791 lines
26 KiB
Python
791 lines
26 KiB
Python
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"""
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Linear algebra
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--------------
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Linear equations
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................
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Basic linear algebra is implemented; you can for example solve the linear
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equation system::
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x + 2*y = -10
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3*x + 4*y = 10
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using ``lu_solve``::
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>>> from mpmath import *
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>>> mp.pretty = False
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>>> A = matrix([[1, 2], [3, 4]])
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>>> b = matrix([-10, 10])
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>>> x = lu_solve(A, b)
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>>> x
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matrix(
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[['30.0'],
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['-20.0']])
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If you don't trust the result, use ``residual`` to calculate the residual ||A*x-b||::
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>>> residual(A, x, b)
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matrix(
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[['3.46944695195361e-18'],
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['3.46944695195361e-18']])
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>>> str(eps)
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'2.22044604925031e-16'
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As you can see, the solution is quite accurate. The error is caused by the
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inaccuracy of the internal floating point arithmetic. Though, it's even smaller
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than the current machine epsilon, which basically means you can trust the
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result.
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If you need more speed, use NumPy, or ``fp.lu_solve`` for a floating-point computation.
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>>> fp.lu_solve(A, b) # doctest: +ELLIPSIS
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matrix(...)
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``lu_solve`` accepts overdetermined systems. It is usually not possible to solve
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such systems, so the residual is minimized instead. Internally this is done
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using Cholesky decomposition to compute a least squares approximation. This means
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that that ``lu_solve`` will square the errors. If you can't afford this, use
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``qr_solve`` instead. It is twice as slow but more accurate, and it calculates
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the residual automatically.
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Matrix factorization
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....................
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The function ``lu`` computes an explicit LU factorization of a matrix::
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>>> P, L, U = lu(matrix([[0,2,3],[4,5,6],[7,8,9]]))
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>>> print(P)
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[0.0 0.0 1.0]
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[1.0 0.0 0.0]
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[0.0 1.0 0.0]
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>>> print(L)
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[ 1.0 0.0 0.0]
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[ 0.0 1.0 0.0]
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[0.571428571428571 0.214285714285714 1.0]
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>>> print(U)
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[7.0 8.0 9.0]
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[0.0 2.0 3.0]
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[0.0 0.0 0.214285714285714]
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>>> print(P.T*L*U)
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[0.0 2.0 3.0]
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[4.0 5.0 6.0]
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[7.0 8.0 9.0]
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Interval matrices
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-----------------
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Matrices may contain interval elements. This allows one to perform
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basic linear algebra operations such as matrix multiplication
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and equation solving with rigorous error bounds::
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>>> a = iv.matrix([['0.1','0.3','1.0'],
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... ['7.1','5.5','4.8'],
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... ['3.2','4.4','5.6']])
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>>>
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>>> b = iv.matrix(['4','0.6','0.5'])
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>>> c = iv.lu_solve(a, b)
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>>> print(c)
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[ [5.2582327113062568605927528666, 5.25823271130625686059275702219]]
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[[-13.1550493962678375411635581388, -13.1550493962678375411635540152]]
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[ [7.42069154774972557628979076189, 7.42069154774972557628979190734]]
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>>> print(a*c)
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[ [3.99999999999999999999999844904, 4.00000000000000000000000155096]]
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[[0.599999999999999999999968898009, 0.600000000000000000000031763736]]
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[[0.499999999999999999999979320485, 0.500000000000000000000020679515]]
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"""
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# TODO:
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# *implement high-level qr()
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# *test unitvector
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# *iterative solving
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from copy import copy
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from ..libmp.backend import xrange
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class LinearAlgebraMethods(object):
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def LU_decomp(ctx, A, overwrite=False, use_cache=True):
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"""
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LU-factorization of a n*n matrix using the Gauss algorithm.
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Returns L and U in one matrix and the pivot indices.
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Use overwrite to specify whether A will be overwritten with L and U.
