No.1294
Dmitri Turevski (Russia) &
René J. Millour (France)

Original Fairy problems
JF-10/2017-3/2018:
October’2017 – March’2018


Definitions: (click to show/hide)


No.1294 Dmitri Turevski
Russia

original – 30.03.2018

Solution: (click to show/hide)

white Ph7 Sd8 Ke8 black Pg6c7c4d4 Kd6 Rd7a1 Sc5h1 Bh3

#21                                           (3+10)
Black Maximummer
White Maximummer


No.1294.1 Dmitri Turevski &
René J. Millour

Russia / France

version of No.1294 – 04.04.2018

Solution: (click to show/hide)

White Pe7 Ka5 Sa4 Black Sh8 Sd7 Pg6 Kc4 Be4 Pb3 Pf3 Rh1

#21                                             (3+8)
Black Maximummer
White Maximummer


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Juraj Lörinc
2 years ago

Nice outmaneuvering of bR and bB by wS, very well done!

Marjan Kovacevic
Marjan Kovacevic
2 years ago

It looks as the first part of solution leads to losing tempo, not gaining, because White gives the order of moves to Black. Nice and joyful play around the corners, with pinning of Sd8 as the core moment to give white the freedom of choice.

Dmitri Turevski
Admin
Dmitri Turevski
2 years ago

Thank you Marjan, I believe you are correct.

dupont
dupont
2 years ago

Very nice problem, but it seems there is something wrong with the obstacle. When the Bh3 stands on square e6, 12.Qxd4+ is illegal because of the longer move 12.Qxh1.

Dmitri Turevski
Admin
Dmitri Turevski
2 years ago
Reply to  dupont

Ah, that’s what happens when you have too many versions of a problem. Thanks, I’ll ask Julia to update the comment.

shankar ram
shankar ram
2 years ago

Could have been possible entry for for H-P.Rehm 80-JT? 😉

Dmitri Turevski
Admin
Dmitri Turevski
2 years ago
Reply to  shankar ram

When’s that, 2022? That’s a motivation to add another preparatory plan for removing black guard from f8 before plaing Ke8-Kf8.

shankar ram
shankar ram
2 years ago

Sorry! It’s his 75th JT!

Dmitri Turevski
Admin
Dmitri Turevski
2 years ago

The comments in the solution were updated to reflect corrections by Marjan and Nicolas.

Paul Rãican
Paul Rãican
2 years ago

The knight c5 seems to be useless. Isn’t it?

Dmitri Turevski
Admin
Dmitri Turevski
2 years ago

There is a very strong cook:
1.Sb7+ Ke5 2.Sd8 Ra8 3. h8=S! and two knignts checkmate quite quickly in double maximummer.
There are several ways to prevent it, bSc5 seemed most economic.

René J. Millour
René J. Millour
2 years ago

Dmitri, this is another view, Popeye 4.79 C+

White Sd8 Ke8 Ph7
Black Pc7 Kd6 Sg5 Bd4 Pc3 Ph3 Ra1 Sh1
Stipulation #21
Condition Blackmaximummer Whitemaximummer

Set play: 1…Ra1-a8 2.h7-h8=Q Ra8-a1 3.Qh8*d4 #
Actual play: 1.Sd8-c6 ! Ra1-a8 + 2.Sc6-b8 Ra8-a1 3.Sb8-a6 Ra1-g1 4.Sa6-b4 Rg1-a1 5.Sb4-a2 Ra1-g1 6.Sa2-c1 Bd4-h8 7.Sc1-b3 Rg1-a1 8.Sb3-a5 Ra1-g1 9.Sa5-c6 Rg1-a1 10.Sc6-b8 Ra1-a8 11.Ke8-f8 Ra8-a1 12.Sb8-a6 Ra1-g1 13.Sa6-b4 Rg1-a1 14.Sb4-a2 Ra1-g1 15.Sa2-c1 Bh8-d4 16.Sc1-b3 Rg1-a1 17.Sb3-a5 Ra1-g1 18.Sa5-c6 Rg1-a1 19.Sc6-d8 Ra1-a8 20.h7-h8=Q Ra8-a1 21.Qh8*d4 #

Forcing Bd4 to play on h8 is paradoxical, preventing the promotion.
Only 11 pieces.
Model mate.

Dmitri Turevski
Admin
Dmitri Turevski
2 years ago

Thank you, René.
Indeed, there are lighter versions possible, and I considered possibilities with the bishop on dark squares, but the Bd4-h8, while visually paradoxical, makes 11.Ke8-f8 the only legal move by white, and eliminates the choice, which was the point of the problem in the first place.

Joost
Joost
2 years ago

White has another legal move (11. Kd8).

René J. Millour
René J. Millour
2 years ago

Thanks, Joost, for mentioning that 11.Ke8-f8 is NOT “the only legal move by White”, as Dmitri asserts.
The choice is NOT eliminated, as asserted. White has 2 legal moves. Among them, 11.Ke8-f8 is a true waiting move.
It is also possible to have a 3rd legal move, wPh3 instead of bP, then 11.h3-h4 fails on 20.h4xg5.

