Edge Coloring

A may contain parallel edges.

Definition 1

Edge coloring of a graph G is an assignment of colors to its edges so that adjacent edges are assigned distinct colors.

Definition 2

The chromatic index, or the edge-chromatic number of a graph G is the smallest integer n for which there is an edge coloring of G which uses n colors.

1 Edge coloring is a special case of vertex coloring.

Definition 3

For a given G = (V,E), the L(G) is a graph H whose set of vertices is E and the set of edges consists of all pairs (ei,e2) where e1 and e2 are adjacent edges in G.

Theorem 1 For a simple graph H, there is a solution G to L(G)= H iff H decomposes into complete subgraphs, with each vertex of H appearing in at most two of these complete subgraphs.

Proposition 1

Every edge coloring is a partition of the edges into matchings.

Proposition 2

There is an edge-coloring of G in n colors iff there is a vertex coloring of L(G) which uses n colors.

2 For a graph G, χ′(G) and ∆(G) denote the chromatic index and the maximum vertex respectively.

Theorem 2 ∀G, χ′(G) ≥ ∆(G). Proof. Obvious.

Theorem 3 ∀G, χ′(G) ≤ 2∆(G) − 1. Proof. Apply the greedy algorithm: color edges one-by-one, using for each edge the smallest positive integer which is available.

Theorem 4 ∀G, if G is bipartite, then χ′(G)=∆(G). Proof. Not obvious.

Theorem 5 (Shannon [1948]) For any multigraph G, 3∆(G) χ′(G) ≤ . 2

Proof. Not obvious.

Theorem 6 (Vizing [1964]) For any graph G, χ′(G) ≤ ∆(G)+1. Proof. Not obvious.

3 Tools for edge-coloring

a b a b a

b a b a b

Definition 4

Given a partial edge-coloring of G, a path P is called 2-colored if all of its edges are colored using two colors. A 2-colored path P is called maximal if no other 2-colored path contains P . Re-coloring of a 2-colored path is an operation which switches the colors of the edges of the path.

Proposition 3

Given a partial edge-coloring of G and a maximal 2-colored path P , re-coloring of P yields a new (legal) edge-coloring of G. Prove it.

4 Practical strategy for edge-coloring: color edges one-by-one, and use recoloring of some maximal 2-colored paths in order to reduce the number of used colors. One edge is not colored; how to recolor the current partial coloring to obtain a complete edge-coloring with 8 colors?

1 4 5 2 3 6

1 2 3 7 8 4 5 6

5 Theorem 7

If all vertices of graph G but one have degree 3 and the remaining vertex has degree 2, then χ′(G)=4.

Theorem 8

Define ω(G) as follows: |E(H)| ω(G) = max⌈ ⌉. H⊆G |V (H)| ⌊ 2 ⌋ Then χ′(G) ≥ ω(G).

Proof. ???

6 Theorem 9

Every graph (multigraph) of degree three can be edge-colored using at most four colors.

a b

Proof. Select four colors, say 1,2,3, and 4, and start coloring edges one-by-one. For every next edge ab to be colored, there are at most four colored edges that may restrict the choice of a color for the edge. See figure above. If one of the colors 1,2,3,4 is not used for the edges adjacent to ab, we color ab using that color and proceed to the next uncolored edge. Now let us consider the “bad” case: all four colors are used “around” ab. It’s easy to see that up to renumbering the colors of the edges are as they are presented on the next figure.

c 1 3

a b

2 4

Consider the maximal (4,1)-colored path starting at vertex b. If this

7 path doesn’t end at a, we re-color the path (change 4 to 1 and 1 to 4). This will make color 4 absent “around” b, so we use it for ab. Finally, if the maximal (4,1)-colored path starting at vertex b ends at a, un-color (b,c) and color 3 edge ab. If either 1 or 2 is missing at c, we use it for (b,c), else we un-color the maximal (4,1)-colored path starting at vertex b. We already know that ends at a, thus after decoloring, color 4 is missing at both b and c, so we us it for (b,c). The result is that we end up with one more colored edge than we started with. 4 1 c 1 4 2

