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MatPlus.Net Forum X-Files: Anticipations Pam-Krabbe-Rochade, aka Extra-Long, Castling-Anticipated By 65 Years!

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Calvet: Captures on the rebirth square are allowed (for ASC this means that the rebirth can take place on the capture square)
Cheylan: Captures on the rebirth square are not allowed (for ASC this means that the rebirth square is always different from the capture square)

@Joost: Knew two castling could be done :-)
BTW, what about normal Circe with telefrag? :-)
(Definition: If X captures Y on the rebirth square,
Y is reborn anyway and captures X. Iterated telefrag
is possible, so with one capture you can turn almost
turn a Fischer 960 into a standard starting position :-)

@Hauke: That's Circe assassin.

QUOTE
@Hauke: That's Circe assassin.

Not really, since Circé Assasin does not have chains of rebirths. This sounds more like Pongracz Circé.

EDIT: This was cooked by Michel Caillaud.
An improved version which eliminates the possibility of this class of cook is given in reply 52 of this thread.

For the record the earlier post was:
rn3bnQ/ppp1r2p/4k3/5pK1/8/1P6/P1PR1PPP/qN4NR PG in 13.5
Here's another old idea refurbished for the modern P-K age. No other fairy conditions, no obvious cooks, and I've run it through Natch at 4 strategic points, together with an a=>b run in Jacobi, but if someone can figure out how to convincingly HC+ this, I would be very appreciative. And maybe it's fun to solve, too.

Ha! ha! not so easy to solve.
Nice, however there seems to be possible interversions in the middle of the game? 9.Kf4 dxc1=R 10.Rd2 Re1 11.b3 or 9.b3 dxc1=R 10.Rd2 Re1+ 11.Kf4

While trying to guess what was the novelty in your problem, another one occured to me :
(= 14+14 )
PG in 12.5
That was a subject in Quartz 49...

Hi Michel - thanks for your persistence with this. For a C?, I think it’s important for me not to have revealed my intended solution right away so that any solver has a fresh mind and can hopefully stumble on any cook. I think you have diverged from me earlier, as in my intended solution e.g. it’s never possible for me to play b3 so late. Can you send me the first 8.0 moves either by note here or by email.

I will do similar honours for you and attempt to solve your 12.5. There are numerous castling themes which can be repurposed for P-K - the issue is that such problems may of necessity be a bit too long for people to enjoy solving manually, and difficult to H+ or HC+.

Francois Labelle agreed yesterday that this will be a cool thing to add to Jacobi, but cannot say when he may be able to implement it, as it needs to await a future re-architecture...

(48) Posted by Rewan Demontay (Real Name: James Malcom) [Friday, Jan 15, 2021 04:50]

Another theme to explore: Valladao. PK takes out 2/3 already, so can the final 1/3 be added in somehow? I won't be surprised if it cannot be achieved in a short manner.

For Michel’s problem, I guess an engine will reveal that PG in 12.5 has no orthodox solution, but PG in 12.0 with suitable retraction has 1 solution. Then since move counting reveals that Wh has only one possible last move, his is HC+. The characteristic raking position of wQ & wB is very suggestive, James... :-)

Another one that hasn't been shown yet, I think: A proofgame that's dualistic in n moves but unique with P-K-castling on move n+0.5.

Ooooh, I see it now Andrew! Thanks. :). That's a nice one Michael, I've got it solved now.

(= 12+12 )
PG in 13.5

Michel cooked the early version (thank you!), so here's an improved version. Still can't check with engine: Popeye (which does the orthodox position at 7.0 in less than a second) has said nothing after 2 hours over the same diagram when I just add the keyword "Rokagogo" - and this is before any P-K would happen! Orthodox play from 7.5 position to the end is C+ under Jacobi in 0.2 sec.

Theme: hidden Wh P-K + fake Bl P-K

Intended solution: 1. e4 d5 2. Bc4 Bf5 3. exf5 dxc4 4. f6 Qd4 5. fxe7 f6 6. b3 Kf7 7. e8=R Qxa1 (C+ to here) 8. P-K c4 9. Kf4 cxd2 10. c3 dxc1=R 11. Qd4 Re1 12. Rb2 Re7 13. Qxa7 Ke6 14. Qxa8

Here are my "lines of thought" (rather than proof)

No ortho sol at 13.5, so there must be one P-K. (Can’t have two in a game.) If we retract experimentally R: 14. ... Qa7xRa8 13. P-K, then again there is no orthodox solution. If Bl genuinely P-Ked then bK blocked earlier Wh promotion. Thus wRb2 is original and we can count 12 Wh moves. So at least 2 Wh pieces were captured on homebase, but that can't include e1, where bR promoted. Some second Bl unit must have marauded the back rank, and I don't see a way.

