28 October 2011

One more time

This is the last part of the little series dealing with twinning in chess problems. I picked some more helpmates featuring quite interesting ways of creating twins.

  1Dieter Müller  
Problem-Echo 03/1998
3rd Prize 16th TT (twomovers)
a) Diagram
b) wBBc4/d4
c) wRRc4/d4

  2Arthur Christopher Reeves  
The Problemist 01/2008
a) Diagram
b) wK ↔ wR
c) wK ↔ wN
d) bK ↔ bR

  3Henk Weenink  
The Problemist 07/2010
a) Diagram
b) all men that move in a)
change colour

1a) 1. Bb7 Nc6 2. Bf4 N×b6#
b) 1. e5 B×b6 2. e4 Bd5#
c) 1. Bf4 Rc7 2. Rg5 Rd8#
This way of twinning demonstrated here was another trigger for me to do some composing. After a short time, I already managed to create a nice h#2 (sorry, I can't reveal details). Based on this, I still try to find improved versions. I guess, you'll see the result(s) in a problem magazine, some day.
2a) 1. Kb3 Kb1 2. Rb4 Re3#
b) 1. Kc3 0-0-0 2. Tb4 Rd3#
c) 1. Rc5 Nb3 2. Rb5 Re4#
d) 1. Kc3 Kb1 2. Kb3 Re3#
I'm not so sure whether the fourth part is necessary — same mate position as in a), but achieved with a tempo move. It fits well with the other parts, though: each time, a king and one of his fellow men change places.
3a) 1. Bc7 Nf3 2. Ra6 Re1 3. Rd6 Re5#
b) 1. Nf3 Bc7 2. Re1 Ra6 3. Re5 Rd6#
Same moves in both solutions with colours reversed. Funny!

21 October 2011

After the key

It was rather by chance that I became curious and searched for examples where the twinning is achieved by taking the position that arises after the key move — another type of continuous problems. I found a lot of compositions! Many of them are so-called perpetua mobilia. This expression perpetuum mobile or Pauly theme is named after the German-Romanian problemist Wolfgang Pauly (1876-1934). There, you have the key move of solution a) that leads to the position b) the key move of which in turn leads to a). As a consequence, the set play of one part is the solution of the other part after the key move (see the following example).

Jan Kubecka
feenschach 01-03/1976
34th TT (B)
a) Diagram
b) after the key

The solution is
a) *1. - f8=Q 2. Kh5 Qh6# and 1. Kg7 f8=N 2. Kh8 Bf6#
b) *1. - f8=N 2. Kh8 Bf6# and 1. Kg6 f8=Q 2. Kh5 Qh6#.

I am not so crazy about those perpetua mobilia, though there are very many of them. Still, there is an abundance of other chess problems left featuring those "after the key" twins. Most of them are helpmates, but I managed to spot some directmate and selfmate compositions, as well. I hope you enjoy the choice I've made.

  1György Bakcsi  
Probleemblad, 1971
2nd Prize
a) Diagram
b)-e) after the key of a)-d)

  2Frank Christiaans  
Die Schwalbe 06/2001
a) Diagram
b) after the key

  3Horst Bäcker  
Problemkiste 12/2008
a) Diagram
b) after the key

1a) 1. a4! Kxc6 2. Be4#
b) 1. a5! Kxc6 2. Qxf3#
c) 1. a6! Kxc6 2. Qc5#
d) 1. a7! Kxc6 2. a8=B/Q#
e) 1. a8=N! Kxc6 2. Qxf3#
That's also a way to demonstrate an Excelsior! The helpmate No. 9 below enhances this effect.
2a) 1. Nd3! (Z) Kxb3 / Kxd3 2. Bd1 / 0-0-0#
b) *1. - Kxb3 / Kxd3 2. Bd1 / 0-0-0#
1. Qe8! (Z) Kxb3 / Kxd3 2. Qa4 / Qe4# (0-0-0??)
The key of a) gives two flights. Position b) provides a retroanalytically motivated dual avoidance. The last move was Kb1-/xc2 which proves that the wR had already moved and thus castling long is forbidden.
3a) 1. Rxh2? Nh3 2. Rxh3 Kd5 3. Rh5+ Ke6 4. ?
1. Bh8! Kd5 2. Qf5+ Kc4 3. Qe4+ Kxc5 / Kb3 4. Qd4 / Qa4#
1. - Ne2 2. Rd1 (thr. Qd7#) Nd4 3. Rxd4 h1=Q 4. Qd7#
b) 1. Ba1? Ne2 2. Rd1 Nd4! 3. ?
1. Rxh2! Nh3 2. Rxh3 Kd5 3. Rh5+ Ke6 / Kc4 4. Re5 / Qc2#
The try of a) is the solution of b), taking back the key move of a) is the try of b).

