Newsgroups: comp.ai.games
Subject: Re: AI to develop strategic play in war games
References: <5en0r2$3ki@dfw-ixnews9.ix.netcom.com> <5epjd1$cf1@sjx-ixn4.ix.netcom.com> <5eran2$lgq$1@nntp-1.io.com> <5es6rt$erg@sjx-ixn6.ix.netcom.com>
Organization: University of Calgary CPSC
From: laneb@cpsc.ucalgary.ca (Brendan Lane)
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Date: 25 Feb 97 17:11:32 GMT
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In article <5es6rt$erg@sjx-ixn6.ix.netcom.com>,
Eric Dybsand  <edybs@ix.netcom.com> wrote:
>
>Russ,
>
>The original point, was whether a player, always making the best
>move possible for a position, could ever lose.  I believe he can.
>I also, thought you had posted, that there was a reference to this
>being possible in the book you keep citing.  Is there?
>
>In addition, your description of the outcome of 2 perfect players
>in chess or go, proved that one perfect player (who always made the
>best move possible for the position) could in fact, lose.  In fact,
>your own tic-tac-toe example above, described X making the best
>move throughout the game and still lost. 
>
>I have no mathmatical proof that this is possible, and I do not 
>need one for my own belief that this can happen.  If you do, then
>I leave that as an exercise for you to undertake.
>
> ... much other stuff deleted ...

Okay, well, I suppose I can come in here and give what I see as a
'proof' (or at least a good argument) why a player always making the
best move will win.  Let's define a position to be winnable for Black
(or player 1, or whoever) if either 1) Black can play and win in one
turn (in chess, for example, if Black can checkmate in one move), or
2) Black has a move to a position that, after any move by White,
becomes a position winnable for Black.  (Okay, the definition isn't
quite perfect - it's possible that, for some games like chess, White
could return the game to a previous state.  Let's assume some kind of
ko rule, like in Go - the board state can never be repeated.)

Now, suppose Black is in a black-winnable position.  Then he must have
some move that either wins the game, or moves the game to a position,
White to play, that White can only move to another black-winnable
position.  Use the ko rule, which guarantees finite games[1], and we
see that Black, starting in a black-winnable position, if he plays the
best move every time, must win.  Similarly, if White starts in a
white-winnable position, then, if he plays the best move every time,
he must win.

Note that, if a position is white-winnable, and it's White's turn,
then _Black has no chance_, given perfect play by White, and
vice-versa.

We thus have several situations.  A given position can be either
black-winnable (B-w), white-winnable (W-w), both black-winnable and
white_winnable (BW-w), or neither black- nor white-winnable (N).

		Player
Pos.\	   Black	|	White
B-w |   black to win	|        ?
W-w |	    ?		|     white to win
BW-w|	black to win	|     white to win
N   |       ?		|        ?

So then, what happens if it's Black's turn to move, and the position
is not winnable for him?  This means that, for any move he makes,
there is a move for White that makes the next position _also_ not
Black-winnable.  We assume that there is at least one move that Black
can make that ensures the position White is faced with is not
white-winnable - otherwise, no matter what move Black makes, White
will be faced with a white-winnable position, and so doom Black.  In
this case, Black's current position is doomed.

But we assume that there is a move for Black that makes the next
position not white-winnable.  Then White is put in the same situation.
If there is no move that makes the next position for Black not
black-winnable, then White is doomed, and putting White in this
situation is the 'best' move for Black.  But otherwise, White is in
the same position as Black.  So she has to put Black in a
non-black-winnable position, from which he tries to put her in a
non-white-winnable position, and so on.  Eventually, one of the two of
them will be put into a position which is winnable for them, or the
game will end in stalemate.  Let's suppose that White is put in a
position which is white-winnable.  Black's best move never involves
this, so Black must have had no choice.  Thus, he was doomed.  But
White put him in this position from a position which Black created,
so, given perfect play, Black must have known that White, playing
perfectly, would put him in a doomed position, and so must have again
had no choice about it.  Thus, he was doomed at this position as well.

We continue this back to find that Black was doomed at every position
from the start.  (Of course, we can do the same for White.)  Thus,
given perfect play, if a player loses, he was doomed from the start.

Well, I hope that was cogent enough to clear things up.  I think that
you can probably get rid of the whole first half, and it'll still make
sense.  Ahem.

Brendan L.
__________
[1] - Well, the ko rule guarantees finite games, but doesn't guarantee
that someone will win.  It _is_ possible that the ko rule will result
in a forced stalemate.

-- 
Words are indeed the most powerful invention  | laneb @ cpsc.ucalgary.ca |
of all; even the slow, stupid, and            |          that's          |
simple-minded can use them to great effect.   | The University of Calgary|
                     - Brendan Lane           |   'The Barbarian West'   |
