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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