AB: First proof of "can castle", without recourse to conventions. (2009-11-22)
A.Buchanan: deadpos doesn't understand about retro logic yet, so does not eliminate R: 1. Ke4xLSBe3? (2022-11-13)
AB: typo in diagram: Be6 is f5. (2001-12-29, hidden) more ...
comment

Shortest known proof game showing Valladao-Monteiro (after 8.5 moves - the extra half-move is to move beyond the check, and is the only move which preserves uniqueness of the PG).
Composed for http://abrobecker.free.fr/chess/IntroductionToPGs.pdf
>>> Agreed with Enzo Minerva that this is EM v AB: 24-Dec-2014. (Original is at P1284755.)
Kevin Begley: Partially anticipated, by Enzo Minerva, Best Problems 2005 (which shows Valladao-Monteiro in as many moves, based upon the same idea). (2014-03-01)
A.Buchanan: Thanks for highlighting this, Kevin. I had searched PDB for prior art before composing this one, but Enzo's problem was (and still is) absent. I feel that my little composition does add something more. In mine, the identity-defining Ceriani-Frolkin bishop move is not a capture. And WQ has potentially three routes to its final square, including the CF promotion square, but the two duals are eliminated by the waiting move d3. So, is Enzo still active? I am thinking this partial anticipation probably a case of "AB after EM" but if Enzo is still around I would like to discuss with him out of courtesy. (2014-05-01)
Kevin Begley: Andrew, at best, I see yours as a version. (2014-05-01)
Kevin Begley: Compare to 'P1284755'. Clearly Andrew's version is better (and not only because of the extra move avoiding check in diagram -- which may be forgiven in a proofgame aiming specifically for brevity), however thematically, they are nearly indistinguishable. I leave it for Andrew and Enzo to collaboratively determine the proper credit here. (2014-05-01)
A.Buchanan: Oh for Pete's sake... there's *another*, now I look in WinChloe:
Alexeï GASPARYAN Oakham PG Tourney,
The Problemist Supplement 2009-10
1° Mention d'Honneur.
rnb2rk1/pp1p1ppp/3p4/2q5/1b6/3p4/PPP1PKPP/RNBQ1BNR
PG in 9.0
1.f3 e5 2.f4 Bb4 3.f5 Ne7 4.f6 00 5.f×e7 e4 6.e8=N Qg5 7.Nd6 c×d6 8.Kf2 Qc5+ 9.d4 e×d3ep+ (C+)
Now the date in WinChloe is wrong: what counts is the closing date for the tournament, 31-12-2009 (although the judging was only held in 07-2010, and publication was in 11-2010). Mine was published in 01-2010 (composed on 10-01-2010. I think I fold at this point, and allow Messrs. Minerva & Gasparan to duke it out, if they care. (2014-05-04)
A.Buchanan: @Kevin: and on the subject of versioning. Do you have a link to anything that gives guidelines to relative attribution. The Codex is sketchy (as usual), and I don't know where else to look. (2014-05-04)
Henrik Juel: On versioning:
When inspired by a published problem by AA to improve it, I would write 'hj after AA' if my contribution is large and 'AA, version by hj' is it is small. Of course, 'improve', 'large', and 'small' are not well defined, so it is up to the improver and the editor what to choose, and up to the judge whether an award is warranted.
In the Minerva-Gasparyan-Buchanan case the later authors surely were not inspired by the first, so here I would suggest the following possibilities: anticipated, partially anticipated, and not anticipated, again leaving the final verdict to the judge (2014-05-04)
A.Buchanan: I don't see any fundamental difference between accidental and deliberate partial anticipation. Ultimately the choice is between labeling the problem as "A,version", "A,version B" or "B, after A". I am totally ok with whatever Minerva/Gasparyan eventually determine. However, I don't think that PDB supports "A, version B" currently. (2014-05-05)
A.Buchanan: Not heard anything from Messrs Minerva or Gasparyan. I should rephrase my earlier statement as: "I don't see any fundamental difference between accidental partial anticipation, and acknowledged deliberate versioning, in terms of the set of possible final outcomes." (2014-12-24)
A.Buchanan: Resolved with Enzo :D! (See above!) (2014-12-24)
A.Buchanan: Resolved with Alexey in Facebook yesterday :D! AB: "I would like to ask you: are you ok with the current attribution of my problem?" AG: "Yes I agree." So there we are. Maybe EM & AG will conduct diplomatic negotiations at some point, but that's not my concern :) (2021-10-05)
Kevin Begley: Partially anticipated, by Enzo Minerva, Best Problems 2005 (which shows Valladao-Monteiro in as many moves). (2014-03-01, hidden) more ...
comment

Objective: Staugaard (Pam-Krabbé) castling as mating move.
Henrik Juel: Hint below...
The problem is probably 'correct', as Natch 3.1 reports no solution in 12.5 moves, and a unique solution for the first 12.0 intended moves

Hint:
The joke happens in the final white move and relies on an unintended reading of the castling rules (2019-12-23, hidden)
Henrik Juel: Searching for 'joke' and 'castling' gives many hits in the PDB, but none in a proof game, so this may be a first
The search also revealed that the joke was invented by a fellow Dane in 1907, P1277302, and re-discovered in 1972, P1277303 (2019-12-23, hidden)
Adrian Storisteanu: Good grief, Andrew, you got the joke indicator all wrong, should be (-: (cf. my comment to P0007559)... :-) (2019-12-24, hidden)
A.Buchanan: Haha I’d forgotten this. What’s the difference in meaning between :-) and (-: (2019-12-24, hidden)
Adrian Storisteanu: For one thing, the common smiley is overloaded (joking, happy, ironic, ..). A less common angle (-: suggests the chess problem itself should be viewed from a different / uncommon angle. Duh. (2019-12-24, hidden)
James Malcom: I like how has its own tag now! I made note of the “anticipation” over on Matplus awhile ago. In the thread, Per Olin shared a h#4 that has Pam-Krabbe castling, but the author and source are unknown. Anyone want to add it in? See here: http://www.matplus.net/start.php?px=1577245612&app=forum&act=posts&fid=anti&tid=2302 (2019-12-25, hidden)
A.Buchanan: Hi James don’t you have edit privileges yourself? Why not try adding it in yourself? If there’s anything out of order, we can sort it out easily (2019-12-26, hidden)
A.Buchanan: In the mean time, if anyone wants to add (-: to their own joke problems that would be great, so it can become a standard going forwards.
There are two problems with jokes: (1) identifying that there is a joke going on at all, rather than misprint, composer error, weird convention (2) identifying which rule is being broken etc if joke is accepted. Adrian’s smiley can at least help with item 1.
The problem is without the smiley, awareness the joke may depend too much on the context in which the problem is being presented - the composer themselves should have a chance to influence this process more if they choose (2019-12-26, hidden)
James Malcom: Thanks for Gerd’s wonderful and much appreciated work, this animation to this PG is now complete at last! (2020-05-03, hidden)
James Malcom: Thanks for Gerd’s wonderful and much appreciated work, this animation to this PG is now complete at last! Also. I’ve gone through and recalculated all problems with Pam-Krabbe castling (2020-05-03, hidden)
A.Buchanan: I don't think it's practical to run a proof game of this length in WinChloe with rockagogo condition. The best line of attack is to focus on the last move. We know there is no orthodox solution, so any solution must include P-M castling. If this was the last move, then we know there is a unique solution for the first 12.0 moves, and we are home. The other candidates for last move are 13. d2-d3 (so wLb4 is promoted: too many Wh moves required) or 13.Te1*e2 (all 13 Wh moves spoken for implies other pieces were all captured on their starting square, but it's not so quick to be able to dismiss this alternative candidate). Does anyone have a quick proof that last move was not 13.Te1-(x)e2# (2021-01-06, hidden)
James Malcom: Andrew, the proof is simple enough, actually.

