Babson task
The Babson task is a task from chess composition . A mutual all-round conversion is required in immediately successive moves by the same pawn , whereby the conversion in all variants must take place on both sides into the same piece.
Origin of the task
The term goes back to the American Joseph Babson (born December 26, 1852 , † December 20, 1929 ). In 1914 he published the first mutual conversion, which, however, was still carried out by three farmers. Babson failed to master this task in the years that followed. That is why the American Powers started a composition tournament under the name Babson Tournament in 1925 and donated $ 20 as a prize (or 25, even if the key move is a subversion ). The tournament was won by Henry Wald Bettmann , but the pieces submitted later by H. August and W. Krämer were almost identical.
Babson Tournament, 1926
1st prize
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Solution:
1. a8L!
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1. ... 2. fxg1D f8d DXF1 3. b5 + Dxb5 matt , or 2. ... dxc5 + 3 + b5 Dxb5 matt
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1.… fxg1T 2. f8T! Rxf1 3.Rxf1 Rxa6 mate
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1.… fxg1L 2. f8L! Bxc5 3rd Bxc5 Rxa6 mate
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1.… fxg1S 2. f8S! Nxh3 3rd Rxh3 Rxa6 mate
After the key move (also a sub-transformation) the newly emerging black figure must be eliminated. When converting to a queen, this is always possible with the exception of the variants mentioned; when converting to rook and bishop , White is prevented from mating himself, and when converting to a knight , the d7-square is defended through which Black could otherwise escape .
After depicting the Babson task in Selbstmatt , its implementation in an orthodox task still seemed impossible. In particular, the Parisian metallurgy engineer Pierre Drumare (1913-2001) systematically researched the Babson for over 20 years, but without finding a correct representation. Only the until then completely unknown composer Leonid Wladimirowitsch Jarosch was able to present a correct task without conversion figures in the starting position in 1983. The problem with number 23 was titled "!!! Jest task Babsona?" (in German "Is it the Babson task?") and later became known worldwide as the problem of the century .
Schachmaty w SSSR, March 1983
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Solution:
1.Rxh4
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1.… cxb1D 2. axb8D Qxb2 3. Qb3 Qc3 4. Qxc3 mate , or 2.… Qe4 3.T / Qxf4 Qxf4 4.D / Rxf4 mate
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1.… cxb1T 2. axb8T! Rxb2 3rd Rb3 Kxc4 4th Rxf4 mate
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1.… cxb1L 2. axb8L! Be4 3. Bxf4 and 4. Be3 or 4. Be5 mate
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1.… cxb1S 2. axb8S! Nxd2 3rd Nc6 + Kc3 4th Rc1 mate
The duals were accepted that way in the first presentation. A few months later, Jarosch created a dual-free version that received 1st prize. In the orthodox problem, the sub- metamorphoses into rook and bishop are motivated by avoidance of stalemate .
In the meantime 16 orthodox problems have been composed with the Babson task, and the Babson task has also been implemented with fairy tale characters and then 5 sub-transformations each. Pierre Drumare also created a correct Babson. However, a challenge remained. So far, only equal conversions had followed one another, and in 2003 Peter Hoffmann started work on the cyclic Babson. In September 2005 the following problem appeared in the German magazine Schach under the number 15.778.
Schach, September 2005
Dedicated to Tim Krabbé
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Solution:
1. Bxc6
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1.… d1D 2. gxf8L! Qd4 + 3. exd4 Kxf6 4. d5 mate
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1.… d1T 2. gxf8S +! Kd6 3. c8S +! Kc7 4. Ne6 mate , or 3.… Kc5 4. Qxc2 mate
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1.… d1L 2. gxf8T! Kd6 3. Qd2 + Kc5 4. Qd4 mate
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1.… d1S 2. gxf8D Nc3 + 3. Kxa5 Ke5 4. Qe7 mate
First representation of the cyclical Babson without conversion figures. A later version of this study received a special award.
The Babson Task in the Study
The Babson task was not completely fulfilled in a study, but it was shown with three (instead of four) conversions . A composition theme tournament held in 2005 by Gerd Wilhelm Hörning, Gerhard Josten and Siegfried Hornecker for the Babson task in the study with a prize fund of 500 euros was not submitted.
Jan Rusinek
Jan Rusinek (born December 2, 1950 in Cracow ) once tried to portray such a "three-quarter Babson", but failed. Presumably his creation inspired Gady Costeff .
Roycroft Jubilee Tourney, EG 1978
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Intended solution : 1. Rb8 – b1 + e2 – e1D 2. Rb1xe1 + , and now:
- 2.… g2 – g1D 3. f7xg8D Qg1xe1 4. Qg8 – a8 + Kh1 – g1 5. Bh8 – d4 + Kg1 – f1 6. Qa8 – f3 + win
- 2.… g2 – g1L 3. f7xg8L Kh1 – g2 4. Bg8 – h7 Kg2 – f2 5. Bh7 – e4 Kf2xe1 6. Ka4xa5 Bg1 – c5 7. Lh8 – e5 win
- 2.… g2 – g1S 3. f7xg8S Kh1 – g2 4. Ng8xe7 h2 – h1D 5. Ne7 – f5 Kg2 – f2 6. Re1 – f1 + Kf2xf1 7. Nf5 – g3 + Kf1 – g2 8. Ng3xh1 win
There are many duals present, moreover the problem according to 1.… g1D 2. fxg8D Qd1 + is unsolvable.
Gady Costeff
Gady Costeff (born May 7, 1961 in Be'er Scheva in Israel ) correctly portrayed the three-quarter Babson.
Magyar Sakkélet, 1984
2nd Special Honorable Mention
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Solution :
1. d6 – d7 Rc8xe8 2. Rb2xb1 + , and now
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2.… e2 – e1D 3. d7xe8D De1xb1 4. Bb5 – d7 Qb1xd3 5. Bd7 – b5 win switchback
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2.… e2 – e1L 3. d7xe8L Win
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2.… e2 – e1S 3. d7xe8S Kf1 – e2 4. Rb1 – b2 + Ke2 – e3 5. Rb2xf2 Ke3xf2 6. Ng8 – f6 Kf2 – e3 7. Nf6xd5 + Ke3 – d2 8. Nd5 – f4 Kd2 – e3 9. Nf4– g2 + profit
First correct representation of a three-quarter Babson.
Chess in Israel, 1997
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Solution :
1.… Bg8 – c4 +! 2. Kd3 – d4 Rc7 – d7 + 3. Kd4 – c5 Nb6 – a4 + 4. Kc5xc4 Ng2 – e3 + 5. Kc4 – b4 Rd7xb7 + 6. Kb4xa4 Ne3 – d1 7. Ra1xd1 , and now:
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7.… e2xd1D 8. e7 – e8D Qd1 – d4 + 9. c3 – c4! Profit
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7.… e2xd1T 8. e7 – e8T! Rd1 – d2 9. Re8 – f8 + Rb7 – f7 10. Rf8xf7 + Kf6xf7 11. Ka4xb3 win
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7.… e2xd1S 8. e7 – e8S + Kf6 – f7 9. Ne8 – d6 + , and 10. Nd6xb7 win
Nice presentation of the subject, especially due to the long introduction.
Web links
- 100 Years: Babson Task in the Orthodox Directmate by Peter Hoffmann
- Leonid Jarosch's website ( Memento from February 18, 2008 in the Internet Archive ) (Russian) (accessed August 12, 2012)
- "Sons of Babson" by Tim Krabbé (English)
- All Babsons with comments on a replayable Javascript board