14 problem(s) found in 4045 milliseconds (displaying 14 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT G='Fairies' AND G='Mathematics'] [download as LaTeX]
1 - P0008784
Niels Høeg
The Chess Amateur 07/1926
(1+1)
Längste BP ohne Schach. Welches war der letzte Zug?
Niels Høeg
The Chess Amateur 07/1926
(1+1)
Längste BP ohne Schach. Welches war der letzte Zug?
R: 5899. Kg2xTh1
The game ends after 50 consecutive moves without captures or pawn moves (loss of castling right is not included here), or when there is not enough material to mate (say K-K or K-KS). There are 30 captures and 96 pawn moves (including 8 pawn captures) available, so the longest game seems to last (30+96-8)x50=5900 moves. This cannot be achieved because of the move-loss when the draw-preventing move shifts between white and black. Niels Høeg believed that 2 moves were lost and stated the solution as 5898.- Kb7xTa8. Karl Fabel later showed that only 1.5 moves need be lost.
Since 1926, there have been some relevant innovations to rules and conventions
(1) 50 move rule applies only to retro compositions, and will trigger automatically (no issue with writing down the move). (Codex 1953?)
(2) Removal of rules about draw by insufficient material (Laws 1997)
(3) Dead position rule introduced (Laws 1997)
(4) 75 move rule introduced (Laws 2014)
(5) Dead position rule applies only to retro compositions (Codex 2015)
(6) Articles 9.2 & 9.3 apply to chess problems - this includes 50 move rule and excludes 75 move rule (Codex 2019)
This is certainly a composition rather than a question about over the board chess. And it is certainly a retro composition. So the 50 move rule will dominate the 75 rule. The standard interpretation of interaction between 50 move rule and Dead Position in compositions is that Dead Position assessment *is* aware of looming automatic draw by 50 moves. (Note there is a similar assessment for interaction between Dead Position and Draw by Repetition.)
So we can argue that the game cannot last to 5898.5 moves, because the final move leads to a mandatory draw: either the king captures the last officer, or the king avoids the capture and the game ends in draw under the 50 move rule. So the position is dead at 5898.0. But even 5898.0 is too long for the diagram position with the kings so far apart. In the alternate reality if the last capture of a rook does not take place, there must be sufficient moves left for the kings to come together so that the rook can deliver checkmate. This will take at least 6.0 moves.
There is also still an ambiguity in the rules as to whether checkmate overrides draw by 50 moves. This is explicitly mentioned in the 75 move rule, but not in 50 move rule. I assume that checkmate *does* take priority.
Or does a valid problem only exist in the context of the rules and conventions that pertained at the time of its composition? The Codex does not opine on this general point.
Compare P1331022
The game ends after 50 consecutive moves without captures or pawn moves (loss of castling right is not included here), or when there is not enough material to mate (say K-K or K-KS). There are 30 captures and 96 pawn moves (including 8 pawn captures) available, so the longest game seems to last (30+96-8)x50=5900 moves. This cannot be achieved because of the move-loss when the draw-preventing move shifts between white and black. Niels Høeg believed that 2 moves were lost and stated the solution as 5898.- Kb7xTa8. Karl Fabel later showed that only 1.5 moves need be lost.
Since 1926, there have been some relevant innovations to rules and conventions
(1) 50 move rule applies only to retro compositions, and will trigger automatically (no issue with writing down the move). (Codex 1953?)
(2) Removal of rules about draw by insufficient material (Laws 1997)
(3) Dead position rule introduced (Laws 1997)
(4) 75 move rule introduced (Laws 2014)
(5) Dead position rule applies only to retro compositions (Codex 2015)
(6) Articles 9.2 & 9.3 apply to chess problems - this includes 50 move rule and excludes 75 move rule (Codex 2019)
This is certainly a composition rather than a question about over the board chess. And it is certainly a retro composition. So the 50 move rule will dominate the 75 rule. The standard interpretation of interaction between 50 move rule and Dead Position in compositions is that Dead Position assessment *is* aware of looming automatic draw by 50 moves. (Note there is a similar assessment for interaction between Dead Position and Draw by Repetition.)
