Find the infimum and supremum of the sum of the dihedral angles in a tetrahedron; see http://math.stackexchange.com/questions/310005.

Question310005.java

public class Question310005 {
    final static int pairs [] [] [] = {{{0,1},{2,3}},{{0,2},{1,3}},{{0,3},{1,2}}};

    public static void main (String [] args) {
        int ntrials = Integer.parseInt (args [0]);

        double min = Double.MAX_VALUE;
        double max = 0;

        double [] [] x = new double [4] [3];

        double [] e = new double [3];
        double [] [] t = new double [2] [3];
        double [] n = new double [2];
        double [] s = new double [2];

        for (int m = 0;m < ntrials;m++) {
            for (int i = 0;i < 4;i++)
                for (int j = 0;j < 3;j++)
                    x [i] [j] = Math.random ();
            double sum = 0;
            for (int i = 0;i < 3;i++)
                for (int j = 0;j < 2;j++) {
                    for (int k = 0;k < 3;k++) {
                        e [k] = x [pairs [i] [j] [0]] [k] - x [pairs [i] [j] [1]] [k];
                        for (int l = 0;l < 2;l++)
                            t [l] [k] = x [pairs [i] [j] [0]] [k] - x [pairs [i] [1 - j] [l]] [k];
                    }
                    double dot = 0;
                    s [0] = s [1] = 0;
                    for (int k = 0;k < 3;k++) {
                        for (int l = 0;l < 2;l++) {
                            n [l] = e [(k + 1) % 3] * t [l] [(k + 2) % 3] - e [(k + 2) % 3] * t [l] [(k + 1) % 3];
                            s [l] += n [l] * n [l];
                        }
                        dot += n [0] * n [1];
                    }
                    sum += Math.acos (dot / Math.sqrt (s [0] * s [1]));
                }
            max = Math.max (max,sum);
            min = Math.min (min,sum);
        }

        System.out.println ("minimum : " + min / Math.PI);
        System.out.println ("maximum : " + max / Math.PI);
    }
}