The back-home fairy condition
(9th of June, 2013)
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It’s interesting that Nicolas has already a first Prize at J.M. Trillon-MT with an astonishing proof game: 1.d4 e5 2.d5 e4 3.d6 e3 4.dxc7 exf2+ 5.Kd2 d5 6.e4 d4 7.e5 d3 8.Ke3! d2 9.Qh5 Bg4 (needs to say that if 8.Qh5? Bg4 then 9.Ke1! is mandatory) 10.e6 Bf3 11.e7 Kd7 12.exf8=Q! (the Q promotion will be justified later) 13.c8=Q+! Kd6 (now, the Knight e7 is pinned, so Sg8 is forbidden) 14.b4 Rxf8 15.b5 Sg8 16.Be2 d1=S+! 17.Bxd1 f5 18.Sd2 f1=S+! 19.Sxf1 Sc6 20.Qg6+ Kc5 21.a4 (now, the Knight c6 is pinned by Qc8 and the Q promotion is justified) 21…Rxc8. Four Schnoebelen promotions. The problem has two co-authors, M. Caillaud and M. Richter.
Very interesting Fairy condition. Just one question: Is there any restriction on the “back-home” move? I presume it can also be a capture.
The only restriction is that a back-home move must be légal. So a capture is allowed, only self-checks (in the back-home sense) are forbidden.
Of course one have to pay attention to the fact that a self-check in the back-home sense generally differs from a self-check in the orthodox sense.
It is perhaps not very clear in the rules, so I add that only a visible piece in the diagram position has a back-home square. It is its initial square, which is even kept after a rebirth. But a new piece (for example a sentinel) doesn’t have any back-home square.
Back-home looks indeed interesting and tricky. Did I get it right? Could this be correct h=2* or at lest, are the final positions stalemates?
White Ke4 Re3 Pf3 Pg3 Pc2 Rf2 Qg2
Black Pf4 Pd2 Pe2 Ph2 Bd1 Ke1
h=2*, Back-Home, 7+6
1… gxf4 2.h1S Qf1=
1.fxg3 f4 2.h1B Rf1=
Setplay – not a stalemate, black may play 3.Kxf2
Solution – I do not see any issue, but WinChloe cannot find the solution. Also, 1…Kd3 seems as good as 1…f4.
I may be wrong, of course…
I woke up suddenly this morning because in sleep I became aware that set-play is not stalemate. 2…Qf1 Kxf2 is indeed a very stupid mistake showing that the scheme is basically wrong. It’s usually wise to sleep BEFORE posting 🙂
1.fxg3 f4 opens the diagonal so after 2.h1B Rf1=, 3.Bxg2+ is illegal because then the back-home move 3…Rf2 would not be legal and bK would be attacked>3.Bxg2-selfcheck?
Of course I may be wrong again (:
No, 1…Kd3 is not good because of 3.Bh1xQg2 which is no more a self-check.
The solution is not a stalemate position because Black can play 3.Bxc2 which is not a self-check. Indeed the game continue with 3…Rf2 4.Bd1.
Of course I see the idea of the problem, 3.Bxg2 is a self-check as it is a check and therefore White has no more its back-home possibility Rf2.
Your problem would be correct, dear Nikola, if Bd1 was also promoted, so that d1 was no more its initial position…
The following scheme is C+:
1.d1=F h5 2.d2 T×g3 3.ç3 Té3 4.h1=C Tf1=
Of course it must be polished, the ultimate goal in this setup would be a Black AUW!
yes stupid me, Bxc2 is not a check because of a back-home obligation. I should have slept twice before posting. I did think about other possibilites but that “false check” by Bd1 seemed as an elegant tool.
Anyway, the idea of 2 antidualistic promotions to B/S, seemingly can’t be realized with a scheme like that. I am personally not much interested in AUW, but in the motivations for particular moves, including the promotions.
The failure with h=2 predicts that h=3 could be only worse (without a solving program). I’ll try once again, with sincere apologies if there are too stupid mistakes again.
At least I hope that the comments of my failures could provide a better understanding of Back-Home condition to some other visitors. A rough attempt:
White Rd8 Pf5 Sg5 Pf4 Kg4 Pa2 Re2 Qf2
Black Pf7 Pe6 Pa3 Pc3 Pf3 Pc2 Pd2 Pg2 Sb1 Rc1 Kd1
1.g1B Sxe6 2.fxe6 Kg3 3.exf5 Re1=
1.fxe2 Sf3 2.exf5+ Kg5 3.g1S Qe1=
No, these are not stalemates:
After 4. deB(or Q) white have to play Kg4, followed by 5. B(Q) d2
Yes, without WinChloe it is difficult to manage back-home traps! I hope that Popeye will soon be able to handle this condition too.
Your solutions are false. I have to find by myself why, as the computer doesn’t answer!
For the second one, I think that 4.KxQe1 is legal as White must reply its back-home move 4…Kg4, which is no more a self-check as Black must at its turn play its back-home move 5.Kd1!
For the first one, this is more or less the same reason: 4.Kxe1 Kg4 5.Kd1+
But WinChloe gives 5 “true” solutions:
1.g1=T f×é6 2.Tg2+ R×f3 3.f5 R×g2=
1.g1=T R×f3 2.Tg2 f×é6 …
1.g1=F f×é6 2.F×f2 R×f3 3.f5 R×f2=
1.g1=F f×é6 2.f5+ R×f3 3.F×f2 …
1.g1=C C×f3 2.C×f3 f6 3.Cé5+ f×é5=
Dear Georgy and Nicolas, thanks for your patience.
The legality of 3 extra half-moves decides about the stalemate, even the reduced choice of promotions 4.d2xe1Q/B. A true challenge is to use these possibilities richly.
Yes, on one hand it should exists even more complicated situations where the status of some move depends on more than 3 extra half-moves. As said in the article:
“It implies that some induction process may appear in order to verify back-home move legality. Nevertheless it seems it never leads to infinite loops, although it may lead to a rather delicate exploration of what can happen in the future of the position. ”
On the other end, it is possible for a side to checkmate (or to stalemate) even when equipped with some back-home move possibility. A very nice example is provided by the series problem of Christian in the article.
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