chapter 2 set-graceful and set-sequential...
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Chapter 2
Set-Graceful and Set-Sequential
Graphs
This chapter contains some foundational results on set valuations of graphs. In section 1,
we establish a necessary condition for a complete graph to be set-graceful. In this section
we prove that a uniform binary tree with one pendant edge added at the root vertex is set-
graceful and establish a necessary and sufficient condition for the graph H +Km to be set-
graceful, where H is a set-graceful graph with n edges and n+1 vertices. We characterize
the complete set-graceful tripartite graph K1,m,n. If Gf is the full augmentation of the
graph G, a set-graceful graph, then we prove that the corona of Gf and K1 is set-graceful.
We also give an embedding of a graph into a set-graceful graph and a cycle into an Eulerian
set-graceful graph. In section 2 we introduce bi set-graceful graphs and establish some
results on bi set-graceful graphs. The main result, in this section is the characterization
of r-regular connected bi set-graceful graphs. Section 3 of this chapter deals with set-
sequential graphs. We characterize complete set-sequential bipartite graphs and identify
some classes of graphs which are not set-sequential. We also prove that G is set-sequential
if and only if G + K1 is set-graceful. If G is a set-graceful graph then we prove that
G∪Kt is set-sequential for some positive integer t and also prove that every graph can be
embedded into a set-sequential graph.
18
Chapter 2 19
2.1 Introduction
Motivated from multiplicity and social relationship Acharya [Ach83]
introduced the idea of assigning subsets of a set to the elements of a
graph. Given a graph G = (V, E) and non-empty sets X, Y, Z any
function
f : V (G) → 2X
f : E(G) → 2Y
f : V (G) ∪ E(G) → 2Z
is called a set-assignment of vertices, edges and elements of G respec-
tively; by simply a set-assignment Acharya defined any of them. Moti-
vated from number valuations of graph, Acharya defined set-indexers,
set-graceful and set-sequential graphs. We find several papers con-
sider with structure of graphs admitting a variety of number valued
functions on their vertex and or edge sets (See, Rosa [Ros67], Golumb
[Gol80] Bloom [Blo77], Slater [Sla81], Grace [Gra82]). The most fa-
mous notion of number valuation of graph is on graceful numberings,
first formulated by Rosa [Ros67]. A graceful labelling of G is an in-
jective function f : V (G) → {0, 1, 2, . . . , q} so that for the induced
function gf : E(G) → N defined from E(G) into the set N of natural
numbers by setting gf(uv) = |f(u) − f(v)| for every uv ∈ E(G), we
have gf(G) = {gf(e) : e ∈ E(G)} = {1, 2, . . . , q}}. A natural set theo-
retical analogue of the ’distance’ |f(u)− f(v)| between numbers f(u)
and f(v) is the symmetric difference f(u)⊕ f(v) when they are taken
to be subsets of set X. Following this idea Acharya defined set-indexer
Chapter 2 20
of a graph G = (V, E) as follows
Definition 2.1.1. [Ach83] Let G = (V, E) be a graph, X be a
nonempty set and 2X denote the set of all subsets of X. A set-indexer
of G is an injective set-valued function f : V (G) → 2X such that the
function f⊕ : E(G) −→ 2X − {∅} defined by f⊕(uv) = f(u)⊕f(v)
for every uv ∈ E(G) is also injective, where ⊕ denotes the symmetric
difference of sets.
Remark 2.1.2. As stated, Definition 2.1.1 can be thought of as
applicable to infinite graphs as well.
Figure 2.1 gives an example of graph with a set-indexer.
Figure 2.1:
He then defined a graph G = (V, E) to be set-graceful if there
exists a non-empty set X and a set-indexer f : V (G) → 2X such that
Chapter 2 21
f⊕(E(G)) = 2X − {∅}, such an indexer being called a set-graceful
labelling of G.
Figure 2.2 gives an example of a set-graceful graph.
Figure 2.2:
The following theorem gives a straight forward necessary condition
for a graph G to be set-graceful.
Theorem 2.1.3. [Ach83] A necessary condition for a graph G =
(V, E) to have a set-graceful labelling with respect to a set X of car-
dinality n is that it be possible to partition V (G) into two subsets Ve
and Vo such that the number of edges joining the vertices of Vo with
those of Ve is exactly 2n−1.
Remark 2.1.4. If a (p, q)-graph is set-graceful then q = 2m − 1 for
some positive integer m. This implies almost all graphs of order p,
and hence almost all graphs are not set-graceful. Further, for every
Chapter 2 22
positive integer m, there exists a set-graceful graph of size q = 2m−1.
However, not all (p, q)-graphs with q = 2m − 1 are set-graceful as, for
instance, it is not difficult to verify that the path P8 is not set-graceful.
More generally the following more results are well known.
Theorem 2.1.5. [MP89] The complete graph Kn is set-graceful if
and only if n ∈ {2, 3, 6}.
Theorem 2.1.6. [MP89] The cycle Cn is set-graceful if and only if
n = 2m − 1, for some integer, m ≥ 2.
Definition 2.1.7. A caterpillar is a tree having a path that contains
at least one vertex of every edge.
Definition 2.1.8. For a graph G the splitting graph S ′(G) is ob-
tained from G by adding for each vertex v of G a new vertex v′ so
that v′ is adjacent to every vertex that is adjacent to v .
Definition 2.1.9. The corona of two graphs G1 and G2 is the graph
G = G1 � G2 formed from one copy of G1 and |V (G1)| copies of G2
where the ith vertex of G1 is adjacent to every vertex in the ith copy
of G2.
Definition 2.1.10. The Line graph L(G) of a graph G is defined
as a graph with E(G) as its vertex set with two vertices of L(G) are
adjacent whenever the corresponding edges of G are adjacent.
Definition 2.1.11. A graph G is said to be bi set-graceful if both
G and its line graph are set-graceful.
Chapter 2 23
Definition 2.1.12. [Ach83] A graph G is said to be set-sequential
if there exists a nonempty set X and a bijective set-valued function
f : V (G)∪E(G) → 2X −{∅} such that f⊕(uv) = f(u)⊕f(v) for every
uv ∈ E(G), where ⊕ denotes the symmetric difference of sets.
Figure 2.3 gives an example of a set-sequential graph, with respect
to a set X = {1, 2, 3, 4}.
Figure 2.3:
Definition 2.1.13. Twig is a graph obtained from a path by at-
taching exactly two pendent edges to each internal vertex of the path.
