝−8)
16
if 𝑝 ≡
0(𝑚𝑜𝑑4),
(𝑝−2)(5𝑝−18)
16
Observation. In a k-regular semigraph of order 𝑝 p and size q, and so
every 1-regular semigraph is of even order.
If G is a 1-regular semigraph, then GUrK2 𝐺𝑈𝑟𝐾2is also a 1-regular semigraph, for any r ϵ N.
𝑟 ∈ 𝑁.
4. CLASSIFICATION OF 1-REGULAR SEMIGRAPHS WITH THREE
M-VERTICES
We classify the 1-regular semigraphs with three m-vertices, according to their structure. In
particular, we classify according to the adjacencies between the three m-vertices.
Let H be the class of all 1-regular semigraphs with exactly three m-vertices (say) x, y and
z. The semigraphs in 𝐻 H can be classified into four types as follows:
Type A : x, y, z are mutually adjacent
Type B : Exactly two pairs of vertices x, y, z are adjacent
Type C 𝐶: Exactly one pair of vertices x, y, z are adjacent
Type D : x, y, z are not adjacent.
Theorem 4.1. If G 1. 𝑙𝑓𝐺is a semigraph of Type i and H is a semigraph of Type j, 𝑖, 𝑗 ∈
{𝐴, 𝐵, 𝐶, 𝐷}, 𝑖 ≠ 𝑗, i,j ϵ{A, B, C, D}, i ≠ j, then G𝐺 is not isomorphic to H.𝐻.
Proof: Since isomorphism preserves the adjacency of the vertices, the result follows easily.
In this paper, we study the semigraphs of Type A in detail.
The semigraphs of Type B, C and D are studied in [9,10].
5. SEMIGRAPHS OF TYPE A
Let 𝒜 denote the class of all semigraphs of Type A.
For every G ϵ 𝒜, x, y, z are mutually adjacent. Semigraphs of Type A are classified
according as
(i) x, y, z lie on the same edge
(ii) x, y, z do not lie on the same edge.
Let 𝒜1, 𝐴1, 𝐴2 𝒜2 denote the class of all semigraphs in 𝒜 𝐴in which (i), (ii) hold respectively.
The next result follows easily and so we state the result without proof.
Theorem 5.1.𝑙𝑓𝐺 ∈ 𝐴1 If G ϵ 𝒜1 and H ϵ 𝒜2𝐻 ∈ 𝐴2, then G𝐺 is not isomorphic to H.
6. SEMIGRAPHS IN 𝒜1𝐴1
The class 𝒜1𝐴1 is further classified into three categories according as
(i) x, y, z lie on a 5-semiedge
(ii) x, y, z lie on a 𝑎44-semiedge
(iii) x, y, z lie on a 3-semiedge.
,
16
)
8
5
)(
4
( −
− p
p
),
4
(mod
0

p ,
16
)
18
5
)(
2
( −
− p
p
).
4
(mod
2

p
2
kp
q =

Let 𝐴11, 𝐴12, 𝐴13 𝒜11, 𝒜12, 𝒜13 denote the class of all semigraphs in 𝒜1 in which (i), (ii), (iii)
hold respectively. By the classification of 𝒜1, the next result follows.
Theorem 6.1. 𝑙𝑓𝐺 ∈ 𝐴1𝑖If G ϵ 𝒜1i and H ϵ 𝒜1j, i,j ϵ {1,2,3}, i ≠ j, then G is not isomorphic to
H.
Let 𝒜′1i 𝐴1𝑖
′
denote the class of all semigraphs in 𝐴1𝑖 𝒜1i of order p𝑝, having no component
of order 2, for i = 1,2,3.
We first analyze the isomorphism between the semigraphs in 𝐴1𝑖
′
𝒜′1i and then we
enumerate the semigraphs in 𝒜1i, for i = 1,2,3.
6.1. Semigraphs in 𝒜′11𝐴11
′
In any semigraph 𝐺 G in 𝒜′11, x, y, z lie on a 5-semiedge. Hence there are edges of the form
(u1,x,y,z,u2), (x,w1), (y,w2) and (z,w3). (Note that the order of 𝐺 G is at least 8). Furthermore, all
the vertices other than u1, u2, w1, w2, w3, x, y and z, lie in a 3-semiedge with a m−vertex x, y or
z.
If there are r (≥ 0) 3-semiedges with x as a m-vertex and s (≥ 0) 3-semiedges with z as a
m-vertex, then there are 3-semiedges with y as a m-vertex.
have 0 ≤ 𝑟 ≤
Since we
𝑝−8
2
, 0 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
, and
p ≥ 8.
semigraph in 𝐴11
′
𝒜′11 can be
Hence any
denoted by , where with
V = {u1, u2, w1, w2, w3, x, y, z, a1, … , a2r, b1, …, b2t, c1, …, c2s} and
X = { (u1,x,y,z,u2), (x,w1), (y,w2), (z,w3) } U { (ai,x,ai+r) | 1 ≤ i ≤ r }
U { (bj,y,bj+t) | 1 ≤ j ≤ t } U { (ck,z,ck+s) | 1 ≤ k ≤
s },
and
Since the role of the m-vertices x and z are interchangeable, it follows that 𝐴𝑟,𝑠,𝑝
(11)
≅ 𝐴𝑠,𝑟,𝑝
(11)
. Hence
we can assume that r ≤ s. Then and
Figure 1
Then, we get the next two results.
0 ≤ 𝑟 ≤ 𝐿
𝑝−8
4
⌋
Lemma 6.2. For
𝑝−8−2𝑟
2
, 𝐴𝑟,𝑠,𝑝
(11)
1
and 𝑟 ≤ 𝑠1, 𝑠2 ≤
is isomorphic to iff s1 = s2.
2
2
2
8 s
r
p −
−
−
,
0
2
2
2
8

−
−
− s
r
p
)
11
(
,
, p
s
r
A )
,
(
)
11
(
,
, X
V
A p
s
r =
2
2
8
0
r
p
s
−
−


2
2
8
0
r
p
s
−
−


.
0
2
2
2
8

−
−
−
=
s
r
p
t





 −


4
8
0
p
r .
2
2
8 r
p
s
r
−
−


)
11
(
14
,
1
,
0
A





 −


4
8
0
p
r ,
2
2
8
2
1
r
p
s
s
r
−
−



)
11
(
,
, 1 p
s
r
A )
11
(
,
, 2 p
s
r
A
2
8
0
−


p
r
,
2
8
0
−


p
r

M.S. Chithra
Proof : Let 𝐴𝑟,𝑠1,𝑝
(11)
= (𝑉, 𝑋)
with
V = {u1, u2, w1, w2, w3, x, y, z, a1, …, a2r, b1, …, 𝑏2𝑡1
, c1, …, 𝑐2𝑠1
} and
X = { (u1,x,y,z,u2), (x,w1), (y,w2), (z,w3) } U { (ai,x,ai+r) | 1 ≤ i ≤ r }
U { (bj,y, 𝑏𝑗+𝑡1
) | 1 ≤ j ≤ t1 } U { (ck,z, 𝑐𝑘+𝑠1
) | 1 ≤ k ≤ s1 }, and
and .
Let
with
V' = {u'1, u'2, w'1, w'2, w'3, x', y', z', a'1, …, a'2r, b'1, …, 𝑏′2𝑡2
, c'1,
…, 𝑐′2𝑠2
} and
X' = { (u'1,x',y',z',u'2), (x',w'1), (y',w'2), (z',w'3) } U { (a'i,x',a'i+r) | 1 ≤ i ≤ r }
(b'j,y', 𝑏′𝑗+𝑡2
) | 1 ≤ j ≤ t2 } U { (c'k,z', 𝑐′𝑘+𝑠2
) | 1 ≤
U {
k ≤ s2 },
and .
If s1 = s2𝑠1 = 𝑠2, the result is obvious. For the converse part, assume the contrary
that 𝐴𝑟,𝑠1,𝑝
(11)
≅ 𝐴𝑟,𝑠2,𝑝
(11)
when 𝑠1 ≠ 𝑠2. In any isomorphism 𝜃 between 𝐴𝑟,𝑠1,𝑝
(11)
and 𝐴𝑟,𝑠2,𝑝
(11)
, we have 𝜃((𝑢1, 𝑥, 𝑦, 𝑧, 𝑢2)) =
(𝑢1
′
, 𝑥′
, 𝑦′
, 𝑧′
, 𝑢2
′
) ; and so {𝜃(𝑥), 𝜃(𝑧)} = {𝑥′
, 𝑧′}. By counting the number of 3-semiedges
with 𝑥, 𝑧, 𝑥′
, 𝑧′
as 𝑚 −vertices, we have {𝑟, 𝑠1} = {𝑟, 𝑠2}, a contradiction.
Theorem 6.3. For p ≥ 8, the number of non-isomorphic semigraphs in 𝐴11 𝒜11 of order 𝑝 p is
given by 𝑇𝐴11,𝑝
= {
𝑝(𝑝−4)(𝑝−5)
96
, 𝑖𝑓 𝑝 ≡ 0(𝑚𝑜𝑑4)
(𝑝−1)(𝑝−2)(𝑝−6)
96
, 𝑖𝑓 𝑝 ≡ 2(𝑚𝑜𝑑4) .
Proof : Using Lemma 6.2, all the semigraphs 𝐴𝑟,𝑠,𝑝
(11)
with and 𝑟 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
0 ≤ 𝑟 ≤ 𝐿
𝑝−8
4
⌋ are non-isomorphic.
Thus for a fixed 𝑟, the number of non-isomorphic semigraphs of the
form 𝐴𝑟,𝑠,𝑝
(11)
is (
𝑝−8−2𝑟
2
− 𝑟 + 1) =
𝑝−6−4𝑟
2
.
Hence the number of non-isomorphic semigraphs of order 𝑝(≥ 8) in 𝐴11
′
is given by
𝑇𝐴11,𝑝
′ = ⌊
𝑝−8
∑4
𝑟=0
⌋
𝑝−6−4𝑟
2
= 𝐿
𝑝−4
4
⌋(
𝑝−6
2
− 𝐿
𝑝−8
4
⌋)
Now, for any 𝐺 ∈ 𝐴11, one of the component of 𝐺 is 𝐴𝑟,𝑠,𝑝
(11)
, with 0 ≤ 𝑟 ≤ ⌊
𝑝−8
4
⌋,
𝑟 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
; and all the other components of 𝐺 are 𝐾2.
)
,
(
)
11
(
,
, 1
X
V
A p
s
r =
2
2
8
1
r
p
s
r
−
−

