ON SD-PRIME CORDIAL GRAPHS

Let G = (V (G), E(G)) be a simple, finite and undirected graph of order n. Given a bijection f : V (G) → {1, . . . , n}, we associate 2 integers S = f(u) + f(v) and D = |f(u) − f(v)| with every edge uv in E(G). The labeling f induces an edge labeling f ′ : E(G) → {0, 1} such that for any edge uv in E(G), f ′(uv) = 1 if gcd(S,D) = 1, and f ′(uv) = 0 otherwise. Let ef ′(i) be the number of edges labeled with i ∈ {0, 1}. We say f is an SD-prime cordial labeling if |ef ′(0) − ef ′(1)| ≤ 1. Moreover G is SD-prime cordial if it admits an SD-prime cordial labeling. In this paper, we investigate the SD-prime cordiality of some standard graphs. AMS Subject Classification: 05C78, 05C25


Introduction
Let G = (V (G), E(G)) (or G = (V, E) for short if not ambiguous) be a simple, finite and undirected graph of order |V | = n and size |E| = m.All notation not defined in this paper can be found in [1].
In 1967, Rosa introduced the first paper on graph labeling.Since then, there have been more than 1500 research papers on graph labeling (see the dynamic survey by Gallian [5]).
In [10,11], the authors introduced the concept of prime graphs and prime cordial graphs.Definition 1.1.A bijection f : V → {1, . . ., n} induces an edge labeling f ′ : E → {0, 1} such that for any edge uv in G, f ′ (uv) = 1 if gcd(f (u), f (v)) = 1, and f ′ (uv) = 0 otherwise.Such a labeling is called a prime labeling if f ′ (uv) = 1 for all uv ∈ E. We say G is a prime graph if it admits a prime labeling.
Several results on prime and prime cordial graphs can be found in [2,3,4,8,9].In [6], Lau and Shiu introduced a variant of prime graph labeling which is defined as follows.
Lau and Shiu then proved the following necessary and sufficient condition for the existence of an SD-prime labeling.
Theorem 1.1.A graph G of order n is SD-prime if and only if G is bipartite and there exists a labeling f : V → {1, 2, . . ., n} such that for each edge uv of G, f (u) and f (v) are of different parity and gcd(f In this paper, we introduce the concept of an SD-prime cordial labeling defined as follows. Definition 1.4.A bijection f : V → {1, . . ., n} induces an edge labeling f ′ : E → {0, 1} such that for any edge uv in G, f ′ (uv) = 1 if gcd(S, D) = 1, and f ′ (uv) = 0 otherwise.The labeling f is called an SD-prime cordial labeling We say that G is SD-prime cordial if it admits an SD-prime cordial labeling.
We shall drop the subscript f ′ if the context is clear.

