Andrew Buchanan, Michel Caillaud
(Singapore / France)

Original Fairy problems
01.01.2021 - 31.12.2022

Definitions / Определения

No. 1615 Andrew Buchanan,
Michel Caillaud

Singapore / France
original - 30.05.2021

white Bf1c1 Ke1 Qd1 Ph2g2f2e2d2c2b2a2 Sg1b1 Rh1a1 black Bf8c8 Ke8 Qd8 Ph7g7f7e7d7c7b7a7 Sg8b8 Rh8a8
SPG 13,5                       13+13
Staugaard castling

Solution: (click to show/hide)

Notify of
Newest Most Voted
Inline Feedbacks
View all comments
shankar ram
shankar ram
May 30, 2021 18:30

Interesting – W and B 0-0-0-0-0-0 are mutually exclusive. Also interesting – A PG with Retro content!

Andrew Buchanan
Andrew Buchanan
June 1, 2021 19:14
Reply to  shankar ram

A proof game is often like a jigsaw: have various fragments of play which need to be consolidated, even after all the chess thinking has been done. Just like a jigsaw where you begin at the edges, so it can be easiest to look at the beginning and the end and work towards the middle. Slot in the bigger fragments first and then the smaller ones must fill in the spaces which remain.

shankar ram
shankar ram
June 1, 2021 19:23

Yes, Andrew, You spoke of this approach when I asked how a PG is solved – from the starting position (forward play) or the final position (backwards play)? My comment here referred to the presence of familiar Retro motifs like W or B can’t castle in a PG. I’m sure there are plenty of other PGs having such content. I’m a PG noob!

andrew buchanan
andrew buchanan
June 10, 2021 16:54

As far as I know, no engine can handle Staugaard castling at the moment. So I would be grateful if you can break this problem manually if you can. Here is the first part of my retro reasoning, with engine support where possible.
(1) Natch-3.3 says no orthodox solutions in 13.5 (or -0.5,1.0,1.5). So someone has castled Staugaard. And only 1 Staugaard can happen per game!
(2) Suppose Black castled, then that’s 6 moves +3 Ke6-f-g-h4 +2 Qd8-d3-c2 + 2 Re7-e3-a3. All Bl moves are accounted for, so bBc8 & a bP were captured at home. The only white captures were on a3, c8 & d/e7. So wPh never captured, and the only Bl move on h file is Kgxh, so that must have captured wPh, and wRh must have exited the file.
If bPe7 was captured at home, it must have been by P, N, or B. The piece was not captured there and moved on, but couldn’t have promoted on d8. Bc1 was only released later by bxa3 If it was wS, then 6 moves taken up Sb1-c3-d5xe7-d5-c3-b1 +2 wK +3 wRh + 1 wPf 1 +1wPh. Except: there’s no way to capture bBc8.
So it was bPd7 that was captured at home, while bPe7 promoted. It must have captured twice, as no other way to clear wPe2. Only candidate is wRh. So wRe2 is promoted. 7 moves wPc-…xd7xc8-e8-e2 + axb f4, h4, wK-f2-d3, wRh1-h3-d/f3. All White moves are accounted for.
cxd7 which releases bQ can only happen *after* castling. So first 6 Bl moves are forced. Wh must produce wR on d/f3 by W3, then extricates wK via f2, so wR was captured on d3. 1. h4 e5 2. Rh3 e4 3. Rd3 exd3 4. f4 dxe2 5. Kf2 e1=R. Pause White can only play 6. c4 then 0-0-0-0-0-0, and we can check A-B in Jacobi for the remaining 7.5 moves C+ 0.091. So that half of the problem is closed reasonably surely.
(3) The other possibility is that White castled. Was castling the final move?
If we retract it, Natch 3.3 finds 12 candidates for the prior PG in 13.0, e.g.: 1.Ph2-h4  Pe7-e5  2.Rh1-h3  Pe5-e4  3.Rh3-d3  Pe4xd3 4.Pe2-e4  Qd8xh4  5.Pe4-e5  Ke8-e7  6.Pe5-e6  Qh4-b4 7.Pe6xd7  Qb4-a3  8.Pb2xa3  Pd3xc2  9.Bc1-b2  Pc2-c1=Q 10.Pd7xc8=R Qc1-c2  11.Bb2-c1  Ke7-f6  12.Rc8-e8  Kf6-g5 13.Pf2-f4  Kg5-h4 *but* all of them imply 12.Rc8-e8 i.e. White couldn’t then castle.
I am reasonably confident of all reasoning so far. I will have a shot at writing down the rest when I am fresh, but this gives an indication to those who would break this that they should concentrate their efforts on demonstrating an earlier White castling. Thank you

Last edited 2 days ago by Andrew Buchanan