Runtime Optimization of sympy code using numpy or scipy

Question

I'm new to programming and it's my first question here. I spend a lot of time on this problem. It's an abstract from a programme I'm writing for my thesis. Maybe it's trivial to some of you experts but I might even lack mathematical knowledge.

I want to solve following set of equations for b1, b2 and c3:

0 = cos(b1)+ cos(b2)- 0.0166
0 = sin(b1)+ sin(b2)+ 0.3077*c3 - 0.6278
0 = cos(b1)- cos(b2)+ 5.4155*c3 - 4.3547

b1 and b2 are angles and should therefore between [0, 2*pi] and c3 should be between [0,1], but this is not necessary. I can filter results later.

I found a solution using sympy:

import sympy as sy
b1 = sy.Symbol('b1',real=True)
b2 = sy.Symbol('b2',real=True)
c3 = sy.Symbol('c3',real=True)
a = sy.cos(b1)+sy.cos(b2)-0.0166
b = sy.sin(b1)+sy.sin(b2)+0.3077*c3-0.6278
c = sy.cos(b1)-sy.cos(b2)+5.4155*c3-4.3547
Erg = sy.solve([a,b,c,],[b1,b2,c3],dict=True)

[{c3: -0.4634, b1: 2.7245, b2: 0.3739}]

The results are reasonable, but unfortunately this leads to a calculation time of 6s which would result in a total calculation time of over 7hrs for my programme. Please help me find a faster solution. I tried scipy

from scipy.optimize import fsolve
import numpy as np

def equations(p):
    b1, b2, c3 = p
    return (np.cos(b1)+np.cos(b2)-0.0166, np.sin(b1)+np.sin(b2)+0.3077*c3-0.6278,np.cos(b1)-np.cos(b2)+5.4155*c3-4.3547)

b1, b2, c3 =  fsolve(equations, (-0.4634, 2.7245, 0.3739))

This results in under 1s in (-9.8418e-14, 5.6621e-15,-7.5495e-14). I don't know which number belongs to which variable, but they don't make any sense anyway. Another option would be optimizing the equation for minimal c3. Please don't hesitate to ask if I haven't been clear about anything.

Answer 1

If I fix the missing parenthesis in your numpy example, I get a different result:

In [40]: b1,b2,c3
Out[40]: (0.21407625679722384, 2.8598524043642226, 0.4463029985826017)

And anyway your input points are not really good, since your variables are mixed up. Your function is defined in the order b1,b2,c3 , yet the starting point corresponds to the symbolic solution of c3,b1,b2 . Changing the order gives me

In [42]: b1,b2,c3
Out[42]: (2.9450733464142882, 0.07278767695320594, 1.1693849677611501)

Note that for a given b1,b2 you'll also have a solution with b1+2*k*pi,b2*2*l*pi where k,l are integers.

I didn't put these numbers into your equations, but I'm pretty sure they solve them. The problem is, there are a bunch of solutions. As I said, your first two variables are periodic, and even then there can be a lot of other solutions with c3 . If you use a bunch of different starting points for fsolve , you'll get a bunch of different solutions.

What's worse: your solution doesn't enforce the 0<c3<1 bound.

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On the mathematics: due to the diversity of possible solutions, you might want to simplify the numerical problem on paper first.

For instance, the first equation is special, as it doesn't contain c3 . You can relate b1 to b2 , albeit in a nonlinear way. For each b1 you'll know that

b2=+-acos(0.0166-cos(b1))+2*k*pi

Which is already a constraint. Furthermore, adding Eq. 1 to Eq. 3 gives you

0 = 2*cos(b1)+ 5.4155*c3 - 4.3547-0.0166

in other words, c3 is linearly dependent on cos(b1) . It might help if you introduce this into Eq. 2, getting a connection between functions of b1 and b2 . Essentially you have variables as b1,b2(b1),c3(b1) , with the single Eq. 2 to solve.

Another note: I'm guessing that the large number of equations comes from a lot of combinations for numerical factors in your equations. Isn't that right? If so, you might want to introduce some symbolic constants in the equations instead of the floating-point literals, and try to solve that. If you succeed, you don't have to solve a huge number of sets of nonlinear equations: you only have to solve one set, then substitute your various parameters.