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Extremal problems on detectable colorings of trees
Discrete Mathematics 308 (2008) 1951 – 1961 www.elsevier.com/locate/disc
Extremal problems on detectable colorings of trees Henry Escuadro, Ping Zhang Department of Mathematics, Western Michigan University, Kalamazoo, MI 49008, USA Received 9 May 2006; received in revised form 11 April 2007; accepted 25 April 2007 Available online 6 May 2007
Abstract Let G be a connected graph of order 3 or more and let c : E(G) → {1, 2, . . . , k} be a coloring of the edges of G (where adjacent edges may be colored the same). For each vertex v of G, the color code of v is the k-tuple c(v) = (a1 , a2 , . . . , ak ), where ai is the number of edges incident with v that are colored i (1 i k). The coloring c is called detectable if distinct vertices have distinct color codes; while the detection number det(G) of G is the minimum positive integer k for which G has a detectable k-coloring. For each integer n 3, let DT (n) be the maximum detection number among all trees of order n and dT (n) the minimum detection number among all trees of order n. The numbers DT (n) and dT (n) are determined for all integers n 3. Furthermore, it is shown that for integers k 2 and n 3, there exists a tree T of order n having det(T ) = k if and only if dT (n) k DT (n). © 2007 Elsevier B.V. All rights reserved. MSC: 05C05; 05C15; 05C70 Keywords: Detectable coloring; Detection number
1. Introduction A fundamental problem in graph theory concerns finding means to distinguish the vertices of a connected graph from one another. One of the earlier methods introduced to distinguish the vertices of a graph is due to Sumner [14] and Entringer and Gassman [7]. In particular, if the equality of the open neighborhoods of every two vertices of a graph G implies that the vertices are the same, then the vertices of G are uniquely determined by their open neighborhoods. If the automorphism group Aut(G) of a graph G is the identity group, then all vertices of G are discernible. On the other hand, if Aut(G) is not the identity group, then G contains at least one nontrivial orbit. The vertices that belong to a common orbit are indistinguishable. Necessarily, vertices that belong to the same orbit have the same degree. Therefore, the vertices of a graph containing a nonidentity automorphism cannot be distinguished by its automorphism group—at least not by its automorphism group alone. Erwin and Harary [8] introduced the idea of selecting a subset S of the vertex set of a graph G such that the subgroup of Aut(G) that fixes every vertex of S is the identity group. Albertson and Collins [2] and Harary [10] introduced the notion of coloring the vertices of G in such a way that the subgroup of color-preserving automorphisms of Aut(G) is the identity group. In these ways, the vertices of a graph G can be distinguished from one another with the aid of certain automorphisms of G.
E-mail address: [email protected] (P. Zhang). 0012-365X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2007.04.048
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H. Escuadro, P. Zhang / Discrete Mathematics 308 (2008) 1951 – 1961
A different way to distinguish the vertices of a connected graph G from one another was introduced by Harary and Melter [11] and Slater [13]. They suggested finding an ordered set W of vertices of G, say W = {w1 , w2 , . . . , wk }, and assigning to each vertex v the ordered k-tuple cW (v) = (a1 , a2 , . . . , ak ), where ai = d(v, wi ) is the distance between v and wi for 1i k. The ordered k-tuple cW (v) is sometimes called the distance code of v. If distinct vertices of G have distinct distance codes, then the vertices of G are distinguishable. In 1985 Harary and Plantholt [12] introduced the idea of distinguishing the vertices of a graph G by assigning colors to the edges of G in such a way that for every two vertices of G, one of the vertices is incident with an edge assigned one of these colors that the other vertex is not. They referred to the minimum number of colors needed to accomplish this as the point-distinguishing chromatic index of G. This manner of distinguishing the vertices of a graph was then studied by Aigner et al. [1] in 1992 and more recently by others as well. There is yet another manner in which differences among the vertices of a connected graph G can be detected that combines many of the features of the foregoing methods. Let G be a connected graph of order n3 and let c : E(G) → {1, 2, . . . , k} be a coloring of the edges of G for some positive integer k (where adjacent edges may be colored the same). The color code of a vertex v of G is the ordered k-tuple c(v) = (a1 , a2 , . . . , ak ) (or simply, c(v) = a1 a2 . . . ak ), where ai is the number of edges incident with v that are colored i for 1 i k. Therefore, k
ai = degG v.
i=1
The coloring c is called detectable if distinct vertices have distinct color codes; that is, for every two vertices of G, there exists a color such that the number of incident edges with that color is different for these two vertices. Thus a detectable coloring of a graph G is a special kind of edge coloring. The detection number det(G) of G is the minimum positive integer k for which G has a detectable k-coloring. Such a coloring is called a minimum detectable coloring. Detectable colorings originated with the work of Burris [3,4] and was studied further in [5,9]. A well-known result from graph theory concerning the degrees of the vertices of a graph is the following. Theorem A. Every nontrivial graph contains at least two vertices having the same degree. One consequence of Theorem A is that the vertices of a nontrivial connected graph cannot be distinguished by their degrees alone. Thus if all edges of a nontrivial graph are assigned the same color, then any two vertices of the same degree will have the same color code. Therefore, the detection number of every connected graph of order 3 or more is at least 2. On the other hand, if all edges of a connected graph G of order at least 3 are assigned distinct colors, then the color codes of the vertices of G are distinct and so this coloring is detectable. Therefore, the detection number is always defined for a connected graph of order at least 3. Furthermore, if G is a connected graph of order n3, then 2 det(G) n. Theorem B below was established in [5], while Theorem C was established in [1,3–5]. Theorem B. A pair k, n of positive integers is realizable as the detection number and the order of some nontrivial connected graph if and only if k = n = 3 or 2 k n − 1, where k = n = 3 only occurs for the complete graph K3 of order 3. Theorem C. If c is a detectable k-coloring of a connected graph G of order at least 3, then G contains at most r+k−1 r vertices of degree r. The detection numbers of cycles, complete graphs, and complete bipartite graphs have been determined and detectable colorings of connected r-regular graphs and trees have been studied as well (see [1,3–5,9]). It is our goal to study some extremal problems concerning detection numbers of trees. In a given tree of order n 3, let ni denote the number of vertices of degree i. In every nontrivial tree, the following equality is known to hold
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(see [6, p. 93], for example): n1 = 2 + n3 + 2n4 + 3n5 + 4n6 + · · · .
