Counting Lattice Points in the $24$ -Cell

Our next example deals with a well-known combinatorial object, the $24$ -cell. Its description is given in the file ``EXAMPLES/24_cell'':

24 5 
2 -1  1 -1 -1
1  0  0 -1  0
2 -1  1 -1  1
2 -1  1  1  1
1  0  0  0  1
1  0  1  0  0
2  1 -1  1 -1
2  1  1 -1  1
2  1  1  1  1
1  1  0  0  0
2  1  1  1 -1
2  1  1 -1 -1
2  1 -1  1  1
2  1 -1 -1  1
2  1 -1 -1 -1
1  0  0  1  0
2 -1  1  1 -1
1  0  0  0 -1
2 -1 -1  1 -1
1  0 -1  0  0
2 -1 -1  1  1
2 -1 -1 -1  1
2 -1 -1 -1 -1
1 -1  0  0  0

Now we invoke the counting function of LattE by typing:

    ./count EXAMPLES/24_cell

The last couple of lines that LattE prints to the screen look as follows:

Total Unimodular Cones: 240
Maximum number of simplicial cones in memory at once: 30

*****  Total number of lattice points: 33  ****

Computation done. 
Time: 0.429686 sec

Therefore, there are exactly $33$ lattice points in the $24$ -cell. We get the same result by using the homogenized Barvinok algorithm:

    ./count homog EXAMPLES/24_cell

The last couple of lines that LattE prints to the screen look as follows:

Memory Save Mode: Taylor Expansion:

****  Total number of lattice points is: 33  ****

Computation done. 
Time: 0.957031 sec

But how many of these $33$ points lie in the interior of the $24$ -cell?

    ./count int EXAMPLES/24_cell

The last couple of lines that LattE prints to the screen look as follows:

Reading .ext file...


*****  Total number of lattice points: 1 ****

Therefore, there is only one of the $33$ lattice points in the $24$ -cell lies in the interior.