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On a chessboard, a piece has a set number of legal moves. It can range from 0 to 27. However, this amount can also restricted. My previous questions have covered n=1 and n=2, it is time for n=3!

Given that:

The entire chess set of 32 pieces is up for grabs.

Construct:

A position in which as many of them as possible have exactly three legal moves.

In the likely advent of a tie, the position containing the shortest proof game wins.

May the best solution be found!

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I'll start the fun:

30 pieces, reachable in 23 1/2 moves.
enter image description here

Proof:

1.b4 g5 2.Bb2 Nf6 3.Nc3 Rg8 4.Nd5 Ne4 5.Bh8 Bg7 6.Rb1 Ba1 7.f4 c5 8.Nf3 Nc6 9.b5 g4 10.bxc6 gxf3 11.Rb6 Rg3 12.Ra6 Rh3 13.g4 b5 14.g5 b4 15.g6 b3 16.Rb6 Rg3 17.Nb4 Ng5 18.Bg2 Bb7 19.d4 e5 20.Qd3 Qc7 21.Kd1 Rc8 22.Qe3 Qd6 23.Rf1 Qb8 24.Qg1 *

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  • $\begingroup$ 32 is possible, but not in a real match. The tie break suggest you look for an achievable position. Do you? $\endgroup$
    – Retudin
    May 22 at 10:32
  • $\begingroup$ @Retudin Yeah, I assumed legal position. Otherwise my guess would be it's too easy, has too many solutions and no tie breaker. $\endgroup$
    – loopy walt
    May 22 at 10:42
  • $\begingroup$ @Retudin Yes, it must be legal, hence my reasoning for the retrograde analysis tag. $\endgroup$ May 22 at 18:48

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