Multicolored Subgraphs in an Edge Colored Graph Hung-Lin Fu Department of Applied Mathematics NCTU, Hsin Chu, Taiwan 30050

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). A 3-edge coloring of 5-cycle

Facts on Edge-Colorings Let G be a simple graph with maximum degree (G). Then, the minimum number of colors needed to properly color G, (G), is either (G) or (G) + 1. (Vizing’s Theorem) G is of class one if (G) = (G) and class two otherwise. Kn is of class one if and only if n is even. Kn,n is of class one.

Rainbow Subgraph Let G be an edge-colored graph. Then a subgraph whose edges are of distinct colors is called a rainbow subgraph of G. It is also known as a heterochromatic subgraph or a multicolored subgraph. Note that the edge-coloring of the edge-colored graph may not be proper. In this talk, all edge-colorings are proper edge-colorings. Therefore, a rainbow star can be found easily.

Rainbow 1-factor Theorem (Woolbright and Fu, JCD 1998) In any (2m-1)-edge-colored K2m where m > 2, there exists a rainbow 1-factor. Conjecture (Transversal) In any n-edge-colored Kn,n, there exists a rainbow 1-factor for each odd integer n.

Room Squares A Room square of side 2m-1 provides a (2m-1)-edge-coloring of K2m such that 2m-1 edge- disjoint multicolored 1-factors exist. 35 17 28 46 26 48 15 37 13 57 68 24 47 16 38 25 58 23 14 67 12 78 56 34 36 45 27 18

Orthogonal Latin Squares A Latin square of order n corresponds to an n-edge-coloring of Kn,n. A Latin square of order n with an orthogonal mate provides n edge-disjoint multicolored 1-factors of Kn,n. 1 2 3 1 2 3 2 3 1 3 1 2 3 1 2 2 3 1

Multicolored Subgraph Conjecture Given an (n-1)-edge-colored Kn for even n > n0, a multicolored Hamiltonian path exists.(By whom?) How about multicolored spanning trees?

Brualdi-Hollingsworth’s Conjecture If m>2, then in any proper edge coloring of K2m with 2m-1 colors, all edges can be partitioned into m multicolored spanning trees.

Multicolored Tree Parallelism K2m admits a multicolored tree parallelism (MTP) if there exists a proper (2m-1)-edge-coloring of K2m for which all edges can be partitioned into m isomorphic multicolored spanning trees.

K6 admits an MTP T1 T2 T3 Color 1: 35 46 12 Color 2: 24 15 36

Two Conjectures on MTP Constantine’s Weak Conjecture For any natural number m > 2, there exists a (2m-1)-edge-coloring of K2m for which K2m can be decomposed into m multicolored isomorphic spanning trees.

Constantine’s Strong Conjecture If m > 2, then in any proper edge coloring of K2m with 2m-1 colors, all edges can be partitioned into m isomorphic multicolored spanning trees. (Three, so far!)

Theorem (Akbari, Alipour, Fu and Lo) For m is an odd positive integer, then K2m admits an MTP. Fact. Constantine’s Weak Conjecture is true.

The Idea An m-total-coloring of a graph G is a mapping from V(G)E(G) into {1,2,3, …, m} such that incident vertices and edges receive distinct images. 1 2 3 2 3 A 4-total-coloring of 5-cycle 1 1 4 3 2

An example: m = 5 Bipartite difference : 1 (Use a color for these edges.)

An example: m = 5 Bipartite difference : 1 and 2

An example: m = 5 Bipartite difference : 1, 2, and 3

An example: m = 5 Bipartite difference : 1, 2, 3 and 4

An example: m = 5 4 2 Tree 1 1 3

An example: m = 5 Tree 2

An example: m = 5 Tree 3

An example: m = 5 Tree 4

An example: m = 5 Tree 5

An example: m = 5 These trees are isomorphic to : Done!

Edge-colored Kn, n is Odd It is well known that Kn is of class 2 when n is odd, i. e. the chromatic index of Kn is n. In order to find multicolored parallelism, each subgraph has to be of size n. The best candidate is therefore the Hamiltonian cycles of Kn.

MHCP K2m+1 admits a multicolored Hamiltonian cycle parallelism (MHCP) if there exists a proper (2m+1)-edge-coloring of K2m+1 for which all edges can be partitioned into m multicolored Hamiltonian cycles.

The Existence Theorem (Constantine, SIAM DM) If n is an odd prime, then Kn admits an MHCP. Conjecture Kn admits an MHCP for each odd integer n.

MHCP Lemma (Fu and Lo, DM) Let v be a composite odd integer and n is the smallest prime which is a factor of v, say v = mn. If Km admits an MHCP, then Km(n) admits an MHCP.