 2
2
2
8 1
1
s
r
p
t
−
−
−
=
,
4
8
0 




 −


p
r
2
2
8
2
r
p
s
r
−
−


2
2
2
8 2
2
s
r
p
t
−
−
−
=
,
4
8
0 




 −


p
r
,
4
8
0 




 −


p
r
.
4
8
4
6
4
4
2
4
6
4
8
0
'
,
11













 −
−
−





 −
=
−
−
= 





 −
=
p
p
p
r
p
T
p
r
A
p
)
11
(
,
,
)
11
(
,
, 2
1 p
s
r
p
s
r A
A 
)
'
,
'
(
)
11
(
,
, 2
X
V
A p
s
r =

Hence the number of non‐isomorphic semigraphs of order 𝑝(≥ 8) in 𝐴11 is given by
𝑇𝐴11,𝑝
= 8 ≤ 𝑛 ≤
𝑝 ∑𝑛𝑒𝑣≤𝑛 𝑇𝐴11,𝑛
′ = 8 ≤ 𝑛 ≤
𝑝 ∑𝑛𝑒𝑣≤𝑛 𝐿
𝑛−4
4
⌋(
𝑛−6
2
− 𝐿
𝑛−8
4
⌋)
On simplification, we get the result
6.2. Semigraphs in 𝐴12
′
In any semigraph 𝐺 in 𝐴12
′
, 𝑥, 𝑦, 𝑧 lie on a 4-semiedge. Hence there are edges of the
form (𝑥, 𝑦, 𝑧, 𝑢1), (𝑦, 𝑤1), (𝑧, 𝑤2) and there is at least one 3-semiedge containing 𝑥 as a 𝑚-
vertex. Furthermore, all the vertices other than 𝑢1, 𝑤1, 𝑤2, 𝑥, 𝑦 and 𝑧, lie in a 3-semiedge with
a 𝑚-vertex 𝑥, 𝑦 or 𝑧.
Now, as in Section 6.1, any semigraph in 𝐴12
′
can be denoted by 𝐴𝑟,𝑠,𝑝
(12)
, where 𝐴𝑟,𝑠,𝑝
(12)
=
(𝑉, 𝑋) with
𝑉 = {𝑢1, 𝑤1, 𝑤2, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑡, 𝑐1, … , 𝑐2𝑠} and 𝑋 =
{(𝑥, 𝑦, 𝑧, 𝑢1), (𝑦, 𝑤1), (𝑧, 𝑤2)} ∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟}
∪ {1 ≤ 𝑗 ≤ 𝑡} ∪ {1 ≤ 𝑘 ≤ 𝑠},
1 ≤ 𝑟 ≤
𝑝−6
2
, 0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
and 𝑡 =
𝑝−6−2𝑟−2𝑠
2
≥ 0.
Note that the role of 𝑥 and 𝑧 cannot be interchangeable.
The next two results can be proved as in Section 6.1.
Figure 2
Lemma 6.4. For 1 ≤ 𝑟 ≤
𝑝−6
2
and 0 ≤ 𝑠1, 𝑠2 ≤
𝑝−6−2𝑟
2
(12)
1
𝑟 ≤ 𝑠1, 𝑠2 ≤
𝑝−8−2𝑟
2
(11)
1
is isomorphic to
iff s1 = s2.𝐴𝑟,𝑠,𝑝
(12)
2
(12)
= (𝑉, 𝑋) 𝐴𝑟,𝑠1,𝑝
(12)
= (𝑉, 𝑋) with
V = {u1,w1,w2,x,y,z,a1, …, a2r,b1, …, 𝑏2𝑡1
,c1, …, 𝑐2𝑠1
} and
X = { (x,y,z,u1), (y,w1), (z,w2) } U { (ai,x,ai+r) | 1 ≤ i ≤ r }
.
4
8
2
6
4
4
8
8
'
,
11
,
11













 −
−
−





 −
=
= 





n
n
n
A
A
T
p
n
neven
p
n
neven
T n
p
)
12
(
12
,
1
,
1
A
,
2
2
6
,
0 2
1
r
p
s
s
−
−


)
12
(
,
, 1 p
s
r
A
)
12
(
,
, 2 p
s
r
A

M.S. Chithra
U { (bj,y, 𝑏𝑗+𝑡1
) | 1 ≤ j ≤ t1 } U { (ck,z, 𝑐𝑘+𝑠1
) |1≤ k ≤ s1 },
and
Let 𝐴𝑟,𝑠2,𝑝
(12)
= (𝑉′, 𝑋′) 𝑉 = {𝑢1, 𝑤1, 𝑤2, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑡, 𝑐1, … , 𝑐2𝑠}with
V' = { u’1,w’1,w’2, x’,y’,z’,a’1, …, a’2r,b’1, …, 𝑏′2𝑡2
,c’1, …, 𝑐′2𝑠2
} and
X' = { (x',y',z',u'1), (y',w'1), (z′,w'2) } U { (a'i,x',a'i+r) | 1 ≤ i ≤ r }
U { (b'j,y', 𝑏′𝑗+𝑡2
) | 1 ≤ j ≤ t2 } U { (c'k,z', 𝑐′𝑘+𝑠2
) | 1 ≤ k ≤ s2 },
𝑉 = {𝑢1, 𝑤1, 𝑤2, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑡, 𝑐1, … , 𝑐2𝑠}
𝑉′
=
{𝑢1
′
, 𝑢2
′
, 𝑤1
′
, 𝑤2
′
, 𝑤3
′
, 𝑥′
, 𝑦′
, 𝑧′
, 𝑎1
′
, … , 𝑎2𝑟
′
, 𝑏1
′
, … , 𝑏2𝑡2
′
, 𝑐1
′
, … , 𝐶2𝑠
′
2}In any isomorphism θ
between and ,𝐴𝑟,𝑠1,𝑝
(11)
= 𝐴𝑟,𝑠2,𝑝
(12)
we have 𝜃((𝑥, 𝑦, 𝑧, 𝑢1)) = (𝑥′
, 𝑦′
, 𝑧′
, 𝑢1
′
)
.
Clearly θ (u1) = u'1; and so θ (z) = z'.
Hence s1 = s2. Converse is obvious
Theorem 6.5. For 𝑝 > 8, the number of non-isomorphic semigraphs in 𝐴12 of order 𝑝 is given
by 𝑇𝐴12,𝑝
=
(𝑝−2)(𝑝−4)(𝑝−6)
48
.
Proof: Using Lemma 6.4, all the semigraphs 𝐴𝑟,𝑠,𝑝
(12)
𝐴𝑟,𝑠2,𝑝
(12)
with 1 ≤ 𝑟 ≤
𝑝−6
2
, 0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
are non-isomorphic.
Thus for a fixed 𝑟, the number of non-isomorphic semigraphs of the form 𝐴𝑟,𝑠,𝑝
(12)
is
(
𝑝−6−2𝑟
2
+ 1) =
𝑝−4−2𝑟
2
.
Hence the number of non-isomorphic semigraphs of order 𝑝( ≥8) in 𝐴12
′
is given by
For any 𝐺 ∈ 𝐴12, one of the component of 𝐺 is 𝐴𝑟,𝑠,𝑝
(12)
, with 1 ≤ 𝑟 ≤
𝑝−6
2
,
0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
Hence, the number of non-isomorphic semigraphs of order 𝑝(≥ 8) in 𝐴12 is given by
On simplification, we get the result
.
0
2
2
2
6 1
1 
−
−
−
=
s
r
p
t
,
2
6
1
−


p
r ,
2
2
6
0 1
r
p
s
−
−


.
0
2
2
2
6 2
2 
−
−
−
=
s
r
p
t
,
2
6
1
−


p
r ,
2
2
6
0 2
r
p
s
−
−


)
12
(
,
, 1 p
s
r
A
)
12
(
,
, 2 p
s
r
A
.
8
)
4
)(
6
(
2
2
4
2
6
1
'
,
12
−
−
=
−
−
= 
−
=
p
p
r
p
A
p
r
T p
.
8
)
4
)(
6
(
8
8
'
,
12
,
12






