Die Schwalbe

19 problem(s) found in 4107 milliseconds (displaying 19 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT A='Carbo y Batlle, Juan' AND G='Mathematics'] [download as LaTeX]

1 - P0008784
Niels Høeg
The Chess Amateur 07/1926
P0008784
(1+1)
Längste BP ohne Schach. Welches war der letzte Zug?
R: 5899. Kg2xTh1
play all play one stop play next play all
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

A.Buchanan: There's a question whether DP rule has visibility of 3Rep & 50M state. The current consensus among most of the tiny group who might care is that for retros, it does have visibility, but for purely forward problems, it does not. This align with the idea that by default 50M & DP rules apply only to retros (2023-09-06)
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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comment
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...
2 - P1000273
Mario Velucchi
Matematika na shahmatnoy doske 1976
P1000273
(0+0)
Stelle 4DD und 1 S so auf das Brett, dass alle nicht besetzten Felder gedeckt sind!
a) 8 symmetries
play all play one stop play next play all
more ...
comment
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...
3 - P1178762
Colin V. Vaughan
9 feenschach 32 01-03/1976
P1178762
(15+2)
14 freiwillige Doppelschach-Matts mit Umwandlungsfiguren
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comment
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...
4 - P1182074
William A. Shinkman
666 The Golden Argosy , p. 286, 1929
P1182074
(5+5)
Pattaufhebung, wodurch Schwarz 32 Züge erhält
1. Lb4
play all play one stop play next play all
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
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...
5 - P1249837
George Burt Spencer
St. Paul Dispatch 1906
P1249837
(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#
play all play one stop play next play all

Duplicate Diagram: P1126102, P1129822, P1145270, P1145376, P1149794, P1159233, P1164692, P1164885, P1175321

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comment
Keywords: Oneliner (16)
Genre: 2#, Mathematics
Computer test: Popeye Windows-64Bit v4.69
FEN: 2RKQbk1/2QB1k1K/KQpk1BPN/RKpp1QPk/Bpkp1KQn/BkNRKBBN/k1BQP1K1/BBK1kN1Q
Input: Nikolai Beluhov, 2012-09-24
Last update: A.Buchanan, 2020-11-27 more...
6 - P1258968
Unto Heinonen
Suomen Tehtäväniekat 12/1993
P1258968
(5+3) C+
ser-h#14. How many solutions?
1. h5 2. h4 3. h3 4. h2 5. h1=D 6. Dg1 7. Da7 8. h5 9. h4 10. h3 11. h2 12. h1=D 13. Dhg1 14. Dgb6 Sb4#
is one example.
play all play one stop play next play all
The first and last moves are forced, but in between there is a 6+6 queue.
Catalan numbers C_n = (2n)!/(n!(n+1)!).
Here there are C_6 = (12!/(6!7!) = 132 solutions.
Problem 1: Suomen Tehtäväniekat Christmas solving contest Dec 1993.
paul: Unique solution with Relay Chess. (2021-08-16)
more ...
comment
Keywords: Path enumeration (Catalan), Seriesmover
Genre: Mathematics, Fairies
Computer test: Gustav 4.1 d gives 132 lines in solution file
FEN: 8/1N5p/k1P4p/8/8/3N4/8/1K3B2
Input: A.Buchanan, 2013-01-11
Last update: Olaf Jenkner, 2021-08-16 more...
7 - P1277937
Mark Kirtley
George P. Sphicas

1801 U.S. Problem Bulletin 05-06/1989
P1277937
(15+6) C+
s#7
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#
play all play one stop play next play all
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
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...
8 - P1312478
Nenad Petrovic
Problem 7, p. 109, 03/1952
P1312478
(9+4)
#2
Projective Chess
Solution:
1. Kg1!
1. ... Ke4,Kf3 2. D-PH#
1. ... d2,Kg4,Kg5 2. D-PD1#
1. ... Sc7~ 2. D-PV#
1. ... Sb6~ 2. D-PD2#

