Die Schwalbe

9 problem(s) found in 4821 milliseconds (displaying 9 problem(s)). [COMMENTDATE>=20200919 AND NOT K='Hilfsrückzüger' AND NOT K='Typ C' AND K='Retro Strategy (RS)'] [download as LaTeX]

1 - P0000195
Frank Christiaans
7167 Die Schwalbe 126 12/1990
P0000195
(13+10)
#3
b) wSe4 nach g8
a) 1. 0-0? droht 2. Tf8+ Kd7 3. Lxe6#
1. ... Lxc5! 2. Sxc5 0-0-0!
1. Thf1! 0-0-0 2. Lxe6+ Td7 3. Tf8#
b) 1. Thf1? Lxc5 2. Sxc5 0-0-0!
1. 0-0! Lxc5 2. Sxc5 ... (0-0-0?) 3. Tf8#
play all play one stop play next play all
a) zuletzt h6-h5 und beide Rochaden sind noch möglich. Schläge von Weiß: Bbxc, Bcxd, Bhxg, Bfxg-g8=L.
b) gegenseitiger Ausschluss der beiden Rochaden. 3 Fälle sind zu betrachten:
b1) zuletzt f7xDe6; Weiß muss 4x geschlagen haben: Bfxexdxc8=L, d.h. s0-0-0 ist unzulässig.
b2) zuletzt d7xDe6; zu den 4 Schlägen von a) wird auch der sBf gebraucht, der sich auf f1 umwandeln muss, also w0-0 unzulässig.
b3) zuletzt h6-h5: wBf zieht über f7 zum UW-Feld g8, g.h. s0-0-0 unzulässig.
Henrik Juel: C+ Popeye 4.61 after analysis (2020-10-30)
A.Buchanan: A very harmonious problem. Suppose that both sides retain castling rights. wLh3 is obtrusive, so wBf/h promoted via g7, and escaped via e6 so sBe6 came from d7 later. Wh has made at least 4 pcs, while Bl has made 2 (D & Lc) dxe6, gxh6. If last move was dxe6, then sD & sLc were captured by officers in cage. Thus sBf was captured by wB, and therefore promoted on f1, disrupting wK. Therefore the last move was h6-h5. In (a) there is no problem retaining both castling rights, but in (b) Sg8 must have reached that square via f6 *after* the promotion, disrupting bK. However in (b) the castling rights are incompatible. In (a) 1. 0-0? Lxc5! 1. Tf1!, while in (b) the reverse.
Neither twin is a PRA problem: in (a) there is no conflict between the castling, while in (b) RS applies: the pre-emptive Wh castling means we are in a reality where Bl cannot castle. (2020-10-31)
comment
Keywords: Castling key (wksg), Obvious promotion (L), Retro Strategy (RS)
Genre: Retro, 3#
FEN: r3k3/ppp1p2p/Nb2p3/P1PP2Pp/4N3/7B/3PPRP1/4K2R
Input: Gerd Wilts, 1995-06-03
Last update: Alfred Pfeiffer, 2022-01-07 more...
2 - P0003417
John Frederick Keeble
2206 The Problemist Fairy Chess Supplement 16, p. 173, 02/1936
P0003417
(4+5) C+
h#2.5 (AP)
1. ... exd6ep 2. 0-0-0 dxe7+ 3. Tf8 exf8=T,D#
play all play one stop play next play all
H.Juel: If Black may castle, last move was not made by Ke8 or Ta8, but by Pd5, so it was d7-d5, and hence White may capture en passant. 1... exd6ep 2.0-0-0 dxe7 3.Tf8 exf8=D,T# where the castling (belatedly) justifies the e.p. capture.
Not 1... exd6ep 2.Kd8? dxe7 3.Kc8 e8=D,T
AB: Another 120 retro tries (=60x2 due to tolerated dual promotion) of general form 1. exd6ep 2. T~ dxc7 3. T~ c8=D,T.
No set play
Nachdruck in "Die Schwalbe" 33 mit Diagrammfehler (wBe6 fehlt).
VL: The first AP-problem! Thematic illegal try:
1. ... exd6ep 2.Kd8? dxe7+ 3.Kc8 e8=T(D)#?? (2007-02-11)
Mario Richter: 46 years between this Problem and P1366663 - perhaps this was not Keeble's first AP-Problem?! (2019-08-29)
Ladislav Packa: Of course I know the idea a posteriori. But let my heresy be forgiven - I do not consider it logical. In this particular example, I consider e.p. for proof that black can castling - not vice versa. (2019-08-29)
