Working On My Knight Moves

There are 8 types of knight moves, corresponding to $(\pm 1, \pm 2 ) , ( \pm 2, \pm1 )$ .

Let's consider how many valid moves there are for the case of $(+1, +2)$ .
We are moving 1 to the right and 2 to the top. Hence, we cannot start in the right most column, nor can we start in the top 2 rows. Given any other starting position, it is clear we can move $(+1, +2)$ . Hence, there are 7 columns $\times$ 6 rows of possible starting positions, which gives us 42 total.

This analysis is the same for each of the 8 moves, and each of them has 42 starting positions. Hence, there are a total of $8 \times 42 = 336$ possible moves that the knight can make.

Note: This analysis generalizes easily to a $n \times m$ board. What is the generalized answer?

Generally, on a $n \times n$ board one can make a translation $(x, y)$ from $(n-|x|)\cdot (n-|y|)$ different positions.

The knight has eight translations as possible moves, with $|x| = 1$ and $|y| = 2$ or vice versa.

Thus there are $8\cdot (n-1)\cdot (n-2)$ possible knight moves on an $n \times n$ board, and with $n = 8$ we get $8\cdot 7\cdot 6 = 336$ .

Note that the $8\times 8$ grid is symmetric about the center and hence, $f(x,y) = f(8-x,y) =f(x,8-y)=f(8-x,8-y)$

Consider a $4\times 4$ grid as shown below

Notice that a knight on the corner ( marked red ) can move to only two squares. A knight on a yellow square can move to 3 squares. A knight on the green squares can move to 4 squares. On the blue, 6 squares, and on the purple, no restriction (all 8 knight moves are possible).

Hence the total will be $4 \times ( 4 \times 8 + 4\times 6 + 5\times 4 + 2 \times 3 + 1 \times 2 ) = 336$

Python 3:

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count = 0
def f(x, y):
    moves = 0
    for dx in (-2,-1,1,2):
        for dy in (-2,-1,1,2):
            if abs(dx) != abs(dy):
                x1, y1 = dx+x, dy+y
                if x1 <= 7 and x1 >= 0 and y1 <= 7 and y1 >= 0:
                    moves += 1
    return moves
for x in range(8):
    for y in range(8):
        count += f(x, y)
print ("Answer:", count)