ON THE LOCATING-CHROMATIC NUMBERS OF SUBDIVISIONS OF FRIENDSHIP GRAPH

Let c be a k-coloring of a connected graph G and let π = {C1, C2, . . . , Ck} be the partition of V (G) induced by c. For every vertex v of G, let cπ(v) be the coordinate of v relative to π, that is cπ(v) = (d(v, C1), d(v, C2), . . . , d(v, Ck)), where d(v, Ci) = min{d(v, x)|x ∈ Ci}. If every two vertices of G have different coordinates relative to π, then c is said to be a locating k-coloring of G. The locating-chromatic number of G, denoted by χL(G), is the least k such that there exists a locating k-coloring of G. In this paper, we determine the locating-chromatic numbers of some subdivisions of the friendship graph Frt, that is the graph obtained by joining t copies of 3-cycle with a common vertex, and we give lower bounds to the locating-chromatic numbers of few other subdivisions of Frt.


INTRODUCTION
The concept of locating-chromatic number was first studied by Chartrand et al. [1] by combining the concept of graph partition dimension and graph coloring. The locating-chromatic numbers of some classes of graphs were studied, especially recently for certain Barbell graphs in [2], Halin graphs in [3], and graphs resulting from certain operations of other graphs, such as join of graphs in [4] and Cartesian product of graphs in [5]. Trees with certain locating-chromatic number were also studied in [6] and [7]. Bounds for locating-chromatic numbers of trees and subdivisions of graph on one edge were also established in [8] and [9], respectively.
Suppose that G = (V, E) is a simple connected graph. Let c be a k-coloring on G and let π = {C 1 , C 2 , ..., C k } be the partition of V = V (G) induced by c. For every vertex v of G, let c π (v) = (d(v, C 1 ), d(v, C 2 ), . . . , d(v, C k )) be the coordinate of v relative to π, where d(v, C i ) = min{d(v, x)|x ∈ C i } is the shortest distance between v and vertices in C i . If every two vertices of G have different coordinates relative to π, then c is said to be a locating k-coloring of G. The locating-chromatic number of G, denoted by χ L (G), is the least k such that there exists a locating k-coloring of G. As shown by Chartrand et al. in [1], if u and v are vertices of G such that d(u, w) = d(v, w) for every w ∈ V − {u, v}, then c(u) = c(v).
In [4], Behtoei and Anbarloei studied the locating chromatic number of friendship graph F r t , which is the graph obtained by joining the complete graph K 1 to the t disjoint copies of K 2 . They showed that χ L (F r t ) = 1 + min{k|t ≤ k 2 }. In this paper, we study some subdivisions of F r t and their locating-chromatic numbers. In general, a subdivision of a graph G is a graph obtained by replacing some edges of G, say 1 , 2 , . . ., r , respectively with paths P 1 , P 2 , . . ., P r of length one or greater, where these paths may differ in length. In particular, when we say a subdivision of a graph on some edges l ≥ 0 times, we are specifying which or how many edges are replaced and ensuring the paths replacing the edges are all of length l + 1. Purwasih et al. [9] showed that χ L (G) ≤ 1 + χ L (H) if G is a subdivision of a graph H on one edge. We investigate the case where H = F r t by determining the locating-chromatic number of any subdivision of F r t on one edge and also the locating-chromatic number of any subdivision of F r t once on each of its cycle. We also give a tight upper bound for any subdivision of F r t . Throughout this paper, for t ≥ 2 we denote the center of F r t , that is the vertex with the largest degree, by z. For every natural number n, we also denote [n] = {1, 2, . . . , n}.

MAIN RESULTS
In this section, we determine the locating-chromatic number of any subdivision of F r t on one edge. We also determine the locating-chromatic number of any subdivision of F r t once on each of its cycle.
2.1. Subdivision of F r t on one edge. Throughout this subsection, let t ≥ 2 and l ≥ 1 be two natural numbers and let G be a subdivision of F r t on one edge l times. For each n ≥ 3, we define d n = n−1 2 + 1. Observe that if t ≥ 3, we have d k−1 < t ≤ d k for some k ∈ {4, 5, 6, . . .}. We begin with the following lemmas. Lemma 1. If 3 ≤ t = d k and l = 2, then χ L (G) = k + 1.
