I am writing a program that requires a root finder of some sort, but every root finder I have used is unsatisfactorily slow. I'm looking for a way to speed this up.
I have used the SymPy's nsolve, and although this produces very precise results, it is very slow (if I do 12 iterations of my program it takes 12+ hours to run). I wrote my own bisection method, and this works much better, but is still very slow (12 iterations takes ~ 1 hour to run). I have been unable to find a symengine solver, or that is what I would be using. I will post both of my programs (with the bisection method and with nsolve). Any advice on how to speed this up is greatly appreciated.
Here is the code using nsolve:
from symengine import *
import sympy
from sympy import Matrix
from sympy import nsolve
trial = Matrix()
r, E1, E = symbols('r, E1, E')
H11, H22, H12, H21 = symbols("H11, H22, H12, H21")
S11, S22, S12, S21 = symbols("S11, S22, S12, S21")
low = 0
high = oo
integrate = lambda *args: sympy.N(sympy.integrate(*args))
quadratic_expression = (H11-E1*S11)*(H22-E1*S22)-(H12-E1*S12)*(H21-E1*S21)
general_solution = sympify(sympy.solve(quadratic_expression, E1)[0])
def solve_quadratic(**kwargs):
return general_solution.subs(kwargs)
def H(fun):
return -fun.diff(r, 2)/2 - fun.diff(r)/r - fun/r
psi0 = exp(-3*r/2)
trial = trial.row_insert(0, Matrix([psi0]))
I1 = integrate(4*pi*(r**2)*psi0*H(psi0), (r, low, high))
I2 = integrate(4*pi*(r**2)*psi0**2, (r, low, high))
E0 = I1/I2
print(E0)
for x in range(10):
f1 = psi0
f2 = r * (H(psi0)-E0*psi0)
Hf1 = H(f1).simplify()
Hf2 = H(f2).simplify()
H11 = integrate(4*pi*(r**2)*f1*Hf1, (r, low, high))
H12 = integrate(4*pi*(r**2)*f1*Hf2, (r, low, high))
H21 = integrate(4*pi*(r**2)*f2*Hf1, (r, low, high))
H22 = integrate(4*pi*(r**2)*f2*Hf2, (r, low, high))
S11 = integrate(4*pi*(r**2)*f1**2, (r, low, high))
S12 = integrate(4*pi*(r**2)*f1*f2, (r, low, high))
S21 = S12
S22 = integrate(4*pi*(r**2)*f2**2, (r, low, high))
E0 = solve_quadratic(
H11=H11, H22=H22, H12=H12, H21=H21,
S11=S11, S22=S22, S12=S12, S21=S21,
)
print(E0)
C = -(H11 - E0*S11)/(H12 - E0*S12)
psi0 = (f1 + C*f2).simplify()
trial = trial.row_insert(x+1, Matrix([[psi0]]))
# Free ICI Part
h = zeros(x+2, x+2)
HS = zeros(x+2, 1)
S = zeros(x+2, x+2)
for s in range(x+2):
HS[s] = H(trial[s]).simplify()
for i in range(x+2):
for j in range(x+2):
h[i, j] = integrate(4*pi*(r**2)*trial[i]*HS[j], (r, low, high))
for i in range(x+2):
for j in range(x+2):
S[i, j] = integrate(4*pi*(r**2)*trial[i]*trial[j], (r, low, high))
m = h - E*S
eqn = m.det()
roots = nsolve(eqn, float(E0))
print(roots)
Here is the code using my bisection method:
from symengine import *
import sympy
from sympy import Matrix
from sympy import nsolve
trial = Matrix()
r, E1, E = symbols('r, E1, E')
H11, H22, H12, H21 = symbols("H11, H22, H12, H21")
S11, S22, S12, S21 = symbols("S11, S22, S12, S21")
low = 0
high = oo
integrate = lambda *args: sympy.N(sympy.integrate(*args))
quadratic_expression = (H11-E1*S11)*(H22-E1*S22)-(H12-E1*S12)*(H21-E1*S21)
general_solution = sympify(sympy.solve(quadratic_expression, E1)[0])
def solve_quadratic(**kwargs):
return general_solution.subs(kwargs)
def bisection(fun, a, b, tol):
