Download Chapter 4 square sum graphs

Chapter 4
square sum graphs
In this Chapter we introduce a new type of labeling of graphs which is closely related to
the Diophantine Equation x2 + y 2 = n and report results of our preliminary investigations
on this new concept. Let G be a (p, q)-graph. G is said to be a square sum graph if there
exist a bijection f : V (G) → {0, 1, ..., p−1} such that the induced function f ∗ : E(G) → N
given by f ∗ (uv) = [f (u)]2 + [f (v)]2 for every uv ∈ E(G) are all distinct. The square
sum labeling f is called a prime square sum labeling if f ∗ (uv) is 1 or a prime number
∀uv ∈ E(G). We prove that for a prime square sum graph G, f (u) and f (v) are relatively
prime ∀e = uv ∈ E(G) with f ∗ (e) 6= 1. Moreover we prove that the complete graph Kn
for n ≤ 5, the cycles Cn for n ≥ 3, trees, cycle cactus, ladders and the complete lattice
grids are square sum. We also proved that every graph can be embedded into a square
sum graph. Also we characterize the class of prime square sum graphs.
4.1
Introduction
Abundant literature exists as of today concerning the structure of
graphs admitting a variety of functions assigning real numbers to their
elements so that certain given conditions are satisfied. Here we are
interested the study of vertex functions f : V (G) → A, A ⊆ N for
78
Chapter 4
79
which the induced edge function f ∗ : E(G) → N is defined as f ∗ (uv) =
[f (u)]2 + [f (v)]2 , ∀e ∈ E(G) are all distinct. As we know that the
notion of prime labeling originated with Entrnger and was introduced
in a paper by Tout, Dabboucy and Howalla [Gal05]. A graph with
vertex set V is said to have a prime labeling if its vertices are labeled
with distinct integers 1, 2, ..., |V | such that for each edge xy the labels
assigned to x and y are relatively prime [Gal05]. This concept leads
to define a prime square sum graph which is a square sum graph with
edge values are only prime numbers together with 1. Our motivation
is to study how the number theoretic problems are related to the
structure of graphs.
Definition 4.1.1.
Let G be a (p, q)-graph. G is said to be a square
sum graph if there exist a bijection f : V (G) → {0, 1, ..., p − 1} such
that the induced function f ∗ : E(G) → N given by f ∗ (uv) = [f (u)]2 +
[f (v)]2 for every uv ∈ E(G) are all distinct. The square sum labeling f
is called a prime square sum labeling if f ∗ (uv) is 1 or a prime number
∀uv ∈ E(G).
Figures 4.1 and 4.2 give two graphs which are square sum and two
graphs which are prime square sum respectively.
Figure 4.1:
Chapter 4
80
Figure 4.2:
The following theorem is a consequence of the definition of a square
sum graph.
Theorem 4.1.2.
For any (p, q)-graph G = (V, E) and for any
square sum labeling f : V (G) → {0, 1, 2, ..., p − 1}, we have
X
f ∗ (e) =
X
[f (u)]2 d(u)
u∈V (G)
e∈E(G)
.
Let G be a square sum graph. Then f ∗ (ei ) = [f (ui )]2 +[f (vi )2 ]
P ∗
P
for all ei = (ui vi ) ∈ E(G) so that
f (e) = [f (ui )]2 + [f (vi )2 ]. In
P ∗
the sum
f (e) each f (ui ) occurs d(ui ) times. Hence
Proof.
X
e∈E(G)
f ∗ (e) =
X
[f (u)]2 d(u)
u∈V (G)
.
Corollary 4.1.3.
If G = (V, E) be a (p, q)-graph which is r-regular
P
and square sum, then 6 f ∗ (e) = rp(p − 1)(2p − 1).
Chapter 4
Proof.
81
By the theorem 4.1.2,
X
f ∗ (e) =
e∈E(G)
X
[f (u)]2 d(u)
u∈V (G)
. Since G is r-regular, d(ui ) = r for all ui ∈ V (G). Hence
∗
f (e) = r
X
[f (u)]2 =
rp(p − 1)(2p − 1)
6
.
Theorem 4.1.4.
Let G = (V, E) be a (p, q)-graph which is square
Q
sum graph with the square sum labeling f . Then f ∗ (e) can be represented as the sum of two squares.
Proof.
Let G be a graph having two edges say e1 = u1 v1 and e2 =
u2 v2 . Then f ∗ (e1 ) = [f (u1 )]2 +[f (v1 )]2 and f ∗ (e2 ) = [f (u2 )]2 +[f (v2 )]2 .
