[math-fun] A visual riddle (hard)
The arrangement in the square grid shown below has four-fold rotational symmetry and none of its squares (apart from the central one) has more than two neighbors. Of course there are very many such arrangements. This one, however, has a very special property, unique within all such(*) arrangements of 289 squares. (*) edge-to-edge connected, four fold symmetry, and fewest possible number of neighbors. Which?
................................................... ................................................... ................................................... ................................................... ..........................#####.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..................#########...#.................... ..................#...........########............. ............###...#..................#............. ............#.#...#..........#########............. ............#.#...#..........#..................... ............#.#...#..........#..................... ............#.#...########...#..................... ............#.#..........#...#...#######........... ............#.#..........#...#...#.....#........... ....#########.#..........#...#...#.....#........... ....#.........########...#...#...#.....#........... ....#....................#.......#.....#........... ....#....................#.......#.....#........... ....########.............#.......#.....#........... ...........#.....#################.....#........... ...........#.....#.......#.............########.... ...........#.....#.......#....................#.... ...........#.....#.......#....................#.... ...........#.....#...#...#...########.........#.... ...........#.....#...#...#..........#.#########.... ...........#.....#...#...#..........#.#............ ...........#######...#...#..........#.#............ .....................#...########...#.#............ .....................#..........#...#.#............ .....................#..........#...#.#............ .............#########..........#...#.#............ .............#..................#...###............ .............########...........#.................. ....................#...#########.................. ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#####.......................... ................................................... ................................................... ................................................... ...................................................
Best regards, jj Hints below, after some several blank lines. Hints: A) 289 = 15^2 + 8^2 B) Consider two copies of the arrangement C) Consider 289 copies of the arrangement P.S.: almost-spoiler at http://jjj.de/visual-riddle/visual-riddle.png
OK, I get it now. Though what continues to baffle me is how you came to be looking for such an object in the first place! The version below makes the symmetry more apparent when viewed without proportional spacing (view source, or drag/drop to text file). WFL
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On 11/6/16, Joerg Arndt <arndt@jjj.de> wrote:
The arrangement in the square grid shown below has four-fold rotational symmetry and none of its squares (apart from the central one) has more than two neighbors.
Of course there are very many such arrangements. This one, however, has a very special property, unique within all such(*) arrangements of 289 squares.
(*) edge-to-edge connected, four fold symmetry, and fewest possible number of neighbors.
Which?
................................................... ................................................... ................................................... ................................................... ..........................#####.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..................#########...#.................... ..................#...........########............. ............###...#..................#............. ............#.#...#..........#########............. ............#.#...#..........#..................... ............#.#...#..........#..................... ............#.#...########...#..................... ............#.#..........#...#...#######........... ............#.#..........#...#...#.....#........... ....#########.#..........#...#...#.....#........... ....#.........########...#...#...#.....#........... ....#....................#.......#.....#........... ....#....................#.......#.....#........... ....########.............#.......#.....#........... ...........#.....#################.....#........... ...........#.....#.......#.............########.... ...........#.....#.......#....................#.... ...........#.....#.......#....................#.... ...........#.....#...#...#...########.........#.... ...........#.....#...#...#..........#.#########.... ...........#.....#...#...#..........#.#............ ...........#######...#...#..........#.#............ .....................#...########...#.#............ .....................#..........#...#.#............ .....................#..........#...#.#............ .............#########..........#...#.#............ .............#..................#...###............ .............########...........#.................. ....................#...#########.................. ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#####.......................... ................................................... ................................................... ................................................... ...................................................
Best regards, jj
Hints below, after some several blank lines.
Hints:
A) 289 = 15^2 + 8^2
B) Consider two copies of the arrangement
C) Consider 289 copies of the arrangement
P.S.: almost-spoiler at http://jjj.de/visual-riddle/visual-riddle.png
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I still don't get it. Can someone make a non-ASCII version? Thanks, Jim On Mon, Nov 7, 2016 at 10:08 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
OK, I get it now. Though what continues to baffle me is how you came to be looking for such an object in the first place!
The version below makes the symmetry more apparent when viewed without proportional spacing (view source, or drag/drop to text file).
WFL
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On 11/6/16, Joerg Arndt <arndt@jjj.de> wrote:
The arrangement in the square grid shown below has four-fold rotational symmetry and none of its squares (apart from the central one) has more than two neighbors.
Of course there are very many such arrangements. This one, however, has a very special property, unique within all such(*) arrangements of 289 squares.
(*) edge-to-edge connected, four fold symmetry, and fewest possible number of neighbors.
Which?
