quasi
Posts:
10,479
Registered:
7/15/05


Re: A New Graph Coloring Conjecture.
Posted:
Jun 28, 2013 5:33 AM


bill wrote: >quasi wrote: >> bill wrote: >> >> >Forced Five Set: A connected subset of five vertices >> >(not isomorphic to K5), that cannot be 4colored. >> >> With the definition as you've stated it, there's no such >> thing as a forced 5set. >> >> More generally, there's no such thing as a forced nset, >> for any positive integer n. >> >> To be precise, the following is true and easily proved: >> >> If a graph G with n vertices is not complete, then G is >> (n1)colorable. >> >> Thus, your stated definitions of forced 4set (F4S) and >> forced 5set (F5S) are badly flawed, since, if interpreted >> literally, there are no such sets. >> >> >In the example above, vertices 0 1 7 8 9 constitute an F5S. >> >> No, not the way you've defined it. >> >> The entire graph (on all 10 vertices) is not 4colorable, >> but the subgraph determined by vertices 7,8,9,0,1 is >> definitely 4colorable. In other words, the subgraph >> determined by the vertices 7,8,9,0,1 and the edges between >> them (inherited from the original graph) _is_ 4colorable. >> >> Let H be the subgraph determined by vertices 7,8,9,0,1. The >> other vertices (vertices 2,3,4,5,6) and their edges are not >> part of H. H has only the 5 vertices 7,8,9,0,1 and has >> exactly 9 edges. In fact, all pairs of distinct vertices of >> H are adjacent except for the pair 7,1. Thus, if we only >> want to color the subgraph H (without considering vertices >> and edges not internal to H), it can be done with 4 colors. >> For example, using colors A,B,C,D, color the vertices >> 7,8,9,0,1 as A,B,C,D,A respectively. >> >H is ******forced***** by vertices 2 3 4 5 6. As a member of >G, H cannot be 4colored. If H is 4colored, then some other >set in G will be a F5S.
Your concept of forced 5set is as unclear as ever.
>Am I the only person who would like to see a short >snappy proof of the 4 CT?
Sure, that would be nice.
>I hope to create a simple proof of the Four Color Theorem. In >this context, I don't think that I will be allowed to presume >that forced sets do not exist.
But you _will_ be expected to give a rigorous mathematical definition of forced sets. You previously said you couldn't do that.
>I am fairly certain that a forced set may be cited as the >primary reason for the more common impasses. > >In this context; an unresolved impasse in the attempted >4coloring of a planar graph might be due to the presence >of a forced set.
If you can't define forced sets in a way that others can understand, there's not much chance that anyone would be able to follow a proposed proof of yours of the 4CT.
quasi