So we can argue that the game cannot last to 5898.5 moves, because the final move leads to a mandatory draw: either the king captures the last officer, or the king avoids the capture and the game ends in draw under the 50 move rule. So the position is dead at 5898.0. But even 5898.0 is too long for the diagram position with the kings so far apart. In the alternate reality if the last capture of a rook does not take place, there must be sufficient moves left for the kings to come together so that the rook can deliver checkmate. This will take at least 6.0 moves.
There is also still an ambiguity in the rules as to whether checkmate overrides draw by 50 moves. This is explicitly mentioned in the 75 move rule, but not in 50 move rule. I assume that checkmate *does* take priority.
Or does a valid problem only exist in the context of the rules and conventions that pertained at the time of its composition? The Codex does not opine on this general point.
Compare P1331022
Duplicate Diagram: P1101148, P1189676, P1191185, P1304589
Keywords: Longest Proof Game, Last Move?, only Kings, Non-Unique Proof Game, Dead Position, 50 move rule, Constrained problem, Type A, Miniature, Golden Age (pre-dead), Aristocrat
Genre: Mathematics, Retro
FEN: k7/8/8/8/8/8/8/7K
Reprints: Schackproblemet 1928
Schach und Zahl 1966
Input: Gerd Wilts, 1997-06-20
Last update: A.Buchanan, 2024-01-18 more...
Genre: Mathematics, Retro
FEN: k7/8/8/8/8/8/8/7K
Reprints: Schackproblemet 1928
Schach und Zahl 1966
Input: Gerd Wilts, 1997-06-20
Last update: A.Buchanan, 2024-01-18 more...
2 - P1000273
Mario Velucchi
Matematika na shahmatnoy doske 1976
(0+0)
Stelle 4DD und 1 S so auf das Brett, dass alle nicht besetzten Felder gedeckt sind!
Mario Velucchi
Matematika na shahmatnoy doske 1976
(0+0)
Stelle 4DD und 1 S so auf das Brett, dass alle nicht besetzten Felder gedeckt sind!
a) 8 symmetries
Keywords: Construction task, Aristocrat, Miniature
Genre: Mathematics
FEN: 8/8/8/8/8/8/8/8
Reprints: 10806 Die Schwalbe 183 06/2000
Input: Gerd Wilts, 2000-08-01
Last update: A.Buchanan, 2023-06-03 more...
Genre: Mathematics
FEN: 8/8/8/8/8/8/8/8
Reprints: 10806 Die Schwalbe 183 06/2000
Input: Gerd Wilts, 2000-08-01
Last update: A.Buchanan, 2023-06-03 more...
3 - P1178762
Colin V. Vaughan
9 feenschach 32 01-03/1976
(15+2)
14 freiwillige Doppelschach-Matts mit Umwandlungsfiguren
Colin V. Vaughan
9 feenschach 32 01-03/1976
(15+2)
14 freiwillige Doppelschach-Matts mit Umwandlungsfiguren
Keywords: Einzügerrekord (B1d4/D), Non-standard material (TTLSSSSS)
Genre: Mathematics
FEN: K2R2B1/3N4/2p1N3/RN1k1NR1/2N1N3/3N4/B2R2Q1/8
Input: HBae, 2010-10-26
Last update: A.Buchanan, 2020-12-13 more...
Genre: Mathematics
FEN: K2R2B1/3N4/2p1N3/RN1k1NR1/2N1N3/3N4/B2R2Q1/8
Input: HBae, 2010-10-26
Last update: A.Buchanan, 2020-12-13 more...
