Approximating Maximum Edge Coloring in Multigraphs Uriel Feige Eran Ofek Udi Wieder Weizmann Institute

The Problem Max Edge t-Coloring Given a multigraph G and a number t, edge color as many edges as possible using t colors. t = 3 Gray edges – uncolored Every color class is a matching 2

Hardness results For fixed t ¸ 2: Max Edge t-Coloring is NP-hard. Implicit in [Holyer ‘81, Leven Galil ’83, Papadimitriou ‘80]. Easily extends to hardness of approximation within a factor of 1 - t. 3

Greedy Algorithm Greedy: Optimum: Repeatedly take a maximum matching from the graph (and remove it). The approximation ratio is at least The ratio may be as bad as ¾ (for every even t): Greedy: Optimum: 4

A Related Problem The chromatic index ’(G): the least number of colors required to edge color all the edges of G. In a simple graph: (G) · ’(G) · (G) +1 [Vizing ‘64]. 5

Algorithm for Simple Graphs Find an edge subgraph with degree bounded by t, containing as many edges as possible [Edmonds and Johnson ’70]. Edge color the subgraph using at most t +1 colors (can be done using Vizing’s algorithm). Take the t largest color classes. Approximation ratio: 6

What About Multigraphs ? Input: multigraph G, number of colors t = 4. Maximum subgraph with degree 4: In each triangle only 4 edges can be colored (using 4 colors). 8 edges are colored. 7

What About Multigraphs ? Input: multigraph G, number of colors t = 4. The optimum solution has 12 edges: The approximation is not better than (can be generalized for every even t). 8

What Went Wrong? The maximum degree  is not enough to predict the chromatic index, the chromatic index can be as large as : Let S be a subset of vertices of size : A matching over the edges in S has size . At least colors are needed. For odd S: it may be as large as . Odd density: 9

Bounds for the chromatic index ’(G) Let G be a multigraph.  is the maximum degree of G and  is the odd density of G. Lower bound: ’(G) ¸ max{, }. Upper bounds: ’(G) · max{1.1 + 0.8, } [Nishizeki, Kashiwagi ’90]. ’(G) · (1 + o(1))max{, }, not constructive [Kahn ‘96]. Conjecture: ’(G) · max{ + 1, } [Seymour ‘79] algorithm [NK ’90] 10

Modifying the Approach for Multigraphs Find a maximum edge subgraph with , · t. Edge color the subgraph using as few colors as possible (can be done using [NK ‘90]). Take the t largest color classes. This yields a approximation. 11

Difficulty The first step of finding a largest edge subgraph with , · t is NP-hard for t=2 (also for t > 2 ??). Solution: Algorithm which “approximates” the first step: OPT= number of edges in an optimal solution (, · t). The algorithm outputs an edge subgraph G’ with: (G’) · t+1, (G’) · t + O(t1/2). The number of edges in G’ is at least OPT. 12

Approximation Using a Linear Program Every edge has value between 0 and 1. The solution has degree and density bounded by t, however it contains fractional edges. Fractional edges are denoted by GF, integer edges denoted by GI. Round the fractional edges of GF (in 3 steps). Return GI plus Round(GF). 13

Step 1 – Parallel Edges in GF Handling parallel fractional edges: GF GI The solution GF + GI remains valid. GF becomes a simple graph. 14

Steps 2 and 3: Round(GF) Round the fractional edges of GF achieving an integral solution with the following properties: Degrees exceed the original fractional degrees by at most 1. The sum of weights of edges does not decrease. 15

Putting All Together Theorem: Adding Round(GF) to GI yields: Degrees are bounded by t+1. Odd densities are bounded by t + O(t1/2). Proof: Consider S µ V : If |S| · t1/2: because GF contributes at most |S| edges per vertex. If |S| ¸ t1/2: follows from degree bound. 16

Step 2 – Rounding GF The degrees of vertices in GF remain the same. Iterating this step eliminates all fractional even cycles in GF. Find a fractional even length cycle in GF. Shift values from red edges to green edges until one edge becomes integral (0/1): Case 1: Case 2: 1 Gray edges: edges which become 0/1. GF GF 17

Step 3 – Tree like structure of fractional GF Odd cycles are vertex disjoint (otherwise an even cycle exists). The fractional edges of GF form a tree like structure. A node of the tree is either a vertex of G or an odd cycle of G: 18

LPval Gives a Good Prediction The value of the LP predicts the optimal solution of Max Edge t-Coloring within a factor of 1+ot(1): Round(LP) has: ,  · t+o(t) and value ¸ LPval . Round(LP) be colored with t+o(t) colors (Kahn ‘96). Thus there exist a solution for Max Edge t-Coloring with value of (1-ot(1))LPval . 19

Summary ratio Red – Greedy algorithm. Blue – LP based algorithm. Green – Special algorithm for t=2 with ratio of 0.77 . t 20