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Aug 15, 2017 at 18:25 comment added Tony Huynh @FedorPetrov Yes, it looks like you are right. I simplified the argument as suggested. Thanks!
Aug 15, 2017 at 18:24 history edited Tony Huynh CC BY-SA 3.0
Simplifications thanks to Fedor Petrov
Aug 15, 2017 at 17:54 comment added Fedor Petrov Do we really need that $\delta_0=\delta_1$ [I did not ubderstand the proof if this in the case when the minimal degree vertuces in $G_0$ have edges to $G_1$]? It looks that it suffices to use that both of them are at least $\delta-1$.
Aug 15, 2017 at 14:54 history edited Tony Huynh CC BY-SA 3.0
edited body
Oct 7, 2011 at 14:38 history edited Tony Huynh CC BY-SA 3.0
fixed typo
Oct 7, 2011 at 11:03 history edited Tony Huynh CC BY-SA 3.0
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Oct 7, 2011 at 8:34 history edited Tony Huynh CC BY-SA 3.0
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Oct 7, 2011 at 5:50 history undeleted Tony Huynh
Oct 7, 2011 at 5:50 history edited Tony Huynh CC BY-SA 3.0
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Oct 6, 2011 at 15:57 history deleted Tony Huynh
Oct 6, 2011 at 8:26 history answered Tony Huynh CC BY-SA 3.0