RADIO AND RADIO ANTIPODAL LABELINGS FOR CIRCULANT GRAPHS G(4k + 2; {1, 2})

Abstract

A radio k-labeling f of a graph G is a function f from V (G) to $Z^+{\cup}\{0\}$ such that $d(x,y)+{\mid}f(x)-f(y){\mid}{\geq}k+1$ for every two distinct vertices x and y of G, where d(x, y) is the distance between any two vertices $x,y{\in}G$. The span of a radio k-labeling f is denoted by sp(f) and defined as max$\{{\mid}f(x)-f(y){\mid}:x,y{\in}V(G)\}$. The radio k-labeling is a radio labeling when k = diam(G). In other words, a radio labeling is an injective function $f:V(G){\rightarrow}Z^+{\cup}\{0\}$ such that $${\mid}f(x)=f(y){\mid}{\geq}diam(G)+1-d(x,y)$$ for any pair of vertices $x,y{\in}G$. The radio number of G denoted by rn(G), is the lowest span taken over all radio labelings of the graph. When k = diam(G) - 1, a radio k-labeling is called a radio antipodal labeling. An antipodal labeling for a graph G is a function $f:V(G){\rightarrow}\{0,1,2,{\ldots}\}$ such that $d(x,y)+{\mid}f(x)-f(y){\mid}{\geq}diam(G)$ holds for all $x,y{\in}G$. The radio antipodal number for G denoted by an(G), is the minimum span of an antipodal labeling admitted by G. In this paper, we investigate the exact value of the radio number and radio antipodal number for the circulant graphs G(4k + 2; {1, 2}).

1. Introduction

Let G be a connected graph with vertex set V(G) and edge set E(G) and let k be an integer, k ≥ 1. A radio k-labeling f of G is an assignment of non negative integers to the vertices of G such that |f(x)−f(y)| ≥ k+1−d(x, y ), where d(x, y ) denotes the distance for every two distinct vertices x and y of G. The span of the function f is max{|f(x) − f(y)| : x, y ∈ V(G)} and denoted by sp(f). The radio k-labeling number of G is the smallest span among all radio k-labelings of G. Chartrand et al. [1] was the first, who studied the radio k-labeling number for paths, where lower and upper bounds were given. These bounds have been improved by Kchikech et al. [7]. The radio k-labeling becomes a radio labeling for k = diam(G). A radio labeling is a function from the vertices of the graph to some subset of non negative integers. The task of radio labeling is to assign to each station a non negative smallest integer such that the disturbance in the nearest channel should be minimized. In 1980 [5], Hale presented this channel assignment for the very first time by relating it to the theory of graphs.

Multilevel distance labeling problem was introduced by Chartrand et al. [4] in 2001. A radio labeling is an injective function f : V(G) → Z+ ∪ {0} satisfying the condition

for any pair of vertices x, y in G. Where d(x, y ) is the distance between any distinct pair of vertices in G, which is the length of the shortest path between them. The largest number that f maps to a vertex of a graph is the span of labeling f. Radio number of G is the minimum span taken over all radio labelings of G and is denoted by rn(G). When k = diam(G) − 1, a radio k- labeling is referred to as a (radio) antipodal labeling, because only antipodal vertices can have the same label. The minimum span of an antipodal labeling is called the antipodal number, denoted by an(G). In [1] and [2], Chartrand et al. were studied the radio antipodal labeling for path and cycle. In [3], Chartrand et al. gave general bounds for the antipodal number of a graph. The exact value of the radio antipodal number of path was found in [9]. Justic and Liu have computed the radio antipodal number of cycles. In [10], by using a generalization of binary Gray codes the radio antipodal number and the radio number of the hypercube are determined.

An undirected circulant graph denoted by G(n; ±{1, 2, ..., j}) where is defined as a graph with vertex set V = {0, 1, 2, ...n − 1} and an edge set E = {(i, j) : |j − i| ≡ s (mod n), s ∈ {1, 2, ..., j}}. For the sake of simplicity, take the vertex set as {v1, v2, ...vn} in clockwise order.

Remark 1.1. The diameter of class of circulant graphs which are going to be discussedin this paper is:

In this paper, radio and radio antipodal numbers for the class of circulant graphs G(4k + 2 : {1, 2}) are computed.

