30 December 2011

Christmas Quizzes (5)

Traditionally, during the week from December 25th to January 1st, ChessBase brings daily chess puzzles selected by John Nunn. The puzzle index page is here. Nunn indicates that the solver should look out for the special moves pawn promotion, en passant capture and castling. All the puzzles deal with them in some way. He also promises some easier puzzles than in previous years. I don't mind them being tough — I am just not too happy when I am to guess (know?) constellations and other stuff related to astronomy. We'll see. So far, solving went smoothly.

For the last time, I have Christmas puzzles for you taken again from the CHESS magazine.


  1Tibor Schönberger  
Népszava, 1923
[8/2N5/8/1p5R/3P4/1P6/k1N5/2K5]
  #3(6+2)  

  2T. R. Dawson  
 
[4k3/2p3p1/2P3P1/2P3P1/2p1B1p1/2Pp1pP1/3PpP2/4K3]
White plays and forces stalemate in 4 moves

  3T. R. Dawson  
 
[8/4p3/2p1p1p1/4K3/2p1R1p1/2PrQrP1/2PPpPP1/4k3]
White takes back his last move and mates in one

[r7/rp1p2QQ/1npR4/3P2p1/2N5/1P3n1q/5P1B/bB6]
4   Place the two kings on the board, so that
a) Black stands legally checkmated.
b) White stands legally checkmated.
c) White can give checkmate in half a move.
d) Black can give checkmate in half a move.


These are the solutions:
11. Na8! Kxb3 2. Rh3+ Kc4/Ka2,4 3. Nb6/Ra3#
1. - b4 2. Na1 ~ 3. Ra5#
21. Bf5! K~ 2. Bxg4 ~ 3. Bh5 ~ 4. g4 and White has stalemated himself!
3White can only retract h2xRg3 and then mate with 1. Qxe2#.
Rectracting h2xPg3 is not legal: the black pawns would have had to capture eight times, but only seven are missing.
If we retract b2xPc3 or b2xRc3, it would have been impossible for the black king to move to the first rank.
We can't retract a king move due to illegal checks.
Retracting a queen or a rook move doesn't leave any possibility to mate in one.
4a) +wKg6, +bKe4
b) +wKb2, +bKd1 (last move was 1.- a1=B#)
c) +wKb4, +bKd4: White already picked up the black pawn and completes the en passant capture d5xe6 e.p.
d) +wKh8, +bKc8: Black completes castling long and puts his rook to d8.

23 December 2011

Christmas Quizzes (4)

The 2011 solving contest of the Stuttgarter Zeitung is there! It was published last Saturday and, of course, you can find a reprint of the puzzles on Harald Keilhack's site. You are confronted with the usual suspects "add a unit", "release the position" and Proca-retractor. It's a must for all fans of that sort of chess problems. As for me, I did have great fun solving them.

I still have some of Hugh Courtney's puzzles left that I want to show you. As always, I struggle with the (lack of proper) references and again I am sorry for that. Anyway, I hope that you'll enjoy my selection and that you don't know all of it already.

  1
 
[8/8/3p4/1P1Pp2P/2BbP1pN/1p3p2/1Ppn1Pp1/r1R1nkqb]
Add white king and white knight, so that White can mate in one

  2Darwin Cabrera  
 
[6N1/4K1pk/6Np/8/8/8/8/8]
White takes back his last move and mates in one

  3Francis C. Collins  
Land and Water, 1879
[4N3/P4p1p/PB5k/PB3KRp/1N4P1/8/PPP4Q/R7]
White takes back his last move and mates in one

[1n1brnB1/ppp1k1pp/2P2pp1/1PKpP/2bRR1P1/QP1P2NP/8/2B5]
4   It's White to move. Here are some statements about the position:
  1. White mates in two.
  2. Black's last move was not legal.
  3. It's mate in one.
  4. Black's last move may have been legal, but the position itself is not legal.
  5. Black's last move was d5.
  6. Black's last move was Ke7.
Which two statements are right and why?

