Dmitrij Baibikov

(v) 1524.1 Dmitrij Baibikov (Israel)

Original Fairy problems
01.01.2019 - 31.12.2020

The problem is withdrawn from the current tournament JF-R2019-20 with an improvement No.1524-2 belonging to the next tournament, JF-R-2021-22.

Definitions / Определения

Some introduction by the Author:

This problem has some background.
In the “ Bulletin” 7/2015 an article “Wenigsteiner in Proca Retractor AntiCirce” was published by Vlaicu Crişan and Eric Huber. The article ended with a call to break the record achieved by Paul Rãican (-18 & # 1). Three months later, a problem by Alex Levit (-19 & # 1) appeared on JF.
Seeing the problem, I immediately realized that it was a new record. Therefore, the first comment on JF wasn’t unexpected. However, the second comment came as a surprise. It turned out that the new record (-21 & # 1) had moved to Problem Paradise. The next surprise was waiting for me when reading the materials of the Wenigsteinerjahrespreis competition for 2016.
The new record (-23 & # 1) by the same author was published in the next issue of PP.
I have long wanted to return the record to JF. I hope that my task is correct and that I succeeded.

And also the hint for solving – the solution includes the following moves:
3.Ke1×Rd1(+wKe1), 9.Ke4×Bf4(+wKe1), 13.Ke3×Pf4(+wKe1), 17.Kh3×Bg4(+wKe1), 33.Kf7×Qe8(+wKe1)

No. 1524 Dmitrij Baibikov
original - 22.07.2020

White Ke1 Black Kc5 Pc2g5
-34 & #1 Proca                       1+3

Solution: (click to show/hide)

No. 1524.1 Dmitrij Baibikov
version of No.1524 - 26.12.2020

White Ke1 Black Kc5 Pg5g6
-33 & #1 Proca                       1+3

Solution: (click to show/hide)

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Alex Levit
Alex Levit
July 24, 2020 12:59

An incredible achievement! I could not even imagine that such a complicated play with 13 (!) uncaptures is possible in Wenigsteiner Proca.

Bojan Bašić
Bojan Bašić
October 21, 2020 04:14

Could this be a cook?
-22.Kf2xRg3(+wKe1) Rd1-d2+ -23.Ke1-f2 Rd2-d1+ -24.Kg5xQg4(+wKe1) Kd8-c7+ -25.Kf4-g5 Ba7-b8+ -26.Ke3-f4 Bb8-a7+ -27.Kf2-e3 Rd1-d2+ -28.Ke1-f2 Rd2-d1+ -29.Kf4xP(B)e4 Ba7-b8+ -30.Ke3-f4 Bb8-a7+ -31.Kf2-e3 Rd1-d2+ -32.Ke1-f2 Rd2-d1+ -33.Kf7×Qe8(+wKe1) ~+ -34.Ke6-f7 & 1.Kd7#

Aleksandr Levit
Aleksandr Levit
December 29, 2020 23:50

And what about

12.Kf2xBg2(+wKe1) Rd1-d2+ 13.Ke1-f2 Rd2-d1+ 14.Kf3xSg4(+wKe1) Bh3-g2+ 15.Kf2-f3 Rd1-d2+ 16.Ke1-f2 Rd2-d1+ 17.Kf2xBg2(+wKe1) Rd1-d2+ 18.Ke1-f2 Rd2-d1+ 19.Ke6×Bd7(+wKe1) Be8-d7+ 20.Ke5-e6 Kc7-c6+ 21.Kd5-e5 Kc8-c7+ 22.Kd6-d5 Ba7-b8+ 23.Kc6-d6 and forward 1.Kc7#

Last edited 2 months ago by Aleksandr Levit
Dmitrij Baibikov
Dmitrij Baibikov
December 30, 2020 21:15

Aleksandr, it’s not a cook.
1) 12.Kf2xBg2(+wKe1) Rd1-d2+ and forward 1…Ba7#
2) Black has a better defense: 20…Kd6-c6+! (or 22…Kc7-c8+!) and no solution.

Bojan Bašić
Bojan Bašić
January 4, 2021 04:44

How about this, seems like a cook in 32?

-1.Ke1×Pd2(+wKe1) d3-d2+ -2.Ke1×Rd1(+wKe1) Rd2-d1+ -3.Kf2×Bg2(+wKe1) Rd1-d2+ -4.Kf1-f2 Ba8-g2++ -5.Ke1-f1 Rd2-d1+ -6.Kf2×Sf1(+wKe1) Rd1-d2+ -7.Ke1-f2 Rd2-d1+ -8.Ke4×Bf4(+wKe1) Kc6-c5+ -9.Ke3-e4 Bb8-f4++ -10.Kf2-e3 Rd1-d2+ -11.Ke1-f2 Rd2-d1+ -12.Ke1×Pf2(+wKe1) f3-f2+ -13.Kh3×Bg4(+wKe1) Bh5-g4+ -14.Kg3-h3 with two variations:

-14…. Kd6-c6+ -15.Kg2-g3 g7-g6+ -16.Kf2-g2 Rd1-d2+ -17.Ke2-f2 Ke7-d6++ -18.Ke1-e2 Rd2-d1+ -19.Kf5xPg6(+wKe1) Kf7-e7+ -20.Kf4-f5 Ba7-b8+ -21.Ke3-f4 Bb8-a7+ -22.Kf2-e3 Rd1-d2+ -23.Ke1-f2 Rd2-d1+ -24.Kf4xBf5(+wKe1) Ba7-b8+ -25.Ke3-f4 Bb8-a7+ -26.Kf2-e3 Rd1-d2+ -27.Ke1-f2 Rd2-d1+ -28.Kg3xBh2(+wKe1) Kf8-f7++ -29.Kf2-g3 Rd1-d2+ -30.Ke1-f2 Rd2-d1+ -31.Kd7xSe8(+wKe1) Be4-f5+ -32.Ke6-d7 & 1.Ke6-f7#;

-14…. Kc7-c6+ -15.Kg2-g3 g7-g6+ -16.Kf2-g2 Rd1-d2+ -17.Ke1-f2 Rd2-d1+ -18.Ke3xPf4(+wKe1) f5-f4+ -19.Kf2-e3 Rd1-d2+ -20.Ke1-f2 Rd2-d1+ -21.Kf2×Se3(+wKe1) Rd1-d2+ -22.Ke1-f2 Rd2-d1+ -23.Kg1xPh2 h3-h2+ -24.Kf2-f1 Rd1-d2+ -25.Ke1-f2 Rd2-d1+ -27.Kg1xQh1(+wKe1) Kd8-c7+ -28.Kf2-g1 Rd1-d2+ -29.Ke1-f2 Rd2-d1+ -30.Kf7xQe8(+wKe1) ~+ -31.Ke6-f7 & 1.Ke6-d7#.

Dmitrij Baibikov
Dmitrij Baibikov
January 6, 2021 22:26
Reply to  Bojan Bašić

Bojan, it’s not a cook in 32…
Forward defense is possible in the second variant: -29.Ke1-f2 Rd2-d1+ & 1…Bg3#. (Small typo present in your notation: -29. is really -28.)
… but the first variant demonstrate that try No.2 is wrong and the problem has dual: 12.Ke1xPf2(+wKe1) f3-f2+ 13.Kh3xBg4(+wKe1) Bh5-g4+ 14.Kg3-h3 Kc7-c6+ (14…Kd6-c6+? 15.Kg2-g3 and quick checkmate in 32 according to the first variant) 15.Kg2-g3 g7-g6+ 16.Kf2-g2 Rd1-d2+ 17.Ke1-f2 Rd2-d1+ 18.Ke3xPf4(+wKe1) f5-f4+ 19.Kf2-e3 Rd1-d2+ 20.Ke1-f2 Rd2-d1+ and further according to solution.
Correction: -bPg5, +wPg5 (2+2). Solution (including tries) is without changes.

Bojan Bašić
Bojan Bašić
January 9, 2021 01:55

You are right about the forward defence in the second variation (sorry about that, I concentrated on the first variation since it was the “more important” one, and did not check the second variation carefully enough). Your proposed correction is a clever and subtle one (it took me a few minutes to undestand why it eliminates the mentioned dual at all). But it brings a new problem: replacing a black pawn by a white piece opens space for some cooks that do not work in the initial problem (solely) because of too many black promoted pieces. Here is one such: -1.Ke1×Pd2(+wKe1) d3-d2+ -2.Ke1×Rd1(+wKe1) Rd2-d1+ -3.Kf2×Bg2(+wKe1) Rd1-d2+ -4.Kf1-f2 Ba8-g2++ -5.Ke1-f1 Rd2-d1+ -6.Ke4xPf4 Kc6-c5+ -7.Ke3-e4 f5-f4+ -8.Kf2-e3 Rd1-d2+ -9.Ke1-f2 Rd2-d1+ -10.Kf3xBg4(+wKe1) Bh3-g4+ -11.Kf2-f3 Rd1-d2+ -12.Kf1-f2 Bg4-h3+ -13.Ke1-f1 Rd2-d1+ -14.Kf3xPf4(+wKe1) Bh3-g4+ -15.Kf2-f3 Rd1-d2+ -16.Ke1-f2 Rd2-d1+ -17.Kd4xPc5(+wKe1) Kd7-c6+ -18.Ke3-d4 Ke7-d7+ -19.Kf2-e3 Rd1-d2+ -20.Ke1-f2 Rd2-d1+ -21.Kf1xPf2(+wKe1) Bg4-h3+ -22.Ke1-f1 f3-f2+ -23.Kh3xRh4(+wKe1) Bh5-g4+ -24.Kg3-h3 g7-g6+ -25.Kf2-g3 Rd1-d2+ -26.Ke1-f2 Rd2-d1+ -27.Kg2xQh3(+wKe1) Kd8-e7+ -28.Kf2-g2 Rd1-d2+ -29.Ke1-f2 Rd2-d1+ -30.Kf2xSe3(+wKe1) Rd1-d2+ -31.Ke1-f2 Rd2-d1+ -32.Kf7xQe8(+wKe1) ~+ -33.Ke6-f7 & 1.Kd7# (I found this one a while ago and discarded it because the final uncapture would be illegal with bPg5, but works in the proposed correction).