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"""
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if not A.rows == A.cols:
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raise ValueError('need n*n matrix')
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# get from cache if possible
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if use_cache and isinstance(A, ctx.matrix) and A._LU:
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return A._LU
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if not overwrite:
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orig = A
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A = A.copy()
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tol = ctx.absmin(ctx.mnorm(A,1) * ctx.eps) # each pivot element has to be bigger
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n = A.rows
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p = [None]*(n - 1)
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for j in xrange(n - 1):
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# pivoting, choose max(abs(reciprocal row sum)*abs(pivot element))
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biggest = 0
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for k in xrange(j, n):
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s = ctx.fsum([ctx.absmin(A[k,l]) for l in xrange(j, n)])
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if ctx.absmin(s) <= tol:
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raise ZeroDivisionError('matrix is numerically singular')
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current = 1/s * ctx.absmin(A[k,j])
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if current > biggest: # TODO: what if equal?
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biggest = current
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p[j] = k
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# swap rows according to p
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ctx.swap_row(A, j, p[j])
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if ctx.absmin(A[j,j]) <= tol:
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raise ZeroDivisionError('matrix is numerically singular')
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# calculate elimination factors and add rows
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for i in xrange(j + 1, n):
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A[i,j] /= A[j,j]
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for k in xrange(j + 1, n):
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A[i,k] -= A[i,j]*A[j,k]
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if ctx.absmin(A[n - 1,n - 1]) <= tol:
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raise ZeroDivisionError('matrix is numerically singular')
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# cache decomposition
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if not overwrite and isinstance(orig, ctx.matrix):
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orig._LU = (A, p)
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return A, p
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def L_solve(ctx, L, b, p=None):
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"""
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Solve the lower part of a LU factorized matrix for y.
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"""
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if L.rows != L.cols:
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raise RuntimeError("need n*n matrix")
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n = L.rows
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if len(b) != n:
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raise ValueError("Value should be equal to n")
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b = copy(b)
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if p: # swap b according to p
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for k in xrange(0, len(p)):
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ctx.swap_row(b, k, p[k])
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# solve
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for i in xrange(1, n):
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for j in xrange(i):
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b[i] -= L[i,j] * b[j]
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return b
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def U_solve(ctx, U, y):
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"""
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Solve the upper part of a LU factorized matrix for x.
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"""
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if U.rows != U.cols:
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raise RuntimeError("need n*n matrix")
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n = U.rows
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if len(y) != n:
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raise ValueError("Value should be equal to n")
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x = copy(y)
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for i in xrange(n - 1, -1, -1):
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for j in xrange(i + 1, n):
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x[i] -= U[i,j] * x[j]
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x[i] /= U[i,i]
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return x
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def lu_solve(ctx, A, b, **kwargs):
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"""
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Ax = b => x
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Solve a determined or overdetermined linear equations system.
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Fast LU decomposition is used, which is less accurate than QR decomposition
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(especially for overdetermined systems), but it's twice as efficient.
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Use qr_solve if you want more precision or have to solve a very ill-
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conditioned system.
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If you specify real=True, it does not check for overdeterminded complex
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systems.
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"""
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prec = ctx.prec
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try:
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ctx.prec += 10
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# do not overwrite A nor b
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A, b = ctx.matrix(A, **kwargs).copy(), ctx.matrix(b, **kwargs).copy()
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if A.rows < A.cols:
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raise ValueError('cannot solve underdetermined system')
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if A.rows > A.cols:
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# use least-squares method if overdetermined
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# (this increases errors)
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AH = A.H
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A = AH * A
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b = AH * b
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if (kwargs.get('real', False) or
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not sum(type(i) is ctx.mpc for i in A)):
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# TODO: necessary to check also b?
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x = ctx.cholesky_solve(A, b)
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else:
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x = ctx.lu_solve(A, b)
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else:
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# LU factorization
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A, p = ctx.LU_decomp(A)
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b = ctx.L_solve(A, b, p)
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x = ctx.U_solve(A, b)
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finally:
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ctx.prec = prec
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return x
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def improve_solution(ctx, A, x, b, maxsteps=1):
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"""
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Improve a solution to a linear equation system iteratively.
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This re-uses the LU decomposition and is thus cheap.
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Usually 3 up to 4 iterations are giving the maximal improvement.
