$L(1,1)-$ Labeling of Direct Product of Cycles

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by  Tayo Charles Adefokun

L(1,1)− LABELING OF DIRECT PRODUCT OF CYCLES
5
1
0 TAYO CHARLES ADEFOKUN1 AND DEBORAH OLAYIDE AJAYI2
2

n
Abstract. AnL(1,1)-labelingofagraphGisanassignmentoflabelsfrom{0,1··· ,k}
a
J totheverticesofGsuchthattwoverticesthatareadjacentorhaveacommonneigh-
9 bor receive distinct labels. The λ11− number, λ11(G) of G is the minimum value k
2 such that G admits an L(1,1) labeling. We establish the λ11− numbers for direct
product of cycles Cm ×Cn for all positive m,n ≥ 3, where both m,n are even or
]
O when one of them is even and the other odd.
C
.
h
t
a
m 1. Introduction
[
The L(h,k)-labeling problem (or (h,k)-coloring problem) is that of vertex labeling

1 of an undirected graph G with non-negative integers such that for every u,v ∈ V(G),
v
uv ∈ E(G), |l(u)−l(v)| ≥ h and for all u,v ∈ V(G),d(u,v) = 2, |l(u)−l(v)| ≥ k.
5
3 The difference between the largest label and the smallest label assigned is called the
3 span. The aim of L(h,k)−labeling is to obtain the smallest non negative integer
7
λk(G), such that there exists an L(h,k)-labeling of G with no label on V(G) greater
0 h
. than λk(G).
1 h
0 Motivated by Hales’ 1980 paper [8], which provided a new model for frequency
5
assignment problems as a graph coloring problem, Griggs and Yeh [7] formulated the
1
: L(2,1) problem to model the channel assignment problem. The general notion of
v
L(h,k)- labeling was first presented by Georges and Mauro [6] in 1995. The topic
i
X
has since then been an object of extensive research for various graphs. Calamonerri’s
r
a survey paper [4] contains known results on L(h,k)-labeling of graphs.
L(1,1)-labeling(or strong labeling condition) of a graphis a labeling ofG such that
vertices with a common neighbor are assigned distinct labels. The usual labeling (or
propervertex coloring)conditionisthatadjacentvertices havedifferent colors, butfor
L(1,1), also all neighbors of any vertex are colored differently. This is equivalent to a
proper vertex-coloring of the square of a graph G. Note that a proper k- coloring of a
graph is a mapping α : V(G) → {1,··· ,k} such that for all uv ∈ E(G) α(u) 6= α(v)
and the square G2 of G has vertex V(G) with an edge between two vertices which
are adjacent in G or have a common neighbor in G. The chromatic number χ(G) of
G is the smallest k for which G admits a k-coloring. Therefore, χ(G2) = λ1(G)+1
1
for a graph G.
Key words and phrases. L(1,1)-labeling,D-2 Coloring,DirectProductofGraphs,CrossProduct
of Graphs, cycle graphs
2010 Mathematics Subject Classification. Primary: 05C35.
1
2 T.C ADEFOKUNAND D.O. AJAYI
Labelingofgraphpowersisoftenmotivatedbyapplicationsinfrequencyassignment
and has attracted much attention [See for example, [1]]. L(1,1)-labeling has applica-
tions in computing approximation to sparse Hessian matrices, design of collision-free
multi-hop channel access protocols in radio networks segmentation problem for files
in a network and drawings of graphs in the plane [3, 13, 15, 16] to mention a few.
For graphs G and H, the direct product G × H have vertex set V(G) × V(H)
where two vertices (x ,x ) and (y ,y ) are adjacent if and only if (x ,y ) ∈ E(G)
1 2 1 2 1 1
and (x ,y ) ∈ E(H). This product is one of the most important graph products with
2 2
potential applications in engineering, computer science and related disciplines. [11].
The L(h,k)- labeling of direct product of graphs was investigated in [2, 5, 9, 12, 14,
18, 19, 20].
In particular, Jha et al [12] gave upper bounds for λk-labeling of the multiple direct
1
product and cartesian product of cycles with some conditions on k and the length of
the cycles. They also presented some cases where we have exact values. In addition
by using backtracking algorithm, they computed λd(C × C ) for 2 ≤ d ≤ 4 and
1 m n
4 ≤ m,n ≤ 10. Since every L(2,1)-labeling is an L(1,1)-labeling, then λ1(G) ≤
1
λ2(G). Therefore, their results for d = 2 provided upper bounds for L(1,1)-labeling
1
of C ×C for 4 ≤ m,n ≤ 10. The only result for λ1-labeling for direct product of
m n 1
two cycles in the paper is that if m,n ≡ 0mod5 then λ1(C ×C ) = 4.