4 - P1182074
William A. Shinkman
666 The Golden Argosy , p. 286, 1929
(5+5)
Pattaufhebung, wodurch Schwarz 32 Züge erhält
William A. Shinkman
666 The Golden Argosy , p. 286, 1929
(5+5)
Pattaufhebung, wodurch Schwarz 32 Züge erhält
1. Lb4
Romantischer Rekord
Romantischer Rekord
A.Buchanan: This kind of problem is traditionally in PDB placed in the mathematics genre, rather than fairy, because it involves, you know, counting stuff (2020-11-22)
comment
comment
Keywords: Construction record
Genre: Mathematics
FEN: 8/8/7K/p7/R3b2k/6rP/4p3/4B1N1
Input: HBae, 2010-12-09
Last update: A.Buchanan, 2020-11-22 more...
Genre: Mathematics
FEN: 8/8/7K/p7/R3b2k/6rP/4p3/4B1N1
Input: HBae, 2010-12-09
Last update: A.Buchanan, 2020-11-22 more...
5 - P1249837
George Burt Spencer
St. Paul Dispatch 1906
(34+15) C+
#2 auf jeder Reihe und jeder Linie
George Burt Spencer
St. Paul Dispatch 1906
(34+15) C+
#2 auf jeder Reihe und jeder Linie
a) 1. Lc2! ... 2. Ld3# 1. ... Kxa1 Lc1#
b) 1. Dd4! ... 2. Dxb4#
c) 1. Dxc6! ... 2. Dxc5#
d) 1. Txd4 ... 2. Txd5 1. ... Kc5 2. Db4# 1. ... Ke5 2. Df4#
e) 1. De4! ... 2. Db1#,Dh1# 1. ... Kd1 2. Db1# 1. ... Kf1 2. Dh1# (Fleck)
f) 1. Ld5+! Ke8 2. Dc8#
g) 1. Dd7! Kf8 2. Df7# 1. ... Kh8 2. Dh7#
h) 1. De4! ... 2. Dg4#
i) 1. Ld4! ... 2. Sg3#
j) 1. Db4! Ka1 2. Da3#,Db1# (dual)
k) 1. Lgd6! Kc2 2. Ld1# 1. ... Kc4 2. Ld5#
l) 1. Dc8+! Kd3 2. Dc2# 1. ... Kd5 2. Dc6#
m) 1. Ta3! ... 2. Th3#
n) 1. Kb7! ... 2. Dxc6#
o) 1. De5! ... 2. Dg7#
p) 1. Tc7! Kh8 2. Dxf8#
b) 1. Dd4! ... 2. Dxb4#
c) 1. Dxc6! ... 2. Dxc5#
d) 1. Txd4 ... 2. Txd5 1. ... Kc5 2. Db4# 1. ... Ke5 2. Df4#
e) 1. De4! ... 2. Db1#,Dh1# 1. ... Kd1 2. Db1# 1. ... Kf1 2. Dh1# (Fleck)
f) 1. Ld5+! Ke8 2. Dc8#
g) 1. Dd7! Kf8 2. Df7# 1. ... Kh8 2. Dh7#
h) 1. De4! ... 2. Dg4#
i) 1. Ld4! ... 2. Sg3#
j) 1. Db4! Ka1 2. Da3#,Db1# (dual)
k) 1. Lgd6! Kc2 2. Ld1# 1. ... Kc4 2. Ld5#
l) 1. Dc8+! Kd3 2. Dc2# 1. ... Kd5 2. Dc6#
m) 1. Ta3! ... 2. Th3#
n) 1. Kb7! ... 2. Dxc6#
o) 1. De5! ... 2. Dg7#
p) 1. Tc7! Kh8 2. Dxf8#
Duplicate Diagram: P1126102, P1129822, P1145270, P1145376, P1149794, P1159233, P1164692, P1164885, P1175321
1. d7! and e.g. d3 2. d8=L b5 3. a8=S gxf5 4. g8=D h2 5. h8=T h1=D 6. Txh1 dxc2 7. De6+ Lxe6#
Author's Note: Black can play d3, b5, gxf5, and h2 in any order, with corresponding White promotions as replies, giving all 24 permutations of White AUW (LSDT, LDST, etc.). If Black plays g5 instead of gxf5, then White replies fxg5 instead of promoting to D. A white pawn added at g5 would eliminate this pesky variation, but the white pawns are all used up!