Non-solutions
1. D-PD2? Sc7a8,Sc7d5!
1. D-PH+? Kg5!
1. D-PD1+? Ke4!
1. Kh1,Kg2? Sb6a8,Sb6d5!
1. Kh3? Kf3!
1. h6? Kg5!
1. a3,a4,a7,d6,f6? Sb6~!
1. Dxb6,Dxc7,Dc6,Df2,Dg1,Db4,Da3? Ke5!
1. De5? Kxe5!
1. De3? Kxe3!
1. Dd4+ Kf3!
1. Db5,Dd5? SxD!
1. Da5? Sc7~! (the reason I think for wBa2a6)
play all play one stop play next play all
In der Lösung bezeichnet 'PH' das projektive Feld in horizontaler Richtung; 'PV' das projektive Feld in vertikaler Richtung; 'PD1' ist das projektive Feld in Richtung der Diagonale a1-h8; 'PD2' ist das projektive Feld in Richtung der Diagonale a8-h1.

A.Buchanan: Together with excellent paradoxical key, and the 4 clean mating lines, a large number of potential cooks are prevented quite economically (see non-solutions, above). wBf6 in particular has many roles. So on this occasion a minor dual at one point later in the solution seems completely acceptable. A very original composition!
Interessanterweise werden in der Lösung alle vier unendlich entfernten Felder des projektiven Schachbretts benutzt.
A.Buchanan: In projective geometry, every family of parallel straight lines intersects at an infinitely distant point. Chess problem composers in the former Yugoslavia have adapted this idea for the chessboard, adding four special squares “at infinity.”

Now a queen on a bare board, for example, can zoom off to the west (or east) and reach a square “at infinity” from which she attacks every rank on the board simultaneously from both directions. She might also zoom to the north (or south) to reach a different square at infinity; from this one she attacks every file simultaneously, again from both directions. Finally she can zoom to the northwest or southeast and attack all the diagonals parallel to a8-h1, or zoom to the northeast or southwest and attack all the diagonals parallel to a1-h8. These four “infinity squares,” plus the regular board, make up the field of play. (2015-11-20)
Henrik Juel: Fine explanation, Andrew, but I still need to see the intended solution (if you know it); this may also clarify the weird stipulation (2015-11-20)
Alfred Pfeiffer: Diese Schachvariante wurde 1952 von Prof. Lav Rajcic und Nenad Petrovic erfunden (D.B.Pritchard: The Encyclopedia of Chess Variants) (2015-11-20)
A.Buchanan: The articulate explanation is not mine: I found it at http://www.futilitycloset.com/?s=chess
I think one needs to add that only line pieces have the privilege to leave the regular board in this way. Which suggests another variant with Nightriders. Now there are another 4 squares accessible only by Nightriders. (2015-11-21)
Henrik Juel: Thanks, Andrew and Alfred
Nice key, but too bad about the trial after 1... Kg3 (2015-11-21)
Henrik Juel: Andrew has pointed out that the flaw after 1... Kg3 is only a dual (2015-11-22)
A.Buchanan: Someone asks me: is there a cook? 1. D-PD2 Sc7a8,Sc7d5 can be answered by 2. Kh3# (2020-11-10)
A.Buchanan: By the way: the squares at infinity are on a straight line, but that cannot be navigated by the queen. For suppose the contrary: then the direction of movement along that line would have to be parallel to some direction of movement of the queen, i.e. by symmetry parallel to *all* directions of movement of a queen, i.e. since lines parallel to one line are parallel to one another, we conclude that all lines on the board are parallel to one another. This is absurd, so the queen cannot navigate the line at infinity (2020-11-10)
comment
Keywords: Projective Chess, no 8x8 board
Genre: Fairies, Mathematics
FEN: 8/2n5/Pn1P1P2/2Q2P1P/2P2k2/3p4/P6K/8
Reprints: Matematika Na Shahmatnoi Doske
Schachmaty i matematika , p. 119, 1983
Schach und Mathematik [Gik] , p. 150, 1986
Input: A.Buchanan, 2015-11-20
Last update: A.Buchanan, 2020-11-10 more...
9 - P1369313
Klaus Funk
8 Die Schwalbe 210, p. 616, 12/2004
P1369313
(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
Keywords: Einzügerrekord, Minimal
Genre: Mathematics
FEN: k7/2QNRNB1/2N3N1/1R2p2R/2N1RK2/2BN4/8/8
Input: HBae, 2019-11-14
Last update: TBr, 2021-04-15 more...
10 - P1382970
James Malcom
MatPlus.net Forum 3/3/2019
P1382970
(15+2)
#1 with double check

