After a capture the capturing piece (Ks included) must immediately be removed to its game array square (necessarily vacant, else the capture is illegal). Captures on the rebirth square are allowed in Anti-Circe Calvet and not allowed in Anti-Circe Cheylan. Game array squares are determined as in Circe: R, B & S go to the square of the same colour as the capture; Ps stay on the file of capture; fairy pieces go to the promotion square of the file of capture.
После взятия берущая фигура (включая королей) должна быть немедленно поставлена на поле, занимаемое ею в расстановке фигур перед началом партии (обязательно свободное, иначе взятие нелегально). Взятия на полях возрождения разрешены в Anti-Circe Calvet и не разрешены в Anti-Circe Cheylan. Расстановка фигур определяется как в Circe: ладьи, слоны и кони возвращаются на поля того цвета, на которых были взятия. Взявшая пешка возвращается на линию, на которой она взяла. Сказочные фигуры – на поля превращений на линии взятия.
stipulations
'Proca Retractor', '-N & #1'
задания
-N & #1:
The stipulation means: retract N moves and than mate in 1.
Задание означает: возврат N ходов, а затем белый матующий ход.
Proca Retractor:
The one of the Defensive Retractors, where White and Black retract (take back) a given number of moves (-n moves), in order to reach a position in which a forward stipulation (e.g. #1, s#1) is met. Given a choice, Black will retract those moves that avoid or delay White’s goal. Unless otherwise stated, Black may defend by mating White through a forward move, if such an opportunity exists after one of Black’s retractions.
The different kinds of Defensive Retractors only differ in the way uncaptures are handled, of course only legal uncaptures are allowed. In Proca Retractors the retracting side decides, if and what man was uncaptured.
Один из защитных ретракторов, где белые и черные берут назад заданное количество ходов (-n ходов), чтобы достичь позиции, в которой соблюдаются форвардные задания (например, #1, s#1).
Если у них есть выбор, черные возвращают те ходы, которые избегают или задерживают цель белых. При возможности выбора, черные могут защищаться, матуя белых прямым ходом, если после взятия назад очередного хода у них возникнет такая возможность.
Различные виды защитных ретракторов отличаются только способом обработки возврата взятий, конечно, разрешены только легальные возвраты взятий. В Proca ретракторах сторона, делающая возврат, решает, была ли взята фигура и какая.
Some introduction by the Author:
This problem has some background.
In the “Chessproblem.ca 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
Israel
original - 22.07.2020
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, 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.
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).
An incredible achievement! I could not even imagine that such a complicated play with 13 (!) uncaptures is possible in Wenigsteiner Proca.
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#
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#
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.
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#.
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.
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).