An interface to Normaliz
PyNormaliz - A python interface to Normaliz
PyNormaliz provides an interface to Normaliz via libNormaliz. It offers the complete functionality of Normaliz, and can be used interactively from python. For a first example, see this introduction by Richard Sieg (Slighty outdated: for the installation follow the instructions below).
- python 2.7 or higher or python 3.4 or higher
- Normaliz 3.8.9 or higher https://github.com/Normaliz/Normaliz/releases
(The current version of PyNormaliz is under construction. Normaliz 3.8.9 not yet released. Use current master of Normaliz for current master of PyNormaliz.) The source packages of the Normaliz realeases contain PyNormaliz.
The PyNormaliz install script assumes that you have executed
within the Normaliz directory. To install PyNormaliz navigate to the Normaliz directory and type
Also see Appendix E of the Normaliz manual.
The command Cone creates a cone (and a lattice), and the member functions of Cone compute its properties. For a full list of input and output properties, see the Normaliz manual.
We assume that you are running python 3.
import PyNormaliz from PyNormaliz import *
To create a simple example, type
C = Cone(cone = [[1,0],[0,1]])
All possible Normaliz input types can be given as keyword arguments.
The member functions allow the computation of the data of our cone. For example,
returns what its name says:
[[0, 1], [1, 0]]
is the matrix of the two Hilbert basis vectors. The ouput matrices of PyNormaliz can be used also in Normaliz input files.
One can pass options to the compute functions as in
C.HilbertSeries(HSOP = True)
Note that some Normaliz output types must be specially encoded for python. Our Hilbert Series is returned as
[, [1, 1], 0]
to be read as follows:  is the numerator polynomial, [1,1] is the vector of exponents of t that occur in the denominator, which is (1-t)(1-t) in our case, and 0 is the shift. So the Hilbert series is given by the rational function 1/(1-t)(1-t). (Aoso see ee this introduction.)
One can also compute several data simultaneously and specify options ("PrimalMode" only added as an example, not because ot is particularly useful here):
C.Compute("LatticePoints", "Volume", "PrimalMode")
with the result
This is the lattice length of the diagonal in the square. The euclidean length, that has been computed simultaneously, is returned by
with the expected value
Floating point numbers are formatted with 4 decimal places and returned as strings (may change). If you want the euclideal volume at the maximum floating point precision, you can use the low level interface which is intermediate between the class Cone and libnormaliz:
Algebraic polyhedra can be computed by PyNormaliz as well:
nf = [ "a2-2", "a", "1.4+/-0.1" ] D = Cone( number_field = nf, cone = [["1/7a+3/2", "-5a"],["4/83a-1","97/81"]])
It is important to note that fractions and algebraic numbers must be encoded as strings for the input.
S = D.SupportHyperplanes() S [['-1470/433*a+280/433', '-1'], ['-32204/555417*a-668233/555417', '-1']]
Very hard to read! Somewhat better:
pretty_print(S) -1470/433*a+280/433 -1 -32204/555417*a-668233/555417 -1
But we can also get floating point approximations:
pretty_print(D.SuppHypsFloat()) -4.1545 -1.0000 -1.2851 -1.0000
By using Python functions, the functionality of Normaliz can be extended. For example,
def intersection(cone1, cone2): intersection_ineq = cone1.SupportHyperplanes()+cone2.SupportHyperplanes() C = Cone(inequalities = intersection_ineq) return C
computes the intersection of two cones. So
C1 = Cone(cone=[[1,2],[2,1]]) C2 = Cone(cone=[[1,1],[1,3]]) intersection(C1,C2).ExtremeRays()
yeilds the result
[[1, 1], [1, 2]]
If you want to see what Normaliz is doing (especually in longer computations) activate the terminal output by
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