Let f (u1 ) = a, f (v1 ) = b, f (u2 ) = c and f (v2 ) = d so that
f ∗ (e1 ).f ∗ (e2 ) = (a2 +b2 ).(c2 +d2 ) = (ac+bd)2 +(ad−bc)2 . By applying
the principle of induction on the number of edges we get the result.♦
As we know that certain numbers like 3, 6, 7 etc.cannot be written as sum of two squares. Hence there are forbidden numbers which
cannot appear as an edge labeling. Hence the interesting question is,
what are those numbers which are not forbidden as edge labeling.
We have the following theorem.
Theorem 4.1.5.
Let G be a square sum graph which admits a
square sum labeling f . Then f ∗ (e) not congruent to 3(mod 4), ∀e ∈
E(G).
Chapter 4
Proof.
82
If possible, let f ∗ (e) ≡ 3(mod 4) for some e ∈ E(G)
Let f ∗ (e) = a2 + b2 where e = (uv), f (u) = a and f (v) = b.
Then a2 + b2 ≡ 3(mod 4). This is not possible. Hence f ∗ (e) not
congruent to 3(mod 4), ∀e ∈ E(G).♦
Theorem 4.1.6.
Let G be a square sum graph which admits a
square sum labeling f . If f ∗ (e) ≡ 1(mod2), ∀e ∈ E(G), then G is
bipartite.
Proof.
Let us suppose that f ∗ (e) ≡ 1(mod 2), ∀e ∈ E(G)
Let f ∗ (e) = a2 + b2 where e = uv ∈ E(G), f (u) = a, f (v) = b.
Then a2 + b2 ≡ 1(mod 2). Since a2 + b2 is an odd number, we have a
and b are of opposite parity.
Let X = {u : f (u) is even } and Y = {v : f (v) is odd }. The
sets X and Y form a disjoint partition of V (G) and any edge with
f ∗ (e) ≡ 1(mod 2) is of one end in X and the other end in Y .♦
proposition 4.1.7.
The following two cases are not possible for a
square sum graph.
(i)f ∗ (e) ≡ 2(mod 4), ∀e ∈ E(G)
(ii)f ∗ (e) ≡ 0(mod 4), ∀e ∈ E(G).
Proof.
(i) Suppose for all edge e = uv ∈ E(G).
f ∗ (e) ≡ 2(mod 4). But f ∗ (e) = [f (u)]2 + [f (v)]2 ≡ 2(mod 4).
Let f (u) = 0. Then [f (v)]2 ≡ 2(mod 4), which is not possible since
the square of an integer is either congruent to 0(mod 4) or congruent
to 1(mod 4). Hence 0 does not belong to f (V (G)).
(ii) Suppose for all edge e = uv ∈ E(G) we have f ∗ (e) ≡ 0(mod 4).
Chapter 4
83
Hence f ∗ (e) = [f (u)]2 + [f (v)]2 ≡ 0(mod 4).
Let f (u) = 1. Then [f (v)]2 ≡ −1( mod 4) , which is not possible since
the square of an integer is either ≡ 0(mod 4) or ≡ 1(mod 4). Hence
1 does not belong to f (V (G)).♦
proposition 4.1.8.
If f ∗ (e) ≡ 1(mod 4), ∀e ∈ E(G) ⇒ G is bi-
partite.
f ∗ (e) ≡ 1(mod 4), ∀e ∈ E(G) ⇒ f ∗ (e) ≡ 1(mod 2), ∀e ∈
Proof.
E(G). Then apply Theorem .♦
4.2
Some classes of square sum graphs
Theorem 4.2.1.
Proof.
The graph G = K2 + mK1 is a square sum graph.
Let V (G) = {u1 , u2 , ...um+2 } where V (K2 ) = {u1 , u2 }. Define
f : V (G) → {0, 1, ..., m + 1} by f (ui ) = i − 1, 1 ≤ i ≤ m + 2. Clearly,
the induced function f ∗ is injective, for if f ∗ (u1 ui ) = f ∗ (u2 uj ), then we
get [f (u1 )]2 + [f (ui )]2 = [f (u2 )]2 + [f (uj )]2 . By taking f (ui ) = x and
f (uj ) = y, we get an equation of the form x2 − y 2 = 1, which implies
either one of x or y is zero. That is either f (ui ) = 0 or f (uj ) = 0,
which is not possible since f is a bijection. Hence G is a square sum
graph.♦
Chapter 4
84
The following figure 4.3 illustrates the theorem for m = 4.
