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Dec 11, 2015 at 7:39 answer added Ali timeline score: 0
Feb 25, 2012 at 3:56 vote accept David Galvin
Feb 25, 2012 at 3:51 comment added darij grinberg PS: By "equivalent", I mean "equivalent by an argument substantially simpler than any proof I know for König's theorem".
Feb 25, 2012 at 3:50 comment added darij grinberg Hmm. It is easy to see that $\chi\left(\overline G\right) = n - \left(\text{number of edges in a maximum matching of }G\right)$ and $\omega\left(\overline G\right) = \left(\text{size of maximal independent subset of }G\right) = n - \left(\text{size of minimal vertex cover of }G\right)$ (because independent subsets are exactly the complements of vertex covers). So this exercise doesn't just follow from König's theorem; it is also equivalent to it...
Feb 25, 2012 at 3:43 answer added Russ Woodroofe timeline score: 3
Feb 25, 2012 at 3:08 history asked David Galvin CC BY-SA 3.0