  4Baruch Lender  
Die Schwalbe 02/1983
a) Diagram
b) after the key

  5Gunter Jordan  
Schach 03/2009
5th Honourable Mention
a) Diagram
b) after the key

4a) 1. Nf2! Qxb3 2. Qe6+ Qxe6#
1. - Qxb1+ 2. Ne4+ Qxe4#
1. - axb3 2. Nxg4 Qxb1#
b) 1. Nxg4! Qxb3 2. Qe6+ Qxe6#
1. - Qxb1+ 2. Rd3+ Qxd3#
1. - axb3 2. Bh2 Qxb1#
Observe the different continuations after Qxb1+ and axb3.
5a) 1. Rh7! Ka7 2. b8=R+ Bd7 3. R1b5! axb5 4. Rb7+ Ka6 5. Qc6+ Bxc6#
b) 1. Rh8! Ka7 2. b8=B+ Kb6! 3. Qxc4 Kb7 4. Qc7+ Ka8 5. Qc6+ Bxc6#
Different underpromotions.

  6Fadil Abdurahmanovic  
Hans Peter Rehm
feenschach 01-04/1989
a) Diagram
b)-d) after the key of a)-c)

  7Fadil Abdurahmanovic  
feenschach, 1997
a) Diagram
b) after the key

  8Christer Jonsson  
Michel Caillaud
StrateGems 10-12/2000
1st Prize
a) Diagram
b)-d) after the key of a)-c)

  9Jaroslav Stun  
The Problemist 11/1988
a) Diagram
b) after the key

  10Anders Lundström  
feenschach 07-09/1986
3rd Prize, 43rd TT
a) Diagram
b) after the key

  11KCharles Mason Fox  
The Chess Amateur 08/1929
a) Diagram
b)after the key

6a) 1. Ke4 c8=N 2. Kf5 Nd6#
b) 1. Kxd5 c8=B 2. Kxc6 Bf3#
c) 1. Kxe6 c8=R 2. Kd7 Bg4#
d) 1. Kf5 c8=Q+ 2. Kf4 Qg4#
Allumwandlung and round trip of the black king. This composition is an ímproved version of a 1st Prize by Abdurahmanovic which was published in feenschach 08/1985.
7a) 1. Rh8 0-0 2. d2 Re2#
b) 1. 0-0 Be6+ 2. Kh8 Rxh7#
Two remarkable things about a): Both black rook and white king have to step aside. Even the passive black knight is of thematic relevance. Solution b) is rather unspectacular.
8a) 1. d5 Nd2 2. d4 Ne4#
b) 1. d4 Bb3 2. d3 Nd5#
c) 1. d3 Nd4 2. d2 Ne2#
d) 1. d2 Bxa4 2. d1=N Rb3#
Impressing — Black moves are only done by the pawn!
9a) 1. Nf4 Nd5 2. Ke4 Nb4#
b) 1. Nd5 Ng4 2. Ke4 Ne5#
Self-pins and batteries.
10a) *1. - Bh4 2. Nf2 Rg3#
1. Nxe3 Bf1 2. Ng4 Rxf5#
b) *1. - Bf1 2. Ng4 Rxf5#
1. Ng2 Be1 2. Nf4 Rg3#
Nice model mates.
11a) 1. Nf3 Rh5+ 2. Kg4 Nf6#
b) 1. Ne5+ Ke2 2. Kxe4 Ng3#
Though the black d-pawns are a little bit suspicious, it's still a small surprise that the checkmating knight of a) is captured in b). Model mate in a) and even ideal mate in b).

14 October 2011

Round and round

Today, I'll show you some continuous helpmates which feature another fascinating way of twinning: except for the initial position, of course, each part of the chess problem results from the previous checkmate position plus an additional alteration.

In April 2010, the British Chess Problem Society (BCPS) organised the 6th European Chess Solving Championship (ECSC) being held in Sunningdale. In conjunction with this event, composing tourneys were being run. The helpmate tourney required twomovers in three or more parts which, taken together as a whole, show one or more round trips. In a round trip a piece leaves a square by a certain route and later in the problem returns to it by a different route. Interestingly, four of the five awarded problems are continuous helpmates. See yourself:

Alexandr J. Semenenko
Waleri J. Semenenko
ECSC Internet Tourney 2010
1st Prize
a) Diagram
b) mate position of a), bK → c7
c) mate position of b), bK → h8
d) mate position of c), bK → g1
e) mate position of d), bK → a3
f) mate position of e), bK → c8
g) mate position of f), bK → h6
h) mate position of g), bK → f3
    a) 1. Nb3 Ne3-c4 2. Na1 Rg2xb2#
b) 1. Ra8 Nc4-d6 2. Kb8 Rb2xb7#
c) 1. Kh7 Nd6-f5 2. Th8 Rb7xg7#
d) 1. g3 Nf5-e3 2. g2 Rg7xg2#
e) 1. a5 Rg2-b2 2. a4 Ne3-c4#
f) 1. Bg5 Rb2-b7 2. Ld8 Nc4-d6#
g) 1. Ng3 Rb7-g7 2. Sh5 Nd6-f5#
h) 1. Kf2 Rg7-g2+ 2. Kf1 Nf5-e3#

Double round trip by the white pieces.

