30 March 2012

Striptease

It was in 2011, I don't remember when exactly, that I saw the following chess problem. It is a highly remarkable demonstration of the so-called striptease theme: in each further twin one more piece is removed.
Udo Marks
The Problemist Supplement, 11/2005
[1b4bn/3p4/1p1P4/BP1p1P2/2kPpP2/pR2P3/KpP5/1B6]
  see below(12+9)  

a) diagram, #10: 1. Kxa3! Bxd6+ (1. Bxb6? Bxd6! [2. - Nf7? 3. Kxa3!] 2. Ba5 Nf7! 3. b6 Nd8 4. ?) 2. Ka4 (3. Rc3#) bxa5 3. b6 Bf7 (3. - Bxf4 4. b7 Bxe3 5. b8=Q Bd2 6. c3+ Bxc3 7. Qc7+ Kxd4 8. Qxc3#) 4. Ba2 b1=Q 5. Bxb1 Be5 (5. - Bc5 6. b7 Bxd4 7. b8=Q Bxe3 8. Qb5+ Kxd4 9. c3#) 6.fxe5 Be8 7. Ra3 (8. Ba2#) d6+ 8. Kxa5 Ba4 9. Kxa4 ~ 10. Ba2#
b) -Nh8, #9: 1. Bxb6! Bxd6 2. Ba5 Bc5 3. b6 Bxd4 (3. - Bxb6 4. Rxb6 ~ 5. Rb4+ Kc3 6. Ra4#) 4. exd4 Kxd4 5. b7 e3 6. c3+! Kc5 7. b8=Q (8. Qc7#) d4 8. Qc7+ Kd5 9. Rb5#
c) -Pc2 in b), #7: 1.Be1! (2. Rb4#) Bxd6 2. Bc2 b1=Q+ 3. Rxb1 Bf7 (3. - Bc5 4. Bd1 Bxd4 5. Be2+ Kc5 6. Bb4#) 4. Bd1 (5. Be2#) Bh5 5. Bxh5 (6. Be2#) Kd3 6. Rb3+ Kc2/Kc4 7. Rc3#/Be2#
d) -Bg8 in c), #6: 1. Be1! (2. Rb4#) Bxd6 2. Bc2 Bc5 3.Bd1 b1=Q+ 4. Rxb1 (5. Bd2#) Bxd4/Kd3 (4. - Bb4 5. Be2#) 5. Be2+/Rb3+ Kc5/Kc4 6. Bb4#/Be2#
e) -Bb8 in d), #5: 1. Kxa3! bxa5 2. b6 a4 3. b7 axb3 4. b8=Q Kc3 5. Qb4#
f) -Pa3 in e), #4: 1. Kxb2! bxa5 2. Kc2 a4 3. Rb2 a3 4. Ba2#
g) -Pb2 in f), #3: 1. Bc2! bxa5 2. Bd1 a4 3. Be2#
h) -Pb6 in g), #2: 1. Rb4+! Kc3 2. Ra4#
i) -Pe4 in h), #1: 1. Bd3#!
j) -Pd5 in i), #3: 1. e4! Kxd4 2. Bb6+ Kc4 3. Bd3#
k) -Pf4 in j), #4: 1. Bb6! Kd5 2. Bc5 Kc4 3. Bd3+ Kd5 4. e4#
l) -Pd4 in k), #5: 1. Be4! Kc5 2. Bc7 Kc4 3. Ka3 Kc5 4. Rc3+ Kxb5 5. Bd3#

Really impressive!

23 March 2012

Bohemian Rhapsody

I just had to choose this title ... Today, I show some early threemovers by Miroslav Havel. I don't think they are published a lot. At most 13 chess pieces are used in each diagram, all problems show beautiful model mates. Some of them have additional variations which are not given in the solutions. These are not considered to be part of the themes and thus are omitted.

