Rainbow Graph Designs Hung-Lin Fu (傅 恒 霖) Department of Applied Mathematics National Chiao Tung University Hsin Chu, Taiwan 30010

Goals The subject we study is to find a special graph design of a properly edge-colored graph; the coloring may be prescribed or arbitrarily given. A good place to start is considering the complete graphs which are properly edge-colored with minimum number of colors, their chromatic indices. If the order of the complete graph is even, this coloring is also known as a 1-factorization coloring.

Preliminaries A (proper) k-edge-coloring of a graph G is a mapping from E(G) into {1, . . ., k } such that incident edges of G receive (distinct) colors. In this talk, all colorings are proper, i. e. incident edges receive distinct colors. A proper 3-edge-coloring of K4

Chromatic Indices It is well-known that the complete graph of order n, Kn, is of Class 1 if and only if n is even. That is, ’(Kn) = n – 1 if n is even and ’(Kn) = n if n is odd. A graph G is of Class 1 if its chromatic index is equal to ∆(G) its maximum degree.

Rainbow subgraph A subgraph H in an edge-colored graph G is a rainbow subgraph of G if no two edges in H have the same color. (A rainbow graph is also known as a multicolored, hetero-chromatic or poly-chromatic graph.) 1 2 3 4 5 1 2 3 4 5 A rainbow 5-cycle in a properly edge-colored K5

Motivation Theorem (Woolbright and Fu, 1998) In any (2m-1)-edge-colored K2m where m > 2, there exists a rainbow 1-factor. This result has been further improved to complete uniform hypergraphs by Saad El-Zanati et al. later. It remains unknown for finding two or more edge-disjoint rainbow 1-factors. Quite recently, Paul Horn and his student claim that they can find many edge-disjoint rainbow 1-factors if m is large enough (by using probabilistic method).

More Rainbow 1-factors If we are allowed to design an edge-coloring for K2m, then we are able to find (2m – 1) rainbow 1-factors inside K2m. This can be done by simply constructing a room square of order 2m, see example of order 8.

Room square of order 8 7 edge-disjoint rainbow matchings! 1,8 5,7 3,4 2,6 3,7 2,8 6,1 4,5 5,6 4,1 3,8 7,2 6,7 5,2 4,8 1,3 2,4 7,1 6,3 5,8 3,5 1,2 7,4 6,8 4,6 2,3 1,5 7,8 7 edge-disjoint rainbow matchings!

Why rainbow? Rainbow is beautiful!

Taipei City with two rainbows

Prescribed coloring Constantine’s Weak Conjecture (2002) For any m > 2, K2m can be (2m－1)-edge-colored in such a way that the edges can be partitioned into m isomorphic multicolored (rainbow) spanning trees. This conjecture has been verified later in 2006. (Work jointly with S. Akbari, A. Alipour and Y. H. Lo.)

Isomorphic rainbow trees in K6 T1 T2 T3 Color 1 x3x5 x4x6 x1x2 Color 2 x2x4 x1x5 x3x6 Color 3 x2x5 x3x4 x1x6 Color 4 x2x6 x1x3 x4x5 Color 5 x1x4 x2x3 x5x6 x1 x2 x3 x4 x5 x6 T1 3 2 4 5 1

On K2m+1 Case Constantine’s Weak Conjecture on odd order (2005) For any m ≥ 2, K2m+1 can be (2m+1)-edge colored in such a way that the edges can be partitioned into m rainbow isomorphic spanning unicyclic subgraphs. Later, we (with Y. H. Lo) verify this conjecture by decomposing K2m+1 (with prescribed coloring) into rainbow Hamilton cycles. All 2m+1 colors occur in a Hamilton cycle.

Remark If the edge-coloring is not designed, then finding rainbow spanning trees is not that easy except the first one!

Conjectures Constantine’s Strong Conjecture (2002) If m > 2, then in any properly (2m－1)-edge-colored K2m, all edges can be partitioned into m isomorphic rainbow spanning trees. Constantine’s Strong Conjecture on odd order (2005) In any properly (2m+1)-edge-colored K2m+1, all edges can be partitioned into m isomorphic rainbow spanning unicyclic subgraphs. Unicyclic graph: a graph with exactly one cycle.