A.Buchanan: Hi Ladislav. If ep is ok, then Black just moved a pawn with double hop. That doesn't stop Black having e.g. moved bK right at the beginning of the game. So you can't prove that Black definitely can castle this way. What am I missing here? (2019-08-29)
Ladislav Packa: Your statement is perfectly fine. However, the AP condition is usually interpreted as castling is evidence of e.p. However, the second party does not have to do castling - the possibility has already been demonstrated by the existence of e.p. (2019-08-29)
A.Buchanan: Hi again Ladislav. Please try again, I am obviously being very stupid. I don't understand how 1. exd6ep is any more indicative that Black can castle than 1. K~. (2019-08-30)
Ladislav Packa: Hi, Andre!
e.p. is just a result of the d7-d5 move. Black can castle only if it was the last move of black (even without wPe5). Therefore, it does not have the logic of joining the e.p. with castling. For example, ask yourself if black is allowed to castle in the position without wPe5. (2019-08-30)
A.Buchanan: @Ladislav: ep is not "proof that black can castle", rather it is a *necessary* condition for Black to be able to castle. The logical implication is the other way round. We can castle because of the optimistic convention, and because sPd might just have moved. Without wPe, there is no solution, but whatever White plays, Black can castle (2020-12-11)
Hans-Jürgen Manthey: was heißt hier Thematisch Illegal ??
Mir völlig wurscht ob ep möglich oder nicht, nach dem weißen Zug ist natürlich Kd8 möglich !!
also ist auch 1. ... exd6ep 2. Kd8 dxe7+ 3. Kc8 e8DT# eine Lösung, sonst müßte auf die erzwungende Rochade in der Aufgabenstellung hingewiesen werden ! (2020-12-11)
A.Buchanan: Can one tolerate “tolerated dual promotion” in Valladao? Well Valladao himself did, so it must be ok (2022-03-26)
more ...
comment
Keywords: a posteriori (AP) (Type Petrovic), En passant as key, Castling (sg), Promotion (D/T), Tolerated dual promotion (D/T), Valladao Task, Retro Strategy (RS)
Genre: h#, Retro
Computer test: Popeye 4.61
FEN: r3k3/2p1p3/2P1P3/2KpP3/8/8/8/8
Reprints: V. Die Schwalbe 33, p. 323, 06/1975
2001 CHM Themes 12/2000
3 Die Schwalbe 215, p. 240, 10/2005
A1 Problemkiste 169 02/2007
402 Eigenartige Schachprobleme , p. 130, 2010
14 Die Schwalbe 241, p. 374, 02/2010
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-06-10 more...
3 - P0003745
Olavi Arvid Riihimaa
2077 Stella Polaris 12/1968
P0003745
(12+12) C+
h#2
Duplex
s) 1. 0-0 Td1 2. Txd8 Txd8#
w) 1. ... 0-0-0 Tf8 2. Txf1 Txf1#
play all play one stop play next play all
a2xLb3, d2xD/Tc3, f2xT/De3, d7xT/Dc6, f7xD/Te6. wLc1 opened no doors.
If both sides can still castle, then dxc6 preceded all white captures, but relies on prior release of wD/T. This could not have happened, so castling mutex.
This composition was marked as PRA, but it is RS.
more ...
comment
Keywords: Retro Strategy (RS), Castling, mutual exclusive, Quasi-symmetrical position
Genre: h#, Retro
Computer test: C+ Popeye 4.61 and analysis
FEN: 3Nk2r/1pp1p1p1/p1p1p2p/6N1/1P2n3/2P1P2P/1PP1P1P1/R3Kn2
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2021-06-30 more...
4 - P0004454
Immo Fuß
Die Schwalbe , p. 463, 03/1939
Allen Teilnehmern herzlich gewidmet
Internationaler Lösungswettkampf 1938
P0004454
(13+11)
#3
1. Tf1? 0-0! Thematic retro try
1. Dxd7+,Sf6+,exd7+,exf7+? Kf8!
1. Dxa8+? Sd8!
1. Sh7? Txh7!

