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})
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