scipy’s linalg module contains two functions, solve_triangular, and cho_solve. The functions can be called by prepending them by scipy.linalg..
scipy: https://docs.scipy.org/doc/scipy/reference/generated/scipy.linalg.cho_solve.html
Solve the linear equations
\begin{equation}
\mathbf{A}\cdot\mathbf{x} = \mathbf{b}
\end{equation}
given the Cholesky factorization of A. As opposed to scipy, the function simply takes the Cholesky-factorised matrix, A, and b as inputs.
# code to be run in micropython
from ulab import numpy as np
from ulab import scipy as spy
A = np.array([[3, 0, 0, 0], [2, 1, 0, 0], [1, 0, 1, 0], [1, 2, 1, 8]])
b = np.array([4, 2, 4, 2])
print(spy.linalg.cho_solve(A, b))array([-0.01388888888888906, -0.6458333333333331, 2.677083333333333, -0.01041666666666667], dtype=float64)
scipy: https://docs.scipy.org/doc/scipy/reference/generated/scipy.linalg.solve_triangular.html
Solve the linear equation
\begin{equation}
\mathbf{a}\cdot\mathbf{x} = \mathbf{b}
\end{equation}
with the assumption that a is a triangular matrix. The two position arguments are a, and b, and the optional keyword argument is lower with a default value of False. lower determines, whether data are taken from the lower, or upper triangle of a.
Note that a itself does not have to be a triangular matrix: if it is not, then the values are simply taken to be 0 in the upper or lower triangle, as dictated by lower. However, a ⋅ x will yield b only, when a is triangular. You should keep this in mind, when trying to establish the validity of the solution by back substitution.
# code to be run in micropython
from ulab import numpy as np
from ulab import scipy as spy
a = np.array([[3, 0, 0, 0], [2, 1, 0, 0], [1, 0, 1, 0], [1, 2, 1, 8]])
b = np.array([4, 2, 4, 2])
print('a:\n')
print(a)
print('\nb: ', b)
x = spy.linalg.solve_triangular(a, b, lower=True)
print('='*20)
print('x: ', x)
print('\ndot(a, x): ', np.dot(a, x))a:
- array([[3.0, 0.0, 0.0, 0.0],
[2.0, 1.0, 0.0, 0.0], [1.0, 0.0, 1.0, 0.0], [1.0, 2.0, 1.0, 8.0]], dtype=float64)
b: array([4.0, 2.0, 4.0, 2.0], dtype=float64) ==================== x: array([1.333333333333333, -0.6666666666666665, 2.666666666666667, -0.08333333333333337], dtype=float64)
dot(a, x): array([4.0, 2.0, 4.0, 2.0], dtype=float64)
With get the same solution, x, with the following matrix, but the dot product of a, and x is no longer b:
# code to be run in micropython
from ulab import numpy as np
from ulab import scipy as spy
a = np.array([[3, 2, 1, 0], [2, 1, 0, 1], [1, 0, 1, 4], [1, 2, 1, 8]])
b = np.array([4, 2, 4, 2])
print('a:\n')
print(a)
print('\nb: ', b)
x = spy.linalg.solve_triangular(a, b, lower=True)
print('='*20)
print('x: ', x)
print('\ndot(a, x): ', np.dot(a, x))a:
- array([[3.0, 2.0, 1.0, 0.0],
[2.0, 1.0, 0.0, 1.0], [1.0, 0.0, 1.0, 4.0], [1.0, 2.0, 1.0, 8.0]], dtype=float64)
b: array([4.0, 2.0, 4.0, 2.0], dtype=float64) ==================== x: array([1.333333333333333, -0.6666666666666665, 2.666666666666667, -0.08333333333333337], dtype=float64)
dot(a, x): array([5.333333333333334, 1.916666666666666, 3.666666666666667, 2.0], dtype=float64)