# Infinite Knight's Tour

Does there exists a one-to-one and onto map $f:\mathbb{N}\rightarrow \mathbb{Z}[i]$ such that $|f(n+1)-f(n)|=\sqrt{5}$?

That is to get from $f(n)$ to $f(n+1)$ you have to move like a knight in chess: move two vertically and then one horizontally or two horizontally and then one vertically.

I know there is a lot written about when a knight tour is possible on a finite board. Check out the wiki and wolfram for some background. I can see how you might be able to do this in a type of infinite outward spiral if you could travel down a corridor of an arbitrary length. It seems that there is already some interest in that. But you would also need to make sure that you "end" your journey on the far side of the corridor in such a way that you could turn to travel down the next corridor.

Motivation: The key part of my question is the infinite part. I have recently been thinking about how we can traverse infinite spaces which is particularly relevant to mathematicians (as opposed to say computer scientists). For example when computing something over the entire set of Guassian integers. Let's say that some function $f$ does exist. Then $$\sum_{z\in \mathbb{Z}[i]} \alpha_z =\sum_{n=1}^\infty \alpha_{f(n)}$$

I am not sure that such a thing could practically be done. But my point is that it's valuable to know different ways we can traverse infinite spaces.

• Shouldn't matter where we start: WLOG $f(0)=0$. – Mason Jun 30 '18 at 20:57
• I would approach it by making tiles, say $8 \times 8$. You can make an infinite spiral of the tiles that covers the whole plane. You need to show two tiles, one that enters on one edge at a space you choose and exits on the opposite edge on the corresponding space, and one that exits on a neighboring edge for the times the spiral turns. I chose $8 \times 8$ just because there is a lot of freedom in knight's tours on a board that size so I am guessing you can find the tiles you need. – Ross Millikan Jun 30 '18 at 21:55
• Got an answer? Lost the answer? What happened? This does look pretty convincing: demonstrations.wolfram.com/AnInfiniteKnightsTour – Mason Jun 30 '18 at 22:51
• @Mason, I thought that the construction in the link provides an example, although I am not sure at this moment... – Sangchul Lee Jun 30 '18 at 22:53
• Indeed it seems fine. The graphics that the web page created confused my eyes, but I indeed confirmed that it misses no cells. Sorry for confusions! :) – Sangchul Lee Jun 30 '18 at 23:06

According to this article, it seems that an infinite knight's tour on $\mathbb{Z}^2$ exists:
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