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ARITHMETIC LABELINGS AND GEOMETRIC LABELINGS OF COUNTABLE GRAPHS

Gurusamy Rengasamy Vijayakumar School of Mathematics

Tata Institute of Fundamental Research Homi Bhabha Road, Colaba, Mumbai 400 005, India

e-mail: vijay@math.tifr.res.in

Abstract

An injective map from the vertex set of a graph G—its order may not be finite—to the set of all natural numbers is called an arithmetic (a geometric) labeling of G if the map from the edge set which assigns to each edge the sum (product) of the numbers assigned to its ends by the former map, is injective and the range of the latter map forms an arithmetic (a geometric) progression. A graph is called arithmetic (geometric) if it admits an arithmetic (a geometric) labeling. In this article, we show that the two notions just mentioned are equivalent—

i.e., a graph is arithmetic if and only if it is geometric.

Keywords: arithmetic labeling of a graph, geometric labeling of a graph.

2010 Mathematics Subject Classification: 05C78, 05C63.

All graphs considered in this article are countable and simple. The set of all positive integers is denoted by N and N = {0} ∪ N. For basic information about graph theory we rely on [7]. Let G = (V, E) be a graph and f be any map from V to N; we associate with f two maps from E to N, denoted by f+and f×: for all uv ∈ E, f+(uv) = f (u) + f (v) and f×(uv) = f (u) × f (v).

If f and f+ are injective and the elements of f+(E) form an arithmetic progression—i.e., if this set can be written as {k+nd : n ∈ N and n < |E|}—

then f is called an arithmetic labeling of G. (For information about this labeling for finite graphs, see [1, 2].) If f and f× are injective and the elements of f×(E) form a geometric progression—i.e., if this set can be

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written as {arn : n ∈ N and n < |E|} where r may not be an integer—

then f is called a geometric labeling of G. (For details about this labeling for finite graphs, we refer the reader to [4, 5]; in this connection, see [3]

also.) If a graph admits an arithmetic (a geometric) labeling then it is called arithmetic (geometric). If σ is an arithmetic labeling of a graph G, then the map: v 7→ 2σ(v) where v ∈ V (G) is a geometric labeling of G.

Thus, if a graph is arithmetic then it is geometric also. The objective of this article is to prove the converse; to this end, we begin with a simple result.

Lemma 1. Letf be a geometric labeling of a graph G = (V, E) and f×(E) = {arn : n ∈ N and n < |E|}. Let H be a component of G and p, q be two vertices ofH. If the distance between p and q is odd, then for some n ∈ Z, f (p)f (q) = arn; otherwise for some n ∈ Z, f (q) = f (p)rn. If H is not a bipartite graph, then for each x ∈ V (H), there is some n ∈ Z such that (f (x))2 = arn.

P roof. If x ∈ N (p), then for some n ∈ Z, f (p)f (x) = arn; if w is a vertex of H such that d(w, p) = 2, then it is easy to verify that f (w) = f (p)rnfor some n ∈ Z. Thus continuing, we obtain the following: If d(p, q) is odd, then for some n ∈ Z, f (p)f (q) = arn; otherwise for some n ∈ Z, f (q) = f (p)rn.

Now suppose that H is not bipartite. Then either A := {x ∈ V (H) : d(x, p) is odd} or V (H) \ A is not an independent set of H; therefore either in A or in V (H) \ A, there exists a pair of adjacent vertices u, v. By the hypothesis, for some k ∈ Z, f (u)f (v) = ark. If u, v ∈ A, then by what has been derived above, for some m, n ∈ Z, f (p)f (u) = armand f (p)f (v) = arn whence by the preceding three equalities, it follows that (f (p))2 = ar` for some ` ∈ Z; if u, v ∈ (V (H) \ A) then for some m, n ∈ Z, f (u) = f (p)rm and f (v) = f (p)rn whence for some ` ∈ Z, (f (p))2 = ar`. Since p is arbitrary, the second part of the conclusion also holds.