1. 0-0! (droht 2. Txf7,Dxa8+) 0-0? illegal
1. ... Lxb2,Lxb4,b5,c6 2. Dxa8+ Sd8 3. Sxc7#
1. ... Kf8 2. Txf7+
1. ... Tg8 2. Dxd7+,Dxa8+
play all play one stop play next play all
Suppose both players can castle, and derive a contradiction.
White captures: a3xb4,fxe,g2xh3,Bc8 & QR in cage a8--d8.
Black captures: a7xb6,fxe/g,Bc1. bPg promoted. bPf promoted or was captured by wPfxe.
So all captures accounted for. Pieces captured by pawns were wRB & bQXX
b6 is dark, so light wBf1 was not captured there. By elimination, a7xRb6, which released bQ.
Was an original officer captured on b4 to release wR?
bQ not yet released
bRh never moved, bRa captured in cage
bB wrong shade, bBc8 captured in cage
bS couldn't escape g1, and two others on board.
So it must have been a promoted officer captured on b4 earlier.
What was captured on h3, to open the line for promotion on g1? Must be original as wPa & bPa have not yet captured. For the same reasons as axb4, we can eliminate all 4 possible officer types.
Contradiction! So at least one player cannot castle.

We apply the Retro Strategy (RS) convention, and White is permitted to castle, while the range of possible histories shrinks to exclude all cases where Black might have castled. This common special case of RS is also known as Mutual Exclusion.
Kees: Only one castling is legal With black castling there's no #3
1. 0-0!
1. ... Kf8 2.Txf7+ Kg8 3 Dxa8#
1. ... c6 2. Dxa8+ Pd8 3. Pc7#

(axNb3, and for f1=N or g1=N wK must move. so bD must pass bK)
Somebody can better explain than me. (2022-02-14)
A.Buchanan: Hi Kees thanks for the solution which grabs the essence - I have used more words, please point out any slips I might have made! :) (2022-02-15)
comment
Keywords: Retro Strategy (RS), Castling, mutual exclusive (wksk)
Genre: Retro, 3#
Computer test: Popeye v4.87 for forward play Non-trivial thinking for retro logic
FEN: n3k2r/1ppppn2/1p2P3/3N2Np/QP6/b6P/1PPPP2P/4K2R
Reprints: (14) Problem 144-147 12/1971
Input: Gerd Wilts, 1995-06-03
Last update: A.Buchanan, 2022-02-15 more...
5 - P0573628
Klaus Wenda
v Mat (Belgrade) 01-02/1979
1. Lob
P0573628
(12+13) C+
h#2
Duplex
s) 1. 0-0 Tf1 2. Sd1 Sf7#
w) 1. ... 0-0 2. Tf8 Tc1 3. bxc1=D/T#
play all play one stop play next play all
If both sides can castle then d2-d3, f2xLe3, g2x?h3, while sBg-g1=?. If this were S/L it couldn't escape without checking wK, but if D/T then it must still be on the board. Yet Th8 is original. So mutex applies. Note this is RS, not PRA.
more ...
comment
Keywords: Castling, mutual exclusive, Tolerated dual promotion, Retro Strategy (RS)
Genre: h#, Retro
Computer test: C+ Popeye 4.61 (except for promotion dual) and analysis C+ Popeye 4.61 (except for promotion dual) and analysis
FEN: 2bqk2r/ppppp2p/2n5/6N1/7B/3PP2P/Pp2PnRP/1B2K2R
Reprints: König & Turm 2000
W172 Dreiklang 2001
Input: Torsten Linss, 1999-04-17
Last update: A.Buchanan, 2021-06-30 more...
6 - P1382808
Ronald Turnbull
Andrew Buchanan

