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 Post subject: d-regular, planar, 2-colored graphPosted: Wed, 29 Sep 2010 22:53:48 UTC
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Joined: Wed, 29 Sep 2010 22:33:04 UTC
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Hey.

Find the values of d for which a d-regular, planar, 2-colored (coloring the vertices) graph exists.

If the graph is planar, then (where m = number of edges, n = number of vertices).
Also, (because of the d-regularity).
From the two above i can say that d < 6 (i'm not so sure of that either).

Can someone help?
thanks

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 Post subject: Re: d-regular, planar, 2-colored graphPosted: Thu, 30 Sep 2010 04:46:55 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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VanDelay wrote:
Hey.

Find the values of d for which a d-regular, planar, 2-colored (coloring the vertices) graph exists.

If the graph is planar, then (where m = number of edges, n = number of vertices).
Also, (because of the d-regularity).
From the two above i can say that d < 6 (i'm not so sure of that either).

Can someone help?
thanks

Right, so what remains is for you to show that each of d=1,2,3,4,5 is possible (or not).

(By the way, if you allow infinite graphs you have to consider too.)

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 Post subject: Posted: Thu, 30 Sep 2010 13:03:36 UTC
 S.O.S. Newbie

Joined: Wed, 29 Sep 2010 22:33:04 UTC
Posts: 2
I think i solved it (it's only for finite graphs), if anyone's intrested:

The d=1 case is easy (just two connected vertices).

For d>1:
We know that , so there must be at least one cycle within. Since the graph is 2-colored, we know that there are no cycles of odd length, so the girth (size of the smallest cycle) is 4 (or greater).

If we put g=4 and in :

Now this tells us that graphs with d>=4 don't exist (We still need to find those with d=2,3).
The 2-regular is a square and 3-regular is a cube (so that there's no intersection between edges).

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