Multicolored Subgraphs in Multicolored Subgraphs in an Edge Colored Graphan Edge Colored Graph

Multicolored Subgraphs in Multicolored Subgraphs in an Edge Colored Graphan Edge Colored Graph

Hung-Lin FuHung-Lin FuDepartment of Applied MathematicsDepartment of Applied Mathematics

NCTU, Hsin Chu, Taiwan 30050NCTU, 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 subgra

ph 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 we may consider the edge-coloring of the edge-colored graph which is not a proper edge-coloring.

• In this talk, all edge-colorings are proper edge-colorings. Therefore, a rainbow star can be found easily.

Rainbow 1-factorTheorem (Woolbright and Fu, JCD 1998) In any (2m-1)-edge-colored K2m where m > 2,

there exists a rainbow 1-factor.

Conjecture (Fu) In any (2m-1)-edge-colored K2m, there exist 2

m-1 edge-disjoint rainbow 1-factors for integers m which are large enough.

Theorem (Hatami and Shor, JCT(A) 2008) In any n-edge-colored Kn,n, there exists a ra

inbow matching of size larger than n – (11.053)(log n)2.

Conjecture (Ryser) In any n-edge-colored Kn,n, there exists a ra

inbow 1-factor if n is odd and there exists a rainbow matching of size n – 1 if n is even.

•What if we can assign the edge-colorings?

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-colore

d Kn for even n > n0, a multicolored Hamiltonian path exists. (By whom?)

• Problem (Fu and Woolbright) Find “a” longest multicolored path in

a (Kn)-edge-colored Kn.

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 Color 3: 25 34 16 Color 4: 26 13 45 Color 5: 14 23 56

2

5 46

13

T1

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, SIAM DM)

For m is an odd positive integer, then K2m admits an MTP.

Fact. Constantine’s Weak Conjecture is true.

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. (Unicyclic spanning subgraphs are also great!)

• 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.

MHCP

• 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.

The Existence

Lemma 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.

Theorem Kv admits an MHCP for each odd integer v.

MHCP (Fu and Lo, DM 2009)

• 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.

Can we do it if the edge-coloring is given?

Observation

• If the edge-coloring is arbitrarily given, then finding MTP is going to be very difficult.

• If we drop “isomorphism” requirement for the above case, then may be we can find many 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.

Joint work with Y.H. Lo• Theorem For any proper (2m-1)-

edge-coloring of K2m, there exist around m1/2 mutually edge-disjoint multicolored spanning trees.

Definitionφ is a (2m-1)-edge-coloring of K2m, and φ(xy) = c.

Define 1.φx-1 (c) = y and φy

-1 (c) = x

2. xy = x‹c› = y‹c› x

y

1

2

3

4

5

= φx-1 (4)

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..

Define T[x1,x2] = T – {e1,e2} + {x1‹c2›, x2‹c1›}.

Definition

3

4 u

v

1

23

4

5

T

u

v

1

2

5

3

4

T[u,v]

Sketch proof1. Pick any two vertices x∞, x1, let T∞ and T1 be the stars

with centers x∞, x1, respectively.

x∞ x1

T∞

2. Pick x2, u, v1, and let the colors be as follows.

x∞ x1

v1x2u

1 2c1

c2

2. Pick x2, u, v1, and let the colors be as follows.

.

x∞ x1

v1x2u

1 2c1

c2

2 1

u1

2. Pick x2, u, v1, and let the colors be as follows.

Find and . Redefine T1 = T1[x2,v1]

x∞ x1

v1x2u2 1

2

11(2)x u

1

1(1)v

u1

T1

2. Pick x2, u, v1, and let the colors be as follows.

Find and . Define

x∞ x1

x2u2

1

22 1T [ , ]xS u u2

11(2)x u

1

1(1)v

u1

T2

c1

c2

2. The structure of T∞ so far.

x∞ x1

T∞

x2

2. The structure of T1 so far.

x∞ x1

T1

x2

2. The structure of T2 so far.

x∞ x1

T2

x2

3. Choose x3, u, v1, v2 and let the colors be as follows.

x∞ x1 x2

v1x3u v2

1 2c1c

2

3 4

3. Choose x3, u, v1, v2 and let the colors be as follows.

Redefine T1 = T1[x3, v1].

x∞ x1

T1

x2

v1x313u v2u1

3. Choose x3, u, v1, v2 and let the colors be as follows.

Redefine T2 = T2[x3, v2].

x∞ x1

T2

x2

v1x3u v2

24

u2

3. Choose x3, u, v1, v2 and let the colors be as follows.

Define

x∞ x1

T3

x2

x334

33 1 2T [ , ]xS u u

u1 u2

• Adjust T1, T2, T3 to add an extra T4.

• ………

• ………until the n-th tree is found.

Property: 1. Ti and Tj are edge-disjoint for i ≠ j ≠ ∞.

2. In each Ti, i ≠ ∞, x∞ is a leaf which is of distant 1 from

its center xi.

3. The number of trees n is determined by way of m.

T1, T2, T3, …, Tn are the desired trees.

Corollary For any proper (2m-1)-edge-coloring of K

2m-1, there exist around m1/2 mutually edge-dis

join multicolored unicyclic spanning subgraphs.

1. Adding an extra vertex, named x∞, to form a (2m-1)-edge-colored K2m.

2. Apply Theorem 1 to construct T∞, T1, T2, …

3. Drop x∞.

4. Adding one specific edge colored the missing color φ(x∞xi) to each Ti, for i =1,2,...,

Sketch proof

Don’t Stop!