14 double check mates

http://matplus.net/start.php?px=1607750786&app=forum&act=posts&fid=gen&tid=2281
HBae: 14 Züge wurden schon 1976 dargestellt, siehe P1178762 (2020-12-13)
comment

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...
11 - P1384831
Lion Xray
Facebook 31/12/2020
P1384831
(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
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...
12 - P1387362
Manfred Zucker
Freie Presse (Chemnitz) 24/06/1977
P1387362
(1+1)
Wie viele verschiedene letzte Züge sind möglich?
29 Lösungen
play all play one stop play next play all
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
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...
13 - P1391117
Andrew Buchanan
Discord Chess Problems & Studies Server 26/06/2021
P1391117
(3+0)
If playing chess on a 5x5 toroidal board, which is abstractly the most powerful: a bishop, a knight or a rook?
[Example pieces are shown in the diagram for visual jollity, particular locations don't matter.]

All 3 pieces can move to 8 squares in 1 move, and any square on the board in 2 moves. Indeed any two directions can be mapped by a linear transformation onto any two directions. I.e. any of these pieces, which can be regarded as moving in two directions, can be transformed to any other, or indeed to a "chimera" e.g. something which moves N-S, or diagonally NW-SE: half-rook and half-bishop.

So given that all of these lines are equivalent, how could one piece be better than another? Bishops and rooks are riders, so can be blocked, even on a torus. So for example Ra1 cannot move to a3 if a2 & a4(or a5) are blocked. However, the knight remained a leaper. Although the 4 squares accessed by a left knight-move (two squares away, then one to the left) do fall on a straight line on the 5x5 Torus, the knight cannot be blocked from accessing any of them. Similarly for right knight-moves.

There are 6 straight line directions on the 5x5 torus, so there are 15 pieces one might conceive which can move on exactly 2 lines. Three are the orthodox R,B,S, and 12 are chimerae. The knight is the most powerful of the 15. The 8 chimerae which share half the knight DNA are the next most powerful, blockable only on one of their two lines, while the 6 with no knight DNA (rook, bishop and 4 chimerae) are the weakest.
Henrik Juel: Your very first sentence is probably right, Andrew, but I am not convinced
I can see that TLS all can move to eight squares in one move
and it is clear that T (and probably L) can reach all 25 squares in two moves
But how does Sc3 reach a1 in two moves? (2021-06-27)
Henrik Juel: Never mind
Sc3-e4-a1 or Sc3-d5-a1 (2021-06-27)
A.Buchanan: A "left-knight" on c3 can move to a2,b5,e4,d1. These look as if they are just scattered, but in fact they are on a straight line with gradient 1/2. It runs from c3 - e4 -b5 - d1 - a2 and back to c3. Similarly the "right-knight" defines an orthogonal line with gradient 2: c3 - d5 etc. A whole knight makes a move on one of these lines, just like a rook makes a move on a file *or* a rank. It's then obvious that the knight can access any square in 2 moves. The catch is that while a rook can be blocked on a line by two pieces (*two* needed because it's a cycle), a knight accesses all the points as a leaper: it's not actually riding *along* the line, it's just mimicking that effect. So it's unblockable. The "stars are right" for this trick only when the torus is 5x5. (2021-06-28)
A.Buchanan: I think this space has 25*24*20 = 12,000 symmetries. Map a1 anywhere (25), then map a2 anywhere else (24), then map b1 anywhere but the 5 points on the line containing a1 & a2 (20). So it's "doubly transitive" on directions - this means any of these 15 piece types can be mapped to any other. (2021-06-28)
Henrik Juel: Your enlightenment is appreciated, Andrew (2021-06-28)
comment
Keywords: Torus (5x5), no 8x8 board
Genre: Fairies, Mathematics
FEN: qqqqqqqq/qqqqqqqq/q5qq/q5qq/q1BNR1qq/q5qq/q5qq/qqqqqqqq
Input: A.Buchanan, 2021-06-27
Last update: A.Buchanan, 2021-06-27 more...
14 - P1401567
Andrew Buchanan
Discord Chess Problems & Studies Server 31/05/2022
P1401567
(4+1)
ser-h#n
How many solutions, including all shorter ones!
(ignore draw by repetition & 50 move rule)
1. Kh8 Dh6#
1. Kf8 2. Ke8 Tg8#
1. Kf8 2. Ke8 3. Kd8 Tg8#
1. Kf8 2. Ke8 3. Kd8,Kf8 4. Ke8 Tg8#