PDB Website 20/01/2012
RT, correction AB
P1382808
(4+12)
h#3
AP
PRA
1. bxc3ep 0-0! (for AP reasons, not 1...Kf2?/Th2? etc) 2. Ta1 Txa1 3. Kc4 Txa4#
1. fxe3ep 0-0! (for AP reasons, not 1...Tf1?/Txh3? etc) b2 Txf3 3. Tc3 Tf4#
play all play one stop play next play all
For AP as for anything else, the default meta-convention is PRA before RS. Here we use AP to lock down wK & wTh1, so the last move was R: 1. c2-c4 or 1. e2-e4. There is a different h#3 in either case, beginning with ep and then 0-0.
Chessically corrects P1012059 following discussions. It's doubtful that AP logic works as simplistically described here, so this may still be unsound logically. Nevertheless, might as well fix the chess!
See P1399112, where the two parts each prove that one side of White castling rights remain. If this is correct, and only *one* part is required per castling, this implies that this Turnbull-Buchanan problem doesn’t work. For one part we have solution c_S and tries c_T1 & c_T2. In the other part, solution e_S and tries e_T1 & e_T2. Any of 5 combinations work:
c_S + e_S
c_S + e_T1
c_S + e_T2
c_T1 + e_S
c_T2 + e_S
because they all confirm that White retains castling rights.
Classify this problem as Golden Age because it’s on the wrong side of the evolving standard.
A.Buchanan: Let ??? Indicate respectively that w00, c ep & e ep are ok. Then YYN, YNY, NYN, NNY, NNN are the contenders.
Under PRA we reduce to YYN, YNY & NNN but the last has no solution so not the right paradigm.
Under SPRA with AP we reduce to YYN, YNY. Each considered as a separate part has kind of AP h#3, but I don’t understand how consolidation would take place over PRA anyway
Under RS with AP, we have 2 solutions assuming again that no consolidation. (2022-03-23)
comment
Keywords: a posteriori (AP) (Type Petrovic - cee), En passant as key (2), Castling (wk), Retro Strategy (RS), Golden Age (AP unconsolidated)
Genre: h#, Retro
Computer test: Forward logic sound by Popeye v4.85 However, seems that AP logic is not sound
FEN: 8/8/8/1nr1n3/ppPkPp2/rp1p1p1p/8/4K2R
Input: A.Buchanan, 2020-12-08
Last update: A.Buchanan, 2023-04-06 more...
7 - P1390054
Paul Buerke
3419 Schach 6, p. 92, 03/1960
P1390054
(13+13)
#2
1. Tad1? 0-0!
1. 0-0-0! ... 2. Td8# (1. ... 0-0? ist illegal)
play all play one stop play next play all
Aus der LB: ... eine der beiden Parteien hat noch das Rochaderecht, aber nur eben eine, die andere nicht! Das verrät eine rückschauende Analyse der Stellung. Weiß muß daher mit 1. 0-0-0! festlegen, daß er und nicht der Gegner zur Rochade berechtigt ist, sonst dürfte der sich (nach 1. Ta-d1?) mit 1. ... 0-0! verteidigen. ... Wenn 1. 0-0-0 eine Lösung ist und daraus folgt, daß Schwarz in der vorliegenden Position nun nicht mehr rochieren darf, dann folgt daraus weiter, daß nunmehr auch 1. Ta-d1! neben 1. 0-0-0 eine richtige (sozusagen eine "sekundäre") Lösung ist.

AB: White missing: QBB
Black missing: QBB
No promotions!
White's captures: cxBb, dxBe, gxQh
Black's captures: cxBb, dxBe, gxQh
(Kingside bishops can only be captured by d-pawns, so to avoid deadlock, g-pawns must capture queens not bishops.)

