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What is the maximum number of black and white pieces that are involved in a checkmate position? That is, if any piece were removed from the board, the position would not be checkmate because that piece is

  • checking the king,
  • blocking a square where the checked king can move,
  • attacking an unoccupied square where the checked king can move, and/or
  • preventing the king or other pieces from capturing or blocking the checking piece.

Note that both kings must also be involved in the position as described above.

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    $\begingroup$ I'm not sure how the title "Checkmate in Last Move" is relevant to the question being asked. Or am I missing something fundamentally obvious? $\endgroup$
    – happystar
    Commented Mar 8, 2021 at 8:12
  • $\begingroup$ there shall be a legal last move to checkmate $\endgroup$
    – TSLF
    Commented Mar 8, 2021 at 8:24
  • $\begingroup$ This feels like it is trivially 9 pieces total? 8 pieces to block the squares around the king and one to do the checking. The 8 pieces could either be the same colour as the king to block the movement or the opposite colour to cover the square to prevent the king moving there. Or of course any combination of the two. (I guess you could also count the king as a tenth piece since it is part of the checkmate process). Am I missing something? $\endgroup$
    – Chris
    Commented Mar 8, 2021 at 10:10
  • $\begingroup$ From the definition "if any piece was removed from the board, checkmate on last move would not be possible", you can include other pieces required to defend white king from being checked, I suppose? $\endgroup$
    – justhalf
    Commented Mar 8, 2021 at 10:53
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    $\begingroup$ I edited the post to make what I see to be the current requirements clearer. If I misinterpreted anything, let me know and I'll try to add it in $\endgroup$
    – HTM
    Commented Mar 8, 2021 at 21:29

5 Answers 5

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If I understand the question correctly,

20 (7 black, 13 white) which if removed would make gxh8B# impossible (sorry board is rotated).
![enter image description here]

Thanks to @aguy for spotting a mistake (now fixed).

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  • $\begingroup$ I think that your understanding of the question and mine certainly differ. My understanding of the question is that it must be a checkmate position (which this doesn't seem to be unless I am being very dense) and that you cannot remove any pieces and it remain a checkmate position. It is very possible though that you are right and I am wrong (given the other answer also doesn't seem to be a checkmate position).. $\endgroup$
    – Chris
    Commented Mar 8, 2021 at 11:20
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    $\begingroup$ @Chris I'd normally agree with you. I was trying to make sense of OP's mention of "last move", so I basically made a position which allows for one checkmating last move (gxh8Q or gxh8B). $\endgroup$
    – loopy walt
    Commented Mar 8, 2021 at 11:28
  • $\begingroup$ @hexomino Yes, which is why they are not counted: 7 black pieces are the 6 directly surrounding their king and the bishop at h8. $\endgroup$
    – loopy walt
    Commented Mar 8, 2021 at 11:34
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    $\begingroup$ In this interpretation of the puzzle, removing either of the black pawns would still make it checkmate as the rooks are defending that square... $\endgroup$
    – TakingNotes
    Commented Mar 8, 2021 at 12:13
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    $\begingroup$ @justhalf Thanks! Your kind comment persuaded me to make a separate question where you can all play by my rules ;-D $\endgroup$
    – loopy walt
    Commented Mar 9, 2021 at 6:33
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With the blocking being ok , and the white king required to be part of it , my improved answer:

22: All 16 white and 6 black (king and adjacent pieces). Note that this position should be achievable by promoting 12 pawns.

enter image description here

(OLD answer)As I understand it, this should do

13 pieces, including the black king (the knight is redundant). Removing any other piece ether does not allow the mate or gives the king an escape. Note: removing 2,4 ,6, or 8 pieces allows the mate for several pairs, to that might invalidate my answer.

