We have our idealised paintbrush $\ [0,1],\ $ and it is living in $\ \mathbb{R}^2.$ We can move it and rotate it, but we cannot lift the brush from the paper. We want to paint a square, let's say $\ [0,9] \times [0,9].$
Painting the square without going over any area more than once is easy if you are - at all- allowed to move the paintbrush in a straight parallel to it's long side. We see this happening when the paintbrush hits the edges of the square:
The paintbrush has not covered any area more than once.
The problem:
If we add the extra condition that the paintbrush is never allowed to move parallel to it's long side, can we paint the square without going over the same area twice?
This question can be extended to any other connected shapes - in particular a circle would be interesting also. But it's one question per question, so let's stick to the square.
At first, I thought the paintbrush would get stuck in a corner, but there is a way to avoid this:
After going round the edges, we end up in the same corner we started in but with the paintbrush in a different position. And the paintbrush can escape from this corner and continue trying to paint the remaining inside of the square. But I'm not sure how to/ if it is possible to paint the square's inside with the rules. Also, it doesn't have to be a $\ 9\times 9\ $ square, and different sized squares might give different answers.
Edit: I'm not sure this works - I think we go over the same area twice, but I'm not certain. The corners do seems to be a problem...
Probably this whole problem can be formalised into a general/geometric topology question, but I'm not sure how to do this.
Has this kind of thing been studied before? Perhaps not if it isn't particularly useful or applicable? But I think it might be interesting from a purely problem-solving perspective.
You can cover a $1$ by $1$ square either by going "straight through", perpendicular to the straight edge of the paintbrush, or you can end up on an adjacent edge of the $1$-by-$1$ square using the following manoeuvre:
So for an $n$ x $m$ rectangle, you can just start on the outside and spiral inwards:
Going "straight through" for the most part, but when changing direction by 90 degrees, using the manoeuvre above, to change to an adjacent edge of the square you are currently on, whilst painting the current square you are on (usually a corner square of what's left).