−
−
=
=
p
n
neven
p
n
neven
n
n
A
A T
T n
p

′
′
, 𝑥, 𝑦, 𝑧 lie on a 3-semiedge. Hence there are edges of the
form(𝑥, 𝑦, 𝑧), (𝑦, 𝑤1) and there is at least one 3-semiedge containing 𝑥 as a 𝑚 −vertex, and
there is at least one 3-semiedge containing 𝑧 as a 𝑚-vertex. Furthermore, all the vertices other
than 𝑤1, 𝑥, 𝑦 and 𝑧, lie in a 3-semiedge with a 𝑚−vertex 𝑥, 𝑦 or 𝑧.
Now, any semigraph in 𝐴13
′
(13)
(13)
= (𝑉, 𝑋) with
V = {w1, x, y, z, a1, …, a2r, b1, …, b2t, c1, …, c2s} and
X = { (x,y,z), (y,w1), (z,w2) } U { (ai,x,ai+r) | 1 ≤ i ≤ r }
U { (bj,y,bj+t) | 1 ≤ j ≤ t
} U { (ck,z,ck+s) | 1 ≤ k
≤ s }
∪ {(𝑐𝑘, 𝑧, 𝑐𝑘+𝑠)|1 ≤ 𝑘 ≤ 𝑠},1 ≤ 𝑟 ≤
𝑝−6
2
, 1 ≤ 𝑠 ≤
𝑝−4−2𝑟
2
and𝑡 =
𝑝−4−2𝑟−2𝑠
2
≥ 0.
Since the role of the 𝑚−vertices 𝑥 and 𝑧 are interchangeable, it follows that
𝐴𝑟,𝑠,𝑝
(13)
≅ 𝐴𝑠,𝑟,𝑝
(13)
. Hence we can assume that 𝑟 ≤ 𝑠. Then 1 ≤ 𝑟 ≤ ⌊
𝑝−4
4
⌋ and
𝑟 ≤ 𝑠 ≤
𝑝 − 4 − 2𝑟
2
.
Figure 3
Lemma 6.6. For 1 ≤ 𝑟 ≤ ⌊
𝑝−4
4
⌋ and 𝑟 ≤ 𝑠1, 𝑠2 ≤
𝑝−4−2𝑟
2
, 𝐴𝑟,𝑠1,𝑝
(13)
is isomorphic to 𝐴𝑟,𝑠2,𝑝
(13)
iff
𝑠1 = 𝑠2.
Theorem 6.7. For 𝑝 ≥ 8, the number of non-isomorphic semigraphs in 𝐴13 of order 𝑝 is given
by
𝑇𝐴13,𝑝
= {
𝑝(𝑝 − 4)(𝑝 − 5)
96
, 𝑖𝑓 𝑝 ≡ 0(𝑚𝑜𝑑4)
(𝑝 − 1)(𝑝 − 2)(𝑝 − 6)
96
, 𝑖𝑓 𝑝
≡ 2(𝑚𝑜𝑑4) .
.
0
2
2
2
4

−
−
−
=
s
r
p
t
,
2
6
1
−


p
r
2
2
4
1
r
p
s
−
−


)
13
(
14
,
2
,
1
A

M.S. Chithra
7. SEMIGRAPHS IN 𝐴2
Let 𝐴2
′
denote the class of all semigraphs in 𝐴2 of order 𝑝, having no component of order 2.
For every 𝐺 ∈ 𝐴2
′
, all the 𝑚-vertices 𝑥, 𝑦, 𝑧 are mutually adjacent but they do not lie on the
same edge. Then there are three edges 𝐸1, 𝐸2, 𝐸3 such that 𝑥, 𝑦 lie in 𝐸1; 𝑦, 𝑧 lie in 𝐸2 and 𝑧, 𝑥
lie in 𝐸3. The vertices 𝑥, 𝑦 and 𝑧 may be end vertices or middle vertices in their corresponding
edges.
We shall categorize the semiedges in 𝐴2
′
according to the positions of 𝑥, 𝑦, 𝑧 in 𝐸1, 𝐸2, 𝐸3.
Let 𝑒, 𝑚 denote the positions ‘end’, ‘middle’ respectively.
7.1. Labelling of semigraphs in 𝐴2
′
We first attach labels to the edges 𝐸1, 𝐸2, 𝐸3 as follows:
The label of an edge is an ordered pair that denotes the positions (m or e) of the two vertices
that lie on them. We fix that the label of the edge 𝐸1 denotes the position of (𝑥, 𝑦) in 𝐸1; the
label of the edge 𝐸2 denotes the position of (𝑦, 𝑧) in 𝐸2; and the label of the edge 𝐸3 denotes
the position of (𝑧, 𝑥) in 𝐸3.
We label a semigraph 𝐺 in 𝐴2
′
, using the labels of the edges 𝐸1, 𝐸2, 𝐸 in 𝐺. Let 𝑙1, 𝑙2, 𝑙3
denote the labels of 𝐸1, 𝐸2, 𝐸3 respectively. Then, the label of 𝐺, denoted by 𝑙(𝐺), is an ordered
triple (𝑙1, 𝑙2, 𝑙3) (Refer Fig.1).
E1 E1
A semigraph with the label A semigraph with the label
((𝑚, 𝑚), (𝑒, 𝑚), (𝑒, 𝑚)).
((𝑚,𝑚),(𝑒,𝑚),(𝑚,𝑒)).
Figure 4
Note that li’s are ordered pairs for 𝑖 = 1,2,3.
There are 26
= 64 possibilities for the labels (𝑙1, 𝑙2, 𝑙3) .
Since 𝐺 is 1-regular, each of the 𝑚-vertices 𝑥, 𝑦 and 𝑧 can be an end vertex in only one
edge.
Hence 𝑥 cannot have the position 𝑒 in both the edges 𝐸1 and 𝐸3; 𝑦 in 𝐸1 and 𝐸2; and 𝑧 in 𝐸2
and𝐸3 E3.
We shall refer this as “𝑥, 𝑦 and 𝑧 cannot have two e’s”.
Next, to count the number of impossible labelings, we apply the Principle of Inclusion and
Exclusion. Suppose that x have two e’s. Then the label of 𝑦 may be `𝑚’ or `e’ in 𝑙1 and 𝑙2; and
similarly the label of 𝑧 may be `𝑚’ or `e𝑒’ in 𝑙2 and 𝑙3.
Hence the number of labelings with 𝑥 or 𝑦 or 𝑧 having two
𝑒’𝑠 = 3 × 2 × 2 × 2 × 2 = 48.
Next, suppose that 𝑥 and 𝑦 have two e’s. Then the label of 𝑧 may be `m’ or 𝑒’ in 𝑙2 and 𝑙3.
Hence the number of labelings with two of the vertices 𝑥, 𝑦, 𝑧, having two
𝑒’𝑠 = 3𝐶2 × 2 × 2 = 12.

Next, there is only one labeling with all 𝑥, 𝑦 and 𝑧 having two 𝑒’𝑆.
Hence the number of impossible labelings= 48 − 12 + 1 = 37.
Let 𝐿 denote the set of all possible labelings of semigraphs in 𝐴2
′
. Then
|𝐿| = 64 − 37 = 27.
We shall show that these 27 possible labelings correspond to 7 non-isomorphic classes of
semigraphs. We partition 𝐿 such that 𝐿 = 𝑈𝑖=1
7
𝐿𝑖, where 𝐿𝑖’s are defined as follows:
𝐿1 = {((𝑒, 𝑒), (𝑚, 𝑚), (𝑚, 𝑚)), ((𝑚, 𝑚), (𝑚, 𝑚), (𝑒, 𝑒)), ((𝑚, 𝑚), (𝑒, 𝑒), (𝑚, 𝑚))}
𝐿2 = {((𝑒, 𝑒), (𝑚, 𝑚), (𝑒, 𝑚 )), ((𝑚, 𝑚), (𝑒, 𝑚), (𝑒, 𝑒 )), ((𝑒, 𝑚), (𝑒, 𝑒), (𝑚, 𝑚)),
((𝑒, 𝑒), (𝑚, 𝑚), (𝑚, 𝑒)), ((𝑚, 𝑚), (𝑚, 𝑒), (𝑒, 𝑒)), ((𝑚, 𝑒), (𝑒, 𝑒), (𝑚, 𝑚))}
𝐿3 = {((𝑚, 𝑚), (𝑒, 𝑚), (𝑚, 𝑒 )), ((𝑒, 𝑚), (𝑚, 𝑒), (𝑚, 𝑚)), ((𝑚, 𝑒), (𝑚, 𝑚), (𝑒, 𝑚))}
𝐿4 = {((𝑚, 𝑚), (𝑚, 𝑚), (𝑒, 𝑚)), ((𝑚, 𝑚), (𝑒, 𝑚), (𝑚, 𝑚 )), ((𝑒, 𝑚), (𝑚, 𝑚), (𝑚, 𝑚)),
((𝑚, 𝑚), (𝑚, 𝑚), (𝑚, 𝑒)), ((𝑚, 𝑚), (𝑚, 𝑒), (𝑚, 𝑚)), ((𝑚, 𝑒), (𝑚, 𝑚), (𝑚, 𝑚))}
𝐿5 = {((𝑚, 𝑚), (𝑒, 𝑚), (𝑒, 𝑚)), ((𝑒, 𝑚), (𝑚, 𝑚), (𝑒, 𝑚)), ((𝑒, 𝑚), (𝑒, 𝑚), (𝑚, 𝑚)),
((𝑚, 𝑚), (𝑚, 𝑒), (𝑚, 𝑒)), ((𝑚, 𝑒), (𝑚, 𝑚), (𝑚, 𝑒)), ((𝑚, 𝑒), (𝑚, 𝑒), (𝑚, 𝑚))}
𝐿6 = {((𝑚, 𝑚), (𝑚, 𝑚), (𝑚, 𝑚))}
𝐿7 = {((𝑒, 𝑚), (𝑒, 𝑚), (𝑒, 𝑚)), ((𝑚, 𝑒), (𝑚, 𝑒), (𝑚, 𝑒))}.
Notation: If 𝑙1 = (𝑖, 𝑗) is a label of an edge, where 𝑖, 𝑗 ∈ {𝑒, 𝑚}, then 𝑙1
−1
denotes the label (𝑗, 𝑖).
Then (𝑙1
−1
)−1
= 𝑙1.
7.2. Isomorphism in 𝐴2
′
Theorem 7.1. Any semigraph in 𝐴2
′
with the labeling (𝑙1, 𝑙2, 𝑙3) (l1,l2,l3) is isomorphic to a
semigraph in 𝐴2
′
with the labeling (l2,l3,l1) (𝑙2, 𝑙3, 𝑙1) and to a semigraph in 𝐴2
′
with the labeling
(l3,l1,l2).
Proof: We prove the result for only one case, and the proof is similar in all the other cases.
Consider a semigraph 𝐺 = (𝑉, 𝑋) in 𝐴2
′
with the labeling (𝑙1, 𝑙2, 𝑙3), where 𝑙1 =
(𝑒, 𝑒), 𝑙2 = (𝑚, 𝑒) and 𝑙3 = (𝑚, 𝑚) .
Let V = {u1,u2,u3,x,y,z,a1,…,a2r,b1,…,b2s,c1,…,c2t} and 𝑋 = 𝑋1𝑈𝑋2𝑈𝑋3 ∪ {𝐸1, 𝐸2, 𝐸3},
where 𝐸1 = (𝑥, 𝑦), 𝐸2 = (𝑢1, 𝑦, 𝑧), 𝐸3 = (𝑢2, 𝑧, 𝑥, 𝑢3), 𝑋1 = {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟},
𝑋2 = {1 ≤ 𝑗 ≤ 𝑠} and 𝑋3 = {1 ≤ 𝑘 ≤ 𝑡}, 0 ≤ 𝑟 ≤
𝑝−6
2
,
0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
and 𝑡 =
𝑝−6−2𝑟−2𝑠
2
≥ 0.
Consider the semigraph 𝐺′
= (𝑉′
, 𝑋′
) in 𝐴2
′
with
𝑉′
= {𝑢1
′
, 𝑢2
′
, 𝑢3
′
, 𝑥′
, 𝑦′
, 𝑧′
, 𝑎1
′
, … , 𝑎2𝑟
′
, 𝑏1
′
, … , 𝑏2𝑠
′
, 𝑐1
′
, … , 𝑐2𝑡
′
} and
𝑋′
= 𝑋1
′
𝑈𝑋2
′
𝑈𝑋3
′
∪ {𝐸1
′
, 𝐸2
′
, 𝐸3
′
}, where 𝐸1
′
= (𝑢1
′
, 𝑥′
, 𝑦 𝐸2
′
= (𝑢2
′
, 𝑦′
, 𝑧′
, 𝑢3
′
), 𝐸3
′
= (𝑧′
, 𝑥E'1
= (u1',x',y'), E'2 = (u2',y',z',u'3), E'3 = (z’,x’),
X1' = {(a'i,z',a'i+r) | 1 ≤ i ≤ r}, X2' = {(b'j,x',b'j+s) | 1 ≤ j ≤s}, 𝑋1
′
= 0 ≤ 𝑟 ≤
𝑝−6𝑖𝑍′
2
, 0 ≤ 𝑠 ≤
{(𝑎′
, 𝑎𝑖+𝑟
′
)|1
≤𝑖≤𝑟𝑝−6−2𝑟
2
𝑎𝑛𝑑𝑡 =
𝑝−6−2𝑟−𝑖𝑠(𝑏𝑗
′
,𝑥′,𝑏′
2
≥ 0} , 𝑋 2
′
= {+𝑆)|1. ≤ 𝑗 ≤ 𝑠}and

