Is it possible to restore both prisoner tallies, or at least their difference, by only looking at the final position (provided both players passed equally often, as in most cases) ?
The answer is, no.
The (trivial) proof:
This (non-seki)
could be (perfect)
0 B, 0 W
as well as (faulty)
0 B, 1 W
as well as (silly)
1 B, 2 W
as well as (crazy)
5 B, 6 W
The problem, of course, is that one does not know who played the last stone:
NB + PB = NW + PW + X <=> PB - PW = NW - NB + X
(with N on board, P off board, and X either 0 or 1)
I wouldn't have cared to add this go triviality if the tally master himself hadn't ignored it once ;-)
-- Robert Pauli