# Only 4 are used!

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• 05-14-2004
jinx
Only 4 are used!
You can color any map using only four or less colors! I can't remember the name of the guy who proved this but its true. Can anyone prove me wrong?

I mean to say that the four colors allow no color to touch the same color on and edge (but the areas may share a vertex point).
• 05-14-2004
Salem
• 05-14-2004
axon
>>Can anyone prove me wrong?

LOL...pft, why would we prove you wrong, if you didn't come up with that.

BTW, in my data structures and discrete math class, our prof said he will give out an A for the course if anyone can prove the theroem wrong...
• 05-14-2004
Zach L.
>> Can anyone prove me wrong?
Given a satisfactory set of axioms, quite easily.
• 05-14-2004
RoD
>>Can anyone prove me wrong?

Yes. Your are wrong because i am right. I cannot be wrong because i am right. There fore if im always right and you are not me, you are wrong and therefore i have proved you to be wrong.
• 05-14-2004
DavidP
try coloring this map with only 4 colors....

(yes the land formation is completely unrealistic, but still!)
• 05-14-2004
jlou
Quote:

Originally Posted by DavidP
try coloring this map with only 4 colors....

(yes the land formation is completely unrealistic, but still!)

Make the yellow and the green areas both be green, then make the uncolored area be yellow.
• 05-14-2004
jlou
Or if you wanted to fill in every area:
• 05-14-2004
Glirk Dient
What about a square divided into 4 equal parts(4 smaller squares) and completely surrounded by the same land.
• 05-14-2004
Zach L.
As I said, you just gotta play around with the axioms a bit. Here is a representation of one which (I am fairly sure, correct me if I am wrong) cannot be colored with only four colors. Its been umm... compressed sorta because I have no artistic skill, and a planar 2D space probably would not work for this kind of graph anyway, but the dots are colorable regions, and the lines between them signify that they share an edge.
• 05-14-2004
DavidP
echo Glirk Dient and Zach L.
• 05-14-2004
JaWiB
I'm assuming the theorem does not allow regions to lie entirely around other ones...If that's not the case then Glirk Dient is probably right. I've tried to represent Zach's here on top and Glirk's on bottom. Zach's doesn't seem to work because one of the colors (in this case, the green) would have to cross over another one. (The green could be light blue)

Edit: I accidentally didn't make the yellow area connect to the green on the top one, but it wouldn't make a difference
• 05-14-2004
jlou
Glirk Dient's won't work, because two areas are allowed to have the same color if they only touch at a point (the areas may share a vertex point).

In the top one for JaWiB, the light blue can be changed to dark green.

In the bottom one, I assume the light blue and the dark blue overlap by more than just a vertex. If so, the red one can be turned yellow (or the yellow can be turned red). If not, then it is like Glirk Dient's comment, and you only need three colors.

I don't see areas in Zach L.'s picture. How does it apply?
• 05-14-2004
DavidP
Zach's is more theoretical. His wont actually work on a single 2d plane, but would require a 3 dimensional representation because he has overlapping lines.
• 05-14-2004
Zach L.
Quote:

Originally Posted by DavidP
Zach's is more theoretical. His wont actually work on a single 2d plane, but would require a 3 dimensional representation because he has overlapping lines.

Precisely! :D

jlou, as for the picture, I just decomposed it into the more graph theoretical representation. The maps can be represented as points (the colored areas), with lines connecting them (indicating that the two regions share an edge). The primary motivation for that particular representation is that I have no artistic skill.
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