1. Introduction
A graph labeling is an assignment of integers to the vertices or edges or both subject to some conditions. Labeled graphs are becoming an increasingly useful family of Mathematical Models from a broad range of applications. The graph labeling problem has a fast development recently. This problem was first introduced by Alex Rosa in 1967. Since Rosa’s article, many different types of graph labeling problems have been defined around this. This is not only due to its mathematical importance but also because of the wide range of the applications arising from this area, for instance, x-rays, crystallography, coding theory, radar, astronomy, circuit design, and design of good radar type codes, missile guidance codes and convolution codes with optimal autocorrelation properties and communication design [2]. All graphs considered here are finite simple and undirected. The number of vertices of a graph G is called order of G, and the number of edges is called size of G. Let G1 and G2 be two graphs with vertex sets V1 and V2 and edge sets E1 and E2 respectively. Then their join G1 + G2 is the graph whose vertex set is V1 ∪ V2 and edge set is E1 ∪ E2 ∪ {uv : u ∈ V1 and v ∈ V2}. Let G1, G2 respectively be (p1, q1), (p2, q2) graphs. The corona of G1 with G2, is the graph G1 ⊙ G2 obtained by taking one copy of G1 and p1 copies of G2 and joining the ith vertex of G1 with an edge to every vertex in the ith copy of G2. In 1987, Cahit introduced the concept of cordial labeling of graphs [1]. Sundaram, Ponraj, Somasundaram [7] have introduced the notion of prime cordial labeling. A prime cordial labeling of a graph G with vertex set V is a bijection f : V → {1, 2, . . . , |V|} such that if each edge uv is assigned the label 1 if gcd (f(u), f(v)) = 1 and 0 if gcd (f(u), f(v)) > 1, then the number of edges labeled with 0 and the number of edges labeled with 1 differ by at most 1. Also they discussed the prime cordial labeling behavior of various graphs. In [5,6], Seoud and Salim gave an upper bound for the number of edges of a graph with a prime cordial labeling as a function of the number of vertices. For bipartite graphs they gave a stronger bound and also they determine all prime cordial graphs of order at most 6. Recently Ponraj et al. [4], introduced the concept of k-difference cordial labeling of graphs and studied the 3-difference cordial labeling behavior of of star, m copies of star etc. Also they discussed the 3-difference cordial labeling behavior of path, cycle, complete graph, complete bipartite graph, star, bistar, comb, double comb, quadrilateral snake, S(K1,n), S(Bn,n). Motivated by these labelings we introduce k-prime cordial labeling of graphs. Also in this paper we investigate the 3-prime cordial labeling behavior of path, cycle, complete graph, star, wheel etc. Let x be any real number. Then ⌊x⌋ stands for the largest integer less than or equal to x and ⌈x⌉ stands for smallest integer greater than or equal to x. Terms not defined here follow from Harary [3].
2. k-prime cordial labeling
Definition 2.1. Let G be a (p, q) graph and 2 ≤ p ≤ k. Let f : V (G) → {1, 2, . . . , k} be a function. For each edge uv, assign the label gcd (f(u), f(v)). f is called a k-prime cordial labeling of G if |vf (i) - vf (j)| ≤ 1, i, j ∈ {1, 2, . . . , k} and |ef (0) − ef (1)| ≤ 1 where vf (x) denotes the number of vertices labeled with x, ef (1) and ef (0) respectively denote the number of edges labeled with 1 and not labeled with 1. A graph with a k-prime cordial labeling is called a k-prime cordial graph.
Example 2.2. An example of a 3-prime cordial labeling of a graph is given in FIGURE 1.
FIGURE 1
Remark 2.1. A 2-prime cordial labeling is a product cordial labeling [2].
Theorem 2.3. Every graph is a subgraph of a connected k-prime cordial graph.
