full friendly index set — i

Discrete Applied Mathematics 161 (2013) 1262–1274

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Discrete Applied Mathematics

journal homepage: www.elsevier.com/locate/dam

Full friendly index set — I✩

Deepa Sinha ∗, Jaspreet KaurCentre for Mathematical Sciences, Banasthali University, Banasthali-304022, Rajasthan, India

a r t i c l e i n f o

Article history:Received 2 September 2012Accepted 4 October 2012Available online 19 November 2012

Keywords:Cordial labelingFriendly labelingFull friendly index set

a b s t r a c t

Let G = (V , E) be a graph, a vertex labeling f : V → Z2 induces an edge labeling f ∗: E →

Z2 defined by f ∗(xy) = f (x)+ f (y) for each xy ∈ E. For each, i ∈ Z2 define vf (i) = |f −1(i)|and ef (i) = |f ∗−1(i)|. We call f friendly if |vf (1) − vf (0)| ≤ 1. The full friendly index setof G is the set of all possible values of ef (1) − ef (0), where f is friendly. In this paper, westudy the full friendly index sets of some standard graphs such as the complete graph Kn,the cycle Cn, fans Fm and F2,m and the Cartesian product of P3 and Pn i.e. P3 × Pn.

© 2012 Elsevier B.V. All rights reserved.

1. Introduction

In this paper, all graphs are assumed to be simple, finite and undirected. All undefined notations and concepts may befound in [4,12,14].

Let G = (V , E) be a graph, and Γ be an abelian group and f : V → Γ be a vertex labeling of G with the elements of Γ .Let f ∗

: E → Γ be defined by

f ∗(xy) = f (x) ◦ f (y), ∀xy ∈ E,

where ‘◦’ is the group operation. For each g ∈ Γ , let vf (g) = |f −1(g)| and ef (g) = |f ∗−1(g)|. We call f a Γ -friendly labelingif

|vf (g1) − vf (g2)| ≤ 1, ∀g1, g2 ∈ Γ , (1)

Γ -edge friendly labeling if

|ef (g1) − ef (g2)| ≤ 1, ∀g1, g2 ∈ Γ (2)

and Γ -cordial if both (1) and (2) hold (cf.: [18]).A graph G is said to be cordial if it admits a Z2-cordial labeling, where Z2 = {0, 1} is the additive group with respect to

modulo 2 addition (cf.: [5]). Cordial labelings of graphs have been studied extensively in [2,7,6,9,17,19,20,22,26].Z2-friendlylabelings are simply called friendly labelings (cf.: [11]). The number if (G) = ef (1) − ef (0) is called the friendly index of f . Thefriendly index set of the graph G, denoted FI(G), is defined as

FI(G) = {|if (G)| : f is a friendly labeling of G}.

The concept was extended by Harris and Kwong [27] to full friendly index set for the graph G, denoted FFI(G), defined as

FFI(G) = {if (G) : f is a friendly labeling of G}.

✩ Research is supported by the Department of Science and Technology (Govt. of India), New Delhi, India under the project SR/S4/MS:409/06.∗ Corresponding author.

E-mail addresses: [email protected] (D. Sinha), [email protected] (J. Kaur).

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D. Sinha, J. Kaur / Discrete Applied Mathematics 161 (2013) 1262–1274 1263

A vertex v is called a k-vertex if f (v) = k, k ∈ {0, 1}. Similarly, an edge e is called a k-edge if f ∗(e) = k, k ∈ {0, 1}. Notethat if f : V (G) → Z2 is a friendly labeling, so is its inverse labeling g : V (G) → Z2 defined by g(v) = 1 − f (v) ∀ v ∈ V (G).Moreover, if (G) = ig(G). Readers interested in friendly labelings and the friendly index set of a graphmay consult a numberof relevant literatures that are mentioned in the reference section, particularly [21,23–25].

In one of the presentations on this particular topic in the intensive interaction session held in the Center forMathematicalSciences, Banasthali University, B.D. Acharya suggested to justify the term ‘‘friendly’’ in the definition of a ‘‘friendly labeling’’invoking the following context: Suppose the label 0 is replaced by +1 and 1 is replaced by −1 in the definition of a friendlylabeling. InterpretG as a social network inwhich individualswith positive (negative) attitudes are labeled as being ‘‘positive’’(respectively, ‘‘negative’’), indicated by assigning the weight +1 (−1), and, similarly, the character of their interpersonalrelationships (as may be evident from the characteristics of their behavioral expressions such as friendship or enmity) beingpositive or negative are taken as being represented by the weights +1 or −1 accordingly. With such a representation of a‘socio-psychological phenomenon’, friendly labelingµ finds justification in the interpretation of the state of the social systemin which the subgroups Vµ(+1) = {u : µ(u) = +1} and Vµ(−1) = V (G)−Vµ(+1) form an equitable partition of the socialgroup V . A partition of a social group in this manner presents many interesting socio-psychological situations. For instance,when seen in terms of their populations vf (0) =: vµ(+1) and vf (1) =: vµ(−1), the ‘‘minority group’’ has a tendency tobecome increasingly argumentative and assertive in its interaction with the ‘‘majority group’’ whereas the latter tends toincreasingly perceive and recognize the threat of the eventual possibility of the reduction in its dominance in wielding itspower over thematters of overall common interest to the entire social system; these tendencies attain their peaks when thepopulations of the minority and the majority groups are equitable, whence the groups are in a state of ‘‘forced friendliness’’with each other in order to make best of the situation and to maintain a delicate balance in order to derive the best of theopportunities provided by the entire social system. Since the overall stability of a social system is well known to dependlargely on the attitudes and behaviors of its individual members, it is important to develop rigorous mathematical analysisof friendly labelings (or, equivalently, ‘equitable 2-partitions’) of the vertices of a graph (that provides a ‘structural model’for a discrete dynamical social system). Hence, a mathematical analysis of friendly labelings is the main aim of this paper.