Proposition 2. Let G = (V, E) be a graph such that the number of its nontrivial components which are bipartite is finite. If G is geometric, then it is arithmetic.

P roof. We can assume that G has no isolated vertices. Suppose that f is a geometric labeling of G. Let U be the union of the vertex sets of all non- bipartite components and (Ai, Bi), i = 1, 2, . . . , n − 1 be the bipartitions of the remaining components. (see Definition 1.2.17 of [7].) Let f×(E) = {ark: k ∈ N and k < |E|}. For each i ∈ {1, 2, . . . , n − 1}, choose a vertex xi ∈ Ai.

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Let v be a vertex in U . By Lemma 1, for some k ∈ Z, (f (v))2 = ark; therefore 2 log f (v) − log a is an integer, viz., k. (In this proof, ‘r’ is the base for every logarithm.) For any i ∈ {1, 2, . . . , n − 1}, let x and y be vertices in Ai and Bi, respectively. Then by Lemma 1, for some j, k ∈ Z, f (x) = f (xi)rj and f (y)f (xi) = ark. Therefore [log f (x)−log f (xi)] = j ∈ Z and [log f (y) + log f (xi) − log a] = k ∈ Z. Let

µ = dmax{2n log a + n, 2n log f (xi) : i = 1, . . . , n − 1}e.

Define a map g : V 7→ Z as follows.

For all x ∈ U , g(x) = n[2 log f (x) − log a] + µ; if i ∈ {1, 2, . . . , n − 1}, then for all x ∈ Ai, g(x) = 2n[log f (x) − log f (xi)] + i + µ and for all x ∈ Bi, g(x) = 2n[log f (x) + log f (xi) − log a] − i + µ.

It is easy to verify that if x ∈ W and y ∈ V \ W where W ∈ {U, Ai, Bi : i = 1, . . . , n − 1}, then g(x) 6≡ g(y) (mod 2n); using this fact, we find that g is injective. By the choice of µ, clearly for each v ∈ V (G), g(v) > 0.

For each e ∈ E, it can be verified that g+(e) = 2n log f×(e) + 2µ − 2n log a.

Therefore g+ is injective and the elements of g+(E) form an arithmetic progression. Thus it follows that g is an arithmetic labeling.

Corollary 3. If a graph is finite and geometric, then it is arithmetic.

For another proof of Corollary 3, see [6]. It is easy to find the graphs—

finite and infinite—which are arithmetic; e.g., it can be verified that every finite path, the infinite path with one leaf and every complete graph whose order is less than 5 are arithmetic and it can be shown that every complete graph with at least 5 vertices—it can be infinite—is not arithmetic. (For more examples of finite arithmetic graphs, see [1, 5].) The next result which yields a class of infinite graphs which are arithmetic, plays a decisive role in settling the main result, viz., Theorem 5.

Theorem 4. LetG = (V, E) be a graph such that its order is countable and the number of its nontrivial components which are bipartite is infinite. Then G is arithmetic.

P roof. First consider the possibility that the number of components of order 2 is infinite. Let H be the union of all other components. Let f be an injective map from V (H) to {2n+3 : n ∈ N}. It can be verified that f+ is injective and J := {n ∈ N : 8n /∈ f+(E(H))} is infinite. We can take

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{Hj : j ∈ J} as the set of all components which are copies of K2. Let f be extended to a map g : V 7→ N so that for each j ∈ J, g(V (Hj)) = {4j − 1, 4j + 1}. Then g and g+ are injective and g+(E) = {8n : n ∈ N}; i.e., g is an arithmetic labeling of G.

So, let us assume that the number of components which are of order 2 is finite. Let V = {v1, v2, v3, . . .}. Let Z be the union of the vertex sets of all components each of which is either non-bipartite or K2 or K1. Inductively we construct for each n ∈ N, an induced subgraph Gn and a map fn : V (Gn) 7→ N such that the conditions (a), (b), . . . , (f) and (g) given below are fulfilled.

