No.1611
Torsten Linß
(Germany)
Original Fairy problems
JF-2021-I:
01.01.2021 - 30.06.2021
Definition provided by author:
reci-hs#N: The stipulation is a straight forward synthesis of the reciprocal helpmate and the helpselfmate — reciprocal helpselfmate: like hs#, but with its final move white can both force a s#1 and deliver a #1. For soundness the “keymoves” of both the s#1 and the #1 should be unique, while the s# may have variations.
No. 1611 Torsten Linß
Germany
original - 18.05.2021
dedicated to Vlaicu Crișan
White Kc4 Rb5g2 Ba6 Pb6
Black Ka4 Be4
reci-hs#9 5+2
Solution: (click to show/hide)
Thank you very much for the dedication, Torsten! Lovely problem with a novel stipulation.
If it was a hs#9 white could play 8.Bb7 & 9.Rb5 in another order and it would be cooked (in fact Rb5 could be played even earlier). It seems like the reci- stipulation only ensures that the rook remains at h5 until the end, or am I missing something?
I guess so 😉
In 8 moves one can easily reach a position like Ka8 Rg7f5 Sb8 – Kb6 Ba6 which solves as a s#1, but not a #1.
reci-hs# requires the construction of a more complex (pre)final position than hs#.
If the Rg7 was on c7 (white Rc7f5 Sb8 Ka8 — black Kb6 Ba6) it would solve as both (1.Rc6# & 1.Rb7+ Bxb7#).
So not really a more complex prefinal position, just a lack of a single halfmove (Rg2-g7-c7).
In a reciprocal helpmate, black, on his last move has the option of either #1 or h#1. The goal is #1.
So, a strict extrapolation would mean that in a reciprocal helpselfmate, black, on his last move should have the option of either s#1 or hs#1. We replace #1 by s#1 as the goal.
In both of the above, it is black who has the option.
In 1611, it is white who has the option, so it would seem like a half-duplex.
But, in addition, the option is that of either #1 or s#1, which is a different combination, compared to the above.
This makes sense. Except “black, on his last move should have the option of either s#1 or hs#1”
s#1 and hs#1 are the same stipulation (WB# – two halfmoves, white starts), the confusion is due to the different sides order in h# and hs#.
What you probably meant was that the black has option either s#1 halfduplex (BW# – two halfmoves) or hs#1.5 (BWB# – three halfmoves) – that indeed would be a full analogy to reci-h#.
Right. hs#1 in this context implies 3 plies: BWB. In the normal sense, hs#1 is equivalent to s#1 (and also equivalent to [semi-]r#1 and h#1 by W).
If by “equivalent stipulation” you mean “any position would have the exact same set of solutions”, then there are three equivalence classes:
(1) s#1 is equivalent to hs#1
(2) semi-r#1 is equivalent to h#1 by W (but not to s#1)
(3) r#1 is not equivalent to any of those two
Why (1) != (2):
white Kb1 black Rc8 Ka3
semi-r#1 – 1 solution (1.Ka1 Rc1#), s#1 – 0 solutions
Why (2) != (3) and (1) != (3):
white Rh5a5 Pb6 Ka6 Bb7 black Kb8 Ba8 (this is Torsten’s problem before move 9)
semi-r#1 and s#1 – Rb5 solves, but in r#1 white is forced to play Rh8#
I meant something like: WKh1, WSf1; BKh3, BPs g3,g4,h2,h4;
s#1 = r#1 = semi-r#1 = hs#1 = h#1 by W = 1.Sd2, g2#
In view of the cumbersome definition, if we interpret a Reci-hs# as an exact analogue of a Reci-h#, I think Torsten’s decision to christen this form as a Reci-hs# is justified. May it live long, and prosper!
One more example by Torsten:
Can’t this be done in the current Popeye brancht with a sstipulation?
Need somebody familar with that sstuff to confirm!
Theoretically, it can, with
sstip white 16hh[1a[#]d & 2ad[#]]
However, it would take too long.
This is correct. I’ve just checked the last 5 moves with Popeye 4.85 as “white 8hh[1a[#]d&2ad[#]]”. This took about 1h.
How does this work with black to move?
This finds the intended solution. But:
this finds no solution.
would have to be