Another interesting context is found in the introductory note [3]: Application of the finite elements method for solvingdifferential equations requires partitioning of the underlying area into simple figures – e.g. squares – and assigning the nodesof the partition to the processors of a parallel computing system. This leads to the problem of constructing an assignment insuch a way that the total amount of computations would be equally distributed among the processors and the informationflow between the processors would be minimized. In other words, one has to partition the vertex set of the underlyinggraph into a certain number of parts (corresponding to the number of processors in the system) consisting of almost equalnumber of vertices so that the number of edges connecting the vertices of different parts is as small as possible (see [1]).Hence, if the number of parts is denoted by k, for k = 2 we have the notion of bisection width of a graph G = (V , E), denotedΘ0

2 (G), defined as the minimum number of edges interconnecting the vertices in the subsets of an equitable bipartition, or asoften called, a bisection, {V1, V2} of V ; here, ‘equitable’ means that the cardinalities of V1 and V2 differ by at most 1. Themin-bisection problem (MBP) is thus to find a bisection of Gwhichminimizes the number of edges connecting the two subsets. Anedge e connects V1 and V2, if one of its end points lies in V1 and the other lies in V2. In general, there is no known polynomialtime algorithm for this problem. Clearly, solving the MBP is equivalent to finding a friendly labeling f with minimum ef (1)value, namely Θ0

2 (G), f ranging over the set FG of all friendly labelings of G. Therefore, one looks forward to solving theproblem for specific classes of graphs that admit a ‘‘good’’ solution, in the sense that there does exist a polynomial timealgorithm to solve the MBP. For example, if the graph is a tree the problem becomes easier in this sense (e.g., see [8]). Onecan go through [1,13,16] for other convincing motivations.

Keeping the above in view, in this paper, we determine the full friendly index sets of some of the standard graphs suchas the complete graph Kn, the cycle Cn, fans Fm and F2,m and the Cartesian product of P3 and Pn i.e. P3 × Pn.

2. Some elementary properties

Now, we present some known and derived results which will be used in our preliminary investigation. Here onwards, bysimply a labeling of a graph G = (V , E) we shall mean any function f : V → Z2.

Lemma 1 ([27]). Let f be a labeling of a graph G that contains a cycle C as its subgraph. If C contains a 1-edge, then the numberof 1-edges in C is a positive even number.

Proof. This follows from the famous Kirchhoff’s Voltage Law (KVL) as stated in [15, Theorem 2, p. 26] for integer values ofvoltages and currents in a graph (which may be regarded essentially as a symmetric digraph). �

We can observe from the proof of Lemma 1 that friendly labelings yield a special class of ‘marked balanced signed graphs’in which a friendly labeling is treated as amarking µ, that assigns to each vertex an element of the involutory multiplicativegroup M2 = {−1, +1}, and the product σ(uv) = µ(u)µ(v) to each edge uv, giving rise to the sets Vµ(+1) = {u : µ(u) =

+1} and Vµ(−1) = V (G) − Vµ(+1) that form an equitable ‘‘Harary bipartition’’ of the vertex-set V (G) of the signed graphSµ = (Gµ, σ ), where by a signed graph we mean a graph in which every edge is designated as being positive or negative(equivalently, as being assigned one of the weights +1 and −1). Further, a signed graph S = (G, σ ) is said to be balancedwhenever every cycle in it contains an even number of negative edges (e.g., see [10]).

1264 D. Sinha, J. Kaur / Discrete Applied Mathematics 161 (2013) 1262–1274

The following is an interesting consequence of Lemma 1.

Corollary 1. Let f be a labeling of an Eulerian graph G. Then, the number ef (1) of 1-edges in G is an even number.

Proof. Since any Eulerian graph can be decomposed into edge-disjoint cycles, the result follows from Lemma 1. �

Lemma 2 ([27]). If G = (V , E) is any graph of order greater than 1, then

FFI(G) ⊆ {2i − q : 1 ≤ i ≤ q}.

Lemma 3 ([27]). If G is a 2-connected graph of order greater than 1, then

FFI(G) ⊆ {2i − q : 2 ≤ i ≤ q}.

Lemma 4 ([27]). If G is a 2-connected bipartite (p, q)-graph with p ≥ 2, then ef (0) = 1. Hence, q − 2 ∈ FFI(G).

Lemma 5 ([27]). For n ≥ 2, FFI(Pn) = {2i − n + 1 : 1 ≤ i ≤ n − 1}.

Lemma 6 ([27]). Let H be a (pH , qH)-graph and K be a (pK , qK )-graph. Then

pK FFI(H) + pHFFI(K) ⊂ FFI(H × K)

where H × K denotes the Cartesian product of H and K .

Lemma 7 ([27]). Let H be a (pH , qH)-graph with a friendly labeling h, and let K be a (pK , qK )-graph. If pH is even, thenih(H)pK − qKpH ∈ FFI(H × K).

3. FFI(Kn)

Theorem 1. The full friendly index set of the complete graph Kn is

FFI(Kn) =

n2

n is even,

n − 12

n is odd.

Proof. Let f be a friendly labeling of Kn, V0 = {v : f (v) = 0} and V1 = {v : f (v) = 1}. The result involves considering thefollowing two cases:Case 1: Let n be even. Clearly, |V0| = |V1| =

n2 . Every edge within V0 or V1 is a 0-edge. So, the number of 0-edges is n

2 (n2 − 1)

and every edge from V0 to V1 will be a 1-edge. So, the number of 1-edges is n2n2 . Therefore, FFI(Kn) =

n2

.

Case 2: Let n be odd. Clearly, either |V0| =n−12 and |V1| =

n+12 or vice versa.