M.S. Chithra
𝑋3
′
= {(𝑐𝑘
′
, 𝑦′
, 𝑐𝑘+𝑡
′
)|1 ≤ 𝑘 ≤ 𝑡}, 0 ≤ 𝑟 ≤
𝑝−6
2
,
0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
and 𝑡 =
𝑝−6−2𝑟−2𝑠
2
≥ 0.
Note that the labeling of 𝐺′
is ((𝑚, 𝑒), (𝑚, 𝑚), (𝑒, 𝑒)) = (𝑙2, 𝑙3, 𝑙1) .
Define a bijection θ : V → V’ by θ(x) = z', θ(y) = x', θ(z) = y', θ(u1) = u1', θ(u2) =
u2', θ(u3) = u3', θ(ai) = ai', 1 ≤ i ≤ 2r, θ(bj) = bj', 1 ≤ j ≤ 2s, θ(ck) = ck', 1 ≤ k ≤ 2t.𝜃(𝑥) =
𝑧≤
′
𝜃(𝑦) = 𝑥′
, 𝜃(𝑧) = 𝑦≤
′
𝜃(𝑢1) = 𝑢1
′
, 𝜃(𝑢2) = 𝑢2
′
, 𝜃(𝑢3) = 𝑢3
′
, 𝜃(𝑎𝑖) = 𝑎𝑖
′
, 1 ≤ 𝑖 ≤
2𝑟, 𝜃(𝑏𝑗) = 𝑏𝑗
′
, 1 ≤ 𝑗 ≤ 2𝑠, 𝜃(𝑐𝑘) = 𝑐𝑘
′
, 1 ≤ 𝑘 ≤ 2𝑡.
Then 𝜃(𝐸1) = (𝑧′
, 𝑥′) = 𝐸3
′
, 𝜃(𝐸2) = (𝑢1
′
, 𝑥′
, 𝑦′) = 𝐸1
′
, 𝜃(𝐸3) = (𝑢2
′
, 𝑦′
, 𝑧′
, 𝑢3
′ ) = 𝐸2
′
,
𝜃((𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)) = (𝑎𝑖
′
, 𝑧′
, 𝑎𝑖+𝑟
′
) , for 1 ≤ 𝑖 ≤ 𝑟, 𝜃((𝑏𝑗, 𝑦, 𝑏𝑗+𝑠)) = (𝑏𝑗
′
, 𝑥′
, 𝑏𝑗+𝑠
′
), for 1 ≤ 𝑗 ≤
𝑠 and ((𝑐𝑘, 𝑧, 𝑐𝑘+𝑡)) = (𝑐𝑘
′
, 𝑦′
, 𝑐𝑘+𝑡
′
), for 1 ≤ 𝑘 ≤ 𝑡.
Hence 𝜃 is an isomorphism and 𝐺 ≅ 𝐺′
.
Repeating the procedure for 𝐺′
, 𝐺 is also isomorphic to a semigraph 𝐺′′
in 𝐴2
′
with the
labeling (𝑙3, 𝑙1, 𝑙2).
Figure 5
Theorem 7.2. Any semigraph𝐺 G in 𝐴2
′
with the labeling (𝑙1, 𝑙2, 𝑙3) is isomorphic to a
semigraph H in 𝐴2
′
with the labeling (𝑙3
−1
, 𝑙2
−1
, 𝑙1
−1
).
Proof: We prove the result for only one case and the proof is similar in all the other cases.
Consider a semigraph 𝐺 = (𝑉, 𝑋) in 𝐴2
′
with the labeling (𝑙1, 𝑙2, 𝑙3) =
((𝑚, 𝑚), (𝑒, 𝑚), (𝑒, 𝑚)).
Let V = {u1, …, u5, x, y, z, a1, …, a2r, b1, …, b2s, c1, …, c2t} and
𝑋 = 𝑋1𝑈𝑋2𝑈𝑋3 ∪ {𝐸1, 𝐸2, 𝐸3, 𝐸4}, where
𝐸1 = (𝑢1, 𝑥, 𝑦, 𝑢2), 𝐸2 = (𝑦, 𝑧, 𝑢3), 𝐸3 = (𝑧, 𝑥, 𝑢4), 𝐸4 = (𝑥, 𝑢5) ,
𝑋1 = {1 ≤ 𝑖 ≤ 𝑟}, 𝑋2 = {1 ≤ 𝑗 ≤ 𝑠} and 𝑋3{(𝑐𝑘, 𝑧, 𝑐𝑘+𝑡)|1 ≤ 𝑘 ≤ 𝑡},
0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
and 𝑡 =
𝑝−8−2𝑟−2𝑠
2
≥ 0.
Consider the semigraph 𝐻 = (𝑉′
, 𝑋′
) in 𝐴2
′
with
𝑉′
= {𝑢1
′
, … , 𝑢5
′
, 𝑥′
, 𝑦′
, 𝑧′
, 𝑎1
′
, … , 𝑎2𝑟
′
, 𝑏1
′
, … , 𝑏2𝑠
′
, 𝑐1
′
, … , 𝑐2𝑡
′
} and
𝑋′
= 𝑋1
′
𝑈𝑋2
′
𝑈𝑋3
′
∪ {𝐸1
′
, 𝐸2
′
, 𝐸3
′
, 𝐸4
′
}, where
𝐸1
′
= (𝑢1
′
, 𝑥′
, 𝑦′), 𝐸2
′
= (𝑢2
′
, 𝑦′
, 𝑧′), 𝐸3
′
= (𝑢3
′
, 𝑧′
, 𝑥′
, 𝑢4
′ ), 𝐸4
′
= (𝑥′
, 𝑢5
′
),
𝑋′
1 = {1 ≤ 𝑖 ≤ 𝑟}, 𝑋′
2 = {1 ≤ 𝑗 ≤ 𝑠} 𝑎𝑛𝑑
𝑋′3 = {(𝑐′
𝑘, 𝑦′, 𝑐′𝑘+𝑡)|1 ≤ 𝑘 ≤ 𝑡}, 0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
and 𝑡 =
𝑝−8−2𝑟−2𝑠
2
≥ 0.