(a) For each x ∈ V (Gn), there exists an integer k ∈ {0, 1, 3, 5, 7} such that fn(x) ≡ k. (8 is the modulus of every congruence occurred in this proof.) (b) Xn∩ Z = ∅ = Yn∩ Z and Zn⊂ Z where Xn = {x ∈ V (Gn) : fn(x) ≡ 1 or 5}, Yn = {x ∈ V (Gn) : fn(x) ≡ 3 or 7} and Zn = {x ∈ V (Gn) : fn(x) ≡ 0}.

(c) fn is injective and fn+(E(Gn)) = {8k : k = 1, . . . , |E(Gn)|}.

(d) If n > 1, then V (Gn−1) ⊆ V (Gn) and fn is an extension of fn−1. (e) There exist a vertex αn ∈ Xn and a vertex βn ∈ Yn, such that they

are adjacent and for each x ∈ Xn, fn(x) 6 fnn), for each x ∈ Yn, fn(x) 6 fnn) and for each x ∈ Zn, fnn) > fn(x) < fnn).

(f) {v1, v2, . . . , vn} ⊆ V (Gn).

(g) If (A, B) is the bipartition of a component of G − Z and x, y belong to V (Gm) ∩ (A ∪ B), then the following hold.

x, y ∈ A or x, y ∈ B ⇒ fn(x) ≡ fn(y) and x ∈ A, y ∈ B ⇒ fn(x) + fn(y) ≡ 0.

We can assume that v1, v2, v3 ∈ Z and v/ 1 is adjacent to both v2 and v3. Let G1 = G[v1, v2, v3]. Set f1(v1) = 1, f1(v2) = 7 and f1(v3) = 15. Let α1 = v1 and β1 = v3. Then (a), . . . , (f) and (g) are satisfied for n = 1. Now suppose for some m ∈ N, there exists an induced subgraph Gm and a map fm : V (Gm) 7→ N satisfying the conditions listed above. Let us construct the subgraph Gm+1, the map fm+1 and set the vertices αm+1, βm+1as described below.

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Case 1. vm+1∈ V (Gm).

Let Gm+1 = Gm, fm+1 = fm, αm+1= αm and βm+1= βm. Case 2. vm+1∈ (Z \ V (Gm)).

Let J = {j ∈ N : vj ∈ V (Gm)∩N (vm+1)}. (J may be empty.) Let L = {1, 2, . . . , 2` − 1, 2`} \ {18fm(vj) : j ∈ J} where ` = 18fmmβm). For each k ∈ L, choose a distinct component Hk of G − Z so that V (Hk) ∩ V (Gm) = ∅ and let xk, yk be two adjacent vertices of Hk. Let V (Gm+1) = V (Gm) ∪ {vm+1, xj, yj : j ∈ L}. For each x ∈ V (Gm), let fm+1(x) = fm(x). Set fm+1(vm+1) = fm+mβm). For each k ∈ L, set fm+1(xk) = fmm) + 4k and fm+1(yk) = fmm) + 4k. Note that Xm+1 = Xm ∪ {xk : k ∈ L}, Ym+1 = Ym∪ {yk : k ∈ L} and Zm+1 = Zm∪ {vm+1}. Let αm+1 = x2` and βm+1 = y2`. It can be verified that |E(Gm+1) − E(Gm)| = |J| + |L| = 2`

and {fm+1+ (e) : e ∈ E(Gm+1) − E(Gm)} = {fm+1+ (vkvm+1) : k ∈ J} ∪ {fm+1+ (xjyj) : j ∈ L} = {fm+mβm) + 8k : 1 6 k 6 2`}.

Case 3. vm+1∈ (Z ∪ V (G/ m)).

Let (S, T ) be the bipartition of the component which contains vm+1. By (g), we can assume that Xm∩ T = Ym∩ S = ∅. Assume that vm+1 ∈ S.

Let J = {j ∈ N : vj ∈ V (Gm) ∩ N (vm+1)}. (J may be empty.) Let the number ρ be chosen in {0, 4} as follows: If V (Gm) ∩ (S ∪ T ) = ∅, ρ = 0.