If both sides retained castling rights, then only these six moves release the capturees, and none can be the first! Contradiction. Therefore, at least one player shifted Pf and K to allow Q to escape. Suppose it was just Black who did this, can White retain castling rights? Yes, and the order of captures must be wPgxQh3 bPdxBe6 wPcxBb3 bPgxQh6 wPdxBe3 bPcxBb6.

This problem was composed according to the Retro Strategy "meta-convention" which was the default at the time. Under this, if castling rights conflict as here, whoever castles first denies the other the right. These days PRA is the default, but when 'a solution is not possible according to the PRA convention' under the Codex "meta-meta-convention" we fail over to RS.

White on the move must castle, and the set of game histories contracts to the one in which Black lost castling rights. 1. Td1? 0-0!

The final point of the German comment by LB translates as: "If 1. 0-0-0 is a solution and it follows from this that Black is no longer allowed to castle in the present position, then it also follows that 1. Ta-d1! next to 1. 0-0-0 is a correct (so to speak a "secondary") solution." Maybe this is a joke: hard to say. I suspect the author is confused about the nature of RS. RS does not allow us to conclude anything fixed about the current position. All it implies is that *after* 1. 0-0-0, the set of game histories collapses to the ones in which White could castle, and in all of these, Black cannot castle. There is no secondary solution.
Henrik Juel: I believe that current thinking leads to
1.Tad1? 0-0!
1.0-0-0! and Black cannot prevent 2.Td8# (2021-05-25)
Henrik Juel: The author surely did not know anything about meta-meta-conventions
He just used common sense
1. White may castle (because there exist game histories where he has not lost this right)
2. The castlings are mutually exclusive
3. and Bob's your uncle
To me point 2 is the main content of the problem (thanks for providing the proof, Andrew) and I see little reason for searching for a better forward play
The last part of the german LB comment makes little sense to me (2021-05-25)
A.Buchanan: Henrik, it was Werner Keym not me who invented the "meta-meta-convention". All I am doing is naming the layers so that people can see what's going on in the opaquely-written Codex.
1) convention (castling, ep)
2) meta-convention (PRA, RS)
3) meta-meta-convention (PRA RS)

Implicit in Werner's change (but never stated) is the idea that the new Code *does* apply to pre-existing compositions. The problem remains sound under the m-m-c. But do we have the right to over-write the way that an old composition should be understood? How should a new generation of solvers be educated to view the spectrum of historical compositions?

"Common sense" is useful in life (I wish I had more), but if you say that is *necessary* for this problem, you close the door on algorithms ever being able to solve simple retros like this, which is absurd. Moreover, if we say that everything is inherently vague and sloppy, relying on "common sense" then we limit the sophistication of retros that can be composed.

What makes it worse is that the c, m-c & m-m-c are written are over-narrowly scoped, excluding fairies and even other orthodox concepts. There is great interest in combining fairies and retros these days, but mostly these are proof games. One reason for this is that the current conventions are so inhospitable to fairy, so people avoid fairy non-PG retros. This was a completely avoidable issue.

Yes I agree that there's no point having better forward play unless it's really good in its own right.

I hadn't read the German before, but the final sentence in the LB comment reveals a fundamental misunderstanding of RS. I am surprised that the editor printed it. (2021-06-29)
more ...
comment
Keywords: Quasi-symmetrical position, Retro Strategy (RS), Castling, mutual exclusive
Genre: Retro, 2#
FEN: n3k2r/pp2p2p/1p2p2p/4Np1N/3R1P1r/1P2P2P/PP2P2P/R3K2n
Input: Felber, Volker, 2021-05-25
Last update: Alfred Pfeiffer, 2021-09-01 more...
8 - P1401566
Jean-Michel Trillon
Siegfried Hornecker
Andrew Buchanan

