If Black promoted on h1, then Black could have reached the position with only one bishop move, Bc8-h3.
White could have reached the position with only two bishop moves, Bf1-e2-d1.
In order for that to have happened, White must have given up at least three pieces along the a8-h1 diagonal, and promoted pawns to get the pieces back.
White could have given up three pieces and two pawns on the diagonal. given that White has two pawns left, and gave up two, that means that three of the remaining four could have been promoted to replace the lost pieces.
So it looks like the minimum number of bishop moves is one for Black and two for White, for a total of three.
I'm sorry. I misunderstood the problem. You're supposed to find the minimum number of moves it takes the bishops to travel on every square on the board.
Originally posted by Bloodnok If Black promoted on h1, then Black could have reached the position with only one bishop move, Bc8-h3.
White could have reached the position with only two bishop moves, Bf1-e2-d1.
In order for that to have happened, White must have given up at least three pieces along the a8-h1 diagonal, and promoted pawns to get the pieces back.
White could have ...[text shortened]... ike the minimum number of bishop moves is one for Black and two for White, for a total of three.
It looks more complicated than that to me - how did the two kings get where they are?
Originally posted by iamatiger Obviously they have to move there... But I can't quite see how the black king gets to where it is without the white bishop moving more than twice.
The original bishop didn't have to move... it could have been captured on its home square with white promoting to a bishop on e8 then moving to d1 via a4.
Retro Problem 1.
01 Jan '12 18:51
Back to Top