Proof. By definition of G, there are exactly d k − 1 number of 3-cycles and a 5-cycle in G. Consider the collection of 2-subsets of 2 , we can denote the elements of [k − 1] 2 by u 1 , u 2 , . . . , and u ( k−1 2 ) . Now, we start by assigning colors to the vertices of G. We immediately assign the color k + 1 to the vertex z. To assign colors to other vertices, observe that since there are d k − 1 = k−1 2 number of 3-cycles, there are 2 k−1 2 vertices other than z that lie on a 3-cycle. We denote these vertices by v 1 , v 2 , . . . , and v 2( k−1 2 ) , where v 2i−1 and v 2i are on the same 3-cycle for each i = 1, 2, . . . , k−1 2 . If we write assign the color a i and b i respectively to v 2i−1 and v 2i . To finish the color assignment, let the 5-cycle in G be zw 1 w 2 w 3 w 4 z. Assign the colors 1, k + 1, k, and 1 respectively to w 1 , w 2 , w 3 , and w 4 . Let c be the obtained coloring. Clearly, c is a well-defined graph coloring since no two adjacent vertices are assigned the same color. We show that c is a locating coloring. Let x and y be two vertices with the same color. If x and y are in the same cycle, then the only possibilities are, without loss of generality, either (x, y) = (z, w 2 ) or (x, y) = (w 1 , w 4 ). However, in both of these cases, the k-th component of the coordinate of x and y differ since 2 = d(x, w 3 ) = d(y, w 3 ) = 1 and w 3 is the only vertex colored k.
Let us now assume that x and y are in different cycles. If both are in different 3-cycles, clearly their coordinates differ since their neighbors other than z have different colors by definition of u 1 , u 2 , . . ., and u ( k−1 2 ) . If, without loss of generality, x is in a 5-cycle and y is in a 3-cycle, then either x = w 1 or x = w 4 since the colors k and k + 1 are not assigned to y. In both cases, however, their neighbors other than z also have different colors. Hence, their coordinates differ. Thus, we have shown that c is a locating (k + 1)-coloring and that χ L (G) ≤ k + 1.
We now show that χ L (G) > k by contradiction. Suppose that there exists a locating k-coloring c for G. Suppose that c (z) = k. Hence, without loss of generality, the pair of vertices {v 2i−1 , v 2i } has to be assigned by the pair of colors Moreover, without loss of generality, let c (w 1 ) = 1. If c (w 2 ) = k, then we let c (w 2 ) = m ∈ {2, 3, ..., k − 1}. However, there are two vertices v p and v p+1 such that (c (v p ), c (v p+1 )) = (1, m). Observe that d(w 1 , w) = d(v p , w) for any vertex w that is assigned by any color other than 1. Hence, w 1 and v p have the same coordinate, contradicting the definition of a locating coloring. Thus, we must have c (w 2 ) = k. However, by the same argument, we must also have c (w 3 ) = k, which contradicts the definition of coloring. Thus, we have shown that χ L (G) > k and we conclude that χ L (G) = k + 1. Proof. Since t = d k and l = 2, there are exactly d k − 1 number of 3-cycles and an (l + 3)-cycle in G. We start by coloring G. Assign the color k to the vertex z and assign colors to the vertices lying in 3-cycles other than z by using the same way used in the proof of the previous lemma. Consider these cases. a. Suppose that the (l+3)-cycle is zs 1 s 2 . . . s 4q−1 z for some q. Assign the color k to s 2 , s 4 , . . . , s 4q−2 . Assign the color 1 to s 1 , s 3 , . . . , s 2q−1 . Assign the color 2 to s 2q+1 , s 2q+3 , . . . , s 4q−1 . Let c be the resulting coloring. Clearly, c is a well-defined coloring. We show that c is a locating coloring. Let x and y be two vertices with the same color. If x and y are in the same cycle, then both have to be in the (l + 3)-cycle. If both are colored k and their neighbors are only vertices of color 1, then their coordinates differ by their distances to a vertex colored 3 since t = 3, or other colors other than 1 and 2. It is also the case when their neighbors are only vertices of color 2. If their neighbors are vertices of color 1 and 2, one of them is z and the other one is s 2q . In this case, their coordinates also differ by their distances to a vertex colored other than 1 and 2. The same argument also applies if the color of x and y are 1 or 2. Moreover, if x is in the (l + 3)-cycle and y is in a 3-cycle without loss of generality, then x and y are not colored k. However, the neighbors of x are only vertices colored k, while some of the neighbors of y are not colored k. Hence, their coordinates differ. Thus, c is a locating k-coloring. b. Suppose that the (l+3)-cycle is zs 1 s 2 . . . s 4q−3 z for some q. Assign the color k to s 2 , s 4 , . . . , s 4q−4 . Assign the color 1 to s 1 , s 3 , . . . , s 2q−1 . Assign the color 2 to s 2q+1 , s 2q+3 , . . . , s 4q−3 . Let c be the resulting coloring. Clearly, c is a well-defined coloring. We show that c is a locating coloring. Let x and y be two vertices with the same color. By using the same argument as in part a, we see that c is indeed a locating k-coloring. c. Suppose that the (l + 3)-cycle is a (2q − 1)-cycle zs 1 s 2 . . . s 2q−2 z. Assign the color k to s 2 , s 4 , . . ., s 2q−6 and s 2q−3 . Assign the color 1 to s 1 , s 3 , . . ., s 2q−5 . Assign the color 2 to s 2q−2 and s 2q−4 . Let c be the resulting coloring. Clearly, c is well-defined. The argument to show that c is indeed a locating k-coloring is similar to part a or part b with minor difference, that is if x and y are two vertices of color k other than z, then their neighbors are either only vertices of color 1 or only vertices of color 2.