1
Narodni Listy, 1899
[3b4/7p/K2pp3/8/1p1k1N2/P4P2/2Q1PP2/8]
  #3(7+6)  

2
Obrazkova Revue, 1900
[8/5K2/8/2pk1B2/N2N2p1/1p2p1Q1/1b2b3/8]
  #3(5+7)  

3
Nove Parizske Mody, 1901
[1Q1K4/p7/p4p2/3k4/1BN2p2/6b1/5P2/1B6]
  #3(5+6)  

4
Romanleser, 1902
[8/1nN5/pK4bp/3p4/4k3/6Q1/2P1BP2/8]
  #3(6+6)  

Solutions:
11. Nd5! (2. Kb5 exd5/~ 3. Qb2/Qe4#)
1. - bxa3 2. Qc3+ Kxd5 3. e4#
1. - exd5 2. f4 ~ 3. e3#
1. - Kxd5 2. Qe4+ Kc5 3. axb4#
Three different mates by the pawns.
21. Qb8! (2. Nb6+ Kxd4 3. Qh8#)
1. - Bc4 2. Qf4 (2. Nb6#) cxd4 3. Be4#
1. - Bxd4 2. Nb6+ Kc6 3. Bd7#
Nice long queen moves. Unfortunately, Nb6+ appears in each of the main variations.
31. Be7! (2. Qd6+ Kxc4 3. Qd3#)
1. - Ke6 2. Ne5 (3. Qd6#) fxe5 3. Qb3#
1. - Kxc4 2. Qb4+ Kd5 3. Qe4#
1. - f3 2. Ne3+ Kd4 3. Qb2# (2. - Kc6/Ke6 3. Be4/Ba2#)
41. Ne8! (2. Ng7 ~/d4 3. Bf3/Qe3#)
1. - Bxe8 2. Bd3+ Kd4 3. Qg7#
1. - Kf5 2. Bg4+ Kg5 3. f4# (2. - Ke4 3. Qe3#)
1. - d4 2. Bd3+ Kd5 3. Nc7#

16 March 2012

Havel's Helpmates

You might regard this as a sequel to last week's Unexpected and call it "Unexpected 2". But these blog posts are not intended to be like the Die Hard movies ... Moreover, I like the alliteratve title. ;-)

Anyway, I was quite astonished to see that Miroslav Havel composed quite a bunch of helpmates, some very nice onces, as well. So, this week, I concentrate on this area and hope you like what I chose for you.

The first one is rather a joke problem and could be presented on April Fools' Day or New Year's Eve. No. 3 makes good use of the available space and surely comes up with a little surprise, doesn't it? I like the twins of No. 5, you too? No. 6 is really curious, though I don't really understand the sense of so many phases — just to accumulate as many as possible?!

1
Narodni osvobozeni, 1938
[8/K1Q5/7N/3k1pP1/3Pp3/8/8/8]
  h#2*(5+3)  

2
Ceskoslovensky Sach, 1937
[2Q5/8/2K5/8/3k4/3pp3/8/8]
  h#2*(2+3)  

3
Sach, 1942
[8/1K1B4/3kP3/8/8/8/8/8]
  h#3*(3+1)  

4
Ceskoslovensky Sach, 1937
[4b3/1nkP2K1/1r6/8/5N2/8/8/8]
  h#2*(3+4)  

5
Fide Revue T., 07-09/1958
[8/8/1B6/2b4K/4NN2/5k2/3P4/8]
  h#2(5+2)  
a) Diagram
b) shift position (without wP)
one square down/left
c) shift position again (without wB) one square down/left