MHCP Example: K9 Starts from K3(3) 1 2 3 3 1 1 3 2 3 2 1 2 1 3 2 2 3 1

MHCP Example: K9 3 3 1 2 2 1

MHCP Example: K9 4 5 6 4 4 6 5 6 5 4 5 4 6 6 5 5 6 4

MHCP Example: K9 7 8 9 9 7 8 8 9 7 7 9 8 9 8 7 8 7 9

MHCP Example: K9 C1 1 2 3 7 Bipartite difference = 0 4 5 8 6 9

MHCP Example: K9 C2 3 9 2 Bipartite difference = 1 8 1 6 7 5 4 Bipartite difference = 2

MHCP Example: K9 C3 1 7 8 9 2 Bipartite difference = 2 3 4 5 6 Bipartite difference = 0

MHCP Example: K9 Use C1 + 3K3 to construct two Hamiltonian cycles :

MHCP Example: K9 C1 1 2 3 7 4 5 8 6 9

MHCP Example: K9 C1 4 5 6 Color set = {4,5,6} 1 2 3 7 8 7 9

MHCP Example: K9 C1’ 5 6 3 7 9 4 8 1 2

MHCP Example: K9 C1” 4 1 2 7 8 5 6 9 3

Not Many Are found in General Let μ be an arbitrary (2m-1)-edge-coloring of K2m. Then there exist three isomorphic multicolored spanning trees in K2m for m > 2. Joint work with Y.H. Lo.

Observation If the edge-coloring is arbitrarily given, then the proof of finding MTP is going to be very difficult. If we drop “isomorphism” for the above case, then can we find m multicolored spanning trees of K2m?

Problem: How many multicolored spanning trees of an edge-colored K2m can we find if m is getting larger? Guess? Of course, the best result is m.

Definition Su φ is a (2m-1)-edge-coloring of K2m, and φ(xy) = c. φx-1 (c) = y. x‹c› denotes the edge xy. x-star : Sx NT(x) u v 1 2 3 4 5 u‹4› or v‹4› Su = φu-1 (4)

Definition Assume T is a multicolored spanning tree of K2m with two leaves x1, x2. Let the edges incident to x1 and x2 be e1 and e2 respectively, and φ(e1) = c1, φ(e2) = c2. Define T[x1,x2] = T – {e1,e2} + {x1‹c2›, x2‹c1›}. u v 1 2 5 3 4 T[u,v] u v 1 2 3 4 5 T 3 4

Definition T is a set of mutually disjoint multicolored spanning trees in a (2m-1)-edge-colored K2m. A multicolored spanning tree T is said to be feasible for T if the union of trees in T and T are disjoint.

Constructing Process T = {T1,…,Tn}, xi is the root of Ti. Let U = Pick u U, and let φ(uxi ) = i . Choose v1 U \ {u} s.t. T1[u,v1] is feasible for T \T1. For 2 ≤ I ≤ n, choose vi U \ {u} s.t. Ti [u,vi] is feasible for T \Ti and (1) …, (2)…, (3)…… Redefine Ti = Ti [u,vi] and let Tn+1 = Su [u1,u2,…,un].

Constructing Process x1 x2 x3 x4 x5 U

Constructing Process x1 x2 x3 x4 x5 u 1 2 3 4 5

Constructing Process x1 x2 x3 x4 x5 1 v1 6 2 v2 7 3 v3 8 4 v4 9 5 v5 10 u5 10 5 u4 9 4 u3 8 3 u2 7 2 u1 1 6 u

Constructing Process u 1 2 3 4 5 6 6 7 8 9 10 10 9 8 7 u5 u4 u3 u2 u1

Note Colors 6~10 are distinct. v1‹7›, v2‹8›, v3‹9›, v4‹10›, v5‹6› are not incident to x1 ~ x5. v1‹7›, v2‹8›, v3‹9›, v4‹10›, v5‹6› can not form a cycle. In order to get vn, |U| ≥ 6n-4.

Counting Let an = |V(K2m) - U|. U = a2 ≤ 7. an ≤ an-1 + 1 + 4(n-1). → an ≤ 2n2-n+1. x1 Xn-1 . . u u1 un-1 . . . u1 u v1 u1 u v2

Theorem (with Y.H. Lo) For m large enough, we can find at least about (2m)1/2 multicolored spanning trees in (2m-1)-edge-colored K2m.

Don’t Stop!

We Have to Stop! Never Ever Give Up!!

Thank you for your attention!

Happy 60th Birthday to 藤原 良教授