Figure 4.3:
Theorem 4.2.2.
The complete graph Kn is a square sum graph if
and only if n ≤ 5.
Proof.
The square sum labeling of the complete graph Kn for n ≤ 5
is given in figures ?? and ??. Now, let Kn , n ≥ 6 and f : V (G) →
{0, 1, 2, ..., n−1} be a vertex function which induces a function f ∗ given
by f ∗ (xy) = [f (x)]2 + [f (y)]2 . Assume n ≥ 6 and since the graph is a
complete graph we get two edges e1 and e2 such that f ∗ (e1 ) = 02 +52 =
25 and f ∗ (e2 ) = 32 +42 = 25. Hence, we conclude that when n ≥ 6, f ∗
is not injective Hence Kn , n ≥ 6 is not a square sum graph.♦
Theorem 4.2.3.
Proof.
Cycles are square sum graphs.
Let Cp be a cycle of length p and let Cp = (u1 u2 ...up u1 ).
Case(i): p is odd.
Define f : V (Cp ) → {0, 1, ..., p − 1} as f (ui ) = i − 1, 1 ≤ i ≤ p. Here f
is an increasing function on V (Cp ), so f ∗ is also an increasing function
on E(Cp ) − {up u1 }. Hence f ∗ (ei ) 6= f ∗ (ej ), i 6= j, for every ei , ej ∈
E(Cp ) − {up u1 }. When p is odd, f (up ) is even. Therefore f ∗ (up u1 )
Chapter 4
85
is even and hence f ∗ (up u1 ) 6= f ∗ (ej ) for every ej ∈ E(Cp ) − {up u1 };
hence f ∗ is injective and f is a square sum labelling on Cp .
Case(ii): when p is even and p 6= 6.
We know that the equation a2 + (a + 1)2 = c2 has only one non-zero
positive consecutive integer solution which are given by a = 3, a+1 = 4
and c = 5. Hence for Cp , p 6= 6 we can have the same labeling as in
Case (i).
For Cp , p = 6, consider the following particular labeling given by,
f : V (Cp ) → {0, 1, 2, 3, 4, 5} as f (ui ) = i − 1, 1 ≤ i ≤ 4, f (u5 ) = 5
and f (u6 ) = 4. Then f ∗ is injective and f is a square sum labeling of
C6 .♦
The square sum labeling of cycles C5 and C6 are given in the figure
4.4.
Figure 4.4:
Theorem 4.2.4.
Proof.
Trees are square sum graphs.
Let T be a tree and let v1 ∈ V (T ) be the vertex with maxi-
mum degree. Start from the vertex v1 , and applying BFS algorithm,
label the vertices of T with 0, 1, 2, ..., p − 1 in the order in which
they are visited. Since f is increasing on the vertex set of T and
Chapter 4
86
f ∗ (ui uj ) = [f (ui )]2 + [f (uj )]2 , we have f ∗ is also an increasing function on the edge set of T . Hence f ∗ is injective and f is a square sum
labeling on T .♦
Remark 4.2.5.
Spanning subgraph H of a square sum graph G
is square sum since V (G) = V (H) and f (V (G)) = f (V (H)) =
{0, 1, ..., p − 1}. But all subgraphs of a square sum graph need not
be square sum as the following figure 4.5 illustrates.
Figure 4.5:
Definition 4.2.6.
A cycle-cactus is a graph which consisting of n
copies of Ck , k ≥ 3 concatenated at exactly one vertex is denoted as
(n)
Ck .
Theorem 4.2.7.
Proof.
(n)
The cycle-cactus Ck
is a square sum graph.
(n)
For a given integer k ≥ 3, let Ck be the cycle-cactus consist-
ing of n copies of the cycle Ck , denoted G1 , G2 , . . . , Gn , all concatenated at exactly one vertex, say z. Let the vertices of Gi , 1 ≤ i ≤ n
(n)
be labelled z, ui1 , ui2 , . . . , ui(k−1) , 1 ≤ i ≤ n. Define f : V (Ck ) →
Chapter 4
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{0, 1, ..., p − 1}. First of all, let f (z) = n(k − 1).
Case (i)
k is odd.
