The Minimum Number Of Coloring Of Knots

This paper mainly studies the minimum number of colorings for all non-trivially 19-colored diagrams of any 19-colorable knot K.Suppose that K is a 19-colorable knot and D is a non-trivially 19-colored diagram of K,where the coloring of a crossing of D is(x/y/z),where x and z corresponds to the color of the two lower arcs and y corresponds to the color of the upper arc.Next suppose there is a color x in the diagram D,then the coloring of crossings in D may be in the following three cases:(x/x/x)?(a/x/b)or(x/a/b),where a and b are not equal to x.Then follow the next steps to remove all arcs colored x.The first step,by Reidemeister move on the local of diagram D,eliminate monochromatic crossing of(x/x/x)-type,and then obtain a new diagram D1;The second step,by Reidemeister move on the local of diagram D1,eliminate crossing of(a/x/b)-type,and then obtain a new diagram D2,it can be seen that D2 no longer contains the upper arcs colored as x,next,only need to consider the lower arc colored as x;The third step,because all arcs colored as x in D2 are lower arcs,then it is obvious that each lower arc colored as x must be connected to two crossings,(x/a/2a-x)-type and(x/b/2b-x)-type,where a and b are not equal to x,next,consider the following two cases:the first is that a is equal to b;and the second is that a is not equal to b.Then by Reidemeister move on the local of diagram D2,and finally a new diagram D3 which no longer contain arcs colored as x can be obtained.Through the above steps,by a serious of Reidemeister move to D,the color x is successfully eliminated.In this paper,by using this method,the following thirteen colors are eliminated successively from nineteen colors:18?17?9?11?6?8?5?1?15?16?4?0?3.It can be seen that for any 19-col orable knot K,at least six colors are enough to color K,that is,the minimum number of 19-col orable knot is six.