Definition 2.1.14. For any integer n ≥ 4, the Wheel Wn is the
n+1-vertex graph obtained by joining a vertex to each of the n vertices
of the cycle Cn.
2.2 Set-Graceful Graphs
In this section we will obtain some new results on set-graceful graphs.
The following theorem gives a necessary condition for the complete
Chapter 2 24
graph to be set-graceful.
Theorem 2.2.1. A necessary condition for a complete graph Kn
to be set-graceful with respect to a set X is that (n − 2) is a perfect
square.
Proof. Assume Kn is set-graceful with respect to a set X of car-
dinality, say m ≥ 2. Then n(n−1)2 = 2m − 1 ⇒ n2−n
2 = 2m − 1 ⇒
2m+1 = n2 − n + 2. By Theorem 2.1.3 there exists a partition of
the vertex set of Kn into two nonempty subsets A and B such that
there exists 2m−1edges having their both ends not belonging to ei-
ther A or B. Let |A| = a and |B| = b. Then a + b = n. Number
of edges between A and B in Kn is ab. Hence ab = 2m−1. Now,
(a− b)2 = n2 − 4.2m−1 = n2 − 2m+1 = n2 − (n2 − n + 2) = n− 2, is a
perfect square. ♣
Remark 2.2.2. The condition is not sufficient as K11 is not set-
graceful, by theorem 2.1.5.
It is interesting to note that every set-graceful (p, q)-graph G =
(V, E) with respect to a set X of cardinality n can be embedded in
a set-graceful (q + 1, q)-graph H. This may be achieved as follows.
Let f be a set-indexer of G. Then 2X − f(V ) has m(G) = 2n − p
elements each of which does not appear as a set assigned to a vertex
in (G, f). Then adjoin m(G) isolated points to G and assign to them
the sets from 2X − f(V ). Every set-graceful graph with q edges and
q + 1 vertices is considered as fully augmented.
We shall soon see that such graphs when they are set-graceful become
Chapter 2 25
amenable to generate an infinite ascending chain of fully augmented
set-graceful graphs each of which contains its predecessor as an in-
duced subgraph.
Theorem 2.2.3. If H is a set-graceful graph with n edges (n ≥ 1)
and n + 1 vertices then the join of H and Km is set-graceful if and
only if m = 2n1 − 1, n1 ∈ N .
Proof. Let G = H + Km. Since H is set-graceful H contains 2k − 1
edges and 2k vertices for some k ∈ N . Take V (H) = {u, u1, u2, . . . , u2k−1}
and V (Km) = {v1, v2, . . . , vn1} ∪ {wl
j, 2 ≤ j ≤ n1, 1 ≤ l ≤ 2j−1 − 1}.
Now |V (Km)| = n1 + 2n1−1 − 1 + 2n1−2 − 1 + · · · + 2n1−(n1−1) − 1 =
1+2+22+· · ·+2n1−1 = 2n1−1. Let X, X1, X2, . . . , Xn1be the sets with
X ⊂ X1 ⊂ X2, . . . ,⊂ Xn1where |X| = k, |X1| = k + 1, . . . , |Xn1
| =
k + n1 and X = {1, 2, . . . , k},
Xi = X ∪ {(k + 1), (k + 2), . . . , (k + i)}, 1 ≤ i ≤ n. Define the la-
belling f : V (G) → 2Xn1 as follows: Label the vertex u by ∅ and ui by
subsets of X in such a way that H is set-graceful. Label v′is and wljs
as
f(vi) = Xi, 1 ≤ i ≤ n1
f(w12) = X2 −X1
label wlj by proper subsets of Xj−X which contains the element k+j,
for j > 2. Clearly f⊕(e), e ∈ E(G) are all distinct and f⊕(E(G)) =
2Xn1 − {∅}. Hence f is a set-graceful labelling of G. Therefore G is
set-graceful.
Conversely suppose G is set-graceful. Then |E(G)| = 2km + 2k − 1 =
Chapter 2 26
2x − 1 for some x ∈ N . That is 2k(m + 1) = 2x ⇒ m + 1 = 2x−k, put
x− k = x1, then m = 2x1 − 1 where x1 ∈ N . ♣
Figure 2.4 illustrates the theorem, when X = {1, 2} and m = 23−1.
Figure 2.4:
Corollary 2.2.4. If Sn denote the star with 2n− 1 spokes and m =
2n1 − 1 for n1 ∈ N , then the join Sn + Km is set-graceful.
Corollary 2.2.5. For any set-graceful tree T , the join T + Kn is
set-graceful for all n = 2x − 1 where x ∈ N .
Theorem 2.2.6. Pn + Km is set-graceful if n ≤ 2 and m = 2n1 − 1
where n1 ∈ N .
Proof. If n = 1 and m = 2n1 −1, then Pn +Km is a star with 2n1 −1
spokes which is set-graceful .
Let n = 2 and m = 2n1−1, n1 ∈ N . The number of edges of Pn+Km is
Chapter 2 27
2(2n1 −1)+1 = 2n1+1−1. Now Consider X = {1, 2, 3, . . . , n1, n1 + 1}.
Let V (Pn) = {u1, u2} and V (Km) = {v1, v2, . . . , v2n1−1}. Define f :
V (Pn +Km) → 2X by saying f(u1) = ∅, f(u2) = X and the remaining
vertices be assigned all the proper subsets of X containing the element
1 in an injective manner. Then edge labels are distinct and f⊕(E(Pn+
Km)) = 2X − {∅}. Hence f is a graceful set-labelling. Therefore
Pn + Km is set-graceful. ♣
Corollary 2.2.7. Pn + Km is not set-graceful for all n 6= 2x1, and
for all m 6= 2x2 − 1 for x1 > 2, x1, x2 ∈ N .
Proof. Suppose Pn + Km is set-graceful. Then |E(Pn + Km)| =
mn+n−1 = 2x−1 for some x ∈ N . That is n(m+1) = 2x ⇒ n = 2x1
and m+1 = 2x2 for some x1, x2 ∈ N . Hence Pn+Km is not set-graceful
if n 6= 2x1 or m 6= 2x2 − 1 , x1, x2 ∈ N . ♣
Corollary 2.2.8. Pn + Km is not set-graceful for n = 22 and m =
2n1 − 1 for n1 ∈ N .