Discord Chess Problems & Studies Server 02/06/2022
J-MT, correction SH & AB
P1401566
(7+8) C+
#4 AP
Maximummer
b) g6->g5
a) 1. Ld6 Td8 2. fxg6 Txd5 3. g7 Th5 4. g8=D#
b) 1. fxg6ep Txd5 2. Le5 0-0-0 3. g7 Th8 4. gxh8=D,T#
play all play one stop play next play all
What did Black play last, and why not b7-b5?
a) Can only be Kxe8 to escape check. Thus castling rights are definitely lost.
b) Also the possibility of g7-g5, so 000 rights can remain. This breaks the earlier solution but allows another solution in which AP retro-justifies ep with 000, which is also necessary for the mate. The solution beginning 1.Ld6 does not work any more, because under RS convention, castling is legal and so must be played.
FYI: here are two non-unique games validated by Jacobi v0.7.5 to prove the legality of the diagram positions. 000 rights remain in b, while in a are retained until the penultimate move.
a) 1.e4 Sf6 2.a4 Sd5 3.exd5 Sc6 4.a5 Sb4 5.h4 Sd3+ 6.Ke2 Sxb2 7.g3 Sc4 8.Bb2 Sa3 9.Be5 Sxc2 10.Bxc7 Sb4 11.Sc3 Sa6 12.Se4 Sb8 13.Sg5 Sa6 14.Se6 Sb8 15.Sxd8 Sa6 16.Bb8 Sb4 17.Sxf7 Sa6 18.Sh6 Sc5 19.Ra4 Sa6 20.Rc4 Sc7 21.Rc6 Sa6 22.Rd6 Sb4 23.Sf3 Sa2 24.Sg5 Sb4 25.Se6 Sa2 26.Sxf8 Sb4 27.Sg6 Sa6 28.Sxe7 Sb4 29.Qc2 Sa6 30.Kd3 Sc7 31.Be2 Sa6 32.f4 Sc7 33.Rb1 Sa6 34.Rb5 Sc7 35.Kc3 Sa6 36.Kb2 Sc7 37.Ka1 Se6 38.Bg4 Sd8 39.Rf6 Sc6 40.Be6 Sd8 41.Sf7 Sc6 42.Sg5 Sd8 43.Qxc8 Rf8 44.Bg4 Rh8 45.Sf5 Rf8 46.Sd4 Rh8 47.Sde6 Rf8 48.Kb2 Rxf6 49.Ka1 Rh6 50.Bh5+ Rxh5 51.Sd4 g6 52.g4 a6 53.Qc1 Sc6 54.f5 Sxd4 55.Rb3 Sxb3+ 56.Ka2 Sxa5 57.Ka1 Sc6 58.Ka2 Se5 59.Ka1 Sf7 60.Ka2 Sxg5 61.Ka1 Sf7 62.Qc8+ Sd8 63.Qxd8+ Kf7 64. Qe8+ Kxe8
b) 51. … a6 52.Qc1 Se6 53.Qc2 Sxg5 54.f5 Sf3 55.g4 Sxd4 56.Rb3 Sxb3+ 57.Ka2 Sxa5 58.Qb1 Sb3 59.Qa1 Sxa1 60.Kxa1 g5
Corrects P1401480.
Henrik Juel: a) 1.fxg6 Txd5 2.Le5 0-0-0 3.g7 Th8 4.gxh8=DT#
b) 1.fxg6ep Txd5 2.Le5 0-0-0 3.g7 Th8 4.gxh8=DT# (2022-06-02)
Henrik Juel: a) is C+ Popeye 4.61, except for the promotion mate dual; no AP here
In b) Popeye with 'opt enp g6' gives the intended solution; here the castling is needed to rule out the possibility of Kf7xDe8 as last move
But how do we know that last move was not f6xg5 or h6xg5? (2022-06-02)
A.Buchanan: Hi Henrik thanks for examining this.
In a), the last move must have been Kxe8 to deal with checking piece. Otherwise black would have played e.g. b7-b5. If you forbid castling in Popeye you will see a completely different solution.
In b) hxg5 or fxg5 is too short to compete with b7-b5. (2022-06-02)
Henrik Juel: Yes, I missed b7-b5
Popeye with 'opt noc a8' gives this solution for a) 1.Ld6 Td8 2.fxg6 Txd5 3.g7 Th5 4.g8=D#
Very nice, with analysis in both parts (2022-06-02)
comment
Keywords: a posteriori (AP) (Type Petrovic), Maximummer, Castling (sg), En passant as key, Promotion (D,D/T), Valladao Task, Retro Strategy (RS)
Genre: h#, Retro, Fairies
Computer test: HC+ Popeye v4.87 for forward logic, Jacobi v0.7.5 to validate demo games & humans for non-trivial retro-logic
FEN: rB2k3/1p1p3p/p5p1/3P1P1r/6PP/8/3P4/K7
Input: A.Buchanan, 2022-06-02
