*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#1orh#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 eithers#1orhs#1. We replace#1bys#1as the goal.In both of the above, it is

blackwho has the option.In

1611, it iswhitewho has the option, so it would seem like a half-duplex.But, in addition, the option is that of either

#1ors#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#1orhs#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) orhs#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.