Proof. Let G be a given (p, q) graph. Take k-copies of complete graph Kp. Let Gi be the ith copy of Kp. Let Let
The vertex set and edge set of the super graph G∗ of G is as follows: Let and Clearly G∗ has kp + m vertices and edges. Next we assign a labeling to the vertices of G∗. Let m = kt+r, 0 ≤ r < k. Assign the label i to all the vertices of Gi (1 ≤ i ≤ k). Then assign the label 1 to v1, v2, . . . , vn, 2 to vt+1, vt+2,. . . , v2t, 3 to v2t+1, v2t+2, . . . , v3t, . . . . . . k to v(k−1)t+1, v(k−1)t+2,. . . , vkt. Finally assign the labels 1, 2, . . . , r to the vertices vkt+1, vkt+2, . . . , vkt+r respectively. The vertex and edge conditions of the above labeling is given below: vf (1) = vf (2) = . . . = vf (r - 1) = vf (r) = p + t + 1, vf (r + 1) = . . . = vf (k) = p + t. This forces f to be a k-prime cordial labeling of G∗.
Hence every graph is a subgraph of a connected k-prime cordial graph. □
Theorem 2.4. If k is even, then the path Pn, n ≠ 3 is k-prime cordial.
Proof. Let Pn be the path u1u2 . . . un. Let n = kt+r, 0 ≤ r < k. Assign the label 2, 4, 6, . . . , k to first path vertices. Then assign again the label 2, 4, 6, . . . , k to the next path vertices. Continue in this way until we reach the vertex that receive the label k.
Case 1. r is even.
Then assign the labels 2, 4, 6, . . . , r to the vertices and 1, 3, 5, . . . , r − 1 to the vertices
Case 2. r is odd.
Here assign the labels 2, 4, 6, . . . , r + 1 to the vertices and 3, 5, . . . , r to the vertices
Now assign the labels 1, 3, 5, . . . , k − 1 to the vertices then label the next t vertices of the path by 1, 3, 5, . . . , k−1. Continue this process until we reach the vertex ukt+r. Obviously this labeling is a k-prime cordial labeling. □
Corollary 2.5. The cycle Cn, n ≠ 3 is k-prime cordial where k is even.
Proof. Let u1u2 . . . unu1 be the cycle Cn. Assign the label to the vertices ui as in theorem 2.4. Then replace the label of by k. Clearly this vertex labeling is a k-prime cordial labeling of Cn. □
Theorem 2.6. The bistar Bn,n is k-prime cordial for all even k.
Proof. Let u, v be the central vertices of Bn,n and ui and vi (1 ≤ i ≤ n) be the pendent vertices which are adjacent to u, v respectively. Let n = kt+r, 0 ≤ r < k. Assign the label 1, 2 respectively to the vertices u, v. Then assign the label to the vertices ui (1 ≤ i ≤ n) as follows: Assign the odd integers 1, 3, 5, . . . , k−1 to the first vertices namely respectively. Then assign the labels 1, 3, 5, . . . , k−1 to the next vertices respectively. Continue this process until we reach the last parts vertices Note that in this process the vertex ukt receive the label k −1. If r is even, then assign the labels 1, 3, . . . , r − 1 to the vertices and to the vertices if r is odd then assign the labels 1, 3, . . . , r to the vertices and to the vertices
Next we move into the other side pendent vertices, namely, vi (1 ≤ i ≤ n). Assign the labels 2, 4, . . . , k to the vertices respectively. Then assign the labels 2, 4, 6, . . . , k to the vertices respectively. Proceed this way until we reach the last part Clearly the vertex vkt received the label k. If r is even, then assign the labels 2, 4, . . . , r to the vertices and to the vertices if r is odd then assign the labels 1, 3, . . . , r −1 to the vertices and to the vertices □
Next we investigate the 3-prime cordial labeling behavior of some graphs.
3. 3-prime cordial labeling
Theorem 3.1. The path Pn is 3-prime cordial if and only if n ≠ 3.
Proof. For n = 3, it is trivial that vf (1) = vf (2) = vf (3) = 1. But ef (0) = 0. This implies |ef (0) − ef (1)| > 1. Assume n ≠ 3. Let Pn be the path u1u2 . . . un.
Case 1. n ≡ 0, 1 (mod 3).
