Any time someone uses the word "any", the word "all" or "some" would be clearer. :-) Jim Propp On Friday, August 4, 2017, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Fun question, Dan!
Part of the fun being trying to figure out whether "any" was existential or universal --- this time around, the example saves the day! WFL
P.S. "or" above was exclusive ...
On 8/4/17, Veit Elser <ve10@cornell.edu <javascript:;>> wrote:
On Aug 4, 2017, at 6:57 AM, Veit Elser <ve10@cornell.edu <javascript:;>>
wrote:
Assuming I understood the puzzle, I found a network of length sqrt(8) based on four edges of the regular octahedron.
You only need three edges, so that brings the length down to 3/sqrt(2).
-Veit
On Aug 3, 2017, at 9:19 PM, Dan Asimov <dasimov@earthlink.net
<javascript:;>> wrote:
The distant planet of Torus is basically the unit cube
C = [0,1]^3
with corresponding points on opposite faces identified:
(0,y,z) ~ (1,y,z), (x,0,z) ~ (x,1,z), (x,y,0) ~ (x,y,1).
Or if you prefer, it's the quotient group of R^3 as an abelian group by the subgroup Z^3. ----- ----- It is desired to create a road network through Torus with two
properties:
I. It must be a connected network,
and
(When the torus is viewed as a cube:)
II. There must be a circuit all around the torus from any point in the plane x = 0 to the corresponding point in x = 1, from any point in the plane y = 0 to the corresponding point in y = 1, and from any point in
z
= 0 to the corresponding point in z = 1. -----
For example, the 3D plus-sign P ("+") — consisting of the 3 segments connecting each face-midpoint of the cube to the corresponding face-midpoint on the opposite face — will do the job. The total length of the plus-sign is 3.
However, it is not the shortest solution. Can you find a network satisfying I. and II., that also is shorter than P ??? How short can it be ???
—Dan
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