Narayan Shankar Ram (India) &
Juraj Lörinc (Slovakia)


Original Fairy problems
JF – 2017(I): January – June

Definitions: (click to show/hide)

No.1224 Narayan Shankar Ram &
Juraj Lörinc

India / Slovakia

original – 05.06.2017

Solutions: (click to show/hide)

White Rf7 Pb5 Pf5 Bg5 Ph5 NHb4 Pf4 NHe3 NHf3 Kg3 NHh3 Pa2 Qf2 Rg1 Black Ke8 Pa6

r#5              b) -wPa2               (14+2)
Nightrider-Hopper b4,e3,f3,h3

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Kjell Widlert
Kjell Widlert
June 7, 2017 22:36

The basic mechanism is indeed simple and well-known: it has been known under the German name “Kontrawechsel” since 1938, and it has been shown with as little as 4 pieces. (There is no established English term, but in Hans Peter Rehm’s new book on logic in fairy chess it is translated as “Interchanged counter”.)

The idea is that Black has two moves (in this problem: a5 and axb5); one requires White to play the moves A and B… (where “…” indicates that there may some further moves connected with B), the other requires White to play A and C… So White must start with A which is common to both variations; if he starts with B or C, Black can choose to continue with the “wrong” variation. With the right first move by White, Black must decide on which variation to play before White has committed himself.

The special feature here is the doubling by twinning, where the white move A is different (a4 or NHh1) and the roles of the two variations are exchanged because in a) but not in b) the bP will make a further capture after the first move, interchanging the promotion squares.

The schemes given by the composers say the same thing as I have done here, but in other ways.

A technical feature I find interesting is the reason why a4 is needed in both variations in a), and NHh1 in both variations in b). In most cases, the reason is that Black cannot be given a choice of capturing or not capturing after the first move (in one variation in b), NHb4 must move in order to allow the bP to move at all).

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