Assign the label 2 to the vertices Then assign the label 3 consecutively to the vertices until we have received the edges with the label 0. If all the 3’s are exhausted then assign the label 1 to the remaining vertices; otherwise consider the non labeled vertex ui such that ui−1 is labeled and assign the labels 1, 3 to the vertices ui, ui+1, ui+2, . . . alternatively until 3’s are exhausted. Finally assign the label 1 to the remaining vertices.
Case 2. n ≡ 2 (mod 3).
As in case 1, assign the labels to the vertices u1, u2, . . . , un. Now, let i be the least positive integer such that the label of ui-1 = the label of ui+1 = 3, and the label of ui = 1. Finally interchange the labels of ui and ui+1. Clearly this vertex labeling satisfies both vertex and edge conditions. □
Corollary 3.2. The cycle Cn is 3-prime cordial if and only if n ∉ {3, 4, 6}.
Proof. One can easily verify that if n ∈ {3, 4, 6}, then Cn is not 3-prime cordial. Consider the case that n ∉ {3, 4, 6}. It is clear that the 3-prime cordial labeling of the path given in theorem 3.1 is also a 3-prime cordial labeling of Cn. □
Theorem 3.3. If T is a 3-prime cordial tree, then T ⊙ K1 is also a 3-prime cordial graph.
Proof. Let the order of T be p. Then the following three cases arise. Let f be a 3-prime cordial labeling of T.
Case 1. p ≡ 0 (mod 3).
Let p = 3t. In this case, vf (1) = vf (2) = vf (3) = t. We now assign the labeling g to the vertices of T ⊙ K1 with the help of the labeling f of T. Assign the label 2 to the vertices of T ⊙ K1 whose support received the label 2 under the labeling f. Then assign the label 3 to the pendent vertices whose supports received the label 3 and assign 1 to all the remaining pendent vertices whose support received the label 3. Now we turn to the pendent vertices whose supports received the label 1. Assign 3 to of these vertices. Finally assign 1 to the remaining vertices. Obviously the labeling given above is a 3-prime cordial labeling of T ⊙ K1 for this case.
Case 2. p ≡ 1 (mod 3).
Let p = 3t + 1. Here the following three subcases arise:
Take the same labeling g of case 1. In the case of (a), there is a non labeled vertex whose support received the label 1. Assign 3 to that vertex. In the case of (b), assign 3 to the non labeled vertex. In the case (c), assign the label 2 to the non labeled vertex. It is easy to verify that this vertex labeling h is a 3-prime cordial labeling of T ⊙ K1 for this case.
Case 3. p ≡ 2 (mod 3).
Let p = 3t + 2. In this case, the following three subcases arise:
In the case of (a), take the labeling h of subcase (b) of case 2. Assign 1 to the non labeled vertex. In the case of (b), take the labeling h of subcase (a) of case 2. Assign 2 to the non labeled vertex. In the case of (c), take the labeling h of subcase (a) of case 2. Assign 3 to the non labeled vertex. It is easy to verify that this labeling is a 3-prime cordial labeling for this case. □
Corollary 3.4. The comb Pn ⊙ K1 is 3-prime cordial.
Proof. The proof follows from theorems 3.1 and 3.3 for n ≠ 3. For n = 3, the 3-prime cordial labeling of P3 ⊙ K1 is given in FIGURE 2.
FIGURE 2
□
Corollary 3.5. The crown Cn ⊙ K1 is 3-prime cordial if and only if n ≠ 3.
Proof. Suppose n = 3. Then vf (1) = vf (2) = vf (3) = 2. Also ef (0) ≤ 2. It follows that, |ef (0) − ef (1)| = 2. Hence C3 ⊙ K1 is not 3-prime cordial. Conversely, one can easily verify that the labeling given in corollary 3.4 is also a 3-prime cordial labeling of Cn ⊙ K1. □
A rooted tree consisting of n branches, where the ith branch is a path of length i is called an olive tree and it is denoted by OTn.
Theorem 3.6. Olive tree OTn is 3-prime cordial.