Let |V0| =n−12 and |V1| =

n+12 . Every edge within V0 and V1 will be 0-edge. So, number of 0-edge will be ( n−1

2 )( n−12 ) and

every edge from V0 to V1 will be 1-edge. So, number of 1-edge will be ( n+12 )( n−1

2 ). Therefore, FFI(Kn) = n−1

2

.

Similarly, we can easily prove if |V0| =n+12 and |V1| =

n−12 .

Thus,

FFI(Kn) =

n2

n is even,

n − 12

n is odd.

�

4. FFI(Cn)

Theorem 2. The full friendly index set of a cycle Cn is

FFI(Cn) =

2i − n : i = 2j; 1 ≤ j ≤

n2

.

Proof. By Lemmas 1 and 3, we have

FFI(Cn) ⊆

2i − n : i = 2j; 1 ≤ j ≤

n2

. (3)


Now, for ef (1) = 2x, 1 ≤ x ≤ n

2

, we define the friendly labeling f of Cn as:

f (vi) =

0 i = 1, 2, . . . ,n2

− x,

i =

n2

− x + 1,

n2

− x + 3, . . . ,

n2

− x + (2x − 1),

1 i =

n2

− x + 2,

n2

− x + 4, . . . ,

n2

− x + 2x,

i =

n2

+ x + 1,

n2

+ x + 2, . . . , n.

Thus, 2i − n : i = 2j; 1 ≤ j ≤

n2

⊆ FFI(Cn). (4)

From (3) and (4), we get

FFI(Cn) =

2i − n : i = 2j; 1 ≤ j ≤

n2

. �

FI(Cn) =

{0, 4, 8, . . . , n} if n ≡ 0 (mod 4),{2, 6, 10, . . . , n} if n ≡ 2 (mod 4),{1, 3, 5, . . . , n − 2} if n is odd.

The friendly index set of Cn coincides with that discussed by Kwong et al. [21].Let Fx denote the set of all friendly labelings f such that ef (1) = x. Now, we will construct the sets for all possible values

of ef (1) of Cn, i.e. F2x, 1 ≤ x ≤ ⌊n2⌋.

Take an alternate sequence of 0’s and 1’s starting with 0 of length 2x. Now insert the remaining (⌊ n2⌋ − x)(or⌈ n

2⌉ − x) 0’sin the neighborhood of any 0 of the sequence in all or in parts and (⌈ n

2⌉ − x)(or⌊ n2⌋ − x) 1’s in the neighborhood of any 1 of

the sequence in all or in parts. All obtained sequences are the friendly labelings such that ef (1) = 2x, 1 ≤ x ≤ ⌊n2⌋. It can

be easily seen that these are the only possible friendly labelings such that ef (1) = 2x as corresponding to each switchingfrom 0 to 1 and 1 to 0 in the sequence we have a 1-edge.

5. FFI(Fm)

Fm, known as the fan is defined as K1 + Pm. We will refer the vertices of the fan Fm by ‘‘u1, u2, . . . , um, v’’ (as shown inFig. 1). In this section, by Pm we will always mean a path formed by the vertices ‘‘u1, u2, . . . , um’’ and by K1,m we will meanthe edges due to central vertex v of the fan Fm and the vertices lying on the path Pm. Clearly, Pm and K1,m are edge disjoint.

Lemma 8. Let f be the friendly labeling of Fm,m ≥ 2, then the minimum and maximum values of ef (1) arem

2

+ 1 andm

2

+ m − 1 respectively.

Proof. Let f be the friendly labeling of Fm and g be the labeling on the path graph Pm such that g(ui) = f (ui), 1 ≤ i ≤ mand h be the labeling on star K1,m such that h(ui) = f (ui), 1 ≤ i ≤ m, h(v) = f (v). There arises following cases:Case 1: Let m be odd. By friendliness, Fm can be labeled in the following ways:

1. When u1, u2, . . . , um are labeled with any permutation of (m+12 ) 0’s and (m−1

2 ) 1’s and vertex v is labeled 1.Clearly, for all possible labelings under defined conditions, the value of eh(1) is (m+1

2 ) and the minimum and maximumvalues of eg(1) are 1 andm− 1 for the permutations (0 0 . . . 0 1 1 . . . 1) and (0 1 0 1 . . . 1 0) respectively. Therefore, theminimum and maximum values of ef (1) on Fm under defined conditions are m+3

2 and m+12 + m − 1 respectively.

2. When u1, u2, . . . , um are labeled with any permutation of (m−12 ) 0’s and (m+1

2 ) 1’s and vertex v is labeled 0.Similarly, amongst all the possible labelings on Fm, the value of eh(1) is (m+1

2 ) and the minimum andmaximum values ofeg(1) are 1 andm− 1 for the permutations (0 0 . . . 0 1 1 . . . 1) and (1 0 1 0 . . . 0 1) respectively. Therefore, the minimumand maximum values of ef (1) on Fm under defined conditions are m+3

2 and m+12 + m − 1 respectively.

Case 2: Let m be even. By friendliness, Fm can be labeled in following ways:

1. When u1, u2, . . . , um are labeled with any permutation of (m2 ) 0’s and (m

2 ) 1’s and vertex v is labeled 0.In this case, amongst all the possible labelings on Fm, the value of eh(1) is (m

2 ) and the minimum andmaximum values ofeg(1) are 1 andm− 1 for the permutations (0 0 . . . 0 1 1 . . . 1) and (0 1 0 1 . . . 0 1) respectively. Therefore, the minimumand maximum value of ef (1) on Fm under defined conditions are m

2 + 1 and m2 + m − 1 respectively.


Fig. 1. Fan Fm .