If it was Wh did the P-K, then probably Bl promoted too, as bRa8 cannot easily have escaped. We can then count 13+12 moves of existing pieces, with just 1 more move for each light-squared bishop. Given the moves are known, I think their order is unique. bQ must be on a1 before P-K.

One doable manual check (which would have picked up the earlier cookage!) is to look at the intended solution and see if there are any different routes for any piece. The fix avoids cooks with wQg4 & bQf6. Now, it seems as if wQ might go via a4 rather than d4. But then wPb must move after wPc, however bQd4-a1 requires wPb to move *before* wPc, so that's not possible. I find this mechanism sweet.

I’d prefer wR on d2 (or c2) rather than b2, but it seems "safer" at the moment on b2.

Any thoughts very welcome...

EDIT: problem extended by 0.5 moves, in order to promote from exact PG to SPG.

And here's one in response to Joost's request:
(= 12+15 )
PG in 11.5 HC+ Natch-3.2 C+
Thematic paradox: if you retract the last two single moves, the position is unsound as a PG in 10.5. (3 bogus variants with no P-K castling rights appear.)

A key observation for dual avoidance given that the engines are of limited use in this joke/fairy context is that Bl has no other retractions apart from the P-K. Surely improvements are possible, but I will post this for now since it does the job! :-)

It's not an SPG, so perhaps add 12. Rgb1?

@Joost: TL;DR Because the matrix isn’t confirmed as fastest for the task, I didn’t have SPG as an objective yet. Added it now at your request. Chessically easy, the lack of an engine complicates the HC+ logic: worth explaining this (sorry for length).

HC+ details
===========

If there are no solutions for 10.5, 10.0 & 9.5, then our unique PG in 11.0 would be SPG. However, in each case, we can only test up to just before the P-K castling itself, and then must extrapolate forward. This is all *independent* of the paradoxical truncation (that the penultimate position has multiple PGs).

If the last move was Bl (11.0, 10.0), then (by deliberate construction) I can't see any alternative to it being the P-K itself. There's just 1 sol for 11.0 (the penultimate position at 10.5 has 4 sols, but only one allows P-K). So up to HC+, 11.0 is unique, and analogously 10.0 has zero solutions.

Last move by Wh (10.5, 9.5) is harder. To retract the Bl P-K, we must retract a Wh move first, which may release other Bl unmoves, not just the P-K. This would force us to retract a further pair of Wh/Bl moves, and so on, until we have unwound to the P-K in all branches! Mercifully, to prove the problem *isn’t* an SPG, it's sufficient to find *one* short solution. R: Sb1-d2 P-K does have a unique solution in 9.5. This does allow P-K, so we can roll forward to a 10.5 solution for the original position, proving this is not SPG.

It's common in such circumstances to extend the proof game by a single move. (Though not every composer likes this: Ronald Turnbull for example.) wRfb1 (coming *after* P-K) is again a good deliberate construction. So let's look at the four lengths:
11.5 Retract R: Rf1-b1 P-K, as nothing else is plausible. Already HC+: 1 sol.
11.0 No candidates Bl moves. P-K would block wKR. 0 sols.
10.5 Analogous to 11.5, 0 sols.
10.0 Analogous to 11.0, 0 sols.
So up to HC+ (not 100% rigorous) this is SPG, IMHO.

I've extended the problem accordingly, truncating now at 1.0 not 0.5. Still interested in a more efficient matrix overall though.

OK the flood of posts is abated now. I plan to post the following into PDB:

post 12: AB+JM+HR+MC (dedicated to C.Staugaard) 10.0 P-K#, nix
post 21: NP 7.5 P-K & 0-0, nix
post 22: AB 7.5 P-K & 0-0-0, nix
post 31: AB 7.5 P-Ked K mated, nix
post 38: MC 8.0 P-K, C-F
post 46: MC 12.5 P-K Valladao, nix
post 52: AB (dedicated to MC) 13.5 P-K real & fake, 2xnix
post 53: AB (dedicated to JDH) 11.5 P-K paradoxical truncation, nix

These all seem plausibly sound as PG & SPG, but I prefer not to mark any of them "C+" or "HC+" for now. The case analysis is tedious and potentially error-prone, and I would rather wait until P-K is implemented efficiently in a search engine. Please post cooks, improvements, claims or other suggestions thanks! Thanks to Michel for further checking of the second fake/hidden position, which seems to have survived for now.

abated but not ended.

dedicated to Hauke Reddmann
(= 11+14 )
PG in 11.5 Circe RI

such problems seem at first sight hard to solve (and they usually are as solvers have not enough elements to grasp) but here :
12 white moves can be counted from the diagram, so missing white units were captured on their initial squares
the position can be tested with Jacobi and no "normal" solution appears...