(1)
In [4] it was shown that for each tree T of order at least 3, √ det(T ) max{n1 , 4.62 n2 , 8} + 1. By Theorem C, if T is a tree with detection number k, then T contains at most k end-vertices and at most n2 k(k+1) 2 .
k(k+1) 2
of degree 2. Thus n1 k and Solving n2 = for k, we obtain k = order at least 3 with n1 end-vertices and n2 vertices of degree 2, then √ −1 + 8n2 + 1 . det(T ) max n1 , 2
√ −1+ 8n2 +1 . 2
k(k+1) 2
vertices
Thus if T is a tree of
Our goal is to determine how large and how small the detection number of a tree of order n can be. For each integer n 3, let DT (n) denote the maximum detection number among all trees of order n and dT (n) the minimum detection number among all trees of order n. That is, if Tn is the set of all trees of order n, then DT (n) = max{det(T ) : T ∈ Tn }, dT (n) = min{det(T ) : T ∈ Tn }. It is easy to see that the star K1,n−1 of order n 3 has detection number n − 1. Therefore, the following is a consequence of Theorem B. Observation 1.1. For each integer n3, DT (n) = n − 1. 2. Minimum detection numbers of trees According to Theorem C, every tree of order n3 having detection number k contains at most k end-vertices and at most k(k+1) vertices of degree 2. It then follows by (1) that 2 n k +
k 2 + 5k − 4 k(k + 1) + (k − 2) = . 2 2
Furthermore, if T is a tree of order n = k +5k−4 with det(T ) = k, then T must contain exactly k end-vertices, exactly 2 k(k+1) vertices of degree 2, and exactly k − 2 vertices of degree 3. We will use the following well-known results (see 2 [6, p. 203–206], for example) in the proof of Theorem 2.1. 2
Theorem D. For each positive integer k, the complete graph K2k can be factored into k − 1 Hamiltonian cycles and a 1-factor. Theorem E. For each positive integer k, the complete graph K2k+1 is Hamiltonian factorable (into k Hamiltonian cycles). We now determine dT (n) for the values of n mentioned above. Theorem 2.1. Let k 2. If n = Proof. Let n = k 2 +5k−4 2
k 2 +5k−4 . 2
k 2 +5k−4 , 2
then dT (n) = k.
First, we show that dT (n)k. Assume, to the contrary, that there exists a tree T of order
such that det(T )k − 1. By Theorem C, T has at most k − 1 end-vertices and at most
k(k−1) 2
vertices of
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T5:
e4,0 e3,0
e2,0
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Fig. 1. A labeling of T5 in Case 1.
degree 2. Therefore, T contains at least k(k − 1) k 2 + 5k − 4 = 2k − 1 − (k − 1) − 2 2 vertices of degree 3 or more. It then follows by (1) that T contains at least 2k + 1 end-vertices, which is impossible. 2 Next, we show that dT (n)k. It suffices to construct a tree of order k +5k−4 with detection number k. We construct 2 having detection number k such that Tk contains exactly k end-vertices, exactly k 2+k a tree Tk of order n = k +5k−4 2 vertices of degree 2, and exactly k − 2 vertices of degree 3. We consider two cases, according to whether k is odd or k is even. Case 1. k is odd. Then k = 2 + 1 for some positive integer . For each integer i with 0 i k − 1, let 2
2
Qi : ui,1 , ui,2 , . . . , ui,+2 be a copy of the path P+2 of order + 2 and let Q : v1 , v2 , . . . , vk−2 be a path of order k − 2. The tree Tk is obtained from Q and Qi (0 i k − 1) by adding the edges u0,1 v1 , uk−1,1 vk−2 2 and ui,1 vi (1i k − 2). Then the order of Tk is k +5k−4 . Since Tk has k end-vertices, det(Tk ) k. It remains to show 2 that det(Tk )k. For each pair r, s of integers with 0 r k − 1 and 1 s + 1, let er,s = ur,s ur,s+1 be the edge joining the vertices ur,s and ur,s+1 , for each integer t with 1 t k − 2, let et,0 = ut,1 vt be the edge joining the vertices ut,1 and vt , and let e0,0 = u0,1 v1 and ek−1,0 = uk−1,1 vk−2 . Such a labeling is shown for T5 (so k = 5 and hence = 2) in Fig. 1. Consider the complete graph K2+1 with V (K2+1 ) = {1, 2, . . . , 2 + 1}. By Theorem E, K2+1 can be factored into Hamiltonian cycles, say H1 , H2 , . . . , H . For each integer i with 1 i , suppose that Hi : 1 = ai,1 , ai,2 , . . . , ai,2+1 , 1, where ai,j (1j 2 + 1) is the jth vertex of Hi . We may assume, without loss of generality, that H1 : 1, 2, . . . , 2, 2 + 1, 1.
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Fig. 2. A detectable 5-coloring of T5 in Case 1.
Observe that for each integer i with 1i , the factor Hi induces a permutation i of V (K2+1 ) = {1, 2, . . . , 2 + 1} where i = (ai,1 ai,2 . . . ai,2+1 ). In particular, 1 = (1 2 . . . 2 + 1) and in general i (ai,j ) = ai,j +1 if 1 j 2 and i (ai,2+1 ) = ai,1 for each integer i with 1 i . We now define a coloring c : E(Tk ) → V (K2+1 ) of the edges of Tk . For each pair r, s of integers with 0 r k − 1 and 0 s + 1, define ⎧ k if r = 0 and s = 0, 1, ⎪ ⎨ c(er,s ) = r if 1r k − 1 and s = 0, 1, ⎪ ⎩ s−1 (c(er,s−1 )) if 0 r k − 1 and 2 s + 1 and define c(vi vi+1 ) = 1 for 1i k − 3. For example, if k = 5, then K5 can be factored into the two Hamiltonian cycles H1 : 1, 2, 3, 4, 5, 1 and H2 : 1, 3, 5, 2, 4, 1, where their corresponding permutations are 1 = (1 2 3 4 5) and 2 = (1 3 5 2 4), respectively. Then the coloring c is illustrated in Fig. 2 for the tree T5 . Observe that: (1) If 0 r = t k − 1 and 0 s + 1, then c(er,s ) = c(et,s ). (2) If e and f are two different edges of Tk each of which is incident to an end-vertex, then c(e) = c(f ). It follows that the end-vertices of Tk have distinct color codes. (3) The color codes of the vertices of degree 3 of Tk , namely, v1 , v2 , . . . , vk−2 are also distinct since c(v1 ) = (2, 0, . . . , 0, 1), c(vk−2 )=(1, 0, . . . , 0, 1, 1, 0) and for each integer i with 2 i k −3, c(vi )=(2, 0, . . . , 0, 1, 0, . . . , 0) where 1 occurs in the ith position. (4) Every vertex of degree 2 of Tk is incident with two edges having a unique pair of colors. Therefore, c is a detectable k-coloring of Tk and so det(Tk ) k. Therefore, det(Tk ) = k. Case 2. k is even. Then k = 2 for some positive integer . Let T2 be the path P5 : v1 , v2 , v3 , v4 , v5 . The edge coloring c defined by c(v1 v2 ) = c(v2 v3 ) = 1 and c(v3 v4 ) = c(v4 v5 ) = 2 is a detectable 2-coloring of T2 . Thus we may assume that k 4. For each integer i with 0 i 2k − 1, let Qi : ui,1 , ui,2 , . . . , ui,+2 be a copy of the path P+2 of order + 2 and for each integer i with 2k i k − 1, let Qi : ui,1 , ui,2 , . . . , ui,+1 be a copy of the path P+1 of order + 1. Also, let Q : v1 , v2 , . . . , vk−2
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Fig. 3. The tree T8 in Case 2.