2. When u1, u2, . . . , um are labeled with any permutation of (m2 + 1) 0’s and (m

2 − 1) 1’s and vertex v is labeled 1.Similarly, amongst all the possible labelings on Fm, the value of eh(1) is (m

2 + 1) and the minimum and maximum valuesof eg(1) are 1 and m − 2 for the permutations (0 0 . . . 01 1 . . . 1) and (0 1 0 1 0 . . . 1 0 0) respectively. Therefore, theminimum and maximum values of ef (1) on Fm under defined conditions are m


3. When u1, u2, . . . , um are labeled with any permutation of (m2 − 1) 0’s and (m

2 + 1) 1’s and vertex v is labeled 0.Amongst all the possible labelings on Fm, the value of eh(1) is (m

2 + 1) and the minimum and maximum values of eg(1)are 1 and m − 2 for the permutations (0 0 . . . 0 1 1 . . . 1) and (1 0 1 0 . . . 1 0 1 1) respectively. Therefore, the minimumand maximum values of ef (1) on Fm under defined conditions are m


4. When u1, u2, . . . , um are labeled with any permutation of (m2 ) 0’s and (m

2 ) 1’s and vertex v is labeled 1.Amongst all the possible labelings on Fm, the value of eh(1) is (m

2 ) and the minimum and maximum values of eg(1) are1 and m − 1 for the permutations (0 0 . . . 0 1 1 . . . 1) and (0 1 0 1 . . . 0 1) respectively. Therefore, the minimum andmaximum values of ef (1) on Fm under defined conditions are m


Thus, for the graph Fm, m ≥ 2, the minimum andmaximum values of ef (1) arem

2

+1 and

m2

+m−1 respectively. �

Theorem 3. The full friendly index set of Fm is

FFI(Fm) =

2i − 2m + 1 :

m2

+ 1 ≤ i ≤

m2

+ m − 1

.

Proof. Using Lemma 3, we have

FFI(Fm) ⊆ {2i − 2m + 1 : 2 ≤ i ≤ 2m − 1}.

By Lemma 8, we get

FFI(Fm) ⊆

2i − 2m + 1 :

m2

+ 1 ≤ i ≤

m2

+ m − 1

. (5)

For all friendly labelings f of K1,m, ef (1) =m

2

. Thus,

FFI(Pm) + FFI(K1,m) ⊆ FFI(Fm)

or {2i − m + 1 : 1 ≤ i ≤ m − 1} + {2i − m : i =m

2

} ⊆ FFI(Fm)

2i − 2m + 1 :

m2

+ 1 ≤ i ≤

m2

+ m − 1

⊆ FFI(Fm). (6)


FFI(Fm) =

2i − 2m + 1 :

m2

+ 1 ≤ i ≤

m2

+ m − 1

. �

6. FFI(F2,m)

A fan F2,m is defined as K2 + Pm. We will refer to the vertices of fan F2,m by ‘‘u1, u2, . . . , um, v1, v2’’ (as shown in Fig. 2). Inthis section, by Pm wewill always mean a path formed by the vertices ‘‘u1, u2, . . . , um’’, by star K1,m wewill mean the edgesdue to vertex v1 of the fan and the vertices lying on the path Pm and by star K ′

1,m wemean the edges due to vertex v2 of thefan and the vertices lying on the path Pm. Clearly, Pm, K1,m and K ′

1,m are edge disjoint.


Fig. 2. Fan F2,m .

Lemma 9. Let f be a friendly labeling of F2,m, m ≥ 4, and E (shown in Fig. 3) be the subgraph of F2,m such that the labeling f ′

is the restriction of f on E. The labeling f ′ can be extended to a labeling g ′ of F (shown in Fig. 3) such that labeling g defined onF2,m+2 as

g(v) =

f (v) v ∈ F2,mg ′(v) v ∈ F2,m

is friendly and eg(1) = ef (1) + 2.

Proof. Since for all the possible friendly labelings of the graph F2,m,m ≥ 4, there exist vertices ui, ui+1 labeled with ‘‘0, 1’’.Consider subgraph E of F2,m as a graph induced on the vertices ‘‘ui, ui+1, v1, v2’’ will be labeled with one of the labelings‘‘0, 1, 0, 0’’, ‘‘0, 1, 0, 1’’, ‘‘0, 1, 1, 1’’ respectively. Now, construct F2,m+2 by introducing the vertices x and y on the edge uiui+1.

Define g ′ as follows:

ui ui+1 v1 v2 x y0 1 0 0 0 10 1 0 1 0 10 1 1 1 0 1

Clearly in each case of defined labeling g ′ of F , the labeling g of F2,m+2 is friendly and eg(1) = ef (1) + 2. �

Lemma 10. For F2,m, there exist friendly labelings f and g such that ef (1) = 2m − 1 and eg(1) = 2m.

Proof. For ef (1) = 2m − 1, the proof can be discussed in the following cases:Case 1: Let m be even. Define the labeling f as:

f (v) =

0 v = v1, ui; i = 1, 3, . . . ,m − 11 v = v2, ui; i = 2, 4, . . . ,m.

Case 2: Let m be odd. Define the labeling f as:

f (v) =

0 v = v1, ui; i = 1, 3, . . . ,m1 v = v2, ui; i = 2, 4, . . . ,m − 1.

Clearly, f is a friendly labeling and ef (1) = 2m − 1.For eg(1) = 2m, the proof can be discussed in the following cases:

Case 1: Let m be even. Define the labeling g as:

g(v) =

0 v = v1, v2, ui; i = 2, 4, . . . ,m − 21 v = um, ui; i = 1, 3, . . . ,m − 1.

Case 2: Let m be odd. Define the labeling g as:

g(v) =

0 v = v1, v2, ui; i = 2, 4, . . . ,m − 11 v = ui; i = 1, 3, . . . ,m.

Clearly, g is a friendly labeling and eg(1) = 2m. �


Fig. 3. Graphs E and F .

Lemma 11. Let f be the friendly labeling of the graph F2,m,m ≥ 2, then the minimum and maximum value of ef (1) are m + 1and 2m respectively.

