Lost Prisoner Tally

    Keywords: Theory

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

Lost Prisoner Tally last edited by MrTenuki on July 9, 2006 - 01:50
RecentChanges · StartingPoints · About
Edit page ·Search · Related · Page info · Latest diff
[Welcome to Sensei's Library!]
Search position
Page history
Latest page diff
Partner sites:
Go Teaching Ladder
Login / Prefs
Sensei's Library