# Chess Construction Challenge #3-The Inverse Babson

Today's challenge is to construct four positions, with White to move and win, such that:

1. In each position, White's move is to promote to a particular piece, for a position per promotion choice.

2. Black's next best move is to promote to a piece, and Black promotes to the same piece that White does.

3. Therefore, White may not promote with check, but Black can

4. The position must be legal, and FIDE laws of chess apply.

The real challenge is that you must use as few pieces as possible for each position. You must also state both White's and Black's motivation for their promotion.

Good luck!

• congrats on 2.5k rep! Sep 23, 2019 at 13:16
• nice problem but I don't think it's "inverse Babson". Not sure what that would be: rather it's "Babson in baby steps". Jun 29, 2020 at 11:48

For promotion to queen, a solution with four pieces:

Best play is

1. b8=Q a1=Q; White could also play Kb2 to win but that takes a move longer. Black should promote since it forces White to win the piece with 2. Qa8+ or Qa7+, giving the black king the chance to escape (temporarily). Black could also promote to a rook, that would be equivalent; I'm not sure if this disqualifies the position, but it will be hard to find one with less pieces.

For promotion to knight, one with nine pieces:

The solution here is

1. e8=N, threatening mate in one; the only way for Black to postpone this is by a check, which can only be done with 1... h1=N+. White will win anyway with 3. Nc7#.

• Well, technically, Black’s best move is to promote to a queen, so promoting to a rook is irrelevant in this case, Sep 23, 2019 at 13:47
• Practically speaking, yes, but technically both moves are best. That is not allowed for some kind of chess problems but you set the rules :) Sep 23, 2019 at 13:49
• Heh, it's hard enough finding positions where promoting to bishop or rook is the strongest move, I'm not a problemist. I might be able to find positions where it doesn't matter (for the distance to mate) if you promote to a bishop or a queen, but I wouldn't be satisfied with those. Sep 24, 2019 at 19:02
• For practical purposes, Black would play 1... a1N+ in the first position to reach a trickier queen v. knight ending, instead of allowing an easy skewer. Oct 10, 2019 at 5:25
• @Cloudy7 I don't think it really matters; after 2. Kb2 the knight will be captured on the next move already. Oct 10, 2019 at 12:48

Promotions to knights, 8 pieces

White to play and mate in 3.

Solution:

1. e8N! f1N! (any other move allows a faster mate with 2.Ng7#) 2. Kh3! a.l. 3. Ng7#/Nf6#

Nobody has a offered a bishop construction yet. Here is one, though imperfect:

White to move and mate in 3

Solution:

1.b8B a1B 2.Bf4 g3 3.xg3#

It is not great because

Although 1...a1B limits White choice, and it would have prevented the #3 in case of 1.b8Q?, it is not technically "better" than 1...a1Q

The difficulty to build a problem with BB or RR promotions is that

We need two stalemating ideas: one that is avoided by White's underpromotion, and a second one which makes Black's underpromotion better than a Q-promotion and allows him to postpone the mate. Most probably, it will require a #4 stipulation to have both underpromotions on the first move.

Promotion to rook, 5 pieces (note this may not qualify)

FEN: 8/7P/8/8/8/K7/1p1N4/k7 w - - 0 1

The strongest move for White is 1.h7-h8=R (since the queen/bishop promotions stalemates Black, and the knight is obviously not enough (actually, after 1.h7-h8=N one of the best moves for Black is 1...b2-b1=N+, forking White's king and knight so forcing an exchange - kind of same-piece promotion in a side variation)). After that, 1...b2-b1=R is one of the best moves for Black, unfortunately alongside with 1...b2-b1=Q (so I write above that it probably does not qualify). Note that 1...b2-b1=B is worse, since it allows immediate checkmate with 2.Nd2-b3#. After 1...b2-b1=Q(R) follows 2.Nd2-b3+ Q(R)b1xb3+ 3.Ka3xb3 Ka1-b1 4.Rh8-h1# (or 1...b2-b1=N+ 2.Nd2xb1 Ka1xb1 3.Rh8-c8 Kb1-a1 4.Rc8-c1#).

I believe this qualifies:

White, in an attempt to smother-mate Black's king, plays 1. e8N. To save the draw, Black then is forced to respond with a check 1... f8N+. (Note that the white king marching to capture the black pawn does not work due to the risk of stalemate.) After the king moves to a square, let's say g2, the game continues 2. Kg2 Nxf3. White is forced to play 3. Kxf3 and take the draw, else lose against a large pawn mass.
Follow-up after my original thoughts: After some thought, White may be able to play 1. e8Q and try to get the queen to h1 or h2. Black then responds 1... f8Q, and will play next turn one of Qg1+, Qg2+, or Qh1+ to draw.

Promotion to knight ,8 pieces, no other pawns

1.d8N (threatens Qa5 Qb7 (and Nc6), only to be postponed 2 moves by:) e1N 2: Kg4 Ne3/f6 Kh3
alternatives like 1.d8Q, Nab6 will take much more time to win

Seeing as how there are no complete answers yet, here are my best efforts as minimization so far. As @Evargalo touched on in their answer, Black's underpromotion to a rook and bishops can never "truly" be the best in a sense. Also, in reply to @Glorfindel's answer, I found a position in which Black's queen promotion defense is genuinely the best.

Queen Promotion

4 Pieces-This is the proven possible minimum. White can only win by queening, and Black's queening provides the longest possible resistance.

Rook Promotion

15 pieces is my best result so far. This is a mate in 3: 1. e8=R bxa1=R 2. Re4 N~ 3. RxN#

Bishops Promotion

I have found quite a few positions with 14 pieces, but this one is the "lightest" so far in terms of piece value. Once more, this is a mate in 3: 1. e8=B h1=B 2. Ba4 Ka2 3. Bc2#

Knight Promotion:

At 7 pieces, this is likely the minimum due to the need to stop 1. Nh6. Lastly, this is also a mate in 3: 1. e8=N f1=N+ 2. Kg1 ~ 3. g7#