Proof. Suppose P4 +Km is set-graceful with respect to a set X. Let
V (P4) = {u1, u2, u3, u4} and V (Km) = {v1, v2, . . . , v2n1−1}. Label ui
by Bi, 1 ≤ i ≤ 4 and vj by Aj, 1 ≤ j ≤ 2n1 − 1, where A′js and B′
is are
subsets of X. Since the symmetric differences of all subsets of the set
X is ∅, we get
(A1⊕B1)⊕(A1⊕B2)⊕(A1⊕B3)⊕(A1⊕B4)⊕
(A2⊕B1)⊕(A2⊕B2)⊕(A2⊕B3)⊕(A2⊕B4)⊕...
(A2n1−1⊕B1)⊕(A2n1−1⊕B2)⊕(A2n1−1⊕B3)⊕(A2n1−1⊕B4)⊕
Chapter 2 28
(B1⊕B2)⊕(B2⊕B3)⊕(B3⊕B4) = ∅. Then B2⊕B3 = ∅, which is a
contradiction to the injectivity of vertex labelling. ♣
The above discussion leads to the following conjecture.
Conjecture 2.2.9. Pn + Km is not set-graceful for all n > 2
Theorem 2.2.10. K1,m,n is set-graceful if and only if m = 2n1 − 1
and n = 2n2 − 1 for all n1, n2 ∈ N .
Proof. Let G = K1,m,n be the complete tripartite graph with one
vertex in the first partition say U = {u}, m vertices in the second
partition say V = {v1, v2, . . . , vm} and n vertices in the third partition
say W = {w1, w2, . . . , wn}. Let m = 2n1 − 1 and n = 2n2 − 1 for
n1, n2 ∈ N . Then |V (G)| = 2n1 + 2n2 − 1 and |E(G)| = 2n1+n2 − 1.
Let X be a set with |X| = n1 + n2. Partition X into 2 sub sets Y and
Z so that X = Y ∪ Z and |Y | = n1 and |Z| = n2. Define the vertex
labelling function f : V (G) → 2X as follows. Label u by ∅ and vi, wj
by subsets of Y and Z respectively in an injective manner.
Clearly f⊕(uvi), f⊕(uwj) and f⊕(viwj), 1 ≤ i ≤ m, 1 ≤ j ≤ n are
all distinct and list all subsets of X except ∅ where f⊕ is the induced
edge function. Hence f⊕(E(G)) = 2X − {∅} and f is a set-graceful
labelling of G. Therefore, K1,m,n is set-graceful.
Conversely, suppose K1,m,n is set-graceful. Then |E(K1,m,n)| = mn +
m + n = 2x − 1 for some x ∈ N . That is mn + m + n + 1 = 2x ⇒
(m+1)(n+1) = 2x which implies m+1 = 2x1 and n+1 = 2x2 for some
x1, x2 ∈ N . Hence, m = 2x1 − 1 and n = 2x2 − 1 for some x1, x2 ∈ N .
♣
Chapter 2 29
Figure 2.5 illustrates the theorem for m = n = 3.
Figure 2.5:
Theorem 2.2.11. Let X be a set with |X| = m. A uniform binary
tree with one pendant edge added at the root vertex having 2m−1 edges
is set-graceful.
Proof. Consider a uniform binary tree Tm with one pendant edge
added at the root vertex having 2m − 1 edges. Let
v11;
v21, v22;
v31, v32, v33, v322;
v41, v42, . . . , v423;...
vm1, vm2, . . . , vm2m−1
be the vertices of Tm at the first , second , . . . , mth levels respectively.
Take X = {1, 2, 3, . . . ,m}, Xm = {3, 4, . . . ,m}, Ym = {2, 3, 4, . . . ,m}.
Chapter 2 30
Now, order all subsets of X by putting A < B for distinct A and
B if and only if either |A| < |B| or |A| = |B| and min(A − B) <
min(B − A). Let 〈Ai〉 and 〈Bi〉 on Xm and Ym respectively be the
increasing sequences obtained in accordance with ordering relation ′ <′
and each containing the element m. Arrange subsets of X containing
m in the orders:
℘m = B1, {1} ∪B1, B2, {1} ∪B2, B3, {1} ∪B3,. . .
<m = A1, {1} ∪A1, {2} ∪A1, {1, 2} ∪A1, A2, {1} ∪A2,{2} ∪A2, . . .
where A′i s and B′
i s are subsets of Xm and Ym respectively.
Define f : V (Tm) → 2X , as f(v) = ∅, where v is the pendant vertex
added at the root vertex and label the root vertex v11 of Tm by {1}.
The vertices in the second and third levels are labelled as follows
f(v21) = {2}; f(v22) = {1, 2};
f(v31) = {2, 3}; f(v32) = {1, 2, 3}; f(v33) = {1, 3}, f(v34) = {3}.
Suppose Tn, n < m is set-graceful. Then, identify the vertices in the
nth level whose labels follow the order of ℘n and rename the vertices
in the nth level in such a way that the label of vni is the ith member
of ℘n. Now, we can define the labelling of the vertices in the n + 1th
level, in general as follows.
Label the vertices adjacent to vni by
f(vni)⊕ Ad i2e
, f(vni)⊕ ({1} ∪ Ad i2e
), when i is odd;
f(vni)⊕ ({2} ∪ A i2), f(vni)⊕ ({1, 2} ∪ A i
2), when i is even, ⊕ denotes
the symmetric difference. Now Tn+1 is set-graceful.
By principle of induction on n, where n < m, Tm, the uniform binary
tree with one pendant edge added at the root vertex having 2m − 1
Chapter 2 31
edges with the above mentioned labelling is clearly set-graceful. ♣
The 5th levels edges are formed as follows:
Figure 2.6:
Chapter 2 32
Figure 2.7 illustrates the theorem for m = 4 .
Let m = 4, ℘3 is {3}, {1, 3}, {2, 3}, {1, 2, 3}.
Figure 2.7:
Now we study some classes of graphs that do not admit set-gracefulness.
Theorem 2.2.12. The splitting graph S ′(G) of a set-graceful graph
G is not set-graceful.
Proof. Let G be a set-graceful graph. Then G has 2n − 1 edges for
some n ∈ N . It’s splitting graph S ′(G) has 3(2n − 1) edges. Suppose
S ′(G) is set-graceful. Then there exists an x in N such that 3(2n−1)=
2x − 1. That is,
(2+1)(2n−1) = 2x−1 ⇒ 2(2n+2n−1−1) = 2x ⇒ 2n+2n−1−1 = 2x−1,
which is a contradiction. Hence, S ′(G) is not set-graceful. ♣
Theorem 2.2.13. S ′(Pn) is not set-graceful for all n.