Since we know PK castling must have taken place, let's count up how many moves we know must have happened from the diagram at a bare minimum: 1 queen move+2 bishops moves+1 knight move+6 pawn moves divided between 5 for wPe2 and 1 for wBd3+1 king move. This takes away 11 out of 13 possible White moves. There are 2 left over, enough for Re2 to e1 back to e2.

However, there is a major flaw-how did White's c2 pawn and f1 bishops get captured then, with neither of them moving by the above argument? The h1 rook and g2 pawn were of course taken by Black's bishop.

Black's king and bishops literally have no time, in 12 Black moves as a limit, to capture them AND get in position, along with the pawn and queen move.

There, the White c2 pawn f1 pawn did move, leaving no time for a rook switchback between e1 and e2,

So the PG is C+ after all. (2021-01-06, hidden)
James Malcom: Furthermore, with the above arguments, the wPc2 couldn't have promoted, as there is then no time to capture wBe2, nor could wPe2 have time to move and be captured to let the Black king pass through. (2021-01-07, hidden)
James Malcom: Lastly, the White pieces couldn't have been captured at home due to simple math. for Black, king moves+1 pawn move+at least 3 moves to take Pg2 & Ph1 takes 3 moves=10/12 Black moves. This leaves nowhere enough time to capture the Pc2 and Bf1 at home. (2021-01-07, hidden)
A.Buchanan: I never doubted that a careful manual retro argument for soundness could be made. But if this implies that the problem is C+ or HC+, then we could reclassify thousands of existing retros as “C+“. I am ok to bend the current notion of C+ a little to include “thinking” but this subjectively goes too far for me. And what to be gain by obsessively classifying this (superseded) problem as “C+”? There is no benefit.
One root issue is that we are missing an overall concept of soundness in PDB. I have a draft email for Gerd about this based on the Codex which is quite good in this area, but I have so many other things to do too. Maybe I could post it in MatPlus but then I fear many hobbyhorses would be taken out of their stables for a gallop: (2021-01-07, hidden)
A.Buchanan: I have removed C+ sticker from this one, not to prejudge the issue, but so that by querying for k
='pam' and not cplus, one can see all 5 P-K problems where there are issues. All the other 14 I think get a clean bill of health: WinChloe rokagogo is sufficient to genuinely validate: all their C+ stickers are 100% merited. Happy if we put back C+ on this one, once we decide what "C+" means. (2021-01-07, hidden) more ...
comment

See P1234039 and P1375777
The joke here is that pawns may promote to enemy officers (but not to pawns or kings)
Henrik Juel: R: 1... 0-0-0+ 2.a7-a8=sT! (2020-10-12)
A.Buchanan: Maybe this is a moderate joke where we allow promotion to Bl officers, while son-of-P1375777 can be the extreme joke where promotion to any Bl unit is allowed (i.e. pawn & king too), including in the history of the game. What would Josef Cumpe think? :-) He and I are both dressed as clerics in out chesscomposers' blogspot pics: http://chesscomposers.blogspot.com/2012/08/august-16th.html. Perhaps he was a real one though? (2020-10-13)
comment

See P1234039, P1375777 and P1380852.
The joke here is that pawns may promote to *any* unit (friend or enemy; officer, king or pawns).
Henrik Juel: R: 1... 0-0-0+ 2.a7-a8=sT
not 2.e7-e8=sK, because then sTa8 would have moved, and castling not allowed (2020-10-14)
A.Buchanan: Exactly, Henrik. Also: sB on 7th rank could retract to 8th rank, so we need sT instead (2020-10-14)
Henrik Juel: A non-joke solution is
R: 1... 0-0-0+ 2.h5-h6 (2020-10-14, hidden)
A.Buchanan: Sorry Henrik typo have fixed (2020-10-14, hidden) more ...
comment

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

Corrects (with a little more functionality) P0001361
Henrik Juel: How does one find the parity of the white moves, with DLSS missing? (2020-12-09)
Henrik Juel: Never mind
If White can castle, then last move was e2-e3, so we know where the missing white men were captured, and the parity is easily found (2020-12-09)
Henrik Juel: Suppose (for contradiction) that both sides may castle
Black has the move, so last move was e2-e3, and [Dd1,Lf1] were captured at home, and [Sb1,Sg1] were captured on e6,d5
The number of white and black moves in the game leading to the diagram are both even, so White should have the move
Conclusion: the castlings are mutually exclusive
In the solution Black must castle to get his king to h8, so White cannot get his Th1 to f1 by castling (2020-12-09)
comment

James Malcom: How does this work exactly? What forbids the rook from just making some random tempi move in a game that leads to this position? (2020-12-12)
Mario Richter: James, think of time pressure. In the diagram, Black has only two retractions (a7xSb6xSa5). The only way to unlock the position is the retraction of e2-e3.
Little hint: castling is considered as a king m,opve ... (2020-12-12)
Henrik Juel: By now there are so many clues revealed for this fine retro that I dare to give a retroplay despite the Byzantine dating, in this case indicating an ongoing solving competition
R: 1.Tf1xPf6 b6xSa5 2.0-0 a7xSb6 3.Lf1-e2 Kd4,Kf4 4.e2-e3+ etc.
Incidentally, TxP is not the first move, but the last... (2020-12-12)
Henrik Juel: I missed the from square in the retraction by the black king, sorry
3... Kd4-e5,Kf4-e5 (2020-12-12)
A.Buchanan: Ha, there is no solving competition for these either. But it's to avoid Gerd's solving competition protection algorithm picking up false positives (2020-12-12)
A.Buchanan: TxP happens to be the last move, but it's also the first effective move of the rook, as it is considered to be a *passenger* in the castling, and not to have moved yet. Otherwise can't get first move of a rook fixed if it capturing (except possibly by DP logic). See the explanatory article: https://www.dieschwalbe.de/download/artikel/Passenger_Rooks.pdf (2020-12-12)
A.Buchanan: Your solution is completely correct, Henrik! (2020-12-12)
comment

Henrik Juel: Oh, I was confused by the LLLLl, which should be TLLLLl, I believe
R: 1.Lg2-f3 b4-b3 2.Tf1xSf6! b5-b4 3.0-0 b6-b5 4.Lf1-g2 Kf4-g5,Kh4-g5 5.g2-g3+ (2020-12-12)
A.Buchanan: Thanks for the catch. I copied the diagrams straight from the article FEN, so should be clean of transcription errors, but then I put in the non-standard material field by hand & eye, so could be other errors as here. (2020-12-12)
A.Buchanan: Your move sequence is again correct! :D (2020-12-12)
Henrik Juel: Is there a diagram error?
White has three rooks (2020-12-12, hidden) more ...
comment