Proof. Let f be the friendly labeling of F2,m, f ′, g and h be the labeling on the path Pm, K1,m and K ′1,m respectively such that

f ′(ui) = f (ui), 1 ≤ i ≤ m, g(ui) = f (ui), 1 ≤ i ≤ m, g(v1) = f (v1), h(ui) = f (ui), 1 ≤ i ≤ m, h(v2) = f (v2). Now, theproof can be done by considering following cases:

1. Let m be even. By friendliness, m2 + 1 vertices must be labeled by 0 and the remaining m

2 + 1 vertices must be labeled by1. There arises following cases:(a) v1 and v2 both are labeled 0. Now, the vertices u1, u2, . . . , um can be labeled with any permutation of (m

2 −1)0′s and(m2 + 1)1′s. Clearly, for all possible permutations, eg(1) = eh(1) = (m

2 + 1) and ef ′(1) is minimum andmaximum forthe permutations (0 0 . . . 0 1 1 . . . 1) and (0 1 0 1 . . . 0 1 0 0) respectively. The corresponding values of ef ′(1) are 1and m − 2 respectively. Therefore, the minimum and maximum values of ef (1) on F2,m under defined condition arem + 3 and 2m respectively.

(b) v1 and v2 both are labeled 1. As in Case 1, we can evaluate the minimum and maximum possible values of ef (1) onF2,m under defined condition which will bem + 3 and 2m respectively.

(c) v1 is labeled 0 and v2 is labeled 1. As in Case 1, we can evaluate the minimum and maximum possible values of ef (1)on F2,m under defined condition which will bem + 1 and 2m − 1 respectively.

(d) v1 is labeled 1 and v2 is labeled 0. As in Case 1, we can evaluate the minimum and maximum possible value of ef (1)on F2,m under defined condition which will bem + 1 and 2m − 1 respectively.

2. Let m be odd. By friendliness, m+12 vertices must be labeled by 0 and the remaining m+3

2 vertices must be labeled by 1 orvice versa. There arises the following cases according to the labeling of u1, u2, . . . , um:(a) m+1

2 0’s and m+32 1’s.

i. v1 and v2 both are labeled 0. Now, the vertices u1, u2, . . . , um can be labeled with any permutation of m−32 0′s and

m+32 1′s. Clearly, for all possible permutations, eg(1) = eh(1) =

m+32 and ef ′(1) is minimum andmaximum for the

permutations (0 0 . . . 0 1 1 . . . 1) and (1 0 1 0 . . . 1 0 1 1 1) respectively. The corresponding values of ef ′(1) are 1and m − 3 respectively. Thus, minimum and maximum possible values of ef (1) on F2,m under defined conditionarem + 4 and 2m respectively.

ii. v1 and v2 both are labeled 1. As in Case 1, we can evaluate the minimum andmaximum possible values of ef (1) onF2,m under defined condition which will bem + 2 and 2m respectively.

iii. v1 is labeled 0 and v2 is labeled 1. As in Case 1, we can evaluate theminimum andmaximum possible value of ef (1)on F2,m under defined condition which will bem + 1 and 2m − 1 respectively.

iv. v1 is labeled 1 and v2 is labeled 0. As in Case 1, we can evaluate theminimum andmaximum possible value of ef (1)on F2,m under defined condition which will bem + 1 and 2m − 1 respectively.

(b) m+32 0’s and m+1

2 1’s.i. v1 and v2 both are labeled 0. Now, the vertices u1, u2, . . . , um can be labeled with any permutation of m+1

2 0′sand m−1

2 1′s. Clearly, for all possible permutations eg(1) = eh(1) =m−12 and ef ′(1) is minimum and maximum

for the permutations (0 0 . . . 0 1 1 . . . 1) and (0, 1, 0, 1, . . . , 0, 1, 0) respectively. The corresponding values are 1andm− 1. Thus, minimum andmaximum possible values of ef (1) on F2,m under defined condition arem+ 1 and2m respectively.

ii. v1 and v2 both are labeled 1. As in Case 1, we can evaluate the minimum andmaximum possible values of ef (1) onF2,m which under defined condition will bem + 3 and 2m respectively.

iii. v1 is labeled 0 and v2 is labeled 1. As in Case 1, we can evaluate the minimum and maximum possible values ofef (1) on F2,m which under defined condition which will bem + 1 and 2m − 1 respectively.

iv. v1 is labeled 1 and v2 is labeled 0. As in Case 1, we can evaluate the minimum and maximum possible values ofef (1) on F2,m which under defined condition which will bem + 1 and 2m − 1 respectively.

Thus, for the graph F2,m,m ≥ 2, the minimum and maximum values of ef (1) arem + 1 and 2m respectively. �


Fig. 4. Graph A.

Theorem 4. The full friendly index set of F2,m,m ≥ 4, is

FFI(F2,m) = {2i − 3m + 1 : m + 1 ≤ i ≤ 2m}.

Proof. By Lemma 3,

FFI(F2,m) ⊆ {2i − 3m + 1 : 2 ≤ i ≤ 3m − 1}.

By Lemma 11, we get

FFI(F2,m) ⊆ {2i − 3m + 1 : m + 1 ≤ i ≤ 2m}. (7)

Now, using induction onm we will prove that

{2i − 3m + 1 : m + 1 ≤ i ≤ 2m} ⊆ FFI(F2,m).

Form = 4, define the labelings f , g, h and j as:

f (v) =

0 v = u1, u2, v11 v = u3, u4, v2

, g(v) =

0 v = u1, u4, v11 v = u2, u3, v2

,

h(v) =

0 v = u1, u3, v11 v = u2, u4, v2

and j(v) =

0 v = u1, u2, u41 v = u3, v1, v2.

Clearly, f , g, h and j are friendly labelings and ef (1) = 5, eg(1) = 6, eh(1) = 7, ej(1) = 8. Thus,

{2i − 11 : i = 5, 6, 7, 8} ⊆ FFI(F2,4).