be a path of order k − 2. The tree Tk is obtained from Q and Qi (0 i k − 1) by adding the edges u0,1 v1 , uk−1,1 vk−2 , 2 and ui,1 vi (1i k − 2). Then the order of Tk is k +5k−4 . Since Tk has k end-vertices, det(Tk ) k. It remains to show 2 that det(Tk )k. For each pair r, s of integers with 0 r k − 1 and 1s , let er,s = ur,s ur,s+1 be the edge joining the vertices ur,s and ur,s+1 , for each integer t with 1t k − 2, let et,0 = ut,1 vt be the edge joining the vertices ut,1 and vt , let e0,0 = u0,1 v1 and ek−1,0 = uk−1,1 vk−2 , and for each integer t with 0 t 2k − 1, let et,+1 = ut,+1 ut,+2 be the edge joining the vertices ut,+1 and ut,+2 . Such a labeling is shown for T8 (so k = 8 and hence = 4) in Fig. 3. Consider the complete graph K2 with V (K2 ) = {1, 2, . . . , 2}. By Theorem D, K2 can be factored into − 1 Hamiltonian cycles, say H1 , H2 , . . . , H−1 , and a 1-factor F. For each integer i with 1 i − 1, suppose that Hi : 1 = ai,1 , ai,2 , . . . , ai,2 , 1, where ai,j (1j 2) is the jth vertex of Hi . We may assume, without loss of generality, that H1 : 1, 2, . . . , 2 − 1, 2, 1. Observe that for each integer i with 1i − 1, the factor Hi induces a permutation i of V (K2 ) = {1, 2, . . . , 2} where i = (ai,1 ai,2 . . . ai,2 ). In particular, 1 = (1 2 . . . 2) and in general i (ai,j ) = ai,j +1 if 1 j 2 − 1 and i (ai,2 ) = ai,1 for each integer i with 1 i − 1. Note that the 1-factor F induces the permutation of V (K2 ) = {1, 2, . . . , 2} given by = (x1 y1 )(x2 y2 ) · · · (xk/2 yk/2 ), where xi and yi are neighbors in F for i = 1, 2, . . . , 2k .
H. Escuadro, P. Zhang / Discrete Mathematics 308 (2008) 1951 – 1961
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Fig. 4. A detectable 8-coloring of T8 in Case 2.
We now define a coloring c : E(Tk ) → V (K2 ) of the edges of Tk . For integers r and s where 0 r k − 1 and 0 s + 1, define ⎧ if r = 0, k − 1 and s = 0, yk/2 ⎪ ⎪ ⎪ ⎪ ⎪ k ⎪ ⎪ ⎪ yr if 1 r − 1 and s = 0, ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ k ⎪ ⎪ ⎪ yr−k/2+1 if r k − 2 and s = 0, ⎪ ⎪ 2 ⎪ ⎪ ⎨ if r = 0 and s = 1, c(er,s ) = xk/2 ⎪ ⎪ k ⎪ ⎪ ⎪ xr if 1 r − 1 and s = 1, ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ k ⎪ ⎪ s−2 (c(er,s−1 )) if 0 r − 1 and 2 s + 1, ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ k ⎩ (c(e r k − 1 and 1 s , s−1 r,s−1 )) if 2 where 0 is the identity permutation, and ⎧ k ⎪ if 1 i − 1, ⎨ xi 2 c(vi vi+1 ) = ⎪ k ⎩ yi−k/2+1 if i k − 3. 2 For example, if k = 8, then K8 can be factored into the three Hamiltonian cycles H1 : 1, 2, 3, 4, 5, 6, 7, 8, 1,
H2 : 1, 3, 5, 2, 7, 4, 8, 6, 1
and
H3 : 1, 5, 7, 3, 8, 2, 6, 4, 1,
where their corresponding permutations are 1 = (1 2 3 4 5 6 7 8),
2 = (1 3 5 2 7 4 8 6),
and
3 = (1 5 7 3 8 2 6 4),
respectively, and a 1-factor F with E(F ) = {17, 24, 36, 58}. The permutation (which corresponds to the 1-factor F) is then given by = (1 7)(2 4)(3 6)(5 8). The coloring c is illustrated in Fig. 4 for the tree T8 . Observe that: (1) If Ai ={c(e0 , i), c(e1 , i), . . . , c(ek/2−1,i ), c(ek/2,i−1 ), . . . , c(ek−1,i−1 )} for i=1, 2, . . . , +1, then Ai =V (K2 )= {1, 2, . . . , 2}. In particular, A1 = {x1 , x2 , . . . , xk/2 , y1 , y2 , . . . , yk/2 }.
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(2) The set A+1 corresponds to the set of colors used for the edges incident to the end-vertices of Tk . It follows that the color codes of the end-vertices are distinct. (3) The color codes of the vertices of degree 3 of Tk , namely, v1 , v2 , . . . , vk−2 , are also distinct since (i) c(v1 ) consists of exactly three 1’s and (2 − 3) 0’s with two of these three 1’s occurring in the y1 th and yk/2 th positions; (ii) c(vi ) consists of exactly three 1’s and (2 − 3) 0’s with vi being the only vertex having 1 in both the xi−1 th and xi th positions for 1i k/2 − 1; (iii) c(vi ) consists of one 1, one 2 and (2 − 3) 0’s with vi being the only vertex having 2 in the yi−k/2+1 th position for 2k i k − 3; and (iv) c(vk−2 ) consists of exactly three 1’s and (2 − 3) 0’s with two of these three 1’s occurring in the yk/2−1 th and yk/2 th positions. (4) Every vertex of degree 2 of Tk is incident with two edges having a unique pair of colors. Therefore, c is a detectable k-coloring of Tk and so det(Tk ) k. Hence det(Tk ) = k.
If T is a tree of order n with det(T ) = 2, then n 5 by Theorem C and (1). It is easy to see that det(Pn ) = 2 for 3 n 5. Thus dT (n) = 2 if 3n 5. With the aid of Theorem 2.1, we now establish the following for k 3. Theorem 2.2. Let k 3. If k 2 + 3k − 6 (k − 1)2 + 5(k − 1) − 4 k 2 + 5k − 4 , = + 1 n 2 2 2 then dT (n) = k. Proof. Let n be an integer such that that there exists a tree T of order n and at most
k 2 −k 2
2 k 2 +3k−6 n k +5k−4 . 2 2
k 2 +3k−6 2
First, we show that dT (n) k. Assume, to the contrary,
such that det(T ) k − 1. By Theorem C, T has at most k − 1 end-vertices
vertices of degree 2. Therefore, T contains at least
n − (k − 1) −
k2 − k k 2 − k k 2 + 3k − 6 − (k − 1) − =k−2 2 2 2
vertices of degree 3 or more. It then follows by (1) that T contains at least k end-vertices, which is impossible. Therefore, dT (n) k. 2 . Assume therefore that n = We next show that dT (n)k. Theorem 2.1 shows that this is true when n = k +5k−4 2 k 2 +5k−4 2
− p where 1p k + 1. We consider two cases according to whether k is odd or even. Case 1. k is odd. Then k = 2 + 1 for some positive integer . There are two subcases, according to whether 1 p k or p = k + 1. 2 − p from the tree Tk described in Theorem 2.1 by Subcase 1.1. 1 p k. Construct the tree T of order k +5k−4 2 suppressing the vertices ui,1 so that the edges ei,0 and ei,1 become the single edge fi where 0 i p − 1. Define an edge coloring c∗ : E(T ) → V (K2+1 ) of T by c(ei,0 ) if e = fi for some i with 0 i p − 1, c∗ (e) = c(e) otherwise. The color codes for the vertices of T are all those of Tk except those (2 + 1)-tuples for which 2 occurs in the kth coordinate and those for which 2 occurs in the ith coordinate for 1 i p − 1. Thus c∗ is a detectable k-coloring of T. Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. 2 . Consider the tree T together with the edge coloring described in Subcase Subcase 1.2. p = k + 1 and so n = k +3k−6 2 1.1 when p = k. We delete the edge v1 v2 , identify the vertices v1 and v2 , and call this new vertex v. This gives us a tree 2 . Observe that the color codes of the vertices of T are those of T except for those of v1 and v2 , T of order n = k +3k−6 2 and that v is the only vertex of T with degree greater than 3. Hence, we have a detectable k-coloring of T . Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. Examples are shown in Fig. 5 for k = 5, n = 20 (where 1 p = 3 k) and k = 5, n = 17 (where p = 6 = k + 1). Case 2. k is even. Then k = 2 for some positive integer 2. There are three subcases, according to whether 1 p 2k , 2k + 1p k, or p = k + 1.