1 m n
In this paper, we solve the L(1,1)-labeling problem for direct product of cycles C ,
m
C , m,n ≥ 3, except form ∈ {16,18,22,26,32,36,46}, n ∈ {14,16,18,26,28,34}and
n
for these outstanding cases we conjecture that λ1(C ×C ) = 5.
1 m n
The paper is organized as follows: We give some preliminaries in Section 2 and
obtain the λ1 labeling numbers for C ×C for m ≥ 3 and n = 4 and 6 and some of
1 m n
their multiples in Section 3. Section 4 deals with labeling of direct product of bigger
cycles.
2. Preliminaries
Let G be a finite simple undirected graph with at least two vertices. For subgraph
V′ ⊆ V(G), we denote by L(V′) the set of L(1,1)−labeling on V′ and for a non-
negative integer, say, k, we take [k(ǫ)] as the set of even integers and zero in [k] while
[k(o)] is the set of odd integers in [k]. Suppose further that v ∈ V(G), we denote d
v
as the degree of v.
The following results, remarks and definitions are needed in the work.
Theorem 2.1. [10] Graph G×H is connected if and only if G and H are connected
and at least one of G and H is non-bipartite.
Remark 2.2. (i) Let G = C ×C , where m,n are even positive integers. Then,
m n
G= G ∪ G , where G and G are the connected components of C × C ,
1 2 1 2 m n
where
{V(G ) = u v : i ∈ [(m−1)(ǫ)],j ∈ [(n−1)(ǫ)] or
1 i j
i ∈ [(m−1)(o)];j ∈ [(n−1)(0)]} and
{V(G ) = u v : i ∈ [(m−1)(ǫ)],j ∈ [(n−1)(o)] or
2 i j
LABELING PRODUCTS OF CYCLES 3
i ∈ [(m−1)(o)];j ∈ [(n−1)(ǫ)]}.
Note that G and G are isomorphic and it is demonstrated in the graph C ×C
1 2 4 6
below
Fig. 1: The components of C4×C6
(ii) Suppose G = C ×C such that G = G′ ∪G′′, where G′,G′′ are components
m n
of G, then, λ1(G) = max{λ1(G′),λ1(G′′)}.
1 1 1
(iii) Let G = C ×C , where m is even and n odd positive integers. Then, G ≡ G ,
m n 1
where G is any of the two connected components of C ×C .
1 m 2n
0,0 0,1 0,2 0,3 0,4 0,0
3,0 3,0
2,0 2,0
1,0 1,0
0,0 0,1 0,2 0,3 0,4 0,0
Fig. 2: C4×C5 isisomorphic to a component of C4×10
Let P be a path of length m−1. The following results are from [2]:
m
Corollary 2.3. For m ≥ 3, λ1(P ×C ) = 5
1 m 6
Lemma 2.4. For m ≥ 5, n ≥ 9, n 6≡ 0 mod 5 λ1(P ×C ) ≥ 5
1 m n
A useful lower bound on L(1,1)-labeling for any graph G is contained in the fol-
lowing Lemma:
Lemma 2.5. [6] If G is a graph with maximum degree △, and G includes a vertex
with △ neighbors, each of which is of degree △, then λ1(G) ≥ △
1
From the lemma, we have for m,n ≥ 3 λ1(C ×C ) = 4
1 m n
4 T.C ADEFOKUNAND D.O. AJAYI
3. Labeling of C ×C ,n = 4,6
m n
In this section, we investigate the λ− numbers of graph product C × C and
m 4
C ×C , where m ≥ 3.
m 6
Let G′ be the connected component of the product graph under consideration.
Lemma 3.1. For m ≥ 4, and even, λ1(C ×C ) ≥ 5.
1 m 4
Proof. Let G′ ⊂ C ×C , where m ≥ 4 and even. Suppose V ,V ,V ⊂ V(G′). Let
m 4 i i+1 i+2
G′ bethesubgraphofG′ inducedbyV ,forallj ∈ [2].Then, V(G′)={u v ,u v ,u v ,
1 i+j 1 i 0 i 2 i+1 1
u v ,u v ,u v }. Now it is clear that the diameter of G′ is 2. Thus for every
i+1 2 i+2 0 i+2 2 1
pair v ,v ∈ V(G′), d(v ,v ) ≤ 2. Thus, l(v ) 6= l(v ) for all v ,v ∈ V(G′). Now,
1 2 1 1 2 1 2 1 2 1
|V(G′)| = 6. Therefore λ1(C ×C ) ≥ 5. (cid:3)
1 1 m 4
Remark 3.2. Note that if G′ ⊂ C × C , m ≥ 4 with m even and v ∈ V , for
m 4 i i
some i, V ⊂ V(G′), such that l(v ) = α ∈ [m], with λ1(G′) = m, then α ∈/
i i i 1 1
L{V V V V }.
i−2 i−1 i+1 i+2
Theorem 3.3. λ1(C ×C ) = 7
1 4 4
Proof. Let G′ ⊂ C ×C and V ⊂ V(G′) for each i ∈ [3]. Clearly, V(G′) = ∪3 V .
