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| (1) Posted by Hauke Reddmann [Friday, Aug 25, 2023 22:37] |
A to B to A
Inspired by Andreas Thoma.
Give two positions A, B such that A->B takes much longer than B->A.
This should be pointless in orthodox chess (either both sides
can take back their moves, and you have 1:3 halfmoves, or there
is an irreversible move, meaning one way is simply impossible).
Just as demo 4:12:
 (= 3+3 )
Circe Rex inclusive
forth:
1...g5 2.fxg6 O-O 3.O-O-O
back:
3...Kh7 4.gxh7(Ke8) Rh8 5.Kd2 Rxh7(Ph2) 6.h4 Rh6 7.h5 Re6 8.Re1 Rxe1(Ra1) 9.Kxe1(Rh8)
The sharp-eyed viewers already see I'd cheated, the White king forfeited castling.
Also, move order isn't completely unique.
Any takers?
You may choose "unique proof game" or "shortest proof game"
and use whatever obscure fairy condition you like.
For starters, can you repair my proof game such that both castling rights get
restored?
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| (2) Posted by Kevin Begley [Saturday, Aug 26, 2023 00:36] |
Circe should work well, but subforms of the Power Transfer condition (including Point Reflection) would seem optimal for this job.
Here's a simple scheme (please pardon the symmetry) which is C+ (Jacobi 2297 seconds):
A:
 (= 5+5 )
]
B:
 (= 5+5 )
Condition: Power Transfer Antipode
A=>B 1.0 : 1.e4 e5
B=>A 11.0 : 1.Kd8 Kd1 2.Kc8 Kc1 3.Kb8 Kb1 4.Ka8 Ka1 5.e3 e6 6.Ka7 Ka2 7.e2 e7 8.Kb8 Kb1 9.Kc8 Kc1 10.Kd8 Kd1 11.Ke8 Ke1
2:22
note: this can likely be extended -- possibly to infinity (providing the larger board keeps a single square of antipode corresponsdence, and with a smarter arrangement of pawns to prevent alternative King paths) -- but it's not quite as easy as I had first imagined; therefore, I leave this as an exercise for the reader. :)
Power Transfer Antipode -
Unit A moves like the unit of either color that stands on the square 4 ranks and 4 files away from the departure square of A.
When that square is vacant, unit A moves normally.
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| (3) Posted by Kevin Begley [Saturday, Aug 26, 2023 01:56] |
The following extension is probably sound (not sure any tool can completely verify this -- but so far Jacobi has found only the intent):
A:
 (= 8+8 )
B:
 (= 8+8 )
Condition: Power Transfer Antipode
A=>B 1.0: 1.e4 e5
B=>A 15.0: 1.Kg8 Kg1 2.Kf8 Kf1 3.Ke8 Ke1 4.Kd8 Kd1 5.Kc8 Kc1 6.Kb8 Kb1 7.Ka8 Ka1 8.e4e3 e5e6 9.Kb8 Kb1 10.Kc8 Kc1 11.Kd8 Kd1 12.Ke8 Ke1 13.Kf8 Kf1 14.Kg8 Kg1 15.Kh8 Kh1
2:30
I suspect it might also be possible to extend this simple scheme to an infinite board (if the Antipode square is recalculated accordingly).
If my guess is correct, the ratio approaches infinity, and the remaining challenge would be to maximize this ratio for finite boards (especially 8x8) or for fairy boards.
Note: this scheme might also be extended to an infinite vertical cylinder board (with the Kings positioned near the halfway point of an infinite vertical cylinder board, the ratio still approaches infinity). With slight modifications (shift the King's ranks, and add more cook-stopping pawns), I suspect this scheme can be extended to an infinite anchor ring.
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| (4) Posted by Kevin Begley [Saturday, Aug 26, 2023 06:35] |
ps: another condition well suited to this task is Degradation (aka Degradierung, aka Relegation Chess): non-royal units which move onto a square where a pawn began the game are immediately demoted to a Pawn.
If you add that condition to your Circe idea, you might achieve a Vallado!?
Just be careful to remember that only moves result in units demoting to pawns (officers are not demoted when rebirthed onto a square where a Pawn began the game).
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| (5) Posted by Joost de Heer [Saturday, Aug 26, 2023 09:31] |
The question isn't pointless in orthodox if the side moving in A isn't the side moving in B.
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| (6) Posted by James Malcom [Saturday, Aug 26, 2023 14:28] |
Unless I'm missing something here, this counts as an orthodox A to B to A.
A
 (= 5+6 )
 (= 5+6 )
A->B: 1. ... Bc7+
B->A: 1. Ke8 Kb7 2. Rd8 Bb8 3. Nc7 Kc6 4. Rc8 Kb7 5. Kd8 Kc6 6. Na8+
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| (7) Posted by Hauke Reddmann [Saturday, Aug 26, 2023 20:02] |
@James: I see what you did :-)
(I forgot that in the first move, a self-pin can
be played, which is impossible to take back immediately.)
