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RSS Feed for Wolfram Community showing any discussions in tag Discrete Mathematics sorted by activeThe inverse of moving from a map to a network topology
http://community.wolfram.com/groups/-/m/t/1358396
Caveat: this may be more of a math problem than a Mathematica-specific problem, but I thought some in this group might have insight.
A solved problem is moving from a map in which there are districts with boundaries to a representation of that map as a network in which the vertices are the districts and edges represent shared boundaries between districts.
fr = UndirectedGraph[
NestGraph[#["BorderingCountries"] &, Entity["Country", "France"],
3]]
But what about the inverse problem: how does one move from a network to a map which is consistent with that network. There are likely to be an infinite number of such maps, but how does one even find a single exemplar. I'm thinking this is actually quite a difficult problem, but perhaps some people here might have insight on the matter and how such an exercise might be tackled using Mathematica.Seth Chandler2018-06-19T13:27:01ZPool Noodle Spikey
http://community.wolfram.com/groups/-/m/t/1356464
I made a [compound of 5 tetrahedra](http://mathworld.wolfram.com/Tetrahedron5-Compound.html). It's surprisingly sturdy.
![noodle spikey][1]
This was built with [pool noodles](https://www.amazon.com/dp/B01BY1S2US/). Noodles are 55" (about 4.5 feet) long with a 2 3/8" diameter. Cut in half these give a Length/Diameter ratio of 11.5789. How close is that to a perfect Length/Diameter ratio?
![pool noodles][2]
First, lets build up a dodecahedron with simple vertices and an edge length of 1.23607 or $\sqrt5-1$.
From that dodecahedron, find the ten tetrahedra with edge length $2 \sqrt2 $ .
Then find five disjoint tetrahedra. The following code works.
tup=Tuples[{1,-1},{3}];
gold=Table[RotateRight[{0, \[Phi], 1/\[Phi]},n],{n,0,2}];
dodec=RootReduce[Union[Join[tup,Flatten[Table[gold[[n]] tup[[m]],{n,1,3},{m,1,8}],1]]]/.\[Phi]-> GoldenRatio];
tetra=FindClique[Graph[#[[1]]\[UndirectedEdge]#[[2]]&/@Select[Subsets[dodec,{2}],Chop[2 Sqrt[2]-EuclideanDistance@@N[#]]==0&]],{4},All];
compounds=FindClique[Graph[#[[1]]\[UndirectedEdge]#[[2]]&/@Select[Subsets[tetra,{2}],Length[Intersection[#[[1]],#[[2]]]]==0&]],{5},All];
Manipulate[Graphics3D[{
Table[{{Yellow,Red,Green,Purple,Blue}[[n]],Tube[#,thickness]&/@Subsets[compounds[[1,n]],{2}]},{n,1,k}],
Table[{{Yellow,Red,Green,Purple,Blue}[[n]],Sphere[#,thickness 2 ]&/@compounds[[1,n]]},{n,1,k}]}, Boxed-> False, SphericalRegion->True,
ViewAngle-> Pi/10, ImageSize-> 650],
Row[{Control@{{k,5, "number shown"},1,5,1, ControlType->Setter },Spacer[15],
Control@{{thickness,.11, "thickness"},.08,.20,.01, Appearance-> "Labeled" }}], SaveDefinitions->True]
![Manipulate 5 tetrahedra][3]
The "perfect" Length/Diameter ratio for rigid tubes seems to be 11.8565. The half-noodle ratio is 11.5789. Since foam is forgiving, I figured that would give a tighter figure, and that turned out to be correct.
For a regular dodecahedron with edge length 1, the inradius and circumradius are 1.11351 and 1.40125.
For an edge length of 1.23607, the inradius and circumradius are 1.37638 and 1.73204.
Based on the sizes of the tetrahedra, we can find the scaling factor of 11.4021.
Height in inches is about 31 inches tall. Distance between vertices is about 14 inches.
The notebook also has a color template. And that's how to build a spikey from pool noodles.
[1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=noodle5tetra.png&userId=21530
[2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=noodles.png&userId=21530
[3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=manipulate5tetra.png&userId=21530Ed Pegg2018-06-15T19:42:04Z