[math-fun] How to make Bill Gosper fall off his chair
Fortunately, I was already lying down. Shouldn't you be able to pack three copies of https://jjj.de/tmp-math-fun/ec3464-curve-tile-BBB-decomp.pdf around its lower left corner, erase the black components, and thereby get a Kochflake from six as well as three of those shapes? Wait, that doesn't add up. You need to put those black things back somehow. --rwg
OK, the curve had its 3-symmetric tile (BBB) seemingly 6-symmetric. The following has BBB obviously 3-symmetric: wild-curve-A-decomp.pdf wild-curve-B-decomp.pdf wild-curve-tile-AAAAAA-decomp.pdf wild-curve-tile-ABAB-decomp.pdf wild-curve-tile-BBB-decomp.pdf 8^) * Bill Gosper <billgosper@gmail.com> [Jul 13. 2018 08:16]:
Fortunately, I was already lying down. Shouldn't you be able to pack three copies of https://jjj.de/tmp-math-fun/ec3464-curve-tile-BBB-decomp.pdf around its lower left corner, erase the black components, and thereby get a Kochflake from six as well as three of those shapes? Wait, that doesn't add up. You need to put those black things back somehow. --rwg _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Jörg, https://jjj.de/tmp-math-fun/ec3464-curve-multi-decomp.pdf contains a self-similar trisection of the Kochflake. How much can you coarsen it? --Bill On Thu, Jul 12, 2018 at 8:34 PM Bill Gosper <billgosper@gmail.com> wrote:
Fortunately, I was already lying down. Shouldn't you be able to pack three copies of https://jjj.de/tmp-math-fun/ec3464-curve-tile-BBB-decomp.pdf around its lower left corner, erase the black components, and thereby get a Kochflake from six as well as three of those shapes? Wait, that doesn't add up. You need to put those black things back somehow. --rwg
Should "coarsen" mean lower iterates, here you go: ec3464-curve-multi-decomp-it1.pdf ec3464-curve-multi-decomp-it2.pdf ec3464-curve-multi-decomp-it3.pdf // and that one is the fourth iterate: ec3464-curve-multi-decomp.pdf Iterate zero is a hexagon. Best regards, jj * Bill Gosper <billgosper@gmail.com> [Jul 14. 2018 08:13]:
Jörg, https://jjj.de/tmp-math-fun/ec3464-curve-multi-decomp.pdf contains a self-similar trisection of the Kochflake. How much can you coarsen it? --Bill
On Thu, Jul 12, 2018 at 8:34 PM Bill Gosper <billgosper@gmail.com> wrote:
Fortunately, I was already lying down. Shouldn't you be able to pack three copies of https://jjj.de/tmp-math-fun/ec3464-curve-tile-BBB-decomp.pdf around its lower left corner, erase the black components, and thereby get a Kochflake from six as well as three of those shapes? Wait, that doesn't add up. You need to put those black things back somehow. --rwg
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On 2018-07-13 23:25, Joerg Arndt wrote:
Should "coarsen" mean lower iterates, here you go: ec3464-curve-multi-decomp-it1.pdf ec3464-curve-multi-decomp-it2.pdf ec3464-curve-multi-decomp-it3.pdf // and that one is the fourth iterate: ec3464-curve-multi-decomp.pdf
Iterate zero is a hexagon.
Best regards, jj No, sorry, I meant fewer fractal pieces. But I was confused . . .
* Bill Gosper <billgosper@gmail.com> [Jul 14. 2018 08:13]:
Jörg, https://jjj.de/tmp-math-fun/ec3464-curve-multi-decomp.pdf contains a self-similar trisection of the Kochflake. Oops, no. You have two different fractal tiles here, in area ratio 3:2. Unless there's an interdissection between them, we still lack the magic trisection: slice up three flakes; reassemble into a bigger flake.
To iterate, Mandelbrot dissects the flake into subflakes, six small and one large, where the large has thrice the area of a small, which fairly begs for an actual trisection into smalls. Redissecting the large produces another small plus six very smalls, 1/3 scale. Areas 6/27 + 7/9 = 1. --Bill
How much can you coarsen it? --Bill
On Thu, Jul 12, 2018 at 8:34 PM Bill Gosper <billgosper@gmail.com> wrote:
Fortunately, I was already lying down. Shouldn't you be able to pack three copies of https://jjj.de/tmp-math-fun/ec3464-curve-tile-BBB-decomp.pdf around its lower left corner, erase the black components, and thereby get a Kochflake from six as well as three of those shapes? Wait, that doesn't add up. You need to put those black things back somehow. --rwg
On Fri, Jul 13, 2018 at 11:52 AM Bill Gosper <billgosper@gmail.com> wrote:
Jörg, https://jjj.de/tmp-math-fun/ec3464-curve-multi-decomp.pdf contains a self-similar trisection of the Kochflake. How much can you coarsen it? --Bill
On Thu, Jul 12, 2018 at 8:34 PM Bill Gosper <billgosper@gmail.com> wrote:
Fortunately, I was already lying down. Shouldn't you be able to pack three copies of https://jjj.de/tmp-math-fun/ec3464-curve-tile-BBB-decomp.pdf around its lower left corner, erase the black components, and thereby get a Kochflake from six as well as three of those shapes? Wait, that doesn't add up. You need to put those black things back somehow. --rwg
Not being sure what exactly to do, here is one A^6 surrounded by 6 of B^3: ec3464-curve-tiles-1-AAAAAA-and-6-BBB.pdf Now that shape might be the same as for A^6 (and B^3 might be that shape as well), I don't know. Even if this is the case, it does not hold for other curves, see wild-curve-tiles-1-AAAAAA-and-6-BBB.pdf This one certainly has a shape different from A^6 and B^3. What I really like about these curves is that instead of one curve being decomposed into smaller copies of itself, now there are two curves decomposed into smaller copies of both curves. See either ec3464-curve-A-decomp.pdf together with ec3464-curve-B-decomp.pdf or wild-curve-A-decomp.pdf together with wild-curve-B-decomp.pdf The grids used before (triangle, square, tri-hex) all had one kind of edge, this one has two. Now guess what I expect to find on grids with k different edges... Hacker question: how do I map the points of the grid to a 2-dim array, preferably using integer coordinates? My best idea so far is to address the points in two steps: address the hexagons by their Eisenstein coordinates, then the six points on them in some (rather arbitrary) way. Another idea would be to distort to the (3^5.6) grid (implicitly dropping some edges there) and use Eisenstein coordinates there, but this feels somewhat of an ugly thing to do. Best regards, jj
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participants (2)
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Bill Gosper -
Joerg Arndt