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Subject: Approximate # of Unique chess positions = (current estimated #)/4

Author: Les Fernandez

Date: 22:25:18 01/17/02


Every so often this subject gets brought up and I have a thought.  Lets say that
the current estimated number of unique positions is designated by N.  When we
say number of positions we need also to take into account some symmetries that
exist.  If symmetry is found in more then 1 position then we actually do not
have N number of unique positions.  With this is mind examine the following 4
diagrams which were generated by my Permutator application:

[D]k1K4R/8/8/8/8/8/8/8 w - -
acd 4; acn 91; acs 1; ce 32762; pv Rg8; id "-XTDa.1-";

[D]R4K1k/8/8/8/8/8/8/8 w - -
acd 4; acn 91; acs 1; ce 32762; pv Rb8; id "-XTDb.1-";

[D]8/8/8/8/8/8/8/K1k4r b - -
acd 4; acn 91; acs 1; ce 32762; pv Rg1; id "-XTDc.1-";

[D]8/8/8/8/8/8/8/r4k1K b - -
acd 4; acn 91; acs 1; ce 32762; pv Rb1; id "-XTDd.1-";

These 4 diagrams, although symmetrically different, are in fact the identical
position and contain the identical solution.  This being true implies that when
estimates for N are performed we need to apply the following 2 rules for
arriving at a much closer estimate.  #1 N/2 applies for positions that have any
type of castling rights and #2 N/4 for positions that have no castling rights.

Although N remains still fairly large we are much closer to the right number
then how it has been estimated in the past.  Whether dividing N by 2 or 4 will
atleast cut the estimate by 1/2 and I suspect there are many more positions with
no castling rights then with.  Perhaps other symmetries exist that we have not
found yet,  and yes I do have some in mind <s>.

Les



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