Last update: A.Buchanan, 2022-06-03 more...
9 - P1401711
Andrew Buchanan
OP008 The Hopper Magazine I01 24/12/2021
"Ak ja, retten, retten; hvad hjælper det, at du har retten, når du ikke har nogen magt?"
P1401711
(3+3) C+
h#2.5 (2 solutions)
Position after Black's 5,696th move
1. ... Se6 2. Th2 Ta8 3. Th7 Txe8#
1. ... Txa2 2. Lg6 Tg2 3. Lh7 Sf7#
not 1. ... 0-0-0 2. Lg6 Tg1 3. Lh7 Sf7#?? (game just ended by 50M)
play all play one stop play next play all
A full solution is really too big for PDB, but it is available at https://www.thehoppermagazine.com/AA084
Henrik Juel: solutions
1...Sg5-e6 2.Ra2-h2 Ra1-a8 3.Rh2-h7 Ra8*e8 #
1...Ra1*a2 2.Be8-g6 Ra2-g2 3.Bg6-h7 Sg5-f7 #
not 1...0-0-0? 2.Be8-g6 Rd1-g1 3.Bg6-h7 Sg5-f7 #
because White has lost his right to castle, as Andrew will explain, I hope... (2023-07-29)
A.Buchanan: White has not necessarily lost the right to castle, but if he can castle then the 50M rule triggers before the mate can be executed. It is pretty complicated, sorry. (2023-07-29)
A.Buchanan: We don't do mottoes much these days. Mrs Baird was a big fan. On page 301 of the July 1916 Chess Amateur P.H. Williams wrote:
"I think it was Mrs Baird who did more to search Shakespeare for accidental (or deliberate) reference to chess, since all her retractors had Shakespearean mottoes, and her knowledge of the Avonian bard was obviously extensive. Her example was followed by other composers of retractors, who considered such positions would be incomplete without some reference to the poet."
But there are other great playwrights, and Asteroid 5696 was named after one such, whose bitter quotation accurately states this problem's theme. (2023-07-31)
Henrik Juel: The citation is by danish philosopher Søren Kierkegaard (1813-1855) and means something like
Oh yes, to be right, to be right; it does not help much that you are right, when you have no power. (2023-07-31)
Henrik Juel: Asteroid 5696 is named Ibsen after the norwegian playwright Henrik Ibsen (1828-1906), who must have borrowed the citation from Søren Kierkegaard (2023-07-31)
A.Buchanan: Thanks for this Henrik - I resolved the question here: https://philosophy.stackexchange.com/questions/101229/did-ibsen-originate-this-statement
Beyond the castling try, the other innovation here is in the solution beginning 1...Se6. Castling convention and 50M I think are trying to combine by Retro Strategy (being optimistic about 50M rights like we are with castling, rather than pessimistic like we are with ep). (2023-08-04)
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Keywords: Aristocrat, Miniature, 50 move rule, Castling, Exchange of roles (T/S, Guard/Mate), Chumakov theme (l/t, simplified), Retro Strategy (RS), Model mate (2), Constrained problem
Genre: h#, Retro
Computer test: HC+ Popeye 4.61 and analysis
FEN: 4b2k/8/8/6N1/8/8/r7/R3K3
Reprints: AA084 The Hopper Magazine I04 13/07/2023
Input: A.Buchanan, 2022-06-09
Last update: A.Buchanan, 2023-08-27 more...
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