Proof. Assign the label 1 to the central vertex. We now consider the path of order n. Assign the label 3 to all the vertices of this path. Note that the number of edges with label 0 is n − 2. Next we move to the next path of order n − 1. Assign the label 3 to the vertices of the path until 3’s are used. Now check the edges with label 0. If it is then we have half of the edges with the label 0. Otherwise we consider the label 2. Assign the label 2 to the next non labeled vertices of this path. (If the label 3 has appeared in the pendent vertex of the path then move to the next path). Proceed this way assign the label 2 to the vertices of the paths until we get the edges with the label zero. It is easy to observe that we have some remaining 2’s. Now assign the labels 1 and 2 alternatively. The edges with vertex labels 1 and 2 together with the edges incident with the central vertex and the edge with the vertex label 3 and 2 contributes 1’s. (Possible if the last used 3 is not a label of the pendent vertex of a path). Clearly the above labeling pattern is a 3-prime cordial labeling. □
The next investigation is about K2 + mK1. Let V (K2 + mK1) = {u, v, ui : 1 ≤ i ≤ m} and E(K2 + mK1) = {uv, uui, vui : 1 ≤ i ≤ m}.
Theorem 3.7. K2 + mK1 is not 3-prime cordial.
Proof. Let u and v be the vertices of K2. Note that the order and size of K2 + mK1 are m + 2 and 2m+ 1 respectively.
Case 1. f(u) = f(v) = 1. Here, all the edges receive the label 1. This implies ef (1) = 2m + 1, a contradiction.
Case 2. f(u) = f(v) = 2.
Subcase 2a. m ≡ 0 (mod 3). Let m = 3t, t ≥ 1. In this case vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = vf (3) = t + 1, vf (2) = t or vf (1) = t, vf (2) = vf (3) = t + 1. Suppose vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = t, vf (2) = vf (3) = t + 1 then ef (0) = 2t − 1 and ef (1) = 4t + 2. This shows that ef (1) − ef (0) = 2t + 3 > 1, a contradiction. If vf (1) = vf (3) = t + 1, vf (2) = t, then ef (0) = 2t−3 and ef (1) = 4t+4. This implies ef (1) − ef (0) = 2t+7 > 1, a contradiction.
Subcase 2b. m ≡ 1 (mod 3). Let m = 3t + 1. In this case, vf (1) = vf (2) = vf (3) = t+1. Then ef (0) = 2t−1, ef (1) = 4t+4. Thus ef (1)−ef (0) = 2t+5 > 1, a contradiction.
Subcase 2c. m ≡ 2 (mod 3). Let m = 3t + 2. Here, vf (1) = t + 2 vf (2) = vf (3) = t + 1 or vf (1) = vf (3) = t + 1, vf (2) = t + 2 or vf (1) = vf (2) = t + 1, vf (3) = t + 2. If vf (1) = t + 2, vf (2) = vf (3) = t + 1 or vf (1) = vf (2) = t + 1, vf (3) = t+2 then ef (0) = 2t−1, ef (1) = 4t+6. Hence ef (1)−ef (0) = 2t+7 > 1, a contradiction. For vf (1) = vf (3) = t + 1, vf (2) = t + 2, ef (0) = 2t + 1, ef (1) = 4t + 4. Therefore ef (1) − ef (0) = 2t + 3 > 1, a contradiction.
Case 3. f(u) = f(v) = 3. Similar to case 2.
Case 4. f(u) = 1, f(v) = 2.
Subcase 4a. m ≡ 0 (mod 3). Let m = 3t, t ≥ 1. In this case vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = vf (3) = t + 1, vf (2) = t or vf (1) = t, vf (2) = vf (3) = t + 1. Suppose vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = t, vf (2) = vf (3) = t + 1 then ef (0) = t and ef (1) = 5t + 1. This shows that ef (1) − ef (0) = 4t + 1 > 1, a contradiction. If vf (1) = vf (3) = t + 1, vf (2) = t, then ef (0) = t − 1 and ef (1) = 5t + 2. This implies ef (1) − ef (0) = 4t + 3 > 1, a contradiction.