Wonderful Michel!

Still working offline with Michel to get the most harmonious representation of the hidden-fake idea. I’ve uploaded all the other listed ones into PDB for future generations to enjoy and marvel at.

In the mean time, I just stumbled across this:

(= 14+14 )
PG in 7.0

which is perhaps as brief as it's going to get for double castling? It’s paradoxical that a speed record includes a tempo move!

Here are some missing solutions.

Problems quoted in post 56 :

post 12: AB+JM+HR+MC (dedicated to C.Staugaard) 10.0 P-K#, nix
1.d4 f5 2.Kd2 f4 3.Kd3 f3 4.Ke4 fxe2 5.Kf5 e1=R 6.Qe2 g5 7.Qxe7 Nxe7 8.Kf6 Bg7 9.Kxg7 Nd5 10.Kxh8 0-0-0-0-0-0#

post 21: NP 7.5 P-K & 0-0, nix
1.e4 Nf6 2.Qf3 Nh5 3.Qf6 exf6 4.e5 Be7 5.exf6 0-0 6.fxe7 Ng3 7.e8=R Nxh1 8.0-0-0-0-0-0

post 22: AB 7.5 P-K & 0-0-0, nix
1.e4 d5 2.Bd3 Bf5 3.exf5 Qd6 4.f6 Nd7 5.Bg6 0-0-0 6.fxe7 Qxh2 7.e8=R Qxh1 8.0-0-0-0-0-0

post 31: AB 7.5 P-Ked K mated, nix
1.d4 e5 2.Bf4 exf4 3.Qd3 f3 4.Qxh7 fxe2 5.f3 g6 6.Kf2 e1=R 7.Qxh8 0-0-0-0-0-0 8.Qe5#

post 38: MC 8.0 P-K, C-F
1.e4 d5 2.e5 Qd6 3.exd6 Kd7 4.dxe7 Kc6 5.e8=R Kb6 6.0-0-0-0-0 Bg4 7.Kf4 Bxe2 8.Kg3 Bc4

post 46: MC 12.5 P-K Valladao, nix
1.e4 h5 2.e5 h4 3.Qh5 h3 4.Be2 hxg2 5.h4 Rh6 6.Rh3 Rb6 7.Ra3 d6 8.Ra6 Kd7 9.Bg4+ f5 10.exf6 e.p.+ Kc6 11.fxe7 bxa6 12.e8=R Kb7 13.0-0-0-0-0-0

post 52: AB (dedicated to MC) 13.5 P-K real & fake, 2xnix
1.e4 d5 2.Bc4 Bf5 3.exf5 dxc4 4.f6 Qd4 5.fxe7 f6 6.b3 Kf7 7.e8=R Qxa1 8.0-0-0-0-0-0 c3 9.Kf4 cxd2 10.c3 dxc1=R 11.Qd4 Re1 12.Rb2 Re7 13.Qxa7 Ke6 14.Qxa8

post 53: AB (dedicated to JDH) 11.5 P-K paradoxical truncation, nix
1.e3 h6 2.Bd3 Rh7 3.Bxh7 e6 4.Bf5 exf5 5.Nf3 f4 6.0-0 fxe3 7.Ne1 exd2 8.Qf3 dxe1=R 9.Be3 b6 10.Nd2 b5 11.Bb6 0-0-0-0-0-0 12.Rfb1

Then
post 57 : MC (dedicated to HR) 11.5 3 white castlings
1.e4 d6 2.e5 Kd7 3.exd6 Kc6 4.dxe7 Bh3 5.e8=R Bxg2 6.0-0-0-0-0-0 Qxd2 7.Bxg2(Bc8) Qxd1 8.Be4 Qxc1 9.Nf3 Qxe3(Ke1) 10.0-0 Qxf2 11.Nbd2 Qxg1(Ke1) 12.0-0-0

post 58 : AB 7.0 double castling, speed record, nix
1.g3 e6 2.Bg2 e5 3.Bxb7 e4 4.Nf3 exf3 5.0-0 fxe2 6.Bxa8 e1=R 7.Bg2 0-0-0-0-0-0

Here is another development, related with post (50) by Joost, concerning games with a distance of 0.5 moves.
We looked for a position where SPG in n is unique with Staugaard castling, and PG in n+0.5 is still unique.
There are 2 forms to show this theoritical case.

Michel Caillaud, Andrew Buchanan
(= 14+14 )
PG in a) 7.0 moves b) 7.5 moves
Staugaard castling

Michel Caillaud
(= 14+16 )
SPG? a) Orthodox b) Staugaard castling