Such a test cannot be linear, due to the fact that the number of colors is even. Similarly all even color tests are non-linear, and thus it is possible to distinguish among knots having identical Alexander polynomials by using these tests.
While one may generate such tests by checking all possible combinations, a far easier way would be the following. One considers a finite group G and let C(G) be a conjugacy class of G. One considers elements of C(G) as the "colors" of the test, and for i,j belonging to C(G), M(i,j)=jij^{-1}. Such a color test would be valid, although it might be reducible.
Due to the fact that the permutation groups and their conjugacy classes are well known and classified, one could easily use these classes to generate color tests. Take for instance P_4, the 4-permutation group, and the conjugacy class 3,1. This class contains permutations mapping (a,b,c,d) to (a,c,d,b). It contains 8 elements and would generate an 8-color test; such a test however splits into two copies of the 4-color test.
Through the use of color tests generated from equivalence classes of P_4 and P_5, as well as the linear color tests, one may distinguish all inequivalent knots with up to 10 crossings (at least). For more information, have a look at a summary of my work.
Charilaos Aneziris, charilaos_aneziris@standardandpoors.com
Copyright 1995
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