Henrik Juel: Good motivation for TxL
R: 1.Tf1xLf8! e4-e3 2.0-0 e5-e4 3.Lf1-g2 e6-e5 4.g2xPh3 h4-h3 and e.g.
5.d4-d5 h5-h4 6.Le4-c6 h6-h5 7.c7-c8=L! h7-h6 8.c6-c7 c7xLb6
f7xTg8=L happened early (2020-12-12)
comment

Hans-Jürgen Manthey: 1. Lg2-a8 c3-c2 2. Tf1xTf6 c4-c3 3. O-O ??? (Rochade nicht möglich wegen sTf6 !)

mögliche Lösung:
1. Tf1xTf6 c3-c2 2. Lf3-a8 c4-c3 3. O-O c5-c4 4. Lg2-f3 c6-c5 5. Lf1-g2 Kf4-g5 6. g2-g3+ Sg5-h7 7. Ld3-g6 Se4-g5 8. Th7-g7 Dg6-f7 9. f7-f8L Tf5-f6 10. Le5-h8 Lf8-h6 11. Lh4-e7 Dg5-g6 12. La7-e5 Tf6-f5 13. Lb8-a7 Th6-f6 14. Lf6-d8 Sc5-e4 15. Lb2-f6 Dd8-g5 16. Lf2-h4 Kg5-f4 17. f6-f7 Sa6-c5 18. Te7-h7 Th8-h6 19. f5-f6 Kh6-g5 20. La3-b2 Kg7-h6 21. Lf7-e8+ Kf6-g7 22. Lc1-a3 Kg7-f6 23. b7-b8L h7-h5 24. La7-f2 Kf6-g7 25. Lb8-a7 Kg7-f6 26. b6-b7 Kf6-g7 27. b7-b8L Kg7-f6 28. Te8-e7 Df6-d8 29. Ta8-e8 g5-g4 30. Le8-f7 f7xSe6 31. Sf4-e6+ Sb8-a6 32. Ta1-a8 Dd8-f6 33. e7-e8L Kf6-g7 34. e6-e7 Ke7-f6 35. b5-b6 g6-g5 36. b6-b7 g7-g6 37. c4xLb5 La6-b5 38. c2-c4 Lc8-a6 39. La6-d3 Ke8-e7 40. Lb7-a6 Ke7-e8 41. La8-b7 Ke8-e7 42. a7-a8L Ke7-e8 43. d5xTe6 Te3-e6 44. a6-a7 Ke8-e7 45. a5-a6 Ke7-e8 46. a4-a5 Ke8-e7 47. a2-a4 Ta3-e3 48. Sh3-f4 Ta8-a3 49. Sg1-h3 b7xSc6 50. Sa7-c6 Sc6-b8 51. Sb5xa7 e7xDd6 52. Dh6-d6 Sb8-c6 53. Dd2-h6 Sc6-b8 54. Dd1-d2 Sb8-c6 55. f4-f5 Sc6-b8 56. f2-f4 Sb8-c6 57. d4-d5 Sc6-b8 58. d2-d4 Sb8-c6 59. Sc3-b5 Sc6-b8 60. Sb1-c3 Sb8-c6 61. b5-b6 Sc6-b8 62. b4-b5 Sb8-c6 63. b2-b4 (2020-12-14)
Henrik Juel: e.g. R: 1.Lg2-a8 c3-c2 2.Tf1xTf6 c4-c3 3.0-0 c5-c4 4.Lf1-g2 Kf4-g5 5.g2-g3+ (2020-12-12, hidden)
A.Buchanan: wLf3 must shield uncastling from sTf6. (2020-12-13, hidden)
Henrik Juel: Yes, of course (2020-12-13, hidden) more ...
comment

Henrik Juel: R: 1.Tf1xDf6 b3xLc2 2.Le4-c2 and e.g. 2... c5-c4 3.Lf3-e4 c6-c5 4.0-0 b4-b3 5.Lg2-f3 b5-b4 6.Lf1-g2 Kf4-g5 7.g2-g3+ (2020-12-12, hidden) more ...
comment

Henrik Juel: White captured TxPf3+, axbxcxd, bxcxd, cxd, fxPe, g5xh6, and h6xg7-g8=L
R: 1.Tf1xPf3+ f4-f3 2,f3xPe4+ e5-e4 3.0-0 Kg2-g3 (2020-12-12)
comment

Henrik Juel: Black has played c3/a3xLb2-b1=T, earlier Kb4-a3-b2-b1, and still earlier Sb3xTa1
R: 1.Tf1xSd1+ Sc3-d1 2.0-0+ (2020-12-12)
Henrik Juel: A very impressive set of retros
Thanks for making them available, Andrew (2020-12-12)
A.Buchanan: Very welcome Henrik: thanks for your invaluable solving! (2020-12-13)
comment

Henrik Juel: R: 1.Tf1xLf8+ f4xg3 2.0-0 e5xf4 3.Lf1-g2 d6xe5 4.g2xDTSh3
and there is still room for White to unpromote via c7 and f7 (2020-12-13, hidden) more ...
comment

Henrik Juel: R: 1... g3xTf2 2.Tf1xTf2 Tf3xTf2 3.0-0 Dh2-h3 and e.g.
4.Tf1-f2 Lg1-e3 5.Sf2-d1 Ke5-f4 6.e3xPd4+
The remaining capture by Black was f5xLg4
[Pa7] was captured by a white officer (2020-12-13)
comment

Minor correction: sBd7->c6 wSb2->wL
Henrik Juel: Not R: 1.Tf1xDf2+ Dg3-f2+ 2.Lf2xPb6 b7-b6 3.0-0 Dh2-g3,
because White can no longer uncapture hxgxh
What was the intention? (2020-12-13)
A.Buchanan: Attempted repair to avoid accidentally retro-blocking c8 (2020-12-23)
Hans-Jürgen Manthey: mögliches Retrospiel: 1. Tf1xDf2# Dg3-f2 2. Lf2xb6+ b7-b6 3. O-O Dh2-g3 4. Se3-g2 Dg2-h2 5. Sc4-e3 Dh2-g2 6. Tg1-g4 Dg2-h2 7. Sa3-c4 Dh2-g2 8. g4-g5 Dg2-h2 9. Lb6-f2 Dh2-g2 10. Ld8-b6 Ke3-f3 11. d7-d8L Kd4-e3 12. Sc3-e2+ Kc5-d4 13. Le2-d1 Kd6-c5 14. Lf1-e2 Tf2-d2 15. Tg2-g1 Tf3-f2 16. Td2-g2 Se2-c1 17. Td1-d2 Te3-f3 18. Tb1-d1 Te6-e3 19. Lc1-b2 Th6-e6 20. f2-f4 Th8-h6 21. g2-g4 Sf4-e2 22. Se2-f3 Sd5-f4 23. Sg1-e2 Ke6-d6 24. Df4-h4 Ta8-h8 25. Da4xBf4 f5-f4 26. Da5-a4 Db8-h2 27. g4xTh5 h4-h3 28. b2-b3 Sb3-a1 29. Dd8xBa5 Sc5-b3 30. Df8-d8 a7-a5 31. Dh8xLf8 f7-f5 32. d6-d7 Kd7-e6 33. e5xBd6 Sf6-d5 34. Dh7-h8 Ke8-d7 35. De4-h7 Sg8-f6 36. Ta1-b1 Dd8-b8 37. Sb1-a3 Th8-h5 38. h3xLg4 h5-h4 39. h2-h3 Lc8-g4 40. De2-e4 h7-h5 41. e4-e5 c7-c6 42. d2-d3 Sa6-c5 43. Dd1-e2 d7-d6 46. e3-e4 Sb8-a6 47. e2-e3 (2020-12-25)
A.Buchanan: Looks like a cook to me heh :D let me check with my esteemed co-author. (2020-12-13, hidden) more ...
comment

