Helpmates in two with grasshoppers and nightriders 8

Older files introducing this combination of stipulation and fairy pieces:

Bedrich Formánek
5979 Probleemblad 3-4 1966

1...Sc5 2.Ga2 Rb3#
1...Se1 2.Ga2 Sc2#

1.Ga2 Sc1 2.Ga4 Rb3#
1.Ka2 Rb3 2.Ga1 Sc1#

Well known quintuple by Rudolf Forsberg is transformed into problem with 2 set plays and 2 solutions by putting a grasshopper on a6. Funny!

h#2 (3+2)
1+0 grasshopper +

Petko A. Petkov
Prize feenschach 1990 (fairy pieces)

a) 1.Nxe4! Bxb6 2.Nf6 Bd4#

b) 1.Rxe4! Nxb6 2.Rf4 Nd5#

c) 1.Bxe4! Rxb6 2.Bb7 Rb3#

Cycles: moving white piece BNR - moving black piece NRB - interfered black piece BNR.

h#2 (7+12)
3+1 nightrider
b) e3 -» b3
c) e3 -» a2

Petko A. Petkov
HM feenschach 1990 (fairy pieces)

1.Ra6 Ga5 2.Rf4! (Nf4?) Ng5#

1.Ra5 Ga4 2.Bf4! (Rf4?) Ge6#

1.Ra4 Ga3 2.Nf4! (Bf4?) f3#

Clearcut strategy: almost all bK flights are safely covered by white, but he needs to free one piece of pf2, Ng7, Gb6 trio to allow mate. For this we have analogic movement on a-file. The square f4 is going to be blocked together with interference of line piece guarding mating square. And now black must be very careful - he mustn't open own line piece. Thus we have cyclical dual avoidance. Very well done!

(Mistake in diagram, reported by Michal Dragoun, corrected.)

h#2 (7+16)
3+1 grasshopper, 1+2 nightrider

bernd ellinghoven
Hans Peter Rehm

4th Comm feenschach 1989 (fairy pieces)

a) 1.Gf7 Gxe4 2.Gd8 Gc4#

b) 1.Nf7 Nc4 Nd8 Nxe4#

Judge Kjell Widlert: "Problems like this one really violate the spirit of twinning: an insignificant change in the position results in a significant change in play. Here the change may be described in very few words (all grasshoppers become nightriders), but there are actually five changes on the board. The trick is to make parts hang together as a whole anyway. That has succeeded admirably, despite the very different nature of G's and N's: there is a strategical correspondence (unpin of g6 and then c8), some half-strategical correspondences (change of function between c8/g6 and b7/g5). The role of d2 in this twinning is also nice. The composers had some bad luck - only 3 pawns are real cook-stoppers, but they had to be placed in the middle of the board (e3-e4-f4) giving the position a rather massive look."

h#2 (8+13)
3+2 grasshopper
b) replace all grasshoppers by nightriders

Peter Bakker
3rd Prize N. Macleod Whisky Tourney 1994

a) 1.Sd7 Qb2 2.Na4 Rh6#

b) 1.Nxf4 Ra2 2.Ng7 Qxf4#

I guess twinning S -> N was thematical and as is already generally well known, in Whisky tourneys not only artistical quality but also solving difficulty counts. That probably explains uneconomical Nd1 in a). On the other hand, analogy is not the worst one...

h#2 (6+12)
1+1 nightrider
b) black Nc5

Comments to Juraj Lörinc.
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