That is, b7 must not be captured on its place.
Look at this:
1.b3 g6 2.b4 g5 3.b5 g4 4.b6 g3 5.hxg3 axb6 6.a3 h6 7.a4 h5 8.a5 h4 9.a6 h3 10.a7 h2 11.Rxh2 Rxa7 12.Rh3 Ra6 13.Rh4 Ra5 14.Rh5 Ra4 15.Rh6 Ra3 16.Rh7 Ra2 17.Rxa2 Rxh7 18.Ra3 Rh6 19.Ra4 Rh5 20.Ra5 Rh4 21.Ra6 Rh3 22.Ra7 Rh2 23.Ra8 Rh1
1.b3 g6 2.b4 g5 3.b5 g4 4.b6 g3 5.hxg3 axb6 6.a3 h6 7.a4 h5 8.a5 h4 9.a6 h3 10.a7 h2 11.Rxh2 Rxa7 12.Rh3 Ra6 13.Rh4 Ra5 14.Rh5 Ra4 15.Rh6 Ra3 16.Rh7 Ra2 17.Rxa2 Rxh7 18.Ra3 Rh6 19.Ra4 Rh5 20.Ra5 Rh4 21.Ra6 Rh3 22.Ra7 Rh2 23.Ra8 Rh1
Ok, black b7 itself might capture white c6. But then c2-c6 are only 4 moves (c3, c4, c5, c6). +7 moves of bishop on c1 - is still less then 12. And don't forget about 4 rooks, from which only half will survive.
However, if you have better suggestions I would like to see it.
However, if you have better suggestions I would like to see it.
to #22: 4 pawns and 4 pieces are "dead", giving only 46 half-moves from 8 * 6 = 48
> to #22: 4 pawns and 4 pieces are "dead", giving only 46 half-moves from 8 * 6 = 48
Mmm? It may be counted differently.
If I understand you correctly, you say that maximum is 46+48=94? May be.
Below is solution in 93 moves.
1.b3 g6 2.b4 g5 3.b5 g4 4.b6 g3 5.hxg3 axb6 6.a3 h6 7.a4 h5 8.a5 h4 9.a6 h3 10.a7 h2 11.Rxh2 Rxa7 12.Rh3 Ra6 13.Rh4 Ra5 14.Rh5 Ra4 15.Rh6 Ra3 16.Rh7 Ra2 17.Rxa2 Rxh7 18.Ra3 Rh6 19.Ra4 Rh5 20.Ra5 Rh4 21.Ra6 Rh3 22.Ra7 Rh2 23.Ra8 Rh1 24.c3 f6 25.c4 f5 26.c5 f4 27.c6 f3 28.exf3 dxc6 29.d3 e6 30.Nd2 Ne7 31.Nb3 Ng6 32.Nd4 Ne5 33.Nf5 Nc4 34.Nh6 Na3 35.Bd2 Be7 36.Be3 Bd6 37.Bf4 Bc5 38.Bg5 Bb4+ 39.Ke2 Kd7 40.Bf6 Bc3 41.Be7 Bd2 42.Nf7 Nc2 43.Nh8 Na1 44.g4 b5 45.g5 b4 46.g6 b3 47.g7 b2 48.g8=N b1=N 49.g3 b6 50.g4 b5 51.g5 b4 52.g6 b3 53.g7 b2 54.f4 c5 55.f5 c4 56.f6 c3 57.f7 c2 58.f8=N+ Kc6 59.Kf3 c1=N 60.Qe2 Qd7 61.Kg4 Kb5 62.f3 c6 63.f4 c5 64.f5 c4 65.f6 c3 66.f7 c2 67.Qf3 Qc6 68.Bg2 Bb7 69.d4 e5 70.d5 e4 71.d6 e3 72.d7 e2 73.d8=N e1=N 74.Ne2 Nd7 75.Ng3 Nb6 76.Ne4 Nd5 77.Ng5 Nb4 78.Nge6 Nbd3 79.Nc7+ Kc4 80.Kf5 Nf2 81.Ne8 Nd1 82.Qg4+ Kd3 83.Bf3 Qb5+ 84.Ke6 Bc6 85.Qg5 Qb4 86.Be4+ Ke2 87.Qg6 Bd5+ 88.Kd7 Qb3 89.Kc8 Kf1 90.Bf5 Bc4 91.Be6 Bd3 92.Bd7 Be2 93.Qh7 Qa2 *
Mmm? It may be counted differently.
If I understand you correctly, you say that maximum is 46+48=94? May be.
Below is solution in 93 moves.
1.b3 g6 2.b4 g5 3.b5 g4 4.b6 g3 5.hxg3 axb6 6.a3 h6 7.a4 h5 8.a5 h4 9.a6 h3 10.a7 h2 11.Rxh2 Rxa7 12.Rh3 Ra6 13.Rh4 Ra5 14.Rh5 Ra4 15.Rh6 Ra3 16.Rh7 Ra2 17.Rxa2 Rxh7 18.Ra3 Rh6 19.Ra4 Rh5 20.Ra5 Rh4 21.Ra6 Rh3 22.Ra7 Rh2 23.Ra8 Rh1 24.c3 f6 25.c4 f5 26.c5 f4 27.c6 f3 28.exf3 dxc6 29.d3 e6 30.Nd2 Ne7 31.Nb3 Ng6 32.Nd4 Ne5 33.Nf5 Nc4 34.Nh6 Na3 35.Bd2 Be7 36.Be3 Bd6 37.Bf4 Bc5 38.Bg5 Bb4+ 39.Ke2 Kd7 40.Bf6 Bc3 41.Be7 Bd2 42.Nf7 Nc2 43.Nh8 Na1 44.g4 b5 45.g5 b4 46.g6 b3 47.g7 b2 48.g8=N b1=N 49.g3 b6 50.g4 b5 51.g5 b4 52.g6 b3 53.g7 b2 54.f4 c5 55.f5 c4 56.f6 c3 57.f7 c2 58.f8=N+ Kc6 59.Kf3 c1=N 60.Qe2 Qd7 61.Kg4 Kb5 62.f3 c6 63.f4 c5 64.f5 c4 65.f6 c3 66.f7 c2 67.Qf3 Qc6 68.Bg2 Bb7 69.d4 e5 70.d5 e4 71.d6 e3 72.d7 e2 73.d8=N e1=N 74.Ne2 Nd7 75.Ng3 Nb6 76.Ne4 Nd5 77.Ng5 Nb4 78.Nge6 Nbd3 79.Nc7+ Kc4 80.Kf5 Nf2 81.Ne8 Nd1 82.Qg4+ Kd3 83.Bf3 Qb5+ 84.Ke6 Bc6 85.Qg5 Qb4 86.Be4+ Ke2 87.Qg6 Bd5+ 88.Kd7 Qb3 89.Kc8 Kf1 90.Bf5 Bc4 91.Be6 Bd3 92.Bd7 Be2 93.Qh7 Qa2 *
It is still seems to me, that your logic sometimes is not correct; however, it may be possible that I do not understand you.
It seems that my 93-moves-example is reorganizable into 94-moves-example.
It seems that my 93-moves-example is reorganizable into 94-moves-example.
#25 with 1 extra halfmove http://de.lichess.org/1nS47By3#187
(moving the queen to b8 instead of h7)
full 94 moves http://de.lichess.org/o1YR6os5#188
Black King to g1 to bring the bishop to f1
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