I have no idea what the "binary icosahedron" should be; and so there is no standard polymake function for this. Please explain.

As far as saving files from a visualization is concerned, please consult our tutorial.

Here is a simple example (not mathematically interesting):

To show the regular icosahedron you can just use icosahedron()->VISUAL; or icosahedron()->VISUAL_GRAPH; if you only want to see the vertices and the edges. What you will get exactly will depend on the configuration of your setup, the default backend being threejs. To show the four planes additionall...

Sorry, the correct homogenization of your example is the following. $quadric = toTropicalPolynomial("max(2*x,1+x+y,2+2*y,1+y+z,2*z,4+2*w)"); $TQuadric = new Hypersurface<Max>(POLYNOMIAL=>$quadric); $TQuadric->VISUAL; My previous explanation is valid: the input needs to be homogeneous, i.e....

polymake uses homogeneous polynomials throughout. Please replace the first line of your code by See this tutorial and these jupyter notebooks for more examples.

Could you please upload your `$my_dual_sub` and `$my_signs`? One way would be to save the `$S_0` object and share the resulting JSON.

I can't promise, but maybe we can do a bit more here.

I still don't understand. Let's postpone this. Concerning your tertatoid, I think I know a way, modulo some experimenting. Here are the steps. (1) Construct the regular tetrahedron as the convex hull of every other vertex of the cube [-1,1]^3. Call this polytope T. (2) From each edge of T construct ...

What is the "perpendicular bisector" of an edge? Do you mean the affine hyperplane which is perpendicular to a given edge, passing through the midpoint of that edge? Your "tetartoid" is given without coordinates, and from the description given on that web page I cannot deduce how...

Of course, we will fix those errors in polymake. However, your specific computation can be rescued as follows: polytope > $C = new Cone(INEQUALITIES =>[[1,0,0,0,0,0,0,0,0,0,0,0],[0,1,0,0,0,0,0,0,0,0,0,0],[0,0,1,0,0,0,0,0,0,0,0,0],[0,0,0,1,0,0,0,00,0,0,0],[0,0,0,0,1,0,0,0,0,0,0,0],[0,0,0,0,0,1,0,0,0,...

OK, so for fixed dimension d, assuming the resulting polytope to be simplicial (with prob 1) is indeed crucial for the algorithms to run in linear time? This is not what I said. Note that in the expression O(mnd) the parameter m (number of facets) depends on n (number of vertices/input points). Rou...