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Fig. 5. Detectable colorings in Case 1.
Subcase 2.1. 1p 2k . Construct the tree T of order k +5k−4 − p from the tree Tk described in Theorem 2.1 by 2 suppressing the vertices ui,2 so that the edges ei,1 and ei,2 become the single edge fi where 0 i p − 1. Define an edge coloring c∗ : E(T ) → V (K2+1 ) of T by c(ei,1 ) if e = fi for some i with 0 i p − 1, c∗ (e) = c(e) otherwise. 2
The color codes for the vertices of T are all those of Tk except those (2)-tuples for which 2 occurs in the xk/2 th coordinate and those for which 2 occurs in the xi th position for 1 i p − 1. Thus c∗ is a detectable k-coloring of T. Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. 2 − p from the tree T in Subcase 2.1 when p = 2k Subcase 2.2. 2k + 1p k. Construct the tree T of order k +5k−4 2 by suppressing the vertices ui,1 so that the edges ei,0 and ei,1 become the single edge fi where 2k i p − 1. Define an edge coloring c : E(T ) → V (K2+1 ) of T by ⎧ ⎨ c∗ (ei,0 ) if e = fi for some i with k i p − 1, 2 c (e) = ⎩ ∗ c (e) otherwise. The color codes for the vertices of T are all those of T except those (2)-tuples for which 2 occurs in the yi−k/2+1 th coordinate for 2k i p − 1. Thus c is a detectable k-coloring of T . Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. 2 . Consider the tree T together with the edge coloring described in Subcase 2.3. p = k + 1. That is, n = k +3k−6 2 Subcase 2.2 when p = k. We delete the edge v1 v2 , identify the vertices v1 and v2 , and call this new vertex v. This gives 2 . Observe that the color codes of the vertices of T are those of T except for those us a tree T of order n = k +3k−6 2 of v1 and v2 , and that v is the only vertex of T with degree greater than 3. Hence, we have a detectable k-coloring of T . Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. Examples are shown in Fig. 6 for (1) k = 8, n = 47 (where 1 p = 3 2k ), (2) k = 8, n = 44 (where 2k + 1 p = 6 k), and (3) k = 8, n = 41 (where p = 9 = k + 1) This completes the proof of Theorem 2.2. By solving for the smallest positive integer n 3 for which n k Proposition 2.3. For each integer n3,
√ −5 + 8n + 41 . dT (n) = 2
2 +5k−4
2
, we obtain the following result.
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By Proposition 2.3, dT (n) ≈ 2n. We are now able to determine all pairs k, n of integers for which there exists a tree T of order n having detection number k. Theorem 2.4. Let k 2 and n 3 be integers. Then there exists a tree T of order n such that det(T ) = k if and only if
√ −5 + 8n + 41 dT (n) = (2) k n − 1 = DT (n). 2 Proof. We have seen that if T is a tree of order n such that det(T )=k, then (2) holds. It remains to verify the converse. For each integer i with i = 0, 1, . . . , n − dT (n) − 1, we construct a tree Gi such that Gi has order n and det(Gi ) = dT (n) + i. Let G0 be the tree of order n described in the proof of Theorem 2.2 together with the detectable dT (n)-coloring, which we denote by c0 here. Then det(G0 ) = dT (n) and G0 has exactly dT (n) end-vertices. Let w0 be a vertex of maximum degree in G0 (there may be more than one choice for w0 ). Since det(G0 ) = n − 1 (as G0 = K1,n−1 ), it follows that there exists a vertex x0 in G0 adjacent to w0 such that degG0 x0 = 1. Construct G1 by deleting the edge w0 x0 , identifying the vertices w0 and x0 , naming the new vertex w1 , introducing a new vertex y1 , and joining y1 to w1 . We note that G1 has order n and has dT (n) + 1 end-vertices. Thus, det(G1 ) dT (n) + 1. We now show that det(G1 ) dT (n) + 1 by giving a detectable (dT (n) + 1)-coloring of G1 . Define c1 : E(G1 ) → {1, 2, . . . , dT (n) + 1} by if e ∈ E(G0 ), c0 (e) c1 (e) = dT (n) + 1 if e = w1 y1 . Then c1 is detectable (dT (n) + 1)-coloring of G1 . This implies that det(G1 ) = dT (n) + 1.
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In general, we construct Gi+1 from Gi and obtain the edge coloring ci+1 from ci , 0 i n − dT (n) − 2, as follows: (1) Let wi be a vertex of maximum degree in Gi . (2) Let xi be a vertex in Gi that is adjacent to wi such that degGi xi = 1. (3) Construct Gi+1 by deleting the edge wi xi , identifying the vertices wi and xi , naming the new vertex wi+1 , introducing a new vertex yi+1 , and joining yi+1 to wi+1 . (4) Define ci+1 : E(Gi+1 ) → {1, 2, . . . , dT (n) + i + 1} by if e ∈ E(Gi ), ci (e) ci+1 (e) = dT (n) + i + 1 if e = wi+1 yi+1 . Observe that for every i = 0, 1, . . . , n − dT (n) − 1: (i) Gi is a tree of order n with dT (n) + i end-vertices; (ii) ci is a detectable (dT (n) + i)-coloring of Gi ; (iii) (i) and (ii) implies that det(Gi ) = dT (n) + i. An example showing the steps in the construction of the trees Gi is given in Fig. 7 for n = 7. This completes the proof. Acknowledgments We are grateful to the referees whose valuable suggestions resulted in an improved paper. References [1] M. Aigner, E. Triesch, Z. Tuza, Irregular assignments and vertex-distinguishing edge colorings of graphs, Combinatorics’ 90, Elsevier Science Pub., New York, 1992 pp. 1–9. [2] M. Albertson, K. Collins, Symmetric breaking in graphs, Electron. J. Combin. 3 (1996) R18. [3] A.C. Burris, On graphs with irregular coloring number 2, Congr. Numer. 100 (1994) 129–140. [4] A.C. Burris, The irregular coloring number of a tree, Discrete Math. 141 (1995) 279–283. [5] G. Chartrand, H. Escuadro, F. Okamoto, P. Zhang, Detectable colorings of graphs, Utilitas Math. 69 (2006) 13–32. [6] G. Chartrand, P. Zhang, Introduction to Graph Theory, McGraw-Hill, Boston, 2005. [7] R.C. Entringer, L.D. Gassman, Line-critical point determining and point distinguishing graphs, Discrete Math. 10 (1974) 43–55. [8] D. Erwin, F. Harary, Destroying automorphisms by fixing nodes, Discrete Math. 306 (2006) 3244–3252. [9] H. Escuadro, F. Okamoto, P. Zhang, On detectable factorizations of cubic graphs, J. Combin. Math. Combin. Comput. 56 (2006) 47–63. [10] F. Harary, Methods of destroying the symmetries of a graph, Bull. Malaysian Math. Sci. Soc. 24 (2001) 183–191. [11] F. Harary, R.A. Melter, On the metric dimension of a graph, Ars Combin. 2 (1976) 191–195. [12] F. Harary, M. Plantholt, The point-distinguishing chromatic index, Graphs and Applications, Wiley, New York, 1985 pp. 147–162. [13] P.J. Slater, Leaves of trees, Congr. Numer. 14 (1975) 549–559. [14] D.P. Sumner, Point determination in graphs, Discrete Math. 5 (1973) 179–187.