Main Results
In [7], the authors show that every bipartite graph is an induced subgraph of an SD-prime graph.Similarly, we have Theorem 2.1.Every (complete) bipartite graph G is an induced subgraph of an SD-prime cordial graph.
Proof.Let the bipartition of G be (A, B) such that |A| = r and |B| = s.Label the vertices in A by integers in X = {1, 3, . . ., 2r − 1}, and the vertices in B by integers in Y = {2 1 , 2 2 , . . ., 2 s }.Now, all the edges of G are labeled 1.Let v ≥ max{2r − 1, 2 s } be sufficiently large.We add v − r − s > 0 new vertices and label them using the integers in {1, 2, . . ., v}\(X ∪ Y ).We can now obtain the required 0-edges by joining new vertices with odd labels to vertices in A or joining new vertices with even labels to vertices in B. Lemma 2.2.Let G be a graph of size m.Consider the conditions (i) f (u) and f (v) are of distinct parity and that gcd(f (u), f (v)) = 1; and (ii) f (u) and f (v) are of same parity.If, for each possible labeling f , the number of (f (u), f (v)) pairs that meets one of the conditions is greater than ⌈ m 2 ⌉, or is less than ⌊ m 2 ⌋, then G is not SD-prime.
Proof.A labeling f that meets condition (i) will have too many or too few 1-edges while one that meets condition (ii) will have too many or too few 0-edges.Hence, f is not an SD-prime cordial labeling.
In the next theorem, we show that all paths and cycles are SD-prime cordial.The approach used turns out to be very useful in getting an SD-prime cordial labeling for the results that follow.We use the notations a cycle in the following constructions using paths and cycles.
We now consider C n .If n ≡ 0, 1, 3 (mod 4), we label the vertices in a way similar to that for P n .Suppose n ≡ 2 (mod 4).For n = 6, label the vertices by 1, 4, 5, 3, 6, 2 consecutively.For n ≥ 10, first define f as in Case (2) above.Note that the labeling gives e(1) − e(0) = 2. Now, switch the vertex label of 3 and 7.It is easy to check that e(1) = e(0) and C n is SD-prime cordial.Definition 2.1.A fan graph of order n ≥ 3, denoted F n , is a graph obtained from a path P n by joining u 1 to u i (3 ≤ i ≤ n).
Theorem 2.8.The fan graph F n is SD-prime cordial for all n ≥ 4.
Proof.By the labeling approach as in the proof of Theorem 2.7, it is clear that F n is SD-prime cordial.Definition 2.2.A double fan graph of order n ≥ 4, denoted DF n , is a graph obtained from a fan graph F n by deleting edge u 1 u 2 and joining u 2 to u i (4 Theorem 2.9.The double fan graph DF n is SD-prime cordial for all n ≥ 4. ); and f (u i ) = i for i ≡ 2, 3 (mod 4).
For n ≡ 0 (mod 4), define f (u i ) as above if i ≤ n − 1 and f (u n ) = n.Clearly, the labeling is SD-prime cordial.
We have joined the two vertices with labels k and k +1, and the two vertices with labels k − 1 and k + 2.An additional six 0-edges and six 1-edges have been added.
Construction C2, to be used when k ≡ 1 (mod 3): We have joined the two vertices with labels k − 1 and k + 1, and the two vertices with labels k and k + 2.An additional six 0-edges and six 1-edges have been added.
Note that the two highest vertex labels are again on the rightmost edge.For a general n, use one of the four simple cases, according to the value of n (mod 4), and use constructions C1 or C2 until P n × P 2 has been attained.Lemma 2.12.The grid P n × P 3 is SD-prime cordial, for all n.
Proof.Assume each copy of P 3 is a vertical path.We first consider the two simplest cases.n = 1: Label consecutively with 1, 3 and 2. n = 2: Label the left P 3 and right P 3 from top to bottom by 1, 3, 2 and 4, 6, 5 respectively.
At this point, we have shown that P n × P 3 is SD-prime cordial, for n = 1 and 2. Furthermore, note that in each of the above labelings, the highest three vertex labels are on the rightmost edge.In the labeling for n = 1, these vertex labels are ≡ 1, 3, and 2 (mod 6), with the highest vertex label in the middle.In the labeling for n = 2, these vertex labels are ≡ 4, 0, and 5 (mod 6), again with the highest vertex label in the middle.
We will continue to add vertices and edges to one of the above grids.At any point, assume that the highest vertex label is k (which is necessarily in the middle of the rightmost edge, and adjacent to the vertices with labels k − 2 and k − 1).Consider adding six vertices, with the vertex labels shown below, to the right.Join the two vertices with labels k − 2 and k + 1, the two vertices with labels k and k + 2, and the two vertices with labels k − 1 and k + 3. Note that the three highest vertex labels are still on the rightmost edge, with the highest vertex label in the middle to get the three rightmost P 3 with labelings as given: For a general n, use one of the simplest cases, according to the value of n (mod 2), and repeatedly add the block shown above until P n × P 2 has been attained.It can be readily verified that, after each step, an additional five 0edges and five 1-edges have been added.Furthermore, in the case when n ≡ 1 (mod 2), the rightmost vertices are ≡ 2, 3, and 1 (mod 6), and in the case when n ≡ 0 (mod 2), the rightmost vertices are ≡ 5, 0, and 4 (mod 6).
Proof.View the m copies of P n as m horizontal paths, with the paths stacked one above another.Label the vertices from left to right, top to bottom.In general, the label of the j-th vertex of the i-th P n is (i − 1)n + j.Clearly, horizontally adjacent labels have opposite parity and are relatively prime, while vertically adjacent labels have the same parity.Thus all horizontal edges have label 1 and all vertical edges have label 0. Among the 2mn − m − n edges, m(n − 1) of them are horizontal and n(m − 1) of them are vertical.Thus there are (n − m) more 1-edges than 0-edges.Since |n − m| ≤ 1, the proof is complete.
Proof.Begin with the vertex labeling as in Lemma 2.13.Since e(1)−e(0) = n − m ≥ 2, we know the labeling is non-SD-prime cordial.Divide each copy of P n into blocks of four vertices, and consider the following modifications to the vertex labeling.
Modification M 1. Changing the top three copies of P n to get: Each of the top three copies of P n has exactly one 0-edge.We claim that the two leftmost edges joining the third and fourth copies of P n have label 1.Consider the leftmost edge with end-vertex labels 2n+2 and 3n+1, and assume that there is a prime p (≥ 3) dividing these two integers.Since their difference is n − 1, which is prime, so p = n − 1.It is obvious that (n − 1) does not divide (2n + 2), a contradiction.Consider the second leftmost edge with end-vertex labels 2n + 1 and 3n + 2, and assume that there is a prime p (≥ 3) dividing these two integers.Since their difference is n + 1, which is prime, so p = n + 1.
It is obvious that (n + 1) does not divide (2n + 1), a contradiction.All the other edge labels are unchanged.Thus, upon applying modification M 1, there are (n − m − 2) more 1-edges than 0-edges.
Modification M 2. Changing the top two copies of P n to get: Each of the top two copies of P n has exactly two 0-edges.We claim that the second and third edges (from the left) joining the second and third copies of P n have label 1.Consider the second edge with end-vertex labels n + 3 and 2n + 2, and assume that there is a prime p (≥ 3) dividing the two vertically adjacent vertices.Since their difference is n − 1, which is prime, so p = n − 1.It is obvious that (n − 1) does not divide (n + 3), a contradiction.Consider the third edge with end-vertex labels n + 2 and 2n + 3, and assume that there is a prime p (≥ 3) dividing the two vertically adjacent vertices.Since their difference is n + 1, which is prime, so p = n + 1.It is obvious that (n + 1) does not divide (n + 2), a contradiction.All the other edge labels are unchanged.Thus, upon applying modification M 2, there are (n − m − 4) more 1-edges than 0-edges.
Modification M 3 i , 1 ≤ i ≤ (n/4) − 1. Changing the top two copies of P n to get: The argument is similar to that in modification M 2. Thus, upon applying modification M 3 i , the difference e(1) − (0) is reduced by 4.
Proof.Label the vertices of the subgrid P n × P m with the original vertex labeling as in Lemma 2.13.
Proof.Label the vertices of the subgrid P n × P m with the original vertex labeling as in Lemma 2.13.
We have in Lemmas 2.14 to 2.17 shown that P n+k × P m is SD-prime cordial for k = 0, 1, 2, 3 and n ≥ m + 2 ≥ 6 if n ≡ 0 (mod 4) and n − 1, n + 1 are primes.We end the paper with the following conjecture: Conjecture 2.1.The grid P n × P m is SD-prime cordial for all n ≥ m ≥ 2.