 

2. Main results

The main theorems of this paper are:

Theorem 2.1. The radio number of the circulant graphs G(4k + 2 : {1, 2}) is given by

Theorem 2.2. The radio antipodal number of the circulant graphs G(4k + 2; {1, 2}) is given by

 

3. Radio number for G(4k + 2; {1, 2})

In this section, we prove the Theorem 1 in two steps. First we provide a lower bound for rn(G(4k + 2; {1, 2})) then define a multilevel distance labeling of (G(4k + 2; {1, 2})) with span equal to the lower bound, thus determining the radio number of (G(4k + 2; {1, 2})).

3.1. Lower bound for G(4k + 2; {1, 2}). The lower bound for the radio number of G(4k + 2; {1, 2}) is determined in following way. First examine the maximum possible sum of the pairwise distance between any three vertices of (G(4k + 2; {1, 2})) and use this maximum sum to compute a minimum possible gap between the ith and (i + 2)nd largest label. Then provides a lower bound for the span of any labeling by using 0 for the smallest label and considering the size of gap into account.

Lemma 3.1. For each vertex on the graph G(4k + 2; {1, 2}) there is exactly one vertex at a distance diameter d, of the graph G.

Proof. We show that d(v1, v2k+2) = k + 1 = d. The path from v1 to v2k+2 is of length k + 1 as v1→ v2(1)+1 → v2(2)+1 → ... → v2(k)+1 → v2(k)+1+1. □

The following Lemma provides a maximum possible sum of the pairwise distances between any three vertices of G(4k + 2; {1, 2}).

Lemma 3.2. For any three vertices u, v, w on the graphs G(4k + 2; {1, 2}),

Proof. By Lemma 3.1, d(v1, v2k+2) = k + 1 = d. Case(i): For odd k.

and a path of length between v2k+2 and v3k+3 is and as

This implies that

Case (ii): For even k.

and a path of length and 1 between v2k+2 and v3k+3 is Also, because Thus, Therefore, for any three vertices u, v, w on the graphs G(4k + 2; {1, 2}),

The minimum distance between every other label (arranged in increasing order) in a multi-level distance labeling (or radio labeling) of G(4k + 2; {1, 2}) is determined by using this maximum possible sum of the pairwise distances between any three vertices of G(4k + 2; {1, 2}) together with the radio condition.

Lemma 3.3.Let f be radio labeling for V (G(4k + 2; {1, 2})), where {xi : 1 ≤ i ≤ 4k + 2} be the ordering of V (G(4k + 2; {1, 2})) such that f (xi) < f (xi+1 ) for all 1 ≤ i ≤ 4k + 1, then

Proof. By definition,

Summing these inequalities yields

Furthermore, by Lemma 4, d(u, v) + d(v, w) + d(w, u) ≤ 2d, so we have

As d = diam(G(4k + 2; {1, 2})) = k + 1, it follows that

Thus

The above Lemma makes it possible to calculate the minimum possible span of a radio labeling of G(4k + 2; {1, 2}).

Theorem 3.4. The radio number of the circulant graphs G(4k + 2; {1, 2}) satisfies

Proof. Let f be a distance labeling for G(4k+2; {1, 2}) and {x1, x2, x3, ..., x4k+2} be the ordering of vertices of G(4k + 2; {1, 2}), such that f (xi) < f (xi+1) defined by f (x1) = 0 and, set di = d(xi, xi+1) and fi = f (xi+1) − f (xi). Then fi ≥ d+1−di for all i. By Lemma 5, the span of a distance labeling for G(4k+2; {1, 2}) is

Thus,

FIGURE 1.Radio labeling and ordinary labeling of G(6; {1, 2})

3.2. Upper bound for rnG(4k + 2; {1, 2}). To complete the proof of Theorem 1, we find upper bound and show that this upper bound is equal to the lower bound for G(4k + 2; {1, 2}). The labeling is generated by three sequences, the distance gap sequence

the color gap sequence

and the vertex gap sequence T

For odd k. The distance gap sequence is given by:

The color gap sequence F is given by:

For even k. The distance gap sequence is given by:

The color gap sequence F is given by:

The vertex gap sequence for all values of k is:

Where ti denotes number of vertices between xi and xi+1.

