25 November 2011


It all started early on the morning of November 26th, 2010. To the day one year ago, my first blog post was published. My original intention was to post every third day. This worked quite well for a while, but I couldn't keep up the pace. Of course, the texts don't always express deep thoughts. Still I aim to maintain a certain quality level and avoid too much blabber. Selecting a subject, thinking about the wording and the layout, etc. — that often takes some time.

I've never advertised this blog a lot as I write just for the fun of it. Hence, the themes I give attention to might not always attract a big audience. But not earning money with the page hits anyway, I don't really care. However, in the course of time, the number of visitors increased and that's surely an encouragement to keep on posting. One interesting fact is that a lot of people are interested in detective chess (top search keywords).

Today, I have a nice composition for you that combines chess and mathematics. It's taken from the book Schach und Zahl by Eero Bonsdorff, Karl Fabel and Olavi Riihimaa.

Dr. Erkki Pesonen
Schach und Zahl, 1966
What is the most probable end of the game?
All choices have the same probability.
a) consider all legal moves
b) first choose the piece, then the move

The key to the solution is this: The fewer moves the higher the probability. A closer look at the position tells that there are two ways to end the game in only three half-moves. Counting moves is all what's left.

a) 1. Nb4 b1=R 2. Nc2#.
Possibilities on half-move 1: Ka4, Kb3, Nb4, Nb8, Nxc5, Nc7 (6)
Possibilities on half-move 2: Kb1, b1=Q, b1=R, b1=B, b1=N, cxb4 (6)
Possibilities on half-move 3: Ka4, Na2, Na6, Nc2, Nc6, Nd3, Nd5 (7)
Thus, the probability is 1/6 * 1/6 * 1/7 = 1/252.
Thematic try: 1. Nxc5? b1=B 2. Nb3# with 1/6 * 1/5 * 1/11 = 1/330.

b) 1. Nxc5 b1=B 2. Nb3#.
Choices on half-move 1: 1. K, N (2) 2. Nb4, Nb8, Nxc5, Nc7 (4)
Choices on half-move 2: 1. K, P (2) 2. b1=Q, b1=R, b1=B, b1=N (4)
Choices on half-move 3: 1. K, N (2) 2. Na4, Na6, Nb3, Nb7, Nd3, Nd7, Ne4, Ne6 (8)
Therefore, the probability is 1/2*1/4 * 1/2*1/4 * 1/2*1/8 = 1/1024.
Thematic try: 1. Nb4? b1=R 2. Nc2# with 1/2*1/4 * 1/3*1/4 * 1/2*1/6 = 1/1152.

18 November 2011

More animals

The fairy pieces I introduce today are less known and not so often used. One reason might be that they are not so flexible and effective on a regular 8x8 board. To exploit their full potential, bigger boards are required. Again, they belong to the family of the leapers: dromedary, antelope, ibis, and flamingo.

Especially the dromedary is quite restricted in its mobility. Compare the following diagrams. On a 8x8 board, it can only go to nine squares no matter where a dromedary is put initially. Using a 10x10 board instead increases the choices just marginally. In that case, there are at most 16 squares available.

Dromedary (DR): (0,3)-leaper

Mobility of dromedaries on a 10x10 board. All potential destination squares after an arbitrary number of moves are marked respectively.

The ibis can only make monochromatic moves, but it can reach all 32 black or white squares of the 8x8 board. It would require an 11x11 board to show an ibis wheel.

Ibis (IB): (1,5)-leaper

The flamingo has a similar flaw when being used on an 8x8 board, for it is only moveable outside the area c3-c6-f6-f3. Finally, there is the antelope which is the most flexible piece of today's quartet.

Flamingo (FL): (1,6)-leaper

Antelope (AN): (3,4)-leaper

After this short but surely sufficient introduction it's time for some compositions where those fairy pieces are used.