Subcase 4b. m ≡ 1 (mod 3). Let m = 3t + 1. In this case, vf (1) = vf (2) = vf (3) = t + 1. Then ef (0) = t, ef (1) = 5t + 3. Thus ef (1) − ef (0) = 4t + 3 > 1, a contradiction.
Subcase 4c. m ≡ 2 (mod 3). Let m = 3t + 2. Here, vf (1) = t + 2 vf (2) = vf (3) = t + 1 or vf (1) = vf (3) = t + 1, vf (2) = t + 2 or vf (1) = vf (2) = t + 1, vf (3) = t + 2. If vf (1) = t + 2, vf (2) = vf (3) = t + 1 or vf (1) = vf (2) = t + 1, vf (3) = t + 2 then ef (0) = t, ef (1) = 5t + 5. Hence ef (1) − ef (0) = 4t + 5 > 1, a contradiction. For vf (1) = vf (3) = t + 1, vf (2) = t + 2, ef (0) = t + 1, ef (1) = 5t + 4. Therefore ef (1) − ef (0) = 4t + 3 > 1, a contradiction.
Case 5. f(u) = 1, f(v) = 3. Similar to case 4.
Case 6. f(u) = 2, f(v) = 3.
Subcase 6a. m ≡ 0 (mod 3). Let m = 3t, t ≥ 1. In this case vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = vf (3) = t + 1, vf (2) = t or vf (1) = t, vf (2) = vf (3) = t + 1. Suppose vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = vf (3) = t+1, vf (2) = t, then ef (0) = 2t−1 and ef (1) = 4t+2. This shows that ef (1) − ef (0) = 2t + 3 > 1, a contradiction. If vf (1) = t, vf (2) = vf (3) = t + 1, then ef (0) = 2t and ef (1) = 4t + 1. This implies ef (1) − ef (0) = 2t + 1 > 1, a contradiction.
Subcase 6b. m ≡ 1 (mod 3). Let m = 3t + 1. In this case, vf (1) = vf (2) = vf (3) = t+1. Then ef (0) = 2t, ef (1) = 4t+3. Thus ef (1)−ef (0) = 2t+3 > 1, a contradiction.
Subcase 6c. m ≡ 2 (mod 3). Let m = 3t + 2. Here, vf (1) = t + 2 vf (2) = vf (3) = t + 1 or vf (1) = vf (3) = t + 1, vf (2) = t + 2 or vf (1) = vf (2) = t + 1, vf (3) = t + 2. If vf (1) = t + 2, vf (2) = vf (3) = t + 1, then ef (0) = 2t, ef (1) = 4t + 5. Hence ef (1) − ef (0) = 2t + 5 > 1, a contradiction. Suppose vf (1) = vf (2) = t + 1, vf (3) = t + 2 or vf (1) = vf (3) = t + 1, vf (2) = t + 2, then ef (0) = 2t + 1, ef (1) = 4t + 4. Therefore ef (1) − ef (0) = 2t + 3 > 1, a contradiction.
Hence K2 + mK1 is not 3-prime cordial. □
Corollary 3.8. K2,n is not 3-prime cordial.
Proof. Since K2,n is obtained from K2 + nK1, by removing the edge uv where deg(u) = deg(v) = n+1 in K2+nK1. Then by theorem 3.7, the proof follows. □
Theorem 3.9. The star K1,n is 3-prime cordial if and only if n ≤ 3.
Proof. It is easy to see that K1,n, n ≤ 3 is 3-prime cordial. On the other hand, if possible, there exist a 3-prime cordial labeling of K1,n, say f. Let u be the vertex with degree n.
Case 1. f(u) = 1. In this case ef (0) = 0 and ef (1) = n. This is a contradiction.
Case 2. f(u) = 2.
