MatPlus.Net

 Website founded by
Milan Velimirović
in 2006

10:09 UTC
ISC 2024
 
  Forum*
 
 
 
 

Username:

Password:

Remember me

 
Forgot your
password?
Click here!
SIGN IN
to create your account if you don't already have one.
CHESS
SOLVING

Tournaments
Rating lists
1-Jan-2024

B P C F





 
 
MatPlus.Net Forum Promenade A little advent puzzle
 
You can only view this page!
(1) Posted by Hauke Reddmann [Friday, Dec 16, 2022 09:43]

A little advent puzzle


Please don't cheat by going to CSE (or writing a program like CSE did).
I'm looking at you, Rewan :-)

Place all 16 (standard) men of White, the black king and some black pawns
on the chessboard. No piece shall be attacked or defended.
Maximize "some". (If this has been done before, which wouldn't surprise
me at all, source please.)
 
(Read Only)pid=24189
(2) Posted by James Malcom [Saturday, Dec 17, 2022 06:20]

I'll leave this one up to the mastes. :-)
 
(Read Only)pid=24191
(3) Posted by seetharaman kalyan [Sunday, Dec 18, 2022 18:48]

Pawns also should be undefended?
 
 
(Read Only)pid=24195
(4) Posted by Hauke Reddmann [Monday, Dec 19, 2022 09:28]

Yes, of course, or I had said "officers".
(This ambiguity of "piece" is a bit annoying;
I don't ever have the problem in German :-)
 
   
(Read Only)pid=24196
(5) Posted by Georgy Evseev [Monday, Dec 19, 2022 18:24]

3 BP, but not feels as maximum
 
   
(Read Only)pid=24199
(6) Posted by Ulrich Voigt [Monday, Dec 19, 2022 23:34]

I have 4 black pawns.
 
   
(Read Only)pid=24200
(7) Posted by Hauke Reddmann [Tuesday, Dec 20, 2022 10:20]

@Ulrich: So does CSE. A proof that 7 pawns can't be surpassed
is obvious by a square counting argument which isn't even that refined,
so methinks this could be a strictly provable optimum.
 
   
(Read Only)pid=24203
(8) Posted by Ulrich Voigt [Tuesday, Dec 20, 2022 12:14]

If you require legal positions, there is a very simple proof that the maximum is 5 black pawns.

Here is a construction with 4 black pawns:

(= 16+5 )


Maybe someone can improve on that...
 
   
(Read Only)pid=24205
(9) Posted by Hauke Reddmann [Wednesday, Dec 21, 2022 14:00]

@Ulrich: Shoot. (The proof. :-) The CSE solution differs
vastly from yours (it surprises you got away with
QBB not all on the border - Qb8 Bg1 Bh1 or so should
be better a priori, because of less squares lost
diagonally).
 
   
(Read Only)pid=24212
(10) Posted by Ulrich Voigt [Wednesday, Dec 21, 2022 18:58]

Are you sure about that? Losing squares on a Bishop or Queen diagonal is a problem only if those squares are not covered by rooks already. Or rather, try to maximize the number of squares that remain available after Queen, Rooks and Bishops:

Qd1, Rb4, Rf2, Ba8, Bg5 (my solution): 18

Qb8, Rc3, Rf5, Bg1, Bh1 (for example): 17

Here is another puzzle that might serve as a basis for solving yours: After placing QRRBB, what is the maximum number of squares that can remain available?
 
   
(Read Only)pid=24213
(11) Posted by Andrew Buchanan [Friday, Dec 23, 2022 00:44]

Is there any virtue in maximizing Black force? E.g. in Ulrich’s example bPa3 might be replacing by bS
 
   
(Read Only)pid=24218
(12) Posted by Hauke Reddmann [Friday, Dec 23, 2022 09:19]

@Andrew: Think that's not so relevant since maximizing count
(not force) was the idea. One could make a different puzzle out of this:

Legal position, maximum material (Q=8,R=5,BS=3,P=1),
no protections or captures. I doubt if 6Q+KK can
be beat then.
 
   
(Read Only)pid=24220
(13) Posted by Andrew Buchanan [Friday, Dec 23, 2022 14:59]

Well I know it's a variant idea. That's why I suggested it. Can't invalidate it just because it's different from your previous concept. Now, I was only suggesting it for the Black force. However your variant's variant of maxing over all force is interesting too. Obviously one can do better than 2K+6Q:

(= 7+2 )
An interesting question is whether a configuration exists for one of the 12 essential solutions to the eight queens problem, under which a *second* vacant square is released by the change of 2 queens to kings. This doesn’t seem to lead to a solution, but possibly there is one where the kings and pawns form a square.

Full white force + 5 Black pawns would be 39+5=44. This new solution would be 54+3=57.

See how fruitful it is to welcome variants, Hauke? :) Happy season's greetings :) :)
 
   
(Read Only)pid=24221
(14) Posted by Hauke Reddmann [Saturday, Dec 24, 2022 10:02]

@Andrew: You know the term "reverse psychology"? I was just too lazy
to find a position where 6Q yield 3 unprotected squares, and you
promptly delivered :P Of course I like variants (e.g. the just published
SCHWALBE construction challenge item B is a close variant to one
we did together here on 16-Oct-09), so Merry Xmas to you too!
 
   
(Read Only)pid=24222
(15) Posted by Andrew Buchanan [Saturday, Dec 24, 2022 10:47]

I’m too lazy to do reverse psychology :-) Chess is easier
 
 
(Read Only)pid=24223

No more posts


MatPlus.Net Forum Promenade A little advent puzzle