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"""
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if A.rows != A.cols:
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raise RuntimeError("need n*n matrix") # TODO: really?
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for _ in xrange(maxsteps):
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r = ctx.residual(A, x, b)
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if ctx.norm(r, 2) < 10*ctx.eps:
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break
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# this uses cached LU decomposition and is thus cheap
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dx = ctx.lu_solve(A, -r)
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x += dx
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return x
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def lu(ctx, A):
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"""
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A -> P, L, U
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LU factorisation of a square matrix A. L is the lower, U the upper part.
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P is the permutation matrix indicating the row swaps.
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P*A = L*U
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If you need efficiency, use the low-level method LU_decomp instead, it's
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much more memory efficient.
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"""
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# get factorization
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A, p = ctx.LU_decomp(A)
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n = A.rows
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L = ctx.matrix(n)
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U = ctx.matrix(n)
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for i in xrange(n):
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for j in xrange(n):
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if i > j:
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L[i,j] = A[i,j]
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elif i == j:
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L[i,j] = 1
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U[i,j] = A[i,j]
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else:
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U[i,j] = A[i,j]
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# calculate permutation matrix
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P = ctx.eye(n)
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for k in xrange(len(p)):
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ctx.swap_row(P, k, p[k])
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return P, L, U
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def unitvector(ctx, n, i):
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"""
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Return the i-th n-dimensional unit vector.
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"""
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assert 0 < i <= n, 'this unit vector does not exist'
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return [ctx.zero]*(i-1) + [ctx.one] + [ctx.zero]*(n-i)
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def inverse(ctx, A, **kwargs):
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"""
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Calculate the inverse of a matrix.
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If you want to solve an equation system Ax = b, it's recommended to use
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solve(A, b) instead, it's about 3 times more efficient.
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"""
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prec = ctx.prec
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try:
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ctx.prec += 10
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# do not overwrite A
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A = ctx.matrix(A, **kwargs).copy()
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n = A.rows
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# get LU factorisation
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A, p = ctx.LU_decomp(A)
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cols = []
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# calculate unit vectors and solve corresponding system to get columns
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for i in xrange(1, n + 1):
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e = ctx.unitvector(n, i)
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y = ctx.L_solve(A, e, p)
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cols.append(ctx.U_solve(A, y))
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# convert columns to matrix
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inv = []
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for i in xrange(n):
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row = []
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for j in xrange(n):
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row.append(cols[j][i])
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inv.append(row)
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result = ctx.matrix(inv, **kwargs)
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finally:
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ctx.prec = prec
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return result
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def householder(ctx, A):
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"""
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(A|b) -> H, p, x, res
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(A|b) is the coefficient matrix with left hand side of an optionally
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overdetermined linear equation system.
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H and p contain all information about the transformation matrices.
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x is the solution, res the residual.
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"""
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if not isinstance(A, ctx.matrix):
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raise TypeError("A should be a type of ctx.matrix")
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m = A.rows
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n = A.cols
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if m < n - 1:
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raise RuntimeError("Columns should not be less than rows")
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# calculate Householder matrix
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p = []
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for j in xrange(0, n - 1):
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s = ctx.fsum(abs(A[i,j])**2 for i in xrange(j, m))
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if not abs(s) > ctx.eps:
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raise ValueError('matrix is numerically singular')
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p.append(-ctx.sign(ctx.re(A[j,j])) * ctx.sqrt(s))
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kappa = ctx.one / (s - p[j] * A[j,j])
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A[j,j] -= p[j]
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for k in xrange(j+1, n):
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y = ctx.fsum(ctx.conj(A[i,j]) * A[i,k] for i in xrange(j, m)) * kappa
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for i in xrange(j, m):
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A[i,k] -= A[i,j] * y
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# solve Rx = c1
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x = [A[i,n - 1] for i in xrange(n - 1)]
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for i in xrange(n - 2, -1, -1):
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x[i] -= ctx.fsum(A[i,j] * x[j] for j in xrange(i + 1, n - 1))
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x[i] /= p[i]
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# calculate residual
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if not m == n - 1:
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r = [A[m-1-i, n-1] for i in xrange(m - n + 1)]
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else:
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# determined system, residual should be 0
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r = [0]*m # maybe a bad idea, changing r[i] will change all elements
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return A, p, x, r
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||
|
|
#def qr(ctx, A):
