Given the following Chess piece relative values and such that both players cooperate, what is the fastest way such that White first has an advantage of at least +10 piece value then secondly, Blacks get the advantage of +10. The latter means that Black gains at least 20 points of value at the end.

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For instance, at first, Black must have captured a Queen and a Pawn with nothing for White, then White captures Black's Queen and a Pawn as well as both Black's rooks.


3 Answers 3


6 Moves

1. d4 b5 2. Bg5 b4 3. Bxe7 b3 4. Bxd8 (+10 w) bxc2 5. Kd2 cxd1=Q+ 6. Ke3 Kxd8 (+11 b)

  • $\begingroup$ Good idea, the promotion! $\endgroup$
    – JKHA
    Commented Jul 19, 2022 at 16:07
  • $\begingroup$ @JKHA one can actually prove that without promotion one cannot do better than 6 full moves. $\endgroup$
    – loopy walt
    Commented Jul 19, 2022 at 16:46

One First proposal. We can surely improve a bit...

1. d4 e5 2. g3 Qh4 3. gxh4 b6 4. dxe5 Bb7 5. e3 Bxh1 6. Ba6 Nxa6 7. Nf3 Bxf3 8. b4 Bxd1

After round 4, black lost Queen + 1Pawn = 10 points

After round 8, white lost Queen + 1 rook, one knight and one bishop = 20 points


Well, fastest +10 for white from start is 3 of his moves - he cannot take any piece in his 1st move, then he can get pawn and queen with his dark square bishop. Fastest +20 for black without promotion is grabbing 9+5+3+3 (or perhaps 9+5+5+1) so 4 of his moves at least, but note we are at 2.5 here, meaning there are 6 full moves minimally. This 6 moves is obviously impossible (just do white's 3 moves and see there is no suitable second capture for black), so we are stuck at least with 7 full moves. I haven't found a really working sequence and I suspect it does not exist, though one almost fine is

  1. d4 g5 2. Bxg5 e5 3. Bxd8 exd4 4. Nc3 dxc3 5. Qd2 cxd2+ 6. Kd1 Kxd8 7. Rc1 dxc1 (here we don't need points from promotion, black has enough without, but it still happens as the pawn reached first row)

With promotion points, black can grab white queen and a piece while making a promotion leading to 9+(9-1)+(3 or 5), requiring 2 of his moves - so, the fastest sequence would be mere 4 full moves; but black pawn requires at least 5 moves to promote so it isn't really possible. It is easy to construct a valid position where you can do it all in just 3 full moves (the fastest from start I found was 1. d4 c5 2. b4 c4 3. Qd3 g5 4. Qc3 e6 5. Qb2 Nh6. This is followed by point grabs: 6. Bxg5 c3 7. Bxd8 cxb2 8. Be7 bxa1=Q).

Can we do it in 5? To do it in 5, black needs to move a pawn in all of his 5 moves. This means white grabs queen without black's help and does it on move 4 by bishop, black's move 4 captures one piece and black's move 5 captures the other. Is that doable? Nope, it is not - for a very simple reason. White's moves until 4 were moving the only piece beyond first row, so black couldn't have taken a piece on his 4th move, just pawns.

There are other ways to make it work in 6 though, one possibility is

  1. d4 g5 2. Bxg5 e5 3. Bxd8 e4 4. Qd3 exd3 5. Nd2 dxc2 6. Rb1 cxb1=Q+

The nice bit about this solution is that white is full +10 ahead after 3rd move (including black's move) while black is full +10 ahead after 6th (after Nd1 or Kd2 from white)


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