|Staugaard castling:||This is a fairy kind of castling, originating in a loophole in the Laws of Chess spotted by the Danish composer Conrad Staugaard in 1907. A promoted rook can be regarded as never having moved. This allows for vertical castling with wKe1 & wRe8 to wKe3 & wRe2 (or analogously for Black). (from Die Schwalbe)|
Some more words about the invention by Andrew Buchanan:
Conrad Staugaard (see the biography) invented the condition later termed Pam-Krabbe. He seems to have been a funny and kind figure: like a Danish Lewis Carroll, but a socially-aware entrepreneur. He deserves to be commemorated with his condition, which is how WinChloe and PDB already identify it. Recently discussed on PDB.
No. 1615 Andrew Buchanan,
Michel Caillaud Singapore / France
original - 30.05.2021
Solution: (click to show/hide)
Fake castling :
From the diagram, it looks as if white has just castled.
In fact, it is a fake castling and black has previously castled!
Some justification for the solution :
Orthodox counting, a minimum of 12 black moves are apparent :
Ke8-h4=4 +Qd8-a3 (for b2×a3)=2 +c1=Q-c2=6.
But h~-h4+ K×h4 implies too many white moves (for wRh1 to reach e2, not through h3).
Same, d7×c2-c1=Q implies too many white moves (white has to capture missing bLc8 and bPe7).
So, instead of a minimal number of 12 black moves, an effective
number of 14 othodox black moves is reached, 1 too many...
Interesting – W and B 0-0-0-0-0-0 are mutually exclusive. Also interesting – A PG with Retro content!
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.
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!
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