4

Anand and Bobby both are members of their school's chess club. And while cleaning up the games room at the end of the day, they saw a board with the following position:

[FEN "rbR1B1Q1/k1n1nN1p/p1pNpBb1/PpKPRqpP/1PP1PrP1/5p2/5P2/8 w - b6 0 1"]

Then the following conversation happened:

  1. Anand claimed that if it were white to move, white has a mate in 1
  2. Bobby then pointed out that if any non-king piece is removed from the board, there is no longer mate in 1
  3. Anand agreed, and further pointed out that in fact there is exactly one combination of two non-king pieces that he can remove from the original position that would still result in a mate in 1

Can you prove each of the statements above?

Source: myself, inspired by my answer to a puzzling question (spoilers in there for the first two questions)

5
  • Thanks Rewan for the edit! As for the image, I noticed that the FEN board display doesn't show on mobile, so I added the image for mobile users. What do you think?
    – justhalf
    Mar 15 at 4:57
  • Ah, in response to that, the third question is not part of the original puzzle (also, the objective is different)! So I believe this is something new =D (and also for those who haven't seen the puzzle yet, that's why I posted it here in different stack)
    – justhalf
    Mar 15 at 4:57
  • In case 2, there might well be mate in 1 after removing a piece, because black's last move could still have been b5. I suggest "proven mate in 1".
    – TonyK
    Mar 15 at 13:47
  • Out of curiosity: How does claim 2 stand if I remove the white pawn on f2?
    – frangge
    Mar 15 at 15:25
  • That's part of the puzzle! See the answer by Remellion below if you want to read an answer.
    – justhalf
    Mar 15 at 15:59
6

The position is 16+15 (16 white units + 15 white units).

Part 1:

Black has no legal retractions except -1...b7-b5, meaning en passant is legal and 1. axb5 ep mates.

Part 2:

Removing anything that is not a king enables a different black retraction (there is one exception), meaning en passant is not legal, and there is no other mate in 1. Aside from obvious removals and retractions, notable cases are:

Remove wPa5; the last move must still be ...b7-b5, but white can't take it en passant.

Remove wPb4; the last move can be ...dxc6 (at some point white played bxc capturing a black pawn).

Remove wPc4; the last move can be ...dxc6 (can be a pawn, for instance) or ...Kb7-a7 (white knight came from c4).

Remove wPf2; the last move can be ...dxe6 (at some point white played fxe capturing a black pawn)

Part 3:

Remove bBb8 and wBe8; 1. Rxc7# is mate.

Removing any unit already invalidates en passant, and so does removing two (there are no interesting interactions here that would keep it legal). So we need to look for a regular mate. The only pieces that can deliver check are the wRc8 and wNd6. The knight can't deliver mate as it is the only thing that could guard b7. So the rook must deliver mate; not on a8 as we can't defend it there with just two removals, so on c7. So we need to remove bBb8 to unguard c7, and wBe8 to guard the b8 square again.

4
  • Nice. Can you give some reasoning why other 2-piece combination doesn't lead to mate in 1? Because Anand's claim is that there is "exactly one".
    – justhalf
    Mar 15 at 5:55
  • 1
    @justhalf sure.
    – Remellion
    Mar 15 at 5:59
  • For the wPb4 and wPf2 case, the white pawn doesn't have to promote in order to be captured by black pawn, right?
    – justhalf
    Mar 15 at 8:59
  • Indeed, it doesn't. I feex.
    – Remellion
    Mar 15 at 9:43

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