Let {2i − 3k + 1 : k + 1 ≤ i ≤ 2k} ⊆ FFI(F2,k).By Lemma 9, we get

FFI(F2,k) + {−2} ⊆ FFI(F2,k+2)

or {2i − 3k − 5 : k + 3 ≤ i ≤ 2k + 2} ⊆ FFI(F2,k+2).

Using Lemma 10, we get

{2i − 3k − 5 : k + 3 ≤ i ≤ 2k + 4} ⊆ FFI(F2,k+2). (8)


FFI(F2,m) = {2i − 3m + 1 : m + 1 ≤ i ≤ 2m}. �

7. FFI(P3 × Pn)

Lemma 12. Let A be the graph shown in Fig. 4 and assume that the vertices u1, v1 and w1 are prelabeled. For c ∈ {−2, 0, 2, 4},the prelabeling can be extended to a labeling g of A such that ig(A) = c and g restricted to the subgraph induced by the verticesu2, v2, w2, u3, v3 and w3 is friendly.


Fig. 5. Graph B.

Proof. The vertices u1, v1, w1 labeled in the following distinct possible ways will establish the result.Case 1: u1 → a, v1 → a, w1 → a; a ∈ {0, 1}.

u2 v2 w2 u3 v3 w3 ig(A)

a + 1 a a a + 1 a + 1 a −2a a + 1 a + 1 a a a + 1 0a + 1 a + 1 a + 1 a a a 2a + 1 a + 1 a a a a + 1 4

Case 2: u1 → a, v1 → a, w1 → a + 1; a ∈ {0, 1}.

u2 v2 w2 u3 v3 w3 ig(A)

a a a a + 1 a + 1 a + 1 −2a + 1 a + 1 a + 1 a a a 0a a + 1 a a + 1 a + 1 a 2a + 1 a + 1 a a a + 1 a 4

Case 3: u1 → a, v1 → a + 1, w1 → a; a ∈ {0, 1}.

u2 v2 w2 u3 v3 w3 ig(A)

a a a a + 1 a + 1 a + 1 −2a + 1 a + 1 a + 1 a a a 0a a a + 1 a + 1 a a + 1 2a a a + 1 a + 1 a + 1 a 4

Case 4: u1 → a, v1 → a + 1, w1 → a + 1; a ∈ {0, 1}.

u2 v2 w2 u3 v3 w3 ig(A)

a + 1 a + 1 a + 1 a a a −2a a a a + 1 a + 1 a + 1 0a a a + 1 a + 1 a + 1 a 2a + 1 a + 1 a a a a + 1 4

Thus, g is restricted to the subgraph induced by the vertices u2, v2, w2, u3, v3 and w3 is friendly and ig(A) = c, c ∈

{−2, 0, 2, 4}. �

Lemma 13. Let B be a graph shown in Fig. 5 and assume that the vertices u1, v1 and w1 are prelabeled (and end vertices ofP3 × P2n). For c ∈ {−1, 1}, the prelabeling can be extended to a labeling g of B such that ig(B) = c and g restricted to thesubgraph induced by u2, v2, w2 is friendly.

Proof. The vertices u1, v1, w1 labeled in the following distinct possible ways would establish the result.Case 1: u1 → a, v1 → a, w1 → a; a ∈ {0, 1}.

u2 v2 w2 ig(B)a a a + 1 −1a a + 1 a 1


Fig. 6. Cartesian Product of P3 and Pn .

Case 2: u1 → a, v1 → a, w1 → a + 1; a ∈ {0, 1}.

u2 v2 w2 ig(B)a a + 1 a + 1 −1a + 1 a a 1

Case 3: u1 → a, v1 → a + 1, w1 → a; a ∈ {0, 1}.

u2 v2 w2 ig(B)a a + 1 a −1a a a + 1 1

Case 4: u1 → a, v1 → a + 1, w1 → a + 1; a ∈ {0, 1}.

u2 v2 w2 ig(B)a a a + 1 −1a + 1 a + 1 a 1

Thus, g restricted to the subgraph induced by the vertices u2, v2 and w2 is friendly and ig(A) = c, c ∈ {−1, 1}. �

Henceforth, we will refer the vertices on three copies of the path Pn (in P3 × Pn) by u1, u2, . . . , un, v1, v2, . . . , vn andw1, w2, . . . , wn respectively, as shown in Fig. 6. We shall call the cycle ui ui+1 vi+1 vi ui the i’th square and the cyclevi vi+1 wi+1 wi vi the (n − 1 + i)’th square.

Observation 5. There exists no friendly labeling f of Pm × Pn such that ef (1) = 2, wherem, n ≥ 3.

Lemma 14. For each integer n ≥ 2, there exists a friendly labeling f of P3 × P2n such that ef (1) = 3, 5, 6, 10n − 3.

Proof. Define the vertex labeling f on P3 × P2n, when k ∈ {0, 1} as:

f (ui) =

k, 1 ≤ i ≤ n + 1k + 1, n + 2 ≤ i ≤ 2n , f (vi) =

k, 1 ≤ i ≤ nk + 1, n + 1 ≤ i ≤ 2n and

f (wi) =

k, 1 ≤ i ≤ n − 1k + 1, n ≤ i ≤ 2n.

Clearly, f is a friendly labeling on P3 × P2n and ef (1) = 5.Define the vertex labeling g on P3 × P2n, when k ∈ {0, 1} as:

g(ui) =

k, i = 2jk + 1, i = 2j − 1,

g(vi) =

k, i = 2j − 1k + 1, i = 2j and

g(wi) =

k, i = 2jk + 1, i = 2j − 1

for 1 ≤ j ≤ n. Clearly, g is a friendly labeling of P3 × P2n and eg(1) = 10n − 3.ConsiderH = P2n, K = P3 in Lemma7. Since 3−2n ∈ FFI(P2n). Let hbe a friendly labeling of P2n such that ih(P2n) = 3−2n,

we get

(3 − 2n)3 − 2(2n) = 9 − 10n ∈ FFI(P3 × P2n),

i.e., there exists a friendly labeling such that ej(1) = 3.