Label the two antipodal vertices of the
cycle G1 by 1 and 2 respectively, of the cycle G2 by 3 and 4 respectively,
. . . and label the two antipodal vertices of the cycle Gn by 2n − 1 and
2n respectively. Next, assign the number 2n + 1 to the unlabeled
vertex adjacent to the vertex labeled 1 in G1 , 2n + 2 to the unlabeled
vertex adjacent to the vertex labeled 2 in G1 , 2n + 3 to the unlabelled
vertex adjacent to the vertex labeled 3 in G2 , 2n + 4 to the unlabeled
vertex adjacent to the vertex labeled 4 in G2 , . . . , 4n − 1 to the
unlabeled vertex adjacent to the vertex labeled 2n − 1 in Gn , and 4n
to the unlabeled vertex adjacent to the vertex labeled 2n in Gn . Next,
assign the number 4n+1 to the unlabeled vertex adjacent to the vertex
labeled 2n + 1 in G1 , 4n + 2 to the unlabeled vertex adjacent to the
vertex labeled 2n + 2 in G1 , 4n + 3 to the unlabeled vertex adjacent to
the vertex labeled 2n+3 in G2 , 4n+4 to the unlabeled vertex adjacent
to the vertex labeled 2n + 4 in G2 , . . . , 6n − 1 to the unlabeled vertex
adjacent to the vertex labeled 4n − 1 in Gn , and 6n to the unlabeled
vertex adjacent to the vertex labeled 4n in Gn . Continuing in this
manner, we end up assigning the number (k − 3)n + 1 to the unlabeled
vertex adjacent to the vertex labeled (k − 5)n + 1 in G1 , (k − 3) + 2
to the unlabeled vertex adjacent to the vertex labeled (k − 5)n + 2 in
G1 , (k − 3)n + 3 to the unlabeled vertex adjacent to the vertex labeled
(k − 5)n + 3 in G2 , (k − 3)n + 4 to the unlabeled vertex adjacent to the
vertex labeled (k − 5)n + 4 in G2 , . . . , (k − 1)n − 1 to the unlabeled
Chapter 4
88
vertex adjacent to the vertex labeled (k − 3)n − 1 in Gn , and (k − 1)n
to the last unlabeled vertex adjacent to the vertex labeled (k − 3)n in
Gn .
Case (ii)
k is even. In this case, label the unique antipodal
vertex of the cycle Gi by i, 1 ≤ i ≤ n. Next, assign the number n + 1
and n + 2 to the unlabeled vertices adjacent to the vertex labeled 1 in
G1 , n + 3 and n + 4 to the unlabeled vertices adjacent to the vertex
labeled 2 in G2 , n + 5 and n + 6 to the unlabeled vertices adjacent to
the vertex labeled 2 in G2 , n + 5 and n + 6 to the unlabeled vertices
adjacent to the vertex labeled 3 in G3 , n+7 and n+8 to the unlabeled
vertices adjacent to the vertex labeled 4 in G4 , . . . , 3n − 1 and 3n to
the unlabeled vertices adjacent to the vertex labeled n in Gn . Next,
assign the number 3n + 1 to the unlabeled vertex adjacent to the vertex labelled n + 1 in G1 , 3n + 2 to the unlabeled vertex adjacent to the
vertex labeled n + 2 in G1 , 3n + 3 to the unlabeled vertex adjacent to
the vertex labeled n + 3 in G2 , 3n + 4 to the unlabeled vertex adjacent
to the vertex labeled n + 4 in G2 , . . . , 4n − 1 to the vertex adjacent
to the unlabeled vertex labeled 3n − 1 in Gn , and 4n to the unlabeled
vertex adjacent to the vertex labeled 3n in Gn . Next, assign the number 4n+1 to the unlabeled vertex adjacent to the vertex labeled 3n+1
in G1 , 4n + 2 to the unlabeled vertex adjacent to the vertex labelled
3n + 2 in G1 , 4n + 3 to the unlabeled vertex adjacent to the vertex
labeled 3n + 3 in G2 , 4n + 4 to the unlabeled vertex adjacent to the
vertex labeled 3n + 4 in G2 , . . . , 6n − 1 to the vertex adjacent to the
unlabeled vertex labeled 4n − 1 in Gn , and 6n to the unlabeled vertex
Chapter 4
89
adjacent to the vertex labeled 4n in Gn . Continuing in this manner,
we end up assigning the number (k − 3)n + 1 to the unlabeled vertex
adjacent to the vertex labeled (k − 5)n + 1 in G1 , (k − 3) + 2 to the unlabeled vertex adjacent to the vertex labeled 2n + 2 in G1 , (k − 3)n + 3
to the unlabeled vertex adjacent to the vertex labeled (k − 5)n + 3
in G2 , (k − 3)n + 4 to the unlabeled vertex adjacent to the vertex
labeled (k − 5)n + 4 in G2 , . . . , (k − 1)n − 1 to the unlabeled vertex
adjacent to the vertex labeled (k − 3)n − 1 in Gn , and (k − 1)n to
the last unlabeled vertex adjacent to the vertex labeled (k −3)n in Gn .