  1Theodor Steudel  
feenschach 12/2005
Dromedary h7

  2Erich Bartel  
Ideal-Mate Review 01-03/1999
Ibis d3

  3Gabriel Nedeianu  
feenschach 12/2004
Flamingo h7

  4Erich Bartel  
Problem Paradise 1998
Antelope h1

11. - e4 2. DRh4! (2. DRe7?) e5 3. Kh7 e6 4. Kh8 e7 5. DRh7 e8=DR#
21. IBe8 g8=IB 2. IBf3 IB×f3 =
1. IBc8 g8=B 2. IBb3 B×b3 =
31. Kg2 FLb6 2. Kf3 FLh5 3. Ke4 FLb4 4. Kd5 FLh3 5. Kc6 FLb2 6. Kb7 FLh1 7. Ka8 Ka6 8. Rb8 FLg7#
41. ANe5 2. ANb1 3. AN×f4 4. ANxb7 5. AN×e3 6. ANh7 7. ANd4 8. ANh1 =
1. K×c2 2. Kd3 3. Ke4 4. K×f4 5. K×e3 6. Kd4 7. Kc5 8. Kb6 =
Antelope star with eight spikes. Either only king moves or only antelope moves.

By now, I've presented quite a bunch of animals. Each of the selected problems have been lightweight examples to concentrate on just one special piece. Now we might feel a little bit more courageous. How about combining several of those fairy pieces in one composition? Look at the last diagram for today — it shows seven different fairy pieces! That takes lots of different rotated pieces, of course.

Krassimir A. Gandev
feenschach 07/1971
Zebra a1,h6
Nightrider a4, b2
Flamingo a8
Ibis c6
Giraffe d6
Camel f6
Grasshopper h7
    We get to see a Super-Allumwandlung:

1. g1=Q 2. Qg4 3. Qe6 Nc3#
1. g1=R 2. Rxg5 3. Re5 Sf4#
1. g1=B 2. Bh2 3. Be5 Se7#
1. g1=S 2. Se2 3. Sd4 e4#
1. g1=G 2. Gxg6 3. Ke5 Rb5#
1. g1=N 2. Ne2 3. Nxc6 Rb5#
1. g1=Z 2. Ze4 3. Zc7 Gb7#
1. g1=CA 2. CAf4 3. CAe7 Gd7#
1. g1=FL 2. FLxh7 3. FLb8 axb8=Z#
1. g1=IB 2. IBxh6 3. Ke4 Nc3#
1. g1=GI 2. GIc2 3. GIxd6 Nf4#

11 November 2011

The Camel and the Grasshopper

We have a special date — and I present a special chess problem. Its author is the Slovak problemist Bedrich Formánek who is well-known for his humorous chess compositions.

Bedrich Formánek
SV CSTV C, 1966
3rd Prize

Let's have a look at the diagram. We know the position has to be legal. So, we ask ourselves what Black's last move was. Obviously, it must have been the move Kg6-/xh5. Consequently, the king on g6 had been in check by the white bishop d3. Now, we reach the crucial point: How did that bishop give a check? Maybe, you didn't realize — but I already gave you some hints!

Yes, the bishop could not give the check directly by moving to d3. It needed the help from another piece. That assistant gave a discovered check by moving to h5 and was then captured by the black king. I can imagine your objection "There is no piece that can accomplish this?!". Yes, you're right — partially. Let me add one word to that statement. There is no orthodox unit that can perform the desired task.

Think about the situation with the black king on g6, once more. What requirements are there for the discovered check? We need a white piece which moved from either e4 or f5 to h5 where it was taken by the black king. Do you know any piece which is capable of doing so? I know you do! A grasshopper (G) can go from f5 to h5. Additionally, we have the camel (CA) that can leap from e4 to h5.

Two remarks: 1. There might be other choices, as well. However, camel and grasshopper are the obvious, since best known, appropriate units. 2. Sure, they are fairy pieces. But you've been alerted with my introductory words, right?

Okay, we now even have two possible last moves for Black. So much for the look into the past. The stipulation is to checkmate in one move. Though we have a nice pawn on g7 being ready to promote, it doesn't help us in any way?! Wrong. Remember, we just have proven that there already has been either a grasshopper or a camel on the board. Do you see something, now?

