From: gwoegi@opt.math.tu-graz.ac.at (Gerhard J. Woeginger)
Subject: Re: outerplanar => 3-colorable
Date: Mon, 29 Jan 2001 17:39:53
Newsgroups: sci.math.research
Summary: 3-colorability of restricted families of graphs
In article Todd Wilson writes:
>Subject: outerplanar => 3-colorable
>From: Todd Wilson
>Date: 28 Jan 2001 18:42:40 -0800
>By the Four Color Theorem, every outerplanar graph is three-colorable.
>Does anyone have a citation for a direct proof of this result?
>
>Todd Wilson
This is folklore and quite easy to prove: Since every outerplanar graph has a
vertex with degree at most 2, an inductive proof works.
Chvatal (A combinatorial theorem in plane geometry, JCT Series B 18, 1975,
39-41) uses three colorability of outerplanar graphs for getting a bound for
the art gallery problem.
___________________________________________________________
Gerhard J. Woeginger (gwoegi@opt.math.tu-graz.ac.at)
==============================================================================
From: David Eppstein
Subject: Re: outerplanar => 3-colorable
Date: Mon, 29 Jan 2001 08:39:09 -0800
Newsgroups: sci.math.research
In article , Todd Wilson
wrote:
> By the Four Color Theorem, every outerplanar graph is three-colorable.
> Does anyone have a citation for a direct proof of this result?
Just remove degree-two vertices one at a time.
You can show there exists a degree-two vertex from the fact that the weak
dual of an outerplanar graph, i.e. planar dual with the vertex
corresponding to the infinite face deleted, is a forest, so it always has
at least one leaf. Alternatively, you can use Euler's formula.
--
David Eppstein UC Irvine Dept. of Information & Computer Science
eppstein@ics.uci.edu http://www.ics.uci.edu/~eppstein/
==============================================================================
From: Fred W. Helenius
Subject: Re: 3-color map theorem
Date: Wed, 21 Feb 2001 10:18:34 -0500
Newsgroups: sci.math
Jan Kristian Haugland wrote:
>Leroy Quet wrote:
>> Jan Willem Knopper wrote:
>> >>Is it true that:
>> >>If you have a flat map, where every country is of a convex shape
>> >>without any holes, then one can always color the map with just 3
>> >>colors such that no two boardering countries are colored the same ?
[snip diagram]
>> >Since all countries are adjacent, no pair can have the same colour.
>>
>> Ah, yes.
>> What if all of the countries are triangles ?
>Isn't there a theorem saying that the chromatic
>number of any cubic planar graph except for the
>tetrahedron is at most 3? That would answer the
>question.
I don't see how. Why should the graph be cubic --
can't a triangle have more than three triangular
neighbors? And you would need to rule out the
tetrahedron...
Regardless, the answer to Leroy's question is no;
four triangular regions can be mutually adjacent:
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--
Fred W. Helenius