play all play one stop play next play all
Solutions in exactly:
1,1,1,2,3,5…
Solutions in up to n:
1,2,3,5,8,13…
and both sequences seem to be more-or-less Fibonacci. What’s going on?

Let's look at something apparently irrelevant: Peg Solitaire (aka Conway's Soldiers). https://en.wikipedia.org/wiki/Peg_solitaire

In this, a peg hops 2 squares orthogonally, capturing the mandatory hurdle. (Is there a fairy chess piece with this movement?) Imagine a long line of them. e.g.
.1234567890abcde.

(We put a space at each end to show the termination.) Split them conceptually into two alternating lines:
.1.3.5.7.9.a.c.e.
&
..2.4.6.8.0.b.d..

No captures can be made, but if we "borrow" one additional peg on the left of each line:
*1.3.5.7.9.a.c.e.
&
.*2.4.6.8.0.b.d..

then these two pegs can hop over all the other pegs in their lines:
................f
&
...............e.

Then combining them again, e can capture f:
.................g

Fibonacci numbers behave exactly the same way when you add them! There is a capturing rule: f(n) = f(n-1) + f(n-2). Then:
sum(k=1...n-1)f(k) = f(n+1) - f(0) - f(1)

The two correction terms correspond to the "*" terms that we added on the left in order to get the cascade started!

So back to chess. Ignoring for now the odd solution with n=1, the number of ser-h#n in EXACTLY n moves is f(n-1), where f(k) is the kth fibonacci number 1,1,2,3,5,8,13,21,34...

[This comes from bK being able to occupy just 4 squares.
A(n) is the number of ways to reach d8 or e8 in n moves, B(n) is the number of ways to reach c8 or f8.
A(n+1) = A(n)+B(n)
B(n+1) = A(n)
You can see that this formula is just the "capturing rule" again.
And to get started: A(1)=0,B(1)=1.
So A(n)=B(n+1)=f(n-1).]

To find all the solutions in UP TO n moves, we need to sum them together:
sum(k=1...n-1)f(k) = f(n+1) - f(0) - f(1)

f(0)+f(1) is just 0+1=1. So this is why we need the extra solution with bKh8.

"Summing it all up", the number of ser-h#n is just f(n+1).

Noam's earlier setting https://pdb.dieschwalbe.de/P1243505 is the first afaik to show the whole Fibonacci sequence, and is one of two he published with the 4-chain (see also P1243525). This problem also uses a 4-chain as the easiest way to get Fib going, but (1) concentrates the mates onto just 2 of the 4 squares in the chain, (2) tots up the cumulative sum from 1 to n, and (3) exploits the curious diversion to h8. And is a seriesmover, as there's not much interest in the two-sided play.

The earlier https://pdb.dieschwalbe.de/P1259074 was the first I think to show a Fibonacci number.

What happens if the BK has more than the 4 free squares? There is a sneaky trick to get the answer quickly: instead of an n-chain, consider a (2n+2)-cycle:

x-x-x-x-x-x =>

x-x-x-x-x-x
|...............|
x..............x
|...............|
x-x-x-x-x-x

n-cycles are more symmetric than n-chains, so easier to analyze.

Counting paths in a graph systematically involves finding the "resonances", termed eigenvalues, which for a graph are always real numbers. The long-term growth rate in the number of paths is dominated by the eigenvalues with largest magnitude. It turns out that the eigenvalues of n-chain & (2n+2)-cycle are mostly the same! The dominant eigenvalues of the (2n+2)-cycle are +-1, but those are the only ones which *don't* carry over to the n-chain. The other eigenvalues are 2*cos(j*pi/(n+1)) for j=1..n, and they *do* carry over. The largest are where j=1 or n, and have opposite sign. We can ignore the sign from the perspective of the overall growth rate, but it's this sign that gives the "parity" in the n-chain paths: they either have odd or even length.