Otherwise, let u ∈ V (Gm) ∩ (S ∪ T ); when u ∈ S, ρ + fmm) ≡ f (u); when u ∈ T , ρ + fmm) + f (u) ≡ 0. (Note that by (g), for all x ∈ S ∩ V (Gm), ρ+fmm) ≡ fm(x) and for all x ∈ T ∩V (Gm), ρ+fmm)+fm(x) ≡ 0.) Let

` = 14[fm+mβm)+ρ]−1. Note that for each j ∈ J, fmm)+fm(vj)+ρ < 8`.

Let L = {1, 2, . . . , `} \ {18[fmm) + fm(vj) + ρ] : j ∈ J}. For each k ∈ L, choose a distinct component Hk of G − Z so that V (Hk) ∩ V (Gm) = ∅ and let xk, yk be two adjacent vertices in Hk. Let y` be chosen so that deg y` >2. Let x`+1 be a neighbour of y`, other than x`. Let V (Gm+1) = V (Gm)∪{vm+1, x`+1}∪{xk, yk: k ∈ L}. For each x ∈ V (Gm), let fm+1(x) = fm(x). Set fm+1(vm+1) = fm+mβm) + fmm) + ρ. For each k ∈ L, set fm+1(xk) = fmm) + 4k and fm+1(yk) = fmm) + 4k. Set fm+1(x`+1) = fm+1(x`) + 8. Note that Xm+1 = Xm ∪ {xk : k ∈ L} ∪ {x`+1, vm+1}, Ym+1 = Ym ∪ {yk : k ∈ L} and Zm+1 = Zm. Let αm+1 = x`+1 and βm+1 = y`. We find that |E(Gm+1) − E(Gm)| = |J| + |L| + 1 = ` + 1 and {fm+1+ (e) : e ∈ E(Gm+1) − E(Gm)} = {fm+1+ (vjvm+1) : j ∈ J} ∪ {fm+1+ (xkyk) : k ∈ L} ∪ {fm+1+ (y`x`+1)} = {fm+mβm) + 8k : 1 6 k 6 ` + 1}.

In each of the three cases, it can be routinely verified that the conditions (a), . . . , (f) and (g) hold when n = m + 1. Since ∪n=1V (Gn) = V (G) and

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for each n ∈ N, fn+1 is the extension of fn, we have an injective map f : V 7→ N such that for each n ∈ N, fn is the restriction of f to V (Gn).

Since f+(E) = {8k : k ∈ N}, f is an arithmetic labeling of G.

Combining Proposition 2 and Theorem 4, we find that if a graph is geo- metric, then it is arithmetic. We have already noted that every arithmetic graph is geometric. Thus we obtain the following.

Theorem 5. A graph is arithmetic if and only if it is geometric.

Acknowledgement

The author expresses his gratitude to the referees for pointing out some errors and for suggesting some modifications.

References

[1] B.D. Acharya and S.M. Hegde, Arithmetic graphs, J. Graph Theory 14 (1990) 275–299.

[2] B.D. Acharya and S.M. Hegde, On certain vertex valuations of a graph, Indian J. Pure and Appl. Math. 22 (1991) 553–560.

[3] L.W. Beineke and S.M. Hegde, Strongly multiplicative graphs, Discuss. Math.

Graph Theory 21 (2001) 63–75.

[4] S.M. Hegde, On multiplicative labelings of a graph, J. Combin. Math. and Com- bin. Comp. 65 (2008) 181–195.

[5] S.M. Hegde and P. Shankaran, Geometric labeled graphs, AKCE International J. Graphs and Combin. 5 (2008) 83–97.

[6] G.R. Vijayakumar, Arithmetic labelings and geometric labelings of finite graphs, J. Combin. Math. and Combin. Comp. (to be published).

[7] D.B. West, Introduction to Graph Theory, Second edition (Printice Hall, New Jersey, USA, 2001).

Received 14 February 2009 Revised 19 October 2009 Accepted 20 October 2009

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