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MatPlus.Net Forum General Forced mutual perpetual
 
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(1) Posted by Hauke Reddmann [Sunday, Sep 26, 2021 21:50]

Forced mutual perpetual


I know there was an article about fairy mutual perpetual check
(where again was the proof that orthodox isn't possible? :-),
but what about the forced version?

A known position, somewhat adapted:
(= 3+3 )

Circe, 1.Rh3+ and so on

I now would be interested if this can be done with (a,b)-riders
and -knights only. Note that the well known camel solution doesn't
work here anymore!
 
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(2) Posted by James Malcom (Rewan Demontay) [Monday, Sep 27, 2021 16:36]

Another classic is easily drained for points. Nightriders, trapped in hell!

(= 5+5 )

 
 
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(3) Posted by Joost de Heer [Monday, Sep 27, 2021 17:10]

1. Kc3 Rd3. Or even simpler: 1. Nd5
 
   
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(4) Posted by Hauke Reddmann [Monday, Sep 27, 2021 19:07]

...which is exactly the point ;-)
With riders-only, the checking riders have to
be pinned, and the pinners also pinned,
probably requiring a large board.
 
   
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(5) Posted by Joost de Heer [Wednesday, Sep 29, 2021 13:32]

https://pdb.dieschwalbe.de/P1281158
 
 
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(6) Posted by seetharaman kalyan [Wednesday, Sep 29, 2021 23:37]

I could not understand (in post 2) why the black rook cannot interpose?
 
   
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(7) Posted by Andrew Buchanan [Thursday, Sep 30, 2021 08:09]

Seetheraman wrote:
 QUOTE 
I could not understand (in post 2) why the black rook cannot interpose?

Yes as others posters noted, that one is cooked. A rare inaccuracy by Rewan.

Meanwhile, I have found a more economical forced perpetual check than post 1, in Circe. It just has 5 pieces, and I believe the loop is unique. (The only variable is what point in the loop one chooses to show in the diagram.) I think the most effective way to test such a claim is to present its construction as a challenge for other folk here. If no-one gets it, I will post my idea in a couple of days.
 
   
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(8) Posted by Jacques Rotenberg [Thursday, Sep 30, 2021 20:27]

Yes Andrew, it seems rather simple in 5 pieces, and nice.

It there a possibility to get a longer loop ?
This question led me to this :
(= 9+8 )
Circe

1.Sh3+ B×a3(Sg1)+ 2.S×a3(Bf8)+ B×h3(Sb1)+ 3.S×h3(Bc8)+
but still... is there a longer loop ?
There should be!
(note that the same knight g1 plays again to h3 only at move 7 (after 6 moves!))
 
   
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(9) Posted by Siegfried Hornecker [Thursday, Sep 30, 2021 20:38]

I composed an endgame study a while ago where a mutual perpetual is forced by check. It was something like White having cannons (Pao in fairy terms) on the entire right side, king on the g-file. Black has rooks on the entire a-file (should have made that d-file, now that I think about it), king on the e-file.

So the kings can wander up and down, but there is no way to escape the perpetual check.

And because I'm a moron, White could checkmate in 1 move in the initial position instead of promoting to a cannon, so the study was incorrect, and I narrowly avoided being banned from feenschach, for being too stupid, by finding it myself and apologizing. But the idea of it fits here.

So here is a corrected version:

(= 9+9 )


= Cannon (Pao)

And if I knew how to make a 5x5 board, that would be the perfect setting probably.
 
 
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(10) Posted by seetharaman kalyan [Thursday, Sep 30, 2021 21:10]

@Jaques
While your idea is fine your setting has a few #1 like Sf3, Sc3/d2. A few wP plugs necessary -- unless we consider a help perpetual !!
 
   
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(11) Posted by Jacques Rotenberg [Thursday, Sep 30, 2021 21:50]

I just wanted to show that such a loop may have a "normal" beginning, put at the beginning the diagram obtained after any of the moves of the loop if you prefer...
 