  5P. Leibovici  
Magyar Sakkvilag, 1931
[8/8/8/7K/8/5p2/4Nk2/5bR1]
  h#3(3+3)  

  6H. Voss  
Aachener Anzeiger, 1933
[8/7k/7r/7r/8/7R/7K/8]
  h#4(2+3)  

In case you should need to have a look at them — here are the solutions:
1+wKd3, +wNh3: 1. Kxd2#. Last moves were e2-e1=N+, Rb1xc1+.
2back f7xBg8=N, then 1. f8=N#
3back 1. Rg8xPg5, then 1. h5xg6 e.p.#
4If Black had just played d5, there would have been no way at all for the black queen's bishop to have emerged from its initial square. And that means that White can't play 1. exd6 e.p.#.
Black's last move was Ke6-e7 as a reaction to White's move f7xg8=B+.
The position is legal: first f7xBg6, then wPf2 to f7, e7xNf6 and finally f7xg8=B+.
It's a mate in two by 1. exf6+ Kxf6 2. g5#.
51. Ke1 Rh1 2. f2 Ng1 3. Be2+ Nf3#
61. Kg6 Kg2 2. Rg5+ Rg3 3. Kh5 Kf3 4. Rhg6 Rh3#
Apparently, this problem was published with reversed colours, so that it was White to move and help Black to mate him. I changed that to make it look like a helpmate as we know it.

16 December 2011

Christmas Quizzes (3)

Welcome to round three. Are you ready? Here we go!

  1E. C. Mortimer  
Chess, April 1952
[rnbqkbnr/ppppp1pp/5p2/8/4P3/8/PPPP1PP1/R1BQKBNR]
  Position after 8,0 moves. Find a game (first moves not unique).

  2unknown author/source  
 
[8/1KP5/8/2p5/1pP5/p7/k7/1R3R2]
  #4(5+4)  

[rnbq1bnr/ppppkppp/8/4R3/4N3/8/1PPPPPPP/2BQKBNR]

3   This puzzle has two parts:
  1. The diagram shows the position after 5. RxPe5#. What were the moves?
  2. Hugh Courtney pointed out that he had found five other sequences of moves wherein the white queen's rook mates the black king in 5 moves. Two of them are quite similar and have the black king on g6. In the other three solutions the king is mated on c6 (twice) and h6, respectively.


  4A. Herbstmann  
Tyovaen Shakki, 1935
1st Prize
[8/8/5P1P/4P3/3p1q2/3bn2K/3k1N2/7Q]
  Win(6+5)  

  5B. J. da Costa Andrade  
unknown source
 
[R7/PPQPPkPP/8/7K/B7/2B5/4r3/8]
  #1(11+2)  

  6Lord Dunsany  
The Times Lit. Supplement, 1922
[3N1rk1/pppp1N1p/5B1P/8/8/RP1P4/rPP3P1/4K2R]
  White to play, can he castle?

  7K. Fabel  
source unknown
[8/8/8/1pNp4/PpbpP3/1bkb4/1N1N4/2K5]
  #3(6+8)  
Solutions:
1There are three ways to make the first three white moves. Apart from that, everything is unique: 1. Nc3 f6 2. h4 Kf7 3. Nh3 Kg6 4. h5+ Kxh5 5. Ng5+ Kxg5 6. e4 Kg6 7. Ne2 Kf7 8. Ng1 Ke8.
21. c8=B! b3 2. Bg4 b2 3. Bd1 Kxb1 4. Bb3#
3These are the six solutions that were given by Courtney. Unfortunately, there are always interchangeable moves. It's still a challenge to find all final positions, isn't it?
a) 1. a3 e5 2. Nc3 Bxa3 3. Ne4 Bf8 4. Ra5 Ke7 5. Rxe5#
b) 1. e4 e6 2. a4 Ke7 3. Ra3 Kf6 4. Qh5 Q,Be7 5. Rf3#
c) 1. e4 e5 2. a4 Ke7 3. Ra3 Kf6 4. Qg4 Q,Be7 5. Rf3#
d) 1. a4 d6 2. a5 Kd7 3. Ra3 Kc6 4. e4 Q,Bd7 5. Rc3#
e) 1. a4 d6 2. a5 Kd7 3. Ra4 Kc6 4. Nc3 Q,Bd7 5. Rc4#
f) 1. e4 f6 2. a4 Kf7 3. Ra3 Kg6 4. Qg4+ Kh6 5. Rh3#
41. Qe1+! Kc2 2. Qc1+ Kb3 3. Qb2+ Kc4 4. Qb4+ Kd5 5. Qd6+ Kc4 6. Qc5+ Kb3 7. Qb4+ Kc2 8. Qb2+ Kxb2 9. Nxd3+. Very nice.
51. d8=N#
6No. If it's White to play, Black's last move must have been 0-0. Therefore, the rook at a2 was promoted from one of the pawns from e7, f7, or g7, any of which would have required either the white king or white rook to have moved. The promotion could not have taken place on a1. Otherwise, the bPe7 would have had to capture four white pieces on black squares — but there are only three available!
71. a5! Bbc2 2. Nb1+ Bxb1 3. Nd1#
1.- Bdc2 2. Nd1+ Bxd1 3. Nb1#
1.- Ba4 2. Nb,cxa4+ (dual) bxa4 3. Nxa4#
This last variation shows why 1. e5? fails: Black plays 1.- Bf5..h7!

