I hope you will forgive my algebraic notation, but I found it easier
to read (since we know every move is a bishop).
White bishops start at b1 and d1,
Black bishops start at b5 and d5.
09. d3-b5 (board is symmetrical)
This was not the first solution that I found, but I liked it because
it was symmetrical. Once you get to step 9 you are basically
With the close of the club championship (congratulations Graham) I
thinking about organising some coaching sessions over the summer (the
latest phase of the blind leading the wossname).
If people would like this, I would like suggestions, and perhaps even
volunteers, for sessions.
Comments on: Martin Loebbing and Ingo Wegener, The Number of Knight's
Tours Equals 33,439,123,484,294 --- Counting with Binary Decision
Comment by the authors, May 15, 1996:
The number of knight's tours given in the paper is incorrect, since
Once upon a time, somewhere in Eastern Europe, there was a great
famine. People jealously hoarded whatever food they could find,
hiding it even from their friends and neighbors. One day a peddler
drove his wagon into a village, sold a few of his wares, and began
asking questions as if he planned to stay for the night.
[No! No! It was three Russian Soldiers! - Lee
[Wait! I heard it was a Wandering Confessor! - Doug Quinn]
White and Black play with piece arrangement, where each
side's pieces are shuffled separately at random subject to two
(1) because castling is such a big part of the game and
adds so much more to planning ("I'll provoke a2-a3 so they won't
castle queen's-side") the possibility of castling on either side
should be preserved. After that, the pieces are randomly shuffled,
(2) there being bishops of either colour square.