27 April 2012


I am sure you've heard about the Logical or New German school — and hopefully it was in connection with chess problems. Searching the Internet in order to get more information about this subject might rather lead you to articles that refer to music. And adding the search word "Problem" will not really be helpful as this gives you links to sites dealing with certain aspects of social science. In fact it's quite hard to get the desired information at all, the more if you have no concrete idea what you're searching for. Anybody interested in Web Content Mining?

The chess problems I have chosen for this post are logical. That term refers to the way in which the solution is structured. In the initial position, White has a so-called main plan, a series of moves with which he wants to checkmate. At first, the execution of this plan fails to a refutation that Black has up his sleeve. Therefore, White first executes a foreplan, whereby Black's defence to the main plan is negated in some way. Roughly speaking, that's what logical problems are about.

Let's start with a lightweight example.

Jörgen Möller
Skakbladet, 1920
    The main plan is to play the queen to the b-file and give checkmate on b8. But the attempt 1. Qb1? Bg3! is premature. Another idea that fails is 1. Qg2+? d5!.

1. Qg7! (2. Qxd7 3. Qc6/Nb6#) Be7 2. Qb2 Bd6 3. Qg2#

The bishop was lured away to another diagonal and thus was forced to replace the good defence Bg3 with the bad defence Bd6. This caused the interference with the pawn d7. This theme is called Interference Roman (translation found here).

Stefan Schneider
Austria-Switzerland Match, 1977
1st Place
1. Re7? with the idea 2. Qxe2+ Bxe2 3. Rxe2+ shows that queen and rook are the wrong way around. Hence, the main plan must be Qe8, followed by Re7 and Rxe2+. But how to do it? 1. Qe8? is much too slow, for the black queen can leave c1 not being bound to defend against Qxd2# anymore. We must find a way to play Qe8 with gain of time — very hard to imagine. Here we go:

1. Ka8!!

Wow! True, an experienced solver might consider this move even without exactly knowing how to proceed thereafter — it's the feel. Otherwise, deep thinking is required.

White now threatens 2. Rh7! 3. Rxh2 4. Ng2#. The point of going to a8 is to avoid 2. Rxh7 being check. In doing so the king has to avoid b8 because of the potential pin by the black bishop coming to h2. Why 1. Kc8? is wrong will be explained later.

1. - Rh8+

Black is so tied down that this is the only way to challenge White's idea. White continues with another incredible move:

2. Qe8!!

2. Rxe8+ 3. Ka7!
Not 3. Kb7? Rb8+ 4. Kxb8 Bh2! This also makes clear that in case of 1. Kc8? the white king would not have been able to escape the h2-b8 diagonal in time.
Black can now no longer guard g2, so that after
3. - Ra8+ 4. Kxa8
there is no way to prevent 5. Ng2# or (when Nf2 moves away) 5. Nd3#.

Alternatively, the rook can only return.

2. - Rh2

By now, we have achieved our goal, so that the main plan can be executed.

3. Re7 ~ 4. Rxe2+ Bxe2 5. Qe2#

Y. Vladimirov
Macleod Memorial Tourney, 1994
1st Prize (Version)
    This one is probably easier to solve despite its length.

White's main plan is Rf8 followed by Rc8#, but Black is stalemated. The foreplan is to build a bishop-rook battery on g1,f2 with the black king on c5 and play Rf8+. But this requires the protection of the pawn d6, so that playing the pawn e2 to e5 is another foreplan. Important things to observe: White always has to a) protect d6 and b) give a check when the king is on c5. See and enjoy how all this is accomplished.

1. Bc1 Kc5 2. Be3+ Kc6 3. Bf4 Kc5 4. Rf5+ Kc6 5. Be5 Kc5 6. Bh2+ Kc6 7. Rf6 Kc5 8. Bg1+ Kc6 9. e3 Kc5 10. e4+ Kc6 11. Bh2 Kc5 12. Rf5+ Kc6 13. e5 Kc5 14. Bg1+ Kc6 l5. Rf2! Kc5 16. Rf8+ Kc6 17. Rc8#

We see a variety of self-interferences employed by White to relieve the stalemate. A very nice puzzle.

