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Approximating Maximum Edge Coloring in Multigraphs

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

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Approximating Maximum Edge Coloring in Multigraphs

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

  2. 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

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

  4. Greedy Algorithm • Repeatedly take a maximum matching from the graph (and remove it). • The approximation ratio is at least • The ratio may be as bad as¾ (for every event): Greedy: Optimum: 4

  5. 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

  6. 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 thet largest color classes. Approximation ratio: 6

  7. 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

  8. 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

  9. 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

  10. 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] We call this algorithm [NK ’90] 10

  11. 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 thet largest color classes. This yields a approximation. 11

  12. 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

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

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

  15. Steps 2 and 3 Round the fractional edges of GF achieving an integral solution with the following properties: • Each degree increases by at most 1. • The sum of weights of edges does not decrease. 15

  16. Putting All Together Theorem: Adding the rounded solution of GF (from steps 2 and 3) 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

  17. 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: 0 1 Gray edges: edges which become 0/1. GF GF 17

  18. 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

  19. 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

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

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