Henrik Juel: R: 1.Tf1xDf3+ Sf2-h1 2.0-0 Se4-f2 and e.g.
3.Sf2-h3 Lh3-g4 4.Lg4-h5 Th5-g5 5.g5-g6 Th6-h5 6.Lh5-g4 (2020-12-13, hidden) more ...
comment

Objective: Staugaard (Pam-Krabbé) castling as mating move.
Henrik Juel: solution
1. Bd2-d4 Bf7-f5 2. Ke1-d2 Bf5-f4 3. Kd2-d3 Bf4-f3
4. Kd3-e4 Bf3xe2 5. Ke4-f5 Be2-e1=T 6. Dd1-e2 Bg7-g5
7. De2xe7 Sg8xe7 8. Kf5-f6 Lf8-g7 9. Kf6xg7 Se7-d5
10. Kg7xh8 0-0-0-0-0-0#
First 9.5 moves are C+ Euclide 1.01
On behalf of the danish inventor of the looong castling, I thank for the dedication (2021-01-21)
James Malcom: After the 12.5 Andrew and I made, this turns out the fastest possible. 9.0 if you go non-unique. (2021-01-21)
A.Buchanan: Welcome, Henrik. Any idea what his initial "C." stands for? I couldn't find it online. (2021-01-22)
Henrik Juel: Unfortunately not, Andrew
In those days the first name was almost always abbreviated to an initial
Christian was a very common first name, but this was usually abbreviated to Chr.
So a wild guess is Carl, but it really could be anything starting with C
In a Skakbladet issue he is referred to as Direktør Staugaard, but this just means that he was in charge of a company (2021-01-22)
A.Buchanan: I have posted an English definition of the keyword. Please check if it suits. (2021-01-22)
Henrik Juel: Searching for direktør staugaard, randers (a danish city) yields
Conrad Staugaard (1852-1922)
so this is my best bet
Conrad Staugaard headed a bank in Randers, founded by him in 1891, but the CV in
http://historiskranders.dk/randers-biografier/
does not mention anything about playing chess or chess problems
Chess playing was very common in the middle class, so the biographer may not have found it worth mentioning (2021-01-22)
Henrik Juel: Your keyword definition seems fine,
except for the spelling of showcase... (2021-01-22)
A.Buchanan: Haha bingo: "Staugaard was a well-known and excellent chess-player, one of the many traits he had in common with Lewis Carroll". MMA SAM - Tidsskrift.dk. Googling for "Conrad Staugaard skak" brings up the pdf for Rama Sama. Please tell us what you make of it, Henrik :) In the mean time, I have updated the author info for him (2021-01-22)
Henrik Juel: Very good, Andrew
Ah, the joys of Google...
In the danish text there is not only your citation on chess, but also one on chess problems
There are also examples of Conrad's excentric behaviour, in line with him inventing the looong castling
That C. Staugaard might be Conrad's younger brother Carl (who surely also knew about chess) is ruled out by the fact that Carl emigrated to South America in 1901, because he experienced disappointments in his love live... (2021-01-22)
A.Buchanan: Yes it said he had lots of amazing concepts. The idea about the banner for his nursemaid to take his children for walks without getting her arms stretched is very funny. This is our man. He also seems to have been a good man, and an effective businessman. And loved enough by his friends for a 120 page book to have been commissioned in his memory. He is reminiscent of other children’s authors like Edward Lear, Dr Seuss, Heinrich Hoffmann... I’m so happy that this problem has brought him back to our awareness (2021-01-23)
comment

Objective: Staugaard (Pam-Krabbé) + queenside castling.
Henrik Juel: solution
1. Be2-e4 Bd7-d5 2. Lf1-d3 Lc8-f5 3. Be4xf5 Dd8-d6
4. Bf5-f6 Sb8-d7 5. Ld3-g6 0-0-0 6. Bf6xe7 Dd6xh2
7. Be7-e8=T Dh2xh1 8. 0-0-0-0-0-0 (2021-01-21, hidden) more ...
comment

Objective: Staugaard (Pam-Krabbé) king is mated.
Henrik Juel: The first 6.5 moves are C+ Euclide 1.01 (2021-05-08)
A.Buchanan: Thanks Henrik. In the advance of an engine supporting Staugaard, here is my purely HC+ proof of soundness. No orthodox solution so 6 Bl moves needed to Staugaard. Other Bl move was g7-g6 or h7xg6. bRh8 & other bPg/bPh never moved - captured by Wh raider(s). Only piece that could have captured wPe was bPe, so it captured twice, other prey being wB, which thus moved once, and could not have been the raider. 6 Wh moves now countable without raiding, so wQ raided, and took 4 moves. All Wh moves now accounted for, so bPe7-e5xf4-f3xe2-e1=T. wQ could only have exited on d file, as diagonal d1-h5 open too late. Thus wQ-d3xh7xh8-e5#. All moves are known, and jigsaw of them moves is unique, ensuring each side can always move. (2021-05-08)
A.Buchanan: I forgot to say that I think my manual proof, while watertight, is too complex to make the status of this problem HC+. I think HC+ is for where the H part is self-evident. (2021-05-08)
Henrik Juel: solution
1. Bd2-d4 Be7-e5 2. Lc1-f4 Be5xf4 3. Dd1-d3 Bf4-f3
4. Dd3xh7 Bf3xe2 5. Bf2-f3 Bg7-g6 6. Ke1-f2 Be2-e1=T
7. Dh7xh8 0-0-0-0-0-0 8. Dh8-e5# (2021-01-21, hidden)
A.Buchanan: Objective: Staugaard (Pam-Krabbé) king is mated (2021-05-08, hidden) more ...
comment

Objective: Play of Staugaard (Pam-Krabbé) castling makes unsound PG sound.
Henrik Juel: I see no castling paradox (2021-01-21)
A.Buchanan: I must complete the stipulation, sry (2021-01-22)
A.Buchanan: Not really paradox in the usual hidden/fake sense. Need a different term for the occasional idea that an earlier point in the proof game is not unique. (2021-01-22)
Henrik Juel: solution
1. Be2-e3 Bh7-h6 2. Lf1-d3 Th8-h7 3. Ld3xh7 Be7-e6
4. Lh7-f5 Be6xf5 5. Sg1-f3 Bf5-f4 6. 0-0 Bf4xe3
7. Sf3-e1 Be3xd2 8. Dd1-f3 Bd2xe1=T 9. Lc1-e3 Bb7-b6
10. Sb1-d2 Bb6-b5 11. Le3-b6 0-0-0-0-0-0 12. Tfb1 (2021-01-21, hidden)
Henrik Juel: The answer to part b is No, e.g.
1. Be2-e3 Bh7-h6 2. Lf1-d3 Th8-h7 3. Ld3xh7 Be7-e6
4. Lh7-f5 Be6xf5 5. Sg1-f3 Bf5-f4 6. 0-0 Bf4xe3
7. Sf3-e1 Be3xd2 8. Dd1-f3 Bd2-d1=T 9. Lc1-e3 Bb7-b5
10. Sb1-d2 Td1xe1 11. Le3-b6 (2021-01-22, hidden) more ...
comment

