Mathematical_Recreations-Kraitchik-2e

Mathematical Recreations
U-4 + 4u-2 + 6 + 4u2 • In the fifth power we must also
omit negative powers greater than the fourth, so we have
5u-3 + IOu-1 + lOu + 4u3 , and so on. The results for the
whole board are given in Figure 145.
The king in chess may move one step in any direction, so
that he may change either his row or column, or both. However,
it is clear that the shortest route to a cell on either
diagonal through his initial position must be along the diagonal.
These two diagonals divide the rest of the board into
four parts such that any shortest route to a cell in one of
these parts must consist of moves in which the row or column
is always changed in the same direction. Hence we may
treat the problem in just one of these quarters of the board
and repeat the results in the other three. If we assume that
he always moves up one row, the alternatives presented to
him at each move are three: either he moves 1 column to the
left or he stays in the same column or he moves 1 column to
the right.
The alternatives may then be represented by
u-1 + 1 + u, and the alternatives resulting from n moves (neglecting
the edges of the board) are given by the terms in the
expansion of (u-1 + 1 + u)". If we use the method of marking
cells we find that each new number is the sum of the three
numbers in the line below that lie immediately below or on
either side (but not including any numbers beyond those in
the diagonals through the initial cell). Figure 146 gives the
table for the first few moves in every direction on an unlimited
board, and Figure 147 gives the table for the White king
on the ordinary board starting from his own cell.
The move of the knight does not permit us to separate the
motion in the horizontal direction from that in the vertical.
Here we may use u for horizontal motion and v for vertical
motion. Then a move 2 columns to the right and 1 row upward
will be represented by u 2 v, and so on. The full set of
alternatives for a single move is thus