Die Schwalbe

4 problem(s) found in 6592 milliseconds (displaying 4 problem(s)). [PROBID='P1258300' OR PROBID='P0003831'OR PROBID='P1004154'OR PROBID='P1176381'] [download as LaTeX]

1 - P0003831
Vasile I. Tacu
RA35 diagrammes 21 05-06/1976
P0003831
(8+8)
Nur die Bauern haben geschlagen; alle 16 Bauern haben sich umgewandelt. Sind alle 4 Rochaden noch zulässig?

Duplicate Diagram: P1004154, P1176381, P1258300

Henrik Juel: Yes, all four castling rights may be preserved, I believe
A feasible proof game seems possible with these 8+8 pawn captures
White axb, bxc, cxb, dxc, exfxg, fxg, hxg
Black axb, dxc, exdxc, fxg, gxhxg, hxg
[Pg2,Pb7,Pc7] promoted without capturing (2018-02-17)
A.Buchanan: I think one capture less for e pawns suffices. The second to move off the file can go to e2 or e7 and promote to minor piece (2018-02-18)
Henrik Juel: Yes, I saw that; but it is stipulated that all captures were made by pawns, so it was natural to change exf to exfxg (2018-02-18)
A.Buchanan: The requirement that all captures are made by pawns seems extraneous. The 16th capture might be made by any of the pawns, so why use it to hide the nice subtlety on the e-file? (2018-02-18)
Henrik Juel: To force the sublety, you could add a black rook on a7, but that would destroy the neat diagram position (2018-02-18)
A.Buchanan: I agree Henrik. There is a curious way to "force the subtlety" in your excellent phrase, by *reversing* the first (unnecessary) condition. It could state instead: "Not every capture was made by a pawn." (2018-02-19)
comment
Keywords: Promotion, Castling (wbsb), Homebase (2), Aristocrat, Symmetrical position
Genre: Retro
FEN: rnbqkbnr/8/8/8/8/8/8/RNBQKBNR
Input: Gerd Wilts, 1995-06-03
Last update: Rainer Staudte, 2019-08-11 more...
2 - P1004154
Bernard Rothmann
R176 Probleemblad 4-2002 08/2002
P1004154a_1.png
(a: 9+9)
P1004154a_2.png
(a: 8+8)
A nach B in 13.0 Zügen
Schwarz beginnt
Circe
1. ... Sg6 Kxg6[+sSg8] 2. Lf5+ Kxf5[+sLc8] 3. Lf4 Kxf4[+sLf8] 4. Txf3[+wBf2]+ Kxf3[+sTa8] 5. Sxf2 Kxf2[+sSb8] 6. Kb2 Tc3 7. Kxc3[+wTa1] Ld4+ 8. Kxd4[+wLc1] Sd5 9. Kxd5[+wSb1] Sd6 10. Kxd6[+wSg1] Lxd7 11. Kxd7[+wLf1] Te8 12. Kxe8[+wTh1] Dxe1[+sTh8]+ 13. dxe1=D[+wDd1]+ Kxe1[+sDd8]
play all play one stop play next play all
Cook: NL
1. ... Sg6 Kxg6[+sSg8] 2. Lxc7[+wSg1] Kg5 3. Txf3[+wBf2]+ Txe6[+sLc8] 4. Ka3 Kg4 5. Kb4 Kxf3[+sTa8] 6. Sxf2 Kxf2[+sSb8] 7. Kb5 Th1 8. Lxb6[+wLc1] Txb6[+sLf8]+ 9. Kxb6[+wTa1] Sd6 10. Kc6 Lxd7+ 11. Kxd7[+wLf1] Se8 12. Kxe8[+wSb1] Dxe1[+sTh8]+ 13. dxe1=D[+wDd1]+ Kxe1[+sDd8] (Ger Blaauw)

Duplicate Diagram: P0003831, P1176381, P1258300

paul: Genre Circe is missing in stipulation.
Intention: 1.Sg6 Kxg6[Sg8] 2.Bf5+ Kxf5[Bc8] 3.Bf4 Kxf4[Bf8] 4.Rxf3[pf2]+ Kxf3[Ra8] 5.Sxf2 Kxf2[Sb8] 6.Kb2 Rc3 7.Kxc3[Ra1] Bd4+ 8.Kxd4[Bc1] Sd5 9.Kxd5[Sb1] Sd6 10.Kxd6[Sg1] Bxd7 11.Kxd7[Bf1] Re8 12.Kxe8[Rh1] Qxe1[Rh8]+ 13.dxe1=Q[Qd1]+ Kxe1[Qd8].

Cooked by Ger Blaauw:
1.Sg6 Kxg6[Sg8] 2.Bxc7[Sg1] Kg5 3.Rxf3[pf2]+ Rxe6[Bc8] 4.Ka3 Kg4 5.Kb4 Kxf3[Ra8] 6.Sxf2 Kxf2[Sb8] 7.Kb5 Rh1 8.Bxb6[Bc1] Rxb6[Bf8]+ 9.Kxb6[Ra1] Sd6 10.Kc6 Bxd7+ 11.Kxd7[Bf1] Se8 12.Kxe8[Sb1] Qxe1[Rh8]+ 13.dxe1=Q[Qd1]+ Kxe1[Qd8].
A correction was published in diagrammes 148/2004. (2010-07-28)
comment
Keywords: A to B, Circe, Homebase (2)
Genre: Retro, Fairies
FEN: 4B2n/1NNp3K/1B2b3/8/8/r4P1R/k2pR2b/3Qr2n
Input: Gerd Wilts, 2002-08-10
Last update: Alfred Pfeiffer, 2019-03-26 more...
3 - P1176381
Ron Fenton
F2838 The Problemist 09/2010
P1176381
(8+8)
BP in 12,5
Der 2. und 3. Zug von Weiß sind Schläge.

