Position analysis puzzle

Given the following position, prove black had promoted at least one of his pawns.

White has made 15 P captures. If black did not promote, then all his pawns must have been sacrificed within the a3-a7-e3 triangle as pawns. But it requires 11 black pawn captures to put them all in there and white is only missing 9 units. Proof by contradiction - black promoted something.

white pawns have made 15 captures, black has 16 pieces total, so all of blacks pieces except for the king managed to move the left to get captured, but there are not enough white pieces for the black h pawn to take in order to get to the other side, unless the white g and h pawns took some black pieces, but all the black pieces were already taken by the 6 pawns pn the a rank, so the h pawn must have promoted.

You can't prove it was the h-pawn; you can only prove that black promoted a pawn.

@qiusi, do you realize that your starting position is illegal?

This reminds me of another position:

I don't expect anyone to find the 34-move game including 7 consecutive black captures to produce this though, although finding both mates in 8 is more reasonable.

You're totally right.

Although it's not completely obvious that 11 black pawn captures are needed.

I thought of it this way:

Number the files from right to left, so h=1, g=2,..., a=8.

A black pawn which starts on file x needs y-x white pieces to capture in order to get to a square on file y.

Taking the eight lowest-file squares in the a3-a7-e3 triangle:

4,5,5,6,6,6,7,7

And the eight black pawns:

1,2,3,4,5,6,7,8

No matter which pawn you pair with each square, the sum of the differences will be the same:

4+5+5+6+6+6+7+7 - (1+2+3+4+5+6+7+8) = 10 >9

Which completes the proof.

As you pointed out this is actually being generous and you in fact need 11 captures.

There are lots of 34-move games that produce this, as shown by the spare time white has to kill in this one.