Observation The above two conjectures are too hard to be verified in my opinion. So, we come back to look at the complete graph with prescribed edge-coloring. Instead of using up all colors, we consider graph designs where the graph is of small order, say a k-cycle where k is less than the order of the complete graph or certain graph with good structure.

Rainbow cycle designs It is easy to see that if Kn (with designed or un-designed edge coloring) can be decomposed into C3’s, then all C3’s are rainbow cycles. But, it is not that trivial for C4-decomposition even we can design the edge coloring. We expect all cycle decompositions of admissible order and a designed edge coloring, a rainbow cycle design can be obtained.(?) This is what I am working now: Rainbow cycle designs.

Examples (Prescribed coloring) A 4-cycle design of order n exists if and only if n ≡ 1 (mod 8). For each odd integer n, there exists an n-edge coloring (Class 2) of Kn. The coloring can be designed as follows. Let V(Kn) = Zn and the edge {x, y} receives the color “i” if x + y is congruent to 2i modulo n. (We use color n for 0.) 1 1 2 3 4 5 2 5 3 4

Cyclic multicolored 4-cycle system Let n = 8k + 1, k is a positive integer. The base cycles are: (3, 4i+1, 4, 4i+3), i = 1, 2, …, k. It is not difficult to check each difference from 1 to 4k occurs exactly once and the colors in the cycle (3, 4i+1, 4, 4i+3) are 2i + 2, 2i + 3, 4k - 2i - 2 and 4k - 2i – 3 which are distinct.

Other cycle-systems? As we have seen rainbow cycle systems of order n do exist when the cycle lengths are 4 and n. Problem: Find a rainbow cycle system for each cycle length larger than 4. Example: For the same edge-coloring, (0, 3, 8, 10, 11, 4) generates a rainbow 6-cycle system of order 13. (Make sure this cycle is a base cycle and a rainbow cycle.)

Recent progress For h, t  N, there exists a rainbow 4h-cycle design of order 8ht + 1. (k = 4h) If k is a prime power of 2 (k > 2), then a rainbow k-cycle design of order n exists if and only if n  1 (mod 2k). For each k  3 (mod 4), there exists a rainbow k-cycle design of order 2k + 1. (*) The idea is to find a suitable labelling of the cycle we consider.

Notation For convenience, we shall use H |R G to denote the existence of a rainbow H-design of G. If G is Kn, we say that a rainbow H-design of order n exists. If H is a path which is not too long comparing to n, then it is not difficult to verify that a rainbow H-design of order n exists as long as n(n-1)/2 is multiple of the length of H. (?)

Rainbow cycle designs of K2m,2m Ch |R Kht,ht for each h  0 or 2 (mod 4), t  N. With the aid of this fact, we can obtain more rainbow cycle designs where the cycle length is odd. This is by the standard construction of combining H | K2m+1 and H | K2m,2m.

Problems I would like to work! Construct a rainbow k-cycle design for each admissible order? Here, the edge-coloring is prescribed. Can we do it if the edge-coloring is arbitrarily given? k = 3 is an exception. We are able to prove that if for each t ≥ 3 a strong rainbow 4-cycle design of order 8t + 1 exists at this moment. (The edge-coloring is arbitrarily given.) How about t = 1 and 2?

Reference [1] S. Akbari, A. Alipour, H. L. Fu and Y. H. Lo, Multicolored parallelisms of isomorphic spanning trees, SIAM J. Discrete Math. 20 (2006), No. 3, 564-567. [2] G. M. Constantine, Multicolored parallelisms of isomorphic spanning trees, Discrete Math. Theor. Comput. Sci. 5 (2002), No. 1, 121-125. [3] G. M. Constantine, Edge-disjoint isomorphic multicolored trees and cycles in complete graphs, SIAM J. Discrete Math. 18 (2005), No. 3, 577-580. [4] H. L. Fu and Y. H. Lo, Multicolored parallelisms of Hamiltonian cycles, Discrete Math. 309 (2009), No. 14, 4871-4876. [5] H. L. Fu and D. E. Woolbright, On the existence of rainbows in 1-factorizations of K2n, J. Combin. Des. 6 (1998), 1-20. [6] H. L. Fu, Rainbow cycle designs, in preprints.

Thank you for your attention!