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Hi Don! First of all, thank you very much for all the Komodo team, testers et al for providing such a great engine and even free. Here is my question: have you planned a release date of a stable compile of Komodo 3 for Windows (all processors and 32&64 bits)? A last thing: are you going to support s...

Hi everybody: Just for curiosity I tried the method of Adam Hair of adjusting odd BF's with polynomials: http://talkchess.com/forum/viewtopic.php?topic_view=threads&p=417416&t=39678 In my case I adjusted with quartic functions (A·x^4 + B·x³ + C·x² + D·x + E) because I splitted in twenty different po...

Hi Sven: Thanks for answer so fast. I wrote 'your' estimates because you posted them and you run them (taking your time...) but I give full credit to Peter and his method (which seems very accurate to my eyes). I also wrote that the estimates seem accurate because the mean values are so close betwee...

Hello everybody: Some people have estimated Perft values for the game of chess up to 20 plies. Here are the links (maybe the list is incomplete): Labelle (2011?) http://wismuth.com/chess/statistics-games.html Muller (10th July): http://talkchess.com/forum/viewtopic.php?topic_view=threads&p=414238&t=...

Hi Sven: It is true: it was with Uri's method the result above 1.997e+18. I misunderstood it, because your corrected value was around 1.979078e+18. Sorry. You also have the reason when you say that it is impossible to say which estimate is 'high' and which 'low': I only say that your fast estimate s...

Hi everybody: In other thread, Mr. Edwards has posted the result of four Perft(11) subtotals of the Perft(13) count: Na3, a6 ---> 1,865,423,942,798,492 Nc3, a6 ---> 2,458,021,022,756,805 b3, a6 ---> 2,110,456,650,953,864 a3, a6 ---> 1,562,327,606,120,748 I hope I do not post typos... I think these ...

Hi everybody: In other thread, Mr. Edwards has posted the result of four Perft(11) subtotals of the Perft(13) count: Na3, a6 ---> 1,865,423,942,798,492 Nc3, a6 ---> 2,458,021,022,756,805 b3, a6 ---> 2,110,456,650,953,864 a3, a6 ---> 1,562,327,606,120,748 I hope I do not post typos... I think these s...

Hi Jesus Thanks for the detailed work. I also tried something similar by taking result of perft(12) and perft(11) with two consecutive moves. At the time I tried it I didn't include the odd-even effect so my result wasn't good. But with your formula which takes care of that effect you should be abl...

Hi Adam: Thanks for clarifying it. If I take the numbers of François Labelle web, the result that you used is 62,846,608,144,842,788 (hoping no typos). Anyway, your estimates are enough good to be considered. I guess that your recalculation should give similar results. I have not checked your equati...

Hi Daniel: I took a look on your new method: http://talkchess.com/forum/viewtopic.php?topic_view=threads&p=417149&t=39678 The Perft count seems very good (it almost fits in my new interval (1.9804681 ± 0.0002024)e+18 , which is a total surprise for me!). I have calculated (I hope there are not erro...