Chapter 2 33
Proof. Let Pn be a path with n vertices say u1, u2, . . . , un. To form
S ′(Pn) add n vertices u′1, u′2, . . . , u
′n respectively. Suppose S ′(Pn) is
set-graceful with respect to a set X and f be the set-graceful labelling
that labels u1, u2, . . . , un by subsets A1, A2, . . . , An and u′1, u′2, . . . , u
′n
by subsets B1, B2, . . . , Bn of X. Since the symmetric difference of all
the nonempty subsets of X is ∅, we get
(A1⊕B2)⊕(B2⊕A3)⊕(A3⊕B4)⊕ . . .⊕(Bn−1⊕An)⊕
(B1⊕A2)⊕(A2⊕B3)⊕ . . .⊕(An−1⊕Bn)⊕
(A1⊕A2)⊕(A2⊕A3)⊕ . . .⊕(An−1⊕An) = ∅, if n is odd.
(A1⊕B2)⊕(B2⊕A3)⊕(A3⊕B4)⊕ . . .⊕(An−1⊕Bn)⊕
(B1⊕A2)⊕(A2⊕B3)⊕ . . .⊕(Bn−1⊕An)⊕
(A1⊕A2)⊕(A2⊕A3)⊕ · · ·+ (An−1⊕An) = ∅, if n is even.
In both the cases we get B1⊕Bn = ∅, which is a contradiction to the
injectivity of the vertex function. Hence, S ′(Pn) is not set-graceful. ♣
Figure 2.8 illustrates the theorem for n = 5.
Figure 2.8:
Theorem 2.2.14. Let G be a (p, q)-graph. S ′(G) is not set-graceful
for q ≡ 0, 2, 3(mod4). Further the only possible values for q so that
S ′(G) could be set-graceful are 21, 85, 341, . . . ,
Chapter 2 34
Proof. Clearly if G is a graph with q edges, then S ′(G) contains 3q
edges. S ′(G) is not set-graceful if 3q 6= 2x − 1 for some x. That is,
3q + 1 6= 2x. If q is even 3q + 1 is odd which cannot be equal to 2x for
any x. Therefore if q ≡ 0, 2(mod 4), then S ′(G) is not set-graceful.
Suppose S ′(G) is set-graceful for q ≡ 3(mod 4) then q − 3 = 4m for
some m which implies q = 4m + 3 ⇒ 3(4m + 3) + 1 = 2x, for some x,
then 12m + 10 = 2x ⇒ 2(6m + 5) = 2x ⇒ 6m + 5 = 2x−1, which is
impossible. Hence S ′(G) is not set-graceful for q ≡ 3(mod 4).
Therefore S ′(G) is not set-graceful for q ≡ 0, 2, 3(mod 4).
Suppose S ′(G) is set-graceful for q ≡ 1(mod4) ⇒ q = 4t + 1
Suppose t is even. Then 3q = 3(4t+1) = 2x−1 ⇒ 12t = 22(2x−2−1) ⇒
3t = 2x−2 − 1, a contradiction.
Suppose t is odd say, t = 2k + 1 and k is odd. Then 3(2k + 1) =
2x−2− 1 ⇒ 6k +3 = 2x−2− 1 ⇒ 2(3k +2) = 2x−2 ⇒ 3k = 2(2x−4− 1),
a contradiction.
Suppose t is odd say, t = 2k + 1 and k = 2n1 for n1 ∈ N
3k = 2(2x−4 − 1) ⇒ 3.2n1 = 2(2x−4 − 1) ⇒ 2n1−1.3 = 2x−4 − 1, which
is a contradiction when n1 > 1.
When n1 = 1 , k = 2, t = 5, q = 21
Suppose t is odd say, t = 2k + 1 and k = 2n2m, m is odd. Then
3k = 2(2x−4 − 1) ⇒ 3m2n2−1 = 2x−4 − 1, a contradiction, for n2 > 1.
When n2 = 1, then t = 2k + 1, and k = 2m
Now we have 3m = 2x−4 − 1. The solution of this equation exist only
for even x ≥ 6 and is of the form m = 40 + 41 + · · · + 4n, n ≥ 0.
Therefore the only possible values of q for which S ′(G) is set-graceful
Chapter 2 35
are 21, 85, 341, . . . . ♣
By discussing about the possible values of q for the (p, q)-graph G,
so that S ′(G) is set-graceful, the following conjecture arises.
Conjecture 2.2.15. In all the possible cases implied by theorem
2.2.14, the graph S ′(G) is set-graceful.
proposition 2.2.16. K3,5 is not set-graceful.
Proof. Suppose K3,5 is set-graceful. Then it is set-graceful with
respect to a set X of cardinality 4, with out loss of generality let
X = {1, 2, 3, 4}. Let Y and Z be a partition of V (K3,5) where
Y = {u1, u2, u3} and Z = {v1, v2, v3, v4, v5}. In the set-graceful la-
belling let n1 labels of Y and m1 labels of Z are odd subsets of X.
Then by theorem 2.1.3 the number of edges with odd set labels is
(3− n1)m1 + (5−m1)n1 = 8 . . . (1)
The possible solutions of (1) are n1 = 1, m1 = 3 and n1 = 2, m1 = 2.
Case 1: n1 = 1 and m1 = 3
Let Ai represents odd subsets and Bj represents even subsets. With
out loss of generality label the vertices u1, u2, u3 by A1, B1, B2 and
v1, v2, v3, v4, v5 by A2, A3, A4, B3, B4 respectively. Then odd subset
edge labels are
A1⊕B3, A1⊕B4, B1⊕A2, B1⊕A3,
B1⊕A4, B2⊕A2, B2⊕A3, B2⊕A4.
Since symmetric difference of all odd subsets of a set X is ∅, we have
(A1⊕B3)⊕(A1⊕B4)⊕(B1⊕A2)⊕(B1⊕A3)⊕
(B1⊕A4)⊕(B2⊕A2)⊕(B2⊕A3)⊕(B2⊕A4) = ∅
Chapter 2 36
That is B3⊕B4⊕B1⊕B2 = ∅ ⇒ B1⊕B3 = B2⊕B4, contradiction to
the injectivity of the edge label. Hence K3,5 is not set-graceful when
n1 = 1
Case 2: n1 = 2 and m1 = 2
With out loss of generality label the vertices u1, u2, u3 by A1, A2, B1
and v1, v2, v3, v4, v5 by A3, A4, B2, B3, B4 respectively.