Author's Note: Black can play d3, b5, gxf5, and h2 in any order, with corresponding White promotions as replies, giving all 24 permutations of White AUW (LSDT, LDST, etc.). If Black plays g5 instead of gxf5, then White replies fxg5 instead of promoting to D. A white pawn added at g5 would eliminate this pesky variation, but the white pawns are all used up!
SCHRECKE: Unlösbar, eventuell fehlt ein wBf4. (2020-10-28)
A.Buchanan: Mark Kirtley wrote to me: George found the original issue of the U.S. Problem Bulletin, showed me, and yes, there is a wPf4 in the original. George and I couldn't remember why we also had a wPa2, but a new computer test (Gustav) showed us the reason: it guards against a major dual Rh1-a1 in some variations.
There is still a minor dual in some variations (of varying length) if the bP promotes to a Q at h1, for then wRh8-h6+, bQh1xh6, and wQe6+ and bQxe6#. (2020-11-11)
A.Buchanan: So I have corrected the confirmed diagram typo (2020-11-11)
A.Buchanan: Have posted a simplified representation of the solution, at the author's request (2021-01-27)
Henrik Juel: So the problem is C+, except for the minor dual mentioned (2021-01-27)
comment
A.Buchanan: Mark Kirtley wrote to me: George found the original issue of the U.S. Problem Bulletin, showed me, and yes, there is a wPf4 in the original. George and I couldn't remember why we also had a wPa2, but a new computer test (Gustav) showed us the reason: it guards against a major dual Rh1-a1 in some variations.
There is still a minor dual in some variations (of varying length) if the bP promotes to a Q at h1, for then wRh8-h6+, bQh1xh6, and wQe6+ and bQxe6#. (2020-11-11)
A.Buchanan: So I have corrected the confirmed diagram typo (2020-11-11)
A.Buchanan: Have posted a simplified representation of the solution, at the author's request (2021-01-27)
Henrik Juel: So the problem is C+, except for the minor dual mentioned (2021-01-27)
comment
Keywords: Allumwandlung, konsekutive Umwandlungen 4 (DTLS)
Genre: s#, Mathematics
Computer test: Gustav 4.1c
FEN: 1BK1N3/P1P3PP/NpkP2p1/5Q2/1Rbp1P2/7p/P1RP4/8
Input: Frank Müller, 2013-09-18
Last update: James Malcom, 2021-01-28 more...
Genre: s#, Mathematics
Computer test: Gustav 4.1c
FEN: 1BK1N3/P1P3PP/NpkP2p1/5Q2/1Rbp1P2/7p/P1RP4/8
Input: Frank Müller, 2013-09-18
Last update: James Malcom, 2021-01-28 more...
7 - P1369313
Klaus Funk
8 Die Schwalbe 210, p. 616, 12/2004
(14+2)
20 erzwungene pattsetzende Schachparaden mit Uf
Klaus Funk
8 Die Schwalbe 210, p. 616, 12/2004
(14+2)
20 erzwungene pattsetzende Schachparaden mit Uf
mit Umwandlungsfiguren
TBr: Diagrammfehler korrigiert (falsch: wLc7) (2021-04-15)
A.Buchanan: Since legality is a (minor) issue in setting up this position, can argue that retro concerns apply, hence DP rule. So the last move was not e6-e5, but must have been d6xe5+ or f6xe5+ with living alternative d5 or f5. So that would account for the last missing White unit (apart from the irrelevant White light-squared bishop). (2021-04-15)
comment
A.Buchanan: Since legality is a (minor) issue in setting up this position, can argue that retro concerns apply, hence DP rule. So the last move was not e6-e5, but must have been d6xe5+ or f6xe5+ with living alternative d5 or f5. So that would account for the last missing White unit (apart from the irrelevant White light-squared bishop). (2021-04-15)
comment
14 double check mates
http://matplus.net/start.php?px=1607750786&app=forum&act=posts&fid=gen&tid=2281
Genre: Mathematics
FEN: 4Q3/2B1N1B1/3R1R2/1QN1k1NQ/5R2/2K1N1B1/4Q3/4n3
Input: James Malcom, 2020-12-13
Last update: James Malcom, 2021-04-19 more...