|
||
|
|
# """
|
||
|
|
# A -> Q, R
|
||
|
|
#
|
||
|
|
# QR factorisation of a square matrix A using Householder decomposition.
|
||
|
|
# Q is orthogonal, this leads to very few numerical errors.
|
||
|
|
#
|
||
|
|
# A = Q*R
|
||
|
|
# """
|
||
|
|
# H, p, x, res = householder(A)
|
||
|
|
# TODO: implement this
|
||
|
|
|
||
|
|
def residual(ctx, A, x, b, **kwargs):
|
||
|
|
"""
|
||
|
|
Calculate the residual of a solution to a linear equation system.
|
||
|
|
|
||
|
|
r = A*x - b for A*x = b
|
||
|
|
"""
|
||
|
|
oldprec = ctx.prec
|
||
|
|
try:
|
||
|
|
ctx.prec *= 2
|
||
|
|
A, x, b = ctx.matrix(A, **kwargs), ctx.matrix(x, **kwargs), ctx.matrix(b, **kwargs)
|
||
|
|
return A*x - b
|
||
|
|
finally:
|
||
|
|
ctx.prec = oldprec
|
||
|
|
|
||
|
|
def qr_solve(ctx, A, b, norm=None, **kwargs):
|
||
|
|
"""
|
||
|
|
Ax = b => x, ||Ax - b||
|
||
|
|
|
||
|
|
Solve a determined or overdetermined linear equations system and
|
||
|
|
calculate the norm of the residual (error).
|
||
|
|
QR decomposition using Householder factorization is applied, which gives very
|
||
|
|
accurate results even for ill-conditioned matrices. qr_solve is twice as
|
||
|
|
efficient.
|
||
|
|
"""
|
||
|
|
if norm is None:
|
||
|
|
norm = ctx.norm
|
||
|
|
prec = ctx.prec
|
||
|
|
try:
|
||
|
|
ctx.prec += 10
|
||
|
|
# do not overwrite A nor b
|
||
|
|
A, b = ctx.matrix(A, **kwargs).copy(), ctx.matrix(b, **kwargs).copy()
|
||
|
|
if A.rows < A.cols:
|
||
|
|
raise ValueError('cannot solve underdetermined system')
|
||
|
|
H, p, x, r = ctx.householder(ctx.extend(A, b))
|
||
|
|
res = ctx.norm(r)
|
||
|
|
# calculate residual "manually" for determined systems
|
||
|
|
if res == 0:
|
||
|
|
res = ctx.norm(ctx.residual(A, x, b))
|
||
|
|
return ctx.matrix(x, **kwargs), res
|
||
|
|
finally:
|
||
|
|
ctx.prec = prec
|
||
|
|
|
||
|
|
def cholesky(ctx, A, tol=None):
|
||
|
|
r"""
|
||
|
|
Cholesky decomposition of a symmetric positive-definite matrix `A`.
|
||
|
|
Returns a lower triangular matrix `L` such that `A = L \times L^T`.
|
||
|
|
More generally, for a complex Hermitian positive-definite matrix,
|
||
|
|
a Cholesky decomposition satisfying `A = L \times L^H` is returned.
|
||
|
|
|
||
|
|
The Cholesky decomposition can be used to solve linear equation
|
||
|
|
systems twice as efficiently as LU decomposition, or to
|
||
|
|
test whether `A` is positive-definite.
|
||
|
|
|
||
|
|
The optional parameter ``tol`` determines the tolerance for
|
||
|
|
verifying positive-definiteness.