Since symmetric difference of all odd subsets of a set X is ∅, we have
(A1⊕B2)⊕(A1⊕B3)⊕(A1⊕B4)⊕(A2⊕B2)⊕
(A2⊕B3)⊕(A2⊕B4)⊕(B1⊕A3)⊕(B1⊕A4) = ∅
That is A3⊕A4⊕A1⊕A2 = ∅ ⇒ A1⊕A3 = A2⊕A4, contradiction to
the injectivity of the edge label.
Hence K3,5 is not set-graceful. ♣
Theorem 2.2.17. Let G be a set-graceful graph. Then the corona of
Gf and K1, that is Gf�K1 is set-graceful if Gf is the full augmentation
of G.
Proof. Let G be a set-graceful graph with respect to a set X of car-
dinality n. With out loss of generality take X = {1, 2, . . . , n}. Let Gf
is the full augmentation of G. Then |V (Gf)| = 2n. To get Gf �K1 we
have to add 2n vertices. Label the new added 2n vertices as follows.
Label the vertex adjacent to the vertex with label ∅ by {1, 2, n + 1}.
Label the vertex adjacent to the vertex with label {x1, x2, . . . , xk, n}
where xi ∈ {1, 2, . . . , (n− 1)}, 1 ≤ i ≤ k. By {x1 + 1, x2 + 1, . . . , xk +
1, n+1}. Label the vertex adjacent to the vertex with label {a1, a2, . . . al}
where aj ∈ {2, . . . , n−1}, 1 ≤ j ≤ l by {1, 2, a1+1, a2+1, . . . , al+1, n+
1}. Label the vertex adjacent to the vertex with label {1, b1, b2, . . . , bt}
Chapter 2 37
where bt ∈ {2, 3, . . . , n− 1} by {1, b1 + 1, b2 + 1, . . . , bt + 1, n + 1}. Ac-
cording to this labelling Gf �K1 is set-graceful. ♣
Figure 2.9 illustrates the theorem.
Figure 2.9:
Hegde [Heg] proved that every graph can be embedded into a con-
nected set-graceful graph. The following theorem gives another em-
bedding of a graph into a connected set-graceful graph.
Theorem 2.2.18. Every finite graph G can be embedded as an in-
duced subgraph of a set-graceful finite graph H.
Chapter 2 38
Proof.
Case 1 : Let G be a finite graph with at least one vertex has degree
≥ 2. Fix that vertex as u and label it by ∅. Label the 2 vertices
u1 and un−1 adjacent to u by{1} and {n − 1} and the other vertices
u2, u3, . . . , un−2 of G by {2}, {3}, ...{n− 2}.
If u and ui , 2 ≤ i ≤ n − 1 are not adjacent add a pendant edge
ui−1vi−1 to ui−1 and label the corresponding vertex vi−1 by {i− 1, i} .
If ui−1 and ui are not adjacent, add an edge between u and a vertex
with label {i− 1, i}.
If ui and uj , j 6= i ± 1 are not adjacent, add an edge to u and label
its vertex by {i, j}.
Then add nc3 + nc4 + · · · + ncn pendant edges to u and label the
corresponding vertices by subsets of {1, 2, . . . , n− 1} with cardinality
≥ 3 in an arbitrary but injective manner. Then the obtained graph
H is set-graceful and G is an induced subgraph of H.
Case 2: Let G be a graph with every vertex have degree ≤ 1. Without
loss of generality, let there is m vertices with degree 1 and n − m
isolated vertices. Define a new graph by taking a new vertex and
joining it to all vertices of the original graph. In this new graph there
are vertices of degree ≥ 2 and hence by case 1 this graph and hence
the original graph can be embedded in a set-graceful graph. ♣
Chapter 2 39
Figure 2.10 illustrates the theorem for n = 5.
Figure 2.10:
We know that a connected finite graph G is eulerian if and only if
all its vertex degrees are even.
Theorem 2.2.19. Every cycle Cn can be embedded into an eulerian
set-graceful graph.
Proof. Let Cn be a cycle of order n, n > 4. Let V (Cn) = {u1, u2, . . . , un}
and X = {1, 2, . . . , n}. Label the vertex ui by the set {i}. Add 2n−1−2
isolated vertices v1, v2, . . . , v2n−1−2.
Define X3, X4, . . . , Xn as follows. X3 = {3, 4, . . . , n}, X4 = {4, 5, . . . , n},
Chapter 2 40
. . . , Xn = {n}. Define Sj, j > 1 as the set of all subsets of X which
contains 1 and j as the minimum elements. Let S1 = S2∪S3∪, . . . , Sn.
Define S as the set S1−{{1, 3, n}, {1, 2, 3}, {1, 3, 4}, . . . , {1, n−1, n}}.
Clearly |S1| = 2n−1−1 and |S| = 2n−1−n. Label the vertices vi where
i > n− 2 by members of S. Join each vi to the vertices u1, and u3.
Case 1 : n is odd.
The n− 2 vertices v1, v2, . . . , vn−2 are joined so that u1, v1, v2,
. . . , vn−2, u3 form a path of length n−1 and label them as follows. La-
bel v1 by {1, 3, n}, v2 by {3, 4, n} and label v3, v4, . . . , vn−2 so that the
corresponding edges v2v3, v3v4, . . . , vn−2u3 have the labels {1, 3, 5, 6},
{1, 3, 6, 7}, . . . , {1, 3, n− 2, n− 1}, {1, 3, n− 1, n}, {1, 3}, {1, 3, 4, 5} re-
spectively.
Case 2: n is even.
The n−2 vertices v1, v2, . . . , vn−2 are joined so that u1, v1, v2, . . . , vn−2, u1
form a cycle of length n − 1 and label them as follows. Label v1 by
{1, 3, n}, v2 by {3, 4, n} and label v3, v4, . . . , vn−2 so that the corre-
sponding edges v2v3, v3v4 , . . . , vn−3vn−2, vn−2u1 have the labels {1, 3, 5, 6},
{1, 3, 6, 7}, . . . {1, 3, n− 2, n− 1}, {1, 3, n− 1, n}, {1, 3}, {1, 3, 4, 5} re-
spectively.
Then clearly every vertex of newly created graph H is of even order
and hence it is eulerian. And also H is set-graceful since all the edge
labels are distinct and the set of edge labels is 2X − {∅}. Therefore
the cycle Cn, n > 4 is an induced subgraph of the set-graceful eulerian
graph G. Hence the theorem. ♣
Chapter 2 41
Figure 2.11 illustrates the theorem for n = 5 .