A.Lobusov & A.Spirin
E.Zepler Memorial Tourney, 1985
1st Prize
    Black to move allowed White to mate on d4 or d5. But there is no waiting move. The goal of the foreplan is to lose a move.

1. Ne8! (threat 2. Nf6 3. Ng4#) Ng8 2. N6c7! Nfe7 3. Ng7!

White now threatens 4. Nxd5+ and 5. Nf5# or vice versa.
3. - Nf6 4. Nce6! and 5. Nxd4# or 5. N(x)f5#.

3. - Nxh6 4. Nge6 Nhf5 5. h6!

There is the waiting move! The diagram position, without the pawn h5, has been repeated. It is Black to move and mate follows on the next move.

Did you notice? During the solution, the two white knights have
swapped places. Moreover, the black knights also performed such a Platzwechsel!

20 April 2012

Not enough

This is a continuation post of Who's missing. Again, I invite you to have a look at some diagrams that ask you to add one unit. All of them are legal positions of the so-called type A, which means that no king is in check and that it is not stated who has the move. Naturally, such chess problems are always a constructional challenge with the main goal to find the most economic position.

  1Andrew Buchanan  
Internet, 2011
  Add one unit(8+6)  

  2Andrew Buchanan  
Internet, 2011
  Add one unit(7+6)  

In No. 1, the white pawns have captured nine black pieces and the black bishop from c8 never left this square. Therefore, the piece to add has to be white. Looking at the pawns on the g- and h-file, we see that the black pawns must have promoted, so that the white pawns could capture them on their way to a6 and a7. The black pawns on the a-file have captured six white pieces, so one white pawn from g2 or h2 must have promoted to be captured, the other had to be captured by a black pawn on these files. The only white piece that could not have been captured is the white rook from a1. It can only be added on a1: +wRa1.

A closer look at the second problem reveals that there have been seven obvious captures by white pawns and at least eight captures by the black pawns. Considering the pawn formations, we see that there must have been two more captures by one side. Counting the missing pieces tells us that this must have been white captures. Two white and one black pawn(s) on the g- and h-file have promoted, the fourth (black) was captured by a white pawn. As all captures are now resolved, we see that only the black f-pawn is left. It has not captured, thus never left its file. The only possibility to add it is: +bPf5.

  3Andrew Buchanan  
Internet, 2011
  Add one unit(6+6)  

  4Thierry Le Gleuher  
Internet, 2011
  Add one unit(6+11)  

As the logic of No. 3 is quite similar to that of the first two, I leave it up to you to prove that the solution is +wBc1.

The last one is a little bit different. The black pawns captured all the missing white pieces. This means that the white h-pawn had to promote and, as a consequence, it had to capture once to bypass the black pawn h7. Moreover, there are the three captures which you immediately can spot by looking at the diagram. Hence, there is one black unit left that can be added. To resolve the position, it is necessary to put this unit on b1. Of course, it's not a black bishop, as all eight black pawns and another bishop on white square are already there. Thus, we have to add a black knight on b1. Then, we can play back for example: 1. - d7xQ,R,B,Nc6 2. ~ Qc1-a3 3. ~ Ka3-b3 etc.

13 April 2012

Exact Echoes

Last year, I already wrote a little bit about echoes and showed examples of a specific type, namely chameleon echoes. This time, I concentrate on something called exact echo.

An exact echo occurs when all the pieces in the two (or more) formations concerned are in the same positions relative to each other. Only the pattern formed by the pieces is echoed, other aspects are ignored.

Before I present some examples of such exactly echoed arrangement of pieces, I want to turn your attention to the following stunning chess problem which has echoed moves.

A. Benedek
The Problemist 5/1977
a) Diagram
b) Rotate 90°

The solution is
a) 1. Kc5 Bb3 2. Bd4 Bxe7#
b) 1. Kc5 Bb3 2. Bd4 Bxe7#
Yes, though the position is rotated, the moves are absolutely identical. Fantastic!

Now, let's have a look at some exact echo problems in which the echoed configurations are related by a shift of one or two ranks.

