Question

In chess, the knight can move in eight directions. It can move two squares vertically followed by
one square horizontally. It can also move two squares horizontally followed by one square vertically.
Since the vertical motion can be up or down and the horizontal motion can be left or right, there
are eight squares on a chessboard where a knight can be positioned after one move.

Suppose that we have an n×n chessboard with a knight placed in one of the corners. (A normal chessboard is 8 × 8.) Prove by induction that for n ≥ 4, the knight in the corner can be placed in any square of the n × n chessboard after a finite sequence of legal knight moves. Note that the knight is not allowed to wander off the board and must stay within the confines of the n × n chessboard. For example, the white knight in the figure above can only move to two places after one move. In your proof by induction, you should start with a large chessboard, then consider one or more smaller chessboards (at least 4 × 4) within it. By the induction hypothesis (and symmetry), you can assume that a knight can move from any corner of the smaller board to any position in the smaller board. Use this to show that a knight can move from any corner of the larger board to any position in the larger board. Some ground rules: • You must make good use of induction. • Do not describe the sequence of moves made by the knight directly. • You must not use iteration instead of induction. (I.e., do not say something like, “Continuing in this manner, the knight can reach any position on the board.”

Answer 1

lution(s):

Suppose we have an chessboard. Label the rows as   from bottom to top of the board.. Label the squares of the row by   from left to right. Now, place a knight at the leftmost bottom square .

Base case:

If the knight has been to some square, we will not revisit it, to avoid repitition.

From , the knight can jump to either or .

From , the knight can jump to either , or .

From , the knight can jump to either or .

From , the knight can jump to either or .

From , the knight can jump to either or .

From , the knight can jump to .

From , the knight can jump to either or .

From , the knight can jump to .

Thus, we have visited all the squares in a chessboard. So, the hypothesis is true for .

Induction hypothesis: Assume that the hypothesis is true for some integer .

Induction step: We will prove that the hypothesis is true for chessboard.

We will first show that the knight can visit all the squares of the row.

From , the knight can jump to and then to .

From (where is such that ), the knight can jump to and then to .

By induction, this shows that the knight can visit the squares for odd integers .

Now, from , the knight can jump to and then to .

From , the knight can jump to and then to .

From (where is such that ), the knight can jump to and then to .

By induction, this shows that the knight can visit the squares for even integers

Combining these two, we have that the knight can visit all the squares of the first row.

By symmetry, the knight can visit all the squares of the first column [Flip the board upside down and rotate it left by degrees]

Now, from ., the knight can jump to , then to and then to .

Note that the square in the chessboard is the leftmost bottom square of the sub-chessboard obtained by deleting the row and the column.

By induction hypothesis, the knight can be placed in any square of this sub-chessboard. So, the knight can be placed in any square of the chessboard. That is to say, the hypotheis is true for chessboard.

Hence, by the principle of mathematical induction, the statement is true for all integers .