Figure 2.11:
Chapter 2 42
Figure 2.12 illustrates the theorem for n = 6.
Figure 2.12:
Chapter 2 43
Embedding of the cycles C3 and C4 in an eulerian set-graceful graph
is as follows:
Figure 2.13:
Figure 2.14:
Chapter 2 44
Theorem 2.2.20. Every star can be embedded as an induced sub-
graph of a set-graceful star.
Proof. Let Sn be a star with |V (Sn)| = n + 1. Let V (Sn) =
{u, u1, u2, . . . , un}, where u is the central vertex. Label u by ∅. If n =
2x− 1 for x ≥ 2, clearly Sn is set-graceful. If 2x− 1 < n < 2x+1− 1 for
x > 2, Then introduce 2x+1−n−1 new vertices, vj, 1 ≤ j ≤ 2x+1−1−n.
Label ui, 1 ≤ i ≤ n and vj, 1 ≤ j ≤ 2x+1 − n − 1 by non-empty
subsets of X, where X = {1, 2, ...x + 1} and join them to the vertex
u. Clearly the newly created graph H will be a set-graceful star and
Sn is a induced subgraph. ♣
Figure 2.15 illustrates the theorem.
Figure 2.15:
Chapter 2 45
2.3 Bi Set-Graceful Graphs
The Line graph of G, denoted by L(G) has E(G) as its vertex set with
two vertices of L(G) are adjacent whenever the corresponding edges
of G are adjacent. A graph G is said to be bi set-graceful if both G
and its line graph are set-graceful.
Lemma 2.3.1. A uniform Binary tree Tn, with one pendant edge
added at the root vertex having 2n − 1 edges is bi set-graceful if and
only if n ≤ 2.
Proof. By theorem 2.2.11 a uniform binary tree Tn, with one pendant
edge added at the root vertex having 2n − 1 edges is set-graceful for
all values of n. For n = 1, L(T1) = K1 is set-graceful. For n = 2,
L(T2) = K3 is set-graceful. Therefore Tn is bi set-graceful for n ≤ 2.
For n ≥ 3, L(Tn) contains 20 + 21 + ... + 2n−2 distinct triangles and
have (2n−1−1)3 edges, which is not set-graceful, since for a set-graceful
graph the number of edges is 2m − 1, where m is a integer. Hence Tn
is bi set-graceful if and only if n ≤ 2. ♣
Every set-graceful cycles are bi set-graceful, Since each cycle is
isomorphic to its line graph. Therefore by theorem 2.1.6, C2n−1 is bi
set-graceful for all n.
Lemma 2.3.2. Star Sn with 2n − 1 spokes is bi set-graceful if and
only if n ≤ 2.
Proof. Star Sn with 2n − 1 spokes is set-graceful for all values of
n. For n = 1, L(S1) = K1 is set-graceful. For n = 2, L(S2) = K3
Chapter 2 46
is set-graceful. Therefore Sn is bi set-graceful for n ≤ 2. For n ≥ 3,
L(Sn) = K2n−1, which is not set-graceful since Kp is set-graceful only
for p = 2, 3, and 6 by theorem 2.1.5. ♣
Lemma 2.3.3. The complete graph Kn on n vertices is bi set-
graceful if and only if n ≤ 3.
Proof. For n = 2, K2 and L(K2) = K1 are set-graceful. For n = 3,
K3 and L(K3) = K3 are set-graceful. By theorem 2.1.5 for n ≥ 4,
Kn is set-graceful only for n = 6. L(K6) contains 60 edges with 15
vertices. Hence it cannot be set-graceful by remark 2.1.4. Therefore
Kn is bi set-graceful if and only if n ≤ 3. ♣
Theorem 2.3.4. r-regular connected (p, q)-graph G is bi set-graceful
if and only if r = 2 and q = 2n − 1 for some n ∈ N .
Proof. Suppose G is a r-regular connected (p, q)-graph and let its
line graph L(G) be a (pl, ql)-graph. Then q = pr2 , L(G) is a (2r − 2)-
regular graph. Then ql = pr2 .2r−2
2 = pr2 (r − 1)
Suppose G is bi set-graceful. Then pr2 = 2n − 1 and ql = 2x − 1 for
some n and x.
Case 1: r is odd
Let r = 2t+1. Then q = p(2t+1)2 = 2n−1. ql = p(2t+1)
2 .2t = p(2t+1)t =
2x − 1 where x ∈ N . That is 2t(2n − 1) = 2x − 1, a contradiction.
Therefore no odd regular graph is bi set-graceful.
Case 2: r is even.
Let r ≡ 0(mod 4). Then r = 4t and q = 4tp2 = 2tp = 2n − 1, a
contradiction.
Chapter 2 47
Let r ≡ 2(mod 4). Then r = 4t + 2 and let t is odd. Then q =
p(4t+2)2 = 2n − 1. Then ql = p(4t+2)
2 (4t + 1) = (2n − 1)(4t + 1) = 2x − 1,
x ∈ N
⇒ 2n+2t + 2n − 22t − 1 = 2x − 1 ⇒ 22(2nt + 2n−2 − t) = 2x ⇒
2nt + 2n−2 − t = 2x−2, a contradiction since t is odd.
Let r = 4t + 2 and let t is even and is of the form 2n1 for n1 ∈ N .
q = (4t+2)p2 = 2n − 1, ql = p(4t+2)
2 (4t + 1) = 2x − 1 , x ∈ N .
⇒ (2n − 1)(4t + 1) = 2x − 1 ⇒ (2n − 1)(2n1+2 + 1) = 2x − 1 ⇒
2n+n1+2 + 2n − 2n1+2 − 1 = 2x − 1 ⇒ 2n(2n1+2 − 2n1−n+2 + 1) = 2x
⇒ 2n1+2 − 2n1−n+2 + 1 = 2x−n, a contradiction
Let r = 4t + 2 and let t is of the form 2n1m, where m is odd. q =
(2t + 1)p = 2n − 1. ql = p(2t + 1)(4t + 1) = 2x − 1.
⇒ (2n − 1)(2n1+2m + 1) = 2x − 1. ⇒ 2n+n1+2m + 2n − 2n1+2m = 2x.
⇒ 2n1+2(m2n + 2n−n1−2 −m) = 2x. ⇒ m2n + 2n−n1−2 −m = 2x−n1−2 ,
a contradiction since m is odd.
