Re: [math-fun] Shortest boundary of tiles in a tessellation of the plane
Yep, you nailed it. I'm guessing Allan did, too, since his 4-digit figure of 3.8637+ agrees with what I got (√2 + √6 = (3.863703305156273+) for the boundary length, using the same non-regular hexagon you mention. However, I don't have a proof handy just now. —Dan William Somsky wrote: ----- Hexagons (not rotationally symmetric) with 120 degree interior angles and a-c = 1 unit, b-c = 1 unit, a-c perpendicular b-c a___ / \ / \c \ / \b____/ On Sun, Aug 9, 2020, 15:30 Allan Wechsler <acwacw@gmail.com> wrote:
A minimum must exist, very roughly, because each tile has unit area, and the minimum circumference for a thingy of unit area is 2 * sqrt(pi), which is 3.5449+.
The circumference achieved by a square tiling is 4. I have a trick to get as low as 3.8637+, and I have a strong feeling that it's the best trick, but am nowhere near a proof.
On Sun, Aug 9, 2020 at 6:03 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
Nit-picking:
"each tile to a tile" perhaps?
Unclear what deformations are implied by "adjust"!
Finally, (why) must such a minimum exist?
WFL
On 8/9/20, Dan Asimov <dasimov@earthlink.net> wrote:
Suppose we tessellate the plane by the unit squares with vertices at the integer points. We'd now like to adjust this tessellation so that
1) The boundary of each tile is a simple closed polygon;
2) The set of translations that carry each tile to another tile is exactly the integer vectors (K,L) ∊ Z^2.
Puzzle: Which tile shape satisfying 1) and 2) is such that each tile's boundary polygon has the shortest possible length?
—Dan
The GLB must exist, but showing it's achieved (rather than approached as a limiting value) needs something like the topological space of tiles being compact. I've no idea if this is true. Rich --- Quoting Dan Asimov <dasimov@earthlink.net>:
Yep, you nailed it. I'm guessing Allan did, too, since his 4-digit figure of 3.8637+ agrees with what I got (?2 + ?6 = (3.863703305156273+) for the boundary length, using the same non-regular hexagon you mention.
However, I don't have a proof handy just now.
?Dan
William Somsky wrote: ----- Hexagons (not rotationally symmetric) with 120 degree interior angles and a-c = 1 unit, b-c = 1 unit, a-c perpendicular b-c a___ / \ / \c \ / \b____/
On Sun, Aug 9, 2020, 15:30 Allan Wechsler <acwacw@gmail.com> wrote:
A minimum must exist, very roughly, because each tile has unit area, and the minimum circumference for a thingy of unit area is 2 * sqrt(pi), which is 3.5449+.
The circumference achieved by a square tiling is 4. I have a trick to get as low as 3.8637+, and I have a strong feeling that it's the best trick, but am nowhere near a proof.
On Sun, Aug 9, 2020 at 6:03 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
Nit-picking:
"each tile to a tile" perhaps?
Unclear what deformations are implied by "adjust"!
Finally, (why) must such a minimum exist?
WFL
On 8/9/20, Dan Asimov <dasimov@earthlink.net> wrote:
Suppose we tessellate the plane by the unit squares with vertices at the integer points. We'd now like to adjust this tessellation so that
1) The boundary of each tile is a simple closed polygon;
2) The set of translations that carry each tile to another tile is exactly the integer vectors (K,L) ? Z^2.
Puzzle: Which tile shape satisfying 1) and 2) is such that each tile's boundary polygon has the shortest possible length?
?Dan
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Equivalently, you can view this problem as drawing a minimum-length finite arrangement of straight lines on the torus R^2 / Z^2 such that removing the lines from the ambient torus results in a simply-connected space. A perturbation argument shows that every vertex where multiple lines meet must necessarily be a degree-3 vertex with three 120-degree angles. Specifically, if you have a vertex A with two outgoing edges that subtend an angle of less than 120 degrees, then let B and C be points on those edges. Then the line segments AB and AC can together be replaced by segments AD, BD, and CD, where D is the Fermat point of the triangle ABC... https://en.wikipedia.org/wiki/Fermat_point ...which necessarily reduces the total length. Consequently, an optimal solution cannot contain any vertices other than degree-3 vertices with three edges meeting at 120 degrees. Returning to the original problem (i.e. thinking of a polygon in the plane instead of a line arrangement on the torus), we can now deduce that every interior angle of the polygon is 120 degrees. So the optimal solution is definitely some equiangular hexagon, and probably the one that William, Dan, and Allen found. -- APG.
Sent: Monday, August 10, 2020 at 1:22 AM From: "Dan Asimov" <dasimov@earthlink.net> To: "math-fun" <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Shortest boundary of tiles in a tessellation of the plane
Yep, you nailed it. I'm guessing Allan did, too, since his 4-digit figure of 3.8637+ agrees with what I got (√2 + √6 = (3.863703305156273+) for the boundary length, using the same non-regular hexagon you mention.
However, I don't have a proof handy just now.
—Dan
William Somsky wrote: ----- Hexagons (not rotationally symmetric) with 120 degree interior angles and a-c = 1 unit, b-c = 1 unit, a-c perpendicular b-c a___ / \ / \c \ / \b____/
On Sun, Aug 9, 2020, 15:30 Allan Wechsler <acwacw@gmail.com> wrote:
A minimum must exist, very roughly, because each tile has unit area, and the minimum circumference for a thingy of unit area is 2 * sqrt(pi), which is 3.5449+.
The circumference achieved by a square tiling is 4. I have a trick to get as low as 3.8637+, and I have a strong feeling that it's the best trick, but am nowhere near a proof.
On Sun, Aug 9, 2020 at 6:03 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
Nit-picking:
"each tile to a tile" perhaps?
Unclear what deformations are implied by "adjust"!
Finally, (why) must such a minimum exist?
WFL
On 8/9/20, Dan Asimov <dasimov@earthlink.net> wrote:
Suppose we tessellate the plane by the unit squares with vertices at the integer points. We'd now like to adjust this tessellation so that
1) The boundary of each tile is a simple closed polygon;
2) The set of translations that carry each tile to another tile is exactly the integer vectors (K,L) ∊ Z^2.
Puzzle: Which tile shape satisfying 1) and 2) is such that each tile's boundary polygon has the shortest possible length?
—Dan
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