# A drunk knight's tour [duplicate]

This question already has an answer here:

Consider an infinite chess board. A knight moves 2 squares forward on one direction, then turn left or right, move 1 square further on. Let's denote this a normal knight, or $$\langle 2,1\rangle$$ knight. And it's known that a knight could reach any square on the board, i.e. denoting the starting square as $$(0,0)$$, a $$\langle 2,1\rangle$$ knight can arrive any square $$(m,n)$$.

Now let's say this knight gets drunk and started to walk $$\langle p,q\rangle$$...

Of course, if both $$p$$ and $$q$$ are even numbers, a $$\langle p,q\rangle$$ knight won't arrive $$(m,n)$$ if both $$m$$ and $$n$$ are odd numbers.

Then brings to the question, in which situation, a drunk $$\langle p,q\rangle$$ knight still could reach any square $$(m,n)$$ ?

## marked as duplicate by user10354138, Ingix, YuiTo Cheng, Yanior Weg, PeterJun 22 at 8:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.