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when toggle format what by license comment
Mar 6, 2015 at 13:03 comment added Fedor Petrov I am afraid that greedy algorithm as is does not provide any better bound, since it may easily happen that each color has about $n(1-\delta)$ components for $\delta=O(1/\sqrt{k})$: partition all vertices onto blocks of $\delta n$ vertices and for any pair of blocs A, B color all edges between A,B in a special color.
Mar 5, 2015 at 18:09 history answered Victor Kleptsyn CC BY-SA 3.0