Proof. We start by coloring the graph G. Assign the color k to z. Assign colors to vertices lying in 3-cycles other than z by using the same way used in the proof of the first lemma, that is by taking different elements in the set [k − 1] 2 as pairs of colors for pairs of vertices in each 3-cycle. Since t < d k , there are less than k−1 2 number of 3-cycles, so that there exist elements in [k − 1] 2 that are not used as a pair of color in any 3-cycle. Denote this element by (g 1 , g 2 ).
It is easy to see, by using the same argument as in the proof of previous lemma, that all vertices colored the same have different coordinates. In this case, vertices colored g 1 and g 2 create the differences. Hence, χ L (G) ≤ k. The proof showing that χ L (G) > k − 1 is similar to the last paragraph of the proof of the last lemma. Thus, we have χ L (G) = k.
From previous three lemmas, we have the following theorem.
Theorem 1. Let t ≥ 3 and l ≥ 1 be two natural numbers. Let G be a subdivision of F r t on one edge l times. For each n ≥ 3, let d n = n−1 2 + 1 and d k−1 < t ≤ d k for some k. Hence, we have χ L (G) = k + 1 if t = d k and l = 2, and χ L (G) = k otherwise.
We treat the case t = 2 separately in the next proposition. Proof. We start by coloring G. Assign the color 4 to the vertex z. Assign the color 1 and 2 to the two vertices lying in the only existing 3-cycle. Now, denote the (l + 3)-cycle in G by zu 1 u 2 . . . u l+2 z.
Assume first that l is even. Assign the colors 1, 3, 1, 3, . . . , 1, 3 respectively to the vertices u 1 , u 2 , u 3 , u 4 , . . . , u l+2 . By doing this, the vertices colored 3 have their coordinates differed by their distances to the vertex colored 4, and the vertices colored 1 have their coordinates differed by their distances to the vertex colored 2. Thus, we obtain a locating 4-coloring.
Assume now that l is odd. Assign the color 4 to the vertex u l+3 2 . Assign the color 1 to each vertex of the form u 1 , u 3 , . . . , u l1 , where l 1 < l+3 2 , and vertex of the form u l2 , . . . , u l−1 , u l+1 , where l 2 > l+3 2 . Assign the color 3 to other remaining vertices. By doing this, the vertices colored 3 have their coordinates differ by their distances to the vertex colored 2, and so do the vertices colored 1. Thus, we obtain a locating 4-coloring.
We have shown that χ L (G) ≤ 4. We now show that χ L (G) > 3. Suppose that there exists a locating 3-coloring on G. Without loss of generality, assume that the 3-cycle in G is colored by 1, 2, and 3, where z is assigned the color 3. Suppose that there exists a vertex colored by 2 in the (l + 3)-cycle in G. Let j be the least index such that u j is colored by 2. If j is odd, then the vertices u 1 , u 3 , . . . , u j−2 have to be colored by 1 since the color of z is 3, and we must also have the vertices u 2 , u 4 , . . . , u j−1 colored by 3. However, the coordinate of u j−1 is equal to the coordinate of z, contradicting the definition of locating coloring. If j is even instead, then the vertices u 1 , u 3 , . . . , u j−1 have to be colored 1 since the color of z is 3, and we must also have the vertices u 2 , u 4 , . . . , u j−2 colored by 3.
However, the coordinate of u j−1 is equal to the coordinate of the vertex colored by 1 on the 3-cycle, contradicting again the definition of locating coloring. Hence, there must not be any vertex colored by 2 on the (l + 3)-cycle. This means that, since z is colored by 3, u 1 , u 3 , . . . , u l+2 have to be colored by 1 and u 2 , u 4 , . . . , u l+1 have to be colored by 3. However, the vertices u 1 and u l+2 have the same color and the same coordinate, contradicting the definition of the locating coloring. Thus, we have χ L (G) = 4.

