26 May 07
Originally posted by geepamoogleThe black bishops can also mate on g2 and h1. The issue is you have to figure out which placement is legal. The bigger issue is that there's only one answer.
I don't see any response by the checked side with any of the following.
White bishop at c6 or d5
Black bishop at d5 or e4
27 May 07
Originally posted by SwissGambitWhite player cannot be in check if it is black's move.. And since neither side is in check, that means the added bishop has to have moved from a location where it does not attack the king's square.
Why does it need to move from a non-checking diagonal?
Now what I would have to consider is if the final move might have been a promotion, in which case the bishop may have been a pawn before the move, and hence did not check the king..
Comments?
27 May 07
Originally posted by geepamoogleIs it true that the "location where it does not attack the king's square" is always off the diagonal with the B and K?
White player cannot be in check if it is black's move.. And since neither side is in check, that means the added bishop has to have moved from a location where it does not attack the king's square.
Now what I would have to consider is if the final move might have been a promotion, in which case the bishop may have been a pawn before the move, and hence did not check the king..
Comments?
Your 2nd point is important. What about the possibility of +bBh1, with the last move ...g2xh1=B+?
30 May 07
Here is the full solution to the Bishop problem. Jirakon was the only one who came close to solving it. All he had left to do is prove that there are no 'spare' captures.
Only B@d5 is a legal mate. B@e4 means that Black captured last move (how else to nullify the check?)
8 captures are needed to promote 16 Bishops:
If all the pawns on the 5th rank capture something, they can all promote.
The problem is that we always get an even number of dark- and light-square Bishops. White needs to promote exactly 7 light square Bishops, and Black needs exactly 5. Thus, the last two available captures must be used to change the square color of a wB and bB.
01 Jun 07
Originally posted by SwissGambitThis puzzle was way over my head. That was a good one SwissGambit 🙂
Here is the full solution to the Bishop problem. Jirakon was the only one who came close to solving it. All he had left to do is prove that there are no 'spare' captures.
Only B@d5 is a legal mate. B@e4 means that Black captured last move (how else to nullify the check?)
8 captures are needed to promote 16 Bishops:
[fen]rnbqkbnr/1p1p1p1p/8/pPpPp ...[text shortened]... Thus, the last two available captures must be used to change the square color of a wB and bB.
01 Jun 07
Originally posted by JirakonActually, it's Black's move.
Oh. Well in that case, wouldn't everything have to be symmetrical? That makes it White's move, so Qf5#.
Ignoring checks and collisions with enemy pieces, the fastest path to the setup for each side is:
R: 6 moves
N: 6 moves (Nc3-e4-g3 and vice versa)
K: 4 moves (castling makes no difference)
Q: 3 moves (Qh1-e4(d5)-d3
B: 5 moves (one extra move by wBf1 to allow wK/Q to use the long diagonal)
P: 2 moves
----
26 moves
But there's no PG in 26.0. There are too many problems with Q or N checks to the enemy King, and too many Bishop collisions on the long diagonal. It is necessary to make one extra move for White just to avoid checks.
1.b4 g5 2.Nc3 Bg7 3.Rb1 Kf8 4.Rb3 b5 5.Ra3 Nc6 6.Nf3 Rb8 7.Nd4 Rb6 8.Nb3 Bd4 9.g4 Kg7 10.Bg2 Ne5 11.Bc6 Kf6 12.Ne4+ Ke6 13.Kf1 Nf6 14.Rg1 Rg8 15.Rg3 Rg6 16.Rh3 Rh6 17.Kg2 Ng6 18.Kg3 [here's the extra move - it allows the wQ to get in place w/o checking] 18...Nd5 19.Qh1 Ra6 20.Qf3 Nb6 21.Qd3 Qh8 22.Kf3 Qe5 23.Ng3 Qd6 24.Bb2 Bf6 25.Ke3 Bb7 26.Bf3 Bc6 27.Bc3
27...Qf4#