................................................... ................................................... ................................................... ................................................... ..........................#####.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..................#########...#.................... ..................#...........########............. ............###...#..................#............. ............#.#...#..........#########............. ............#.#...#..........#..................... ............#.#...#..........#..................... ............#.#...########...#..................... ............#.#..........#...#...#######........... ............#.#..........#...#...#.....#........... ....#########.#..........#...#...#.....#........... ....#.........########...#...#...#.....#........... ....#....................#.......#.....#........... ....#....................#.......#.....#........... ....########.............#.......#.....#........... ...........#.....#################.....#........... ...........#.....#.......#.............########.... ...........#.....#.......#....................#.... ...........#.....#.......#....................#.... ...........#.....#...#...#...########.........#.... ...........#.....#...#...#..........#.#########.... ...........#.....#...#...#..........#.#............ ...........#######...#...#..........#.#............ .....................#...########...#.#............ .....................#..........#...#.#............ .....................#..........#...#.#............ .............#########..........#...#.#............ .............#..................#...###............ .............########...........#.................. ....................#...#########.................. ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#####.......................... ................................................... ................................................... ................................................... ...................................................
Best regards, jj
Hints below, after some several blank lines.
Hints:
A) 289 = 15^2 + 8^2
B) Consider two copies of the arrangement
C) Consider 289 copies of the arrangement
P.S.: almost-spoiler at http://jjj.de/visual-riddle/visual-riddle.png
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_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Thanks Fred. To "get" the 289-omino, it helps to think of the underlying grid as a square grid, and each # as being instead a square that fills the square space available to it. On my Mac I had to copy Fred's depiction to a *text* file, so that it got rid of the vertical line that Mac Mail uses for quoting, and so that each line does not wrap around. —Dan P.S. But what Fraid sed: How did Joerg come to be looking for such a weird tile in the first place? And for that matter: What is known about such tiles in general? Can we assume that the only ways that it tiles the plane are periodic? I seem to recall that a non-periodic connected monotile is knot nown. What about non-periodic monotiles that are *not* connected?
On Nov 7, 2016, at 9:03 AM, James Propp <jamespropp@gmail.com> wrote:
I still don't get it. Can someone make a non-ASCII version?
Thanks,
Jim
On Mon, Nov 7, 2016 at 10:08 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
OK, I get it now. Though what continues to baffle me is how you came to be looking for such an object in the first place!
The version below makes the symmetry more apparent when viewed without proportional spacing (view source, or drag/drop to text file).
WFL
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On 11/6/16, Joerg Arndt <arndt@jjj.de> wrote:
The arrangement in the square grid shown below has four-fold rotational symmetry and none of its squares (apart from the central one) has more than two neighbors.
Of course there are very many such arrangements. This one, however, has a very special property, unique within all such(*) arrangements of 289 squares.
(*) edge-to-edge connected, four fold symmetry, and fewest possible number of neighbors.
Which?
................................................... ................................................... ................................................... ................................................... ..........................#####.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..................#########...#.................... ..................#...........########............. ............###...#..................#............. ............#.#...#..........#########............. ............#.#...#..........#..................... ............#.#...#..........#..................... ............#.#...########...#..................... ............#.#..........#...#...#######........... ............#.#..........#...#...#.....#........... ....#########.#..........#...#...#.....#........... ....#.........########...#...#...#.....#........... ....#....................#.......#.....#........... ....#....................#.......#.....#........... ....########.............#.......#.....#........... ...........#.....#################.....#........... ...........#.....#.......#.............########.... ...........#.....#.......#....................#.... ...........#.....#.......#....................#.... ...........#.....#...#...#...########.........#.... ...........#.....#...#...#..........#.#########.... ...........#.....#...#...#..........#.#............ ...........#######...#...#..........#.#............ .....................#...########...#.#............ .....................#..........#...#.#............ .....................#..........#...#.#............ .............#########..........#...#.#............ .............#..................#...###............ .............########...........#.................. ....................#...#########.................. ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#####.......................... ................................................... ................................................... ................................................... ...................................................
Best regards, jj
Hints below, after some several blank lines.
Hints:
A) 289 = 15^2 + 8^2
B) Consider two copies of the arrangement
C) Consider 289 copies of the arrangement
P.S.: almost-spoiler at http://jjj.de/visual-riddle/visual-riddle.png
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On Mon, Nov 7, 2016 at 4:08 PM, Dan Asimov <asimov@msri.org> wrote:
What about non-periodic monotiles that are *not* connected?
https://en.wikipedia.org/wiki/Socolar%E2%80%93Taylor_tile Andy
Thanks for that link. Fascinating! And discovered only as recently as 2011. —Dan
On Nov 8, 2016, at 8:01 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Mon, Nov 7, 2016 at 4:08 PM, Dan Asimov <asimov@msri.org> wrote:
What about non-periodic monotiles that are *not* connected?
non-ascii version (of the ascii) here: https://flic.kr/p/N3JBzg On Mon, Nov 7, 2016 at 11:03 AM, James Propp <jamespropp@gmail.com> wrote:
I still don't get it. Can someone make a non-ASCII version?
Thanks,
Jim
On Mon, Nov 7, 2016 at 10:08 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
OK, I get it now. Though what continues to baffle me is how you came to be looking for such an object in the first place!
The version below makes the symmetry more apparent when viewed without proportional spacing (view source, or drag/drop to text file).
WFL
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On 11/6/16, Joerg Arndt <arndt@jjj.de> wrote:
The arrangement in the square grid shown below has four-fold rotational symmetry and none of its squares (apart from the central one) has more than two neighbors.