Duplicate Diagram: P0003831, P1004154, P1258300

Ron Fenton: Solution can be made unique by forbidding en passant captures. (2018-02-17)
A.Buchanan: Hi Ron: do you want me to shift that e.p. condition into the stipulation for a version that will be unique proof game. Is it C+? (I.e. find all the solutions by program, then eliminate those manually which don't satisfy the conditions.) (2018-02-17)
Henrik Juel: What was the intended solution, Andrew? (2018-02-17)
Ron Fenton: Mr.Buchanan
That would be great but let me run it past Popeye.exe one more time (it's been running for 20 hours now with no solution). If I remember correctly, it eventually shows the intended solution with duplicates that are eliminated by the 'must capture on moves 2&3 + no e.p. captures stipulations' but it could also be that I gave up and proofed it manually. I can't remember.

BTW: what do I click on to change my P/W? I keep having to log in using the cryptic one provided but I'd like to simplify. (2018-02-17)
A.Buchanan: I am halfway through check of 12.0 moves. There are no shorter solutions in 10.5, 11.0 or 11.5 moves. If you finish the chunky computation for 12.5 then I suggest you examine all solutions to see if there is a single elegant condition which can give uniqueness at once. Something like “maximise number of en passant captures.
To change password, go to settings. (2018-02-18)
Ron Fenton: Sorry, it's probably just me but I don't see 'change password' under Settings so I'm having to request a new one each time I log in. Lost in translation?

I gave my laptop 30+ hours @ depth 12.5 and all Popeye did was blink at me and crash other programs so I finally gave up (an over-sensitive mouse-pad accidentally killed it is #therealstory).

A C+ proof may prove impractical but playing over the solution just now I'm convinced the combined stipulation of...

'NO e.p. captures and White's 2nd and 3rd MUST be captures'

...results in a PG12.5 that is logically sound and unique. (2018-02-18)
A.Buchanan: I see that "Settings" occurs twice(!) on the left hand side. Once as an unclickable header above language selection. Then, AFTER you log in, it appears as something clickable just under the title PDB Account. It's this second occurrence that will allow you to change the password.
I am using Jacobi to test 12.0 moves. It is about 15 hours in, and says it's about 65% through, so I expect about 23 hours in total. For 12.5, I am guessing that it might take about 20 times that, so maybe about 3 weeks on my computer. One good thing about Jacobi is that it gives a lot of feedback about progress, and also allows easy pause/resume. In Firefox, it seems that Jacobi continues at full speed even when it's not the front window.
By the way, since the existence of a solution in x.y moves implies a solution in (x+2).y moves, finding no solution in 10.5, 11.0, 11.5 & 12.0 is a good foundation for the serious quest of 12.5.
If you have been able to rigorously prove by mental logic that your 12.5 solution is unique, then my hat is off to you: you must be a very strong solver. (2018-02-18)
Ron Fenton: I would never consider myself a strong solver, I just know this problem inside-out because it took a while to construct. If I remember correctly, it takes other aristocrats a move (or maybe a half-move) longer to do the job (with Knights a little longer than that) due the unavoidable checking of the Kings. Ironically, if you know the first move by each side it almost solves itself. Sorry to go on and on... (2018-02-18)
A.Buchanan: Jacobi says no solution in 12.0, 67734 secs, 17518 nodes/sec (2018-02-18)
Ron Fenton: Are you running Jacobi @ 12.5? (2018-03-06)
A.Buchanan: No, too much work for my computer (2018-03-06)
comment
Keywords: Constrained problem, Non-Unique Proof Game, Homebase (2), Aristocrat, Symmetrical position
Genre: Retro
FEN: rnbqkbnr/8/8/8/8/8/8/RNBQKBNR
Input: Gerd Wilts, 2010-09-11
Last update: A.Buchanan, 2018-02-18 more...
4 - P1258300
Matti Arvo Myllyniemi
7603 Schach 12/1974
P1258300
(8+8) cooked
s#6
Längstzüger
1. Th2 Lh3 2. De2 Le6 3. Dd3 La3 4. Lxa3 Txh2 5. Sc3 Ta2 6. Dd2 Dxd2#
play all play one stop play next play all
Cook: NL
1. Dd6 Lh3 2. Ta2 Lc8 3. Lg2 Lh3 4. Txh3,Sxh3 Txa2 5. De7 Sxe7 6. Kf1 Dd1#
u.a.

Duplicate Diagram: P0003831, P1004154, P1176381

t/One-Liner 8-feldrig a1h1 und a8h8
comment
Keywords: Maximummer, Homebase (2), Aristocrat
Genre: Fairies
FEN: rnbqkbnr/8/8/8/8/8/8/RNBQKBNR
Input: Erich Bartel, 2013-01-05
Last update: A.Buchanan, 2018-02-17 more...
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The problems of this query have been registered by the following contributors:

Gerd Wilts (3)
Erich Bartel (1)