Subcase 2a. n ≡ 0 (mod 3). Let n = 3t. In this case, vf (1) = vf (3) = t, vf (2) = t + 1 or vf (1) = t + 1, vf (2) = vf (3) = t or vf (1) = vf (2) = t, vf (3) = t + 1. Suppose vf (1) = vf (3) = t, vf (2) = t + 1 then ef (0) = t, ef (1) = 2t. Therefore |ef (1) − ef (0)| = t > 1, a contradiction. If vf (1) = t + 1 vf (2) = vf (3) = t or vf (1) = vf (2) = t, vf (3) = t + 1 then ef (0) = t − 1, ef (1) = 2t + 1. Hence |ef (1) − ef (0)| = t + 2 > 1, a contradiction.
Subcase 2b. n ≡ 1 (mod 3). Let n = 3t + 1. Here, vf (1) = t vf (2) = vf (3) = t + 1 or vf (1) = vf (3) = t + 1, vf (2) = t or vf (1) = vf (2) = t + 1, vf (3) = t. If vf (1) = t, vf (2) = vf (3) = t+1 or vf (1) = vf (2) = t+1, vf (3) = t then ef (0) = t, ef (1) = 2t+1. Hence |ef (1) − ef (0)| = t+1 > 1, a contradiction. For vf (1) = vf (3) = t + 1, vf (2) = t, ef (0) = t − 1, ef (1) = 2t + 2. Therefore |ef (1) − ef (0)| = t + 3 > 1, a contradiction.
Subcase 2c. n ≡ 2 (mod 3). Let n = 3t + 2. In this case, vf (1) = vf (2) = vf (3) = t + 1. Then ef (0) = t, ef (1) = 2t+ 2. Thus |ef (1) − ef (0)| = t + 2 > 1, a contradiction.
Case 3. f(u) = 3. Similar to case 2.
Hence K1,n is 3-prime cordial if and only if n ≤ 3. □
Theorem 3.10. The complete graph Kn is 3-prime cordial if and only if n < 3.
Proof. If n < 3, the proof is trivial. Assume n ≥ 3. If possible, let f be a 3-prime cordial labeling of Kn.
Case 1. n ≡ 0 (mod 3).
Let n = 3t. In this case, vf (1) = vf (2) = vf (3) = t. Then and This implies a contradiction.
Case 2. n ≡ 1 (mod 3).
Let n = 3t + 1. In this case, vf (1) = t + 1 vf (2) = vf (3) = t or vf (1) = vf (3) = t, vf (2) = t + 1 or vf (1) = vf (2) = t, vf (3) = t + 1. If vf (1) = t + 1 vf (2) = vf (3) = t then This forces a contradiction. Suppose vf (1) = vf (3) = t, vf (2) = t + 1 or vf (1) = vf (2) = t, vf (3) = t + 1 then Hence a contradiction.
Case 3. n ≡ 2 (mod 3).
Let n = 3t+2. In this case, vf (1) = vf (2) = t+1, vf (3) = t or vf (1) = vf (3) = t + 1, vf (2) = t or vf (1) = t vf (2) = vf (3) = t + 1. If vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = vf (3) = t + 1, vf (2) = t then and hence a contradiction. For vf (2) = vf (3) = t + 1, vf (1) = t, This implies a contradiction.
Hence Kn is not 3-prime cordial. □
Theorem 3.11. The wheel Wn is not 3-prime cordial.
Proof. Suppose there exist a 3-prime cordial labeling of Wn, say f. Let u be the vertex with degree n.
Case 1. f(u) = 1.
Subcase 1a. n ≡ 0 (mod 3). Let n = 3t. Here, vf (1) = t + 1 vf (2) = vf (3) = t or vf (1) = vf (2) = t, vf (3) = t+1 or vf (1) = vf (3) = t, vf (2) = t+1. Suppose vf (1) = t+1, vf (2) = vf (3) = t then ef (0) ≤ (t−1)+(t−1) = 2t−2 and ef (1) ≥ 6t−(2t−2) = 4t+2. Therefore ef (1)−ef (0) ≥ 2t+4, a contradiction. If vf (1) = vf (2) = t, vf (3) = t + 1 or vf (1) = vf (3) = t, vf (2) = t + 1 then ef (0) ≤ (t − 1) + t = 2t − 1 and ef (1) ≥ 6t − (2t − 1) = 4t + 1. Hence ef (1) − ef (0) = 2t + 2, a contradiction.