Extremal problems on detectable colorings of trees Henry Escuadro, Ping Zhang Department of Mathematics, Western Michigan University, Kalamazoo, MI 49008, USA Received 9 May 2006; received in revised form 11 April 2007; accepted 25 April 2007 Available online 6 May 2007
Abstract Let G be a connected graph of order 3 or more and let c : E(G) → {1, 2, . . . , k} be a coloring of the edges of G (where adjacent edges may be colored the same). For each vertex v of G, the color code of v is the k-tuple c(v) = (a1 , a2 , . . . , ak ), where ai is the number of edges incident with v that are colored i (1 i k). The coloring c is called detectable if distinct vertices have distinct color codes; while the detection number det(G) of G is the minimum positive integer k for which G has a detectable k-coloring. For each integer n 3, let DT (n) be the maximum detection number among all trees of order n and dT (n) the minimum detection number among all trees of order n. The numbers DT (n) and dT (n) are determined for all integers n 3. Furthermore, it is shown that for integers k 2 and n 3, there exists a tree T of order n having det(T ) = k if and only if dT (n) k DT (n). © 2007 Elsevier B.V. All rights reserved. MSC: 05C05; 05C15; 05C70 Keywords: Detectable coloring; Detection number
1. Introduction A fundamental problem in graph theory concerns finding means to distinguish the vertices of a connected graph from one another. One of the earlier methods introduced to distinguish the vertices of a graph is due to Sumner [14] and Entringer and Gassman [7]. In particular, if the equality of the open neighborhoods of every two vertices of a graph G implies that the vertices are the same, then the vertices of G are uniquely determined by their open neighborhoods. If the automorphism group Aut(G) of a graph G is the identity group, then all vertices of G are discernible. On the other hand, if Aut(G) is not the identity group, then G contains at least one nontrivial orbit. The vertices that belong to a common orbit are indistinguishable. Necessarily, vertices that belong to the same orbit have the same degree. Therefore, the vertices of a graph containing a nonidentity automorphism cannot be distinguished by its automorphism group—at least not by its automorphism group alone. Erwin and Harary [8] introduced the idea of selecting a subset S of the vertex set of a graph G such that the subgroup of Aut(G) that fixes every vertex of S is the identity group. Albertson and Collins [2] and Harary [10] introduced the notion of coloring the vertices of G in such a way that the subgroup of color-preserving automorphisms of Aut(G) is the identity group. In these ways, the vertices of a graph G can be distinguished from one another with the aid of certain automorphisms of G.
E-mail address: [email protected] (P. Zhang). 0012-365X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2007.04.048
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A different way to distinguish the vertices of a connected graph G from one another was introduced by Harary and Melter [11] and Slater [13]. They suggested finding an ordered set W of vertices of G, say W = {w1 , w2 , . . . , wk }, and assigning to each vertex v the ordered k-tuple cW (v) = (a1 , a2 , . . . , ak ), where ai = d(v, wi ) is the distance between v and wi for 1i k. The ordered k-tuple cW (v) is sometimes called the distance code of v. If distinct vertices of G have distinct distance codes, then the vertices of G are distinguishable. In 1985 Harary and Plantholt [12] introduced the idea of distinguishing the vertices of a graph G by assigning colors to the edges of G in such a way that for every two vertices of G, one of the vertices is incident with an edge assigned one of these colors that the other vertex is not. They referred to the minimum number of colors needed to accomplish this as the point-distinguishing chromatic index of G. This manner of distinguishing the vertices of a graph was then studied by Aigner et al. [1] in 1992 and more recently by others as well. There is yet another manner in which differences among the vertices of a connected graph G can be detected that combines many of the features of the foregoing methods. Let G be a connected graph of order n3 and let c : E(G) → {1, 2, . . . , k} be a coloring of the edges of G for some positive integer k (where adjacent edges may be colored the same). The color code of a vertex v of G is the ordered k-tuple c(v) = (a1 , a2 , . . . , ak ) (or simply, c(v) = a1 a2 . . . ak ), where ai is the number of edges incident with v that are colored i for 1 i k. Therefore, k
ai = degG v.
i=1
The coloring c is called detectable if distinct vertices have distinct color codes; that is, for every two vertices of G, there exists a color such that the number of incident edges with that color is different for these two vertices. Thus a detectable coloring of a graph G is a special kind of edge coloring. The detection number det(G) of G is the minimum positive integer k for which G has a detectable k-coloring. Such a coloring is called a minimum detectable coloring. Detectable colorings originated with the work of Burris [3,4] and was studied further in [5,9]. A well-known result from graph theory concerning the degrees of the vertices of a graph is the following. Theorem A. Every nontrivial graph contains at least two vertices having the same degree. One consequence of Theorem A is that the vertices of a nontrivial connected graph cannot be distinguished by their degrees alone. Thus if all edges of a nontrivial graph are assigned the same color, then any two vertices of the same degree will have the same color code. Therefore, the detection number of every connected graph of order 3 or more is at least 2. On the other hand, if all edges of a connected graph G of order at least 3 are assigned distinct colors, then the color codes of the vertices of G are distinct and so this coloring is detectable. Therefore, the detection number is always defined for a connected graph of order at least 3. Furthermore, if G is a connected graph of order n3, then 2 det(G) n. Theorem B below was established in [5], while Theorem C was established in [1,3–5]. Theorem B. A pair k, n of positive integers is realizable as the detection number and the order of some nontrivial connected graph if and only if k = n = 3 or 2 k n − 1, where k = n = 3 only occurs for the complete graph K3 of order 3. Theorem C. If c is a detectable k-coloring of a connected graph G of order at least 3, then G contains at most r+k−1 r vertices of degree r. The detection numbers of cycles, complete graphs, and complete bipartite graphs have been determined and detectable colorings of connected r-regular graphs and trees have been studied as well (see [1,3–5,9]). It is our goal to study some extremal problems concerning detection numbers of trees. In a given tree of order n 3, let ni denote the number of vertices of degree i. In every nontrivial tree, the following equality is known to hold
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(see [6, p. 93], for example): n1 = 2 + n3 + 2n4 + 3n5 + 4n6 + · · · .
