|(1) Posted by Rewan Demontay (Real Name: James Malcom) [Sunday, Aug 4, 2019 06:39]; edited by Rewan Demontay (Real Name: James Malcom) [19-08-04]|
A Small Idea
I had an idea awhile ago, and after some refinement, I finally nailed it! I'm here to share it today.
It's White to move and win. How many winning moves are there?
(= 16+1 )
|(2) Posted by Jakob Leck [Sunday, Aug 4, 2019 10:55]|
105, all of which include the necessity to travel back in time to where stalemate was actually a win. ;-)
|(3) Posted by Rewan Demontay (Real Name: James Malcom) [Sunday, Aug 4, 2019 11:49]; edited by Rewan Demontay (Real Name: James Malcom) [19-08-04]|
Precisely my friend!
I set out to specifically engineer a legal position, with no promoted pieces, in which no matter what move White plays, Black is always stalemated. The goal was to optimize that number.
105 is what I have found. Even if I decided let it so fhat stalemate was optional, I still don't think that you could be possiblly do better than 105 possible stalemating moves.
|(4) Posted by Olaf Jenkner [Sunday, Aug 4, 2019 12:10]|
This problem is 50 years old and shows 113 moves:
|(5) Posted by Joost de Heer [Sunday, Aug 4, 2019 13:21]; edited by Joost de Heer [19-08-04]|
It's a different task: 113 voluntary stalemates vs 105 forced stalemates.
|(6) Posted by Olaf Jenkner [Sunday, Aug 4, 2019 13:30]|
Yes, 50 years ago there were 109 forced stalemates:
|(7) Posted by Jakob Leck [Sunday, Aug 4, 2019 22:06]|
Wer kann was Dummes, wer was Kluges denken... ;-)
It's no surprise that such a record already exists. 105 moves is pretty close to Dickens' 109 and probably still quite an achievement as you have chosen the form of a black minimal, which might be harder.
What was a surprise to me is the fact that my half-joking answer should be correct. According to current rules and with the given stipulation your problem should just be marked as unsolvable. It should be marked as a record problem/task like the Dickens problem '109 erzwungene Patterhaltungszüge mit Umwandlungszügen' ('109 forced moves keeping the stalemate including promotions'), where actually I think the word 'Patterhaltungszüge' ('moves keeping the stalemate') is ill-chosen because in the diagram position it's not black's turn, so there is no existing stalemate.
|(8) Posted by seetharaman kalyan [Sunday, Aug 4, 2019 23:26]; edited by seetharaman kalyan [19-08-04]|
Why it cant be black to move? because of the stipulation?
|(9) Posted by Nikola Predrag [Monday, Aug 5, 2019 01:46]|
Actually, White is 'stalemated', wK is not attacked and White has no legal move due to the 'dead position' (5.2.2 FIDE Laws) :-)
|(10) Posted by Rewan Demontay (Real Name: James Malcom) [Monday, Aug 5, 2019 03:31]; edited by Rewan Demontay (Real Name: James Malcom) [19-08-05]|
Well Jakob, doesn't a joke answer to joke problem make a complete circle of satisfaction? What goes around, comes around!
Thanks for the congratulations on it being be an achievement! I suppose the correct title for it would be "Black Minimal, 105 Forced Stalemating Moves With Promotions,"
In all honesty, I just felt like introducing it as if it were a throwaway joke!
Also, what a record for most forced and unforced stalemating moves with promoted pieces?
I took one of the positions that has maximum piece movement in a position, and switched two pieces, and did the math.
205 Voluntary Stalemating Moves
(= 16+3 )
Additionally, just because I can, here is my first successful attempt of what eventually became my 105. You clearly the see the improvements that I've made
95 Voluntary Stalemating Moves
(= 16+1 )
In fact, the fact that my 105 ending up being a non-voluntary one was a complete be accident!
|(11) Posted by Geir Sune Tallaksen Østmoe [Monday, Aug 5, 2019 09:04]|
I was going to suggest that the original diagram could be upside-down, but you’ve got that covered too because of the dark-squared bishop.
|(12) Posted by Siegfried Hornecker [Tuesday, Aug 6, 2019 00:40]|
You might want to remove Bh8 in your 205 move problem. It loses two bishop moves but wins 3 moves (1 each by the 2 queens e5 and h4, 1 by Ra8)
|(13) Posted by Rewan Demontay (Real Name: James Malcom) [Tuesday, Aug 6, 2019 01:42]|
Siegfried, that would not work at all, because then if the WQ on e5 leaves than the BP on b2 is no longer pinned.
|(14) Posted by Olaf Jenkner [Tuesday, Aug 6, 2019 02:19]|
Your "records" were outbid 50 years ago, 211 with one piece less instead of 205,
102 with one piece less instead of 95, see
|(15) Posted by Rewan Demontay (Real Name: James Malcom) [Tuesday, Aug 6, 2019 02:57]|
Thanks for that 211!
However, Olaf, the 102 I have already beat. See my original 105 at the top. Below it in the replies is a voluntary 113 and an involuntary 109.
The 95 I showed was a demonstration of what eventually became my 105. Jakob notes that my 105 is an achievement for involuntary, Black minimal.
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