ConsiderH = P2n, K = P3 in Lemma7. Since 5−2n ∈ FFI(P2n). Let hbe a friendly labeling of P2n such that ih(P2n) = 5−2n,we get

(5 − 2n)3 − 2(2n) = 15 − 10n ∈ FFI(P3 × P2n),

i.e., there exists a friendly labeling such that ek(1) = 6. �

Lemma 15. For each n ≥ 1, there exists a friendly labeling f of P3 × P2n+1 such that ef (1) = 4, 10n − 2, 10n + 2.

Proof. Define the vertex labeling f of P3 × P2n+1, k ∈ {0, 1} as:

f (ui) = f (vi) =

k, 1 ≤ i ≤ nk + 1, n + 1 ≤ i ≤ 2n , f (wi) =

k, 1 ≤ i ≤ n + 1k + 1, n + 2 ≤ i ≤ 2n.

Clearly, f is a friendly labeling of P3 × P2n+1 and ef (1) = 4.Define the vertex labeling g of P3 × P2n+1 as:

g(ui) =

k, i = 1, 2, 2j; 2 ≤ j ≤ nk + 1, i = 3, 2j + 1; 1 ≤ j ≤ n,

g(vi) =

k, i = 2, 2j + 1; 1 ≤ j ≤ nk + 1, i = 1, 2j; 2 ≤ j ≤ n and

g(wi) =

k, i = 1, 2j; 2 ≤ j ≤ nk + 1, i = 2, 2j + 1; 1 ≤ j ≤ n.

Clearly, g is a friendly labeling of P3 × P2n+1 and eg(1) = 10n − 2.Define the vertex labeling h of P3 × P2n+1 as:

h(ui) =

k, i = 2j; 1 ≤ j ≤ nk + 1, i = 2j − 1; 1 ≤ j ≤ n + 1,

h(vi) =

k, i = 2j − 1; 1 ≤ j ≤ n + 1k + 1, i = 2j; 1 ≤ j ≤ n and

h(wi) =

k, i = 2j; 1 ≤ j ≤ nk + 1, i = 2j − 1; 1 ≤ j ≤ n + 1.

Clearly, h is a friendly labeling of P3 × P2n+1 and eh(1) = 10n + 2. �

Lemma 16. For n ≥ 1, there exists no friendly labeling f of P3 × P2n+1 such that ef (1) = 3.

Proof. Suppose there exists a friendly labeling f of P3 × P2n+1 with ef (1) = 3. In accordance to Lemma 1, the three 1-edgesmust occur in two adjacent squares say i’th, (i+1)’th or (2n+i)’th, (2n+i+1)’th or i’th, (2n+i)’th or (i+1)’th, (2n+i+1)’thsquares. Now, we consider the following cases

Case 1: All the 1-edges are horizontal.Then these edges must lie in i’th and (2n + i)’th or (2n + i)’th and (2n + i + 1)’th squares. For this, we have

for k ∈ {0, 1} and for i ∈ 1, 2n

f (uj) = f (vj) = f (wj) =

k, j ≤ ik + 1, j > i.

Clearly, labeling f is not friendly for any value of i ∈ 1, 2n.

Case 2: One 1-edge is vertical and two 1-edges are horizontal.Due to symmetry, we may assume that the vertical edge is uivi. Now, there arises the following cases.

1. i = 1 or 2n + 1.In this case, two horizontal 1-edgesmust be v1v2, w1w2 (or v2nv2n+1, w2nw2n+1). For this, we have f (v1) = f (w1) = k+1(or f (v2n+1) = f (w2n+1) = k+1) and the remaining vertices are labeled by k, where k ∈ {0, 1}. Clearly, f is not a friendlylabeling.

2. 2 ≤ i ≤ 2n.In this case, two horizontal 1-edges must lie in (i− 1)’th and i’th squares and they will be ui−1ui and uiui+1, respectively.For this, we have f (ui) = k + 1 and the remaining vertices are labeled by k, where k ∈ {0, 1}. Clearly, f is not friendlylabeling.


Case 3: If two 1-edges are vertical and one 1-edges are horizontal.By Lemma 1, the two vertical 1-edges must lie in one of the 1’st, 2n’th, (2n+ 1)’th or 4n’th square. Due to symmetry, say

vertical 1-edges lie in the 2n’th square, then the corresponding possible vertex labeling f will be defined as:

f (v) =

k, v = u2n+1, u2nk + 1, v = u2n+1, u2n.

Clearly, f is not friendly.Case 4: If all three 1-edges are vertical.

Due to symmetry, wemay assume that the vertical edgemay lie in i’th and (i+1)’th square then clearly, (i+2)’th squarewill have one 1-edge contradicting Lemma 1. Thus, there exists no friendly labeling for this case also. �

Lemma 17. For n ≥ 1, there exists no friendly labeling f of P3 × P2n such that ef (1) = 4.

Proof. It can be done as in Lemma 16. �

Lemma 18. For n = 2, there exists a friendly labeling of P3 × Pn with ef (0) = 2 if and only if n is odd.

Proof. Lemma 1 asserts that the two 0-edges must lie in the same square. If the two 0-edges are non-adjacent, then therewill be a neighboring square with only three 1-edge. Hence the two 0-edges must be incident to a corner vertex. If we labelthis corner vertex k, then its neighbors will be labeled k as well. Since all other edges are 1-edges, the remaining verticesmust be labeled alternately with k + 1 and k. It is easy to verify that

vf (k) − vf (k + 1) =

1 if n is odd,2 if n is even.