(1)
In [4] it was shown that for each tree T of order at least 3, √ det(T ) max{n1 , 4.62 n2 , 8} + 1. By Theorem C, if T is a tree with detection number k, then T contains at most k end-vertices and at most n2 k(k+1) 2 .
k(k+1) 2
of degree 2. Thus n1 k and Solving n2 = for k, we obtain k = order at least 3 with n1 end-vertices and n2 vertices of degree 2, then √ −1 + 8n2 + 1 . det(T ) max n1 , 2
√ −1+ 8n2 +1 . 2
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vertices
Thus if T is a tree of
Our goal is to determine how large and how small the detection number of a tree of order n can be. For each integer n 3, let DT (n) denote the maximum detection number among all trees of order n and dT (n) the minimum detection number among all trees of order n. That is, if Tn is the set of all trees of order n, then DT (n) = max{det(T ) : T ∈ Tn }, dT (n) = min{det(T ) : T ∈ Tn }. It is easy to see that the star K1,n−1 of order n 3 has detection number n − 1. Therefore, the following is a consequence of Theorem B. Observation 1.1. For each integer n3, DT (n) = n − 1. 2. Minimum detection numbers of trees According to Theorem C, every tree of order n3 having detection number k contains at most k end-vertices and at most k(k+1) vertices of degree 2. It then follows by (1) that 2 n k +
k 2 + 5k − 4 k(k + 1) + (k − 2) = . 2 2
Furthermore, if T is a tree of order n = k +5k−4 with det(T ) = k, then T must contain exactly k end-vertices, exactly 2 k(k+1) vertices of degree 2, and exactly k − 2 vertices of degree 3. We will use the following well-known results (see 2 [6, p. 203–206], for example) in the proof of Theorem 2.1. 2
Theorem D. For each positive integer k, the complete graph K2k can be factored into k − 1 Hamiltonian cycles and a 1-factor. Theorem E. For each positive integer k, the complete graph K2k+1 is Hamiltonian factorable (into k Hamiltonian cycles). We now determine dT (n) for the values of n mentioned above. Theorem 2.1. Let k 2. If n = Proof. Let n = k 2 +5k−4 2
k 2 +5k−4 . 2
k 2 +5k−4 , 2
then dT (n) = k.
First, we show that dT (n)k. Assume, to the contrary, that there exists a tree T of order
such that det(T )k − 1. By Theorem C, T has at most k − 1 end-vertices and at most
k(k−1) 2
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degree 2. Therefore, T contains at least k(k − 1) k 2 + 5k − 4 = 2k − 1 − (k − 1) − 2 2 vertices of degree 3 or more. It then follows by (1) that T contains at least 2k + 1 end-vertices, which is impossible. 2 Next, we show that dT (n)k. It suffices to construct a tree of order k +5k−4 with detection number k. We construct 2 having detection number k such that Tk contains exactly k end-vertices, exactly k 2+k a tree Tk of order n = k +5k−4 2 vertices of degree 2, and exactly k − 2 vertices of degree 3. We consider two cases, according to whether k is odd or k is even. Case 1. k is odd. Then k = 2 + 1 for some positive integer . For each integer i with 0 i k − 1, let 2
2
Qi : ui,1 , ui,2 , . . . , ui,+2 be a copy of the path P+2 of order + 2 and let Q : v1 , v2 , . . . , vk−2 be a path of order k − 2. The tree Tk is obtained from Q and Qi (0 i k − 1) by adding the edges u0,1 v1 , uk−1,1 vk−2 2 and ui,1 vi (1i k − 2). Then the order of Tk is k +5k−4 . Since Tk has k end-vertices, det(Tk ) k. It remains to show 2 that det(Tk )k. For each pair r, s of integers with 0 r k − 1 and 1 s + 1, let er,s = ur,s ur,s+1 be the edge joining the vertices ur,s and ur,s+1 , for each integer t with 1 t k − 2, let et,0 = ut,1 vt be the edge joining the vertices ut,1 and vt , and let e0,0 = u0,1 v1 and ek−1,0 = uk−1,1 vk−2 . Such a labeling is shown for T5 (so k = 5 and hence = 2) in Fig. 1. Consider the complete graph K2+1 with V (K2+1 ) = {1, 2, . . . , 2 + 1}. By Theorem E, K2+1 can be factored into Hamiltonian cycles, say H1 , H2 , . . . , H . For each integer i with 1 i , suppose that Hi : 1 = ai,1 , ai,2 , . . . , ai,2+1 , 1, where ai,j (1j 2 + 1) is the jth vertex of Hi . We may assume, without loss of generality, that H1 : 1, 2, . . . , 2, 2 + 1, 1.
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Observe that for each integer i with 1i , the factor Hi induces a permutation i of V (K2+1 ) = {1, 2, . . . , 2 + 1} where i = (ai,1 ai,2 . . . ai,2+1 ). In particular, 1 = (1 2 . . . 2 + 1) and in general i (ai,j ) = ai,j +1 if 1 j 2 and i (ai,2+1 ) = ai,1 for each integer i with 1 i . We now define a coloring c : E(Tk ) → V (K2+1 ) of the edges of Tk . For each pair r, s of integers with 0 r k − 1 and 0 s + 1, define ⎧ k if r = 0 and s = 0, 1, ⎪ ⎨ c(er,s ) = r if 1r k − 1 and s = 0, 1, ⎪ ⎩ s−1 (c(er,s−1 )) if 0 r k − 1 and 2 s + 1 and define c(vi vi+1 ) = 1 for 1i k − 3. For example, if k = 5, then K5 can be factored into the two Hamiltonian cycles H1 : 1, 2, 3, 4, 5, 1 and H2 : 1, 3, 5, 2, 4, 1, where their corresponding permutations are 1 = (1 2 3 4 5) and 2 = (1 3 5 2 4), respectively. Then the coloring c is illustrated in Fig. 2 for the tree T5 . Observe that: (1) If 0 r = t k − 1 and 0 s + 1, then c(er,s ) = c(et,s ). (2) If e and f are two different edges of Tk each of which is incident to an end-vertex, then c(e) = c(f ). It follows that the end-vertices of Tk have distinct color codes. (3) The color codes of the vertices of degree 3 of Tk , namely, v1 , v2 , . . . , vk−2 are also distinct since c(v1 ) = (2, 0, . . . , 0, 1), c(vk−2 )=(1, 0, . . . , 0, 1, 1, 0) and for each integer i with 2 i k −3, c(vi )=(2, 0, . . . , 0, 1, 0, . . . , 0) where 1 occurs in the ith position. (4) Every vertex of degree 2 of Tk is incident with two edges having a unique pair of colors. Therefore, c is a detectable k-coloring of Tk and so det(Tk ) k. Therefore, det(Tk ) = k. Case 2. k is even. Then k = 2 for some positive integer . Let T2 be the path P5 : v1 , v2 , v3 , v4 , v5 . The edge coloring c defined by c(v1 v2 ) = c(v2 v3 ) = 1 and c(v3 v4 ) = c(v4 v5 ) = 2 is a detectable 2-coloring of T2 . Thus we may assume that k 4. For each integer i with 0 i 2k − 1, let Qi : ui,1 , ui,2 , . . . , ui,+2 be a copy of the path P+2 of order + 2 and for each integer i with 2k i k − 1, let Qi : ui,1 , ui,2 , . . . , ui,+1 be a copy of the path P+1 of order + 1. Also, let Q : v1 , v2 , . . . , vk−2
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Fig. 3. The tree T8 in Case 2.