In the above two cases we have
f (u11 ) < f (u21 ) < f (u31 ) < ... < f (un1 ).
f (u12 ) < f (u22 ) < f (u32 ) < ... < f (un2 ).
f (u13 ) < f (u23 ) < f (u33 ) < ... < f (un3 ).
.............................................
..........................................
f (u1(k−1) ) < f (u2(k−1) ) < f (u3(k−1 ) < ... < f (un(k−1) ) and f (z) =
n(k − 1).
(n)
Hence the induced function on the set of edges of Ck
is injective so
that the assignment f defined above in each case is a square sum la(n)
beling of Ck .
Corollary 4.2.8.
Proof.
(n)
The friendship graph C3
Let k = 3 in the above theorem.♦
is a square sum graph.
Chapter 4
Corollary 4.2.9.
90
For the star K1,p−1 , if we label the central vertex
with any integer between 0 and p − 1 and the remaining values are
assigned to the vertices of unit degree in a one to one manner, we get
a square sum labeling.
Theorem 4.2.10.
Ladder Ln is a square sum graph.
Proof. Let V (Ln ) = {a1 , a2 , . . . , an , b1 , b2 , . . . , bn } and E(Ln ) = {ai bi :
1 ≤ i ≤ n} ∪ {ai ai+1 : 1 ≤ i ≤ n − 1} ∪ {bi bi+1 : 1 ≤ i ≤ n − 1}. Define
f : V (Ln ) → {0, 1 . . . , p − 1} as follows:
f (ai ) = 2i − 2, 1 ≤ i ≤ n and f (bi ) = 2i − 1,
1 ≤ i ≤ n so
that f ∗ (ai bi ) = (2i − 2)2 + (2i − 1)2 , 1 ≤ i ≤ n. Next, f ∗ (ai ai+1 ) =
(2i − 2)2 + (2i)2 , 1 ≤ i ≤ n − 1, f ∗ (bi bi+1 ) = (2i − 1)2 + (2i + 1)2 , 1 ≤
i ≤ n − 1; since f ∗ (ai bi ) is an odd number for 1 ≤ i ≤ n and
f ∗ (ai ai+1 ).Also f ∗ (bj bj+1 ), 1 ≤ i ≤ n − 1, 1 ≤ j ≤ n − 1 are both
even numbers, we have f ∗ (ai bi ) 6= f ∗ (ai ai+1 ) and f ∗ (ai bi ) 6= f ∗ (bi bi+1 ).
Moreover, f is an increasing function on V (G) and so f ∗ (ai bi ) 6=
f ∗ (aj bj ), i 6= j and f ∗ (ai ai+1 ) 6= f ∗ (aj aj+1 ), i 6= j, 1 ≤ i ≤ n − 1, 1 ≤
j ≤ n − 1. Also, f ∗ (bi bi+1 ) 6= f ∗ (bj bj+1 ) and f ∗ (ai bi ) 6= f ∗ (bj bj+1 ), 1 ≤
i ≤ n − 1, 1 ≤ j ≤ n − 1, since these numbers form an increasing
sequence of even numbers. Hencef ∗ is injective and f is a square sum
labeling.♦
Chapter 4
91
The following figure 4.6 gives a square sum labeling of Ln for n = 5.
Figure 4.6:
Theorem 4.2.11.
The complete lattice grids Lmn = Pm × Pn are
square sum graphs.
Proof.
Let G be the grid with V (G) = {u11 , u12 , ..., u1n ,
u21 , u22 , ..., u21 , ..., um1 , um2 , ..., umn }.
We arrange the set of vertices of G into different levels as follows:
u11
u21 , u12
u31 , u22 , u13 , ..., umn .
That is V (G) = {uij : i + j = 2, 3, ..., m + n, i ≤ m, j ≤ n}.
Define f : V (G) → {0, 1, ..., m.n − 1}, by
f (u11 ) = 0, f (u21 ) = 1, f (u12 ) = 2, f (u31 ) = 3, f (u22 ) = 4, f (u13 ) =
5.
Chapter 4
92
After labelling the ith level vertices we will label the (i + 1)th level of
vertices and so on. Then the induced edge labeling given by
f ∗ (e) = [f (u)]2 + [f (u)]2 , ∀e ∈ E(G), is injective,
for if let e1 = u1 v1 and e2 = u2 v2 be any two edges of G.
