I've been thinking about the fact that an n×n square grid has as many cells as a triangular lattice with n triangles on each side.
Is there a nice continuous bijection between them? When n is odd, you can at least keep one line of symmetry throughout.
Here's the same thing for a 2×2 grid. The triangle in the middle has to go one way or the other, so you have to break the vertical line of symmetry
@christianp maybe if you rotate the square 45deg so blue is at the top you could adjust it to be symmetric again?
@christianp I'm sorry, but the colours matched perfectly... https://www.reddit.com/r/PoliticalCompassMemes/comments/f2ibgk/the_real_political_compass/?ref=share&ref_source=link
@christianp There are more or less systematic ways to do it, but deciding what property to attempt to preserve along the way is a kicker.
@christianp I think it may be impossible to do 5x5 while preserving at least 1 original adjacency on each element.
@christianp Use the decomposition of n² into 1+3+5+..., which appears naturally in both lattices starting from a vertex.
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