Define a bijection 𝜃: 𝑉 → 𝑉′
by 𝜃(𝑥) = 𝑥,
′
𝜃(𝑦) = 𝑧′
, 𝜃(𝑧) = 𝑦,
′
𝜃(𝑢1) = 𝑢4
′
, 𝜃(𝑢2) =
𝑢3
′
, 𝜃(𝑢3) = 𝑢2
′
, 𝜃(𝑢4) = 𝑢1
′
, 𝜃(𝑢5) = 𝑢5
′
, 𝜃(𝑎𝑖) = 𝑎𝑖
′
, 1 ≤ 𝑖 ≤ 2𝑟, 𝜃(𝑏𝑗) =
𝑏𝑗
′
, 1 ≤ 𝑗 ≤ 2𝑠, 𝜃(𝑐𝑘) = 𝑐𝑘
′
, 1 ≤ 𝑘 ≤ 2𝑡.
Then 𝜃(𝐸1) = (𝑢4
′
, 𝑥′
, 𝑧′
, 𝑢3
′ ) = 𝐸3
′
, 𝜃(𝐸2) = (𝑧′
, 𝑦′
, 𝑢2
′ ) = 𝐸2
′
, 𝜃(𝐸3) = (𝑦′
, 𝑥′
, 𝑢1
′ ) =
𝐸1
′
, 𝜃(𝐸4) = (𝑥′
, 𝑢5
′ ) = 𝐸4
′
, 𝜃((𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)) = (𝑎𝑖
′
, 𝑥′
, 𝑎𝑖+𝑟
′
) , for 1 ≤ 𝑖 ≤ 𝑟,
𝜃((𝑏𝑗, 𝑦, 𝑏𝑗+𝑠)) = (𝑏𝑗
′
, 𝑧′
, 𝑏𝑗+𝑠
′
), for 1 ≤ 𝑗 ≤ 𝑠 and 𝜃((𝑐𝑘,𝑧,𝑐𝑘+𝑡)) = (𝑐𝑘
′
,𝑦′, 𝑐𝑘+𝑡
′
), for
1 ≤ 𝑘 ≤ 𝑡.
Thus 𝐺 ≅ 𝐻. Also note that 𝑙(𝐻) = ((𝑚, 𝑒), (𝑚, 𝑒), (𝑚, 𝑚)) = (𝑙3
−1
, 𝑙2
−1
, 𝑙1
−1).
Figure 6
Let 𝐴2𝑖
′
= {𝐺 ∈ 𝐴2
′
|𝑙(𝐺) ∈ 𝐿𝑖}, for 𝑖 = 1,2, … ,7.
Applying Theorems 7.1 and 7.2 to the partitions 𝐿𝑖, we note that
𝐴21
′
= {𝐺 ∈ 𝐴2
′
|𝑙(𝐺) = ((𝑒, 𝑒), (𝑚, 𝑚), (𝑚, 𝑚)}
𝐴22
′
= {𝐺 ∈ 𝐴2
′
|𝑙(𝐺) = ((𝑒, 𝑒), (𝑚, 𝑚), (𝑒, 𝑚))}
𝐴23
′
= {𝐺 ∈ 𝐴2
′
|𝑙(𝐺) = ((𝑚, 𝑚), (𝑒, 𝑚), (𝑚, 𝑒))}
𝐴24
′
= {𝐺 ∈ 𝐴2
′
|𝑙(𝐺) = ((𝑚, 𝑚), (𝑚, 𝑚), (𝑒, 𝑚))}
𝐴25
′
= {𝐺 ∈ 𝐴2
′
|𝑙(𝐺) = ((𝑚, 𝑚), (𝑒, 𝑚), (𝑒, 𝑚))}
𝐴26
′
= {𝐺 ∈ 𝐴2
′
|𝑙(𝐺) = ((𝑚, 𝑚), (𝑚, 𝑚), (𝑚, 𝑚))}
and 𝐴27
′
= {𝐺 ∈ 𝐴2
′
|𝑙(𝐺) = ((𝑒, 𝑚), (𝑒, 𝑚), (𝑒, 𝑚))}.
Now the next result follows easily
Theorem 7.3. 𝑙𝑓 𝐺 ∈ 𝐴2𝑖
′
and 𝐻 ∈ 𝐴2𝑗
′
, 𝑖, 𝑗 ∈ {1, … , 7, 𝑖 ≠ 𝑗, then 𝐺 is not isomorphic to 𝐻.
′
′
, there are edges of the form (𝑥, 𝑦), (𝑢1, 𝑦, 𝑧, 𝑢2), (𝑢3, 𝑧, 𝑥, 𝑢4) and
(𝑧, 𝑤1). Furthermore, all the vertices other than 𝑢1, 𝑢2, 𝑢3, 𝑢4, 𝑤1, 𝑥, 𝑦 and 𝑧, lie in a 3-semiedge
with a 𝑚−vertex 𝑥, 𝑦 or 𝑧.
If there are 𝑟(≥ 0) 3-semiedges with 𝑥 as a 𝑚-vertex and 𝑠(≥ 0) 3-semiedges with 𝑦 as a
𝑚 −vertex, then there are
𝑝−8−2𝑟−2𝑠
2
3-semiedges with 𝑧 as a 𝑚 −vertex. Since
𝑝−8−2𝑟−2𝑠
2
≥
0, we have 0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
and 𝑝 ≥ 8. Hence any semigraph in 𝐴21
′
can be
denoted by 𝐴𝑟,𝑠,𝑝
(21)
(21)
= (𝑉, 𝑋) with
𝑉 = {𝑢1, … , 𝑢4, 𝑤1, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑠, 𝑐1, … , 𝑐2𝑡} and 𝑋 =
{(𝑥, 𝑦), (𝑢1, 𝑦, 𝑧, 𝑢2), (𝑢3, 𝑧, 𝑥, 𝑢4), (𝑧, 𝑤1)} ∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟}
∪ {(𝑏𝑗, 𝑦, 𝑏𝑗+𝑠)|1 ≤ 𝑗 ≤ 𝑠} ∪ {(𝑐𝑘, 𝑧, 𝑐𝑘+𝑡)|1 ≤ 𝑘 ≤ 𝑡},

M.S. Chithra
0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
and 𝑡 =
𝑝−8−2𝑟−2𝑠
2
≥ 0.
Since the role of the 𝑚−vertices 𝑥 and 𝑦 are interchangeable, it follows that 𝐴𝑟,𝑠,𝑝
(21)
≅
𝐴𝑠,𝑟,𝑝
(21)
𝑝−8
4
⌋ and 𝑟 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
.
𝑭𝒊𝒈𝒖𝒓𝒆. 𝟕 𝐴1,1,12
(21)
Lemma 7.4. For 0 ≤ 𝑟 ≤ ⌊
𝑝−8
4
⌋ and 𝑟 ≤ 𝑠1, 𝑠2 ≤
𝑝−8−2𝑟
2
(21)
(21)
iff
𝑠1 = 𝑠2.
Theorem 7.5. For 𝑝 ≥ 8, the number of non‐isomorphic semigraphs in 𝐴21 of order 𝑝 is given
by 𝑇𝐴21,𝑝 = {
𝑝(𝑝−4)(𝑝−5)
96
, 𝑖𝑓 𝑝 ≡ 0(𝑚𝑜𝑑4)
(𝑝−1)(𝑝−2)(𝑝−6)
96
, 𝑖𝑓 𝑝 ≡ 2(𝑚𝑜𝑑4) .
′
In any semigraph G𝐺 in 𝐴22
′
, there are edges of the form (𝑥, 𝑦), (𝑢1, 𝑦, 𝑧, 𝑢2), (𝑧, 𝑥, 𝑢3).
Furthermore, all the vertices other than 𝑢1, 𝑢2, 𝑢3, 𝑥, 𝑦 and 𝑧, lie in a 3-semiedge with a
𝑚−vertex 𝑥, 𝑦 or 𝑧.
If there are 𝑟(≥ 0) 3-semiedges with 𝑥 as a 𝑚-vertex and 𝑠(≥ 0) 3-semiedges with 𝑦 as a
𝑚 −vertex, then there are
𝑝−6−2𝑟−2𝑠
2
3-semiedges with 𝑧 as a 𝑚 −vertex.
Since
𝑝−6−2𝑟−2𝑠
2
≥ 0, we have 0 ≤ 𝑟 ≤
𝑝−6
2
, 0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
and 𝑝 ≥ 6. Hence any
semigraph in 𝐴22
′
(22)
(22)
= (𝑉, 𝑋) with
𝑉 = {𝑢1, 𝑢2, 𝑢3, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑠, 𝑐1, … , 𝑐2𝑡}and 𝑋 =
{(𝑥, 𝑦), (𝑢1, 𝑦, 𝑧, 𝑢2), (𝑧, 𝑥, 𝑢3)} ∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟}
0 ≤ 𝑟 ≤
𝑝−6
2
, 0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
and 𝑡 =
𝑝−6−2𝑟−2𝑠
2
≥ 0.
Since the role of the 𝑚−vertices 𝑥 and 𝑦 are not interchangeable, it follows that 𝐴𝑟,𝑠,𝑝
(22)
≇
𝐴𝑠,𝑟,𝑝
(22)
. Then we get the next two results.

