Plotting linear inequalities in Mathematica

Answer 1

A triplet chosen from your set of inequalities will generally determine a point obtained by solving the corresponding triplet of equations. I believe that you want the convex hull of this set of points. You can generate this like so.

cons = randomCons;  (* Your function *)
eqs = Apply[Equal, List @@@ Subsets[cons, {3}], {2}];
sols = Flatten[{x, y, z} /. Table[Solve[eq, {x, y, z}], {eq, eqs}], 1];
pts = Select[sols, And @@ (NumericQ /@ #) &];
ComputationalGeometry`Methods`ConvexHull3D[pts]

Of course, some triplets might actually be underdetermined and lead to lines or evan a whole plane. Thus the code will issue a complaint in those cases.

This appeared to work in the few random cases that I tried but, as Yaro points out, it doesn't work in all. The following picture will illustrate exactly why:

{p0, p1, p2, 
   p3} = {{1, 0, 0, 0, 0, 0, 0, 0}, {1, 1/2, -(1/2), 0, -(1/2), 0, 
    0, -(1/2)}, {1, 0, 1/2, 1/2, 0, 0, -(1/2), 1/2}, {1, -(1/2), 1/2, 
    0, -(1/2), 0, 0, -(1/2)}};
hadamard = KroneckerProduct @@ Table[{{1, 1}, {1, -1}}, {3}];
invHad = Inverse[hadamard];
vs = Range[8];
m = mm /@ vs;
section = 
  Thread[m -> 
    p0 + {x, y, z}.Orthogonalize[{p1 - p0, p2 - p0, p3 - p0}]];
cons = And @@ Thread[invHad.m >= 0 /. section];
eqs = Apply[Equal, List @@@ Subsets[cons, {3}], {2}];
sols = Flatten[{x, y, z} /. Table[Solve[eq, {x, y, z}], {eq, eqs}], 
    1]; // Quiet
pts = Select[sols, And @@ (NumericQ /@ #) &];
ptPic = Graphics3D[{PointSize[Large], Point[pts]}];
regionPic = 
  RegionPlot3D[cons, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, 
   PlotPoints -> 40];
Show[{regionPic, ptPic}]

Thus, there are points that are ultimately cut off by the plane defined by some other constraint. Here's one (I'm sure terribly inefficient) way to find the ones you want.

regionPts = regionPic[[1, 1]];
nf = Nearest[regionPts];
trimmedPts = Select[pts, Norm[# - nf[#][[1]]] < 0.2 &];
trimmedPtPic = Graphics3D[{PointSize[Large], Point[trimmedPts]}];
Show[{regionPic, trimmedPtPic}]

Thus, you could use the convex hull of trimmedPts. This ultimately depends on the result of RegionPlot and you might need to ramp of the value of PlotPoints to make it more reliable.

Googling about a bit reveals the concept of a feasibility region in linear programming. This seems to be exactly what you're after.

Mark

Answer 2

Here is a small program that seems to do what you want, although I have doubts about my understanding because your math and Mathematica skills seem much better than mine.

Anyway, please point out my errors "nicely" :)

The code is:

rstatic = randomCons;                    (* Call your function *)
randeq = rstatic /. x_ >= y_ -> x == y;  (* make a set of plane equations 
                                            replacing the inequalities by == *)

eqset = Subsets[randeq, {3}];            (* Make all possible subsets of 3 planes *)

                                         (* Now find the vertex candidates
                                            Solving the sets of three equations *)
vertexcandidates =      
    Flatten[Table[Solve[eqset[[i]]], {i, Length[eqset]}], 1];

                                         (* Now select those candidates 
                                            satisfying all the original equations *)          
vertex = Union[Select[vertexcandidates, rstatic /. # &]];

                                         (* Now use an UNDOCUMENTED Mathematica
                                            function to plot the surface *)

gr1 = ComputationalGeometry`Methods`ConvexHull3D[{x, y, z} /. vertex];
                                         (* Your plot follows *)
gr2 = RegionPlot3D[rstatic,
        {x, -3, 3}, {y, -3, 3}, {z, -3, 3},
         PerformanceGoal -> "Quality", PlotPoints -> 50]

Show[gr1,gr2]   (*Show both Graphs superposed *)

The result is:

Downside: the undocumented function is not perfect. When the face is not a triangle, it will show a triangulation:

Edit

There is an option to get rid of the foul triangulation

 ComputationalGeometry`Methods`ConvexHull3D[{x, y, z} /. vertex,
                                Graphics`Mesh`FlatFaces -> False]

does the magic. Sample:

Edit 2

As I mentioned in a comment, here are two sets of degenerate vertex generated by your randomCons

 {{x -> Sqrt[3/5]}, 
  {x -> -Sqrt[(5/3)] + Sqrt[2/3] y}, 
  {x -> -Sqrt[(5/3)], y -> 0}, 
  {y -> -Sqrt[(2/5)], x -> Sqrt[3/5]}, 
  {y -> 4 Sqrt[2/5],  x -> Sqrt[3/5]}
 }

and

{{x -> -Sqrt[(5/3)] + (2 z)/Sqrt[11]}, 
 {x -> Sqrt[3/5], z -> 0}, 
 {x -> -Sqrt[(5/3)], z -> 0}, 
 {x -> -(13/Sqrt[15]), z -> -4 Sqrt[11/15]}, 
 {x -> -(1/Sqrt[15]),  z -> 2 Sqrt[11/15]}, 
 {x -> 17/(3 Sqrt[15]), z -> -((4 Sqrt[11/15])/3)}
}

Still trying to see how to cope gently with those ...

Edit 3

This code is not general enough for the full problem, but eliminates the cylinder degenerancy problem for your sample data generator. I used the fact that the pathogenic cases are always cylinders with their axis paralell to one of the coordinate axis, and then used RegionPlot3D to plot them. I'm not sure if this will be useful for your general case :(.

For[i = 1, i <= 160, i++,
 rstatic = randomCons;
 r[i] = rstatic;
 s1 = Reduce[r[i], {x, y, z}] /. {x -> var1, y -> var2, z -> var3};
 s2 = Union[StringCases[ToString[FullForm[s1]], "var" ~~ DigitCharacter]];

 If [Dimensions@s2 == {3},

  (randeq = rstatic /. x_ >= y_ -> x == y;
   eqset = Subsets[randeq, {3}];
   vertexcandidates = Flatten[Table[Solve[eqset[[i]]], {i, Length[eqset]}], 1];
   vertex = Union[Select[vertexcandidates, rstatic /. # &]];
   a[i] = ComputationalGeometry`Methods`ConvexHull3D[{x, y, z} /. vertex, 
            Graphics`Mesh`FlatFaces -> False, Axes -> False, PlotLabel -> i])
  ,

   a[i] = RegionPlot3D[s1, {var1, -2, 2}, {var2, -2, 2}, {var3, -2, 2},
             Axes -> False, PerformanceGoal -> "Quality", PlotPoints -> 50, 
             PlotLabel -> i, PlotStyle -> Directive[Yellow, Opacity[0.5]], 
             Mesh -> None]
  ];
 ]

GraphicsGrid[Table[{a[i], a[i + 1], a[i + 2]}, {i, 1, 160, 4}]]

Here you can find an image of the generated output, the degenerated cases (all cylinders) are in transparent yellow

HTH!