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MatPlus.Net Forum Selfmates A Forced Selfmate In 16 Ply
 
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(1) Posted by Rewan Demontay (Real Name: J. Malcom) [Friday, May 24, 2019 22:07]; edited by Rewan Demontay (Real Name: J. Malcom) [19-05-25]

A Forced Selfmate In 16 Ply


I define a forced selfmate as “the player has no other option after his first move but to go along with the selfmate that they started.” The joke is that the player goes for a selfmate on his first move, but he cannot change his mind afterwards, assuming that the sequence is 4 or more ply long.

I say 4 ply because in order to have a “choice” after his first move, the sequence must be at least three ply long. But since their opponent must mate, one more ply must added on. The colors of the opponents can be switched of course.

Under my set definition, here is an original (to my knowledge) forced selfmate in 16 ply that I constructed myself. Try to solve it, and good luck to those who do!

NOTE: Due to a cook that I just noticed, I had to fix it. It cost be 2 ply, but I see no way to regain those 2 ply at the moment. I also found ways to shave off a piece for each side.

NOTE 2: I found a way to get my 16 ply back! As a bonus, I was also able shave off a white piece! I have heard that composers should always try to minimize their piece/pawn usage.

White To Move-Forced Selfmate In 16 Ply

(= 12+13 )

 
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(2) Posted by Frank Richter [Saturday, May 25, 2019 10:21]; edited by Frank Richter [19-05-25]

It solves in 7 with all white knights on f3+ and 7.- Q:f3#
So may be I misunderstood your intention?

Ps.: .... and please provide a proof game to show, that all pawns (white and black) really could have promoted ;)
 
 
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(3) Posted by Nikola Predrag [Saturday, May 25, 2019 13:23]

It seems that the main problem is guessing a possible meaning of the stipulation, using the information about the selfmate and 16 ply.
Maybe something like "find a key for a longest selfmate (not shortest)", with 1.f4+ in mind.
 
   
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(4) Posted by Joost de Heer [Saturday, May 25, 2019 13:44]; edited by Joost de Heer [19-05-25]

-
 
   
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(5) Posted by Rewan Demontay (Real Name: J. Malcom) [Saturday, May 25, 2019 15:25]; edited by Rewan Demontay (Real Name: J. Malcom) [19-05-25]

Here’s a proof game like you asked for: 1. d4 a5 2. b4 c5 3. dxc5 axb4 4. e4 e6
5. e5 d5 6. Be2 f5 7. Bg4 Bd6 8. exd6 fxg4 9. Bg5 Ra3 10. Qc1 Rf3 11. gxf3 h6 12. Qd1 hxg5

And then pawns on e6 and f2 will be the ones that you see in my diagram. All of other pawns go on to promote their respective pieces that are seen in my diagram.
 
   
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(6) Posted by Rewan Demontay (Real Name: J. Malcom) [Saturday, May 25, 2019 15:58]; edited by Rewan Demontay (Real Name: J. Malcom) [19-05-25]

Frank, while Nf3+ is technically a selfmate move, that is a forced selfmate in 14 ply, not 16 ply as stipulated.

Additionally, here is an example of why a forced selfmate must be at least 4 ply long.

(= 3+4 )


While White can simply win with 1. Qa7#, they go for 1. Qb7+, as it is a selfmate move. But after Black’s queen recaptures, White has no choice but to complete their selfmate: 1. Qb7+ Qxb7+ 2. Qxb7+ Qxb7#. Thus, it becomes a forced selfmate. This also proves that a forced selfmate must be at least 4 ply long.
.
 
   
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(7) Posted by Joost de Heer [Saturday, May 25, 2019 17:53]; edited by Joost de Heer [19-05-27]

Entry 267 on https://timkr.home.xs4all.nl/chess2/diary_14.htm can be changed really easy to a s#11:

(= 12+14 )

s#11
EDIT oops, 1. Ref4 is a short solution. Black queen on f3 added, should still be legal (white PbxPa, PdxPc, PfxPe and black PgxPh results in 7 white promotions (2 on a8, c8, e8 and one on g8), 5 black promotions (2 on h1 and 1 on b1, d1, f1)).
 
   
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(8) Posted by Rewan Demontay (Real Name: J. Malcom) [Saturday, May 25, 2019 17:58]; edited by Rewan Demontay (Real Name: J. Malcom) [19-05-25]

I know. I’ve already done that. But the one I’ve presented has a specific key/theme to it.

You’re set up wouldn’t work because of this line: 1. Rf3+ Ne2f4+ 2. Rfxf4+ Ndxf4+ 3. Rexf4+ Nexf4+ 4. Rdxf4+ N6xf4+ 5. Rbxf4+ N5xf4+ 6. Raxf4+ Nhxf4+ 7. Rxf4+ Nxf4+ 8. Rxf4+ Kxf4+9. Qe5+ Rgxe5#. That’s a s#9, not a s#11.

Here is one that does work.

(= 12+13 )


And after the BK captures the last WR, now we have a true s#11 setup, as the game will be extended two moves more than yours.
 
   
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(9) Posted by Joost de Heer [Sunday, May 26, 2019 08:19]

In a selfmate, short variations are just poor defences, and ignored.
 