enter image description here

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  • $\begingroup$ As far as I understand the question, there's not only the black knight redundant but also the white king. Since there should be no idle (i.e. redundant pieces), I suggest removing the black knight and one of the pawns and moving the white king to b3/f3 respectively, thus making guarded the square formerly occupied by the removed pawn. $\endgroup$
    – trolley813
    Commented Mar 8, 2021 at 20:24
  • $\begingroup$ I dont think you can have two bishops on the same color. One has to be on black and the other on white. You have them on A1 and A7. That's not valid. Similarly, you haver them on C5 and E5. $\endgroup$
    – Joe Ferndz
    Commented Mar 9, 2021 at 3:36
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    $\begingroup$ @JoeFerndz You can PROMOTE a pawn to the second bishop. It is perfectly valid and allowed within the question's rules. $\endgroup$
    – Rewan Demontay
    Commented Mar 9, 2021 at 23:32
  • $\begingroup$ You can add more $\endgroup$
    – TSLF
    Commented Mar 10, 2021 at 3:14
  • $\begingroup$ My understanding per the rules is that this has many pieces that could be removed and not affect the checkmate. A8, for example. If I'm incorrect and these all count, I'd like to know. $\endgroup$
    – Joel Rondeau
    Commented Mar 11, 2021 at 21:25
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My interpretation of the question is that you want a checkmate position where none of the pieces can be removed and it remain a checkmate position. In that case I present the following:

enter image description here

In this setup the black king is in check by the white knights which cannot be captured and the king has no valid moves due to being blocked by black pieces or the white knight. Each square adjacent to the king is only blocked by one piece so if that piece is removed then the king would have a way to escape. There is no way to check the king without that checking piece also covering one of the squares adjacent to the king so I don't think you can involve any more pieces than this

It is possible to change the balance of white and black pieces if you wanted to have more white pieces and less black. eg you could remove the black piece on C4 and put a white pawn on B3 and the effect will be the same.

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    $\begingroup$ rot13(Lbh pna qb ng yrnfg bar orggre - ercynpr gur q6 ebbx jvgu n ovfubc naq cynpr n juvgr ovfubc ba s7. Ynfg zbir jbhyq or n qbhoyr purpx naq zngr.) $\endgroup$
    – Cloudy7
    Commented Mar 8, 2021 at 19:30
  • $\begingroup$ ohg gur bcra purpxvat ovfubc vf erzbinoyr fb vg jnf erzbirq $\endgroup$
    – TSLF
    Commented Mar 9, 2021 at 5:48
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    $\begingroup$ V guvax nabgure jnl gb vaibyir bar zber cvrpr vf ol univat gur zngr qryvirerq ol n cnja, juvpu arrqf gb or cebgrpgrq sebz gur xvat. Sbe rknzcyr, erzbir gur juvgr xavtug, svyy r6 jvgu nalguvat oynpx (dhrra sbe rknzcyr), naq chg juvgr cnjaf ba R4 naq S3 $\endgroup$
    – Pepper
    Commented Mar 10, 2021 at 7:58
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So given the other answers and the comments on them, I have a solution with:

16 pieces all required for the checkmate, both kings included. It does require promoting 3 pawns, so I made them queens to be more obvious.

The board:

enter image description here

The move:

1. e7xd8#

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  • $\begingroup$ not quite the most number..note that the white king is considered 'not involved' even it must be on board and unchecked for legality $\endgroup$
    – TSLF
    Commented Mar 10, 2021 at 3:48
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4: (can probably be improved by a ton)

enter image description here

Chris in the comments claims to describe an 11-piece solution but I don't fully understand his comment so I hope he posts a solution with his answer. I'll update this answer in the future if I find a better answer (which I may soon)...

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  • $\begingroup$ I think that our interpretations of the question differ - I've added an answer myself with a comment on how I interpreted the question. That having been said I'm not sure I understand how you interpreted the question based on this answer! $\endgroup$
    – Chris
    Commented Mar 8, 2021 at 11:39
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    $\begingroup$ Ah! My interpretation of the question is that white is definitely going to win in this situation. However, if you remove any of the non-king pieces, then it will end up in a draw. $\endgroup$
    – TakingNotes
    Commented Mar 8, 2021 at 11:44

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