So the growth rate of an n-chain is 2*cos(pi/(n+1)). This "explains" why the golden ratio (which contains _/5) is the growth rate when n=4, which doesn't seem to have anything to do with 5.

Here's a few values of growth rates from n=1 upwards:
1-chain: 0 (no movement can happen)
2-chain: 1 (just deterministic pendulum)
3-chain: _/2 ~= 1.4142
4-chain: golden ratio = (1+_/5)/2 ~= 1.618
5-chain: _/3 ~= 1.7321
6-chain: ~= 1.8019
7-chain: ~= 1.8477
etc.

Some other random observations. As n increases, so does the growth rate, but it will always be less than 2 because the n-chain is a subgraph of the n-cycle (just missing 1 edge), and the n-cycle always has a growth rate of 2. The growth rate increases with n, because the missing edge is increasingly negligible. There is also a recursive formula for the polynomial of degree n, p_n(t) which contain the eigenvalues as roots:
p_n = - t*p_(n-1) - p_(n-2)
which is reminiscent of both Fibonacci & Pascal, although is more general.
Previous problems of this kind were "exact", excluding short solutions. Here's one where, for a given n, you need to tot up *all* the solutions with m less than or equal to n into a nice sum.
Henrik Juel: n=1, 1.Kh8 Dh6#, total t=1
n=2, 1.Kf8 2.Ke8 Tg8#, t=2
n=3, 1.Kf8 2.Ke8 3.Kd8 Tg8#, t=3
n=4, t=5
After this it gets a little hairy, but Andrew will give the nice sum next week... (2022-06-03)
A.Buchanan: Nice sum supplied :) (2022-06-08)
Henrik Juel: Thanks, Andrew
So the nice sum is t=f(n+1), where f are the Fibonacci numbers given by
f(1)=f(2)=1, f(n)=f(n-1)+f(n-2) for n=3,4,..
n 1 2 3 4 5 6
f 1 1 2 3 5 8
t 1 2 3 5 8 (2022-06-08)
more ...
comment
Keywords: Path enumeration (Fibonacci), Aristocrat, Miniature, Rex solus (s)
Genre: Fairies, Mathematics
FEN: 6k1/6R1/N3K3/8/8/4Q3/8/8
Reprints: MatPlus.net Forum 31/05/2022
Input: A.Buchanan, 2022-06-02
Last update: A.Buchanan, 2022-06-08 more...
15 - P1407174
Noam D. Elkies
Retros mailing list 01/01/2023
P1407174
(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
play all play one stop play next play all
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
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...
16 - P1407176
Andrew Buchanan
Retros mailing list 01/01/2023
P1407176
(16+0)
ser-PG in 19
How many solutions?
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
play all play one stop play next play all
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
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...
17 - P1407177
Andrew Buchanan
OP032 The Hopper Magazine I03 31/12/2022
P1407177
(2+15) C+
ser-#15
How many solutions?

2022 Lösungen
C(14,9)+C(6,3) = 2002+20
Hans-Jürgen Manthey: Es sind genau 2022 Lösungen sagt Popeye !
Das zum Abschluß des Jahres 2022 ist schon erstaunlich ! (2023-01-15)
comment
Keywords: Path enumeration (year), Seriesmover, Minimal
Genre: Fairies, Mathematics
Computer test: 2022 Lösungen sagt Popeye
FEN: 7r/7q/3pnb2/2p2p2/P4pp1/K2p2n1/4pp2/3r3k
Input: A.Buchanan, 2023-01-08
Last update: Miguel Ambrona, 2023-07-26 more...
18 - P1410557
Andrew Buchanan
1002 Mathematics at DEC 30/12/1988
P1410557
(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)
comment
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...
19 - P1414063
Andrew Buchanan
The Puzzling Side of Chess 228 31/10/2023
P1414063
(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 ...
comment
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...
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