Assuming that the last move was Kg6xGh5, we can play 1. gxh8=G#. Similarly, if the last move was Kg6xCAh5, 1. g8=CA# delivers checkmate. Yeah, we solved it! I like this funny fairy retro a lot. How about you?

04 November 2011

A visit to the zoo

Several months have passed without a post mentioning anything related to fairy chess. So, it's high time!

The introduction of new chess pieces is one of the characteristics of fairy chess. I already showed some chess problems with grasshopper and nightrider which are well-known and very popular. Of course, there a many, many more fairy pieces. One category of them are the so-called leapers. A leaper is a piece which moves (leaps) from one square to another by a fixed amount of squares in horizontal and vertical direction. For that reason, you need two integers specifying this vector from the start square to the arrival square and you write (m,n)-leaper.

The term leaper indicates that such a piece cannot be blocked as it jumps over any other piece. Therefore, a check by a leaper can't be parried by interposing, either. You already know this behaviour from the knight in orthodox chess which is — surprise, surprise — the name of the (1,2)-leaper. Interestingly, neglecting checking and castling, the king can be described as a combined leaper. That means it has the movement ability of at least two leapers. The king combines wazir and fers; the wazir is the (0,1)-leaper and the fers the (1,1)-leaper.

As you've already seen, for reasons of convenience, those leapers are assigned names often being taken from fauna and flora. In general, chess diagrams show leapers as a knight rotated by 90 or 270 degrees. Returning to the (m,n)-leapers, I introduce the camel, the giraffe and the zebra (see the following diagrams).

Camel (CA): (1,3)-leaper

Giraffe (GI): (1,4)-leaper

Zebra (Z): (2,3)-leaper

Now, let's have a look at some chess problems featuring those animals. Enjoy!

  1Peter Heyl  
Die Schwalbe 10/2002
Camel a5

  2Erich Bartel  
feenschach 07-09/1979
Camels e3,h6

  3Erich Bartel  
Duplex-Vierer 07/1968
Giraffe e6

  4Erich Bartel  
mpk-Blätter 06/2004 (Corr.)
Giraffe d6

  5Erich Bartel  
Duplex-Vierer 07/1968
Zebra f2

  6Theodor Steudel  
feenschach 12/1955
  h#3*  (4+3)  
Zebra a4

  7Gunter Jordan  
Ideal-Mate Review 10-12/1991
a) Diagram
b) bPd3 → e3
Zebra a4

  8Daniel Novomesky  
harmonie 12/2001
Zebra b5

11. a1=CA Bf6 2. exf6 e7 3. CAb4 e8=B 4. CAa7 Bc6#
1. a1=R CAd6 2.exd6 e7 3. Rc1 e8=Q+ 4. Tc8 Qxc8#
Miniature with two model mates, mixed promotions (sort of an Allumwandlung) and role change of bishop and camel.
21. Rg8+! hxg8=CA 2. Rc8 dxc8=CA=
The black rook/rook clearance results in an attractive mirror stalemate (all squares of the king's field are vacant) by the four camels.
31. d1=GI Kf1 2. GIh2 GId2#
1. Kf1 Kh2 2. GIf2 d1=Q#
41. - Kf6 2. 0-0 Ke7 3. Kh8 Kxf8=
1. - GIe2 2. Rh4 GIa3 3. Re4+ GIxe3=
51. d1=R Zh5 2. Rg1 Ze3#
1. Zd5 d1=Q 2. Zg3 Qh5#
6*1. - Bd4 2. Kb5 Zd2+ 3. Kxa5 Bc3#
1. Kxc5 Bxb6+ 2. Kb4 Zd6 3. Ka3 Bc5#
Echo and changing sacs.
7a) 1. e1=R Kg5 (Kg3?) 2. Re3 Zc1#
b) 1. e1=N Kg3 (Kg5?) 2. Nd3 Zc7#
Key moves with changing promotions, dual avoidance and ideal mates.
81. - Ze7 2. Kb3 Zc4 3. Ka2 Zf6 4. Ka1 Kc2 5. Ba2 Zd3#
1. - Kd1 2. Kb2 Ze7 3. Ka1 Kc2 4. Be4+ Kb3 5. Bb1 Zc4#