Subcase 1b. n ≡ 1 (mod 3). Let n = 3t + 1. In this case, vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = vf (3) = t + 1, vf (2) = t or vf (1) = t vf (2) = vf (3) = t + 1. If vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = vf (3) = t + 1, vf (2) = t then ef (0) ≤ t+(t−1) = 2t−1 and ef (1) ≥ 6t+2−(2t−1) = 4t+3. Hence ef (1) − ef (0) = 2t + 4, a contradiction. For vf (2) = vf (3) = t + 1, vf (1) = t, ef (0) ≤ t + t = 2t and ef (1) ≥ 6t + 2 − 2t = 4t + 2. This forces ef (1) − ef (0) = 2t + 2, a contradiction.
Subcase 1c. n ≡ 2 (mod 3). Let n = 3t + 2. In this case, vf (1) = vf (2) = vf (3) = t + 1. Then ef (0) ≤ t + t = 2t, ef (1) ≥ 6t + 4 − 2t = 4t + 4. Thus |ef (1) − ef (0)| ≥ 2t + 4, a contradiction.
Case 2. f(u) = 2.
Subcase 2a. n ≡ 0 (mod 3). Let n = 3t. Here, vf (1) = t + 1 vf (2) = vf (3) = t or vf (1) = vf (2) = t, vf (3) = t+1 or vf (1) = vf (3) = t, vf (2) = t+1. If vf (2) = vf (3) = t, vf (1) = t+1 then ef (0) ≤ (t−2)+(t−1)+(t−1) = 3t−4 and ef (1) ≥ 6t − (3t − 4) = 3t + 4. Hence ef (1) − ef (0) ≥ 8, a contradiction. If vf (1) = vf (2) = t, vf (3) = t + 1 then ef (0) ≤ (t − 2) + (t − 1) + t = 3t − 3 and ef (1) ≥ 6t − (3t − 3) = 3t + 3. This forces ef (1) − ef (0) ≥ 6, a contradiction. For vf (2) = t + 1 vf (1) = vf (3) = t, ef (0) ≤ (t − 1) + (t − 1) + t = 3t − 2 and ef (1) ≥ 6t − (3t + 2) = 3t + 2. Therefore ef (1) − ef (0) ≥ 4, a contradiction.
Subcase 2b. n ≡ 1 (mod 3). Let n = 3t + 1. Here, vf (1) = vf (2) = t + 1, vf (3) = t or vf (1) = vf (3) = t + 1, vf (2) = t or vf (1) = t vf (2) = vf (3) = t + 1. If vf (1) = vf (2) = t+1, vf (3) = t then ef (0) ≤ (t−1)+(t−1)+t = 3t−2 and ef (1) ≥ 6t+2−(3t−2) = 3t+4. This implies ef (1)−ef (0) ≥ 6, a contradiction. If vf (1) = vf (3) = t + 1, vf (2) = t then ef (0) ≤ (t − 2) + (t − 1) + t = 3t − 3 and ef (1) ≥ 6t+2−(3t−3) = 3t+5. Hence ef (1)−ef (0) ≥ 8, a contradiction. Suppose vf (2) = vf (3) = t+1, vf (1) = t then ef (0) ≤ (t−1)+t+t = 3t−1 and ef (1) ≥ 6t+2−(3t−1) = 3t+3. This implies ef (1)−ef (0) ≥ 4, a contradiction.
Subcase 2c. n ≡ 2 (mod 3). Let n = 3t + 2. In this case, vf (1) = vf (2) = vf (3) = t+1. Then ef (0) ≤ t+(t−1)+t = 3t−1, ef (1) ≥ 6t+4−(3t−1) = 3t+5. Hence ef (1) − ef (0) ≥ 6, a contradiction.
Case 3. f(u) = 3. Similar to case 2.
Hence Wn is not 3-prime cordial. □
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