### Solutions for Mod 2 Problems
If each unit is on its start square (the "game array") it can
only be White to move. The number of moves to reach that configuration
must be even for each side, but for a position with Black to move, the
total number of moves must be odd. Whimsically, these positions might be
termed **vampires**, since they have no mirror image under the
transformation which reverses the board, the colour and who has the move.
A vampire's ancestors (i.e. its prior positions) are also vampiric.
These problems are light, but my interest was piqued by trying to patch
Dr
Julio Sunyer's 1946 composition from *Fairy Chess Review* (edited
by Thomas Rayner Dawson). This led me to produce {A}. [1]
**{A}** 1. Nc3 0-0? (parity) 1...Rf1! 2. Ne4 fxe4 3. fxg3 Rxg8#. The
solution in Sunyer's original h#2.5 has a dual by 2 Ne3 fxe3.
**{B}** Since e.p. is legal, White does have the move, and Black's
last was the double Pawn jump. In the prior position, Black has lost castling
rights, by parity. This is not quite trivial. If Black can castle, the
Black queen is hemmed in, and the Black f pawn must have captured two
knights to reach d5 since it couldn't promote. The White pawn then
captured two knights and bPd5 to reach its current square. But the
parity is wrong, so Black can't castle. The solution is: 1...dxe6ep 2. Kf8 e7+ 3. Kg8 e8=Q#.
Tries include:
1...dxe6ep 2. O-O? exd7 3. Re8 dxe8=Q#
1...dxe6ep 2. O-O? e7 3. Kh8 exf8=Q/R#
1...R-a4/xf6 2. O-O? Rf4/6 3. Kh8 Rxf8#
Parity problems commonly involve questions of castling, but this is perhaps the nearest
one can get to e.p. The difficulty is that the position is vampiric only
if the e.p. is legal, and this cannot be proved. It is fortunate that this
natural diagram admits a unique solution.
**{C}** Set play: 1...Nc3 2 0-0 Nd5 3 Nh8 Nxe7#. However, this does
not work in the actual play, since by parity Black cannot castle. Instead:
1 Kf8 Nc3 2 Kg8 Nd5 3 Nf8 Nxe7#. This was dedicated "in memoriam
Gábor Cseh", since although it matched the pattern for the Memorial
Tourney (set play with an identical chess-problem motif demonstrated in
each phase) it was h#3* not, as required, h#2*.
**{D}** By parity Black cannot castle, so: 1.O-O e3 2.Kh8
Bd3 3 Rg8 Nf7# is a try while 1 Nf5 h4 2 g5 hxg5 3 hxg5
Rxh8# is the solution. This was posted on www.france-echecs.com
to illustrate the concept of parity.
**{E}** It appears that there are two solutions.
1. b6 axb6 2. Ba6 bxa7 3. O-O-O a8=Q#
1. b6 a6 2. Bb7 axb7 3. Kd8 bxa8=Q/R#
But how to use parity to eliminate the castling cook? The White a-Pawn
could have moved two or three times to reach its current position. It
hinges on the Rook odds. Most commonly if White gives these odds, he will
begin the game with Pa3, since the Pawn otherwise is undefended. See here,
for example. Yes but shouldn't the stronger player be able to cope with
the weakness of an undefended pawn? Well I don't know. And how do you *know*
this was the way the Rook odds were being played? Hey, it's only a joke...
Amazingly and disappointingly, this obscure idea was anticipated by Wladimir
Korolkow. However, I find I do prefer mine since it *uses* the
trick to get some clean forward play, not otherwise accessible in this
kind of problem.
**{F}** 1. 0-0 Rf1 (not 1. ...0-0? by parity) 2. Rxf2 Rxf2 3. Kh8
Rf8#. The original problem, here,
admits the cook 1. Kd8 d3 2. Re8 Be3 3. c6 Bxb6# (and two other solutions
that permute the Black moves). This can be fixed by shifting the two Black
units on the a-file, but I have chosen also to return some other units to
their start squares, since I find it more aesthetic. Another
cooked problem by Korolkow admits many mates ending 3...Rh8#, and I
cannot see any way to repair it.
**An open challenge**
Is it possible to have a *duplex* parity h#3? I.e. there is one
well-defined try for each side that depends upon castling, which is
illegal by parity. [Nicolas Dupont]
**An open question**
Is the total number of legal chess positions odd or even? Obviously,
this reduces immediately to a question about vampires. The FIDE
Law concerning draw by repetition defines the concept of position.
Thus castling rights, e.p. status and who has the move can distinguish
otherwise identical positions. However, if two identical pieces are
permuted, that does not by itself change the position.
[1] I asked George
Jelliss, librarian of the British Chess Problem Society, about the
history of the Sunyer composition. He replied:
I looked up the Sunyer problem 6724 in my (now BCVS) collection of [*Fairy
Chess Review*] (mainly photocopies). It occurs on p.30 of the April
1946 issue, but it has the black QR at b8 not a8. [corrected in the
solution (June 1946): this issue is separate from the dual.]
The original Sunyer problem appears as part of an article, Number 99
in the series of *"Five Minute Papers"*; that has the
title *"Knightly Opposition"*. Eight problems are given
and the following text [by Dawson] reads:
We have not previously had in FCR any problems like 6274-8, because
this is the elementary retro-opposition theme on which very many
beginners start, and I have 36 published examples in my files all very
much like these and like one another. However, this is really doing
FCR contributors an injustice, for I see other examples appearing
quite regularly elsewhere. So here is the accumulated group --
J.Sunyer, 1925; H.H.Cross 1941; J.E.H.Creed 1942; and R.Van den Bergh
1945. They show nice variation in the forward play at any rate.
Nos. 6729-31 reveal much more sophisticated examples of S-opposition
retraction.
From this it seems that Dawson had been holding onto the Sunyer
problem for 21 years before publication!
Sunyer composed some superb problems (e.g. a classic bare
kings retraction 1923, and the minimal
setting known for the 50-move rule 1958).
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