a3 b 4 1 4 1 2 1 4

4 4 1 1 4 1 1 4

8 Theorem 10

For any bipartite multigraph G, χ′(G)=∆(G). Proof. Select colors 1,..., ∆(G) and start coloring edge by edge. Given an edge uv, if there is a color i which was not used for any edge incident with either u or v, use it for uv. Suppose the next edge uv is such that every color i which is not used for the edges incident with u is used for that of v, and vise versa. Pick any color i which is not used at v and any color j not used at u. Let uu1 be an edge colored i and let P be a maximal (i,j)-colored path containing uu1.

Claim: P does not contain v. Indeed, otherwise we would have an odd cycle, which is not possible in a . un-coloring P . The resulting edge-coloring does not use i either at u, or at v. Assign color i to uv. Continue until all edges are colored.

9 Two Theorems in Edge Coloring

Theorem 11 (Shannon, 1948)

For every multigraph G with the maximal vertex degree ∆, 3 χ′(G) ≤ ∆ 2

Proof. We describe an algorithm for coloring G using at most 3∆/2 colors. The colors to be used by the algorithm are 3 1, 2,..., ⌊ ∆⌋ 2 The algorithm colors the edges of G one by one starting with G without colored edges. For every next edge ab, we consider the edges incident to a and let A to be the set of colors that are not used (missing) at a; similarly, let B be the set of colors missing b. If A ∩ B 6= ∅, then any color from the intersection can be used to color ab. Thus, let A ∩ B = ∅ and consider any α ∈ A; let bc be the edge colored with α. Let also C be the set of colors missing at c.

α a b c

10 If B ∩ C 6= ∅ and δ ∈ B ∩ C, then the algorithm un-colors bc, uses α to color ab and color δ edge bc. Hence, we can assume that in addition to A ∩ B = ∅, we have C ∩ B = ∅. However, since 3 |A| ≥ ⌊ ∆⌋− ∆+1= ⌊∆/2⌋ +1, 2 3 |B| ≥ ⌊ ∆⌋− ∆+1= ⌊∆/2⌋ +1, 2 3 |C| ≥ ⌊ ∆⌋− ∆= ⌊∆/2⌋, 2 there must be a color γ ∈ A∩C. Take such a γ and any β ∈ B, and consider the maximal (γ, β)-path starting at b. If this path does not end at a, re-color the path and color ab γ. If this path does not end at c, de-color bc, re-color the path, color ab α, and color bc γ.

α α γ a b c a b c β γ γ β γ β β γ β γ γ β

11 Theorem 12 (Vizing [1964]) For every graph G without parallel edges, χ′(G) ≤ ∆(G)+1. Proof. Similar to the proof for Shannon’s theorem, we present an algorithm which colors the edges one by one using just ∆(G)+1 colors. Let the set of available colors be {1, 2,..., ∆+1}. Given a partial edge-coloring of G, let Ω(x) denote the set of colors from {1, 2,..., ∆+1} that are missed at x. For every edge ab, if Ω(a) ∩ Ω(b) 6= ∅, select the color from the intersection and color edge ab with this color. Otherwise, let α1 ∈ Ω(a) and ba1 be the edge colored α1. If α2 ∈ Ω(a1)∩Ω(b) 6= ∅, color α1 edge ab and color α2 edge a2b, Otherwise, let α2 ∈ Ω(a1) and ba2 be the edge colored α2.

Again, if α3 ∈ Ω(a2) ∩ Ω(b), color α1 edge ab, color α2 edge a1b, and α3 edge a2b. Otherwise, pick α3 ∈ Ω(a2) and let edge ba3 be colored α3.

α Recoloring β 1 β α β α 2 α 1 α α 1 α β 1 α 1 2 3 β α α 1 β α a b 3 a b 4

α α 4 5 α α 5 1

Continuing this way, we create a sequence of colors

α1,α2,...,αk and a sequence of edges ab, a1b,...,akb,

12 for which aib is colored αi (i =1, 2,...,k). Obviously, there is k > 0 and i

13