Hence the only possible value of r for which G is bi set-graceful is
r = 2. Since cycles with 2n−1 edges are set-graceful, 2-regular graphs
with 2n − 1 edges are bi set-graceful. Therefore r-regular connected
(p, q)-graph G is bi set-graceful if and only if r = 2 and q = 2n− 1 for
some n ∈ N . ♣
However we could not succeed to characterize the classes of graphs
which are bi set-graceful. Since the number of edges in a set-graceful
graph to be 2m−1, for some m ∈ N , we have to trace those graphs for
which its and its line graph have the size of the form 2m− 1, for some
m ∈ N . We pause the following problem for further investigations.
Chapter 2 48
Problem 2.3.5. Characterize bi set-graceful graphs.
2.4 Set-Sequential Graphs
A graph G is said to be set-sequential if there exists a nonempty set
X and a bijective set-valued function f : V (G) ∪ E(G) → 2X − {∅}
such that f(uv) = f(u)⊕f(v) for every uv ∈ E(G). In this section we
recognize the fact that the notion of set-sequential graphs and that of
set-graceful graphs are closely related as already evident from theorem
1.1.20, this is reinforced in the following results.
Theorem 2.4.1. If G + K1 is set-graceful then G is set-sequential.
Proof. Let f be a set-graceful labelling of G + K1 with respect to
the set X of cardinality m and let G be a (p, q)-graph, then G + K1 is
a (p + 1, p + q)-graph. Let V (G) = {u1, u2, . . . , up} and V (K1) = {v}.
Define a labelling f ∗ : V (G) → 2X − {∅} as f ∗(ui) = f(v) ⊕ f(ui),
where ⊕ denotes the symmetric difference of sets. Then f ∗ is a set-
sequential labelling on G and hence G is set-sequential. ♣
By theorem 1.1.20, we have if G is a connected set-sequential graph
then G + K1 is set-graceful. Here the restriction connected is not
necessary. Then the following corollary is immediate.
Corollary 2.4.2. A graph G is set-sequential if and only if G+K1
is set-graceful.
G + K1 is set-sequential does not imply G is is set-graceful and
G is is set-graceful does not imply G + K1 is set-sequential, for the
Chapter 2 49
complete graph K5 = K4 + K1 is set-sequential but K4 is not set-
graceful and the complete graph K6 is set-graceful but K7 = K6 + K1
is not set-sequential.
Theorem 2.4.3. If G is set-graceful then G ∪Kt is set-sequential
for some positive integer t.
Proof. Let f be a set-graceful labelling of G with respect to the
set X and let |X| = m. Take a new symbol y and include it in
every subset of X assigned by f to a vertex of G. Then the new
sets f1(u) = f(u) ∪ {y} so assigned to the vertices u ∈ V are all
different from the nonempty subsets of X assigned to the edges of G
by f⊕, since clearly, in this process, the sets assigned to the edges of G
remain unaltered. Now, augment t = 2m − p new vertices to G. Then
for each nonempty set A ∈ 2X such that A 6∈ {f(v) : v ∈ V }, define
ZA = A ∪ {y}. Label the new vertices by ZA, A ∈ 2X , resulting is the
augmented graph H = G ∪ Kt. Clearly, H is a set-sequential graph
with f1 as a set-sequential labelling. ♣
Theorem 2.4.4. Set-gracefulness and set-sequentialness are equiv-
alent for a tree.
Proof. To prove the theorem first we prove all set-graceful trees are
set-sequential and viceversa.
Let T be a set-graceful tree with respect to a set X of cardinality n and
let the vertices of T are labelled by A1, A2, . . . , A2n where Ai is a subset
of X. Define Y = X∪{n + 1}. Consider a labelling h : V (T ) → 2Y so
that h(vi) = Ai∪{n+1} where vi ∈ V (T ). Then clearly h is injective,
Chapter 2 50
h(V (T ) ∪ E(T )) = 2Y − {∅} and h(V (T ) ∩ E(T )) = ∅. Hence T is
set-sequential.
Let T1 be a set-sequential tree with respect to a set X of cardinality
n. Let V (T1) = {u1, u2, . . . , u2n−1}. Let the vertex label of ui be
Ai, 1 ≤ i ≤ 2n−1, a subset of X. For Ai = {x1, x2, . . . , xj}, j ≤ n, take
Bi = {x1 − 1, x2 − 1, . . . , xj − 1} when xi 6= 1
Bi = {x1 − 1, x2 − 1, ..., xi−1 − 1, xi+1 − 1, ..., xj − 1} when xi = 1. If
Ai = {1} take Bi = ∅. Since T1 is set-sequential Ai is always not equal
to ∅. Label the vertices of T1 by the labelling f defined by f(ui) = Bi.
Then clearly f is a set-graceful labelling on T1 and T1 is set-graceful.
Hence the theorem. ♣
In theorem 2.2.11, we expose an important class of graphs, a uni-
form binary tree with one pendant edge added at the root vertex is
set-graceful. Hence by theorem 2.4.4, it is set-sequential.
Theorem 2.4.5. Km,n is set-sequential if and only if m = 2x1 − 1
and n = 2x2 − 1 for some x1, x2 ∈ N .
Proof. Suppose Km,n is set-sequential with respect to a set X of
cardinality l. Then mn+m+n = 2l− 1. That is (n+1)(m+1) = 2l,
which implies m = 2x1 − 1 and n = 2x2 − 1 for some x1, x2 ∈ N , where
x1 + x2 = l.
To prove the converse suppose m = 2x1 − 1 and n = 2x2 − 1 for some
x1, x2 ∈ N . Let A = {1, 2, ...x1} and B = {x1 + 1, x1 + 2, ..., x1 +
x2}. Label the m vertices of the first partition of Km,n by non-empty
subsets of A and n vertices of the second partition of Km,n by non-
Chapter 2 51
empty subsets of B in an injective manner. This labelling gives a
set-sequential labelling and hence Km,n is set-sequential. ♣
The following results give some classes of graphs which are not
set-sequential.
Theorem 2.4.6. Fans Pn + K1 are not set-sequential for all n.
Proof. Let G = Pn+K1 with V (K1) = v1 and V (Pn) = {u1, u2, ...un}.
