Answer
There are six base cases, for the cases when i + j ≤ 2.
The knight is already at (0, 0) to start, so the empty sequence of
moves reaches that square.
Work Step by Step
-We use the notation (i, j ) to mean the square in row i and column j and use induction on i +j to show that
every square can be reached by the knight.
--Basis step: There are six base cases, for the cases when i + j ≤ 2.
The knight is already at (0, 0) to start, so the empty sequence of
moves reaches that square.
To reach (1, 0), the knight moves
(0, 0)→(2, 1)→(0, 2)→(1, 0).
Similarly, to reach (0, 1),the knight moves (0, 0) → (1, 2) → (2, 0) → (0, 1).
-Note
that the knight has reached (2, 0) and (0, 2) in the process. For the last basis step there is (0, 0) → (1, 2) → (2, 0) →(0, 1) → (2, 2) → (0, 3) → (1, 1).
-- Inductive step: Assume
the inductive hypothesis, that the knight can reach any square
(i, j ) for which i + j = k, where k is an integer greater
than 1.
We must show how the knight can reach each square (i, j )when i+j=k+ 1.
Because k + 1 ≥ 3, at least one of i and j is at least 2. If i ≥ 2, then by the inductive hypothesis,
there is a sequence of moves ending at (i − 2, j + 1),
because i − 2 + j + 1 = i + j − 1 = k; from there
it is just one step to (i, j ); similarly, if j ≥ 2.