b) R: 1. d7-d8=S# c2-c1=L

Some notes on Bichromatic chess: Bishops cannot move. Castlings are forbidden. Queens are Rooks with
a different shape (!?). Pawns cannot capture neither make double steps, thus they cannot change column
and there is at most one promotion per file (and both Pawns must be absent).
Nd8 attacks both K and has no square to come from, so thge last move is n.d7-d8=N+, thus WNd8.
The Ps have the natural colour, ie WPa4, WPb2, WPe2, WPf2, WPg2, WPh4 and BPa6, BPb6, BPe6,
BPf7, BPg7, BPh7. We can deduce BBf8 since there was no promotion on this column. But then the
BRh8 was not able to go out and has been captured in its homecage.
Moreover, if there was a promotion on the c-file it’s a promotion to B since there is at most one promotion
per file and the extremities are occupied by Bs. Thus there as been no promotion to R (or Q) during
the game, so there remains 1 BR and 2 WRs, and we can deduce the colours of the Ks and of the pieces
touching them: BRb8, BKb7, BQa7, WKc6, WRd6, WRc5, WNe7. We deduce also WQa8.
If WBc8 then it has been promoted, then WBc1 (one promotion per file) but then black is retro-stalemate
before n.d7-d8=N#. Thus our hypothesis is wrong and we have BBc8.
If WBc1 then black is retro-stalemate before n.d7-d8=N+, so we have BBc1 and the last black move is
n-1... c2-c1=B.
(This was the small idea of this problem, a light trap with a promoted Bishop).