9 - P1384831
Lion Xray
Facebook 31/12/2020
(9+9)
In how many ways can you add a White Queen and a Black Queen to this diagram to obtain a legal position?
Lion Xray
Facebook 31/12/2020
(9+9)
In how many ways can you add a White Queen and a Black Queen to this diagram to obtain a legal position?
2021 solutions
A.Buchanan: beautiful! (2021-01-01)
Henrik Juel: Yes, beautiful simplicity; finally one I can handle...
There are 64-9-9 = 46 empty squares, so the queens can be added in 46x45 = 2070 ways
The only illegality is that both kings are in check, which happens in 7x7 = 49 ways
Answer: 2070-49 = 2021 (2021-01-01)
A.Buchanan: Oh the other two today aren't hard, Henrik. But this one is a real gift from the Gods (2021-01-01)
comment
Henrik Juel: Yes, beautiful simplicity; finally one I can handle...
There are 64-9-9 = 46 empty squares, so the queens can be added in 46x45 = 2070 ways
The only illegality is that both kings are in check, which happens in 7x7 = 49 ways
Answer: 2070-49 = 2021 (2021-01-01)
A.Buchanan: Oh the other two today aren't hard, Henrik. But this one is a real gift from the Gods (2021-01-01)
comment
Keywords: Add pieces (Dd), Symmetrical position, Kindergarten Problem
Genre: Retro, Mathematics
FEN: 4k3/3ppp2/ppp3pp/8/8/PPP3PP/3PPP2/4K3
Reprints: Retros mailing list 01/01/2021
Input: A.Buchanan, 2021-01-01
Last update: A.Buchanan, 2021-01-01 more...
Genre: Retro, Mathematics
FEN: 4k3/3ppp2/ppp3pp/8/8/PPP3PP/3PPP2/4K3
Reprints: Retros mailing list 01/01/2021
Input: A.Buchanan, 2021-01-01
Last update: A.Buchanan, 2021-01-01 more...
10 - P1387362
Manfred Zucker
Freie Presse (Chemnitz) 24/06/1977
(1+1)
Wie viele verschiedene letzte Züge sind möglich?
Manfred Zucker
Freie Presse (Chemnitz) 24/06/1977
(1+1)
Wie viele verschiedene letzte Züge sind möglich?
29 Lösungen
Gefragt war natürlich nach dem letzten Einzelzug. Nun kann zuletzt sowohl der weiße als auch der der schwarze König gezogen haben. Der weiße König kann von drei verschiedenen Feldern (a2,b1,b2) gekommen sein, und er kann auf a1 eine schwarze Dame, einen schwarzen Turm, einen schwarzen Läufer, einen schwarzen Springer oder nichts geschlagen haben, das sind 3x5=15 Möglichkeiten. Dasselbe trifft auf den schwarzen König zu, insgesamt ergeben sich so 30 Möglichkeiten für den letzten Zug - und doch ist diese Lösung nicht richtig! Die Stellung wKb2-sLa1 ist partiemöglich (letzte Züge: 1. ... a2-a2=L+! 2. Kxa1), die Stellung sKg2-wLh1 jedoch nicht, denn der weiße Läufer kann (bei schwarzer Königsstellung auf g2) auf keine Weise nach h1 gekommen sein! Diese partieunmögliche Stellung war also zu subtrahieren, das richtige Ergebnis mußte "29" lauten.