4 4 i i=0 i
Let G′ be a subgraph of G′ such that G′ is induced by V ,V ,V . By the proof of
1 1 0 1 2
Lemma 3.1, |V(G′)| = 6 and suppose α′,α′′,∈ L(V ), then by remark 3.2,α′,α′′,∈/
1 k k 3 k k
L(V(G′)). Thus there exists α′,α′′,∈/ [5] such that {α′,α′′} = L(V ), and α′ 6= α′′
1 k k k k 3 k k
since d(v′,v′′) = 2 for v′,v′′ ∈ V . Thus, |L(∪3 V )|=|L(V(G′))| = 6+2. Therefore,
3 3 3 3 3 i=0 i
λ1(C ×C ) = λ1(G′) = 7. (cid:3)
1 4 4 1
Next we present the necessary and sufficient condition under which λ1(C ×C ) is
1 m 4
5.
Theorem 3.4. For m ≥ 4,m even, λ1(C ×C ) = 5 if and only if m ≡ 0 mod 6.
1 m 4
Proof. Let m = 6n, n ∈ N. By Lemma 3.1, λ1(G) ≥ 5. Therefore, λ1(G′) ≥ 5, where
1 1
G′ ⊂ C × C . Let G′′ be the connected component of C ×C . By Corollary 3.2,
m 4 6 4
L(V )∩L(V ) = ∅, L(V )∩L(V ) = ∅, L(V )∩L(V ) = ∅. Now, set L(V ) = L(V ),
0 1 1 2 2 0 0 3
L(V ) = L(V ),L(V ) = L(V ). But L(V ) ∩ L(V ) = ∅, L(V ) ∩ L(V ) = ∅. Thus,
1 4 2 5 5 0 5 1
λ1(G′′) ≤ 5 and λ1(C ×C ) = 5. Thus by re-occurrence along C and C , m = 0
1 1 6 4 n m
mod 6 implies λ1(G) = 5.
1
Conversely, suppose λ1(G) = 5. Let G′ be a connected component of G = C ×
1 m
C , m ≥ 4,m even. Then, λ1(G′) = 5. Now, assume that m 6≡ 0 mod 6 , then
4 1
m = 6n′ +2 or m = 6n′ +4 where n′ ∈ N∪0. For n′ = 0, G = C ×C , for which
4 4
λ1(G′) = 7 by Theorem 3.3.
1
Case i: For m = 6n′ + 2, n′ ∈ N, let V ,V ,V be subsets of V(G′). By Corollary
0 1 2
3.2, L(V ) ∩ L(V ) = ∅, L(V ) ∩ L(V ) = ∅ and L(V ) ∩ L(V ) = ∅. Now let G′ be
0 1 1 2 2 0 1
the subgraph of G′ induced by V ,V ,V . Since L(V ) ∩ L(V ) = ∅ for V ⊂ V(G′)
0 1 2 3 2 3
and λ1(G′) = 5, then L(V ) = L(V ). Let V ⊂ V(G′). Then L(V )∩L(V ) = ∅ and
1 3 0 4 4 3
L(V )∩L(V ) = ∅. Thus L(V ) = L(V ). Let V ⊂ V(G′). Then L(V )∩L(V ) = ∅
4 2 4 1 5 5 4
and L(V )∩L(V ) = ∅ and therefore L(V ) = L(V ). The scheme continues in such a
5 3 4 1
LABELING PRODUCTS OF CYCLES 5
way that L(V ) = L(V +3) for all i ∈ [m−1], that is L(V ) = L(V ) = L(V ) = ··· =
i i 0 3 6
L(V6n′), L(V1) = L(V4) = L(V7) = ··· = L(V6n′+1), L(V2) = L(V5) = L(V8) = ··· =
L(V0). Now, for all va ∈ V0 and vb ∈ V6n′, d(va,vb) = 2. For all vc ∈ V1, vd ∈ V6n′+1,
d(v ,v ) = 2 and finally, L(V )∩L(V ) = ∅. Thus a contradiction.
c d 0 2
Case ii: For m = 6n′ +4, n′ ∈ N, similar argument as in m = 6n+2 applies. Thus,
λ1(C ×C ) = 5 if and only m ≡ 0 mod 6 (cid:3)
1 m 4
Corollary 3.5. Let m ≡ 0 mod 6 and n ≡ 0 mod 4. Then, λ1(C ×C ) = 5
1 m n
Proof. The claim follows from Theorem 3.4 and the re-occurrence of the optimal
labeling of C6n′ ×C4, n′ ∈ N. (cid:3)
Corollary 3.6. For all m 6≡ 0 mod 6, λ1(C ×C ) ≥ 6.