𝑭𝒊𝒈𝒖𝒓𝒆. 𝟖 𝐴1,1,10
(22)
Lemma 7.6. For 0 ≤ 𝑟 ≤
𝑝−6
2
and 0 ≤ 𝑠1, 𝑠2 ≤
𝑝−6−2𝑟
2
(22)
(22)
iff
𝑠1 = 𝑠2.
(22)
= (𝑉, 𝑋) with
𝑉 = {𝑢1, 𝑢2, 𝑢3, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑠1
, 𝑐1, … , 𝑐2𝑡1
} and
𝑋 = {(𝑥, 𝑦), (𝑢1, 𝑦, 𝑧, 𝑢2), (𝑧, 𝑥, 𝑢3)} ∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟}
∪ {(𝑏𝑗, 𝑦, 𝑏𝑗+𝑠1
)|1 ≤ 𝑗 ≤ 𝑠1} ∪ {(𝑐𝑘, 𝑧, 𝑐𝑘+𝑡1)|1 ≤ 𝑘 ≤ 𝑡1},
0 ≤ 𝑟 ≤
𝑝−6
2
, 0 ≤ 𝑠1 ≤
𝑝−6−2𝑟
2
and 𝑡1 =
𝑝−6−2𝑟−2𝑠1
2
≥ 0.
(22)
= (𝑉′
, 𝑋′) with
𝑉′
= {𝑢1
′
, 𝑢2
′
, 𝑢3
′
, 𝑥′
, 𝑦′
, 𝑧′
, 𝑎1
′
, … , 𝑎2𝑟
′
, 𝑏1
′
, … , 𝑏2𝑠2
′
, 𝑐1
′
, … , 𝑐2𝑡2
′
} and
𝑋′
= {(𝑥′, 𝑦′), (𝑢′
1, 𝑦′, 𝑧′, 𝑢′
2), (𝑧′, 𝑥′, 𝑢′3)} ∪ {(𝑎′
𝑖, 𝑥′, 𝑎′𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟}
∪ {(𝑏′
𝑗, 𝑦′, 𝑏′
𝑗+𝑠2
)|1 ≤ 𝑗 ≤ 𝑠2} ∪ {(𝑐′
𝑘, 𝑧′, 𝑐′
𝑘+𝑡2)|1 ≤ 𝑘 ≤ 𝑡2},
0 ≤ 𝑟 ≤
𝑝−6
2
, 0 ≤ 𝑠2 ≤
𝑝−6−2𝑟
2
and 𝑡2 =
𝑝−6−2𝑟−2𝑠2
2
≥ 0.
In any isomorphism 𝜃 between 𝐴𝑟,𝑠1,𝑝
(22)
(22)
, we have 𝜃((𝑥, 𝑦)) = (x’, 𝑦′); and so
{𝜃(𝑥), 𝜃(𝑦)} = {𝑥′
, 𝑦′}.
Now, by counting the number of 3‐semiedges with 𝑥, 𝑦, 𝑥′
, 𝑦′
as 𝑚‐vertices, we have
{𝑟, 𝑠1} = {𝑟, 𝑠2}. Hence 𝑠1 = 𝑠2. Converse is obvious.
by 𝑇𝐴22,𝑝
=
𝑝(𝑝−2)(𝑝−4)
48
.
(22)
𝑤𝑖𝑡ℎ 0 ≤ 𝑟 ≤
𝑝−6
2
, 0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
are
non‐isomorphic.
Thus for a fixed 𝑟, the number of non‐isomorphic semigraphs of the form 𝐴𝑟,𝑠,𝑝
(22)
is
(
𝑝−6−2𝑟
2
+ 1) =
𝑝−4−2𝑟
2
.
Hence the number of non‐isomorphic semigraphs of order 𝑝(≥ 6) in 𝐴′22 is given by
𝑇𝐴′22,𝑝
= ∑
𝑝−6
2
𝑟=0
𝑝−4−2𝑟
2
=
(𝑝−4)(𝑝−2)
8
.
For any 𝐺 ∈ 𝐴22, one of the component of 𝐺 is 𝐴𝑟,𝑠,𝑝
(22)
, with 0 ≤ 𝑟 ≤
𝑝−6
2
,
0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
; 𝑎𝑛𝑑 all the other components of 𝐺 are 𝐾2.

M.S. Chithra
𝑇𝐴22,𝑝
= ∑𝑛 𝑒𝑣𝑒𝑛 6≤𝑛≤𝑝 𝑇𝐴22,𝑛
′ = ∑𝑛 𝑒𝑣𝑒𝑛 6≤𝑛≤𝑝
(𝑛−4)(𝑛−2)
8
.
On simplification, we get the result.
′
′
, there are edges of the form (𝑢1, 𝑥, 𝑦, 𝑢2), (𝑦, 𝑧, 𝑢3), (𝑢4, 𝑧, 𝑥)
and (𝑧, 𝑤1). Furthermore, all the vertices other than 𝑢1, 𝑢2, 𝑢3, 𝑢4, 𝑤1, 𝑥, 𝑦 and 𝑧, lie in a 3‐
semiedge with a 𝑚−vertex 𝑥, 𝑦 or 𝑧.
If there are 𝑟(≥ 0) 3‐semiedges with 𝑥 as a 𝑚‐vertex and 𝑠(≥ 0) 3‐semiedges with 𝑦 as
a 𝑚 −vertex, then there are
𝑝−8−2𝑟−2𝑠
2
3‐semiedges with 𝑧 as a 𝑚 −vertex. Since
𝑝−8−2𝑟−2𝑠
2
≥
0, we have 0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
′
can be
(23)
(23)
= (𝑉, 𝑋) with
𝑉 = {𝑢1, … , 𝑢4, 𝑤1, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑠, 𝑐1, … , 𝑐2𝑡} and
𝑋 = {(𝑢1, 𝑥, 𝑦, 𝑢2), (𝑦, 𝑧, 𝑢3), (𝑢4, 𝑧, 𝑥), (𝑧, 𝑤1)} ∪ {1 ≤ 𝑖 ≤ 𝑟}
∪ {(𝑏𝑗, 𝑦, 𝑏𝑗+𝑠)|1 ≤ 𝑗 ≤ 𝑠} ∪ {(𝑐𝑘, 𝑧, 𝑐𝑘+−)|1 ≤ 𝑘 ≤ 𝑡},
0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
and 𝑡 =
𝑝−8−2𝑟−2𝑠
2
≥ 0.
Since the role of the 𝑚−vertices 𝑥 and 𝑦 are interchangeable, it follows that 𝐴𝑟,𝑠,𝑝
(23)
≅
𝐴𝑠,𝑟,𝑝
(23)
𝑝−8
4
⌋ and 𝑟 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
.
Then we get the next two results as in Section 6.1.
𝑭𝒊𝒈𝒖𝒓𝒆. 𝟗 𝐴1,1,12
(23)
Lemma 7.8. For 0 ≤ 𝑟 ≤ ⌊
𝑝−8
4
⌋ and 𝑟 ≤ 𝑠1, 𝑠2 ≤
𝑝−8−2𝑟
2
(23)
(23)
iff
𝑠1 = 𝑠2.
by 𝑇𝐴23,𝑝 = {
𝑝(𝑝−4)(𝑝−5)
96
, 𝑖𝑓 𝑝 ≡ 0(𝑚𝑜𝑑4)
(𝑝−1)(𝑝−2)(𝑝−6)
96
, 𝑖𝑓 𝑝 ≡ 2(𝑚𝑜𝑑4) .
′
′
, there are edges of the form (𝑢1, 𝑥, 𝑦, 𝑢2), (𝑢3, 𝑦, 𝑧, 𝑢4), (𝑧, 𝑥, 𝑢5),
(𝑥, 𝑤1) and (𝑦, 𝑤2). Furthermore, all the vertices other than 𝑢1, 𝑢2, 𝑢3, 𝑢4, 𝑢5, 𝑤1, 𝑤2, 𝑥, 𝑦 and
𝑧, lie in a 3‐semiedge with a 𝑚−vertex 𝑥, 𝑦 or 𝑧.
If there are 𝑟(≥ 0) 3‐semiedges with 𝑥 as a 𝑚‐vertex and 𝑠(≥ 0) 3‐semiedges with 𝑧 as
𝑝−10−2𝑟−2𝑠
2
3‐semiedges with 𝑦 as a 𝑚 −vertex. Since

𝑝−10−2𝑟−2𝑠
2
≥ 0, we have 0 ≤ 𝑟 ≤
𝑝−10
2
, 0 ≤ 𝑠 ≤
𝑝−0−2𝑟
2
and 𝑝 ≥ 10. Hence any semigraph
in 𝐴24
′
(24)
(24)
= (𝑉, 𝑋) with
𝑉 = {𝑢1, … , 𝑢5, 𝑤1, 𝑤2, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑡, 𝑐1, … , 𝑐2𝑠} and
𝑋 = {(𝑢1, 𝑥, 𝑦, 𝑢2), (𝑢3, 𝑦, 𝑧, 𝑢4), (𝑧, 𝑥, 𝑢5), (𝑥, 𝑤1), (𝑦, 𝑤2)}
∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟} ∪ {(𝑏𝑗, 𝑦, 𝑏𝑗+𝑡)|1 ≤ 𝑗 ≤ 𝑡}
∪ {1 ≤ 𝑘 ≤ 𝑠}, 0 ≤ 𝑟 ≤
𝑝 − 10
2
, 0 ≤ 𝑠 ≤
𝑝 − 10 − 2𝑟
2
and 𝑡 =
𝑝−10−2𝑟−2𝑠
2
≥ 0.
Since the role of the 𝑚−vertices 𝑥 and 𝑧 are not interchangeable, it follows that 𝐴𝑟,𝑠,𝑝
(24)
≇
𝐴𝑠,𝑟,𝑝
(24)
. Then we get the next two results.
Figure 10 𝐴1,0,14
(24)
Lemma 7.10. For 0 ≤ 𝑟 ≤
𝑝−10
2
and 0 ≤ 𝑠1, 𝑠2 ≤
𝑝−10−2𝑟
2
(24)
(24)
iff
𝑠1 = 𝑠2.
(24)
= (𝑉, 𝑋) with
𝑉 = {𝑢1, … , 𝑢5, 𝑤1, 𝑤2, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑡1
, 𝑐1, … , 𝑐2𝑠1
} and
𝑋 = {(𝑢1, 𝑥, 𝑦, 𝑢2), (𝑢3, 𝑦, 𝑧, 𝑢4), (𝑧, 𝑥, 𝑢5), (𝑥, 𝑤1), (𝑦, 𝑤2)}
∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟} ∪ {(𝑏𝑗, 𝑦, 𝑏𝑗+𝑡1
)|1 ≤ 𝑗 ≤ 𝑡1} ∪ {1 ≤ 𝑘 ≤ 𝑠1},
0 ≤ 𝑟 ≤
𝑝−10
2
, 0 ≤ 𝑠1 ≤
𝑝−10−2𝑟
2
and 𝑡1 =
𝑝−10−2𝑟−2𝑠1
2
≥ 0.
(24)
= (𝑉′, 𝑋′) with
𝑉′
= {𝑢′
1, … , 𝑢′
5, 𝑤′
1, 𝑤′
2, 𝑥′
, 𝑦′
, 𝑧′
, 𝑎′
1, … , 𝑎′
2𝑟, 𝑏′
1, … , 𝑏′
2𝑡2
, 𝑐′
1, … , 𝑐′
2𝑠2
}and
𝑋′ = {(𝑢′
1, 𝑥′, 𝑦′, 𝑢′
2), (𝑢′
3, 𝑦′, 𝑧′, 𝑢′
4), (𝑧′, 𝑥′, 𝑢′
5), (𝑥′, 𝑤′
1), (𝑦′, 𝑤′2)}
∪ {(𝑎′
𝑖, 𝑥′, 𝑎′
𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟} ∪ {(𝑏′
𝑗, 𝑦′, 𝑏′𝑗+𝑡2
)|1 ≤ 𝑗 ≤ 𝑡2}
∪ {(𝑐𝑘, 𝑧, 𝑐𝑘+𝑠)|1 ≤ 𝑘 ≤ 𝑠2}, 0 ≤ 𝑟 ≤
𝑝 − 10
2
, 0 ≤ 𝑠2 ≤
𝑝 − 10 − 2𝑟
2
and 𝑡2 =
𝑝−10−2𝑟−2𝑠2
2
≥ 0.
(24)
(24)
, we have 𝜃((𝑧, 𝑥, 𝑢5)) = (𝑧′
, 𝑥′
, 𝑢5
′
) .
Hence 𝜃(𝑥) = 𝑥′
and 𝜃(𝑧) = 𝑧′
. Now, by counting the number of 3‐semiedges with 𝑧, 𝑧′
as 𝑚‐vertices, we have 𝑠1 = 𝑠2. Converse is obvious.

