# positive dimensional solution sets¶

The modules sets, cascades, factor, and diagonal provide some functionality to work with positive dimensional solution sets. In particular, the solve function of the factor module computes a numerical irreducible decomposition of the solution set of a polynomial system. Also polynomials that have variables raised to negative powers, so-called Laurent polynomials are supported.

For isolated solutions, the main outcome of the numerical solver is a list of points, given as tuples of values for the coordinates. For positive dimensional solutions, with numerical homotopy continuation methods we can compute a numerical irreducible decomposition of the solution set. Such a decomposition has two layers:

1. For every dimension of the solution set, we have as many generic points as the degree of the solution set of that dimension.
2. For every dimension of the solution set, those generic points that belong to the same irreducible factor are stored in the same list.

A generic point on a d-dimensional solution set is computed as a solution of the given system of polynomial equations, augmented with d linear equations with randomly generated complex coefficients.

Generic points occur as solutions in the data structure that is called a witness set. With embeddings and cascades, we define homotopies in a top down calculation of a numerical irreducible decomposition. Diagonal homotopies define a bottom up construction of a numerical irreducible decomposition. The application of monodromy loops leads to a factorization of a pure dimensional solution set into irreducible components. Even without having explicit equations for the irreducible factors, with a homotopy membership test we can determine whether any given point belongs to any given factor in the decomposition.

## witness sets¶

A witness set is a data structure to represent a positive dimensional solution set, which is stored as a tuple of two items:

1. An embedding of the polynomial equations that define the solution set, augmented with as many generic linear equations as the dimension of the solution set. To every linear equation corresponds one slack variable.
2. Witness points are solutions in the intersection of the original polynomial equations and the generic linear equations. For generic coefficients of the added linear equations, we obtain generic points on the solution set. The number of witness points equals the degree of the solution set.

In the example below we consider the twisted cubic:

>>> twisted = ['x^2 - y;', 'x^3 - z;']
>>> from phcpy.sets import embed
>>> e = embed(3,1,twisted)
>>> e[0]
'x^2 - y + (-8.23538851649530E-01-5.67259869745581E-01*i)*zz1;'
>>> e[1]
'x^3 - z + (9.35464826338593E-01-3.53419805165623E-01*i)*zz1;'


The last equation of the embedded system is a linear equation with randomly generated complex coefficient. The zz1 denotes the slack variable. Continuing the session:

>>> terms = e[-1].split(')*')
>>> for t in terms: print(t)
...
+ (-8.85038627286137E-01 + 4.65517591731472E-01*i
x + (-2.12324313395875E-02 + 9.99774566519578E-01*i
y + (-9.52478263619098E-01-3.04606561539880E-01*i
z + (-9.59619713608467E-01 + 2.81300560351385E-01*i
zz1+(-3.24025444378001E-01 + 9.46048366308847E-01*i);
>>> from phcpy.solver import solve
>>> s = solve(e, silent=True)
>>> len(s)
3
>>> for sol in s: print(sol)
...
t :  1.00000000000000E+00   0.00000000000000E+00
m : 1
the solution for t :
x : -4.06510360753325E-01   5.24436731997959E-01
y : -1.09783212468901E-01  -4.26377930233570E-01
zz1 : -4.27642353614751E-50  -2.73691106313441E-48
z :  2.68236261633139E-01   1.15752697061077E-01
== err :  3.693E-16 = rco :  1.041E-01 = res :  1.804E-16 =
t :  1.00000000000000E+00   0.00000000000000E+00
m : 1
the solution for t :
x :  7.97989261058868E-01   1.37578286563110E+00
y : -1.25599163259885E+00   2.19571990464483E+00
zz1 :  1.44884468793274E-32  -5.03715049801216E-33
z : -4.02310165732919E+00   2.41891366942435E-02
== err :  1.240E-15 = rco :  1.463E-02 = res :  2.120E-15 =
t :  1.00000000000000E+00   0.00000000000000E+00
m : 1
the solution for t :
x : -1.07164436617733E-01  -9.41488516596475E-01
y : -8.74916410407434E-01   2.01788172926252E-01
zz1 :  0.00000000000000E+00   0.00000000000000E+00
z :  2.83741171803972E-01   8.02099237512644E-01
== err :  9.857E-17 = rco :  8.220E-02 = res :  1.110E-16 =


The variable zz1 is an artificial slack variable. Adding the slack variable via an embedding is a general technique to make overdetermined polynomial systems square, that is: having as many equations as unknowns. Only solutions with zero slack variables matter.

There are four homotopies which involve witness sets.