1 m 4
Theorem 3.7. For C ×C , λ1(C ×C ) = 7
8 4 1 8 4
Proof. Suppose G′ is a connected component of C ×C and suppose that λ1(G′) = 6.
8 4 1
By Corollary 3.2, |L{V ,V ,V }| = 6. Likewise, for all α ∈ L(V ), α ∈/ L(V ) and
0 1 2 k 0 k 7
α ∈/ L(V ). Also, for all α ∈ L(V ), α ∈/ L(V ). Suppose L(V ) = L(V ), then by
k 6 j 1 j 7 2 7
Corollary 3.2, if α ,α ∈ L(V ), then α ,α ∈/ L{V ,V ,V ,V }. Since λ1(G′) = 6,
a b 2 a b 3 4 5 6 1
then there exists only five members of [6] that labels V ,V ,V ,V . However, this
3 4 5 6
contradicts Lemma 3.1. Thus, L(V ) 6= L(V ). Now suppose one of α ,α ∈ L(V ),
2 7 a b 2
say α , labels some vertex v ∈ L(V ), then there exists some α′ ∈ [6] such that
a 1 7 a
α′ ∈/ L{V ,V ,V } such that α′ = l(v ) ∈ V , with v 6= V . Now let α ,α ∈
a 0 1 2 a 2 7 1 2 c d
L(V ). Suppose L(V ) = L(V ). Then by Corollary 3.2, α ,α ∈/ L(V ,V ,V ), which
0 3 0 a b 4 5 6
contradicts Lemma 3.1 since |L(V ,V ,V )| = 6 and [6]\2 = 5. Then, α′ ∈ L(V )
4 5 6 a 3
and also also one of α ,α′ ∈ L(V ). Further, by Corollary 3.2, α ,α′ ∈/ L(V ,V ,V ).
a b 3 a b 4 5 6
Thus, λ1(G′) ≥ 7. Conversely, λ1(G′) ≤ 7 follows directly from re-occurrence of the
1 1
labeling of C ×C . Thus, λ1(G′) = λ1(C ×C ) ≥ 7. (cid:3)
4 4 1 1 8 4
The next result focuses on the λ1−number of C × C , for m ≥ 9. Theorem 3.6
1 m 4
have already established the lower bound for λ1−number of C ×C to be 6 if m is
1 m 4
not a multiple of 6. So we only need to label C ×C with [6] such that it combines
10 4
perfectly with the labeling of C ×C with [5] to establish general bound for all cases
6 4
except when m = 14 which is dealt with separately.
0 1 2 0 0 1 5 1 2 0
5 3 4 5 6 2 6 4
3 4 5 3 3 4 0 3 5 3
2 0 1 2 3 4 0 1
0 1 2 0 0 1 5 1 2 0
Fig.3: 5−L(1,1)-labelingofC4×C6 Fig.4: 6−L(1,1)-labelingofC4×C10
Theorem 3.8. Let m′,m′′ ∈ N ∪ 0, with 10m′ + 6n′′ not a multiple of 6. Then
λ11(C10m′+6m′′ ×C4) = 6.
Proof. ByCorollary3.6, λ1(C ×C ) ≥ 6forallmnotmultipleof6. Theclaimfollows
1 m n
requiredcombinationsofFigures3and4abovewhichshowsthatλ11(C10m′+6m′′×C4) ≤
6 for 10m′ +6m′′ not a multiple of 6. (cid:3)
6 T.C ADEFOKUNAND D.O. AJAYI
Clearly, every even number m ≥ 10, m 6= 14 can be obtained from 10m′ + 6m′′
defined above. Therefore, we can conclude that for all m ≥ 9, λ1(C ×C ) = 6 for
1 m n
all m that is not a multiple of 6 if we can establish that the λ1−number of C ×C
1 14 4
is 6. We show this in the next result.
Theorem 3.9. λ1(C ×C ) = 6
1 14 4
Proof. λ1(C ×C ) = 6 (cid:3)
1 14 4
We have now completely determined the λ1−numbers of (C ×C ) for all m ≥ 3.
1 m 4
In what follows, we investigate the values of λ1(C ×C ).
1 m 6
Proposition 3.10. Let G′ be a connected component of C ×C , m ∈ N,m ≥ 3. Let
m 6
V ⊆ V(G′), i ∈ [m−1]. Then,
i
(i) λ1(C ×C ) ≥ 5.