Gefragt war natürlich nach dem letzten Einzelzug. Nun kann zuletzt sowohl der weiße als auch der der schwarze König gezogen haben. Der weiße König kann von drei verschiedenen Feldern (a2,b1,b2) gekommen sein, und er kann auf a1 eine schwarze Dame, einen schwarzen Turm, einen schwarzen Läufer, einen schwarzen Springer oder nichts geschlagen haben, das sind 3x5=15 Möglichkeiten. Dasselbe trifft auf den schwarzen König zu, insgesamt ergeben sich so 30 Möglichkeiten für den letzten Zug - und doch ist diese Lösung nicht richtig! Die Stellung wKb2-sLa1 ist partiemöglich (letzte Züge: 1. ... a2-a2=L+! 2. Kxa1), die Stellung sKg2-wLh1 jedoch nicht, denn der weiße Läufer kann (bei schwarzer Königsstellung auf g2) auf keine Weise nach h1 gekommen sein! Diese partieunmögliche Stellung war also zu subtrahieren, das richtige Ergebnis mußte "29" lauten.
Duplicate Diagram: P1207607
Dazu schreibt die FP: "Wie viele verschiedene letzte Züge sind möglich? Noch ein kleiner Hinweis: Beachten Sie beim Zählen bitte auch, daß die Könige zuletzt geschlagen haben können.".
(MZ433)
.
Henrik Juel: solution
2x3x5 - 1 = 29
R: Kg2xLh1 is illegal (2021-03-12)
A.Buchanan: This preceded the DP rule by 20 years, so under “Golden Age” principle, that rule does not apply, although it would be sound either way. Under DP rule there would have been 2x2x3=12 possible last moves, as a rook or queen must have just been captured (2021-03-13)
A.Buchanan: Is an only kings position an aristocrat or a kindergarten? :) Is it a one liner? :) I think the answer should be no, no and no (2021-03-13)
comment
(MZ433)
.
Henrik Juel: solution
2x3x5 - 1 = 29
R: Kg2xLh1 is illegal (2021-03-12)
A.Buchanan: This preceded the DP rule by 20 years, so under “Golden Age” principle, that rule does not apply, although it would be sound either way. Under DP rule there would have been 2x2x3=12 possible last moves, as a rook or queen must have just been captured (2021-03-13)
A.Buchanan: Is an only kings position an aristocrat or a kindergarten? :) Is it a one liner? :) I think the answer should be no, no and no (2021-03-13)
comment
Keywords: Aristocrat, Miniature, only Kings, Golden Age, Last Move?, Symmetrical position, Asymmetrical solution
Genre: Retro, Mathematics
FEN: 8/8/8/8/8/8/8/K6k
Input: Felber, Volker, 2021-03-12
Last update: A.Buchanan, 2021-03-13 more...
Genre: Retro, Mathematics
FEN: 8/8/8/8/8/8/8/K6k
Input: Felber, Volker, 2021-03-12
Last update: A.Buchanan, 2021-03-13 more...
11 - P1407174
Noam D. Elkies
Retros mailing list 01/01/2023
(16+16) C+
PG in 10.0
How many solutions?
Noam D. Elkies
Retros mailing list 01/01/2023
(16+16) C+
PG in 10.0
How many solutions?
1. e4 e5 2. Ke2 Df6 3. Kd3 Db6 4. h4 d6 5. Th3 Sd7 6. Kc4 Le7 7. Te3 Sf8 8. Kd5 Lf6 9. Lc4 Ld7 10. Se2 Se7# for example
2023 solutions
2023 solutions
Henrik Juel: Surely, there are 2023 solutions
But how do you compute it? (2023-01-08)
Moldenhauer: Computerprüfung: C+ Jacobi v0.7.6 beta1 Lösungen: 2023 sind bestätigt!