A.Buchanan: Without German KBP in the stip the animation won't work (2021-02-12)
comment

Objective: fastest Staugaard (Pam-Krabbé) + another castling (happens to be kside)
A.Buchanan: Now we know more about Conrad Staugaard and his life, I enthusiastically agree with Henrik's earlier suggestion that we change this theme to be "Staugaard", which would also align with WinChloe. I can make necessary changes in PDB as a batch, and retain also "Pam-Krabbe" as a deprecated keyword, pointing to Staugaard. Pam & Krabbé are famous already, and only popularized Staugaard's funny idea. Any objections? (2021-05-08)
A.Buchanan: OK thank you: I've made the move. Please check the definitions, and let me know anything that needs changing (2021-05-09)
Henrik Juel: Very good, Andrew; the Staugaard fan club (both of us) thanks you
You might want to add that after a white Staugaard castling, wK is on e3 and promoted wR is on e2
I wonder whether the solution notation should be changed to 0-0-0-0-0 with 5 zeroes, so that the number of zeroes is one less than the length of the rook displacement for all three castlings (2021-05-09)
Henrik Juel: Forget my very last remark, please
Short and long castling has rook displacement = number of zeroes,
so Staugaard should remain with 6 zeroes, of course (2021-05-09)
A.Buchanan: Ok listed the target squares as requested. (2021-05-09)
A.Buchanan: Curious that this economy record must begin with a tempo move (2021-05-20)
A.Buchanan: Staugaard seems a much better basis than Chess960 for further generalization of castling. Basically move an unchecked royal king 2 unattacked square towards the rook, and the rook hops the otherside. This works whatever the gap between the two. This is more similar to rokagogo. (2021-05-21)
Henrik Juel: No... (2021-05-08, hidden)
Henrik Juel: solution
1. Bg2-g3 Be7-e6 2. Lf1-g2 Be6-e5 3. Lg2xb7 Be5-e4
4. Sg1-f3 Be4xf3 5. 0-0 Bf3xe2 6. Lb7xa8 Be2-e1=T
7. La8-g2 0-0-0-0-0
The first 6.5 moves are C+ Euclide 1.01 (2021-05-08, hidden)
Rainer Staudte: I agree to renaming. (2021-05-08, hidden) more ...
comment

Corrects P1323157.
Henrik Juel: If both White and Black can castle, there is no way to explain sTh7
If only White can castle, both black rooks are original
If only Black can castle, sTh7 is a promoted pawn
So the castlings are mutually exclusive, and the solution
1.0-0-0 thr. 2.Txb7 thr. 3.Tb8# is C+ Popeye 4.61 (2021-05-11)
Henrik Juel: I had a hard time finding the correct Popeye input, which is 'opt cas a1h8'
Good work, Andrew; if you continue like this, you may expect a peerage letter from the british queen, provided Scotland stays in the union... (2021-05-11)
A.Buchanan: Thanks Henrik: yes I can never remember these parameters, but Olive gave me a hint after I got it wrong :) I replaced wSa7 by wLa5 for an extra variation. These problems were just Christmas puzzles for Dunsany every year in the main Literary magazine of the time. So I fix it in that spirit, but I'm sure I've seen other more intricate problems where both sides have one side boxed in by possibly cross-capturing pawns depending on capture-counting (2021-05-11)
Henrik Juel: The additional variation is 1... Kd8 2.Txa6+ b6,Kc8,Ke8 3.Ta8#
Still C+ Popeye 4.61 (2021-05-11)
comment

Corrects P0000914.
Black to move.
Missing White unit captured on h-file. So wPa promoted, hence 11 necessary captures by White, within budget.
Henrik Juel: C+ Popeye 4.61 (2021-10-10, hidden) more ...
comment

See P0574383.

With the ink still wet on this correction, I would like to remove sD. Sound by Popeye & Retractor, but needs an adapted proof game for C+.

Proof Game: 1. c4 a6 2. g3 Nc6 3. Bg2 Na5 4. Bc6 h5 5. Bb5 axb5 6. g4 c5 7. d4 cxd4 8. e4 f5 9. Be3 fxe4 10. f4 g5 11. Qf3 Nb3 12. Na3 hxg4 13. Nc2 gxf3 14. Rd1 Na1 15. Rd2 Ra3 16. Rg2 Rc3 17. a4 d3 18. Bf2 e3 19. Bg3 e5 20. Re2 e4 21. a5 d5 22. a6 d4 23. a7 Qd5 24. a8=R Qe5 25. Rf2 Kd7 26. Re2 Nf6 27. Rd2 Nd5 28. cxd5 Kd6 29. Rf2 Kc5 30. Re2 b6 31. Rd2 Kc4 32. fxe5 Kb3 33. Re2 fxe2 34. Rb8 Bg4 35. Rb7 Bf3 36. Rc7 Ka2 37. Bf4 gxf4 38. Rc6 Bb4 39. Rc5 Rh3 40. Na3 Ba5 41. Rc4 bxc4 42. Nc2 Ra3+ 43. b4
A.Buchanan: 6n1/8/1q6/b7/1Ppppp2/r2ppp1r/k1N1p2P/n3K1NR has one less piece again, but I prefer the current version as retro logic is more interesting. (2021-11-26)
James Malcom: Enjoy an optimal PG Andrew. The total sum of the queen & knight journeys is 8 no matter which knight ends up on a1. (2021-11-27)
A.Buchanan: :-) thanks James (2021-11-27)
Henrik Juel: Black captured 7 white men with pawns and one more capture
White captured cxd (to explain the pawn inversion on the d-file)
Retroplay 1.b2-b4 Tc3-a3+ 2.Sa3-c2
1.c4*b3ep+ Sc2-b4+ 2.Ka2-b1 Sg1*e2 3.b3-b2 0-0#
HC+ Popeye 4.61 (2021-11-24, hidden) more ...
comment

The cooked P0006972 can be fixed many ways. I wanted to find a single touch that would keep the intended solution, the intended try, and exactly one of the cooks which seemed worth saving, while eliminating all the other cooks.
Henrik Juel: Black captured [Lf1] and the remaining 4 missing white men with exdxcxTb2-b1=L and gxh
1.Bb1*a2 Ra1*a2 2.0-0-0 Ra2*a7 3.Rd8-d7 Ra7-a8 #
1.Qg8-g3 Ra1*b1 2.Qg3*b3 Rb1*b3 3.0-0 Rb3-g3 #
Not 1.Bb1*a2 0-0-0? 2.Qg8*g2 Rd1-g1 3.0-0 Rg1*g2 # (White may not castle, as he has moved his king)
Neat birthday greeting (2021-11-27)
James Malcom: Thanks for the warm gift Andrew. Huzzah! (2021-11-28)
comment

Corrected version of P0003264, with extra tightness.
Henrik Juel: HC+ Popeye 4.61 (2022-01-07, hidden)
Henrik Juel: 1.bxc3ep+ d3 2.0-0 Tf4 3.Tf8 Txf8# (2022-01-07, hidden) more ...
comment

