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| (1) Posted by James Malcom [Sunday, Jun 25, 2023 04:39] |
Fewest Black pieces requiring all 16 White pieces in a checkmate position
In a checkmate position, a piece can be required for any number of reasons. It is possible to have all 16 from one side. It gets interesting to try to reduce and reduce many pieces the checkmated side needs for all 16 to be needed.
I've done it with 6 Black pieces.
 (= 16+6 )
Without wRc7, there is no legal last move; without wPb2, the position is illegal. All other pieces guard, block, or pin. Promoted pieces are allowed, but I managed without.
Five feels impossible, but I've no proof.
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| (2) Posted by Sarah Hornecker [Sunday, Jun 25, 2023 07:53] |
Can you please explain to me what prevents d5-d6 from being the last move without Rc7?
Also, would this work?
 (= 16+5 )
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| (3) Posted by James Malcom [Sunday, Jun 25, 2023 08:45] |
I misplaced it--have shuffled it back to d7.
Unfortunately, it does not work; wPd4 can be removed.
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| (4) Posted by Sarah Hornecker [Sunday, Jun 25, 2023 10:24] |
My bad, we need to shuffle some pieces around.
 (= 16+5 )
Sc7 is needed for the last move only now.
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| (5) Posted by Joost de Heer [Sunday, Jun 25, 2023 10:48] |
cxd7 can be last move I think?
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| (6) Posted by James Malcom [Sunday, Jun 25, 2023 11:42] |
Thanks Joost.
I believe this remedies that, seeing as how White needs exactly 11 pawns captures leftward.
 (= 16+5 )
Thus, it is also retro economic.
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| (7) Posted by Frank Richter [Sunday, Jun 25, 2023 11:46] |
Hm ... where the black a-Pawn was captured?
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| (8) Posted by James Malcom [Sunday, Jun 25, 2023 11:56] |
The literal edge case.
Even with only 10 needed captures, cxd7 is clearly illegal, requiring 12 captures.
 (= 16+5 )
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| (9) Posted by Andrew Buchanan [Sunday, Jun 25, 2023 13:00] |
Very nice this latest one.
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| (10) Posted by Joaquim Crusats [Sunday, Jun 25, 2023 13:43] |
Does this one work?
 (= 16+4 )
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| (11) Posted by Olaf Jenkner [Sunday, Jun 25, 2023 15:35] |
In a legal position a white king is required.
So there is no reason to let him guard a flight of the black king.
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| (12) Posted by Andrew Buchanan [Sunday, Jun 25, 2023 15:37] |
I think Joaquim's one doesn't need Sh2, because Ba8 covers that square.
Shifted Rc7 to g6, so that last move is unique.
Here's a new version. Removing Kh8 leaves a position no less illegal then removing Rg6.
 (= 16+4 )
EDIT: Oo I see Olaf has just posted along the same lines
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| (13) Posted by James Malcom [Sunday, Jun 25, 2023 15:41] |
I had meant to exactly specify the White king must be actively be in the mate, such that moving (not removing) it would destroy the mate/render others pieces useless.
With and without White king participation sets two different types, therefore.
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| (14) Posted by Joaquim Crusats [Sunday, Jun 25, 2023 15:41] |
It's really pleasant to see that inputs from all posters converge into the last diagram.
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| (15) Posted by Mu-Tsun Tsai [Thursday, Jul 13, 2023 20:14] |
 (= 16+3 )
By Mu-Tsun Tsai and Dream Yeh
Using only 3 black units.
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| (16) Posted by Dream Yeh [Thursday, Jul 13, 2023 20:54] |
(= 16+3 )
This position only uses three black pieces.
The only possible explanation is that the last move was
-1. d5xPe6 e.p. e7-e5
-2. d4-d5
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| (17) Posted by Mu-Tsun Tsai [Friday, Jul 14, 2023 07:44] |
 (= 16+3 )
Quick fixing the position. Previous design leads to a surprising cook: remove both d2 and f2, and retract -1. fxe6ep# e7 -2. Rd4+!
Proof that this position is minimal:
Suppose that S is a subset of the White units such that it forms a legal mate position with the three given Black units. The wK is of course in S, and since mutual check is illegal, wRf1 is also in S. Since only the wBa1 covers e5 and g7 squares, it must be in S.
If wNf2 is not in S, the Black king will then be under double check by the bishop and the rook. This is only possible by the last move being -1. fxe6ep#, but still, White would have no legal retraction two single moves earlier, a contradiction. So wNf2 is in S.
The squares g6, g5, f5, e6, e7 is guarded by wRh6, wPh4, wPg4, wNc7 and wPd6 only, respectively, so these 5 units must be in S. If wPg6 is not in S it will again be an illegal double check, so it is.
Now only wBa1 is still checking the Black king, so in order for this position to be a mating position, wBa4, wPb5, wQd1, wPd2 and wPe2 must all be in S.
Finally, wPe6 must be in S to provide legal retraction.
This proves that S is in fact the entire White units.
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| (18) Posted by Joaquim Crusats [Friday, Jul 14, 2023 09:56] |
Very nice. The sublte cook suggests that the idea of "essential white units" can be suitable for interesting "add white units" problems with good tries: e.g. add the missing white units to get a (16,3) position in which all white units are essential for a legal position with Black in checkmate. The less the white units on the initial board still leading to a unique solution, the better. With more black units on the board, even the addition of all 16 white units may be required.
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| (19) Posted by Mu-Tsun Tsai [Saturday, Jul 15, 2023 04:11] |
I just realized that the cooked version can actually be used as a pretty good puzzle: "White checkmates Black in this position. However, we don't really need all White units to do that (legally). Can you see how?"
Update: Ah, not really, the solution won't be unique then. Of course one can require "remove the fewest number of White units..." but that's a bit ugly IMO.
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| (20) Posted by Joaquim Crusats [Friday, Jul 28, 2023 08:57] |
Mu-Tsun Tsai
Dream Yeh
Mat Plus Forum, Jul 14, 2023
(= 16+3 )
#0 with all white units required
Maybe placing the wK in g3 has some advantages:
 (= 9+3 )
Add pieces to get a position in which all 16 white units are essential for a legal mate to the black king.
The position is quite constrained so one imagines that the same approach will be required. f7 is already doubly guarded, the wPa3 has to be essential, and a unit on the first rank to shield the mating unit from the bQ is also needed. So the wBa1 still seems to be a good idea. Now we need a wR or wQ in f2 to avoid the check to the wK and to make essential yet another white unit to avoid the check to the bK. The good Try here is that solutions with a wRf2 will fail because of the ep capture from the wrong side (after removing the convenient white units so that this works as in the cook reported before). And then all the other officers seem to fall in a unique place with the added difficulty that the wRs need to be placed somewhere.
I wonder if the following solution is correct and whether it is unique or not:
 (= 16+3 )
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