  1aMichael McDowell  
US Problem Bulletin 1/1986 (Version)

  1bM. Havel  
Zlatá Praha, 1912

  2aAndrew Kalotay  
Ideal Mate Review #8 3-4/1984

  2bM. Havel  
Zlatá Praha, 1912

1a*1. - Kf4 2. d3 Qf5+ 3. Kh6 Ne5 4. e3#
1. g3 Nd4 2. f3 Ne6 3. d4 Rg5 4. e4#
1b1. Qd6!
1. - e5 2. Qd7+ Kf6 3. Nf2 e4 4. Ng4#
1. - Kf6 2. Kh6 Kf7 3. Qd8 e5 4. Ng5#
2a1. Bf4 Nb4 2. Rd4 Kf1 3. Ke3 Ke1 4. Re4 Nc2#
1. Bf6 Nb2 2. Ke5 e4 3. Rd6 Ke3 4. Re6 Nc4#
2b1. Qf7! with the thematic lines
1. - Kh5 2. Kf4 Kh6 3. h4 g5+ 4. h×g5#
1. - g5 2. Kf2 g4 3. Qf5 g3+ 4. h×g3#

If you bother to examine the different types of exact echoes, you'll find out that there are only five: identity, translation, rotation, reflection and glide reflection. I won't further elaborate on that, you'll surely find appropriate information yourself.

06 April 2012

Retro mix

Today, I have two retro chess problems for you. Enjoy!

T. R. Dawson
Cas, 1922
Add one man so that White to play can mate in two

We know that White is to move in the diagram. Obviously, something must be wrong, since White already has a mate in two by 1. Ne8 and 2. Nc7. So, it seems the addition of a further piece is unnecessary.

Let's start the retro-analysis to find out what's wrong. White has made at least five pawn captures on the kingside. As only six black men are missing, there's at most one capture left for White on the queenside. This means that the white pawn on b6 has never made a capture. Otherwise, that required another capture due to the pawns on a7 and c4, respectively.

As a consequence, the black pawn on b4 has made a capture which implies a second pawn capture by Black on the queenside. Together with the four on the kingside, all black captures are resolved. Therefore, the missing piece must be black. This means that actually only five black men are missing and this fits exactly with what we already found out.

Let's check where White captured. The h6-pawn started from d2 and took pieces on e3, f4, g5 and h6. The h3-pawn must have arrived via the capture g4xh3, so the history of moves in the bottom right comer must have been either g4xh3, then g2-g4 and h2xg3, or g4xh3, then h2xg3, g3-g4 and g2-g3. In either case, White captured on g3. A closer look reveals that all White captures took place on black squares. So, the black bishop from c8 hasn't been taken by White. This is the piece that is to be added.

Where can we put the bishop without disturbing White's mate in two by 1. Ne8? There are just three candidate squares, namely a2, e8 and h5. Not e8, because Black has no previous move. And h5? If so, Black's last move must have been b5-b4 or a5xb4. As White hasn't made a pawn capture on the queenside, the following sequence must have occurred: WPb2 to b5 or b6, Black's a-pawn captured to go behind the b6 pawn, WPa2 to a7 and finally Black played b7xa6. So the last move could not have been a5xb4, since this happened before the white pawn advanced to a7. Therefore, it must have been b5-b4, but in this case the two Black queenside pawn captures took place at b5 and a6. These squares, like e6, f5, g4 and h3, are white squares. Black could not have captured White's queen's bishop.

We have to add a black bishop on a2.

R. Kofman
Shakhmaty Bulletin, 1958
White retracts his last move and then mates in 3

If Black was unable to castle, White could mate in two from the diagram by 1. d2xc3. So, we must retract a move that proves that Black cannot castle. We might try taking back 0-0-0. Then White's king has never moved. This means that the white rook d3 is a promoted pawn. If the pawn promoted at g8 White must have made seven pawn captures, which is too many. The promotion must have taken place at a8, b8, c8 or d8, or f8. In each case the black king must have moved, thus 0-0 is illegal.

We still have the task of mating in three, so we must prevent gxf2+. This can be accomplished by playing 1. 0-0-0! (threat 2. dxc3), the very move we just took back. Funny!