09 December 2011

Christmas Quizzes (2)

I continue with a second installment of this quiz series. Once again, I stick to the original text from Chess as close as it seems appropriate. Unfortunately, Courtney wasn't very generous revealing details on his sources for the puzzles or at the least naming the composers.

[3b4/3p3Q/1BB1kp2/R2bn3/5n2/3p4/3KN3/4R3]
White to play and mate in 4 moves
1   I think this one is quite well-known:
According to legend, a certain German doctor once made a bargain with the devil, and this was the strange agreement that they made: the devil agreed to give the doctor super-human skill at chess for a period of four years, and in return the physician agreed to give the devil his soul upon the expiry of this four year term, BUT on one condition! The proviso was that the devil should first of all solve a 4-move problem which would be specially composed for him by the doctor at the end of the four years. The devil at once agreed to that 'puerile condition' (as he called it), and with a flash and a bang he was gone. The devil duly fulfilled his part of the contract, and the fame of the physician's chess prowess became famous world-wide. However, the day of reckoning came: "Where is this silly problem?" demanded his satanic majesty. "Here it is, all set up ready for you," replied the doctor. The devil made the first three moves on the board accurately and without a moment's hesitation; but the fourth move (obvious to all the bystanders) seemed to baffle him, and, with a diabolical oath, he disappeared in a huff and a puff of smoke, leaving behind him the inevitable odour of brimstone and, of course, a highly delighted German physician!
Why did the doctor's fairly simple four-mover defeat the devil?!

  2unknown author/source  
 
[8/r7/k7/p7/N7/1p3q2/1K6/QR5b]
  h#2(4+6)  

  3A. Baker  
British Chess Magazine, 1916
[3k4/3p4/3p4/p2P2p1/P2P2P1/3P4/3K4/8]
  Win(6+5)  

[6R1/7k/2r5/7B/2q5/8/4r3/5K1R]
4   Composed by Geoff Chandler of Holyrood Court, Edinburgh.
The position occured towards the end of a game between two rival, barbarian chieftains in the year 1560. The player of the white pieces was Gunter, and he had a smile on his face because it was his move and he could see that he had a forced win. What is more the winner of the game was to have the other chieftain's daughter as a prize. The barbarian who had Black was called Bunzer, by the way, and we must assume that his daughter was very pretty: otherwise why was Gunter smiling? So, that is your problem: White to play and win.
Hint: Read the story again very carefully, after which look equally carefully at the squares on which the pieces are placed!

Solutions:
1The devil played the following moves very quickly: 1. Nd4+ Kd6 2. Qxd7+ Nxd7 3. Rxd5+ Nxd5 and only then did he realise that if he played the fourth move of the solution (4. Re6#) then he would have completed the sign of the cross ... foiled again!
21. Qa8 Rg1 2. Bb7 Qf1#. I like it.
31. Kd1!! Ke7 2. Ke2 Kd8 3. Kd2 Ke8 (3. - Kc8 4. Ke3) 4. Kc3 Kd8 5. Kc4 Kc7 6. Kb5 Kb7 7. Kxa5 Ka7 8. Kb5 Kb7 9. a5 Ka7 10. a6 Ka8 11. Kb6 Kb8 12. a7+ Ka8 13. Kc7 1-0. Not too difficult, right?
4The key to this puzzle is that two barbarians are playing: so take hold of the white king and jump it over the black rook which is next to it, keeping hold of the white king now jump it over the black queen, and to conclude White's first move keep hold of the king and jump it over Black's remaining rook: remove the two black rooks and the queen from the board. Black's only move available is Kg6, whereupon the white bishop jumps over the black king which is removed from the board. White has won. Only barbarians, of course, would play Draughts with Chessmen ... didn't you know?!

