OEIS knows the answer…
Yes, and there’s an interesting story behind that entry — more later!
I looked up the OEIS sequence. The relationship to sums of non-negative integers is fascinating!
And I just realized why its true.
Walking along the paths counterclockwise there is an constant number of right turns in each group of diagrams, increasing from group to group.
Indeed! What about the total number of turns, both right & left?
The number of edges increases by 2 each group.
That’s right. Now, how do these two observations relate?
To do a full turn I have to do 4 right turns, plus an equal number of additional left and right turns (assuming we don’t want loops.). So the number of right turns determines the number of left turns.
Took me a couple of looks but since the area and perimeter weren’t constant, there was a clear metric to check next. This would be an interesting challenge to automate, with removal of reflections and rotations and length variations and collision avoidance.
“This would be an interesting challenge to automate” — that’s a very perceptive comment. In fact, the image you see above *was* generated in an automated way, and indeed, it was quite interesting! In a sense, I have been trying (off and on) to figure out how to generate this image for more than a year. I’ll explain more in future posts.
Can’t wait! 🙂
Consider a regular 2n+2 gon with n-1 painted edges where n=1,2,3,4….
The various such painted polygons are represented by your shapes.
It’s just a very difficult game of Tetris 🙂
Start with a square. We bite off one of the corners to get an L shape (see second image). We can aslo bite into an edge to get a C shape (see third image). The corner biting is worth 1 bite and the edge biting is worth 2 bites. For any right angled polygon, edge biting and corner biting is defined the same way. Your shapes are the sets of n-bitten squares where n=0,1,2,3,4.
Pingback: Orthogonal polygons | The Math Less Traveled
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