   
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(10) Posted by Rewan Demontay (Real Name: J. Malcom) [Sunday, May 26, 2019 13:07]

Makes sense. I personally prefer ones without shorter lines myself regardless.
 
   
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(11) Posted by Rewan Demontay (Real Name: J. Malcom) [Sunday, May 26, 2019 15:02]

Hey Nikola,I just noticed your reply, and sorry about that. You are correct that f4+ is the key move. But I would prefer a full solution please!
 
   
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(12) Posted by Nikola Predrag [Sunday, May 26, 2019 15:53]

Rewan, I just tried to guess your intention as questioned by Frank in post No.2.
However, the stipulation is still mysterious and I don't see a solution. After the key, White is not obliged to play for a selfmate(??) so, for instance:
1.f4+ exf3ep. 2.Sh2xf3+ Qh3xf3+ 3.Sh4xf3+ Bxf3+! 4.Kf1! - refutes?!

I'm sorry, that's too much for my guessing-abilities :-)
 
   
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(13) Posted by Rewan Demontay (Real Name: J. Malcom) [Sunday, May 26, 2019 16:35]; edited by Rewan Demontay (Real Name: J. Malcom) [19-05-26]

Crap. Well that’s cooked. That’s a well spotted refutation Nikola! I didn’t see that there!

Here’s a (hopefully!) fixed version. This is legal: 1. a4 d5 2. c4 b5 3. axb5 dxc4 4. e4 e6 5. e5 Bd6 6. exd6 f5 7. Bd3 g5 8. Be4 fxe4 9. Ra3 h5 10. Rg3 Rh6 11. Rg4 hxg4 12. g3 Rh4. 13. gxh4. The pawns on e6 and f2 are the pawns seen in my diagram, and all of the other pawns promote.

The problem is pretty much self-solving once you find the key.

(= 13+13 )

 
   
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(14) Posted by Nikola Predrag [Sunday, May 26, 2019 16:49]; edited by Nikola Predrag [19-05-26]

And what about 6...Bxf3+ 7.Kf1 ?
14 ply 1.Se1-f3+!
 
   
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(15) Posted by Rewan Demontay (Real Name: J. Malcom) [Sunday, May 26, 2019 16:57]; edited by Rewan Demontay (Real Name: J. Malcom) [19-05-26]

BLANK
 
   
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(16) Posted by Rewan Demontay (Real Name: J. Malcom) [Sunday, May 26, 2019 17:35]; edited by Rewan Demontay (Real Name: J. Malcom) [19-05-26]

This is my last resort to restore my glorious 16 ply. Since 7 black queens is way impossible, the last idea that I have is putting the black bishop that I had earlier trimmed off and putting it down on f1 as a light-squared bishop. That way the WK won’t have an escape square for when the BB comes along.

Any cooks?

PG: 1. a4 d5 2. c4 b5 3. axb5 dxc4 4. e4 e6 5. e5 Bd6 6. exd6 f5 7. d4 f4 8. Ra3 h5 9. Rg3 Rh6 10. Rg4 hxg4 11. Be3 Rh3 12. gxh3 fxe3

(= 13+13 )

 
   
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(17) Posted by Olaf Jenkner [Monday, May 27, 2019 23:49]

7 solutions in 7 moves. So I'd think there are at least 6 cookes.
 
   
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(18) Posted by Rewan Demontay (Real Name: J. Malcom) [Tuesday, May 28, 2019 03:52]; edited by Rewan Demontay (Real Name: J. Malcom) [19-05-28]

Those are not even moves under the stipulations, so thus they are are not cooks. The stipulation is for white to selfmate in 16 ply, not 14 ply, which is what would happen if you used a knight on the first move.

They key is here is 1 f4, which forces black to capture en passant. Then white must capture the black pawn with a knight, and the problem solves itself from here. White forces black to mate him, aka a selfmate. Previous versions have been cooked due to White King being able escape to f1 when a black bishop checks. But my last version has f1 filled in, so I don’t see any cooks in it yet,

The joke here is that white goes for a selfmate on their first move, but they cannot changed their minds afterwards, making it a “forced selfmate,” as I coin it.

Also, also, to others, do you think that my cooked version with the black bishop on a8 would make a good “don’t checkmate white” puzzle?
 
   
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(19) Posted by Hauke Reddmann [Tuesday, May 28, 2019 11:49]

Just for the terminology: There is such a thing as a "help-compel mate"
(German: Hilfszwingmatt) which is, how do I formulate it, a help play
to a #1 selfmate. I.e. "mixed" forms of direct and help play exist anyway.
It would be tempting to coin a third form "forced" - i.e., no side can
deviate after the dominoes begin to fall. (Your problem.)

As a table:
Direct form: White aims for goal, Black can't avoid goal
Help form: White and Black aim for goal
Forced form: White and Black can't avoid goal

with possible mixed forms. (If the third form is stipulation from the
very start, it's usually called a "Problem without Words".)

Hauke
 
   
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(20) Posted by Rewan Demontay (Real Name: J. Malcom) [Tuesday, May 28, 2019 13:38]

Thanks for that bag of knowledge Hauke! Also, do you think that my newset 16 ply is solid?
 
   
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MatPlus.Net Forum Selfmates A Forced Selfmate In 16 Ply