02 December 2011

Christmas Quizzes (1)

You know it, right? Yeah, it's quiz time! Lots of web pages, newspapers and (chess) magazines offer their readers brainteasers for their recreation (or desperation). I've chosen some chess puzzles from Hugh Courtney’s famous Christmas Quizzes that were published in the magazine Chess. Have fun!

1What is it that a king, rook, bishop, knight and pawn can all do, but a queen cannot?
2What (if anything) is wrong with the following conversation:
"The Vicar gave Jock a fright last night when he put Jock in double check from both his queens!" "Yes, I saw it, the Vicar had two queens, two pawns and king, and Jock had queen, rook, bishop, knight, three pawns and king: quite a position!" "That's right, and the Vicar's face was quite a picture when Jock's move in reply to the double check actually put the Vicar into checkmate!!" ?
3Is this true or false: Neither a knight nor a pawn can give check without first moving away from its original square. ?

  4G. Kortnog  
source unknown
[5Q2/1Ppn4/B3p3/1K1k4/8/8/8/8]
  h#2(4+4)  

These are the solutions:
1First asked by Arthur C. Moseley (CHESS, October 1959). A queen is the only piece that cannot make a move giving discovered check.
2Nothing is wrong. Consider the following possible position from the game Jock (White) vs. the Vicar (Black). White is in check and play continues 1. e4 fxe3 e.p.+ 2. Kxe3#
[4Q2N/5q2/7p/R2q3k/5p1P/5KP1/4P3/3B4]
3False! Although true of a pawn, a knight may actually give check from its original square — if a pawn promotes. For example, 1. d8=N+ puts a black king at b7 into check. The square d8 is that knight's original square!
41. e5 Ka4,a5,b4 2. Ke6 Bc4#
It took me quite a long time to discover the solution. Yes, the first white move is not unique and you'd consider this helpmate as cooked. But it's still a nice and tricky puzzle!

25 November 2011

Anniversary

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
[8/8/N7/2p5/2P5/K7/1p6/k7]
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.

[3c4/8/8/c2*1N2c1/8/8/3c4/8]
Dromedary (DR): (0,3)-leaper

[X2X2X2X/2x2x2x1/10/X2*1N2X2X/2*1n2x2x1/10/X2X2X2X/2x2x2x1/10/X2X2X2X]
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.

[4c1c4/11/11/11/c9c/5*1N5/c9c/11/11/11/4c1c4]
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.

[7c/1*1N6/7c/8/8/8/8/c1c5]
Flamingo (FL): (1,6)-leaper

[7c/8/8/3*1N4/8/8/7c/c5c1]
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
 
[5K2/7*1n/6k1/8/8/8/4P3/8]
  h#5
  0.1...
(2+2)  
Dromedary h7

  2Erich Bartel  
Ideal-Mate Review 01-03/1999
Commendation
[8/6P1/8/8/8/k1K*1n4/8/8]
  h=2
  2.1...
(2+2)  
Ibis d3

  3Gabriel Nedeianu  
feenschach 12/2004
[8/7*1N/8/K7/8/8/8/1r5k]
  h#8
  
(2+2)  
Flamingo h7

  4Erich Bartel  
Problem Paradise 1998
[8/1p6/8/8/5b2/4p3/2p4p/2K4*1N]
  ser-=8
  2.1...
(2+5)  
Antelope h1

Solutions
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
[*1B4K2/P6*2Q/1R*1R*3B1*3NN*1N/3k2Pp/*2N7/4Pp2/p*2N3Pp1/*1N7]
  ser-h#3
  11.1...
(16+5)  
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
[7b/6P1/7B/6Pk/2P5/3B3K/2P1P3/8]
  #1(8+2)  

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).


[2c1c3/8/c5c1/3*1N4/c5c1/8/2c1c3/8]
Camel (CA): (1,3)-leaper

[8/8/7c/3*1N4/7c/8/8/2c1c3]
Giraffe (GI): (1,4)-leaper

[1c3c2/c5c1/8/3*1N4/8/c5c1/1c3c2/8]
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
[k7/4p3/4P3/*1N7/3B4/8/pK6/8]
  h#4
  2.1...
(4+3)  
Camel a5

  2Erich Bartel  
feenschach 07-09/1979
[rr6/3P3P/4k2*1N/8/8/4*1N3/8/6K1]
  h=2
 
(5+3)  
Camels e3,h6

  3Erich Bartel  
Duplex-Vierer 07/1968
[8/8/4*1N3/8/8/8/3p1K2/7k]
  h#2
  Duplex
(2+2)  
Giraffe e6

