RE: computer-go: Scoring

[Jean-Pierre Vesinet <jpvesinet@xxxxxxxxxx>]

  Here is another counter-proposal to the Taylor-Tromp integrists.
  There are many logical scoring rules. Taylor-Tromp is not the only one. My
personal preference goes to the scoring method I use in my own Java program
- of course :-). It is a useful shortcut to avoid trying many stupid endgame
moves during search. The Benson algorithm recognizes "absolutely"
uncapturable stones (strings impossible to capture even if the opponent
plays an unlimited number of moves in a row), then some regions are defined
as "absolutely" safe territory if they take part in the Benson criteria or
have a little enough potential eye space to allow the opponent to make life.
Note - suicide is not allowed. A player's score is the area covered by his
"absolutely" safe stones and territory.
  This scoring has nice properties:
  - there is a static algorithm to recognize all "absolutely" safe stones
and territory, including capturable alive stones (see example below) and
uncapturable dead stones (for example a one and a half eyed group with a
double ko seki elsewhere is uncapturable if basic ko is in use - see
http://www.msoworld.com/mindzine/news/orient/go/history/nyobutsu.html if you
don't believe me),
  - it does not require capturing all dead stones - most of them are
recognised long before capture.
  Depending on how other rules (superko / basic ko + no result, how seki is
scored), this scoring method may be exactly equivalent to Tromp-Taylor
"flood-fill" scoring, except for position where suicide or seki matters. It
is much more complex, too, so I don't think it is a good rule except for
computer programs.  

For example the static algorithm (no search) gives these results:
5 . . . . . . . . . . . . .
4 O O O O O O O O . . . . .
3 . . . . . . . O . . . . .  The exterior white stones are "absolutely"
uncapturable.
2 X X X X X X . O O O O O .  The two white stones C1 and D1 are capturable
alive.
1 X . O O . X . O . O . O .  All black (X) stones are dead.
  A B C D E F G H J K L M N

There is no need to explicitely capture the black stones. White can safely
ignore any move played by Black anywhere in this corner. This is true even
if there is a ko fight?
I believe this algorithm correctly recognizes all sorts of stones and
territory whose status cannot be changed by further moves.

I would like to undestand your method better because it looks very
interesting - similar or better than mine. Please see questions below.

> From : T.R. Christie [mailto:trc23@xxxxxxxxxxxxxxxxx]
> So, here is the redefined algorithm...
> 
> 
> 
> 
> 
> a black score area is a string of [white OR empty] intersections

Do you mean a _maximum_ set of connected white or empty intersections?

> 
> a white score area is a string of [black OR empty] intersections
> 
> two score areas can be said to intersect iff:

Do you mean two scoring areas of different color (one black, one white)?

>   they share an intersection  OR
>   there is an intersection in one score area that is
>   immediatly adjacent to an intersection in the other score area
> 
> 
> 1. the status of all score areas is initially CONTESTED
> 
> 2. any score area that intersects with one and only one
>      opposing score area has it's status set to SCORING
> 
> 3. any score area that intersects with a SCORING score area
>      has its status set to NON-SCORING

If A intersects B, then B intersects A, therefore SCORING score areas would
also need to be set to NON-SCORING ?

> 
> 4. any score area that only intersects with NON-SCORING score areas
>      has its status set to SCORING
> 
> 
>   calculating the final score:
> 
> 6. any stone in a SCORING score area is removed as a prisoner
> 
> 7. white scores 1 for each intersection contained by a 
> SCORING white score
>    area, plus 1 for each white intersection (Chinese) OR 1 for each
>    prisoner (Japanese).
> 
> 8. black scores 1 for each intersection contained by a 
> SCORING black score
>    area, plus 1 for each black intersection (Chinese) OR 1 for each
>    prisoner (Japanese).
> 
> 9. add komi to white's score if necessary
> 
> 
> 
> a fix, to cope with seki:
> 
> any score area consisting of exactly 2 empty intersections,
> and at least one coloured intersection is always CONTESTED,
> and may not be changed by steps 2, 3 or 4.

Too simple. Sekis can have 1 liberty in common, or 2 or 3, or even more!
This includes the "final" state of a seki, where neither player can add a
stone.

--
Jean-Pierre Vesinet - hopeless tsumego addict and part-time go programmer