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Third Option - Black: Any irrep - Blue: The adjoint irrep
This looks like the correct linear combination which at least
partially answers my problem:
a) In the classic limit, the P term ...

In Kauffmans "abstract tensor" approach, you can translate the Reidemeister moves to tensor equations. (This works equally well for directed and undirected case.) Of course, solving n^6 cubics and ...

Provided you have some "base data" (closed formulae for quadratic Casimirs and
quantum dimensions, for instance), for "low" irreps the quantum 6j symbols
might be calculated recursively using ...

Dear Bertrand & Ross - could you check the following? I neither speak French nor Math :-)
but I computed the same skein relations with magic and trickery (so my results
are unproven, of course). ...

Don't you have to rule out pawn checks as well? A trivial example would be
Ka6 Pb7 - Ka8. (A) Black plays Kb8. (B). Clearly there is no way back
from B to A. The pawn moved only in retroanalysis :-)
...

This surely isn't what you had in mind either, but just one word: Collatz.
(OK, more than one word: Assume that the 3n+1|n/2 process is completely
random. Odd->even, even->1/2*even+1/2*odd. Now it is ...

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