(i)u1 = u2 ⇒ f (v1 ) < f (v2 ) or f (v1 ) > f (v2 ),
then f ∗ (e1 ) 6= f ∗ (e2 ).
(ii) If u1 6= u2 , then f (u1 ) < f (u2 ) ⇒ f (v1 ) < f (v2 ).
Hence f ∗ (e1 ) 6= f ∗ (e2 ) so that f ∗ is a square sum labeling of G.
Hence G is a square sum graph.♦
The following figure 4.7 illustrates the theorem for m = 4, n = 5.
Figure 4.7:
Next we are looking for bipartite graph which are square sum. The
following theorem gives the idea that there exists plenty of graphs
which are not square sum.
Theorem 4.2.12.
if m ≤ 4, ∀n.
The complete bipartite graphs Km,n is square sum
Chapter 4
Proof.
93
Let V (Km,n ) = X ∪ Y 3 X ∩ Y = φ.
Stars K1,n , ∀n are square sum by theorem [3.9].
The different sets of partitions of the vertex sets of Km,n , 2 ≤ m ≤ 5
are given below:
Case (i) m = 2.
X = {x1 , x2 } and Y = {y1 , y2 , ..., yn }, define
f (x1 ) = 0, f (x2 ) = 1 and f (yi ) = i + 1, 1 ≤ i ≤ n.
Case (ii) m = 3.
X = {x1 , x2 , x3 } and Y = {y1 , y2 , ..., yn }, define
f (x1 ) = 0, f (x2 ) = 1, f (x3 ) = 2 and f (yi ) = i + 2, 1 ≤ i ≤ n.
Case (iii) m = 4.
X = {x1 , x2 , x3 , x4 } and Y = {y1 , y2 , ..., yn }, define
f (x1 ) = 0, f (x2 ) = 1, f (x3 ) = 2f (x4 ) = 3 and f (yi ) = i + 3, 1 ≤ i ≤ n.
consider the following sets of equations.
x2 − y 2 = 1, x2 − y 2 = 4, x2 − y 2 = 9, x2 − y 2 = 16, x2 − y 2 =
4, x2 − y 2 = 8, x2 − y 2 = 5, x2 − y 2 = 12, x2 − y 2 = 7.
From the solutions of these nine equations we conclude that f ∗ is
injective for Km,n , m ≤ 4, ∀n. Hence Km,n is square sum if m ≤
4, ∀n.♦
Problem 4.2.13. Characterize the the class of bipartite square sum
graphs.
We have already give some graphs which are not square sum. Therefore our next attempt is to embed such graphs into square sum graphs.
Chapter 4
94
When we succeed in this effort then try to make embeddings into an
eulerian, bipartite, hamiltonian embeddings.
4.3
Embeddings of square sum graphs
Theorem 4.3.1.
Every (p, q)-graph G can be embedded in a con-
nected square sum graph H with 5p−1 + t + q − p edges and 5p−1 + 1
vertices where t is the number of connected components of G.
Proof.
Let G be a graph with vertex set V (G) and let V (G) =
{u1 , u2 , ..., up }. We will embed the graph G in a graph H with |V (H)| =
5p−1 + 1 and |E(H)| = 5p−1 + t + q − p where t is the number of connected components of G. Consider the set of p integers 0, 5, 52 , ..., 5p−1 .
Label the vertices of G with the above p numbers by
f (u1 ) = 0, f (ui ) = 5i−1 , 2 ≤ i ≤ p.
Introduce isolated vertices
v1 , v2 , ..., vn where n = 5p−1 + 1 − p. Labeled these n isolated vertices by
1, 2, 3, 4, 6, 7, ..., 24, 26, 27, ..., 124, 126, ..., 5p−1 − 1. (Note that here we
are omitting the numbers 5, 52 , ..., 5p−1 ).
If G is connected join u1 to all vk , 1 ≤ k ≤ n.
If G is disconnected and having t components, say C1 , C2 , ..., Ct ,
join u1 to exactly one of the vertex of Ci , say ui1 , 2 ≤ i ≤ t, where
the vertex u1 belongs to the component C1 . Then for the resulting
graph H, f is a bijection from V (H) → {0, 1, ..., 5p−1 }. The induced
edge labeling f ∗ is injective, for if assume there exists edges e1 and e2
such that f ∗ (e1 ) = f ∗ (e2 ).