   
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(12) Posted by Jacques Rotenberg [Thursday, Sep 30, 2021 22:40]

if fact, the stipulation I had in mind was something like "get a loop", but if you say that the stipulation may be "white plays and draw", it seems that you can even get a better position with less pieces.
(= 8+5 )

circe
White plays and draw

1.Sa3+ B×h3(Sb1)+ 2.S×h3(Bc8)+ B×a3(Sg1)+ 3.S×a3(Bf8)+
it seems to work
 
   
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(13) Posted by Joost de Heer [Friday, Oct 1, 2021 08:00]

With a fairy piece, 4 pieces can be achieved:
(= 2+2 )

Circe
= Fers+Alfil ((1,1)+(2,2) leaper)
1. Rh1+ Qd1+ 2. Rd1(+Qd8)+ Qd1(+Rh1)+ etc
 
 
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(14) Posted by Andrew Buchanan [Friday, Oct 1, 2021 09:23]

(= 2+3 )
Circe

Here is what Hauke & I'm pretty sure Jacques spotted. Thanks for confirmation!

What is the most economical "Type B" Circe inexorable mutual perpetual check? "Type B" means that the position does not contain a check, but we are told who has the move? That's quite easy - I have found one with 6 units. And what about "Type A", where there is no check, but we can deduce who has the move? Or "Type D" = Duplex, where the position heads inexorably towards mutual perpetual check whoever is on move?

I like inexorability, so I prefer Jacques' original and beautiful SSSBB loop. I don't have a problem with the couple of extra pawns this implies, and it's nice that the only non-standard material is the thematic S.
 
   
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(15) Posted by Hauke Reddmann [Friday, Oct 1, 2021 10:21]

@Andrew: Forget the PM, hadn't read the preceding post then. And I have the 6-piecer. Really tricky, a Bc4 protecting itself :-)

(= 4+5 )


An alternate possibility to force White's entry into the loop.

(= 7+5 )


A mixture of the above and Jacques idea. (WTM again) Maybe this could be
used for Andrew's task - this *almost* it:

(= 10+11 )


WTM is stalemated, BTM must enter the loop.

Finally, by the fairy mantra, "If in doubt, cheat" :P,

(= 7+13 )


Either Diagram Circe or Fischer starting BQ...any.
 
   
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(16) Posted by Andrew Buchanan [Friday, Oct 1, 2021 18:22]

(= 2+4 )
Circe. White to move.

Here's the 6-piece Type B I had in mind (and I think Hauke guessed a colour-flipped form). It's not unique, the bBc5 could equally, but less interestingly, be on e3, f2 or g1.
 
   
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(17) Posted by Hauke Reddmann [Saturday, Oct 2, 2021 10:33]

Yup. (Had a Qa2 instead of Ra4, in that case the B must be there.)
 
   
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(18) Posted by Andrew Buchanan [Sunday, Oct 3, 2021 11:13]

Here is the "Type D" case:
(= 12+4 )
Circe. Result? Duplex.

Could alternatively have wSf8, but then the *non-checking* Sf8xd7[+Bc8] entry to the true mutual perpetual check loop can be postponed indefinitely, which doesn't seem in the spirit of the "forced mutual perpetual check". Or could have wBf8, but I now feel that standard material is more attractive than strict economy, for this particular challenge.
 
   
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(19) Posted by Jacques Rotenberg [Sunday, Oct 3, 2021 19:49]

@ Andrew

I had not your very nice 5 pieces B/Q.
I thought I had a R/Q feature, but it was a mistake!
 
   
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(20) Posted by Hauke Reddmann [Monday, Oct 4, 2021 09:40]

@Jacques: Had QR also first, also overlooked something...I know I'm telepathic but I didn't know my powers reach THAT far :-)

@Andrew: I see you already edited the post, saving me the one or other nagging. :-)
But one comment: Since the stipulation is "Results", one could troll the solver
a bit: e.g. add wPc7,wSg8,wPa5,bPa6 and a bP tempo. Now White could desperately try to free
the Q before the inevitable happens...
 
   
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MatPlus.Net Forum General Forced mutual perpetual