|
||
|
|
|
||
|
|
**Examples**
|
||
|
|
|
||
|
|
Cholesky decomposition of a positive-definite symmetric matrix::
|
||
|
|
|
||
|
|
>>> from mpmath import *
|
||
|
|
>>> mp.dps = 25; mp.pretty = True
|
||
|
|
>>> A = eye(3) + hilbert(3)
|
||
|
|
>>> nprint(A)
|
||
|
|
[ 2.0 0.5 0.333333]
|
||
|
|
[ 0.5 1.33333 0.25]
|
||
|
|
[0.333333 0.25 1.2]
|
||
|
|
>>> L = cholesky(A)
|
||
|
|
>>> nprint(L)
|
||
|
|
[ 1.41421 0.0 0.0]
|
||
|
|
[0.353553 1.09924 0.0]
|
||
|
|
[0.235702 0.15162 1.05899]
|
||
|
|
>>> chop(A - L*L.T)
|
||
|
|
[0.0 0.0 0.0]
|
||
|
|
[0.0 0.0 0.0]
|
||
|
|
[0.0 0.0 0.0]
|
||
|
|
|
||
|
|
Cholesky decomposition of a Hermitian matrix::
|
||
|
|
|
||
|
|
>>> A = eye(3) + matrix([[0,0.25j,-0.5j],[-0.25j,0,0],[0.5j,0,0]])
|
||
|
|
>>> L = cholesky(A)
|
||
|
|
>>> nprint(L)
|
||
|
|
[ 1.0 0.0 0.0]
|
||
|
|
[(0.0 - 0.25j) (0.968246 + 0.0j) 0.0]
|
||
|
|
[ (0.0 + 0.5j) (0.129099 + 0.0j) (0.856349 + 0.0j)]
|
||
|
|
>>> chop(A - L*L.H)
|
||
|
|
[0.0 0.0 0.0]
|
||
|
|
[0.0 0.0 0.0]
|
||
|
|
[0.0 0.0 0.0]
|
||
|
|
|
||
|
|
Attempted Cholesky decomposition of a matrix that is not positive
|
||
|
|
definite::
|
||
|
|
|
||
|
|
>>> A = -eye(3) + hilbert(3)
|
||
|
|
>>> L = cholesky(A)
|
||
|
|
Traceback (most recent call last):
|
||
|
|
...
|
||
|
|
ValueError: matrix is not positive-definite
|
||
|
|
|
||
|
|
**References**
|
||
|
|
|
||
|
|
1. [Wikipedia]_ http://en.wikipedia.org/wiki/Cholesky_decomposition
|
||
|
|
|
||
|
|
"""
|
||
|
|
if not isinstance(A, ctx.matrix):
|
||
|
|
raise RuntimeError("A should be a type of ctx.matrix")
|
||
|
|
if not A.rows == A.cols:
|
||
|
|
raise ValueError('need n*n matrix')
|
||
|
|
if tol is None:
|
||
|
|
tol = +ctx.eps
|
||
|
|
n = A.rows
|
||
|
|
L = ctx.matrix(n)
|
||
|
|
for j in xrange(n):
|
||
|
|
c = ctx.re(A[j,j])
|
||
|
|
if abs(c-A[j,j]) > tol:
|
||
|
|
raise ValueError('matrix is not Hermitian')
|
||
|
|
s = c - ctx.fsum((L[j,k] for k in xrange(j)),
|
||
|
|
absolute=True, squared=True)
|
||
|
|
if s < tol:
|
||
|
|
raise ValueError('matrix is not positive-definite')
|
||
|
|
L[j,j] = ctx.sqrt(s)
|
||
|
|
for i in xrange(j, n):
|
||
|
|
it1 = (L[i,k] for k in xrange(j))
|
||
|
|
it2 = (L[j,k] for k in xrange(j))
|
||
|
|
t = ctx.fdot(it1, it2, conjugate=True)
|
||
|
|
L[i,j] = (A[i,j] - t) / L[j,j]
|
||
|
|
return L
|
||
|
|
|
||
|
|
def cholesky_solve(ctx, A, b, **kwargs):
|
||
|
|
"""
|
||
|
|
Ax = b => x
|
||
|
|
|
||
|
|
Solve a symmetric positive-definite linear equation system.
|
||
|
|
This is twice as efficient as lu_solve.