1 m 6
(ii) Given v ,v ∈ V , d(v ,v ) ≤ 2.
a b i a b
(iii) For all V ⊆ V(G′), |L(V )| = 3
i i
(iv) suppose α ∈ L(V ), then α ∈/ L(V ).
k i k i+2
Proof. The proof of the claims above are as follows:
(i) C ×C contains P ×C . Now from Corollary 2.3, λ1(P ×C ) = 5. Therefore
m 6 m 6 1 m 6
λ1(C ×C ) ≥ 5.
1 m 6
(ii) Let V ⊆ V(G′). V = {u v ,u v ,u v }, where j ∈ {0,1}. Now, since C is a
i i i j i j+2 i j+4 m
cycle,then d(u v ,u v ) = 2. Clearly, d(u v ,u v ) = 2, d(u v ,u v ) = 2 and
i j i j+4 i j i j+2 i j+2 i j+4
thus the claim.
(iii) This is quite obvious.
(iv) It is obvious that for all u v ∈ V and u v ∈ V , d(u v ,u v ) = 2.
i j i i+2 k i+2 i j i+2 k
Therefore, L(V )∩L(V ) = ∅.
i i+2
(cid:3)
The next result describes a property of L(1,1)-labeling of C ×C
m 6
Lemma 3.11. Let α ∈ L(V ), i ∈ [m−1], V ⊆ V(G′), then α labels some vertex
k i i k
v ∈ V . In other words, L(V ) labels V .
i+1 i+1 i i+1
Proof. Let{u v ,u v ,u v } = V and{u v ,u v ,u v } = V Clearly
i j i j+2 i j+4 i i+1 j+1 i+1 j+3 i+1 j+5 i+1
d(u v ,u v ) = d(u v ,u v ) = d(u v ,u v ) = 3. Therefore, suppose
i j i+1 j+3 i j+2 i+1 j+5 i j+4 i+1 j+1
α = l(v ), for some v ∈ V , then, there exists some unique v ∈ V such that
k i i i i+k i+1
l(v ) = l(v ),with |(i−(i+k))| = 3. (The uniqueness of v results from Proposi-
i i+k i+k
(cid:3)
tion 3.10(b).)
Corollary 3.12. If L(V ) = L(V ), then, L(V )∩L(V ) = ∅ and L(V )∩L(V ) =
i i+1 i+2 i i i+3
∅.
It is obvious from Proposition 3.10(d).
Corollary 3.13. λ1(C ×C ) = 5 if and only if m ≡ 0 mod 4
1 m 6
LABELING PRODUCTS OF CYCLES 7
Proof. Let m ≡ 0 mod 4. For m = 4, clearly λ1(C ×C ) = 5, which is obtained from
1 4 6
Corollary 3.6. Now in the case of the general m ≡ 0 mod 4, by re-occurrence of the
labeling of C ×C along C , it follows that λ1(C ×C ) = 5. Conversely, suppose
4 6 m 1 m 6
that λ1(C ×C ) = 5. We show that m ≡ 0 mod 4. Let L(V ) = {α ,α ,α }. By
1 m 6 i i j k
Proposition 3.10 (c),α 6= α 6= α 6= α , that is, |L(V )| = 3. Suppose L(V ) = L(V )
i j k i i 0 1
by Lemma 3.11, then by Corollary 3.12, L(V ) ∩L(V ) = ∅ and L(V ) ∩L(V ) = ∅.
2 1 3 1
Since λ1(C × C ) = 5, Then L(V ) = L(V ) = [5]\L(V ). This scheme continues
1 m 6 2 3 0
such that L(V ) = L(V ) = L(V ); L(V ) = L(V ) = L(V )··· = L(V ) = L(V ) =
0 4 5 2 6 7 0 m−4
L(V ); and L(V ) = L(V ) = L(V ) = ··· = L(V ), n ∈ N, where 4n = (m−1)+
m−3 0 4 8 4(n)
1 = m since C is a cycle. Thus m ≡ 0 mod 4. (cid:3)
m
The implication of the last result is that the lower bound for the λ1−number of
1
graph product C ×C , m ≥ 3 is 6 except for when m ≡ 0 mod 4, in which case the
m 6
optimal λ1−number reduces by 1.
1
Now we consider particular cases where the lower bound is strictly greater than 6.
Theorem 3.14. λ1(C ×C ) = 8
1 6 6
Proof. Supposethatλ1(C ×C ) = 7. Let{V } ⊆ V(G′),foralli ∈ [5]. Byproposition
1 6 6 i
3.10 (d), L(V ) ∩ L(V ) = ∅; L(V ) ∩ L(V ) = ∅ and L(V ) ∩ L(V ) = ∅. Now,
0 2 0 4 4 2
L(V ) ⊆ [7]\L(V ) and L(V ) ⊆ [7]\L(V ). Note that |[7]\L(V )| = 5. Now set
2 0 4 0 0
[7]\L(V ) = [A′]. L(V ) ⊆ [A′]\L(V ) since L(V )∩L(V ) = ∅. Now, |[A′]\L(V )| = 2.