NMax = 100000
f = Lambdify(E, fun)
FA = f(a)
for n in range(NMax):
p = (b+a)/2
FP = f(p)
if FP == 0 or abs(b-a)/2 < tol:
return p
if FA*FP > 0:
a = p
FA = FP
else:
b = p
print("Failed to converge to desired tolerance")
def H(fun):
return -fun.diff(r, 2)/2 - fun.diff(r)/r - fun/r
psi0 = exp(-3*r/2)
trial = trial.row_insert(0, Matrix([psi0]))
I1 = integrate(4*pi*(r**2)*psi0*H(psi0), (r, low, high))
I2 = integrate(4*pi*(r**2)*psi0**2, (r, low, high))
E0 = I1/I2
print(E0)
for x in range(11):
f1 = psi0
f2 = r * (H(psi0)-E0*psi0)
Hf1 = H(f1).simplify()
Hf2 = H(f2).simplify()
H11 = integrate(4*pi*(r**2)*f1*Hf1, (r, low, high))
H12 = integrate(4*pi*(r**2)*f1*Hf2, (r, low, high))
H21 = integrate(4*pi*(r**2)*f2*Hf1, (r, low, high))
H22 = integrate(4*pi*(r**2)*f2*Hf2, (r, low, high))
S11 = integrate(4*pi*(r**2)*f1**2, (r, low, high))
S12 = integrate(4*pi*(r**2)*f1*f2, (r, low, high))
S21 = S12
S22 = integrate(4*pi*(r**2)*f2**2, (r, low, high))
E0 = solve_quadratic(
H11=H11, H22=H22, H12=H12, H21=H21,
S11=S11, S22=S22, S12=S12, S21=S21,
)
print(E0)
C = -(H11 - E0*S11)/(H12 - E0*S12)
psi0 = (f1 + C*f2).simplify()
trial = trial.row_insert(x+1, Matrix([[psi0]]))
# Free ICI Part
h = zeros(x+2, x+2)
HS = zeros(x+2, 1)
S = zeros(x+2, x+2)
for s in range(x+2):
HS[s] = H(trial[s]).simplify()
for i in range(x+2):
for j in range(x+2):
h[i, j] = integrate(4*pi*(r**2)*trial[i]*HS[j], (r, low, high))
for i in range(x+2):
for j in range(x+2):
S[i, j] = integrate(4*pi*(r**2)*trial[i]*trial[j], (r, low, high))
m = h - E*S
eqn = m.det()
roots = bisection(eqn, E0 - 1, E0, 10**(-15))
print(roots)
As I said, they both work as they are supposed to, but they do so very slowly.
Here are some optimizations for your code,
Lambdify(E, fun, cse=True) to make use of Common Subexpression Eliminationpi = sympify(sympy.N(pi)) to use a numeric value of pi. Keeping pi as symbolic hurts because of large expressions..simplify calls to .expand calls.integrate(r**n * exp(-p*r), (r, 0, inf) which can be integrated easily.In [21]: var("n, r, p", positive=True)
Out[21]: (n, r, p)
In [22]: integrate(q*r**n*exp(-p*r), (r, 0, oo))
Out[22]: p**(-n)*q*gamma(n + 1)/p
You can get the advantage of this using a hack like this below. (Ideally sympy should be able to do this faster, but sympy doesn't do a good job on this. I ran into the same issue last summer when trying to solve Dirac and Schrödinger equations symbolically to debug my numeric code. I assume you are trying to do something similar)
def integrate(*args):
args = list(args)
expr = args[0].expand()
r = sympy.S(args[1][0])
limits = args[1][1:]
p = sympy.Wild("p")
n = sympy.Wild("n")
q = sympy.Wild("q")
pattern = q * r**n * sympy.exp(p*r)
terms = expr.args
if not expr.is_Add:
terms = [expr]
result = 0
for arg in terms:
d = sympy.S(arg).match(pattern)
if d is None:
result += sympy.N(sympy.integrate(arg, args[1]))
continue
if d[p].is_number and d[q].is_number and d[n].is_number:
ex = d[q]*(-d[p])**(-d[n])/d[p]*sympy.lowergamma(d[n]+1, -d[p]*r)
result += sympify(sympy.factorial(d[n])*d[q]/(-d[p])**(d[n]+1))
else:
result += sympy.N(sympy.integrate(arg, args[1]))
return result
These 4 changes reduces the time to 16 seconds for me.