Moldenhauer: Computerprüfung: C- oder Cooked Stelvio 1.4 in 1 Sekunde.
Die Diagrammstellung ist verschieden erreichbar.
Hier 2 Möglickeiten:
1.Sa3 d5 2.Sb1 Kd7 3.f3 De8 4.Kf2 e5 5.Kg3 De6 6.Kh4 Dg6
7.Kh3 e4 8.fxe4 f6 9.e5 Ke8+
1.Sa3 d5 2.Sb1 Dd6 3.f3 Da6 4.Kf2 Sd7 5.Kg3 Dg6+ 6.Kh4 e5
7.Kh3 e4 8.fxe4 f6 9.e5 Sb8+ (2023-07-29)
A.Buchanan: Hi Moldenhauer this problem is sound. You are forgetting the constraint: "Drawn game". (2023-07-29)
Moldenhauer: Hi Andrew, ich habe deshalb nur die Diagrammstellung angegeben.
Ich hoffe Andrew ist erlaubt! Mein Vorname ist Klaus. (2023-07-29)
A.Buchanan: Hi Klaus nice to meet you, thanks for all your work. The point of HC+ is that it combines human & computer logic intelligently. It makes zero sense to solve robotically for 9.0. It is definitely valuable to solve this kind of stipulation (1) as PG in 5.0, and then (2) as PG A-B for 2.0. Apparent duals can then be resolved through reasoning about castling, e.p. and premature draw. (2023-07-29)
comment

cheeky retro-active version of P0001136
A.Buchanan: Looking at the original, I am wondering why wBh cannot be on h5. It clearly cannot retract to h3, but how many times is it needed to retract if White retracts from the critical position 1Q3rk1/2pppppp/8/2p4P/2P5/PPRP4/RBpP1PP1/N1K5 (2022-02-07)
A.Buchanan: Maybe the distinguished composer felt that the more spare White moves has, the more creditable is the fact that Black had no way to avoid prior castling. The problem is still sound either way. Of course, with the h#1 version, wBh6 is mandatory. wD location was pretty much arbitrary in the original, but now there is only one wD location for soundness: g3. (e5 & g4 are unsound). (2022-02-07)
A.Buchanan: The solution given in Keym’s wonderful “Eigenartige Schachprobleme” uses an unnecessary move by wPh. (2022-02-23)
A.Buchanan: In offering this version, I do not think that pure retros are in anyway inferior to forward problems, which was perhaps the motivation for forward stipulations in the days of yore. Rather, the aesthetic tension between past and future never tires for me. It is not an issue to me whether the balance of interest in a retro-active problem be equally divided between the two directions. Rather, it’s essential that the position have two readings: that the pieces have two purposes. This formal requirement of double meaning which is ubiquitous and quotidian in a cryptic crossword should also find a place near to the hearth in chess problems. So Morse might welcome Dexter. (2022-02-23)
A.Buchanan: I also like it when the locations of all the units are uniquely determined. This humble version finally nails down wQ & wPh (2023-08-15, hidden) more ...
comment

This is an important schema for AP. Previous cooked problems P0002476 & P0004295 (together with versions of each lacking sBe3) had multiple chess cooks that do not involve e.p. as key. This chessically sound composition allows us to focus on the fundamental theoretical issue of AP.

The double e.p. idea cannot work as an AP retro strategy problem, because one can never prove that one double pawn hop rather than the other occurred. PRA+AP seems potentially feasible, but I see no clear statement of how the normal PRA method should be extended to this case. All prescriptive suggestions are welcome.

In WinChloe, I found a correction from 2002 by Valery Liskovitz, just posted in PDB as P1401449. However it used three extra pawns, rather than 2, so the current version still has merit, but I will add Valery as co-corrector.

See also P1399178.
A.Buchanan: Marking this pattern as Type Petrovic - ccee (2022-02-17, hidden) more ...
comment

Corrects P0000899. See P1399806, a companion problem.
A.Buchanan: I guess can further remove sSb1 & shift wBg3 to g4 (2023-03-31, hidden)
A.Buchanan: I guess can further remove sSb1 & shift wBf3 to g4 (2023-03-31, hidden) more ...
comment

Corrects P0003736. See P1399805, a companion problem.
A.Buchanan: Oops this one doesn’t quite work I will fix it in the morning (2022-03-15)
Mario Richter: Hi Andrew, it would be nice to know the cook. Can you please give it - either here or if you prefer, via email? (2022-03-16)
A.Buchanan: Hi Mario: thanks for your interest. The transient version 4k2r/N1p2p1R/p1pB1ppp/npP5/5P2/5PPP/2P2P2/R3K3 a) allowed wPaxb=R then entering the cage disrupting b0-0 rights. That's 5 pawn captures by White. In the mean time, Black only has two visible pcs, with 4 captures available, so wRh1 (and another White officer) can be captured by officers without disrupting w0-0-0 rights. wSa7 is unhelpful in trying to fix it, but despair turned to joy when I realized that wPb6 & bPc7 both cover c7, and when the latter is missing, the former is not a promotion risk. Hence the new version which the Discord community provisionally thinks are sound. (2022-03-16)
comment

Corrects P0005275
Henrik Juel: C+ Popeye 4.61 and analysis (2022-04-23)
Henrik Juel: solutions
a) 1.cxd4 Lxc4 2.Dd5 Lxc2#
not 1.cxd3ep? Txd1 2.dxc2 Lxc2, because last move could be 0-0
b) 1.cxd3ep cxd3+ 2.Kf3 Lxd1# (2022-04-23)
Mario Richter: In a), why not 1. cxd3ep?
Andrew: "because last move was 0-0."
Henrik: "because last move could be 0-0"
Mario: "because d2-d4 definitively was not White's last move" (2022-04-24)
Henrik Juel: You are right, Mario
But my argument suffices to rule out the ep capture, which is allowed only if you can show that last move was d2-d4 (2022-04-24)
A.Buchanan: Thanks Mario. I think that our opinions are consistent. I wanted to emphasise the castling because of Valladao. In b), h8=L is accurate, even if we never know the result of e8=X in either twin. (2022-04-25)
A.Buchanan: Thanks Mario. I think that our opinions are consistent. I wanted to emphasise the castling because of Valladao. Prior to the castling in a) Black must have played hxg1=D,S. I wish it was possible to determine which. However h8=L in b) is accurate, even if we never know the result of e8=X in either twin. (2022-04-24, hidden)
A.Buchanan: On reflection, in a) sBhxg3 or earlier would also work, without promotion on g1, if White just castled. But I don’t think this affects the count or the significant logic (2022-04-24, hidden)
A.Buchanan: Thanks Mario. I think that our opinions are consistent. I wanted to emphasise the castling because of Valladao. Prior to the castling in a) Black must have played hxg1=D,S. In b), h8=L is accurate, even if we never know the result of e8=X in either twin. (2022-04-25, hidden) more ...
comment

Author: Kardos' cooked P0003269 was from 1956 just after AP compositions began to appear, and I can't decide whether he intended to include AP but it seems to fit this corrected version.
Henrik Juel: C+ Popeye 4.61 and analysis (2022-05-07, hidden)
Henrik Juel: solution
1.gxf3ep+ e3 2.0-0-0 Tb4 3.Tg4 Tb8#
The castling serves two purposes: justifying the ep capture and enabling the mate (2022-05-07, hidden) more ...
comment

