3) My comment about the geometric mean was really jsut frivolous speculation. I have no real reason to suppose that the smallest graph with one of two specified edges mono will have somewhere near that number of vertices. Please ignore that.

4) The cases where the lower bound is greater than the upper bound are simply those that lead directly to the proof that the CNP is greater than 4. Remember that the table is for 4-colourings: I think the only things we know about higher-order colourings are that p_1 is 0 and p_d is less than 1/2 for all d greater than 1/2.

1) The only reason for weakening the statement is that then it is likely that the graph can be smaller – maybe much smaller. One of the main goals of Polymath16 is to find a proof that the CNP is at least 5 that does not rely on computers to verify any properties of any graph. Domotor’s cool realisation is that we would obtain such a proof if we could find a reasonably small graph that has at least one of two specified vertex pairs (both of which must be more than 1/2 apart, of course) the same colour in any 4-colouring. The smaller the graph, the shorter the proof that it has that property; therefore, the hope is that we can find a small enough graph that the proof is human-derivable (in a publishable number of pages!).

2) The probabilities are actually not dependent on each other, because they are not defined in terms of any particular location or orientation of any particular graph in the plane – they are defined in terms of all such locations and graphs. The statement “p_d is less than 1/2” simply means that in any 4-colouring of the plane, fewer than half of the pairs of points at distance d from each other are the same colour – no graphs mentioned yet. So then if p_d1 and p_d2 are both less than 1/2 and we have a graph that has vertices A and B that are d1 apart and also two vertices C and D that are d2 apart, and then we colour the plane, then we know that out of all the infinitely many places and orientations that we can place our graph in the plane, there will be some in which A and B are different colours and C and D are also different colours, because fewer than half of those infinitely many copies of our graph have A and B the same colour and also fewer than half have C and D the same colour. (I’m playing fast and loose with fractions of infinities, of course, but that’s OK in this case, trust me.) So if we have a graph in which either A is the same colour as B or C is the same colour as D in any 4-colouring, we have a contradiction, which means CNP must be greater than 4 (because the P_d’s are all defined in terms of 4-colourings of the plane). Note that d1 and d2 can be equal; in the case I’m suggesting you explore, they are both 8/3.

]]>All three pairs of vertices with distances of 8/3 in the 409-vertex graph are monochromatic. Why do we need to weaken this statement and say that at least one of them is mono? What does this give?

Does the probabilistic argument take into account that if we take several non-unit distances, then, generally speaking, the probability of one event depends on the probability of another? What if, for example, in the presence of one monochromatic pair, the probability of the formation of another sharply increases? (Or is it impossible or does not change the overall logic?)

It is not clear how the geometric mean appears.

Is it normal that some lower bounds (on a wiki page, for example, for 8/3) are greater than upper bounds?

]]>*[Done! – M.]*

So now, in reality we do not have any such bound (though I wonder if we might get it for ). What we do have, though, is that p_d is less than 1/2 for all d greater than 1/2. Therefore, the equivalent logic works if we have a graph in which there are two specific vertex pairs, A,B and C,D, such that in any 4-colouring of the graph either A and B are the same colour or C and D are the same colour. Obviously we already have such graphs, since in your graphs we can dispense with C,D. But with that relaxed condition, it should be possible to remove vertices. The question is, how many? It is tantalising that the geometric mean of 5 and 409 is only 45… if we could find such a graph with under 100 vertices, it’s a fair bet that we could prove the property without a computer.

The next question is where to begin. I think the most promising line of attack would be to use two of the 8/3 pairs in your 409er, so let’s say and . They start out both needing to be mono, and the relaxed property is that one or other of them is mono.

If this works it will be so cool. It was one of he main original goals of the project, and it’s already the result of substantial input from several of us.

]]>I did everything I could to thin the graph while maintaining symmetry. I do not think that it is possible to significantly reduce the number of vertices due to the loss of symmetry. ]]>