Wie kommt man auf sowas? Interessiert auch mich. (2023-01-09)
A.Buchanan: White & Black play are almost independent. Because Black's first move is e5, wK route is determined. It's nearly C(10,3), but wRd3 would block e2-e4, so subtract one solution to make 120-1 = 119 = 7x17. Black's play is nearly as simple: C(9,2) fails due to collision on f8 3 times, and due to collision on f6 1 time, so 21-3-1 = 17. Note the final move is checkmate, a signature flourish that Noam likes to apply (2023-01-09)
comment
But how do you compute it? (2023-01-08)
Moldenhauer: Computerprüfung: C+ Jacobi v0.7.6 beta1 Lösungen: 2023 sind bestätigt!
Wie kommt man auf sowas? Interessiert auch mich. (2023-01-09)
A.Buchanan: White & Black play are almost independent. Because Black's first move is e5, wK route is determined. It's nearly C(10,3), but wRd3 would block e2-e4, so subtract one solution to make 120-1 = 119 = 7x17. Black's play is nearly as simple: C(9,2) fails due to collision on f8 3 times, and due to collision on f6 1 time, so 21-3-1 = 17. Note the final move is checkmate, a signature flourish that Noam likes to apply (2023-01-09)
comment
Keywords: Path enumeration (year), Non-Unique Proof Game, Capture-free
Genre: Retro, Mathematics
Computer test: C+ Jacobi v0.7.6 beta1 Lösungen: 2023 sind bestätigt!
FEN: r3kn1r/pppbnppp/1q1p1b2/3Kp3/2B1P2P/4R3/PPPPNPP1/RNBQ4
Input: A.Buchanan, 2023-01-08
Last update: A.Buchanan, 2024-01-02 more...
Genre: Retro, Mathematics
Computer test: C+ Jacobi v0.7.6 beta1 Lösungen: 2023 sind bestätigt!
FEN: r3kn1r/pppbnppp/1q1p1b2/3Kp3/2B1P2P/4R3/PPPPNPP1/RNBQ4
Input: A.Buchanan, 2023-01-08
Last update: A.Buchanan, 2024-01-02 more...
1. a4 2. Ta3 3. d4 4. h4 5. Thh3 6. Thf3 7. Tf6 8. f4 9. f5 10. Lf4 11. e3 12. Lc4 13. Ld5 14. c4 15. Db3 16. Db6 17. b4 18. Tb3 19. b5 for example
2023 solutions
2023 solutions
Joost de Heer: See Problemas april 2023 for a detailed explanation on how to calculate the number of solutions. (2023-04-04)
A.Buchanan: Hi Joost thanks for the comment. This problem basically is about "triangle + triangle = square", but the article uses a different mechanism (Elastic Belt) to reach the same number (2023-04-05)
comment
A.Buchanan: Hi Joost thanks for the comment. This problem basically is about "triangle + triangle = square", but the article uses a different mechanism (Elastic Belt) to reach the same number (2023-04-05)
comment
Keywords: Path enumeration (year), Seriesmover, Non-Unique Proof Game
Genre: Retro, Mathematics
FEN: 8/8/1Q3R2/1P1B1P2/P1PP1B1P/1R2P3/6P1/1N2K1N1
Input: A.Buchanan, 2023-01-08
Last update: A.Buchanan, 2023-08-27 more...
Genre: Retro, Mathematics
FEN: 8/8/1Q3R2/1P1B1P2/P1PP1B1P/1R2P3/6P1/1N2K1N1
Input: A.Buchanan, 2023-01-08
Last update: A.Buchanan, 2023-08-27 more...
13 - P1410557
Andrew Buchanan
1002 Mathematics at DEC 30/12/1988
(15+4)
White to move. This position has a certain curious property. What is it?
Andrew Buchanan
1002 Mathematics at DEC 30/12/1988
(15+4)
White to move. This position has a certain curious property. What is it?
If one counts the number of legal moves for each white pieces, one gets exactly 0 through 14.