M.S. Chithra
Theorem 7.11. For 𝑝 > 10, the number of non‐isomorphic semigraphs in 𝐴24 of order 𝑝 is
=
(𝑝−4)(𝑝−6)(𝑝−8)
48
.
(24)
with 0 ≤ 𝑟 ≤
𝑝−10
2
and 0 ≤ 𝑠 ≤
𝑝−10−2𝑟
2
non‐isomorphic.
Thus for a fixed 𝑟, the number of non‐isomorphic semigraphs of the form 𝐴𝑟,𝑠,𝑝
(24)
is
(
𝑝−10−2𝑟
2
+ 1) =
𝑝−8−2𝑟
2
.
Hence the number of non‐isomorphic semigraphs of order 𝑝(≥ 10) in 𝐴′24 is 𝑇𝐴′24,𝑝 =
∑
(𝑝−10)
2
𝑟=0
(𝑝−8−2𝑟)
2
=
(𝑝−8)(𝑝−6)
8
.
Now, for G ∈ 𝐴24, 𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐺 𝑖𝑠 𝐴𝑟,𝑠,𝑝
(24)
, with with 0 ≤ 𝑟 ≤
𝑝−10
2
and
0 ≤ 𝑠 ≤
𝑝−10−2𝑟
2
𝑇𝐴24,𝑝 = ∑𝑛≤𝑒𝑣𝑒𝑛 10≤𝑛≤𝑝 𝑇𝐴24,𝑛
′ = ∑𝑛𝑒𝑣𝑒𝑛 10≤𝑛≤𝑝
(𝑛−8)(𝑛−6)
8
.
′
′
, there are edges of the form (𝑢1, 𝑥, 𝑦, 𝑢2), (𝑦, 𝑧, 𝑢3) and (𝑧, 𝑥, 𝑢4)
and (𝑥, 𝑤1). Furthermore, all the vertices other than 𝑢1, 𝑢2, 𝑢3, 𝑢4, 𝑤1, 𝑥, 𝑦 and 𝑧, lie in a 3‐
semiedge with a 𝑚−vertex 𝑥, 𝑦 or 𝑧.
If there are 𝑟(≥ 0) 3‐semiedges with 𝑥 as a 𝑚‐vertex and 𝑠(≥ 0) 3‐semiedges with 𝑦 as
𝑝−8−2𝑟−2𝑠
2
3‐semiedges with 𝑧 as a 𝑚 −vertex. Since
𝑝−8−2𝑟−2𝑠
2
≥
0, we have 0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
′
can be
(25)
(25)
= (𝑉, 𝑋) with
𝑉 = {𝑢1, … , 𝑢4, 𝑤1, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑠, 𝑐1, … , 𝑐2𝑡}and
𝑋 = {(𝑢1, 𝑥, 𝑦, 𝑢2), (𝑦, 𝑧, 𝑢3), (𝑧, 𝑥, 𝑢4), (𝑥, 𝑤1)} ∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟}
∪ {(𝑏𝑗, 𝑦, 𝑏𝑗+𝑠)|1 ≤ 𝑗 ≤ 𝑠} ∪ {(𝑐𝑘, 𝑧, 𝑏𝑘+𝑡 )|1 ≤ 𝑘 ≤ 𝑡},
0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠 ≤
𝑝−8−2𝑟
2
and 𝑡 =
𝑝−8−2𝑟−2𝑠
2
≥ 0.
Since the role of the 𝑚−vertices 𝑥 and 𝑦 are not interchangeable, it follows that 𝐴𝑟,𝑠,𝑝
(25)
≇
𝐴𝑠,𝑟,𝑝
(25)
.
Then we get the next two results.

Figure 11 𝐴1,1,12
(25)
Lemma 7.12. For 0 ≤ 𝑟 ≤
𝑝−8
2
and 0 ≤ 𝑠1, 𝑠2 ≤
𝑝−8−2𝑟
2
(25)
(25)
iff
𝑠1 = 𝑠2.
Proof: Let 𝐴𝑟,𝑠1,𝑝
(25)
= (𝑉, 𝑋) with
𝑉 = {𝑢1, … , 𝑢4, 𝑤1, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑠1
, 𝑐1, … , 𝑐2𝑡1
} and
𝑋 = {(𝑢1, 𝑥, 𝑦, 𝑢2), (𝑦, 𝑧, 𝑢3), (𝑧, 𝑥, 𝑢4), (𝑥, 𝑤1)} ∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟}
∪ {(𝑏𝑗, 𝑦, 𝑏𝑗+𝑠1
)|1 ≤ 𝑗 ≤ 𝑠1} ∪ {(𝑐𝑘, 𝑧, 𝑏𝑘+𝑡1
)|1 ≤ 𝑘 ≤ 𝑡},
0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠1 ≤
𝑝−8−2𝑟
2
and𝑡 =
𝑝−8−2𝑟−2𝑠1
2
≥ 0.
(25)
= (𝑉′, 𝑋′) with
𝑉′ = {𝑢′
1, … , 𝑢′
4, 𝑤′
1, 𝑥′, 𝑦′, 𝑧′, 𝑎′1, … , 𝑎′2𝑟, 𝑏′1, … , 𝑏′2𝑠2
, 𝑐1, … , 𝑐2𝑡2
} and
𝑋′ = {(𝑢′
1, 𝑥′, 𝑦′, 𝑢′
2), (𝑦′, 𝑧′, 𝑢′
3), (𝑧′, 𝑥′, 𝑢′
4), (𝑥′, 𝑤′
1)} ∪ {(𝑎′
𝑖, 𝑥′, 𝑎′𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟}
∪ {(𝑏𝑗
′
, 𝑦′
, 𝑏𝑗+𝑠2
′
)|1 ≤ 𝑗 ≤ 𝑠2} ∪ {(𝑐𝑘
′
, 𝑧′
, 𝑐𝑘+𝑡2
′
)|1 < 𝑘 ≤ 𝑡2},
0 ≤ 𝑟 ≤
𝑝−8
2
, 0 ≤ 𝑠2 ≤
𝑝−8−2𝑟
2
and 𝑡2 =
𝑝−8−2𝑟−2𝑠2
2
≥ 0.
(25)
(25)
, we have
𝜃((𝑢1, 𝑥, 𝑦, 𝑢2)) = (𝑢1
′
, 𝑥′
, 𝑦′
, 𝑢2
′
) ; and so {𝜃(𝑥), 𝜃(𝑦)} = {𝑥′
, 𝑦′}.
Now, by counting the number of 3‐semiedges with 𝑥, 𝑦, 𝑥′
, 𝑦′
as 𝑚 −vertices, we have
{𝑟, 𝑠1} = {𝑟, 𝑠2}. Converse is obvious.
Proceeding as in Theorem 6.5, we get the next result.
Theorem 7.13. For 𝑝 > 8, the number of non‐isomorphic semigraphs in 𝐴25 of order 𝑝 is given
by 𝑇𝐴25,𝑝 =
(𝑝−2)(𝑝−4)(𝑝−6)
48
.
′
′
, there are edges of the form (𝑢1, 𝑥, 𝑦, 𝑢2), (𝑢3, 𝑦, 𝑧, 𝑢4),
(𝑢5, 𝑧, 𝑥, 𝑢6), (𝑥, 𝑤1), (𝑦, 𝑤2) and (𝑧, 𝑤3). Furthermore, all the vertices other than
𝑢1, 𝑢2, 𝑢3, 𝑢4, 𝑢5, 𝑢6, 𝑤1, 𝑤2, 𝑤3, 𝑥, 𝑦 and 𝑧, lie in a 3‐semiedge with a 𝑚−vertex 𝑥, 𝑦 or 𝑧.
Let there be 𝑟(≥ 0) 3‐semiedges with 𝑥 as a 𝑚 −vertex, 𝑠(≥ 0) 3‐semiedges with 𝑦 as a
𝑚‐vertex and 𝑡(≥ 0) 3-semiedges with 𝑧 as a 𝑚 −vertex. Since
𝑝−12−2𝑟−2𝑠
2
≥ 0, we have 0 ≤
𝑟 ≤
𝑝−12
2
, 0 ≤ 𝑠 ≤
𝑝−12−2𝑟
2
′
can be denoted
by 𝐴𝑟,𝑠,𝑡,𝑝
(26)
, where 𝐴𝑟,𝑠,𝑡,𝑝
(26)
= (𝑉, 𝑋) with

