1 problem(s) found in 5893 milliseconds (displaying 1 problem(s)). [PROBID='P1243719'] [download as LaTeX]
1 - P1243719
Andrew Buchanan
Richard Stanley
8574 feenschach 144 11-12/2001
(a: 6+10)
(a: 6+9)
White to move. A to B in
(1) 5.5
(2) 6.0
(3) 6.5
(4) 7.0
(5) 7.5 etc...
Andrew Buchanan
Richard Stanley
8574 feenschach 144 11-12/2001
(a: 6+10)
(a: 6+9)
White to move. A to B in
(1) 5.5
(2) 6.0
(3) 6.5
(4) 7.0
(5) 7.5 etc...
(1,5,9...) 1. g3+ Kh5 2. g4+ Kh6 3. g5+ Kh7 4. fxg6+ Kg8! 5. f5 b1=L+ 6. Kd1!
(2,6,10...) 1. g3+ Kh5 2. g4+ Kh6 3. g5+ Kh7 4. fxg6+ Kh8! 5. f5 b1=L+ 6. Kd1! Kg8
(3,7,11...) 1. g3+ Kh5 2. g4+ Kh6 3. g5+ Kh7 4. fxg6+ Kh8! 5. f5 b1=L+ 6. Kc1! Kg8 7. Kd1
(4,8,12...) 1. g3+ Kh5 2. g4+ Kh6 3. g5+ Kh7 4. fxg6+ Kg8! 5. f5 b1=L+ 6. Kc1! Kh8 7. Kd1 Kg8 etc.
The point of the above problem is that it has infinitely many stipulations! Moreover, they are of the form A->B in k single moves, k + 1 single moves, k + 2 single moves... There was an attempt to make k as large as possible (with no solution in less than k single moves). This has the paradoxical effect that it is harder to make a problem which has a smaller set of valid stipulations!
It is much also easier to compose such a problem if the required
number of moves is always the same mod 2: k, k + 2, k + 4...
(2,6,10...) 1. g3+ Kh5 2. g4+ Kh6 3. g5+ Kh7 4. fxg6+ Kh8! 5. f5 b1=L+ 6. Kd1! Kg8
(3,7,11...) 1. g3+ Kh5 2. g4+ Kh6 3. g5+ Kh7 4. fxg6+ Kh8! 5. f5 b1=L+ 6. Kc1! Kg8 7. Kd1
(4,8,12...) 1. g3+ Kh5 2. g4+ Kh6 3. g5+ Kh7 4. fxg6+ Kg8! 5. f5 b1=L+ 6. Kc1! Kh8 7. Kd1 Kg8 etc.
The point of the above problem is that it has infinitely many stipulations! Moreover, they are of the form A->B in k single moves, k + 1 single moves, k + 2 single moves... There was an attempt to make k as large as possible (with no solution in less than k single moves). This has the paradoxical effect that it is harder to make a problem which has a smaller set of valid stipulations!
It is much also easier to compose such a problem if the required
number of moves is always the same mod 2: k, k + 2, k + 4...
Keywords: A to B, Path enumeration
Genre: Retro, Mathematics
FEN: 8/3pP1p1/6p1/5P2/2p2P1k/2p2p2/PpK2pP1/r7
Reprints: www.jacobi.com 30/10/2017
Input: A.Buchanan, 2012-07-01
Last update: A.Buchanan, 2020-01-03 more...
Genre: Retro, Mathematics
FEN: 8/3pP1p1/6p1/5P2/2p2P1k/2p2p2/PpK2pP1/r7
Reprints: www.jacobi.com 30/10/2017
Input: A.Buchanan, 2012-07-01
Last update: A.Buchanan, 2020-01-03 more...
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https://pdb.dieschwalbe.de/search.jsp?expression=PROBID%3D%27P1243719%27
The problems of this query have been registered by the following contributors:
A.Buchanan (1)
Alfred Pfeiffer: Andrew, there is a trick to do this:
1) Go into the edit mode with any other problem;
2) Replace in the command line of the browser the actual probid by that of the A=B-problem
(eg "http://pdb.dieschwalbe.de/edit.jsp?id=P1224189&positions=true" becomes to
"http://pdb.dieschwalbe.de/edit.jsp?id=P1243719&positions=true"), then click ENTER;
3) Edit the position as usual. (2017-09-13)
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