perft(19) has at least 9 possible stalemates(both white and black has 3 options to change the order of moves at moves 1-3)
1.c4 2.h4 3.Qa4
1.h4 2.c4 3.Qa4
1.c4 2.Qa4 3.h4
1...h5 2...a5 3...Ra6
1...a5 2...h5 3...Ra6
1...a5 2...Ra6 3...h5
1.c4 h5 2.h4 a5 3.Qa4 Ra6 4.Qxa5 Rah6 5.Qxc7 f6 6.Qxd7+ Kf7 7.Qxb7 Qd3 8.Qxb8 Qh7 9.Qxc8 Kg6 10.Qe6 1/2-1/2
I wonder if people can calculate the number of stalemates in perft(19) and estimate this number for bigger n and the same for checkmates and I wonder if there is some n when the number of stalemates in perft(n) is bigger than the number of checkmates in perft(n).
estimating the number of possible stalemates in perft(n)
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Estimating the number of possible stalemates in Perft(n).
Hello Uri:
I understand that you say that 19 is the minimum number of plies where a stalemate can happen. I have no idea about that, the same as the estimate, that should be clearly difficult: Perft(19) will be around 2.462e+19 according to François Labelle (his estimates are very accurate) and maybe one have to go through each position for verifying if a stalemate happens (only a guess). A huge task for everybody and an impossible task for me. Obviously, I also can not say anything about the comparison between checkmates and stalemates in Perft(n).
In conclusion: no help from mine, but I only can point out that c3 is also valid in your fabulous mainline. Well done!
Regards from Spain.
Ajedrecista.
Have you discovered those stalemates by yourself, without computer help? You are a genius! But it surprises me that, once done the most difficult task (by far), which is finding the mainline, then you forget that c3 also does the task (not only c4). So, the minimum number of stalemates for Perft(19) are not 9, but 18, happening with two different final positions: white pawn in c3 and white pawn in c4. I am simply unable of looking for more mainlines, I am not a genius...Uri Blass wrote:perft(19) has at least 9 possible stalemates(both white and black has 3 options to change the order of moves at moves 1-3)
1.c4 2.h4 3.Qa4
1.h4 2.c4 3.Qa4
1.c4 2.Qa4 3.h4
1...h5 2...a5 3...Ra6
1...a5 2...h5 3...Ra6
1...a5 2...Ra6 3...h5
1.c4 h5 2.h4 a5 3.Qa4 Ra6 4.Qxa5 Rah6 5.Qxc7 f6 6.Qxd7+ Kf7 7.Qxb7 Qd3 8.Qxb8 Qh7 9.Qxc8 Kg6 10.Qe6 1/2-1/2
I wonder if people can calculate the number of stalemates in perft(19) and estimate this number for bigger n and the same for checkmates and I wonder if there is some n when the number of stalemates in perft(n) is bigger than the number of checkmates in perft(n).
I understand that you say that 19 is the minimum number of plies where a stalemate can happen. I have no idea about that, the same as the estimate, that should be clearly difficult: Perft(19) will be around 2.462e+19 according to François Labelle (his estimates are very accurate) and maybe one have to go through each position for verifying if a stalemate happens (only a guess). A huge task for everybody and an impossible task for me. Obviously, I also can not say anything about the comparison between checkmates and stalemates in Perft(n).
In conclusion: no help from mine, but I only can point out that c3 is also valid in your fabulous mainline. Well done!
Regards from Spain.
Ajedrecista.
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Re: Estimating the number of possible stalemates in Perft(n)
I did not find the game that lead to the stalemate by myself and it is a known stalemate for many years.
I believe that I heard about this stalemate many years ago in a chess lecture that happened after a tournament that I played(I think that it was more than 30 years ago but I am not sure)
I did not remember the moves but found them in some chess site today.
http://www.ichess.co.il/viewboard.html#!id=1032
You are right that c3 also achieve the same effect and that there are at least 18 stalemates with 19 plies.
I believe that I heard about this stalemate many years ago in a chess lecture that happened after a tournament that I played(I think that it was more than 30 years ago but I am not sure)
I did not remember the moves but found them in some chess site today.
http://www.ichess.co.il/viewboard.html#!id=1032
You are right that c3 also achieve the same effect and that there are at least 18 stalemates with 19 plies.
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Re: Estimating the number of possible stalemates in Perft(n)
Instead c4, White can play d3 with the idea Qd2, Qxa5 instead Qa4, Qxa5. Later Black will be playing 7...Qxd3 instead of 7...Qd3.
And then there is 1.e3 a5 2.Qh5 Ra6 3.Qxa5 h5 4.h4 ... etc.
Note that White can delay h4 until after Qxc7 before playing h4 in all of these.
And then there is 1.e3 a5 2.Qh5 Ra6 3.Qxa5 h5 4.h4 ... etc.
Note that White can delay h4 until after Qxc7 before playing h4 in all of these.
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Re: Estimating the number of possible stalemates in Perft(n)
Fixed last sentence that got mangled.rjgibert wrote:Instead c4, White can play d3 with the idea Qd2, Qxa5 instead Qa4, Qxa5. Later Black will be playing 7...Qxd3 instead of 7...Qd3.
And then there is 1.e3 a5 2.Qh5 Ra6 3.Qxa5 h5 4.h4 ... etc.
Note that White can delay h4 until after Qxc7 in all of these.