M.S. Chithra
𝑉 = {𝑢1, … , 𝑢6, 𝑤1, 𝑤2, 𝑤3, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑠, 𝑐1, … , 𝑐2𝑡} and
𝑋 = {(𝑢1, 𝑥, 𝑦, 𝑢2), (𝑢3, 𝑦, 𝑧, 𝑢4), (𝑢5, 𝑧, 𝑥, 𝑢6), (𝑥, 𝑤1), (𝑦, 𝑤2), (𝑧, 𝑤3)}
∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟} ∪ {(𝑏𝑗, 𝑦, 𝑏𝑗+𝑠)|1 ≤ 𝑗 ≤ 𝑠} ∪ {1 ≤ 𝑘 ≤
𝑡}, 0 ≤ 𝑟 ≤
𝑝−12
2
, 0 ≤ 𝑠 ≤
𝑝−12−2𝑟
2
and 𝑡 =
𝑝−12−2𝑟−2𝑠
2
≥ 0.
Since the role of the 𝑚−vertices 𝑥, 𝑦 and 𝑧 are interchangeable, it is easy to note that, if
𝑟, 𝑠, 𝑡 are non‐negative integers with 𝑟 ≤ 𝑠 ≤ 𝑡 and 𝑟 + 𝑠 + 𝑡 =
𝑝−12
2
, then, upto isomorphism,
there is exactly one semigraph 𝐴𝑟,𝑠,𝑡,𝑝
(26)
in 𝐴26
′
.
𝑭𝒊𝒈𝒖𝒓𝒆 𝟏𝟐 𝐴0,1,1,16
(26)
Theorem 7.14. For 𝑝 ≥ 12, the number of non‐isomorphic semigraphs in 𝐴26 of order 𝑝 is
= {
𝑝3−15𝑝2+60𝑝
288
, 𝑖𝑓 𝑝 ≡ 0(𝑚𝑜𝑑12)
𝑝3−15𝑝2+60𝑝−68
288
, 𝑖𝑓𝑝 ≡
2(𝑚𝑜𝑑12)
𝑝3−15𝑝2+60𝑝−64
288
, 𝑖𝑓𝑝 ≡ 4(𝑚𝑜𝑑12)
𝑝3−15𝑝2+60𝑝−36
288
, 𝑖𝑓𝑝 ≡
6(𝑚𝑜𝑑12)
𝑝3−15𝑝2+60𝑝−32
288
, 𝑖𝑓 𝑝 ≡ 8(𝑚𝑜𝑑12)
𝑝3−15𝑝2+60𝑝−100
288
, 𝑖𝑓𝑝 ≡ 10(𝑚𝑜𝑑12) .
Proof: Any semigraph in 𝐴26
′
is of the form 𝐴𝑟,𝑠,𝑡,𝑝
(26)
, with 𝑟 ≤ 𝑠 ≤ 𝑡.
Without loss of generality let 𝑟 = 𝑛1, 𝑠 = 𝑛1 + 𝑛2 and 𝑡 = 𝑛1 + 𝑛2 + 𝑛3, where 𝑛1, 𝑛2, 𝑛3 ≥
0.
Since 𝑟 + 𝑠 + 𝑡 =
𝑝−12
2
, 3𝑛1 + 2𝑛2 + 𝑛3 =
𝑝−12
2
. Hence 0 ≤ 𝑛1 ≤
𝑝−12
6
.
For a fixed 𝑛1, 2𝑛2 + 𝑛3 =
𝑝−12
2
− 3𝑛1 and so 0 ≤ 𝑛2 ≤ ⌊
𝑝−12
2
−3𝑛1
2
⌋.
For a fixed 𝑛1 and 𝑛2, 𝑛3 =
𝑝−12
2
− 2𝑛2 − 3𝑛1.
So, for a fixed 𝑛1, the number of semigraphs in 𝐴26
′
is ⌊
𝑝−12
2
−3𝑛1
2
⌋ + 1.
Then the number of non‐isomorphic semigraphs in 𝐴26
′
= 𝑇𝐴26,𝑝
′ = ∑
⌊
𝑝−12
6
⌋
𝑛1=0 ⌊
𝑝−8−6𝑛1
4
⌋.
Hence the number of non‐isomorphic semigraphs in 𝐴26 = 𝑇𝐴26,𝑝
= ∑12≤𝑛≤𝑝 𝑛 𝑒𝑣𝑒𝑛 𝑇𝐴26,𝑛
′ .
′
′
, there are edges of the form (𝑥, 𝑦, 𝑢1), (𝑦, 𝑧, 𝑢2) and (𝑧, 𝑥, 𝑢3).
Furthermore, all the vertices other than 𝑢1, 𝑢2, 𝑢3, 𝑥, 𝑦 and 𝑧, lie in a 3‐semiedge with a
𝑚−vertex 𝑥, 𝑦 or 𝑧.

Let there be 𝑟(≥ 0) 3‐semiedges with 𝑥 as a 𝑚 −vertex, 𝑠(≥ 0) 3‐semiedges with 𝑦 as a
𝑚‐vertex and 𝑡(≥ 0) 3-semiedges with 𝑧 as a 𝑚 −vertex. Since
𝑝−6−2𝑟−2𝑠
2
≥ 0, we have
0 ≤ 𝑟 ≤
𝑝−6
2
, 0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
′
can be denoted
by 𝐴𝑟,𝑠,𝑡𝑝
(27)
, where 𝐴𝑟,𝑠,𝑡,𝑝
(27)
= (𝑉, 𝑋) with
𝑉 = {𝑢1, 𝑢2, 𝑢3, 𝑥, 𝑦, 𝑧, 𝑎1, … , 𝑎2𝑟, 𝑏1, … , 𝑏2𝑠, 𝑐1, … , 𝑐2𝑡} and
𝑋 = {(𝑥, 𝑦, 𝑢1), (𝑦, 𝑧, 𝑢2), (𝑧, 𝑥, 𝑢3)} ∪ {(𝑎𝑖, 𝑥, 𝑎𝑖+𝑟)|1 ≤ 𝑖 ≤ 𝑟}
0 ≤ 𝑟 ≤
𝑝−6
2
, 0 ≤ 𝑠 ≤
𝑝−6−2𝑟
2
and 𝑡 =
𝑝−6−2𝑟−2𝑠
2
≥ 0.
Since the role of the 𝑚−vertices 𝑥, 𝑦 and 𝑧 are interchangeable, it is easy to note that, if
𝑟, 𝑠, 𝑡 are non-negative integers with 𝑟 ≤ 𝑠 ≤ 𝑡 and 𝑟 + 𝑠 + 𝑡 =
𝑝−6
2
, then, upto isomorphism,
there is exactly one semigraph 𝐴𝑟,𝑠,𝑡,𝑝
(27)
in 𝐴27
′
.
Figure 13 𝐴1,1,1,12
(27)
Theorem 7.15. For 𝑝 ≥ 6, the number of non‐isomorphic semigraphs in 𝐴27 is given by
𝑇𝐴27,𝑝 = {
𝑝3
+ 3𝑝2
− 12𝑝
288
, 𝑖𝑓 𝑝 ≡ 0(𝑚𝑜𝑑12)
𝑝3
+ 3𝑝2
− 12𝑝 + 4
288
, 𝑖𝑓 𝑝
≡ 2(𝑚𝑜𝑑12)
𝑝3
+ 3𝑝2
− 12𝑝 − 64
288
, 𝑖𝑓𝑝
≡ 4(𝑚𝑜𝑑12)
𝑝3
+ 3𝑝2
− 12𝑝 + 36
288
, 𝑖𝑓𝑝
≡ 6(𝑚𝑜𝑑12)
𝑝3
+ 3𝑝2
− 12𝑝 − 32
288
, 𝑖𝑓𝑝
≡ 8(𝑚𝑜𝑑12)
𝑝3
+ 3𝑝2
− 12𝑝 − 28
288
, 𝑖𝑓𝑝 ≡ 10(𝑚𝑜𝑑12) .
Proof : Any semigraph in 𝐴27
′
is of the form 𝐴𝑟,𝑠,𝑡,𝑝
(27)
, with r ≤ 𝑠 ≤ 𝑡.
Without loss of generality, let 𝑟 = 𝑛1, 𝑠 = 𝑛1 + 𝑛2 and 𝑡1 = 𝑛1 + 𝑛2 + 𝑛3, 𝑛1, 𝑛2, 𝑛3 ≥ 0.
Since𝑟 + 𝑠 + 𝑡 =
𝑝−6
2
, 3𝑛1 + 2𝑛2 + 𝑛3 =
𝑝−6
2
. Hence 0 ≤ 𝑛1 ≤
𝑝−6
6
.
For a fixed 𝑛1, 2𝑛2 + 𝑛3 =
𝑝−6
2
and so 0 ≤ 𝑛2 ≤ ⌊
𝑝−6
2
−3𝑛1
2
⌋.
For a fixed 𝑛1 𝑎𝑛𝑑 𝑛2, 𝑛3 =
𝑝−6
2
− 2𝑛2 − 3𝑛1.

M.S. Chithra
So, for a fixed 𝑛1, the number of semigraphs in 𝐴27
′
is ⌊
𝑝−6
2
−3𝑛1
2
⌋ +1.
Then the number of non-isomorphic semigraphs in 𝐴27
′
= 𝑇𝐴′
27,𝑝 = ∑
(𝑝−6)
6
𝑛1=0 ⌊
(𝑝−2−6𝑛1)
4
⌋.
Hence the number of non-isomorphic semigraphs in 𝐴27
′
= 𝑇𝐴′
27,𝑝 = ∑0≤𝑛≤𝑃 𝑛 𝑒𝑣𝑒𝑛 𝑇𝐴′
27,𝑝.
8. CONCLUSION
In this paper, we have studied the structure of 1-regular semigraphs with three m-vertices. In
particular, we have enumerated the 1-regular semigraphs with three mutually adjacent m-
vertices.
We have already enumerated 1-regular semigraphs with three m-vertices, that are not
mutually adjacent in [9,10]. Moreover, we have enumerated 1-regular semigraphs with one or
two m-vertices in [8].
Further research can be carried out for 1-regular semigraphs with (k≥4) m-vertices.
Moreover, in this paper, we have studied one type of regular semigraphs. Similar work can be
extended for regular semigraphs with edge degree, adjacent degree or consecutive adjacent
degree.
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