Let π : {1, 2, 3, ..., 4k + 2} → {1, 2, 3, ..., 4k + 2} be defined by π(1) = 1 and

Let xi = u π(i) for i = 1, 2, 3, ..., 4k + 2. Then x1, x2, x3, ..., x4k+2 is an ordering of the vertices of G, assuming f (x1) = 0, f (xi+1) = f (xi) + fi. Then for i = 1, 2, 3, ..., 2k + 2,

and for i = 1, 2, ...2k + 2,

We will show that each of the sequences given above, the corresponding π are permutations. For odd k, g.c.d.(4k + 2, k) = 1 and 3k + 2 ≡ −k (mod 4k + 2) implies that (3k + 2)(i − i ′) ≡ k(i ′ − i) ≢ 0 (mod 4k + 2). Because if it does so then k(i ′ − i) ≡ k.0 (mod 4k + 2) and i ′ − i ≡ 0 (mod 4k + 2) which is impossible when Therefore π(2i − 1) ≠ π(2i ′ − 1), if i ≠ i ′. Similarly π(2i) ≠ π(2i ′), if i ≠ i ′. If π(2i) = π(2i ′ − 1), then we get

As k is odd and g.c.d.(4k + 2, k) = 1 it follows that i ′ − i ≡ 0 (mod 4k + 2). This implies that 4k + 2 divides i ′ − i < 2k + 1, which is not possible.

When k is odd, then span of f is equal to:

For even k, g.c.d.(4k + 2, k) = 2 and 3k + 2 ≡ −k (mod 4k + 2) implies that (3k + 2)(i − i ′ ) ≡ k(i ′ − i) ≢ 0 (mod 4k + 2). Because if it does so then k(i ′ − i) ≡ k.0 (mod 4k + 2) and which is impossible when Therefore π(2i − 1) ≠ π(2i ′ − 1), if i ≠ i ′. Similarly π(2i) ≠ π(2i ′), if i ≠ i ′. If π(2i) = π(2i ′ − 1), then

As k is even and g.c.d.(4k + 2, k) = 2 it follows that Which is not possible.

When k is even, then span of f is equal to:

FIGURE 2.Radio labeling and ordinary labeling of G(10; {1, 2})

 

4. Radio antipodal number for G(4k + 2; {1, 2})

In this section, the lower and upper bound for the radio antipodal number are determined and have shown that these bounds are equal.

4.1. Lower bound for an(G(4k + 2; {1, 2}). The technique for finding the lower bound for an(G(4k + 2; {1, 2}) is analogous to that of rn(G(4k + 2; {1, 2}).

Lemma 4.1. Let f be radio antipodal labeling for V (G(4k + 2; {1, 2})), where {xi : 1 ≤ i ≤ 4k + 2} be the ordering of V (G(4k + 2; {1, 2})) such that f (xi) ≤ f (xi+1 ) for all 1 ≤ i ≤ 4k + 1, then

Proof. By definition, f (xi+1) − f (xi) ≥ d − d(xi+1, xi), f (xi+2) − f (xi+1) ≥ d − d(xi+2, xi+1) and f (xi+2) − f (xi) ≥ d − d(xi+2, xi). Summing up these three in-equalities and by Lemma 4, we get

Thus,

Theorem 4.2. The radio antipodal number of the circulant graphs G(4k + 2; {1, 2}) is given by

Proof. Let f be a distance labeling for G(4k+2; {1, 2}) and {x1, x2, x3, ..., x4k+2} be the ordering of vertices of G(4k + 2; {1, 2}), such that f (xi) ≤ f (xi+1) defined by f (x1) = 0 and, set di = d(xi, xi+1) and fi = f (xi+1)−f (xi). Then fi ≥ d−di for all i. By Lemma 7, the span of a distance labeling for G(4k + 2; {1, 2}) is

Thus,

FIGURE 3.Radio antipodal labeling and ordinary labeling of G(6; {1, 2})

4.2. Upper bound for an(G(4k+2; {1, 2}). To complete the proof of Theorem 2, we find upper bound and show that this upper bound is same as the lower bound for an(G(4k + 2; {1, 2})). The technique for an upper bound of an(G(4k + 2; {1, 2}) is analogous to that of rn(G(4k + 2; {1, 2}), with replacing the color gap sequence.

For odd k. The color gap sequence F is given by:

For even k. The color gap sequence F is given by:

When k is odd, then span of f is equal to:

When k is even, then span of f is equal to:

FIGURE 4.Radio antipodal labeling and ordinary labeling of G(10; {1, 2})