Suppose G is set-sequential with respect to a set X having the la-
belling f where f(v1) = A1 and f(ui) = Bi, 1 ≤ i ≤ n, where A1
and Bi are subsets of X. Then f(V (G) ∪ E(G)) = 2X − {∅} and
f(V (G)∩E(G)) = ∅. Since the symmetric difference of all non-empty
subsets of any set is the empty set we have A1⊕B1⊕B2⊕ . . .⊕Bn⊕
(A1⊕B1)⊕(A1⊕B2)⊕ . . .⊕(A1⊕Bn)⊕
(B1⊕B2)⊕ . . .⊕(Bn−1⊕Bn) = ∅ , which implies B1⊕Bn = ∅, if n is
odd and A1⊕B1⊕Bn = ∅, if n is even.
That is B1 = Bn, if n is odd and A1⊕B1 = Bn, if n is even.
Both cases contradicts the injectivity of f . Hence Fans are not set-
sequential. ♣
Observation 2.4.7. For a (p, q)-graph G to be set-sequential, p + q
should be odd. In the case of cycles p and q are the same. Hence p+ q
is even. Therefore cycles are not set-sequential.
Lemma 2.4.8. P4 is not set-sequential.
Proof. Suppose P4 is set-sequential, with vertex labels A1, A2, A3
and A4. Since P4 is set-sequential, A1 ⊕ A2 ⊕ A3 ⊕ A4 ⊕ (A1 ⊕ A2)⊕
Chapter 2 52
(A2 ⊕ A3) ⊕ (A3 ⊕ A4) = ∅, since symmetric difference of all subsets
of a set is empty. Which implies A2 ⊕ A3 = ∅ ⇒ A2 = A3, which is a
contradiction to the injectivity of the vertex labelling. ♣
proposition 2.4.9. Wheel Wn of any order is not set-sequential.
Proof. Let Wn be a wheel of order n + 1 and suppose Wn is set-
sequential.
Case 1 : n is odd.
Since Wn is set-sequential, 3n + 1 = 2x − 1 for some x, which implies
3n + 2 = 2x for some x, a contradiction.
Case 2 : n is even, say n = 2m
Label the central vertex by A and the other vertices by B1, B2,
. . . , B2m. Since Wn is set-sequential and symmetric difference of all
subsets of a set is empty, we get A⊕B1⊕B2⊕· · ·⊕+B2m⊕A⊕B1⊕
· · · ⊕ A ⊕ B2m ⊕ B1 ⊕ B2 ⊕ B2 ⊕ B3 · · · ⊕ B2m ⊕ B1 = ∅ ⇒ A = ∅, a
contradiction. Hence Wn for all n is not set-sequential. ♣
Lemma 2.4.10. A twig t of order less than 16 is not set-sequential.
Proof. For a set-sequential graph, |V (G)|+|E(G)| = 2n−1, for some
n. Therefore the only possible set-sequential twig of order less than
16 is t8, a twig of order 8. Consider t8 and label the vertices on the
path of t8 by A, B, C,D and of the other vertices by A1, A2, A3 and A4.
Suppose t8 is set-sequential. Since the symmetric difference of all the
subsets of a set is ∅, we have A⊕B⊕C⊕D⊕A1⊕A2⊕A3⊕A4⊕(A⊕
B)⊕(B⊕C)⊕(C⊕D)⊕(A1⊕B)⊕(A2⊕C)⊕(A3⊕B)⊕(A4⊕C) = ∅.
Chapter 2 53
Which implies B ⊕ C = ∅, a contradiction to the injectivity of vertex
labelling. Hence the result. ♣
For any integer t ≥ 3, let f2t denote the graph obtained from the
path P2t = (u1, u2, . . . , u2t) by adjoining the new edges of the form
u1+iu2t−i for every i, 0 ≤ i ≤ t− 2.
Lemma 2.4.11. f2t, t ∈ N is not set-sequential.
Proof. Suppose f2t is set-sequential with respect to a set X. Let
V (f2t) = {u1, u2, . . . , u2t} and the vertex label of ui be Ai, a subset
of X. Since the symmetric difference of all subsets of a set is ∅, we
have A1 ⊕ A2 ⊕ · · · ⊕ A2t ⊕ (A1 ⊕ A2) ⊕ (A2 ⊕ A3) ⊕ · · · ⊕ (A2t−1 ⊕
A2t) ⊕ (A1 ⊕ A2t) ⊕ (A2 ⊕ A2t−1) ⊕ · · · ⊕ (At−1 ⊕ At+2) = ∅. Then
A1 ⊕ A2t ⊕ At ⊕ At+1 = ∅ Which implies A1 ⊕ A2t = At ⊕ At+1, a
contradiction to the injectivity of edge labels. Hence f2t is not set-
sequential. ♣
Theorem 2.4.12. Every star can be embedded as an induced sub-
graph of set-sequential star.
Proof. Let Sn be a star with V (Sn) = {u, u1, u2, . . . , un}, where u
is the central vertex. If n = 2x − 1 for some x ∈ N , then clearly Sn
is set-sequential. If 2x − 1 < n < 2x+1 − 1, let X = {1, 2, ...n + 1}.
Label u by {1} and ui by the singleton set {i + 1}. Add 2n − n − 1
new vertices and label them by subsets of X with cardinality > 2 and
containing the element 1 in an injective manner. Then join them to
the vertex u. The newly created graph H will be a set-sequential star
with Sn an induced sub graph. ♣
Chapter 2 54
Figure 2.16 illustrates the theorem.
Figure 2.16:
Hegde [Heg] proved that every graph can be embed into a connected
set-sequential graph. The following theorem gives another embedding
of a graph into a set-sequential graph.
Theorem 2.4.13. Every graph G can be embedded into a set- se-
quential graph H.
Proof. Let G be a graph of order n. Let V (G) = {u1, u2, ...un} and
X = {1, 2, ...n}. Label ui by the singleton set {i}. Define Sj as the
set of subsets of X of cardinality j which contains the element 1. if
ui is not adjacent with uj , where i, j 6= 1 introduce a new vertex,
label it by {1, i, j} and join it to the vertex u1. If u1 is not adjacent to
uk, introduce a vertex, label it by {1, 2, k} and join to the vertex u2.
Then introduce |S4|+ |S5|+ · · ·+ |Sn| new vertices and label them by
Chapter 2 55
the elements of S4, S5, . . . , Sn in an injective manner. The subsets of
cardinality 3 which was not an vertex label is obtained as edge labels
by joining the vertices which was labelled by elements of S4 to u1 and
appropriate elements of S4 to ui, 1 ≤ i ≤ n as required. Then join the
vertices which was labelled by elements of S5, S6, . . . , Sn to u1. The
obtained graph H is set-sequential. ♣
Figure 2.17 illustrates the theorem.
Figure 2.17:
Chapter 2 56
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