I’m still working my way through *A Stitch In Line* but ended up spending rather a lot of time at the start of chapter 10, which discusses “a remarkable image showing Fibonacci snowflakes emerging in the fully packed loop diagram corresponding to a particular 32 × 32 permutation matrix”: http://www.slabbe.org/blogue/2010/10/fibonacci-tiles-can-appear-in-a-fully-packed-loop-diagram/

I didn’t want to stitch the sampler specified for this chapter because I didn’t think it would fit in well with my plan for making a hanging from all my samplers once I’ve finished the book. So I decided to stitch the “remarkable image” instead, but I wanted to understand it first, and this took me on something of a journey! How does a permutation matrix turn into a loop diagram?

The reference given in *A Stitch In Line* is de Gier J, “Fully packed loop models on finite geometries”, in Guttman AJ (Ed.), *Polygons, Polyominoes and Polycubes*, pp. 317–346, Springer; 2009, which is available at https://arxiv.org/abs/0901.3963 — it took me a while to get my head around it, but I think I got it in the end (and will explain in a separate post because of character limit).

Take a permutation matrix, i.e. a matrix where all entries are either 0 or 1, and each column and row have only a single 1 in them.

Make a square lattice with the same number of points in each row and column as the matrix has. The points are referred to as vertices and the lines between them as bonds.

Draw an arrow on each bond of the lattice so (a) each vertex has 2 arrows pointing toward it and 2 pointing away, (b) the vertices corresponding to non-zero entries in your permutation matrix have their “toward” arrows pointing vertically and their “away” arrows pointing horizontally (see de Gier Fig. 4), and (c) at the boundaries of the lattice, all horizonal arrows point into the lattice and all vertical arrows point away.

Look again at Fig. 4; it should be clear that the arrows across a given row or column of the lattice can only change direction at the points corresponding to non-zero entries in the permutation matrix, so (a)–(c) uniquely determine the arrow placement across the lattice.

Now divide the lattice into “even” and “odd” sublattices so the top left vertex is in the even sublattice, its neighbours are in the odd sublattice, their neighbours are in the even sublattice, etc. So e.g. going along the top row, the vertices will be in the even, odd, even, odd, etc etc sublattices, and along the 2nd row, they’ll be in odd, even, odd, even, etc.

Finally!!! Highlight each bond where the arrow points toward a vertex on the even sublattice. This is your loop diagram.

Further thoughts on the pretty picture. While I was stitching it, I was wondering if there were any other permutation matrices that would produce Fibonacci snowflakes. There are indeed — after I finished it, I turned it over and there was the order 3 snowflake in the centre. (The photo below is less clear than the front view because of the knots where I had to join thread, but hopefully it’s clear enough.) This corresponds to taking our original permutation matrix “backwards”, i.e. [17, 18, 20, 21, 14, 11, 24, 10, etc].

I like this because it’s a result that’s trivial to see once you’ve stitched it but is not at all obvious from any other way of rendering it (e.g. drawing it on paper or digitally).

#MathArt #MathsArt #Maths #Math #Embroidery #Sashiko

ALT

@kake
Inspired by your post.
https://mastodon.me.uk/@bbcmicrobot/114569840904150009

@geoffl Oh, very nice! So now we have two ways of making the result obvious :)

I had been pondering writing a renderer in order to make it easier to play around with ideas for finding matrices for the higher-order snowflakes. I have a hunch that the matrices will need to have fourfold symmetry like this one does. I suppose really though the first step should be emailing Sébastien Labbé to find out how they discovered this one.

@kake I'd programned things in BASIC to generate Hitomezashi stitch patterns before and was interested in this variation when I saw you post about it. As it has 4 fold symetry you really only need 1/4 of the probability matrix to genarate it. Unlike the normal matrix where a location only excludes any others in the same row or column this requires each cell chosen in the complete matrix to also have 3 others, as 90° rotations, also selected.

@geoffl Indeed, yes, that’s what I meant by my hunch. However: the snowflakes all have fourfold symmetry but I don’t think it *necessarily* follows that the matrices also have to have it. It might be possible to construct a non-symmetric matrix that has a symmetric snowflake in the middle.

@kake With a bigger grid to stitch across you can have a matrix that has larger offsets on one side than the other without breaking the snowflake (as no points of the probability matrix lay on, or need to lay next to, the boundry of the snowflake) but I can't tell if you can fill the rest of the probability matrix without breaking it. Feels like it should work.

@kake Higher order Fibonacci snowflake.

It's a 4th order snowflake generated using a 68x68 permutation matrix with rotational symmetry.

ALT