[Black]Maximummer: Black must play the geometrically longest move or may choose from among longest moves of equal length, distances being measured from the center of each square. Diagonal and oblique distances are measured from the orthogonal coordinates by using Pythagora’s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply. White Maximummer: Only White must play the geometrically longest moves, Black plays orthodoxal.
(While BlackMaximummer and Maximummer mean the same thing, in Popeye indication of Maximummer cancels out WhiteMaximummer, so to have both BlackMaximummer should be used)

No.1294Dmitri Turevski Russia original – 30.03.2018

white Ph7 Sd8 Ke8
black Pg6c7c4d4 Kd6 Rd7a1 Sc5h1 Bh3

#21 (3+10)
Black Maximummer
White Maximummer

{
The main plan doesn't work because it's white to move: }
1... Ra1-a8 2.h7-h8=Q Ra8-a1 3.Qh8*d4#
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 Bh3-e6! {(4 < 3*sqrt(2) < 5)} 7.Sc1-b3 Rg1-a1 8.Sb3-a5 Ra1-g1 9.Sa5-c6 Rg1-a1 {
tempo is lost, but now there's a new obstacle:
10.Sd8? Ra8 11.h8=Q Ra1 12.Qxh1
} 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 Be6-h3 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# {
(C+ by Popeye 4.79)}

In 2010 I have published a cooked(sic!) problem with similar device and impure logic.
This C+ problem is dedicated to Juraj Lorinc, a proponent of fairies with direct
stipulations, the editor of "Conflictio". (Author)

No.1294.1Dmitri Turevski &
René J. Millour Russia / France version of No.1294 – 04.04.2018

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.

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.

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.

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.

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.

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?

“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.

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.

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!

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.

Glad you’ve accepted it for your tournament! Thank you!
Organizing Latvian solving championship tomorrow.. very busy these days. Will start to publish my “waiting list” from Sunday, 8th of April.

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.

Many thanks, Julia, for the new diagram + photo!
.
At http://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!

Dmitri Turevski Jul 15, 17:57 on No.1422 (AO) { The thing with Alexey's position is that nothing really interesting/original/hs#-specific happens until exf2+. In your... }

Sergey Shumeiko Jul 15, 11:53 on No.1422 (AO) { >>What is the idea of the first two half-moves? The idea is: - the critical... }

Aleksey Oganesjan Jul 13, 19:55 on No.1422 (AO) { >> What is the idea of the first two half-moves? 1) In my opinion, additional... }

Dmitri Turevski Jul 13, 12:03 on No.1422 (AO) { 4.Qa1-g1 is a critical move (exploiting the interference), not anti-critical (avoiding the interference). What is... }

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

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.

Thank you Marjan, I believe you are correct.

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.

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

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

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

Sorry! It’s his 75th JT!

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

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

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.

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.

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.

White has another legal move (11. Kd8).

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.

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?

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.

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.

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!

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.

Julia, Please put a diagram for 1294.1!

Glad you’ve accepted it for your tournament! Thank you!

Organizing Latvian solving championship tomorrow.. very busy these days. Will start to publish my “waiting list” from Sunday, 8th of April.

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.

Many thanks, Julia, for the new diagram + photo!

.

At http://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!

Yes, of course, the problem 1294.1 is added to the list at http://juliasfairies.com/problems/jf-1017-0318/ .

Thank you, René!