|
||
|
|
|
||
|
|
Typical use cases:
|
||
|
|
* A.T*A
|
||
|
|
* Hessian matrix
|
||
|
|
* differential equations
|
||
|
|
"""
|
||
|
|
prec = ctx.prec
|
||
|
|
try:
|
||
|
|
ctx.prec += 10
|
||
|
|
# do not overwrite A nor b
|
||
|
|
A, b = ctx.matrix(A, **kwargs).copy(), ctx.matrix(b, **kwargs).copy()
|
||
|
|
if A.rows != A.cols:
|
||
|
|
raise ValueError('can only solve determined system')
|
||
|
|
# Cholesky factorization
|
||
|
|
L = ctx.cholesky(A)
|
||
|
|
# solve
|
||
|
|
n = L.rows
|
||
|
|
if len(b) != n:
|
||
|
|
raise ValueError("Value should be equal to n")
|
||
|
|
for i in xrange(n):
|
||
|
|
b[i] -= ctx.fsum(L[i,j] * b[j] for j in xrange(i))
|
||
|
|
b[i] /= L[i,i]
|
||
|
|
x = ctx.U_solve(L.T, b)
|
||
|
|
return x
|
||
|
|
finally:
|
||
|
|
ctx.prec = prec
|
||
|
|
|
||
|
|
def det(ctx, A):
|
||
|
|
"""
|
||
|
|
Calculate the determinant of a matrix.
|
||
|
|
"""
|
||
|
|
prec = ctx.prec
|
||
|
|
try:
|
||
|
|
# do not overwrite A
|
||
|
|
A = ctx.matrix(A).copy()
|
||
|
|
# use LU factorization to calculate determinant
|
||
|
|
try:
|
||
|
|
R, p = ctx.LU_decomp(A)
|
||
|
|
except ZeroDivisionError:
|
||
|
|
return 0
|
||
|
|
z = 1
|
||
|
|
for i, e in enumerate(p):
|
||
|
|
if i != e:
|
||
|
|
z *= -1
|
||
|
|
for i in xrange(A.rows):
|
||
|
|
z *= R[i,i]
|
||
|
|
return z
|
||
|
|
finally:
|
||
|
|
ctx.prec = prec
|
||
|
|
|
||
|
|
def cond(ctx, A, norm=None):
|
||
|
|
"""
|
||
|
|
Calculate the condition number of a matrix using a specified matrix norm.
|
||
|
|
|
||
|
|
The condition number estimates the sensitivity of a matrix to errors.
|
||
|
|
Example: small input errors for ill-conditioned coefficient matrices
|
||
|
|
alter the solution of the system dramatically.
|
||
|
|
|
||
|
|
For ill-conditioned matrices it's recommended to use qr_solve() instead
|
||
|
|
of lu_solve(). This does not help with input errors however, it just avoids
|
||
|
|
to add additional errors.
|
||
|
|
|
||
|
|
Definition: cond(A) = ||A|| * ||A**-1||
|
||
|
|
"""
|
||
|
|
if norm is None:
|
||
|
|
norm = lambda x: ctx.mnorm(x,1)
|
||
|
|
return norm(A) * norm(ctx.inverse(A))
|
||
|
|
|
||
|
|
def lu_solve_mat(ctx, a, b):
|
||
|
|
"""Solve a * x = b where a and b are matrices."""
