No.1297
Sergey Shumeiko (Russia)

Original Retro & PG problems
JF – R2017-18


No.1297 Sergey Shumeiko
Russia

original – 15.04.2018

Solution: (click to show/hide)

white Bf1c1 Ke1 Qd1 Ph2g2f2e2d2c2b2a2 Sg1b1 Rh1a1 black Bf8c8 Ke8 Qd8 Ph7g7f7e7d7c7b7a7 Sg8b8 Rh8a8

PG 7                                      (13+13)


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Vlaicu Crisan
Vlaicu Crisan
April 16, 2018 19:18

This should have been presented with unknown pieces and the stipulation “White is mated. PG in 7.0” with exactly the same solution.

C+ Jacobi in 1300s.

dupont
dupont
April 16, 2018 20:18

It looks even more suitable presented with PG 6.5 unknown pieces and #1. Indeed the position remains at home but the economy is reinforced with one more piece on the board. This is still C+ Jacobi:

stipulation PG 6.5
forsyth X1XXX1XX/XXXXXX1X/8/8/8/8/XXXXXXX1/XX1XXXXX ColorThePieces
stipulation #1

1.h4 Sc6 2.h5 Sd4 3.h6 Sxe2 4.hxg7 Sxc1 5.gxf8=B Sd3+ 6.Ke2 Se1 7.Bxe7 Qxe7#

Sergey Shumeyko
Sergey Shumeyko
April 16, 2018 21:38
Reply to  dupont

Option PG 7.5:
2XXXXXX/X1XXXXXX/8/8/8/8/XXXXXXXX/1XXXXX1X
ColorThePieces stipulation #1
1.a4 Sc6 2.a5 Sd4 3.a6 Sxe2 4.ab Sxg1 5.bcS Sf3+ 6.Ke2 Se1 7.Sxe7 Rc8 8.Ra2 & Qxe7#
Phoenix (wS). Ceriani. Excelsior.

Sergey Shumeyko
Sergey Shumeyko
April 16, 2018 21:13

Сергей Шумейко
X1XXXX1X/XXXXXXX1/8/8/8/8/XXXXXX1X/XXXXXX1X
PG 6.5 & #1
1.g4 Sc6 2.g5 Sd4 3.g6 Sxe2 4.gh Sxg1 5.hgS Sf3+ 6.Ke2 Se1 7.Sxe7 & Qxe7#
Phoenix (wS). Ceriani. Excelsior.

dupont
dupont
April 16, 2018 21:25

Yes, this variation is even more homogeneous with involved Knights from both sides.

Sergey Shumeyko
Sergey Shumeyko
April 16, 2018 22:04
Reply to  dupont

Ещё одна симпатичная картина, с забавной игрой коней:
Сергей Шумейко
P1PPPP1P/PPPPPPPP/8/8/8/8/PPPPPPPP/P1PPPP1P
PG 6.0 ColorThePieces stipulation & #1
1.Sc3 Sc6 2.Sd5 Sd4 3.Sxe7 Sxe2 4.Sxg8 Sxg1 5.Sf6+ Ke7 6.Se8 Se2 & Qxe2# 🙂

dupont
dupont
April 16, 2018 23:32

The “visible” problem, dear Julia, is too light, but becomes interesting with unknown pieces. Among the 3 variations Sergey proposed, my favorite is the last one in 6.0 with harmonious holes on the 4 Knight squares. But of course this is to the author to decide.

Sergey Shumeyko
Sergey Shumeyko
April 17, 2018 00:12
Reply to  dupont

Мне последняя версия PG 6.0 тоже нравится больше. Красивая диаграммная картинка. Задание можно придумать с юмором: PG 6.0 & #1. Where are the horses? 🙂
Есть другие варианты?
Если Юлия согласится поставить задачу под No.1297.1 со скрытыми фигурами X1XXXX1X/XXXXXXXX/8/8/8/8/XXXXXXXX/X1XXXX1X, буду доволен.
P.S. Sorry, I don’t speak English.
Popeye Windows-32Bit v4.59 (1024 MB)
1.Sb1-c3 Sb8-c6 2.Sc3-d5 Sc6-d4 3.Sd5*e7 Sd4*e2 4.Se7*g8 Se2*g1 5.Sg8-f6 + Ke8-e7 6.Sf6-e8 Sg1-e2 dia
r1bqNb1r/ppppkppp/8/8/8/8/PPPPnPPP/R1BQKB1R
& 7.Qd1xe2#
solution finished. Time = 0.560 s
BeginProblem
Stipulation dia6.0
Option Variation NoBoard SetPlay Defence 1
Pieces
white Bf1c1 Ke1 Qd1 Ph2g2f2d2c2b2a2 Se8 Rh1a1
black Bf8c8 Ke7 Qd8 Ph7g7f7d7c7b7a7 Se2 Rh8a8
EndProblem

dupont
dupont
April 17, 2018 00:07

The last version is cooked:

1.Sc3 e5 2.Sf3 Sa6 3.Sd5 Sb8 4.Sxe5 Sa6 5.Sxc7+ Sxc7 6.Sxf7 Se7 7.Sd6#

Sergey Shumeyko
Sergey Shumeyko
April 17, 2018 00:18
Reply to  dupont

Thanks. The task in the basket.

Sergey Shumeyko
Sergey Shumeyko
April 17, 2018 00:40
Reply to  dupont

X1XXXX1X/XXXXkXXX/8/8/8/8/XXXXXXXX/X1XXXX1X
PG 6.0 & #1. Where are the horses?

François Labelle
François Labelle
April 17, 2018 00:55

One way to get an animated solution *and* undefined pieces is to end with a dummy move that turns the last position into what we want. For example, here’s the p2w-solution for a PG 6.5 and #1, which I think was the intention:

1.h2-h4 Sb8-c6 2.h4-h5 Sc6-d4 3.h5-h6 Sd4*e2 4.h6*g7 Se2*c1 5.g7*f8=B Sc1-d3 + 6.Ke1-e2 Sd3-e1 7.Bf8*e7 Qd8*e7 + 8.Ra1-a1 [+nOa1][+nOb1][+nOd1][+nOe1][+nOf1][+nOg1][+nOh1] [+nOa2][+nOb2][+nOc2][+nOd2][+nOe2][+nOf2][+nOg2]
[+nOa7][+nOb7][+nOc7][+nOd7][+nOe7][+nOf7][+nOh7] [+nOa8][+nOc8][+nOd8][+nOe8][+nOg8][+nOh8]

The drawback is that the dummy move will appear as part of the solution. To get something perfect we would need to make a feature request to Dmitri Turevski.

Paul Raican
April 14, 2020 14:24

This problem is Jacobi+:
cond #r
stip dia 6.5 forsyth r1bqSb1r/ppppkppp/8/8/8/8/PPPP1PPP/R1B1KB1R

Unique solution: 1.Sc3 Sc6 2.Sd5 Sd4 3.Sxe7 Sxe2 4.Sxg8 Sxg1 5.Sf6+ Ke7 6.Se8 Se2 7.Qxe2 [-e2]
Who dares to test with undefined pieces?

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