0 4 2 4 2 2
However, by Proposition 3.10 (c), |L(V )| = 3. Therefore a contradiction and hence
4
λ(C ×C ) ≥ 8. The labeling in Figure 6 confirms that λ1(C ×C ) ≤ 8, and thus,
6 6 1 6 6
λ1(C ×C ) = 8.
1 6 6
0 1 2 0
8 6 7
6 7 8 3
5 3 4
3 4 5 3
2 0 1
0 1 2 0
Fig.6: 8−L(1,1)-labelingofC6×C6
(cid:3)
Theorem 3.15. λ1(C ×C ) = 7
1 10 6
Proof. Let G′ be a connected component of C ×C . and suppose that λ1(G′) = 6.
10 6 1
Let V ⊂ V(G′) such that L(V ) ⊂ [6]. By Proposition 3.10 (c), L(V ) ∩L(V ) = ∅.
0 0 0 2
Therefore L(V ) ⊂ [6]\L(V ), where |L(V )| = 3. Thus, |[6]\L(V )| = 4. Now, for
0 0 0 0
all v ∈ V and v ∈ V , V ,V ⊂ V(G′), d(v ,v ) = 2, since C is a cycle of
0 0 8 8 0 8 0 8 10
length 10. Therefore, L(V ) ⊂ [6]\L(V ). Now, suppose that, L(V ) = L(V ), by
8 0 8 2
Proposition 3.10, then there exists α ∈ [6] such that α ∈/ L(V ), andα ∈ L(V ).
k k 0 k 2
Thus L(V ) ⊂ L(V ) ∪ α and L(V ) ⊂ L(V ) ∪ α . Now, |L(V )∪α | = 4. By
4 0 k 6 0 k 0 k
8 T.C ADEFOKUNAND D.O. AJAYI
Proposition 3.10, (d), L(V ) ∩ L(V ) = ∅. Thus |L(V )∪L(V )| = 6, which is a
6 4 6 4
contradiction.
Now, suppose L(V ) 6= L(V ) then it is not difficult to see that there exists
8 0
α ,α ∈ [6]\L(V ) such that L(V ) ∩ L(V ) = {α ,α }. Thus L(V ) = {α ,α ,α }
a b 0 8 2 a b 8 a b c
and L(V ) = {α ,α ,α } such that L(V ) ∪ L(V ) = [6]\L[V ]. Now by Proposition
2 a b d 2 8 2
3.10 (d) still, L(V ) ⊆ [6]\L(V ) = L(V ) ∪ α , such that α ∈/ L(V ), α ∈ L(V ).
4 2 0 k k 0 k 2
L(V ) ⊆ [6]\L(V ) = L(V ) ∪ α for α ∈/ L(V ), α ∈/ L(V ), α 6= α . Thus,
0 8 0 j j 2 k 0 j k
|L(V )∪{α ∪α }| = 5. By |L(V ∪L(V ))| = 6, and for all v ∈ V , and v ∈ V ,
0 j k 6 4 6 6 4 4
d(v ,v ) = 2. Thus a contradiction and hence λ(G′) ≥ 7.
4 6
Conversely, we consider the 7−L(1,1)-labeling of C ×C in Figure 5 below.
10 6
0 6 1 2 6 0
1 2 7 5 7
2 3 5 0 3 2
4 0 1 4 0
1 7 4 7 5 1
5 3 6 3 2
0 6 1 2 6 0
Fig.7: 7−L(1,1)-labelingofC10×C6
(cid:3)
Theorem 3.16. λ1(C ×C ) = 6
1 14 6
Proof. Since 14 is not a multiple of 4 and by Corollary 3.13, λ1(C ×C ) ≥ 6.
1 14 6
0 6 5 2 4 6 3 0
1 3 0 3 0 1 4
2 4 1 6 3 2 5 2
0 6 5 2 4 6 3
1 3 0 3 0 1 4 1
2 4 1 6 3 2 5
0 6 5 2 4 6 3 0
Fig.8: 5−L(1,1)-labelingofC14×C6
(cid:3)
We can conclude that for C ×C , if m is even, and m 6≡ 0 mod 4, then λ1(C ×
m 6 1 m
C ) ≥ 6. For m ≥ 14, this class of direct product graphs can be obtained from
6
C14+4m′, where m′ is a non-negative integer.
