Four-Color Problem - Part III

The last post described how when the configuration on the left in a minimal non-colorable map with one face removed could be be modified as shown in the figure. Briefly, the a-c region containing face A in the fugure on the left must also contain face C so the b-d region containing face B is surrounded thus allowing colors b and d in that region to be exchanged leading to the figure on the right. A similer argument applied to the figure on the right
allows the d-a region containing D to have its colors exchanged so face D is now colored a. This process can be repeated as show in the following table:

After 20 exchanges we have the original colors back around our five-edged face. However, it does not necessarily follow that the whole map has its original coloration. Enouch exchanges must bring it back to its original state as the map is finite and has a finite number of states and each exchange is reversable. The number of exchanges which will do this has to be a multiple of 20. [Note that if states are equated if they can be made identical by a permutation of colors this number is reduced to 5.]

One can draw paths (lines) between the faces that have to be connected by a region. Where two regions containing one different color (say a-b and a-c) the intersection must be the common color (a in this example). I drew many such lines in my diagrams labeling them and their intersections with the colors they represented trying to show that they would requiring additional faces in the map. It turned out tthis was not the case and I found a counter example. (The counter example could, of course, but the coloration cannot be found by the method described here.) I will illustrate this counter example and discuss other possible Kempe chains after I return home from Ann Arbor. Is anybody listening?