Post without words #22

Posted on June 9, 2018 by Brent

About Brent

Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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13 Responses to Post without words #22

  1. DEB JYOTI MITRA says:
    June 9, 2018 at 6:01 pm

    Impressive in deed.

  2. Will says:
    June 9, 2018 at 8:50 pm

    It’s a geometric realization of \mathbf{F}_7. Equivalence classes of cells represent the field elements. Arithmetic in the field can be understood geometrically. For example, translating one cell to the left represents adding 1. Translating one cell along the 60^\circ lines heading northeast and northwest represent adding 4 and 5 respectively. Rotating 60^\circ clockwise about a 0 cell represents multiplication by 5, which is a primitive element of the field, and so on.

    • Brent says:
      June 9, 2018 at 9:06 pm

      Thank you! I knew there had to be some algebra hiding in there but I wasn’t sure what it was. This is actually quite different from how I was thinking about creating the image in the first place.

      • Will says:
        June 15, 2018 at 5:45 am

        I want to add that there’s some interesting finite geometry in this picture. Interpret equivalence classes of cells, that is, field elements, as “points”. Consider two adjacent cells in a row, 0 and 1 for example, and the cell directly below, 3 in this example. The centers of these three cells form a downward-pointing triangle. Call a set of three cells in such a configuration a “line”. There are seven “lines” in this figure: {0,1,3}, {1,2,4}, {2,3,5}, {3,4,6}, {0,4,5}, {1,5,6}, and {0,2,6}. By definition, every line contains three points, but since a given point can be in any of the three positions of a downward-pointing triangle, each point is contained in three lines. Furthermore, every cell has six neighbors, and with each of those it forms a distinct “line”. Hence two “points” determine a unique “line”. In addition, any two downward-pointing triangles share one cell, and therefore any two “lines” intersect in a unique “point”. We conclude that our system of equivalence classes of cells and downward-pointing triangles forms a finite projective plane of order 2.

        Once also gets a finite projective plane of order 2 using upward-pointing triangles.

      • Will says:
        June 15, 2018 at 7:03 am

        A second additional remark: if one uses a square tessellation instead of a hexagonal one, and labels consecutive squares in every row with the repeating pattern …, 4, 3, 2, 1, 0, 4, 3, 2, 1, 0, …, with 2 lying directly above 0 in the row above, then one gets a geometric realization of \mathbf{F}_5. (In the arithmetic of \mathbf{F}_5, multiplication by the primitive element 2 corresponds to 90^\circ clockwise rotation.)

        The finite fields \mathbf{F}_5 and \mathbf{F})7 are the only two that are realized in this way as periodic tessellations of the plane. One could say that they live on the torus. In contrast, \mathbf{F}_4 lives on the sphere and corresponds to the tetrahedron. Finite fields \mathbf{F}_q with q\ge 8 correspond to Platonic tessellations of the hyperbolic plane.

  3. Denis says:
    June 11, 2018 at 1:40 pm

    It’s a good step toward the chromatic number proof but I think there’s something missing. I’m tempted to blow off work for a bit this afternoon to provide a “response without words”. (Also this doesn’t seem to be tagged as a PWW.)

      • Brent says:
        June 11, 2018 at 3:17 pm

        Nice! And I completely agree re: something missing; I wasn’t intending for this PWW in and of itself to be a chromatic number proof. Your image gets closer but I think there’s still a lot missing. In particular, how do we know that this pattern can tile the whole plane? How do we know that every color is “the same” in the sense of there being a translation that will send any color to any other? “Just look at it, it’s obvious” somehow doesn’t sit well with me in this case.

        • Denis says:
          June 11, 2018 at 3:31 pm

          Thanks for the kind words 🙂 I’m not sure what the best method of making it clear is. I feel like your image is a clearly repeating pattern and method to fully color the plane but I’m less connected to academic math so I don’t know what the proof requirements are.

          (BTW: shared just my image with some friends as a puzzle and they got to “Something about the number of distinct colors needed so you can’t poke another white region with your 10-foot pole.” within 15 minutes. I then explained and pointed them to your blog.)

          • Brent says:
            June 12, 2018 at 10:16 pm

            I suppose you’re right about the repeating pattern being convincing, especially if you show as many repetitions as I did (most of the pictures I see around the web relating to the Hadwiger-Nelson problem show just a few hexagons, hardly enough to get a general sense for the pattern). But for me this pattern raises more questions that I want to understand: is it possible to do something similar with other numbers of colors? What is special about 7 here? How would you communicate the pattern in a precise way to someone else (such as a computer program for creating a drawing!) without showing them a picture?

            You have clearly found good friends if they are willing to think about a mathematical puzzle for 15 minutes. =)

  4. Pingback: Some words on PWW #22 | The Math Less Traveled

  5. Trevor Hawkes says:
    June 28, 2018 at 1:17 pm

    I may have missed it, but was it specified how to colour the boundaries between the hexagons?

    • Brent says:
      June 28, 2018 at 3:14 pm

      Ah, very good question! No, I didn’t specify. But if you’re considering it in a purely combinatorial sense it doesn’t matter. And if you want to use it as an construction showing that the plane can be colored with seven colors, there is plenty of “wiggle room” so it doesn’t matter there either. So it doesn’t really matter what you pick — just use one of the two/three adjacent colors for each boundary point.

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