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Subject: Need thoughts about my approach to speeding up EGTB generation (Eugene ?

Author: Les Fernandez

Date: 10:17:50 01/01/03


I am interested in hearing comments about an approach that I think has the
ability of saving time in EGTB generation.  I call it a substitution technique
where we substitute various pieces (of the opposing side) at the target square.
Target square is defined as the final location of the sliding piece. Let me
expalin.  If the pv for a particular epd position is Re4 we have 2 possible
scenarios.  Either e4 is blank in which case we can substitute all of the
following pieces (P-N-B-R-Q)onto the target square.  Although I have included
the pawn it may be best to forget about them in this substitution technique
since there appears to be to many rules involved as to whether or not the
position is reachable or legal.  The above technique implies the following.
Take for example the following 3 piece EGTB position assume no piece on target
square and that the pv=Re4 as the solution:

[D]2k5/8/8/8/8/8/4R3/K7 - - w

Now using the above technique you would generate 5 new positions for use in the
4 piece EGTB where the only difference is that captures are involved:

[D]2k5/8/8/8/4p3/8/4R3/K7 - - w
[D]2k5/8/8/8/4n3/8/4R3/K7 - - w
[D]2k5/8/8/8/4b3/8/4R3/K7 - - w
[D]2k5/8/8/8/4r3/8/4R3/K7 - - w
[D]2k5/8/8/8/4q3/8/4R3/K7 - - w

Notice that in all 5 cases above the target square is unchanged but only its
contents.

Had the target square been occupied by any piece then we do the same thing with
a minor difference.  Say for example that the e4 square is occupied by a black
rook, then the substitution we would do is replace that piece by all of the
following:  blank-p-n-b-q

Notice that additon of the blank square which again does not change the results.

By applying this substitution method (even if we do not include pawns) we would
be indirectly generating 4 new positions for the n+1 EGTB where n=number of
pieces in the EGTB.  So if we are generating positions for the 3 piece EGTB for
every position we generate for this set we are actually generating 4 new
positions for the n+1 EGTB.

Be interested in hearing all your thoughts about this apporach and perhaps other
techniques might be found.

Thanks,

Les



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