
Re: 4 colors problem
Posted:
Mar 4, 2014 4:12 PM


According to Prof. Augustus De Morgan:
"A student of mine [Guthrie] asked me to day to give him a reason for a fact which I did not know was a fact  and do not yet. He says that if a figure be any how divided and the compartments differently colored so that figures with any portion of common boundary 'line' are differently colored  four colors may be wanted but not more"
The original letter did not say "adjacent countries" or "neighboring countries", instead it said, "figures with any portion of common boundary 'line'". When two countries are completely apart, one color can be used to color them both. When two countries are connected by a common point, the modern convention for 4 color theorem says that one color can be used to color them both. Without such a convention concerning a common point, we would have a "pie map" situation where a pie sliced into 8 pieces would need 8 different colors instead of 2. By the same convention, a chess board is colored with 2 colors, black and white. But it is but a convention, an agreedupon point (no pun intended) for the sake of having everyone be on the same page. The mathematicians working on this problem could very well have made it a convention to have a point shared by two countries as being the same as a line requiring two different colors to color such two countries, not one same color for both countries. But with so many countries sharing huge borderlines that one has to resolve for 4 color theorem who has time to care about a mere common point between two countries? So a convention was created that if ever there are two countries that does not share a same boundary line but shares a mere common point, you may consider the two countries as same one country and color them both using one color only. And this convention was accepted and followed by all because, after all, how many countries are there that share a mere point with its adjacent country? Not many, one would assert. So, nobody really cared about this point. Even Utah/New Mexico do not have a common point connecting them, but there are some common borderlines between the two, but so miniscule as to be negligible. Same goes with Arizona/Colorado. Negligible enough to be called a point. Another convention for the sake of convenience and having everyone be on the same page doing the same thing as everyone else so there wouldn't be any differential points to iron out later on. Who has time for that?
But in this map of imagined world where each country is shaped as a sqaure of equal size, lined up in a row, row after row, represented by matrix A, B, C... When things are oversimplified like this, should the same convention that work supremely well for the real world be applied too? Real world map has countries with all kinds of zigzagging squiggly borderlines between them. Rarely would there be two countries sharing nothing but a point between the two, allowing the cartographer to color both countries with single color. So in this simplified version of the world map, could then a convention about a common point be modified to say that a common point may be considered the same way as the regular borderlines are, thus requiring two colors to color such countries connected with only a common point? (after all, without considering a common point to be the same as a regular borderline, no two diagonally situated countries would ever share a common borderline, unlike in the real world)
What would the student of Prof. Augustus De Morgan say if he knew that in "a figure be any how divided and the compartments differently colored so that figures with any portion of common boundary line are differently colored", the figure was to be divided into equalsized squares, lined up one after another row after row such that diagonally situated squares are only connected by a common point mimicking borderlines between them, to mimick the real world map?
Anyway, in this imagined world, the convention concering common points (unlike in the realworld examples of 4 color solutions) is that such a common point may be considered the same as borderline, and under such convention, 4 color problem succeeds for matrix B for odd numbers, fails for even numbers.
But now that it is agreed that conventions are rather arbitrary and they are there to make things easier for us, the convention for this imagined world can be changed so that a common point between diagonallypositioned squared cannot be considered adjacent anymore. Then maybe that will lead to the right conclusion for matrix B that everyone can agree upon.
So, let's see...

