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So I came up with a decision procedure for 3SAT which would seem to be completeable in a polynomial amount of time. Naturally, I am assuming it is incorrect, but I don't know where the mistake is.

Ok, so given an instance of 3SAT, reduce it to 3 colorability. A polynomial time method for doing this can be found here.

A necessary condition for 3 colorability is that it doesn't contain the complete graph on 4 vertices as a minor.

Hadwiger's conjecture implies that this is also sufficient for 3 colorability on loopless graphs, (maybe here is the problem; did I misunderstand something?) and it has apparently been proven for this case.

So in this case, our graph has one forbidden minor which stays constant over all 3SAT instances created in this manner. Testing if a minor is in a given graph is fixed parameter tractable (with a constant that depends superpolynomially on the minor) , and given that this parameter is fixed, our problem is solvable in a polynomially bounded amount of time.

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    +1 for trying to prove that NP=P *and* realizing that the proof is probably flawed :-)2012-01-23

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There are 3-colorable graphs with $K_4$ as a minor.

For instance:

enter image description here

So your assumption about necessity is incorrect. The sufficiency is true (and not a conjecture).

(Also, there is really no need to bring 3SAT into the picture, 3-colorability (of loopless graphs) is 'classically' known to be NP-Hard).

For a graph with no $K_4$ (as subgraph) which requires $4$-colours:

enter image description here

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    Yes, exactly, it does. Sorry.2012-01-24