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* added sol3.py for problem_20 * added sol4.py for problem_06 * ran `black .` on `\Python`
72 lines
2.1 KiB
Python
72 lines
2.1 KiB
Python
import numpy as np
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def qr_householder(A):
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"""Return a QR-decomposition of the matrix A using Householder reflection.
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The QR-decomposition decomposes the matrix A of shape (m, n) into an
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orthogonal matrix Q of shape (m, m) and an upper triangular matrix R of
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shape (m, n). Note that the matrix A does not have to be square. This
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method of decomposing A uses the Householder reflection, which is
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numerically stable and of complexity O(n^3).
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https://en.wikipedia.org/wiki/QR_decomposition#Using_Householder_reflections
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Arguments:
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A -- a numpy.ndarray of shape (m, n)
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Note: several optimizations can be made for numeric efficiency, but this is
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intended to demonstrate how it would be represented in a mathematics
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textbook. In cases where efficiency is particularly important, an optimized
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version from BLAS should be used.
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>>> A = np.array([[12, -51, 4], [6, 167, -68], [-4, 24, -41]], dtype=float)
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>>> Q, R = qr_householder(A)
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>>> # check that the decomposition is correct
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>>> np.allclose(Q@R, A)
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True
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>>> # check that Q is orthogonal
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>>> np.allclose(Q@Q.T, np.eye(A.shape[0]))
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True
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>>> np.allclose(Q.T@Q, np.eye(A.shape[0]))
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True
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>>> # check that R is upper triangular
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>>> np.allclose(np.triu(R), R)
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True
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"""
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m, n = A.shape
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t = min(m, n)
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Q = np.eye(m)
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R = A.copy()
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for k in range(t - 1):
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# select a column of modified matrix A':
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x = R[k:, [k]]
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# construct first basis vector
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e1 = np.zeros_like(x)
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e1[0] = 1.0
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# determine scaling factor
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alpha = np.linalg.norm(x)
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# construct vector v for Householder reflection
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v = x + np.sign(x[0]) * alpha * e1
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v /= np.linalg.norm(v)
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# construct the Householder matrix
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Q_k = np.eye(m - k) - 2.0 * v @ v.T
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# pad with ones and zeros as necessary
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Q_k = np.block([[np.eye(k), np.zeros((k, m - k))], [np.zeros((m - k, k)), Q_k]])
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Q = Q @ Q_k.T
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R = Q_k @ R
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return Q, R
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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