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Fischer Random Chess. Play from a random setup. (8x8, Cells: 64) (Recognized!)[All Comments] [Add Comment or Rating]
Brad Hoehne wrote on Wed, Jun 20, 2001 12:00 AM UTC:
Hi,

I have worked out a slightly different method of setting up Fischer random
chess positions with a single six-sided die.   It's fairly easy to
memorizem because it follows logically from the positional rules of the
game.   As far as I can tell it will create all possible positions. Here it
is:

All die rolls are counted from the left side of the board from white's
point of view and apply to remaining empty and 'legal' squares only.

Because the king must be between both rooks, it can only occupy the central
six squares on each side.  Roll a die and place the king on one of the six
'central' squares.

Now place the rooks.   Roll a die for the left rook.  If the number exceeds
the number of squares on the left side of the king, roll again.    Repeat
for the right rook.  If there is only one square to the right or left of
the king, skip the rolls and simply place the rook.

Now place the Bishops.   Place the first bishop based on a die roll.   If
the roll value exceeds the number of remaining squares, roll again.   Place
the second bishop in a similar manner counting only the available squares
of the opposite color of the already placed bishop.

Place the queen with a die roll.  If the die number is 4-6 then subtract 3
from its value (to minimize the number of rolls necessary.)

Place the two knights on the last two squares.

I have yet to study this method in detail to determine if it favors certain
positions.


A modification of the die roll procedure to minimize re-rolls is as
follows:   If there are 2-3 'legal' squares for the rooks or the second
bishop take the remainder of the die in the 'modula' of the number of
remaining squares.  For example, if there are two legal squares for the
left rook, and one rolls a 5, one counts this as a '1', as 1 is the
remainder when one divides 5 by 2.  If the roll had been a '4' one would
count this as a '2'.   In the case of 3 empty squares, one a '5' would
count as a '2'.  A '6' would count as a '3' and a '4' would count  as a '1'
(as in the queen roll, which will always have 3).   

This method will not work without bias when there are 4-6 legal squares
remaining, and re-rolls must be employed.  However, statistically speaking,
fewer rolls will be necessary in such a case anyway.   

It is possible, though highly improbable, that one might require a very
large number of rolls to finally 'nail down' a position for the rooks and
bishops.  But once they are placed, only 1 roll remains.

What do you think?

Brad Hoehne- Columbus, Ohio.