Hmmmm, this cries for record-breaking! Scheme:
 (= 7+7 )
I'm a mathematician, I can't count :-) - can someone check this
with A->B condition (or tell me how Popeye does it)?
I think at least 12 halfmoves are needed for a full cycle:
1.Kc1 Kb3 2.Qd4 Ka3 3.Kd1 Qb3+ 4.Kc1 Qc3+ 5.Kb1 Qc4 6.Qa1+ Qa2
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| (8) Posted by James Malcom [Saturday, Aug 26, 2023 22:44] |
Nein, e.g.: 1.Kc1 Kb3 2.Kd1 Ka3 3.Qd4 Qa1+ 4.Kc2 Qc3+ 5.Kb1 Qc4 6.Qa1+ Qa2+
Here a C+ one of mine.
 (= 9+11 )
Sol: 1. Ka6 Bh7 2. Nd6 Bg8 3. Nb7 Bh7 4. Qa8 Bg8 5. Rb8 Bh7 6. Nd6+ Qb7+ 7. Ka5 Bg8 8. Nb5 Bh7 9. Nc7 Bg8 10. Rc8 Bh7 11. Qb8 Bg8 12. Nb5+ Qc7+
Code: EnglishN
forsyth 1QRRrbb1/P1qpppp1/2k1N1P1/KNP5/ppP2p2/8/8/8
stipulation a=>b12.0 forsyth 1QRRrbb1/P1qpppp1/2k1N1P1/KNP5/ppP2p2/8/8/8
Perhaps more matrixes can be found in Ortho-reconstruction and Tempoverlustspiel.
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| (9) Posted by Hauke Reddmann [Saturday, Aug 26, 2023 23:12] |
For whatever reason I had to add Option WhiteToPlay. <shrug>
Of course, now it calculates until kingdom come...
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| (10) Posted by James Malcom [Sunday, Aug 27, 2023 07:47] |
Hauke, Popeye starts with BTM, so WTM must be specified; Jacobi starts with WTM, so BTM must be specified.
If you don't care about uniqueness, much longer is naturally possible.
 (= 15+10 )
Sol: 1... Qc7+ 2. Kd4 Bg8 3. Ke4 Bh7 4. Rh8 Bg8 5. Rh5 Bh7 6. Rf5 Bg8 7. Rf3 Bh7 8. Rg3 Bg8 9. Rg1 Bh7 10. Re1 Bg8 11. Re2 Bh7 12. Rc2 Bg8 13. Rc1 Bh7 14. Ra1 Bg8 15. Ra5 Bh7 16. Rb5 Bg8 17. Rb7 Bh7 18. Qa8 Bg8 19. Rcb8 Qa5 20. Bc7 Qa6 21. Rc8 Qa5 22. Qb8 Qa6 23. Rb5 Qb6 24. Ra5 Qa6 25. Ra1 Qb7 26. Rb1 Qa6 27. Rc1 Qb7 28. Rc2 Qa6 29. Rd2 Qb6 30. Re2 Qb5 31. Re1 Qb6 32. Rh1 Qb5 33. Ke5 Qb6 34. Rh3 Qb5 35. Rf3 Qb7 36. Rf5 Qb5 37. Rh5 Qb6 38. Rh8 Bh7 39. Rf8 Qb7 40. Bd8+
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| (11) Posted by James Malcom [Sunday, Aug 27, 2023 13:20] |
Back to uniqueness, here is a new record length with no fairy conditions needed; it is indeed a legal position. C+ by Jacobi.
A:
 (= 15+12 )
B:
 (= 15+12 )
A=>B 1.0: 1. Qc7+ Rf8
B=>A: 24.0: 1. Bh7 Rh8 2. Bg8 Rh5 3. Bh7 Rg5 4. Bg8 Rg3 5. Bh7 Rh3 6. Bg8 Rh1 7. Bh7 Rf1 8. Bg8 Rf2 9. Bh7 Rd2 10. Bg8 Rd1 11. Bh7 Rb1 12. Bg8 Rb2 13. Bh7 Ra2 14. Bg8 Ra5 15. Bh7 Rb5 16. Bg8 Rb7 17. Bh7 Qba8 18. Bg8 Rcb8 19. Qb6 Bc7 20. Bh7 Re8 21. Bg8 Rbb8+ 22. Qb7 Rbc8 23. Bh7 Q8b8 24. Bg8 Bd8+
2:48
In simpler terms, we are looking for an A=>B that loops back around the same position with the same side to move, sorta of an anti/opposite ortho-reconstruction.
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| (12) Posted by Hauke Reddmann [Monday, Aug 28, 2023 21:47] |
Makes indeed more sense (since you can split the cycle anywhere -
even after a single half-move).
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| (13) Posted by Andrew Buchanan [Tuesday, Aug 29, 2023 04:45] |
If ortho-reconstruction (ugly uninformative word) is like a Mobius strip, then what we are looking for is either a simple loop to break at some point, or maybe twice round the Mobius.
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