2.2.
Subdivision of F r t once on one edge of each cycle. We now determine the locating-chromatic number of the subdivision of F r t once on one edge of each cycle. This means that each cycle of the graph is a 4-cycle. Let G be such graph, where t ≥ 2. for i = 1, 2, . . . , k − 1, by noting that the components are calculated under modulo k − 1. Observe that in W j , there is no entry that is equal to k. Observe also that W i and W j never equal to each other since their second entries differ for any i and j. By definition, we also see that W i1 and W i2 differ for any different i 1 and i 2 , and that W j1 and W j2 differ for any different j From the definition of W , clearly c is a k-coloring. We now show that c is a locating coloring. Let x and y be two different vertices with the same color in G. If x = z, then y = u i,2 for some i such that c(u i,2 ) = k, if it exists. However, since t ≥ 2 and by definition of c, the vertex x is adjacent to vertices with colors other than c(u i,1 ) and c(u i,3 ), while y is only adjacent to vertices with these colors. Hence, the coordinates of x and y differ, so we assume that x and y are not z. Now, let our x and y be in the 4-cycles C(i 1 ) and C(i 2 ), respectively, where }. There are some cases to consider. For the first case, if x = u i1,2 and y = u i2,2 , then clearly they have different coordinates by looking at the colors of their neighbors. For the next case, if x = u i1,2 and y = u i2,1 (or y = u i2,3 without loss of generality), then y is adjacent to z, which is a vertex colored k, but x is not adjacent to any vertex colored k, so we know that their coordinates differ. For the last case, if x = u i1,1 and y = u i2,1 (without loss of generality), then, by definition of c, the colors of u i1,2 and u i2,2 differ, so that the colors of the neighbors of x and y also differ, and hence x and y have different coordinates. Let } (without loss of generality). Thus, we have, say x = u i1,1 , and y = u i2,2 or y = u i2,1 . However, both neighbors of x are colored k, but only one of the neighbors of k is colored k. Hence, their coordinates differ.
Thus, we have shown that c is a locating k-coloring, so that χ L (G) ≤ k.
We now show that χ L (G) > k−1. Suppose that there exists a locating (k−1)coloring on G, which we denote by c . Assume that c (z) = k − 1. We divide all of the 4-cycles into k − 1 types. Type a consists of all 4-cycles C(i) = zu i,1 u i,2 u i,3 z with c (u i,2 ) = a. Observe that if there exist two 4-cycles of type k −1, say C(i) and C(j), where c (u i,1 ) = c (u j,1 ) without loss of generality, then the coordinates of u i,1 and u j,1 must be the same since both are adjacent only to two vertices colored k − 1 and d( ,1 , x) for each vertex x that is not u i,1 , u i,2 , u i,3 , u j,1 , u j,2 , u j,3 . Hence, since u i,1 and u j,1 must not be colored k − 1, there are at most k−2 number of 4-cycles of type other than k − 1. Thus, by combining with the previous paragraph, there are at most e k−1 number of 4-cycles in G. This contradicts the assumption on t. We conclude that χ L (G) > k − 1, so that χ L (G) = k.

2.3.
Upper bound for arbitrary subdivision of F r t . We now study the upper bound for arbitrary subdivision of F r t . It is known that χ L (G) ≤ 1 + χ L (H) if G is a subdivision of a graph H on one edge. For H = F r t , this bound is strengthened.
Observe that two adjacent vertices in G lie in a C(i). By definition of c, those two vertices have different colors. Thus, c is a k-coloring.
Next, to show that c is locating coloring, let z 1 and z 2 be two different vertices having the same color in G, that is c(z 1 ) = c(z 2 ) = a 0 . If one of z 1 or z 2 is z, say z 1 , then we know that, by definition of c and the fact that t ≥ 2, z 1 is adjacent to at least 4 vertices which are two vertices in the cycle where z 2 belongs and two other vertices in another cycle, and that three of these four vertices have different colors. However, z 2 is only adjacent to at most two vertices with different colors. Hence, the coordinates of z 1 and z 2 differ. Now, let z 1 and z 2 be vertices other than z. Assume that both are in different cycles, say C(i 1 ) and C(i 2 ), respectively. If a 0 = k, then C(i 1 ) and C(i 2 ) are cycles of even length by definition of c. Again, by definition of c, z 1 and z 2 are vertices that have their distances to z the greatest in the cycles containing them, so that z 1 is adjacent to two vertices colored a i1 and b i1 , and z 2 is adjacent to two vertices colored a i2 and b i2 , but u i1 = u i2 . Thus, the coordinates of z 1 and z 2 are different. If a 0 = k, then, since no cycle is a 4-cycle and each of z 1 and z 2 has a neighbor z 1 and z 2 , respectively, that c(z 1 ) = c(z 2 ) by definition, the coordinates of z 1 and z 2 are different. Now, let z 1 and z 2 be in the same cycle C(i), and both are not z. By definition of c, we have a 0 = k. Again by definition of c and the fact that t ≥ 2, there exists a vertex z colored a 0 outside of C(i) and no vertex in C(i) is colored a 0 . By the numbering of C(i), we have d(z 1 , z ) = d(z 1 , z) + d(z, z ) = d(z 2 , z) + d(z, z ) = d(z 2 , z ). Hence, the coordinates of z 1 and z 2 differ. Thus, we have shown that c is a locating coloring and that χ L (G) ≤ k.