Henrik Juel: Here are the 15 pieces, in increasing order of number of moves:
Pd3, Pe6, Ph2, Pc2, Pc7, Sd2, Sf4, Kc5, Pd7, Ld4, Lf3, Tg5, Pf7, Ta1, Da8 (2023-06-27)
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Pd3, Pe6, Ph2, Pc2, Pc7, Sd2, Sf4, Kc5, Pd7, Ld4, Lf3, Tg5, Pf7, Ta1, Da8 (2023-06-27)
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Keywords: Construction task
Genre: Mathematics
FEN: Q3b1n1/2PP1P1k/4P3/2K3R1/3B1N2/1p1P1B2/2PN3P/R7
Input: Mu-Tsun Tsai, 2023-06-27
Last update: A.Buchanan, 2023-06-28 more...
Genre: Mathematics
FEN: Q3b1n1/2PP1P1k/4P3/2K3R1/3B1N2/1p1P1B2/2PN3P/R7
Input: Mu-Tsun Tsai, 2023-06-27
Last update: A.Buchanan, 2023-06-28 more...
14 - P1414063
Andrew Buchanan
The Puzzling Side of Chess 228 31/10/2023
(14+2)
Dead position, with check, maximizing the number of legal but unplayable moves.
Andrew Buchanan
The Puzzling Side of Chess 228 31/10/2023
(14+2)
Dead position, with check, maximizing the number of legal but unplayable moves.
20 legal but unplayable moves.
Black just played c6xd5+ or exd5+, not d6-d5? which would have ended the game in prior dead position.
Henrik Juel: The position is displayed by clicking on one of the five animation symbols and then moving the cursor (2023-12-14)
Henrik Juel: ... and then moving the cursor up (2023-12-14)
A.Buchanan: I think the phrase "legal but unplayable" is happier than "impossible", so I will use the former (2023-12-14)
A.Buchanan: Have reverted to the mode in which PDB shows the diagram directly, rather than having to click animation. (2023-12-16)
more ...
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Henrik Juel: ... and then moving the cursor up (2023-12-14)
A.Buchanan: I think the phrase "legal but unplayable" is happier than "impossible", so I will use the former (2023-12-14)
A.Buchanan: Have reverted to the mode in which PDB shows the diagram directly, rather than having to click animation. (2023-12-16)
more ...
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Keywords: Construction record, Minimal, Dead Position, Type C, Non-standard material (TTLSSSS)
Genre: Retro, Mathematics
FEN: B5Bk/2NRN3/1N3N2/R2p2R1/2KR1N2/4N3/6Q1/8
Input: A.Buchanan, 2023-12-14
Last update: A.Buchanan, 2023-12-16 more...
Genre: Retro, Mathematics
FEN: B5Bk/2NRN3/1N3N2/R2p2R1/2KR1N2/4N3/6Q1/8
Input: A.Buchanan, 2023-12-14
Last update: A.Buchanan, 2023-12-16 more...
Show statistic for complete result. Show search result faster by using ids.
https://pdb.dieschwalbe.de/search.jsp?expression=COMMENTDATE%3E%3D20200919+AND+NOT+K%3D%27Hilfsr%C3%BCckz%C3%BCger%27+AND+NOT+G%3D%27Fairies%27+AND+G%3D%27Mathematics%27
The problems of this query have been registered by the following contributors:
Gerd Wilts (2)HBae (3)
Nikolai Beluhov (1)
Frank Müller (1)
James Malcom (1)
A.Buchanan (4)
Felber, Volker (1)
Mu-Tsun Tsai (1)
A.Buchanan: The old intended interpretation is protected under the Golden Age principle. Suppose we do apply modern rules, codex & clarifications. If Black just played Kb7xRa8, then the alternative leading to mate takes 7.0 moves, so this must have been Black's 5892nd move at the latest. If White just moved, then it was White's 5892nd at latest. So to maximize the length of the game, Black moved last. (2023-09-06) edit (2023-09-06)
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