Corrects P1401496
Henrik Juel: analysis
Black captured [Pa2] with an officer and cxb, dxe, exf, and f3xg2
White captured g4xLh5 and h2xDg3, so if he may castle, Black may capture ep now (2022-05-30)
A.Buchanan: A detail I hadn't noticed: bPh never captured, so wPh5 must have arrived on h-file *above* the light square h3. If it was gxh4, then there was no light square to capture light bB, so it was gxh5. But the capture might have been h3xBg4xQh5. The problem is still sound: the last move can't have been h4-h5. (2022-05-30)
Henrik Juel: HC+ Popeye 4.61 and analysis (2022-05-30, hidden)
Henrik Juel: set play and solution
1...Ke1-d2 2.e4-e3 + f2*e3 #
1.b4*c3 ep. b2-b3 2.Kd4*d3 0-0-0 #
The castling is needed both for the mate and for the legitimization (2022-05-30, hidden)
A.Buchanan: A detail I hadn't noticed: bPh never captured, so wPh5 must have arrived on h-file *above* the light square h3. If it was gxh4, then there was no light square to capture light bB, so it was gxh5. But the capture might have been h3xBg4xQh5. The problem is still sound: the last move can't have been h4-h5.d (2022-05-30, hidden) more ...
comment

Economizes P1080354 by 2 units, adding a solution.
Henrik Juel: C+ Popeye 4.61 (2022-05-30, hidden)
Henrik Juel: solutions
1.Ra3-b3 Ba6*c4 2.Rb3*b4 Ra1*a2#
1.Ka4*b4 Ra1-b1+ 2.Kb4-a4 Ba6-b5#
1.c4*b3ep. 0-0-0 2.e5-e4 Rd1-d4# (not 1... Td1, of course) (2022-05-30, hidden) more ...
comment

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

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)
Henrik Juel: C+ Popeye 4.61 and analysis (2023-07-29, hidden)
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."
So here's one, that clearly states the theme. Curiously, Asteroid 5696 is named after the author! (2023-07-31, hidden)
A.Buchanan: Thanks for this, Henrik. Ibsen (who wrote in Danish I think) wrote this in Act 2 of "Enemy of the People" 1882, a play whose themes are sadly very relevant today. Have you got a precise source for Kierkegaard please? I like him a lot, but I didn't find this. Honestly, it sounds more like Ibsen because it's political & about social commentary. Kierkegaard is more existential, in my limited experience.

I found it translated first as "Oh yes, right—right. What is the use of having right on your side if you have not got might?" which makes a bit more sense from a chess perspective, but I preferred to have it in the original Danish (2023-07-31, hidden) more ...
comment

After P0004346
Henrik Juel: Not completion of 0-0-0, because Black has no last move
Completion of f5xg6ep is feasible, and then 1.Lxb3 Ta8#
Stipulation should be h#1.25, I think (2022-07-10)
A.Buchanan: Hi Henrik: it’s intended to be h#1.75 :-) Can black complete any Valladao move? And if not then what do we do? :-) (2022-07-10)
Henrik Juel: Thanks Andrew
So Black must complete a move, but his four men are stuck, and white pawns captured the remaining 12 black men; so there is nothing for Black to complete
Looks like a weird joke... (2022-07-10)
A.Buchanan: Sorry you don't like the Valladao joke Henrik. "1.75 minus 0.5 = 1.25" is a novel way to combine "unfinished move" with Article 15. I also like the fact that all the cases of pawn capture counting are exact, with no "retro dressing" unit purely to contribute to a balance.s (2022-07-10)
comment

Objective: co-record shortest achievement of task, followed by artistic continuation.
Also version rnbqBbnr/pp2p1Bp/3P1pp1/4N2k/4P3/1Q5P/PP3P1P/RN1R2K1 PG in 14.0 mentioned in article. This offers different artistic continuation (with Pronkin instead of Ceriani-Frolkin).
Henrik Juel: C+ Stelvio 1.0 (2023-03-19, hidden)
Henrik Juel: 1.Pe2-e4 Pc7-c5 2.Bf1-b5 Pc5-c4 3.Sg1-f3 Pc4-c3 4.O-O Pc3xd2
5.Pc2-c4 Pf7-f6 6.Qd1-b3 Pd2-d1=B 7.Bc1-h6 Ke8-f7 8.Pc4-c5+ Pd7-d5
9.Pc5xd6ep+ Kf7-g6 10.Sf3-e5+ Kg6-h5 11.Bb5-e8+ Pg7-g6 12.Bh6-g7 Bc8-h3
13.Pg2xh3 Bd1-g4 14.Pf2-f3 Kh5-h4 15.Pf3xg4 (2023-03-19, hidden) more ...
comment

Objective: co-record shortest achievement of task, followed by artistic continuation.
Henrik Juel: C+ Stelvio 1.0 (2023-03-19, hidden)
Henrik Juel: 1.Pe2-e4 Pb7-b5 2.Pe4-e5 Pb5-b4 3.Bf1-b5 Pb4-b3 4.Sg1-e2 Pb3xc2
5.O-O Pc2xb1=S 6.Qd1-c2 Pf7-f5 7.Qc2xc7 Ke8-f7 8.Qc7-c4+ Pd7-d5
9.Pe5xd6ep+ Kf7-f6 10.Qc4xg8 Bc8-e6 11.Pd6-d7 Qd8-c8 12.Pd7-d8=S Sb8-d7 (2023-03-19, hidden) more ...
comment

Objective: co-record shortest achievement of task, followed by artistic continuation.
Henrik Juel: C+ Stelvio 1.0 (2023-03-19, hidden)
Henrik Juel: 1.Pd2-d3 Pe7-e5 2.Sb1-d2 Pe5-e4 3.Sd2-f3 Pe4xf3 4.Pb2-b4 Pf3xe2
5.Bc1-a3 Pe2xf1=S 6.Qd1-g4 Ke8-e7 7.O-O-O Ke7-d6 8.Pb4-b5+ Pc7-c5
9.Pb5xc6ep+ Kd6-c7 10.Qg4xd7+ Kc7-b6 11.Pg2-g4 Sf1-g3 12.Ph2xg3 (2023-03-19, hidden) more ...
comment

Objective: record achievement of task. Promotion can be to either minor officer, but prefer to keep the "stem cell" to avoid arbitrary choice of non-standard material.
Henrik Juel: with wSd8 C+ Stelvio 1.0 (2023-03-19, hidden)
Henrik Juel: with wSd8
1.Pe2-e4 Sg8-f6 2.Ke1-e2 Sf6-d5 3.Pe4xd5 Pg7-g5 4.Pd5-d6 Bf8-h6 5.Pd6xc7 O-O 6.Pc7xb8=S Qd8-a5 7.Ke2-e3 Pg5-g4+ 8.Pf2-f4 Pg4xf3ep+ (2023-03-19, hidden) more ...
comment