6
Parallele 50, 1951
[8/8/8/3K4/kN6/1n6/8/2nN4]
  h#2(3+3)  
see solution

Solutions
1Here, the usual conventions regarding the en passant capture as key move are simply ignored — consistenly!
*1. - gxf6 e.p. 2. Ke6 Qc6#
1. exd3 e.p. Qc4+ 2. Ke5 Nf7#
I liked it very much!
2*1. - Kd6 2. Ke4 Qg4#
1. d2 Kb5 2. Kd3 Qc4#
Beautiful simplicity.
3*1. - e7 2. Kxd7 Kb6 3. Kc8 e8=Q#
1. Kc5 e7 2. Kb4 e8=Q 3. Ka5 Qe1#
How obvious is this march to a5?
4*1. - d8=R 2. Bc6 Ne6#
1. Nd6 dxe8=Q 2. Rb7 Nd5#
5a) 1. Bf2 Be3 2.Bg3 Ng5#
b) 1. Be1 Kf4 2. Bf2 Nc1#
c) 1. Bb4 Nb3 2. Be1 Ne3#
Another interesting way to create twins.
6a) Diagram: 1. Na5 Nc2 2. Nb3 Nc3#
b) bKa4 → a3: 1. Nd3 Nc3 2. Nb2 Nc2#
c) further bNc1 → d3: 1. Nb2 Nc2+ 2. Ka2 Nc3#
d) further bNd3 → c2: 1. Ka4 Nc6 2. Na3 Nc3#
e) further bK → a1: 1. Kb1 Nd3 2. Na1 Nc3#
f) further wNb4 → e3: 1. Ka2 Nc4 2. Na1 Nc3#
g) further bNb3 → d2: 1. Nc4 Nc3 2. Nb2 Nxc2#
h) further wNe3 → e5: 1. Kb1 Nc3+ 2. Kc1 Nd3#
i) further bK → f1: 1. Ke2 Ng6 2. Kd3 Nf4#
j) further bNd2 → f2: 1. Nd4 Nf3 2. Ne2 Ne3#
k) further bK → e1: 1. Nd4 Ne3 2. Ne2 Nf3#
l) further bK → h1: 1. Ne1 Nf3 2. Ng2 Nxf2#
m) further bNc2 → g3: 1. Kg1 Ne3 2. Nh1 Nf3#
n) further wNd1 → h5: 1. Kh2 Nf4 2. Nh1 Nf3#
o) further bNf2 → g6: 1. Nh4 Nf3 2. Ng2 Nxg3#
p) further bK → h6: 1. Nf5 Nf6 2. Ng7 Nf7#
q) further wNe5 → h4: 1. Ne4 Nf6 2. Ng5 Nf5#
r) further bK → h8: 1. Nf5 Nf6 2. Ng7 Nxg6#
s) further bNg3 → f7: 1. Kh7 Nf5 2. Nh8 Nf6#
t) further wNh4 → d8: 1. Kg8 Ne6 2. Nh8 Nf6#
u) further bNg6 → h4: 1. Nf5 Nf6 2. Ng7 Nxf7#
v) further bK → f5: 1. Ng5 Nf7 2. Ng6 Nh6#
w) further bK → e8: 1. Ng6 Nc6 2. Nf8 Nf6#
x) further wK → f4: 1. Nf5 Ne6 2. Ne7 Nf6#
y) further bK → f8: 1. Nf5 Nf6 2. Ne7 Ne6#
z) further wNd8 → h8: 1. Nf5 Nf6 2. Ng7 Ng6#
aa) further bK → g8: 1. Nf5 Ng6 2. Ng7 Nf6#
ab) further bNh4 → d8: 1. Kf8 Ng6+ 2. Ke8 Nf6#
A twinning monster.

09 March 2012

Unexpected

I have only one diagram today as I am in deep time trouble, sorry. It's a threemover by a famous member of the Bohemian School, Miroslav Kostal who used the pseudonym Miroslav Havel. What's quite puzzling: it's a retro!

Miroslav Havel
Cas, 1922
[r3k1BQ/b1pp1Rpp/1p3pp1/2p5/P5P1/5P2/1PP2P1P/R3K3]
  #3(12+11)  

I closely follow the analysis T. R. Dawson provided.

In chess problems in which the king and rook of one colour are on their original squares, we make the assumption that castling is legal unless proven otherwise. So, we assume that both sides have the right to castle long and therefore Ke1, Ke8, Ra1, Ra8 have never moved.

Captured black men: Q, R, B, N, N.
Captured white men: B, N, N, d-Pawn.

Black has made four captures: f7xg6, e7xf6, a7xb6xc5, accounting for all captured While men. If Pb7-b6xc5 and Pa7xb6, then the Ba7 has no mode of entry. Thus, the captures were Pa7xb6xc5.

The white d-Pawn (or its equivalent after promotion) was captured and it must have made at least one capture to c5 (a second if it promoted anywhere except e8).

As the white Rh1 has moved out, either to f7 or to be captured, the Pg4 has not recently come from g3 or g2, nor the Pf3 from g2. Otherwise, white couldn't castle anymore as the white king had moved.

Let's inspect the north-east corner of the position and see how the men arrived there. As Black has 8 Pawns on the board, the Ba7 came originally from f8. Hence it must return to f8 before Pf6 returns to e7. Supposing that done, we see Rf4-f7, Bc4-g8, Qd5-g8-h8 are a simple system of moves which get the men f7, g8, h8 safely into position without disturbing the Black King. The black Ra8 was evidently captured by the White Q on g8 or h8 in this process.

It follows from this inspection that Black must release White, if at all, by uncapturing Pb6xc5 to give White some liberty. This necessitates Pb6 back to b7, Ba7 out to a5, and then Pb6xc5 may return. But we are confronted here by the fact that the Ba5 checks e1. To retract B to a5 needs a White king move, or needs an interception on b4, c3 or d2. The only man we can possibly use to intercept the line a5-e1 will be the Black man that reappears on f3, and so we necessarily
arrive at the "try" to free White by:

1. Pb7-b6 (the Bc8 does not matter, as there are plenty of Black men for White's few captures), Pe2xNf3! 2. Nd2-f3+ Pa3-a4 3. Bb6-a7 Pa2-a3 4. Ba5-b6, successfully preparing the way for Pb6xc5, but failing by a single move. At this point, White has used all his "spare" moves, and may retract only moves by his king, the Ra1, or the Pg4, meaning that Castling is illegal in each case.