Therefore a friendly labeling with ef (0) = 2 exists if and only if n is odd. �

We will now give the full friendly index set of P3 × Pn for some special values of n.

Example 1.

FFI(P3 × P2) = FFI(P2 × P3) = {−1, 1, 3, 7} [27].

Example 2. In case of P3 × P3, Lemma 3 implies that FFI(P3 × P3) ⊆ {2i− 12 : 2 ≤ i ≤ 12} = {−8, − 6, − 4, − 2, 0, 2,4, 6, 8, 10, 12}. By Lemma 4, 10 ∈ FFI(P3 ×P3). By Lemma 13, we have {−1, 1, 3, 7}+{−1, 1} = {−2, 0, 2, 4, 6, 8} ⊆

FFI(P3 × P3). Thus, by Observation 5, Lemmas 15, 16 and 18, we have

FFI(P3 × P3) = {2i − 12 : 4 ≤ i ≤ 12, i = 11}.

Example 3. In case of P3 ×P4, Lemma 3 implies that FFI(P3 ×P4) ⊆ {2i−17 : 2 ≤ i ≤ 17} = {−13, −11, −9, −7, −5,− 3, − 1, 1, 3, 5, 7, 9, 11, 13, 15, 17}. By Lemma 4, we have 15 ∈ FFI(P3 × P4). By Lemma 12, we have

{−1, 1, 3, 7} + {−2, 0, 2, 4} = {−3, − 1, 1, 3, 5, 7, 9, 11} ⊆ FFI(P3 × P4). Thus, by Observation 5, Lemmas 14, 17,18, we have

FFI(P3 × P4) = {2i − 17 : 3 ≤ i ≤ 17, i = 4, 15, 16}.

Example 4. In case of P3×P5, Lemma3 implies that FFI(P3×P5) ⊆ {2i−22 : 2 ≤ i ≤ 22} = {−18, −16, −14, −12, −10,− 8, − 6, − 4, − 2, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22}. By Lemma 4 we have, 20 ∈ FFI(P3 × P5). By Lemma 13we have, {−11, − 7, − 5, − 3, −1, 1, 3, 5, 7, 9, 11, 13, 17} + {−1, 1} ⊆ FFI(P3 × P5). Thus, by Observation 5,Lemmas 15, 16 and 18, we have

FFI(P3 × P5) = {2i − 22 : 4 ≤ i ≤ 22, i = 21}.

Example 5. In case of P3×P6, Lemma3 implies that FFI(P3×P6) ⊆ {2i−27 : 2 ≤ i ≤ 27} = {−23, −21, −19, −17, −15,− 13, − 11, −9, − 7, − 5, − 3, − 1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27}. By Lemma 4, we have 25 ∈

FFI(P3 ×P6). By Lemma 12, we have {−11, −7, −5, −3, −1, 1, 3, 5, 7, 9, 11, 13, 17}+{−2, 0, 2, 4} ⊆ FFI(P3 ×P6).Thus, by Observation 5, Lemmas 14, 17 and 18, we have

FFI(P3 × P6) = {2i − 27 : 3 ≤ i ≤ 27, i = 4, 25, 26}.

Theorem 6. For n > 1, FFI(P3 × P2n) = {2i − 10n + 3 : 3 ≤ i ≤ 10n − 3, i = 4, 10n − 5, 10n − 4}.


Proof. We use induction on n. The assertion is true for n = 2, 3. Assume it is true for n − 1, where n > 1.Now by Lemma 12, we have

{2i − 10(n − 1) + 3 + 2c : 3 ≤ i ≤ 10(n − 1) − 3, i = 4, 10(n − 1) − 4, c ϵ {−1, 0, 1, 2}} ⊂ FFI(P3 × P2n)or {2i − 10n + 3 : 7 ≤ i ≤ 10n − 6} ⊆ FFI(P3 × P2n).

Thus, by using Observation 5, Lemmas 14, 17 and 18, we have

FFI(P3 × P2n) = {2i − 10n + 3 : 3 ≤ i ≤ 10n − 3, i = 4, 10n − 5, 10n − 4}. �

Theorem 7. For n ≥ 1, FFI(P3 × P2n+1) = {2i − 10n − 2 : 4 ≤ i ≤ 10n + 2, i = 10n + 1}.

Proof. We use induction on n. The assertion is true for n = 1, 2. Assume it is true for n − 1, where n ≥ 1.Now by Lemma 13, we have

{2i − 10(n − 1) + 3 : 3 ≤ i ≤ 10(n − 1) − 3, i = 4, 10(n − 1) − 5, 10(n − 1) − 4} + {−1, 1}⊆ FFI(P3 × P2n+1)

= {2i − 10n − 2 : 5 ≤ i ≤ 10n − 3, 10n − 1, 10n} ⊆ FFI(P3 × P2n+1).

Thus, by using Observation 5, Lemmas 15, 16 and 18, we have

FFI(P3 × P2n+1) = {2i − 10n − 2 : 4 ≤ i ≤ 10n + 2, i = 10n + 1}. �

Remark. Clearly, FI(P3 × P2n) = {10n − 3 − 2i : i = 0, 3, 4, 5, . . . , 5n − 1}, FI(P3 × P2n+1) = {10n + 2 − 2i : i =

0, 2, 3, 4, . . . , 5n + 1}. The above results coincide with the results by Salehi and Bayot [24].

Acknowledgments

The authors express their gratitude to B.D. Acharya and M. Acharya for encouraging us to obtain more results on thefriendly labeling of graphs as well as for their constant readiness for discussions, which almost invariably yielded deeperinsight. We are also thankful to S.M. Hegde who was visiting Centre for Mathematical Sciences, Bansthali University forintensive interaction in graph labelings. Their rigorous efforts in going through the paper have helped the authors producethe paper in its present form.

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