Objective: record achievement of task. Promotion can be to any officer, but prefer to keep the "stem cell" to avoid arbitrary choice of non-standard material.
Henrik Juel: with wSb8 C+ Stelvio 1.0 (2023-03-19, hidden)
Henrik Juel: with wSb8
1.Pe2-e4 Sg8-f6 2.Ke1-e2 Sf6-d5 3.Pe4xd5 Pg7-g5 4.Pd5-d6 Bf8-h6
5.Pd6xc7 O-O 6.Pc7xb8=S Qd8-a5 7.Ke2-e3 Pg5-g4+ 8.Pf2-f4 Pg4xf3ep+ (2023-03-19, hidden) more ...
comment

Objective: record achievement of task. Promotion can be to any officer, but prefer to keep the "stem cell" to avoid arbitrary choice of non-standard material.
Henrik Juel: with wSc8 C+ Stelvio 1.0 (2023-03-19, hidden)
Henrik Juel: with wSc8
1.Pd2-d4 Sg8-f6 2.Ke1-d2 Sf6-e4+ 3.Kd2-e3 Se4-c5 4.Pd4xc5 Pg7-g5 5.Pc5-c6 Bf8-h6 6.Pc6xb7 O-O 7.Pb7xc8=S Pg5-g4+ 8.Pf2-f4 Pg4xf3ep+ (2023-03-19, hidden) more ...
comment

Objective: record achievement of task. Promotion can be to either minor officer, but prefer to keep the "stem cell" to avoid arbitrary choice of non-standard material.
Henrik Juel: with wSd8 C+ Stelvio 1.0 (2023-03-19, hidden)
Henrik Juel: with wSd8
1.Pd2-d4 Sb8-c6 2.Ke1-d2 Sc6-e5 3.Pd4xe5 Sg8-f6 4.Pe5xf6 Pg7-g5 5.Pf6xe7 Bf8-h6 6.Pe7xd8=S O-O 7.Kd2-e3 Pg5-g4+ 8.Pf2-f4 Pg4xf3ep+ (2023-03-19, hidden) more ...
comment

Objective: record achievement of task. Promotion can be to either minor officer, but prefer to keep the "stem cell" to avoid arbitrary choice of non-standard material.
Henrik Juel: with wSd8 C+ Stelvio 1.0 (2023-03-19, hidden)
Henrik Juel: with wSd8
1.Pf2-f4 Pe7-e6 2.Pf4-f5 Bf8-c5 3.Pf5-f6 Sg8-e7 4.Pf6xe7 Bc5-b6 5.Pe7xd8=S Pc7-c5 6.Ke1-f2 O-O 7.Kf2-e3 Pc5-c4+ 8.Pd2-d4 Pc4xd3ep+ (2023-03-19, hidden) more ...
comment

Objective: record achievement of task. Promotion can be to either minor officer, but prefer to keep the "stem cell" to avoid arbitrary choice of non-standard material.
Henrik Juel: With wSf8, C+ Stelvio 1.0 (2023-03-19, hidden)
Henrik Juel: With wSf8
1.Pf2-f4 Pd7-d5 2.Pf4-f5 Bc8-d7 3.Pf5-f6 Bd7-c6 4.Pf6xe7 Sb8-d7 5.Pe7xf8=S Qd8-g5 6.Ke1-f2 O-O-O 7.Kf2-f3 Pd5-d4+ 8.Pe2-e4 Pd4xe3ep+ (2023-03-19, hidden) more ...
comment

Corrects P0000615.
A.Buchanan: corrects P0000615 (2023-05-20, hidden) more ...
comment

A.Buchanan: Test (2023-12-05, hidden) more ...
comment

See P0003666
A.Buchanan: Apart from having removed wBe2/sBe3, I don't think this is an improvement. In the original, sBh6 gave and took. The version is more economical but there's two pawns difference between the twins. (2023-12-06)
Henrik Juel: a) 1.Ta5 Kd1 2.Ta2 Kc2# (not 1.Ta7? Lxc3 2.Ta2 0-0, because White may not castle)
b) 1.Ta7 Lxc3 2.Ta2 0-0# (last move could be h7-h8=L) (2023-12-05, hidden) more ...
comment

Supersedes P0003659.
"The noted Cornish problemist, Dr. Maurice Jago was most prolific during the war when he was a lieutenant in the Royal Army Medical Corps. He became increasingly interested in the more flamboyant and unusual kinds of positions and problems." (www.keverellchess.com, now closed)
Yuri Bilokin: perhaps such a version does not harm the author's intention –wNa7, -wNa8, -wBb8, -wPd3, -bRc8, -bPc7, -bPe7 8/8/1Pqp4/2bb4/1p6/n1p2rP1/1PP1P2k/R3KQ1n (8+12) h#2 b) bPb4-b5 (2024-03-10)
Yuri Bilokin: slip of the pen -wNa7, -wNa8, -wBb8, -wPd3, -bRc8, -bPc7, -bPe7, -bPa4, -bPg6 8/8/1Pqp4/2bb4/1p6/n1p2rP1/1PP1P2k/R3KQ1n (8+10) h#2 b) bPb4-b5 (2024-03-10)
A.Buchanan: Hi Yuri thanks yes it’s more than intention here. In b need to ensure that White can’t have just played axb6, fxg3 or g2-g3, in order to eliminate the castling try: the extra pieces are all to ensure this (2024-03-11)
Yuri Bilokin: Hi Andrew. I apologize, I didn't go there, although the white horse on the a8 square still bothers me. (2024-03-11)
A.Buchanan: WP cap: axb BP caps: fxexdxc. bPh must have promoted, so either bPh3xg2-g1= or wPh was waylaid. In any case, all captures are accounted for. wPf promoted. Easy to see that White cannot have just played g2-g3 or f2xg3 or d2-d3 or axb6. So must have bPb4 rather than b5, allowing for White to have just played Sb5-a7 or b5-b6. If any of the "retro-dressing" White pieces e.g. Sa8 are removed, then White might have just played g2-g3 or d2-d3. At least the retro-dressing units have no possible last move (& wSa8 wBb8 have not just promoted), so there is some craft to their deployment. (2024-03-11)
A.Buchanan: bPg6 blocks a cook so must stay there, but can shift bPe7 to d7 (allowing the removal of wSa8 or wBb8) or to b3/b7 (allowing the removal of all three White officers: wSa7a8 & wBb8.) Maybe that's worth doing? (2024-03-11)
Yuri Bilokin: Thank you for the detailed analysis. Perhaps my poor view will allow you to get an optimal starting position. (2024-03-11)
Henrik Juel: a) 1.Sh1-f2 Qf1*f2 + 2.Kh2-h1 0-0-0 #
b 1.Sa3-b1 Ra1*a4 2.Sh1*g3 Ra4-h4 # (2024-03-11)
Henrik Juel: C+ Popeye 4.61 and analysis (2024-03-11)
A.Buchanan: I had wanted to keep to the spirit of P0003659, in which may be the author wanted to have a bunch of White retro-dressing that could not retract. But going back to the first version P0003659, the White position is much cleaner. So following Yuri being bothered by a horse, I am going to change the diagram here from NBr5/N1p1p3/1Pqp2p1/2bb4/pp6/n1pP1rP1/1PP1P2k/R3KQ1n by removing 4 white units. Hope this is agreeable to all (2024-03-13)
A.Buchanan: bPg6 blocks a cook so must stay there, but can shift bPe7 to d7 (allowing the removal of one of wSa8, wSa7 or wBb8) or to b3/b7 (allowing the removal of all three White officers.) Maybe that's worth doing (2024-03-11, hidden) more ...
comment