Theorem 3.17. Form = 14+4m′, where m′ is a non-negativeinteger, λ1(C ×C ) =
1 m 6
6
LABELING PRODUCTS OF CYCLES 9
Proof. Since for any non-negative integer m′, m = 14 + 4m′ 6≡ 0 mod 4, then by
3.13, λ1(C ×C ) ≥ 6. By combining the labeling in Figure 8 and m′−multiple of
1 m n
the labeling in Figure 3, we have that λ1(C ×C ) ≤ 6 and the result follows. (cid:3)
1 m n
Note that the following result was established in [2]
Theorem 3.18. Let m′,n′ ≡ 0mod10 and A = {12,14,16,18}. Then, for all k ∈
Aandm,n, λ11(Cm′ ×Ck+n′) = 5. Also, let m,n ≡ 0mod5, then λ11(Cm ×Cn) = 4.
4. Labeling of C ×C , n ≥ 8
m n
In this section, we obtainthe λ1−numbers of graphproduct C ×C , where n,m ≥
1 m n
8.
Now we establish the λ1−number for C ×C . Since labeling of product graphs is
1 m 8
commutative, we restrict our work in this section to m ≥ 8 since the cases for smaller
graphs have been taken care of in the last sections.
The following result are helpful to reveal some useful properties of L(1,1)- labeling
of C ×C .
m 8
Lemma 4.1. Let G′ be a connected component of C × C , m ≥ 4. Suppose there
m 8
exist v ,v ∈ V such that α = l(v ) = l(v ) ∈ [p], p ∈ N. then α ∈/ L(V ∪V ).
a b i k a b k i+1 i+2
Furthermore, α ∈/ L(V ∪V ).
k i−1 i−2
Proof. Claim: Let α = l(v ) = l(v ), v ,v ∈ V , V ⊆ V(G′). Then, d(v ,v ) = 4.
k a b a b i i a b
Reason: Clearly, |V | = 4 for i ∈ [m − 1]. Let V = {u v ,u v ,u v ,u v }. So,
i i i 0 i 2 i 4 i 6
d(u v ,u v )=d(u v ,u v )=d(u v ,u v )=2. also, d(u v ,u v ) = 2 since C is a cy-
i 0 i 2 i 2 i 4 i 4 i 6 i 6 i 4 8
cle. However, d(u v ,u v )=d(u v ,u v ) = 4. Thus v = v v and v = v v or
i 0 i 4 i 2 i 6 a i 0 b i 4
v = v v and v = v v . Now, suppose v = u v and v = u v . Let V =
a i 2 b i 6 a i 0 b i 4 i+1
{u v ,u v ,u v ,u v }. Then that d(u v ,u v ) = 1 = d(u v ,u v ) fol-
i+1 1 I+1 3 i+1 5 i+1 7 i 0 i+1 1 i 0 i+1 7
lows from the definition of C ×C . Likewise, d(u v ,u v ) = 1 = d(u v ,u v ).
m n=8 i 4 i+1 3 i 4 i+1 5
Therefore, α ∈ L(V ). Also, let V = {u v ,u v ,u v ,u v }. Then
k i+1 i+2 i+2 0 i+2 2 i+2 4 i+2 6
d(u v ,u v ) = 2=d(u v ,u v ) and d(u v ,u v )=2=d(u v ,u v ) since C
i 0 i+2 0 i 0 i+2 2 i 0 i+2 6 i 0 i+2 2 m=8
is a cycle. Now d(u v ,u v ) = 2 and therefore α ∈/ L(V ). This argument is
i 4 i+2 2(4,6) k i+2
valid for V and V . (cid:3)
i−1 1−2
The consequence of Lemma 4.1 is that if a label is assigned to two vertices on
V ⊂ V(G′), then the label could no longer be assigned to another vertex on the
i
vertex sets two step above or below it. The next result is similar.
Proposition 4.2. Suppose v ∈ V and v ∈ V such that α = l(v ) = l(v ),
i i i+2 i+1 k i i+1
then, d(v ,v ) = 3.
i i+1
Proof. Suppose v = u v without loss of generality, then d(v ,u v ) = 2. Now.
i i 0 i i+1 1(7)
d(v ,u v ) = 3 and d(v ,u v ) = 3. (cid:3)
i i+1 3 i i+1 5
Lemma 4.3. Suppose V ,V ⊂ V(G′) where G′ is a connected component of C ×
1 i+1 m
C . Let α ∈ L(V )∩L(V ) then α ∈/ L(V )∪L(V ).