Chapter 4
95
Case (i). Let e1 = uk ul and e2 = ui uj where 1 ≤ i, j, k, l ≤ p .If
f ∗ (e1 ) = f ∗ (e2 ) which implies (5i )2 + (5j )2 = (5k )2 + (5l )2 .
(i) Let no end vertex of e1 and e2 be labeled as zero and since i, j, k, l 6=
0, divide throughout by (5i )2 where i be the least among them. Hence
1 + 5j−i = 5k−i + 5l−i which is a contradiction.
(ii) Let one of the end vertex of ei or ej be labeled as zero. Then,
f ∗ (ei ) = f ∗ (ej ) ⇒ (5i )2 = (5k )2 + (5l )2 , again contradiction.
(iii) Let one end vertex of ei and ej be labeled as zero. Then
f ∗ (e1 ) = f ∗ (e2 ) ⇒ (5i )2 = (5k )2 , which implies i = j, which is not
possible.
Case (ii). Let e1 = u1 vi and e2 = u1 vj where 1 ≤ i ≤ n and
1 ≤ j ≤ n. Since f (u1 ) = 0 and if we assume f (vi ) = xi and f ∗ (vj ) =
xj , then f ∗ (e1 ) = f ∗ (e2 ) ⇒ f ∗ (u1 vi ) = f ∗ (u1 vj ), which implies that
02 + xi 2 = 02 + xj 2 , not possible, hence f ∗ (u1 vi ) 6= f ∗ (u1 vj ), ∀i 6= j.
Case (iii). Let e1 = u1 vi and e2 = u1 uk where 1 ≤ i ≤ n and
2
1 ≤ k ≤ n. Then f ∗ (e1 ) = f ∗ (e2 ) ⇒ xi 2 = (5k ) . Hence we get
xi = 5k , not possible since f is a bijection.
Case (iv). Let e1 = uk ul and e2 = u1 vi , then f ∗ (e1 ) = f ∗ (e2 ) ⇒
(5k )2 + (5l )2 = xi 2 , since the left side is even ,xi must be even.
Hence xi 2 ≡ 0(mod4) but (5k )2 + (5l )2 ≡ 2(mod4), which is a contradiction.
From the above four cases we conclude that the induced function f ∗
is a square sum labeling of H and hence the graph H is a square sum
graph which contains G as a subgraph.♦
Chapter 4
96
When the given graph is connected the following figure 4.8 represents the embedding of the graph into a connected square sum graph.
Figure 4.8:
When the given graph is disconnected the following figure 4.9 represents the embedding of the graph into a connected square sum graph.
Figure 4.9:
Corollary 4.3.2.
If G is planar , so the square sum graph H.
Chapter 4
97
Corollary 4.3.3.
Every eulerian graph can be embedded into an
eulerian square sum graph.
Proof. Let G be the given eulerian graph. By the theorem 4.3.1, we
can embed G into a square sum graph. In that embedding if we
construct distinct cycles joining the vertices with labeling given by
5 → 4 → 1 → 2 → 3 → 6 → ... → 24 → 5 Similarly cycles with vertex
labeling given by 5 → 26 → 27 → ... → 124 → 5. Here each cycle is
centered at the vertex with labeling 5. Then the embedding becomes
an eulerian embedding.♦
Corollary 4.3.4.
Every bipartite graph can be embedded into a bi-
partite square sum graph.
Proof.
Since each cycle in the corollary 4.3.3 centered at the vertex
with labeling 5 is of even length we have if the given graph is bipartite
then the embedding also bipartite.♦
Corollary 4.3.5.
In a similar way one can show that very Hamil-
tonian graph can be embedded into a Hamiltonian square sum graph.
4.4
Prime square sum graphs
As we know that prime numbers play an important role in number
theory, We investigate the structure of square sum graphs with edge
labeling prime numbers.
Theorem 4.4.1.
Let G be a prime square sum graph. Then f ∗ (e) ≡
1(mod4), ∀e ∈ E(G).
Chapter 4
98
Proof. If f ∗ (e) = 1, then obviously f ∗ (e) ≡ 1(mod4). Otherwise
assume f ∗ (e) 6= 1. Since f ∗ (e) is a prime number either f ∗ (e) ≡
1(mod4) or f ∗ (e) ≡ 3(mod4). But by theorem 4.1.5, f ∗ (e) cannot
be congruent to 3(mod4), hence f ∗ (e) ≡ 1(mod4).♦
Observation 4.4.2.
All prime square sum graphs are bipartite
graph.
The following corollary is immediate.
Corollary 4.4.3.
