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[Jesús Muñoz]

Hi again:

I will use some numbers that were posted here:

http://talkchess.com/forum/viewtopic.php?topic_view=threads&p=416331&t=39678

Sorry for insisting so much with my polynomial extrapolation. I have done linear extrapolation (instead of quadratic) for Perft(11) estimate, to see if increasing (or decreasing) degrees in [alpha(n)]* and [beta(n)]* affects so much in range width and relative error; in fact, it seems that I was right this time (almost, because there is an issue at the end of the post). I also have done linear and quadratic extrapolations for Perft(13) and range width is reduced if the degree of the adjusting polynomial is increased (as I expected). I can say nothing yet about relative error in Perft(13) estimate because Perft(13) is still unknown. Here are the numbers (rounding up to the closest integer):

Perft(11) estimate using linear extrapolation (remember that n = 2x + 3):

[Alpha(x)]* = 2.2591 - 0.4519x
[Beta(x)]* = 2.1489 - 0.4189x

[Alpha(n = 11)]* = 0.4515 (the true one is more less 0.265031).
[Beta(n = 11)]* = 0.4733 (the true one is more less 0.262037).

{Perft(11), [alpha(n = 11)]*} ~ 2,106,537,035,152,254 ~ [true Perft(11)] + 0.4236%
{Perft(11), [beta(n = 11)]*} ~ 2,107,833,647,107,925 ~ [true Perft(11)] + 0.4854%

Taking the arithmetic average of two values above:

[Perft(11), linear extrapolation]* ~ 2,107,185,341,130,090 ± 0.0308%

The central value of this interval is more less +0.4545% than Perft(11) = 2,097,651,003,696,806. As everybody can see, both range width and relative error are greater than the ones with quadratic extrapolation:

Range width: |±0.0308%| > |±0.0187%|
Relative error: |+0.4545%| > |+0.145%|

So it seems than increasing the degree of the adjusting polynomial leads to less range width and less relative error; for Perft(13) I have found that range width is also reduce when increasing the degree of the polynomial:

[Alpha(x) linear]* = 2.3522 - 0.5078x
[Beta(x) linear]* = 2.2545 - 0.4823x

[Alpha(n = 13) linear]* = - 0.1868 (the true one is still unknown).
[Beta(n = 13) linear]* = - 0.157 (the true one is still unknown).

{Perft(13), [alpha(n = 13) linear]*} ~ 1,992,015,814,368,140,589
{Perft(13), [beta(n = 13) linear]*} ~ 1,992,883,788,621,946,743

Taking the arithmetic average of two values above:

[Perft(11)]* ~ 1,992,449,801,495,043,666 ± 0.0218%

===============================================

[Alpha(x) quadratic]* = 2.0908 - 0.2464x - 0.0523x²
[Beta(x) quadratic]* = 1.9516 - 0.1793x - 0.0606x²

[Alpha(n = 13) quadratic]* = - 0.4487 (the true one is still unknown).
[Beta(n = 13) quadratic]* = - 0.4599 (the true one is still unknown).

{Perft(13), [alpha(n = 13) quadratic]*} ~ 1,984,069,630,911,749,172
{Perft(13), [beta(n = 13) quadratic]*} ~ 1,983,623,854,355,425,180

Taking the arithmetic average of two values above:

[Perft(11), quadratic extrapolation]* ~ 1,983,846,742,633,587,176 ± 0.0112%

===============================================

Range width: |±0.0218%| > |±0.0112%|

But when a cubic function is used for adjusting the data, range width is |±0.013%| > |±0.0112%|. This last issue is what I did not expect... so I could not conclude anything. Anyway, less than |±0.03%| in range width seems pretty accurate for me. And I think that relative errors below |0.3%| are quite good. I can only wait and see.

Good luck with your methods and estimators (biased, unbiased, etc.). It seems great stuff to me (viewing how people implement such methods and obtain such good results!).

Regards from Spain.

Ajedrecista.
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