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Original Problems →Previous ; →Next ; →List 2015(I) Please send your original fairy problems to: julia@juliasfairies.com |
No.716 
No.716 – Cornel Pacurar – Wenigsteiner Series Retractor with three sets of SBRQ uncaptures! (JV)
No.716.1 – Cornel Pacurar – An improved version to No.716! (JV)
Definition:
Series Retractor: In the retro phase, White or Black retracts a series of legal moves.
-n [w|b] & forward stipulation: White and/or Black retracts (takes back) a given number of moves (-n moves), in order to reach a position in which a forward stipulation is met.
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No.716 Cornel Pacurar |
Solutions: (click to show/hide) |
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white kf7
black kd4 pf4
-2b & h=1 (1+2) |
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No.716.1 Cornel Pacurar |
Solutions: (click to show/hide) |
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white kd8
black kh4 pg7
-2b & h=1 (1+2) |
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Great Cornel!
Is it a computer-helped problem or did you searched at hand every possibility of twin?
Do you know if there is some possibility for Popeye to be able, a day or another, to handle such a problem with backward play?
Btw, a small “-” is missing in the last solution, dear Julia.
It would certainly be good to see Popeye (and WinChloe, for that matter) handle retractors in the future! I heard that rawbats might be able to solve some type of retractors, I am yet to hear back from Mario Richter.
This is a very rich matrix and the six twins above do not exhaust all the possibilities offered. For instance:
Kb5 / Kd4 f4
-2b & h=1
2 Solutions
– 1.e5xBf4 -2.d6xRe5 & 1.d6-d5 Re5-e3=
– 1.Ke3xSd4 -2.Ke4xQe3 & 1.Ke4-d5 Qe3xf4=
While here we have another set of SBRQ uncaptures, it doesn’t fit into the problem’s intended single-solution twinning scheme. And another interesting one, more like a separate composition altogether:
Ka7 / Kd4 f4
-2b & h=1
3 Solutions
– 1.Kc3xRd4 -2.Kb4xSc3 & 1.Kb4-a5 Rd4xf4=
– 1.Kc4xSd4 -2.Kb4xQc4 & 1.Kb4-a5 Sd4-f3=
– 1.Kc5xRd4 -2.e5xQf4 & 1.Kc5-c6 Qf4xe5=
There are six different ways of picking two pieces out of the four QRBS, so one could imagine having all the pairs in one solution each. This is not so here, as we have RS and QB twice.
The full set of six pairs actually exists:
Branko Koludrovic & Hans Gruber, 2. Prize Springaren 2002
Ka7 Pc7 d2 d6 g2 g5 g7 – Ke4 Se8 Pd3 g3 g4,
-2b & h=1, 3 solutions, b: Ke4>f4
a)
-1.Kf5xRe4 -2.Kg6xQf5 & 1.Kxg7 Rxe8=
-1.Kd5xSe4 -2.Kc6xRd5 & 1.Sf6 Sxf6=
-1.Sf6xSe8 -2.Sd5xQf6 & 1.Sxc7 Sxc7=
b)
-1.Sf6xSe8 -2.Se4xBf6 & 1.Sxd6 Sxd6=
-1.Sf6xBe8 -2.Sd5xQf6 & 1.Ke4 Bc6=
-1.Sf6xRe8 -2.Sd7xBf6 & 1.Se5 Rxe5=
Fine twinning … but four times as many pieces as in 716!
Kjell, thanks for showing Branko and Hans’ fine problem!
I don’t think that this is possible to achieve with just the kings and a black pawn, in any acceptable twining form. I already knew that RQ can be added to this very matrix (e.g. Ka5 / Kd4 f4 -1.Kc5xRd4 -2.e5xQf4 & 1.e5-e4 Qf4xe4=) , the problem is with SB (or BS). While possible as single-solution, that requires a corner stalemate (e.g. Kc3 / Kc1 e7 -1.Kb1xSc1 -2.Ka1xBb1 & 1.e7-e5 Bb1-e4=) which effectively kills any real chance of getting all six.
If you (or anyone else) believe otherwise and manage them all, I would be more than happy to offer co-authorship! Until then, I am still quite happy with 716 (the two RS are, in fact, RS and SR, and the two BQ are completely different).
You have every reason to be happy with your creation! And yes, the solutions are quite sufficiently different.
Of course my showing of the BK/HG problem was not intended to detract from your own fine work, just to show how the same basic idea can be done quite differently. An interesting comparison, I think.
If I happen to find a way to do SB or BS, I’ll be in touch 🙂
Kjell, I have tell you that your messages were not only appreciated, but also very motivating! In the end, we should never really be very happy with our own compositions, should we?! So, I went back to the drawing board with new motivation and inspiration, and after burning the midnight oil I found a way to avoid the SB corner-stalemate and to connect all six pairs. Not in a perfect form, mind you, but in a quite acceptable one, I think. Please see the version below:
Kd8 / Kh4 g7
-2b & h=1
-1.Kg3xSh4 -2.Kh2xQg3 & 1.Kh2-h1 Sh4-g6=
(SQ)
b) Kh4->c2
-1.Kc3xBc2 -2.Kb2xQc3 & 1.Kb2-a2 Bc2-g6=
(BQ)
c) Kh4->g8
-2b & h=1
-1.Kf7xRg8 -2.Kg6xQf7 & 1.Kg6-h6 Rg8xg7=
(RQ)
d) Kh4->a6
-2b & h=1
-1.Kb7xSa6 -2.Ka7xRb7 & 1.Ka7-a8 Rb7xg7=
(SR)
e) Kh4->h8
-2b & h=1
-1.Kh7xBh8 -2.Kg8xRh7 & 1.Kg8-f8 Rh7xg7=
(BR)
+f) Kd8->c8
-2b & h=1
-1.Kg8xSh8 -2.Kf8xBg8 & 1.Kf8-e8 Sh8-g6=
(SB)
The uncapture of a promoted bishop in e) doesn’t bother me and I see it as a plus (Schnoebelen!), not minus. I am curious, though, what others think about this. I do regret the fact that twin f) cannot be obtained by repositioning the black king, as it is the case with all the others.
Oh, I didn’t expect this! Congratulations!
I agree – of course – that the last twin is less than perfect, but it is fully acceptable. And the uncaptured Bh8 may not really be a plus, but it is definitely not a minus! As long as it is legal, it is all right in this kind of problem.