Hello,

This thought is not new and certainly has been
discussed before but it's nice to think about
it again.

How about a match of:

(master & his computer) vs. (master & his computer)

for example consider this table :

master1
master2
master3
master4

software1
software2
software3
software4

I'm not sure if i count correct, but i think
considering playing both as black and white we have

(4!)x(4^2)= 384 games

(The point is that the number of games will be too high)


If we only have 2 masters and 2 softwares
then we have a total of 8 games. 1 day rest
between each match, and we have a nice 2 week
chess event. Both good for chess, promotion
of chess software and it's abilities and flaws
for the users and developers.
The downside is that the master could just sit
there and make the software moves on the board
which most likely results in a draw game.
Or he can reject the computer move and go on with
his own moves. In that case since everything is
already logged, the one who made most self moves
and not lost the game is the winner.
Since a draw this way is worth more than a computer
draw. However there is a flaw. How about nobody
uses the computer and make only his own moves?
well, in that case we already have a
(master vs. master) game just as usual.
A regular chess event by another word.

I probably made some wrong assumptions and
conclusions. All theoretical, mathematical,
practical corrections & (your suggestions)
are welcome.


regards,
kave

"Kave" wrote in message
om


(master & his computer) vs. (master & his computer)


If we only have 2 masters and 2 softwares
then we have a total of 8 games.

Correct.

I'm not sure if i count correct, but i think
considering playing both as black and white we have

(4!)x(4^2)= 384 games


There are n(n - 1)/2 ways to choose 2 teams from n teams.

Hence 4n(n - 1) games will be necessary with n teams.

eg, with 4 teams this equals 48 games (or 24 with no double
round-robin).



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