For the next case, suppose that there are q number of 4-cycles in G. We show that χ L (G) ≤ k. Write q = (k − 1)m + r where r and m are unique integers satisfying 0 ≤ r < k − 1 and m ≥ 0 by the division algorithm. Let the 4-cycles be denoted by Q 1 , Q 2 , . . . , Q q . Consider the complete graph H on the set [k − 1].
Assume that k is odd. We must have m ≤ k−1 2 − 1, otherwise we would have q > k−1 2 , which is a contradiction. Since k − 1 is even, by decomposing H to obtain its Hamiltonian cycles and its 1-factors, there exist subgraphs a Hamiltonian cycle and H i and H j are edge-disjoint subgraphs for each different i and j, and that E i is a complete graph on two vertices and E i and E j are edge-disjoint subgraphs for each different i and j.
For the case r > k−1 2 +1, we have m < k−1 2 −1, so that there exist a Hamiltonian cycle H m+1 that have not yet been associated with the 4-cycles on the previous paragraph. We define the coloring c on the 4-cycles Q (k−1)m+1 , Q (k−1)m+2 , . . ., Q (k−1)m+r associated with the subgraph H m+1 . Similar to the previous paragraph, by changing the role of p with m + 1 and j ∈ [r], and when j = r, we set c(v m+1,j,2 ) := k, we see that adjacent vertices on these cycles have different colors.
Note that for the case that k is even, we obtain Hamilton cycles H 1 , H 2 , . . ., H k−2 2 of H. Again, this time we must have m ≤ k−2 2 . The coloring is done by using the similar way to the case that k is odd, except that the case r > k−1 2 + 1 is replaced with the case r = 0, and the case r ≤ k−1 2 is not needed. Next, color the remaining t − q cycles by using the similar coloring used to color cycles before there was any 4-cycle, by noting that the pair {a i , b i } that is used is the label of two adjacent vertices that have not been used to color the 4-cycles on the above decomposition. Observe that there are exactly t − q pairs of such labels. Thus, we have shown that c is a k-coloring.
We show that c is indeed a locating coloring. Let x and y be two different vertices in G with c(x) = c(y). The cases for x and y that must be considered are: (1) One of them is z (2) Both are not z and are in the same 4-cycle (3) Both are not z and are in the same cycle that is not a 4-cycle (4) Both are not z, x is in a 4-cycle, and y is in a cycle that is not a 4-cycle (5) Both are not z and are in different 4-cycles (6) Both are not z and are in different cycles, but these cycles are not 4-cycles.
The first four cases are easily verified. For the fifth case, if both x and y are colored k, then, since both are not z, their neighbors have to be only two vertices that have the color pair from some labels E i and E j , respectively, that are edgedisjoint subgraphs. Hence, the colors of the neighbors of x and y differ. If both are not colored k, then the colors of the neighbors of x and y also differ since each two 4-cycles have their vertices colored based on the labels of the subgraphs of H that are edge-disjoint subgraphs and since x and y are different vertices. In fact, if the colors of the neighbors of x and y are only k, then, since they belong to different 4-cycles, the colors in both of these cycles must be based on E i and E j that are edge-disjoint subgraphs. This is impossible if x and y are different vertices.
For the sixth case, the same argument also applies by observing the possibilities of the position of x and y and the labels used. Thus, we have shown that c is a locating k-coloring so that χ L (G) ≤ k.
Lastly, for the case t < k−1 2 , the (t + 1)-th, (t + 2)-th, . . ., and so on that have been colored from the coloring on the case t = k−1 2 before is removed so that there are t < k−1 2 cycles remaining, and the above cases can be verified again the similar way. Thus, the theorem is proved.