1. Given a witness set and a point, a homotopy membership test decides whether the point lies on the solution set represented by the witness set.
2. Given all solutions with nonzero values for the slack variables of an embedded system, a cascade homotopy takes those solutions as the start points of solution paths leading to generic points on lower dimensional solution sets.
3. Given a witness set, a monodromy homotopy separates the generic points in the witness set according to the irreducible factors of the solution set.
4. Given two witness sets, a diagonal homotopy computes witness set representations for all components of the intersection of the two given witness sets.

## homotopy membership test¶

Given a witness set and a point, with a homotopy we can decide whether the point belongs to the algebraic set represented by the given witness set. We illustrate this membership test on the cyclic 4-roots problem. First we compute a witness set.

>>> from phcpy.families import cyclic
>>> c4 = cyclic(4)
>>> from phcpy.sets import embed
>>> c4e1 = embed(4, 1, c4)
>>> from phcpy.solver import solve
>>> sols = solve(c4e1)
>>> from phcpy.solutions import filter_zero_coordinates as filter
>>> genpts = filter(sols, 'zz1', 1.0e-8, 'select')
>>> for sol in genpts:
...     print(sol)


Because there are four solutions that satisfy the original cyclic 4-roots problem and a hyperplane with randomly generated coefficients, there is a one dimensional solution set of cyclic 4-roots.

The function membertest takes as input the witness set, represented by the polynomials in c4e1 and the generic points in genpts, and a point. The point is given as a list of doubles, with the real and imaginary parts of all coordinates. The point (1, -1, 1, -1) is thus given as the list [1, 0, -1, 0, 1, 0, -1, 0]. The four extra zeroes are the zero imaginary parts of the four coordinates.

>>> point = [1, 0, -1, 0, 1, 0, -1, 0]
>>> from phcpy.sets import membertest
>>> membertest(c4e1, genpts, 1, point)
residual is  4.00000000000000E+00
point does not lie on the component, as residual >  1.000E-06
False


The function membertest returns False as the residual of the evaluation of the point at the equations does not satisfy the default tolerance.

Testing the point (-1, -1, 0, 0) proceeds as follows. The ... below stands for omitted output.

>>> point = [-1, 0, -1, 0, 1, 0, 1, 0]
>>> membertest(c4e1, genpts, 1, point)
residual is  0.00000000000000E+00
point satisfies the equations, as residual <=  1.000E-06
===========================================================================
== 1 =  #step :  48 #fail :  0 #iter :  61 = regular solution ==
t :  1.00000000000000E+00   0.00000000000000E+00
m : 1                  Length of path :  3.20529115248889E-02
the solution for t :
x0 : -9.99999989106364E-01   7.42569305742753E-09
x1 : -1.00000001089364E+00  -7.42569325135928E-09
x2 :  1.00000000001044E+00   7.19523989696616E-12
x3 :  9.99999999989557E-01  -7.19544395567596E-12
zz1 : -3.89859685757465E-16  -1.56365136224798E-16
== err :  2.235E-08 = rco :  2.239E-09 = res :  1.079E-15 ==
...
match with generic point 1, as difference is  1.074E-08 <=  1.000E-06
Point lies on a solution component.
True


The point passes the residual test. The test continues with the computation of new generic points for a hyperplane that passes through the test point. If the test point is among the new generic points, then the test point belongs to the positive dimensional solution set represented by the witness set. For this example we see that the point (-1, -1, 1, 1) is a singular point on the curve, as can be seen from the estimate for the inverse condition number, rco :  2.239E-09. The default tolerance of 1.0e-6 is high enough in this case for the point to satisfy the membership test.

If the tolerance 1.0e-6 is deemed too sloppy, then we can allow for a stronger tolerance and execute the homotopy membership test in double double precision. More zeroes must be inserted in the test point for the second part (the least significant double) in the double double representation for the real and imaginary parts of the coordinates:

>>> ddpoint = [-1, 0, 0, 0, -1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0]


Instead of 1.0e-6, the new tolerance is 1.0e-12:

>>> membertest(c4e1, genpts, 1, ddpoint, memtol=1.e-12, precision='dd')
residual is 0.00000000000000000000000000000000E+00
point satisfies the equations, as residual <=  1.000E-06
...
== 3 =  #step :  43 #fail :  9 #iter : 119 = singular solution
t : 1.00000000000000000000000000000000E+00  0.00000000000000000000000000000000E+00
m : 1                  Length of path :  2.58366452243257E+01
the solution for t :
x0 : -1.00000000000000000000460097514793E+00  -5.03546515557825836758428944198701E-21
x1 : -9.99999999999999999995399044465921E-01  5.03544121195596798412885036927768E-21
x2 : 9.99999999999999999999995602319048E-01  -4.84887236589678161814305345115216E-24
x3 : 1.00000000000000000000000441729480E+00  4.87281598848246759732685155434612E-24
zz1 : 3.21588798303478472454372240227796E-34  -4.61270682182747444272903697352050E-34
== err :   1.525E-13 = rco :   1.343E-14 = res :   8.687E-26 ==
...
match with generic point 3, as difference is 4.183E-17 <=  1.000E-12
Point lies on a solution component.
True


In double double precision, the condition number estimate for the inverse condition number drops to 1.343E-14 (see the rco field).

To perform the membership test in quad double precision, invoke membertest with precision='qd'.

For solution sets of large degree, the homotopy membership test will run faster in its multitasked version. To run the membership test with 8 tasks, add tasks=8 as last argument of the call to the function.

With a cascade of homotopies, we separate generic points on one equidimensional component from another equidimensional component of the solution set. A cascade starts at the top dimension. We consider an illustrative example:

>>> pols = ['(x^2 + y^2 + z^2 - 1)*(y - x^2)*(x - 0.5);', \
'(x^2 + y^2 + z^2 - 1)*(z - x^3)*(y - 0.5);', \
'(x^2 + y^2 + z^2 - 1)*(z - x*y)*(z - 0.5);']


The polynomials in pols are defined in factored form so for this illustrative example we may read of the equidimensional components of the solution set, which contain the two dimensional sphere, the one dimensional twisted cubic, and the isolated point (0.5, 0.5, 0.5).

To initialize the cascade, we must have solved an embedded polynomial system. With embed(3, 2, pols) we make an embedding of the 3-dimensional system in pols adding two linear equations with random complex coefficients. Two slack variables zz1 and zz2 are added to make this overdetermined system square.

>>> from phcpy.sets import embed
>>> topemb = embed(3, 2, pols)
>>> from phcpy.solver import solve
>>> topsols = solve(topemb, silent=True)


The list topsols contains two types of solutions: those with nonzero values for the slack variables, and those with zero slack variables, which thus satisfy the original equations in pols and the two added linear equations with random complex coefficients. The solutions with zero values for the slack variables define generic points on the two dimensional solution set. We filter the solutions, as follows:

>>> from phcpy.solutions import filter_zero_coordinates as filter
>>> topsols0 = filter(topsols, 'zz2', 1.0e-8, 'select')
>>> topsols1 = filter(topsols, 'zz2', 1.0e-8, 'remove')
>>> print('generic points on the two dimensional surface :')
>>> for sol in topsols0:
...     print(sol)


The solutions with nonzero values for the slack variables are called nonsolutions. These solutions are regular and serve as start solutions in a cascade to compute generic points on the lower dimensional components of the solution set.

>>> from phcpy.cascades import cascade_step
>>> lvl1sols = cascade_step(2, topemb, topsols1)


After the filtering, we must drop variables, coordinates, and hyperplane for the next level in the cascade.

>>> from phcpy.sets import drop_variable_from_polynomials as drop1poly
>>> from phcpy.sets import drop_coordinate_from_solutions as drop1sols
>>> lvl1emb = drop1poly(topemb, 'zz2')
>>> lvl1emb = lvl1emb[:-1]  # dropping the last polynomial
>>> lvl1solsdrop = drop1sols(lvl1sols, len(topemb), 'zz2')
>>> lvl1sols0 = filter(lvl1solsdrop, 'zz1', 1.0e-8, 'select')
>>> lvl1sols1 = filter(lvl1solsdrop, 'zz1', 1.0e-8, 'remove')


Among the solutions at the end of the paths defined by the cascade homotopy are solutions that belong to the two dimensional sphere. These solutions are singular and we filter then away based on threshold for the estimate of the inverse condition number.

>>> from phcpy.solutions import filter_regular as regfilt
>>> reglvl1sols0 = regfilt(lvl1sols0, 1.0e-8, 'select')
>>> for sol in reglvl1sols0:
...     print(sol)


To find the isolated solutions, another cascade homotopy is applied, tracking the paths starting at the nonsolutions at the end of the previous cascade.

>>> lvl2sols = cascade_step(1, lvl1emb, lvl1sols1)
>>> lvl2solsdrop = drop1sols(lvl2sols, len(lvl1emb), 'zz1')
>>> for sol in reglvl2solsdrop:
...     print(sol)


To perform the filtering of the solutions properly, we apply a membership test, defined in the sets module.