  4Erich Bartel  
mpk-Blätter 06/2004 (Corr.)
[4k2r/8/3*1NK3/8/8/8/8/8]
  h=3
  0.2...
(2+2)  
Giraffe d6

  5Erich Bartel  
Duplex-Vierer 07/1968
[8/8/8/8/8/7K/3p1*1N2/7k]
  h#2
  Duplex
(2+2)  
Zebra f2

  6Theodor Steudel  
feenschach 12/1955
[8/8/ppk5/B1B5/*1N7/8/8/K7]
  h#3*  (4+3)  
 
 
Zebra a4

  7Gunter Jordan  
Ideal-Mate Review 10-12/1991
[8/8/8/3pp3/*1N2pk1K1/3p4/4p3/8]
  h#2(2+6)  
a) Diagram
b) bPd3 → e3
Zebra a4

  8Daniel Novomesky  
harmonie 12/2001
[8/8/8/1*1N1b4/8/2k5/8/1K6]
  h#5
  0.2...
(2+2)  
 
Zebra b5

Solutions
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#

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)
[kb4r1/8/1p2p3/1p6/1KNN2rq/8/6b1/8]
  h#2(3+9)  
a) Diagram
b) wBBc4/d4
c) wRRc4/d4

  2Arthur Christopher Reeves  
The Problemist 01/2008
 
[8/8/8/8/1kr5/8/1N6/K3R3]
  h#2(3+2)  
a) Diagram
b) wK ↔ wR
c) wK ↔ wN
d) bK ↔ bR

  3Henk Weenink  
The Problemist 07/2010
 
[3q4/3b4/8/bPpk2p1/1p6/rppK3p/6p1/6NR]
  h#3(4+12)  
a) Diagram
b) all men that move in a)
change colour

Solutions
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
Commendation
34th TT (B)
[8/5P1P/6k1/6B1/8/6K1/8/8]
  h#2*(4+1)  
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
[3R4/2pP3p/2Pp3R/3k4/1P4N1/3B1p2/P3pQ2/4K3]
  #2(10+6)  
a) Diagram
b)-e) after the key of a)-d)

  2Frank Christiaans  
Die Schwalbe 06/2001
 
[2Q5/6Pp/5N1p/2NP3B/2P5/1P6/1Pk2P2/R3K3]
  #2*(12+3)  
a) Diagram
b) after the key

  3Horst Bäcker  
Problemkiste 12/2008
 
[8/2p4Q/p1p1k3/K1P5/8/5p2/5P1p/B5nR]
  #4(6+7)  
a) Diagram
b) after the key

Solutions
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
 
[8/2p2Qp1/R1bk2Pp/4pK1P/pP4p1/pR1N2B1/qp6/bN6]
  s#2(10+12)  
a) Diagram
b) after the key

  5Gunter Jordan  
Schach 03/2009
5th Honourable Mention
[2Q5/1PR5/pk6/pb6/P1p5/2p5/2P2r2/1R4BK]
  s#5(8+7)  
a) Diagram
b) after the key

Solutions
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
 
[8/n1P5/2N1N3/Pp1PPk1P/8/4p3/K3B3/8]
  h#2(9+4)  
a) Diagram
b)-d) after the key of a)-c)

  7Fadil Abdurahmanovic  
 
feenschach, 1997
 
[3rkr2/7n/8/8/8/3p3B/7R/4K2R]
  h#2(4+5)  
a) Diagram
b) after the key

  8Christer Jonsson  
Michel Caillaud
  
StrateGems 10-12/2000
1st Prize
[8/3p4/8/1P4P1/p7/p1k1NN2/p1B5/rR4K1]
  h#2(7+6)  
a) Diagram
b)-d) after the key of a)-c)


  9Jaroslav Stun  
The Problemist 11/1988
 
[B4R2/2b1r1B1/4nN2/5P2/7R/prPkp3/bp6/1n3K2]
  h#2(8+10)  
a) Diagram
b) after the key

  10Anders Lundström  
feenschach 07-09/1986
3rd Prize, 43rd TT
[8/8/8/3p1pRP/K1BPp3/4Pk2/5B2/3n4]
  h#2*(7+5)  
a) Diagram
b) after the key

  11KCharles Mason Fox  
The Chess Amateur 08/1929
 
[8/8/8/3p1kn1/3pNp1R/3K4/8/7N]
  h#2(4+5)  
a) Diagram
b)after the key

Solutions
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).