Virgil
Posts:
9,012
Registered:
1/6/11


Re: 4 colors problem
Posted:
Mar 5, 2014 7:09 PM


In article <5b66e1d607774171864bdda23f4a3025@googlegroups.com>, swtchwrds88@gmail.com wrote:
> "The 4color theorem doesn't claim that every partial 4coloring > of a planar graph can be completed to a full 4coloring." > > The unit pattern (in the previous post I called it the unit function, but > meant "unit pattern") guarantees a full 4coloring, so it's reliable. > > "The claim of the theorem is that of _all_ possible valid > colorings, at least one of them uses at most 4 colors." > > The unit pattern guarantees to use no more than 4 colors. > > "Thus, to show that a given planar graph is a counterexample to > the 4color theorem, it's not sufficient to produce a partial > 4coloring which can't be completed to a full coloring." > > The unit pattern uses a partial 4coloring which completes to a full coloring > for anysized map and for any number of countries. > > "You would have to show that _all_ possible valid colorings use > more than 4 colors, not just the one that you think makes sense." > > What does this mean? > > With a=blue, b=green, c=red, d=yellow, > Color the following map accordingly, > > c d a b > a b c d > c d a b > a b c d > c d a b > > which fulfills the promise of the 4 color theorem. > Now, let's say the second and third countries on the third row unites, > so they have to share the same color. > Now we have, > > c d a b > a b c d > c d d b > a b c d > c d a b > > How would you color this to fulfill the 4 color theorem using any coloring > order?
Assuming each coloring colors a rectangle, with the 'dd' in the center being a single rectangle, the coloring are acceptable as they stand.
Note that it is allowable for two regions to share one or more isolated points of common boundary, but not any interval of positive length as common boundary 