Of course there are very many such arrangements. This one, however, has a very special property, unique within all such(*) arrangements of 289 squares.
(*) edge-to-edge connected, four fold symmetry, and fewest possible number of neighbors.
Which?
................................................... ................................................... ................................................... ................................................... ..........................#####.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..........................#...#.................... ..................#########...#.................... ..................#...........########............. ............###...#..................#............. ............#.#...#..........#########............. ............#.#...#..........#..................... ............#.#...#..........#..................... ............#.#...########...#..................... ............#.#..........#...#...#######........... ............#.#..........#...#...#.....#........... ....#########.#..........#...#...#.....#........... ....#.........########...#...#...#.....#........... ....#....................#.......#.....#........... ....#....................#.......#.....#........... ....########.............#.......#.....#........... ...........#.....#################.....#........... ...........#.....#.......#.............########.... ...........#.....#.......#....................#.... ...........#.....#.......#....................#.... ...........#.....#...#...#...########.........#.... ...........#.....#...#...#..........#.#########.... ...........#.....#...#...#..........#.#............ ...........#######...#...#..........#.#............ .....................#...########...#.#............ .....................#..........#...#.#............ .....................#..........#...#.#............ .............#########..........#...#.#............ .............#..................#...###............ .............########...........#.................. ....................#...#########.................. ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#...#.......................... ....................#####.......................... ................................................... ................................................... ................................................... ...................................................
Best regards, jj
Hints below, after some several blank lines.
Hints:
A) 289 = 15^2 + 8^2
B) Consider two copies of the arrangement
C) Consider 289 copies of the arrangement
P.S.: almost-spoiler at http://jjj.de/visual-riddle/visual-riddle.png
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* James Propp <jamespropp@gmail.com> [Nov 08. 2016 18:45]:
I still don't get it. Can someone make a non-ASCII version?
Will do (not today). Alternative: print two (or more!) of the ASCII pictures on transparencies. These things give a lattice tile: delta-x = 16 and delta-y = 1. No turning is needed, so unequal lengths in x and y do not matter. It's math-fun, I promise! Best regards, jj
Thanks,
Jim
[...]
Done, see http://jjj.de/visual-riddle/ visual-riddle.txt : riddle in ASCII format (essentially my original message) visual-riddle.png : magnified image as requested visual-riddle.pdf : magnified image as requested, as pdf visual-riddle-hint.png : Hint for solution (formerly visual-riddle.png) Enjoy! Best regards, jj * Joerg Arndt <arndt@jjj.de> [Nov 09. 2016 10:32]:
* James Propp <jamespropp@gmail.com> [Nov 08. 2016 18:45]:
I still don't get it. Can someone make a non-ASCII version?
Will do (not today).
Alternative: print two (or more!) of the ASCII pictures on transparencies. These things give a lattice tile: delta-x = 16 and delta-y = 1. No turning is needed, so unequal lengths in x and y do not matter.
It's math-fun, I promise!
Best regards, jj
Thanks,
Jim
[...]
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* Fred Lunnon <fred.lunnon@gmail.com> [Nov 07. 2016 17:26]:
OK, I get it now. Though what continues to baffle me is how you came to be looking for such an object in the first place!
Answer below your message (SPOILER).
The version below makes the symmetry more apparent when viewed without proportional spacing (view source, or drag/drop to text file).
Thanks for that one, hope it encourages other to look harder!
WFL
[...]
It is a tile: the thing tiles the plane as a square does. Any such (4-symmetric) tile corresponds to plane-filling curves on the square grid (this special tile corresponds to just one curve). The search space for curves of order 257 is about 10^70, so way out for direct search (as done before). Looking for the tiles does speed up the search immensely (e.g., by the factor of about 10^7 for not-too-big orders). Now I thought "it's just backtracking, that cannot be hard". It is backtracking, but in 2 dimensions. So I did not yet find an efficient way to obtain all tiles (an avoid duplication). This one is in the class of tiles I currently can find: there is a purely linear way from the center to the out-most points. I computed all those tiles for all orders (and their decompositions into two squares) up to 300. They look (IMO) stunning. Drawback: smaller orders tend to have visual elements of swastikas in them, so I picked one that does NOT look that way. Best regards, jj P.S.: for a numeration system with base (15 + 8*i) take as digits the 257 marked squares, then the generalization of the unit square looks (about) like shown in the image. P.P.S: Is backtracking in 2D really nontrivial, or am I sitting on my brain? I need to find all 4-symmetric edge-to-edge connected arrangements for orders n = 4*k + 1. (Adding the tiling condition is trivial). P.P.P.S: Here is another one, for order 281 = 16^2 + 5^2. As before, the x and y are the shifts for the four adjacent tiles, take (+x, +y), (+y, -x), (-x, -y), and (-y, +x).
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participants (6)
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Andy Latto -
Dan Asimov -
Fred Lunnon -
James Buddenhagen -
James Propp -
Joerg Arndt