be a path of order k − 2. The tree Tk is obtained from Q and Qi (0 i k − 1) by adding the edges u0,1 v1 , uk−1,1 vk−2 , 2 and ui,1 vi (1i k − 2). Then the order of Tk is k +5k−4 . Since Tk has k end-vertices, det(Tk ) k. It remains to show 2 that det(Tk )k. For each pair r, s of integers with 0 r k − 1 and 1s , let er,s = ur,s ur,s+1 be the edge joining the vertices ur,s and ur,s+1 , for each integer t with 1t k − 2, let et,0 = ut,1 vt be the edge joining the vertices ut,1 and vt , let e0,0 = u0,1 v1 and ek−1,0 = uk−1,1 vk−2 , and for each integer t with 0 t 2k − 1, let et,+1 = ut,+1 ut,+2 be the edge joining the vertices ut,+1 and ut,+2 . Such a labeling is shown for T8 (so k = 8 and hence = 4) in Fig. 3. Consider the complete graph K2 with V (K2 ) = {1, 2, . . . , 2}. By Theorem D, K2 can be factored into − 1 Hamiltonian cycles, say H1 , H2 , . . . , H−1 , and a 1-factor F. For each integer i with 1 i − 1, suppose that Hi : 1 = ai,1 , ai,2 , . . . , ai,2 , 1, where ai,j (1j 2) is the jth vertex of Hi . We may assume, without loss of generality, that H1 : 1, 2, . . . , 2 − 1, 2, 1. Observe that for each integer i with 1i − 1, the factor Hi induces a permutation i of V (K2 ) = {1, 2, . . . , 2} where i = (ai,1 ai,2 . . . ai,2 ). In particular, 1 = (1 2 . . . 2) and in general i (ai,j ) = ai,j +1 if 1 j 2 − 1 and i (ai,2 ) = ai,1 for each integer i with 1 i − 1. Note that the 1-factor F induces the permutation of V (K2 ) = {1, 2, . . . , 2} given by = (x1 y1 )(x2 y2 ) · · · (xk/2 yk/2 ), where xi and yi are neighbors in F for i = 1, 2, . . . , 2k .
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We now define a coloring c : E(Tk ) → V (K2 ) of the edges of Tk . For integers r and s where 0 r k − 1 and 0 s + 1, define ⎧ if r = 0, k − 1 and s = 0, yk/2 ⎪ ⎪ ⎪ ⎪ ⎪ k ⎪ ⎪ ⎪ yr if 1 r − 1 and s = 0, ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ k ⎪ ⎪ ⎪ yr−k/2+1 if r k − 2 and s = 0, ⎪ ⎪ 2 ⎪ ⎪ ⎨ if r = 0 and s = 1, c(er,s ) = xk/2 ⎪ ⎪ k ⎪ ⎪ ⎪ xr if 1 r − 1 and s = 1, ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ k ⎪ ⎪ s−2 (c(er,s−1 )) if 0 r − 1 and 2 s + 1, ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎪ k ⎩ (c(e r k − 1 and 1 s , s−1 r,s−1 )) if 2 where 0 is the identity permutation, and ⎧ k ⎪ if 1 i − 1, ⎨ xi 2 c(vi vi+1 ) = ⎪ k ⎩ yi−k/2+1 if i k − 3. 2 For example, if k = 8, then K8 can be factored into the three Hamiltonian cycles H1 : 1, 2, 3, 4, 5, 6, 7, 8, 1,
H2 : 1, 3, 5, 2, 7, 4, 8, 6, 1
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respectively, and a 1-factor F with E(F ) = {17, 24, 36, 58}. The permutation (which corresponds to the 1-factor F) is then given by = (1 7)(2 4)(3 6)(5 8). The coloring c is illustrated in Fig. 4 for the tree T8 . Observe that: (1) If Ai ={c(e0 , i), c(e1 , i), . . . , c(ek/2−1,i ), c(ek/2,i−1 ), . . . , c(ek−1,i−1 )} for i=1, 2, . . . , +1, then Ai =V (K2 )= {1, 2, . . . , 2}. In particular, A1 = {x1 , x2 , . . . , xk/2 , y1 , y2 , . . . , yk/2 }.
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(2) The set A+1 corresponds to the set of colors used for the edges incident to the end-vertices of Tk . It follows that the color codes of the end-vertices are distinct. (3) The color codes of the vertices of degree 3 of Tk , namely, v1 , v2 , . . . , vk−2 , are also distinct since (i) c(v1 ) consists of exactly three 1’s and (2 − 3) 0’s with two of these three 1’s occurring in the y1 th and yk/2 th positions; (ii) c(vi ) consists of exactly three 1’s and (2 − 3) 0’s with vi being the only vertex having 1 in both the xi−1 th and xi th positions for 1i k/2 − 1; (iii) c(vi ) consists of one 1, one 2 and (2 − 3) 0’s with vi being the only vertex having 2 in the yi−k/2+1 th position for 2k i k − 3; and (iv) c(vk−2 ) consists of exactly three 1’s and (2 − 3) 0’s with two of these three 1’s occurring in the yk/2−1 th and yk/2 th positions. (4) Every vertex of degree 2 of Tk is incident with two edges having a unique pair of colors. Therefore, c is a detectable k-coloring of Tk and so det(Tk ) k. Hence det(Tk ) = k.
If T is a tree of order n with det(T ) = 2, then n 5 by Theorem C and (1). It is easy to see that det(Pn ) = 2 for 3 n 5. Thus dT (n) = 2 if 3n 5. With the aid of Theorem 2.1, we now establish the following for k 3. Theorem 2.2. Let k 3. If k 2 + 3k − 6 (k − 1)2 + 5(k − 1) − 4 k 2 + 5k − 4 , = + 1 n 2 2 2 then dT (n) = k. Proof. Let n be an integer such that that there exists a tree T of order n and at most
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such that det(T ) k − 1. By Theorem C, T has at most k − 1 end-vertices
vertices of degree 2. Therefore, T contains at least
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vertices of degree 3 or more. It then follows by (1) that T contains at least k end-vertices, which is impossible. Therefore, dT (n) k. 2 . Assume therefore that n = We next show that dT (n)k. Theorem 2.1 shows that this is true when n = k +5k−4 2 k 2 +5k−4 2
− p where 1p k + 1. We consider two cases according to whether k is odd or even. Case 1. k is odd. Then k = 2 + 1 for some positive integer . There are two subcases, according to whether 1 p k or p = k + 1. 2 − p from the tree Tk described in Theorem 2.1 by Subcase 1.1. 1 p k. Construct the tree T of order k +5k−4 2 suppressing the vertices ui,1 so that the edges ei,0 and ei,1 become the single edge fi where 0 i p − 1. Define an edge coloring c∗ : E(T ) → V (K2+1 ) of T by c(ei,0 ) if e = fi for some i with 0 i p − 1, c∗ (e) = c(e) otherwise. The color codes for the vertices of T are all those of Tk except those (2 + 1)-tuples for which 2 occurs in the kth coordinate and those for which 2 occurs in the ith coordinate for 1 i p − 1. Thus c∗ is a detectable k-coloring of T. Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. 2 . Consider the tree T together with the edge coloring described in Subcase Subcase 1.2. p = k + 1 and so n = k +3k−6 2 1.1 when p = k. We delete the edge v1 v2 , identify the vertices v1 and v2 , and call this new vertex v. This gives us a tree 2 . Observe that the color codes of the vertices of T are those of T except for those of v1 and v2 , T of order n = k +3k−6 2 and that v is the only vertex of T with degree greater than 3. Hence, we have a detectable k-coloring of T . Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. Examples are shown in Fig. 5 for k = 5, n = 20 (where 1 p = 3 k) and k = 5, n = 17 (where p = 6 = k + 1). Case 2. k is even. Then k = 2 for some positive integer 2. There are three subcases, according to whether 1 p 2k , 2k + 1p k, or p = k + 1.