Now Black has no other way whatever of "freeing" White if we maintain our initial assumption that Black may castle long. Therefore, IF BLACK MAY CASTLE, WHITE MAY NOT CASTLE. On other hand, if we insist that White can castle, then Black must have moved his king, whereby the Rf7, etc., obtain "freedom". Thus, IF WHITE MAY CASTLE, BLACK MAY NOT CASTLE.

If Black may castle long, then there is NO MATE IN THREE. But if Black may not castle long, then White easily mates in two, by 1. Rxg7 or 1. Qxg7. Of course, we prevent Black's castling by playing 1. 0-0-0! and are able to mate in three.

02 March 2012

Looking back again

First, some lines about the yearly solving contests I participated in once more. I mentioned them here and here but haven't written about my results yet.

This year, the ChessBase contest was quite easy. And when I have almost no trouble spotting the solution, it's very likely that I become careless. So, in puzzle #2, I overlooked the better defense 1. - Kh4! and chose 1. - Kxg4. As I was so focussed on those knight promotions, I continued exactly with the moves of the solution, although 11. d8=Q would have won quicker. Naturally, I highly enjoyed the helpmate and the serieshelpmate. Normally, I have great respect for proof games with more than seven moves and am rather not touching them. With the one that was to be solved on December 31st it was totally different, though. I can't say why nor how, but in a matter of seconds it was all so clear to me. Okay, almost ... there was just this detail about how/where to get rid of the wBf1. In case you haven't seen this retro yet, I give you the diagram (the lazy/impatient can peek here for the solution):

Bernd Gräfrath
Die Schwalbe April 2009
Dedicated to Cedric Lytton
[rnb2rk1/ppp1nppp/3bp3/8/3p4/4PP2/qPPP2PP/2KR1R2]
  SPG in 12,0 moves(10+15)  

To sum up, it was a nice set of chess problems to solve. On the other hand I was not too sad to not get a prize (you know why). Clearly trickier puzzles challenged the solvers of the Stuttgarter Zeitung (surprise, surprise). It's incredible, I made nearly the same experience as in the other contest. Many experts failed to crack puzzle D. This and also the other chess problems were not a stumbling block to me. Oddly enough, I overlooked one tiny detail: a pawn to prevent retro-stalemate. Despite this mental blackout, I received a book prize. It was this retro, a Proca retractor, because of which I missed the 100 percent by a narrow margin. Could it have been closer?

Werner Keym
Stuttgarter Zeitung Dezember 2011
[2kr2NK/6P1/3P4/2P5/8/8/8/8]
Retract 3 moves, then mate in one.

The solution goes like this: Take back (Black has no choice) 1. e5xd6 e.p. d7-d5 2. Nh6-g8 0-0-0+ 3. Nf5xPh6!, then 1. g8=Q#. Inevitably, you have to un-capture a pawn on h6, so that, for instance, h7-h6 was possible before, for both Ke8 and Ra8 never moved in order to castle! Once again, Keym presented his pet theme, the Valladao-Task (promotion + castling + en passant capture), this time in an orthodox miniature.

A third look back concerns my post One more time. There, as in all October 2011 posts, I had written about various twinning methods in helpmates. Especially diagram #1 had fascinated and inspired me. This week, on Wednesday, I received the latest issue of Probleemblad with the two originals I had sent to Gerard Smits in October and November 2011. Here they are (I won't give the solution to not spoil the fun):

Gerson Berlinger
Probleemblad 1/2012
[8/6PP/4n2P/4k1p1/2p3P1/5p2/3K3b/6r1]
  h#2(5+7)  
a) Diagram
b) wNNg7/h7
c) wBBg7/h7
d) wRRg7/h7

Gerson Berlinger
Probleemblad 1/2012
[8/7p/1P2p2P/1P2k2p/4P2K/4rb2/1PPP2P1/8]
  h#2(9+6)  
a) Diagram
b) wNNd2/g2
c) wBBd2/g2
d) wRRd2/g2

The first diagram was only the second correct version of what I wanted to show. The second diagram shows an improvement. From time to time, I work on further versions and there is already another one ready with sort of an extension.