Dmitri Turevski
Admin
Dmitri Turevski
2 years ago

True, I have missed Kd8, please excuse me. (Funny how many mistakes I make with this scheme, first publishing cooked problem in 2010, then all the variety of fallacies in сщььутеы).

I wonder if it is possible to keep the choice between a tempo AND promotion while having a model mate in the end?

René J. Millour
René J. Millour
2 years ago

Dmitri, please have a look on this other view, Popeye V4.79 C+

White Pe7 Ka5 Sa4
Black Sh8 Sd7 Pg6 Kc4 Be4 Pb3 Pf3 Rh1
Stipulation #21
Condition BlackMaximummer WhiteMaximummer

Set play: 1…Rh1-a1 2.e7-e8=Q Ra1-h1 3.Qe8*e4 #
Actual play: 1.Sa4-c3 ! Rh1-a1 + 2.Sc3-a2 Ra1-h1 3.Sa2-c1 Rh1-h7 4.Sc1-e2 Rh7-h1 5.Se2-g1 Rh1-h7 6.Sg1-h3 Be4-a8 7.Sh3-f2 Rh7-h1 8.Sf2-d1 Rh1-h7 9.Sd1-c3 Rh7-h1 10.Sc3-a2 Rh1-a1 11.Ka5-a6 Ra1-h1 12.Sa2-c1 Rh1-h7 13.Sc1-e2 Rh7-h1 14.Se2-g1 Rh1-h7 15.Sg1-h3 Ba8-e4 16.Sh3-f2 Rh7-h1 17.Sf2-d1 Rh1-h7 18.Sd1-c3 Rh7-h1 19.Sc3-a4 Rh1-a1 20.e7-e8=Q Ra1-h1 21.Qe8*e4 #

“I wonder if it is possible to keep the choice between a tempo AND promotion while having a model mate in the end?”. You should be satisfied: here 11.e7-e8=Q is possible, but premature, failing on 12.Qe8-e1.

Only 11 pieces.
Model mate.

Dmitri Turevski
Admin
Dmitri Turevski
2 years ago

I think this is an excellent improvement: 11.e8=Q is premature, and 11.Ka4? selfblocks – we have the best of two worlds and a model mate!
Can we submit this as a joint version (of course only if you are ok with it) to the most recent tourney of JF? That can be considered unfair by the other participants as the end date has passed.

René J. Millour
René J. Millour
2 years ago

When I saw your 1294, I was a bit disturbed by Sc5 (just anti-cook), by Rd7 (heavy) and by “10.Sd8? Ra8 11.h8=Q” not refuted by “Be6 guarding d5” (as you first said) but by 12.Qxh1.
I immediately tried to find a more economical position. But I must say this was not easy at all: I encountered lots of cooks in lots of positions. Thus, because I spent a long time on the matrix, I agree with your so nice proposal to have a joint version, for JF as you mention.
Really glad to see your last enthusiastic comments, I am now myself very satisfied of the result.
And I feel honored to sign with you!
Thanks for all!

Paul Rãican
Paul Rãican
2 years ago

Here is however a non-unique set play: 1…Rh1-a1 2.e7-e8=S! Ra1-h1 3.Sb2+ Kd5 4.Sc7+ Kc6 and checkmate at move 16. But I guess it isn’t actually a weakness.

shankar ram
shankar ram
2 years ago

Julia, Please put a diagram for 1294.1!

René J. Millour
René J. Millour
2 years ago

Paul, I agree with you, this sequence is not a weakness.
.
1) This sequence is virtual and of course not a cook, as Popeye confirms.
2) It contains so many moves that it is not to be compared with THE set play in only 3 moves, on which the problem is built, and it does not interfere with these 3 moves.
3) It is not at all connected to the actual play and has in fact nothing to do with what is shown in this work [a cook has not either something to do with what is shown in a work, but I repeat: this sequence is not a cook].
.
What is shown in this work is a main plan in only 3 moves (expressed in the set play), which fails to a precise reason (it is White to play)… and which finally works after a long foreplan based on a “tempo manoeuvre” allowing a return close to the starting position, with Black to play… this return itself temporarily fails to another precise reason: the bB is now on a8 instead of e4… this needs a second execution of the “tempo manoeuvre” in order to bring back the B on e4… at this stage at last, the main plan works! This is also combined with wK waiting move, promotion and model mate.
This independent virtual sequence neither adds nor subtracts anything to this logical content.

René J. Millour
René J. Millour
2 years ago

Many thanks, Julia, for the new diagram + photo!
.
At https://juliasfairies.com/problems/jf-1017-0318/, I guess “problem, date, author, country, pieces” should be now “1294.1, 2018-04-04, Turevsky & Millour, Russia/France, 3+8”.
.
As 1294.1 is in the Meredith form, the statistics are slightly changed. “Meredith (>7, 12): 19” becomes “Meredith (>7, 12): 18”.
.
Thanks for all!