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Subcase 2.1. 1p 2k . Construct the tree T of order k +5k−4 − p from the tree Tk described in Theorem 2.1 by 2 suppressing the vertices ui,2 so that the edges ei,1 and ei,2 become the single edge fi where 0 i p − 1. Define an edge coloring c∗ : E(T ) → V (K2+1 ) of T by c(ei,1 ) if e = fi for some i with 0 i p − 1, c∗ (e) = c(e) otherwise. 2
The color codes for the vertices of T are all those of Tk except those (2)-tuples for which 2 occurs in the xk/2 th coordinate and those for which 2 occurs in the xi th position for 1 i p − 1. Thus c∗ is a detectable k-coloring of T. Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. 2 − p from the tree T in Subcase 2.1 when p = 2k Subcase 2.2. 2k + 1p k. Construct the tree T of order k +5k−4 2 by suppressing the vertices ui,1 so that the edges ei,0 and ei,1 become the single edge fi where 2k i p − 1. Define an edge coloring c : E(T ) → V (K2+1 ) of T by ⎧ ⎨ c∗ (ei,0 ) if e = fi for some i with k i p − 1, 2 c (e) = ⎩ ∗ c (e) otherwise. The color codes for the vertices of T are all those of T except those (2)-tuples for which 2 occurs in the yi−k/2+1 th coordinate for 2k i p − 1. Thus c is a detectable k-coloring of T . Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. 2 . Consider the tree T together with the edge coloring described in Subcase 2.3. p = k + 1. That is, n = k +3k−6 2 Subcase 2.2 when p = k. We delete the edge v1 v2 , identify the vertices v1 and v2 , and call this new vertex v. This gives 2 . Observe that the color codes of the vertices of T are those of T except for those us a tree T of order n = k +3k−6 2 of v1 and v2 , and that v is the only vertex of T with degree greater than 3. Hence, we have a detectable k-coloring of T . Since T has k end-vertices, it follows that det(T ) = k. Consequently, dT (n) = k. Examples are shown in Fig. 6 for (1) k = 8, n = 47 (where 1 p = 3 2k ), (2) k = 8, n = 44 (where 2k + 1 p = 6 k), and (3) k = 8, n = 41 (where p = 9 = k + 1) This completes the proof of Theorem 2.2. By solving for the smallest positive integer n 3 for which n k Proposition 2.3. For each integer n3,
√ −5 + 8n + 41 . dT (n) = 2
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By Proposition 2.3, dT (n) ≈ 2n. We are now able to determine all pairs k, n of integers for which there exists a tree T of order n having detection number k. Theorem 2.4. Let k 2 and n 3 be integers. Then there exists a tree T of order n such that det(T ) = k if and only if
√ −5 + 8n + 41 dT (n) = (2) k n − 1 = DT (n). 2 Proof. We have seen that if T is a tree of order n such that det(T )=k, then (2) holds. It remains to verify the converse. For each integer i with i = 0, 1, . . . , n − dT (n) − 1, we construct a tree Gi such that Gi has order n and det(Gi ) = dT (n) + i. Let G0 be the tree of order n described in the proof of Theorem 2.2 together with the detectable dT (n)-coloring, which we denote by c0 here. Then det(G0 ) = dT (n) and G0 has exactly dT (n) end-vertices. Let w0 be a vertex of maximum degree in G0 (there may be more than one choice for w0 ). Since det(G0 ) = n − 1 (as G0 = K1,n−1 ), it follows that there exists a vertex x0 in G0 adjacent to w0 such that degG0 x0 = 1. Construct G1 by deleting the edge w0 x0 , identifying the vertices w0 and x0 , naming the new vertex w1 , introducing a new vertex y1 , and joining y1 to w1 . We note that G1 has order n and has dT (n) + 1 end-vertices. Thus, det(G1 ) dT (n) + 1. We now show that det(G1 ) dT (n) + 1 by giving a detectable (dT (n) + 1)-coloring of G1 . Define c1 : E(G1 ) → {1, 2, . . . , dT (n) + 1} by if e ∈ E(G0 ), c0 (e) c1 (e) = dT (n) + 1 if e = w1 y1 . Then c1 is detectable (dT (n) + 1)-coloring of G1 . This implies that det(G1 ) = dT (n) + 1.
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In general, we construct Gi+1 from Gi and obtain the edge coloring ci+1 from ci , 0 i n − dT (n) − 2, as follows: (1) Let wi be a vertex of maximum degree in Gi . (2) Let xi be a vertex in Gi that is adjacent to wi such that degGi xi = 1. (3) Construct Gi+1 by deleting the edge wi xi , identifying the vertices wi and xi , naming the new vertex wi+1 , introducing a new vertex yi+1 , and joining yi+1 to wi+1 . (4) Define ci+1 : E(Gi+1 ) → {1, 2, . . . , dT (n) + i + 1} by if e ∈ E(Gi ), ci (e) ci+1 (e) = dT (n) + i + 1 if e = wi+1 yi+1 . Observe that for every i = 0, 1, . . . , n − dT (n) − 1: (i) Gi is a tree of order n with dT (n) + i end-vertices; (ii) ci is a detectable (dT (n) + i)-coloring of Gi ; (iii) (i) and (ii) implies that det(Gi ) = dT (n) + i. An example showing the steps in the construction of the trees Gi is given in Fig. 7 for n = 7. This completes the proof. Acknowledgments We are grateful to the referees whose valuable suggestions resulted in an improved paper. References [1] M. Aigner, E. Triesch, Z. Tuza, Irregular assignments and vertex-distinguishing edge colorings of graphs, Combinatorics’ 90, Elsevier Science Pub., New York, 1992 pp. 1–9. [2] M. Albertson, K. Collins, Symmetric breaking in graphs, Electron. J. Combin. 3 (1996) R18. [3] A.C. Burris, On graphs with irregular coloring number 2, Congr. Numer. 100 (1994) 129–140. [4] A.C. Burris, The irregular coloring number of a tree, Discrete Math. 141 (1995) 279–283. [5] G. Chartrand, H. Escuadro, F. Okamoto, P. Zhang, Detectable colorings of graphs, Utilitas Math. 69 (2006) 13–32. [6] G. Chartrand, P. Zhang, Introduction to Graph Theory, McGraw-Hill, Boston, 2005. [7] R.C. Entringer, L.D. Gassman, Line-critical point determining and point distinguishing graphs, Discrete Math. 10 (1974) 43–55. [8] D. Erwin, F. Harary, Destroying automorphisms by fixing nodes, Discrete Math. 306 (2006) 3244–3252. [9] H. Escuadro, F. Okamoto, P. Zhang, On detectable factorizations of cubic graphs, J. Combin. Math. Combin. Comput. 56 (2006) 47–63. [10] F. Harary, Methods of destroying the symmetries of a graph, Bull. Malaysian Math. Sci. Soc. 24 (2001) 183–191. [11] F. Harary, R.A. Melter, On the metric dimension of a graph, Ars Combin. 2 (1976) 191–195. [12] F. Harary, M. Plantholt, The point-distinguishing chromatic index, Graphs and Applications, Wiley, New York, 1985 pp. 147–162. [13] P.J. Slater, Leaves of trees, Congr. Numer. 14 (1975) 549–559. [14] D.P. Sumner, Point determination in graphs, Discrete Math. 5 (1973) 179–187.