Re: computer-go: unmake move?

["Mark Boon" <tesuji@xxxxxxxxx>]

It is a ladder indeed. Davids approach surprises me a little. I had assumed
all programs would start off with a 'ladder' search. Only when it turns out
it can't be captured by the 'ladder' routine would it need to do a more
extensive tactical search, which in its turn calls the 'ladder' routine to
do the last two liberties.

Of course there are advantages doing it the way David does. Most of the
times it's better to play a geta instead of a long ladder. On the other
hand, often it's also better to play the solution that minimises the number
of liberties a chain can get, instead of minimising the search-depth. For
example, if there had been a 'O' stone at 'i', connecting at 'b' in the
third sequence would have worked. But in this case it's clearly an inferior
solution because the '@' move above 'i' would become sente. (This is maybe
not a good example, but in general minimising the number of liberties also
minimises the number of forcing moves of the opponent.)

In my experience, a two-liberty search (or 'ladder' search) can be highly
optimised, way beyond what's possible with more general tactical searches.
So the main reason I'm surprised is that I find it hard to see how Davids
method can compete in terms of performance. A simple 'ladder' routine would
have 50% chance of starting with 'a' and 50% with 'c'. Starting with 'a'
would take 22 nodes while starting with 'c' would take 39 nodes (Average of
30.5, if I counted right). With a specialised 'ladder' routine that can
search ten times faster than a more general tactical routine, it would
finish in 15% of the time or less.

    Mark

P.S. David, your program doesn't try 'k' after 'j' failed in the third
sequence because of some heuristic?


----- Original Message -----
From: Martin Mueller <mueller@xxxxxxxxxxxxxxxxx>
To: <computer-go@xxxxxxxxxxxxxxxxx>
Sent: Friday, June 16, 2000 8:04 AM
Subject: Re: computer-go: unmake move?


>
>     + + @ @ @ @ @ + |
>     + + + O O O @ i |
>     + @ + O @ @ O O k
>     + @ O b O @ @ c h
>     + @ + d @ O a g j
>     _ _ _ f e _ _ _ _
>
> I may be missing something, but: isn't this simply a ladder problem?
>
> Martin
>