Finding Singular Solutions to Polynomial Systems with Perturbed Regeneration
Daniel Bates, Brent R. Davis, Eric Hanson, David Eklund, Chris Peterson
Department of Mathematics, Colorado State University, Fort Collins, CO
What is regeneration and what’s the problem?
Problem
Broad: Develop parallalizable algorithms to find all numerical solutions to a system
of polynomials equations coming from science and engineering.
Focus: Find all isolated regular solutions and singular solutions.
Main Algorithm: Perturbed Regeneration
Regeneration is an equation-by-equation based solving method to find all isolated
solutions to f (z) = 0.
Construct a system
f (z) + f (y )
for some random
complex vector y
Consider solving the system of equations:
#
"
Specific: A powerful method called regeneration only finds regular solutions and
we also want singular solutions.
Suppose a polynomial system of equations f (z) = 0 has at least one isolated
solution.
y (x −2)2
=0
f (x, y ) =
x(y −3)
Since solutions are
all regular, solve
f (z) − f (y ) = 0
using regeneration
Track solutions
f (z)−f (y ) → f (z)
using homotopy
continuation
which has a regular solution (0, 0) and a singular solution (2, 3).
What are regular and singular solution?
Example: Closest points between two curves using software Bertini
The first equation has degree 3 so solve the 3 random linear systems:
"
Suppose z ∗ ∈ CN is an isolated solution of f (z) = 0.
Regular Solutions:
A solution z ∗ is regular if the
Jacobian matrix Jf (z) at z ∗ is full
rank.
Singular Solutions:
A solution z ∗ is singular if the
Jacobian matrix Jf (z) at z ∗ is not
full rank.
#
Linear
or
Doesn’t converge
Quadratic
α3x +β3y +γ3
α2x +β2y +γ2
α1x +β1y +γ1
= 0.
= 0,
= 0,
a1x +b1y +c1
a1x +b1y +c1
a1x +b1y +c1
Now we know the
solutions to:
"
#
(α1x +β1y +γ1)(α2x +β2y +γ2)(α3x +β3y +γ3)
=0
a1x +b1y +c1
What does a Bertini input file
look like?
CONFIG
USEREGENERATION:1;
USERHOMOTOPY: 1;
END;
Now regenerate in the first equation by numerically tracking 3 solutions:
"
#
"
The system
Homotopy Continuation:
Choose a polynomial system g
that is “similiar” to f but is easier to
solve
"
(α1x +β1y +γ1)(α2x +β2y +γ2)(α3x +β3y +γ3)
y (x −2)2
→
a1x +b1y +c1
a1x +b1y +c1
What is polynomial homotopy continuation?
2
#
#
y (x −2)
= 0 has two singular solutions.
a1x +b1y +c1
We discard the singular solutions and proceed by numerically tracking:
"
Solve g and form the homotopy
function given by
2
#
"
2
y (x −2)
y (x −2)
→
a2x +b2y +c2
a1x +b1y +c1
H(z; t) = (1−t)f (z)+tg(z),
so that
Illustration of homotopy continuation and
numerical predictor-corrector methods.
As t → 0, track
Every unknown solution of
H(z;0) = f(z) = 0
Why use regeneration?
#
y (x −2)2
= 0.
(a1x +b1y +c1)(a2x +b2y +c2)
Finally we regenerate the second equation by numerically trackings the 2 regular
solutions:
"
#
"
#
y (x −2)2
y (x −2)2
→
(a1x +b1y +c1)(a2x +b2y +c2)
x(y −3)
to recover the regular solution (0, 0) but miss the singular solution (2, 3).
What is a perturbed system?
Consider the polynomial system of equations f (z) = 0 with isolated possibly singular
solutions.
“By revealing the structure of the solution sets of subsets of the polynomials in the
system, these incremental methods eliminate paths in the later, more expensive
stages of homotopy. This tends to save overall computation.” – J. Hauenstein
For a randomly chosen vector y , the perturbed system:
Flexibility in regeneration:
Order of equations
Equation grouping
Minimal support hyperplanes
turns
INPUT
function L,f,dfx,dfy,
g,dgz,dgw;
variable group x,y,z,w,
l0,l1,l2;
pathvariable t;
parameter s;
s = t;
Two algebraic curves and a representation of the many local
solutions.
#
Thus we have 2 regular solutions to:
"
Solutions of
H(z; 1) = g(z) = 0
Given two algebraic set, it may be difficult to guarantee their nearest (or farthest)
points between them since their are many local solutions. In this case, there are
242 possibilities.
#
"
by consider all pairs from the previous 3 linear systems.
Convergence of Newton’s Method:
H(z; 0) = f (z)
H(z; 1) = g(z)
#
"
f̂ (z) = f (z)−f (y )
y1 = -0.6737-0.7390*I;
...
y7 = 0.5484+0.8362*I;
r0 = 0.9830-0.1837*I;
...
r2 = -0.9958+0.0918*I;
f
=
g
=
dfx =
dfy =
dgz =
dgw =
L
=
END;
paths
timing
part 1 5995 1min 3 sec
part 2 650
10 sec
Paths and timing for
part 1 and 2 of
perturbed
regeneration
#R
regular 0
singular 0
total
0
# C # R+ # C
162
162
120
120
282
282
Number of solutions
using perturbed
regeneration.
yˆ4-(xˆ3-2*x+1)ˆ2-s*y1;
(w-0.7)ˆ4-(z-(z+1.3)ˆ3+1.3)ˆ2-s*y2;
(x-z)*l0-2*l1*(3*xˆ2-2)*(xˆ3-2*x+1)-s*y3;
(y-w)*l0+4*l1*yˆ3-s*y4;
(z-x)*l0+2*11*(3*(z+1.3)ˆ2-1)*(z-(z+1.3)ˆ3+1.3)-s*y5;
(w-y)*l0+4*l1*(w-0.7)ˆ3-s*y6;
r0*l0+r1*l1+r2*12-1-s*y7;
level
0
1
2
3
4
5
6
paths reg. endpoints total discarded sing. endpoints inf. endpoints
1
1
0
0
0
6
6
0
0
0
36
26
10
0
10
104
52
52
0
52
312
312
0
0
0
1872
916
956
0
956
3664
650
3014
0
3014
Regeneration summary for part 1 of perturbed regeneration
Acknowledgement
isolated
singular solutions of f (z) = 0
multiple
regular solutions of f̂ (z) = 0
This material is based upon work supported by the National Science Foundation under Grants No. DMS-1115668 and
DMS-1025564. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the
authors and do not necessarily reflect the views of the National Science Foundation.