2017 Nottingham #2 Award (in full)

Written by Michael Lipton

All three entries composed “onsite”, and one online, are commended, in alphabetical order without ranking.

Rauf Aliovsadzade (USA)




1.Bb2? (2.Ra1) 1...Bxb2!
1.Bc3? (2.Ra1) 1...Bxc3!
1.Bd4? (2.Ra1) 1...Bxd4!
1.Be5? (2.Ra1) 1...Bxe5!
1.Bf6? (2.Ra1) 1...Bxf6!
1.Bg7? (2.Ra1) 1...Bxg7!
1.Bxh8? stalemate

1.d4! (2.Sd2)

D. Stojnic (2004, after P. Ivanic: Jeremy Morse, Tasks and Records 3rd ed., #267) showed eight separate BB refutations to WB tries, five on the diagonal, with, as here, obvious capture refutations. Stojnic needed twelve units; here, seven suffice for a sixfold duel on one diagonal, the theoretical maximum. It seems new. The miniature setting is attractive, but concurrency is a dull way to show separate refutations. Very differently, Hammarström’s 1994 miniature (see p. 4) also shows six WB tries and six BB refutations, though only a fivefold B v B duel – but it adds fivefold Vladimirov mates – as does Banaczec (1996, below, after Rice and Emmerson 1994), an astonishing R v R counterpart of Rauf’s entry, even down to the stalemate try.

Barry Barnes




1.Bc5!          (2.Rxg1)

1...f2/Rb1       2.Qe4/Qh2
(1...Qf1+/e1/xd1 2.Rxf1/xe1/Qxd1)

Eight units seem the minimum needed to combine mates due to “lets-leave-line interference” and pole-of-line unguard. Though out of play, the WB’s key, line-cutting to stop a ruinous dual 1...f2 2.Qc6, is perfectly in keeping. Clear, good construction, if rather slight.

Stephen Emmerson




1.Sb7+?A  Bxb7! a
1.Sc6+?B? Bxc6! b


1...Bb7 a 2.Sxb7 A
1...Bc6 b 2.Sxc6 B

These capture-Vladimirovs lack the paradox associated with non-capture refutations, but this is a light, clear example. However, of several miniatures among the 241 Vladimirovs in the Albrecht-Leiss-Degener database, see Hammarström’s and Banaczec’s fivefold examples.

John Rice
after Yuri Marker, 1977 (and version of YM by Michael Lipton 2017)





1...Sc4  2.Sc5
1...Se4  2.Sb2

1.Qxa5? (2.Qd2)

1...Sc4  2.Qf5
1...Se4  2.Qxa6
1...Bc3  2.Qxc3

1.Qg5?     (2.Qd2)

1...Bc3,Be3 2.Q(x)e3
(1...Sc4    2.Qf5)

1.Qe7?  (block)

1...Sc4,Se4 2.Q(x)e4
1...Bc3,Be3 2.Q(x)e3

1.Qh4!     (2.Rd2)

1...Bc3     2.Sc5
1...Be3     2.Sb2
1...Sc4,Se4 2.Q(x)e4			

If you didn’t know the backstory, you’d ask: why not the top prize? There are three WQ thematic tries, each refuted differently. Set, 1.Qxa5?, 1.Qg5? and 1.Qh4 seem to show three different replies to 1...Bc3, Be3, Sc4 and Se4. (Analytically we can forget 1.Qe7? for a moment.) That would mean 3x4 changes (Zagoruyko), but (a) 1.Qxa5? and 1.Qg5? each fail to a thematic defence (a plus, but preventing it from featuring in a change or transfer); (b) after 1.Qg5 (as after the key) two thematic defences yield the same mate. Yet the nine genuinely distinct variations can, at a stretch, be read as a double Zagoruyko plus ideal Rukhlis – in Meredith! However, as John stated on his entry, the problem improves an essentially similar 10-unit schema by Marker, and my 9-unit version (see p. 4), which have no extra try 1.Qe7? but the same change content. So why, in an extreme-economy tourney, commend a 12-unit follow-up? Because 1.Qe7? and John’s reconstruction greatly enrich the problem, with a good refutation, block strategy – and (by lessening the symmetry) interest and attractiveness. These are often absent from extreme-economy efforts; John shows how to achieve them. The backstory, however, prevents a higher award. Despite the excellent construction, I suspect a Letztform still awaits; speed tourney conditions seldom produce these.

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