The number of vertices p of a prime square sum
graph is even if and only if |X| = |Y | and if the number of vertices p
is odd if and only if |X| ∼ |Y | = 1 .
Theorem 4.4.4.
If G is a prime square sum graph, then it will
have at least one pendant vertex.
Proof.
Let G be a prime square sum graph and let the vertex u be
such that f (u) = 0. Then the vertex u can be adjacent only with
the vertex whose labeling is 1. Hence the degree of the vertex u with
f (u) = 0 is always 1.♦
Remark 4.4.5.
The condition given in Theorem 4.4.4 is not sufficient as the following figure 4.10 illustrates.
Chapter 4
99
Figure 4.10:
Theorem 4.4.6.
For any prime square sum graph f ∗ (e) ≡ 1, 5( mod
8), ∀e ∈ E(G).
Proof.
Let e = uv ∈ E(G). Since G is prime square sum, either
f (u) is even and f (v) is odd or vice versa. But the square of an odd
integer is always congruent to 1( mod 8) and square of an even integer
is congruent to 0, 4(mod 8).
Hence [f (u)]2 + [f (v)]2 ≡ 1, 5(mod 8).♦.
Theorem 4.4.7.
For any prime square sum graph G, f (u) and f (v) are relatively
prime for all e ∈ E(G) except the edge e = uv with f ∗ (e) = 1
Proof. Let G be a prime square sum graph and e = uv ∈ E(G) such
that f ∗ (e) 6= 1.Hence f (u) 6= 0 and f (v) 6= 0. Let us assume f (u)
and f (u) are not relatively prime and let d be the greatest common
divisor of f (u) and f (v). We have f ∗ (e) = f ∗ (uv) = [f (u)]2 + [f (v)]2
and since d divides both f (u) and f (v), it must divide f ∗ (e) also;
which is a contradiction to the fact that f ∗ (e) is a prime. Hence f (u)
and f (v) are relatively prime for all e ∈ E(G) except the edge e = uv
Chapter 4
100
with f ∗ (e) = 1.♦
Now we give an interesting result. We have already mentioned the
definition of a prime graph. A graph G = (V, G) is said to be a prime
graph if its vertices are labelled with distinct integers 1, 2, ..., |V | such
that for each edge xy, the labels assigned to x and y are relatively
prime [?]. The following theorem is an immediate result obtained
from theorem 4.4.7.
Theorem 4.4.8.
Proof.
All prime square sum graphs are prime graphs.
Let G be a prime square sum graph. Then there exists a
function f : V (G) → {0, 1, ..., p − 1} such that the induced function
f given by f ∗ (uv) = [f (u)]2 + [f (v)]2 = {1, p1 , p2 , ..., pq−1 } where pi
are prime numbers. Then by theorem 4.4.4, there exists a vertex
u1 in V (G) such that f (u1 ) = 0 and d(u1 ) = 1. Define a function
f1 : V (G) → {1, 2, ..., p} such that f1 (u1 ) = p and f1 (ui ) = f (ui ), 1 ≤
i ≤ p − 1. Since d(u1 ) = 1, the vertex ui which is adjacent to u1 is
such that f (ui ) = 1. Hence f (ui ) and f (u1 ) are relatively prime. For
all the other edges uv ∈ E(G), f1 (ui ) and f1 (u1 ) are relatively prime
by theorem 4.4.7. Hence f1 is a prime labeling of G so that G is a
prime graph.♦
Corollary 4.4.9.
The converse of the theorem 4.4.8 is not always
true. Figure 4.11 gives a prime graph which is not prime square sum.
Chapter 4
101
Figure 4.11:
Observation 4.4.10.
The complete graph Kn is prime square sum
for n ≤ 2.
Theorem 4.4.11.
The star K1,n is prime square sum if and only if
n ≤ 2.
Proof.
For n = 1, 2, the prime square sum labeling for K1,n is given
in the figure 4.13. Let n > 2. By the theorem 4.4.4, if u ∈ V (K1,n )
is such that f (u) = 0,then d(u) = 1. If we label the central vertex of
K1,n as any one of the numbers 1, 2, ..., n we get the edge values are
not all prime numbers. Hence the star K1n is not prime square sum
for n > 2.♦
We pose the following conjectures and problems:
Conjecture 4.4.12.
Every bipartite graph can be embedded into
a prime square sum graph.
Conjecture 4.4.13.
Every tree can be embedded into a prime
square sum graph.
Problem 4.4.14.
Characterize the class of prime square sum trees.
Chapter 4
103
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