The function run_cascade() takes on input the number of variables in the polynomials and the top dimenion of the solution set. Starting at the top dimension, a witness set representation for each pure dimensional component of the solution set is computed.

## factoring into irreducibles¶

A witness set consists of two parts. The first part of a witness set is a polynomial system with as many added linear equations with random coefficients as the dimension. The number of slack variables (variables that start with the name zz) equals the dimension of the witness set. The second part of a witness set is a list of solutions of the first part. Because the added linear equations have random coefficients, the solutions are generic points on the positive dimensional algebraic set.

Given a witness set, applying monodromy loops those points in a witness set that lie on the same irreducible factor are joined. The application of monodromy is a probabilistic method with unknown probability of failure because it relies on the unknown distribution of the singular solutions.

Below is a simple example, given already in factored form:

>>> p = '(x+1)*(x^2 + y^2 + 1);'


To construct a witness set we import witness_set_of_hypersurface from phcpy.sets:

>>> from phcpy.sets import witness_set_of_hypersurface as wh
>>> (w, s) = wh(2, p)
>>> len(s)


Because the degree of p is three, we see 3 as the outcome of len(s).

>>> from phcpy.factor import factor
>>> f = factor(1, w, s)
>>> f


The result in f is a a list of tuples:

[([1, 2], 8.537360146292391e-15), ([3], 2.1316282072803006e-14)]


The factorization joined the first two solutions of s as they represent the quadratic factor. A generic point for the linear factor is in the second tuple. The second floating point number in each tuple is the residual obtained via the linear trace test, used as stop criterion in the running of monodromy loops.

For polynomials of higher degrees, double double or even quad double could be required to obtain accurate results. The following two commands illustrate how to apply monodromy respectively in double double and quad double precision:

>>> f = factor(1, w, s, precision='dd')
>>> f = factor(1, w, s, precision='qd')


The witness set (w, s) should also have been computed in double double and quad double precision.

The function decompose() takes the output of the run_cascade() function of the cascades module and factors every witness set for the pure dimensional components into irreducible factors. The functions run_cascade() and decompose() lead to a numerical irreducible decomposition of the solution set.

## numerical irreducible decomposition¶

Consider the polynomials defined by the list pols as follows:

>>> pols = ['(x1-1)*(x1-2)*(x1-3)*(x1-4);', \
'(x1-1)*(x2-1)*(x2-2)*(x2-3);', \
'(x1-1)*(x1-2)*(x3-1)*(x3-2);', \
'(x1-1)*(x2-1)*(x3-1)*(x4-1);']
>>> from phcpy.factor import solve, write_decomposition
>>> deco = solve(4, 3, pols, verbose=False)
>>> write_decomposition(deco)


We see the common factor x1-1 which defines a three dimensional solution plane. The factor x1-2 leads to a two dimensional solution plane, with the additional factor x2-1. Furthermore, the system in pols has twelve lines as solutions and four isolated solution points.

The first argument 4 in the solve(4, 3, pols, verbose=False) is the number of variables in the polynomials in pols. The second argument 3 equals the top dimension of the solution set. The write_decomposition() confirms there is one three dimensional linear component, one two dimensional linear component, twelve lines, and four isolated solutions.

## diagonal homotopies¶

Given two witness sets, with diagonal homotopies we can compute generic points on the intersection of the algebraic sets represented by the witness sets, and thus obtain a witness set of the intersection. This section illustrates the intersection of the unit sphere with a cylinder. This intersection defines a quartic curve.

>>> sph = 'x^2 + y^2 + z^2 - 1;'
>>> cyl = 'x^2 + y - y + (z - 0.5)^2 - 1;'


Observe the + y - y line in the assignment to cyl. With this trick we initialize the symbol table for the witness set computation, ensuring that y is present.

Next, we compute a witness sets for the sphere and the cylinder:

>>> from phcpy.sets import witness_set_of_hypersurface as witsurf
>>> sphwit = witsurf(3, sph)
>>> spheqs, sphpts = sphwit
>>> cylwit = witsurf(3, cyl)
>>> cyleqs, cylpts = cylwit


Once we have two witness sets, we call the diagonal_solver method to compute a witness set for the intersection:

>>> from phcpy.diagonal import diagonal_solver as diagsolve
>>> quawit = diagsolve(3, 2, spheqs, sphpts, 2, cyleqs, cylpts)
>>> quaeqs, quapts = quawit
>>> for pol in quaeqs:
...     print(pol)
>>> for sol in quapts:
...     print(sol)