Here is a pair of chess problems to kick off the new year. Your task is to solve both problems and work out what they have in common.

Left: Both sides cooperate to reach the position shown after White’s tenth move (starting from the usual position)

Right: Both sides cooperate to checkmate Black in two moves. Black moves first.

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EDIT: This is for an on-going competition. If you have a solution, please refrain from posting until after 31st December 2020 midnight (it is probably wise to wait an extra day to account for time-zone differences since the site does not mention the time-zone)

Source: Andrew Buchanan, ChessBase India Christmas-New Year Solving 2020 (Left)

Source: Original (Right)


For the proof of game:

Model solution:

1.f4 Nc6 2.f5 Nb4 3.f6 Nxa2 4.fxg7 Nxc1 5.gxf8=B Nb3 6.Bxe7 Nxd2 7.Bh4 c6 8.Bf2 Qb6 9.e3 Qb3 10.Ne2 *


We'll count permutations later but first let's find actually different moves that can be used.
1. black can move their N through a6 instead of c6
2. white can do fxg7 xf8=B xe7 -h4 -f2 ("geh") or fxg7 xf8=B xe7 -c5 -f2 ("gec") or fxe7 xf8=B xg7 -d4 -f2 ("egd")


The white and black paths do not interfere; we can count them separately and multiply the results.
1. black: we have two submaneuvers: the 6 N moves and c6 Qb6 -b3. Qb3 must come last
1a. with Na6 we have 8-choose-2 = 28 ways of interleaving
1b. with Nc6 this reduces to 6-choose-2 = 15
2. white: also two submaneuvers: e3 Ne2 and the pawn promotion and return to f2
2a. "gec" and "egd": e3 -Ne2 must come last, therefore only 2 variations
2b. "geh" 10-choose-2 = 45 ways of interleaving

Grand total:

(28+15) x (45+1+1) = 2021

Happy new year!


For the helpmate, I think White can only mate by

promoting the g- or h-pawn to queen or rook, and removing the bishop on g8. This can be done in a couple of ways:
- first, Bg8-f7, -e6 or xd5 and then either of the four promotions (12 possibilities)
- first, h7-h8=R or =Q and then Bg8-f7, -e6, xd5 or -h7 (8 possibilities)

Black should not be able to

interpose something on the 8th line, which can be arranged by either
- blocking the c-file with Bb1-a2-c4 or Ng4-e3/e5-c4 (to c6 doesn't help of course) (3 possibilities) - moving both rooks from the c-file; there are seven files to choose from for both rooks, and we can either move Rc3 or Rc2 first, so that's 7x7x2 = 98 possibilities.

Together, this yields

(98 + 3) x (12 + 8) = 2020 possible helpmates.

So given that this is a New Year puzzle, there might be one that I'm missing, and @Sid found the last one:

1. Rc8 Bb8 2. Rxb8 Ra5#

which makes 2021 helpmates, exactly the amount of proof games. So that's what they have in common.

  • 1
    $\begingroup$ I think White can also mate by: ...Rc8 Bb8 Rxb8 Ra5# (if I am not wrong). So, that would make it 21 possibilities $\endgroup$ – Sid Jan 2 at 20:12
  • $\begingroup$ @Sid thanks a lot! $\endgroup$ – Glorfindel Jan 2 at 20:16
  • $\begingroup$ @Glorfindel the answer is correct, but the conventional notation for helpmates is 1. Rc8 Bb8 2. Rxb8 Rxa5# instead of 1 ... Rc8 2. Rb8 Rxb8 3. Rxa5#. Well done :) $\endgroup$ – happystar Jan 2 at 22:07

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