Consider a game of chess where White delivers checkmate in their $X^{th}$ move. For all the pieces that remain on the board at the end of the game, consider their Manhattan Distances from their initial positions and take the sum. Let's say this value is $Y$.

What is the minimum possible value of $X+Y$? How many different sequences of moves can achieve this value?


3 Answers 3


With the rules clarified by OP I see at least 80 variations achieving a score of


Main branches:

1.a4/c4 b5 2.ab/cb a6/a5/c6/c5 3.ba/bc Bb7 4.ab/cb Qc8 5.bcR#/bcQ#

For a subtotal of 16 variations.

And the branch found by @Sneftel:

1.d4 c5 2.dc d6/d5 3.cd X 4.de Y 5.edQ# where X can be any one of 14 "time wasting moves" taken back by Y in the next move (28 variations total) 1.d4 e5 2.de d6/d5 3.ed X 4.dc Y 5.cdQ# where X can be any one of 18 time wasting moves (36 variations total)

For a subtotal of 64 variations.


There is also the almost solution:

1.f4/h4 g5 2.fg/hg Nf6/h6 3.gf/gh Bg7 4.fg/hg and now, unfortunately, black has to move and spoil it otherwise 5.ghR#/ghQ# would also achieve 5.

  • $\begingroup$ Nice. Seems I highly underestimated the variety of solutions that would work out. +1 $\endgroup$ Nov 6, 2022 at 8:07

Here's a score of 5.

1. d4 e5 2. dxe5 d6 3. exd6 Qe7 4. dxc7 Qd8 5. cxd8=Q#

The trick is,

under the standard rules, promotion of a pawn results in the pawn being removed from the board and a new piece being placed in that location, rather than the pawn "becoming" the new piece (FIDE 3.7e). So the promoted queen is in its initial position, and thus Y=0.

Black has several options to burn tempo on turns 3 and 4. There's probably a couple of other lines that could clear the necessary black material. So I suspect there's quite a few tied sequences, taking this tack.

  • $\begingroup$ Would be a bit hard on tournament hosts asking them to provide actual shape shifting pawns. $\endgroup$
    – loopy walt
    Nov 3, 2022 at 22:15
  • 4
    $\begingroup$ @loopywalt Tell me you don't want to own a pawn with a clever system of gears to unfold into a queen. $\endgroup$
    – Sneftel
    Nov 3, 2022 at 22:20
  • $\begingroup$ Isn't the new queen 7 squares away from its origin? $\endgroup$
    – JLee
    Nov 3, 2022 at 22:34
  • $\begingroup$ @JLee No, see the second bit. $\endgroup$
    – Sneftel
    Nov 3, 2022 at 22:37
  • 1
    $\begingroup$ @ManishKundu If we accept this interpretation of the rules then there are at least 3 different first moves for white. $\endgroup$
    – loopy walt
    Nov 5, 2022 at 21:59

I get 10 from

1 e3 e5 2 Qh5 Ke7 3 Qxe5#. $X=3, Y=1+5+1=7; X+Y=10$.

  • $\begingroup$ Can you prove that this is minimal, and also whether there are any other ways to get the same value or not? $\endgroup$ Nov 2, 2022 at 18:56
  • 3
    $\begingroup$ Variation with the same score: 1. d3 e5 2. d4 exd4 3. Qxd4 Ke7 4. Qe5# $\endgroup$
    – loopy walt
    Nov 2, 2022 at 19:29

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