8 k i i+1 k i−1 i+2
10 T.C ADEFOKUNAND D.O. AJAYI
Proof. By Proposition 4.2, suppose that α = l(u v ) and that α ∈ L(V ) then,
k i j k i+1
α = l(u v ) or α = l(u v ). Without loss of generality, suppose that in fact,
k i+1 j+3 k i+1 j+5
α = l(u v +5). Thend(u v ,u v ) = 2. Meanwhile, d(u v ,u v ) =
k i+1 j i j i+2 j(j+2,j+6) i+1 j+3(j+5) i+2 j+4
1. Thus, α ∈/ L(V ). Similar argument holds for α ∈/ L(V ). (cid:3)
k i+2 k i−1
By to Lemma 4.3, it is quite clear that if α belongs L(V ) ∩ L(V + 1), then α
k i i k
does not belong to L(V ∪Vi+2). A similar result is as follows:
i−1
Lemma 4.4. Suppose α ∈ L(V )∩L(V ) ⊂ V(G′), where G′ is a connected com-
k i i+2
ponent of C ×C , m ≥ 4, then α ∈/ L(V )
m 8 k i+1
Proof. Let,v ∈ V beu v . Notethat,d(u v ,u v ) = 2sinceC isacycle. Then,
i i i 0 i 0 i+2 0(2,6) 8
the remaining vertex v ∈ V such that l(v ) = α is u v andd(u v ,u v ) =
i+2 i+2 i+2 k i+2 4 i 0 i+2 4
4. Now d(v ,u v ) = 1 and d(u v ,u v ) = 1. Thus, α ∈/ L(V ). (cid:3)
i i+1 1(7) i+2 4 i+1 3(5) k i+1
The consequence of Lemma 4.4 is that if two vertices on V and V share the same
i i+2
label, then that label can not be shared by another vertex on V given that V ,V
i+1 i i+1
and V are all in V(G′).
i+2
Next we establish the lower bound of λ1(C ×C ) where m ≥ 8 and m ≡ 2 mod 6.
1 m 8
We require the following definition.
Let G′ be a connected component of G. Then, V is the class of all vertices on
αk
V(G′) labeled α .
k
Lemma 4.5. For m ≥ 8,m ≡ 2 mod 6, λ1(C ×C ) ≥ 6.
1 m 8
Proof. Case 1: Let α ∈ L(V(G′)) such that if α ∈ L(V ), V ⊂ V(G′), i ∈ [m−1],
k k i i
then there exist v′,v′′ ∈ V such that l(v′) = α = l(v′′). Let V¯ be a class of all V ∈
i i i i k i i
V(G′)suchthatα ∈ L(V ). Nowsuppose, without lossofgenerality, thatV ∈ V¯. By
k i 0
thisandLemma 4.1, andbyassuming thatα labelsV(G′) optimally, suppose V ∈ V¯,
k i
then i ≡ 0 mod 3,i 6= m−2. Since m ≡ 2 mod 6, then there exists n′ ∈ N, such
that m = 6n′+2. Note that m−5 = (6n′+2)−5 = 3(2n′−1). Thus, V ∈ V¯. By
m−5
Lemma 4.1 and since V ,V ∈ V¯, then α ∈/ L(V ∪V ∪V ∪V ). Thus
0 m−5 k m−4 m−3 m−2 m−1
V¯ = {V ,··· ,V }. Set V¯′ = V¯\{V }. Since V¯′ = 2n′ −1, then |V′| = 2n′. Now,
0 m−5 0
(cid:12) (cid:12)
|V(G′)| = (6n′ +2)4. Clearly |V | = 2(2n′) = 4(cid:12)n′.(cid:12)Hence, |V(G′)|=6n′+2 > 6.
αk |Vαk| n′
Case 2: Suppose that for all triple V ,V V ⊂ V(G), α ∈ L(V ∩V ) and by
i i+1 i+2 k i i+1
Lemma 4.3 α ∈/ V . Without loss of generality, we select the initial triple to be
k i+2
V ,V ,V , such that α ∈ L(V ∩V ), α ∈/ V ; (and α ∈ L(V ∩V ),α ∈/ L(V )···).
0 1 2 k 0 1 k 2 k 3 4 k 5
Therefore, α ∈/ V for all i ∈ [m − 1] such that i + 1 ≡ 0 mod 3. Now, m ≡ 2
k i
mod 6 implies there exists n′ ∈ N such that m = 6n′+2. Thus, m−2 ≡ 0 mod 3
and hence α ∈/ L(V ). Now, since α ∈ L(V ) ∩ L(V ), then α ∈/ L(V ) by
k m−3 k 0 1 k m−1
Lemma 4.3 and since C is a cycle. By Lemma 4.4, it is possible for α ∈ L(V )
m k m−2
since α ∈/ L(V ). Thus we, for maximality, assume that α ∈ L(V ). Now,
k m−3 k m−2
Let V¯ ={V ,··· ,V } ⊂ V(G′). Then V¯ = (m−3)+1 = 2(6n′) = 4n′, where
αk 0 m−3 (cid:12) αk(cid:12) h 3 i 3
n′ ∈ N. Thus, for all V ∈ V(G′), |V |(cid:12) = (cid:12)4n′ + 1 since α ∈ L(V ). Thus
i αk k m−2
|V(G′)| = (6n′+2)4 > 6.
|Vαk| 4n′+1