Michael McDowell
ECSC Internet Tourney 2010
2nd Prize
a) Diagram
b) mate position of a), bK → e4
c) mate position of b), bK → e6
d) mate position of c), bK → c6
    a) 1. Nb6-a4 Ne7-c8 2. Na4-c3 Nc8-b6#
b) 1. Nc3-e2 Nb6-a4 2. Ne2-f4 Na4-c3#
c) 1. Nf4-g6 Nc3-e2 2. Ng6-e7 Ne2-f4#
d) 1. Ne7-c8 Nf4-g6 2. Nc8-b6 Ng6-e7#

Only knight moves! The white knight chases the black one, each time mating on the latter's diagram square.

Michael Barth
ECSC Internet Tourney 2010
3rd Prize
a) Diagram
b) mate position of a), bK → c6
c) mate position of b), bK → d3
    a) 1. Qe3-e5 e8=N 2. Re4 Ng7#
b) 1. Qe5-c5 Ne6 2. Rd5 Nd8#
c) 1. Qc5-e3 Nc6 2. Bd4 Nb4#

Round trip by a "magnetic" black queen with a colleague following her to provide blocks. Judge Chris Feather wondered whether the composer missed the chance to include an additional white knight round trip.

Alexandr J. Semenenko
ECSC Internet Tourney 2010
2nd Honourable Mention
a) Diagram
b) mate position of a), bK → a7
c) mate position of b), bK → g8
d) mate position of c), bK → h2
    a) 1. Rh8-a8 Nf4-d3 2. Ra8-a1 Rg2-b2#
b) 1. Rh1-h8 Nd3-c5 2. Rh8-a8 Rb2-b7#
c) 1. Ra1-h1 Nc5-e6 2. Rh1-h8 Rb7-g7#
d) 1. Ra8-a1 Ne6-f4 2. Ra1-h1 Tg7-g2#

Round trips by both white pieces and by the black rooks. It's a neat construction without pawns, though it requires additional black material to keep the white king unter control. But the four solutions are "quite the same".

07 October 2011

Turn! Turn! Turn!

Most probably you know this tune by Pete Seeger which became a No. 1 hit in the mid-60s when it was covered by The Byrds. This title (not necessarily the lyrics, though) presents an appropriate transition from the Oktoberfest time and the joke problems back to the more serious chess compositions. I guess I don't have to further elaborate on that, right?

I like chess problems where the twinning is done by rotating the position on the board. I learnt that there was a composing tournament with that theme in Messigny (France) as part of the 31ème RIFACE (Rencontre Internationale en France des Amateurs de la Composition Échiquéenne), that's an annual French meeting for chess composition enthusiasts. Here are some of the awarded problems for you (rotation is always done clockwise):

  1Mario Parrinello  
Messigny 2010
1st Prize
a) Diagram
b) Rotate 90°

  2Marcos Roland  
Ricardo de Mattos Vieira
Messigny 2010
2nd Prize
a) Diagram
b) Rotate 270°

  3Carlos Lago  
Messigny 2010
3rd Prize
a) Diagram
b) Rotate 90°
c) Rotate 270°

  4Menachem Witztum  
Messigny 2010
2nd Special Prize
a) Diagram
b) Rotate 270°

  5Dieter Müller  
Messigny 2010
1st Honourable Mention
a) Diagram
b) Rotate 90°

  6Franz Pachl  
Dieter Müller
Messigny 2010
3rd Honourable Mention
a) Diagram
b) Rotate 90°

1a) 1. Nc8+ bxc8=Q 2. Ke4 Qc2#
b) 1. Nf8 gxf8=Q 2. Ke4 Qxe7#
Harmonious solutions with knight sacs and self-pins.
2a) 1. Rd4 Nb7 2. Rbd2 f4#
b) 1. Qe4 f7 2. Rg5 Ne3#
Rather random moves by White compared to the homogeneous play of Black, but nice model mates.
3a) 1. Bd6 Kf3 2. Qe5 e4 3. Kd4 Ke2 4. Bc5 Nb5#
b) 1. Be4 Na7 2. Bd5 c6 3. Kd6 cxd5 4. Qe7 Nb5#
c) 1. Bb3 f6 2. Qg7 fxg7 3. Bd1 g8=Q 4. Be2 Qb3#
Impressive miniature. The third solution is not consistent with the other two, though. It's surely difficult for a composer to just not mention it, but it would provide a better picture.
4a) 1. f1=B Rd1 2. Bf2 Bxf2#
b) 1. g5 Rh6 2. gxf4 Re6#
Very nice. Just this black rook ...
5a) 1. Bg6 Nxc5+ 2. Kf5 e4#
b) 1. Kd5 Nf7 2. Bd4 e4#
Beautiful problem with model mates. Black pieces follow each other, the white pawns on e2 and g5 reciprocally take a flight and checkmate.
6a) 1. Kb6 Rd6+ 2. Ka7 Rh7#
b) 1. b5+ Rxb5 2. 0-0-0 Rc1#

Diagram 6 inspired me to experiment a little bit searching for a more economical version. Finally, I arrived at the following result.

Gerson Berlinger
a) Diagram
b) Rotate 90°
        The solutions are

a) 1. Bd3 Rxb2 2. 0-0-0 Rc5#
b) 1. b5 Re3 2. Rh5 Rb4#

In both phases, the white king is protected from a check by the black rook. The white rooks swap roles.