+ my $col = $x[10];
+ my $nep = (qw(a b c d e f g h))[$col];
+
+ if ($x[9] eq 'B') {
+ $nep .= "3";
+ } else {
+ $nep .= "6";
+ }
+
+ #
+ # Showing the en passant square when actually no capture can be made
+ # seems to confuse at least Rybka. Thus, check if there's actually
+ # a pawn of the opposite side that can do the en passant move, and if
+ # not, just lie -- it doesn't matter anyway. I'm unsure what's the
+ # "right" thing as per the standard, though.
+ #