|
||
|
|
r = ctx.matrix(a.rows, b.cols)
|
||
|
|
for i in range(b.cols):
|
||
|
|
c = ctx.lu_solve(a, b.column(i))
|
||
|
|
for j in range(len(c)):
|
||
|
|
r[j, i] = c[j]
|
||
|
|
return r
|
||
|
|
|
||
|
|
def qr(ctx, A, mode = 'full', edps = 10):
|
||
|
|
"""
|
||
|
|
Compute a QR factorization $A = QR$ where
|
||
|
|
A is an m x n matrix of real or complex numbers where m >= n
|
||
|
|
|
||
|
|
mode has following meanings:
|
||
|
|
(1) mode = 'raw' returns two matrixes (A, tau) in the
|
||
|
|
internal format used by LAPACK
|
||
|
|
(2) mode = 'skinny' returns the leading n columns of Q
|
||
|
|
and n rows of R
|
||
|
|
(3) Any other value returns the leading m columns of Q
|
||
|
|
and m rows of R
|
||
|
|
|
||
|
|
edps is the increase in mp precision used for calculations
|
||
|
|
|
||
|
|
**Examples**
|
||
|
|
|
||
|
|
>>> from mpmath import *
|
||
|
|
>>> mp.dps = 15
|
||
|
|
>>> mp.pretty = True
|
||
|
|
>>> A = matrix([[1, 2], [3, 4], [1, 1]])
|
||
|
|
>>> Q, R = qr(A)
|
||
|
|
>>> Q
|
||
|
|
[-0.301511344577764 0.861640436855329 0.408248290463863]
|
||
|
|
[-0.904534033733291 -0.123091490979333 -0.408248290463863]
|
||
|
|
[-0.301511344577764 -0.492365963917331 0.816496580927726]
|
||
|
|
>>> R
|
||
|
|
[-3.3166247903554 -4.52267016866645]
|
||
|
|
[ 0.0 0.738548945875996]
|
||
|
|
[ 0.0 0.0]
|
||
|
|
>>> Q * R
|
||
|
|
[1.0 2.0]
|
||
|
|
[3.0 4.0]
|
||
|
|
[1.0 1.0]
|
||
|
|
>>> chop(Q.T * Q)
|
||
|
|
[1.0 0.0 0.0]
|
||
|
|
[0.0 1.0 0.0]
|
||
|
|
[0.0 0.0 1.0]
|
||
|
|
>>> B = matrix([[1+0j, 2-3j], [3+j, 4+5j]])
|
||
|
|
>>> Q, R = qr(B)
|
||
|
|
>>> nprint(Q)
|
||
|
|
[ (-0.301511 + 0.0j) (0.0695795 - 0.95092j)]
|
||
|
|
[(-0.904534 - 0.301511j) (-0.115966 + 0.278318j)]
|
||
|
|
>>> nprint(R)
|
||
|
|
[(-3.31662 + 0.0j) (-5.72872 - 2.41209j)]
|
||
|
|
[ 0.0 (3.91965 + 0.0j)]
|
||
|
|
>>> Q * R
|
||
|
|
[(1.0 + 0.0j) (2.0 - 3.0j)]
|
||
|
|
[(3.0 + 1.0j) (4.0 + 5.0j)]
|
||
|
|
>>> chop(Q.T * Q.conjugate())
|
||
|
|
[1.0 0.0]
|
||
|
|
[0.0 1.0]
|
||
|
|
|
||
|
|
"""
|
||
|
|
|
||
|
|
# check values before continuing
|
||
|
|
assert isinstance(A, ctx.matrix)
|
||
|
|
m = A.rows
|
||
|
|
n = A.cols
|
||
|
|
assert n >= 0
|
||
|
|
assert m >= n
|
||
|
|
assert edps >= 0
|
||
|
|
|
||
|
|
# check for complex data type
|
||
|
|
cmplx = any(type(x) is ctx.mpc for x in A)
|
||
|
|
|
||
|
|
# temporarily increase the precision and initialize
|
||
|
|
with ctx.extradps(edps):
|
||
|
|
tau = ctx.matrix(n,1)
|
||
|
|
A = A.copy()
|
||
|
|
|
||
|
|
# ---------------
|
||
|
|
# FACTOR MATRIX A
|
||
|
|
# ---------------
|
||
|
|
if cmplx:
|
||
|
|
one = ctx.mpc('1.0', '0.0')
|
||
|
|
zero = ctx.mpc('0.0', '0.0')
|
||
|
|
rzero = ctx.mpf('0.0')
|
||
|
|
|
||
|
|
# main loop to factor A (complex)
|
||
|
|
for j in xrange(0, n):
|
||
|
|
alpha = A[j,j]
|
||
|
|
alphr = ctx.re(alpha)
|
||
|
|
alphi = ctx.im(alpha)
|
||
|
|
|
||
|
|
if (m-j) >= 2:
|
||
|
|
xnorm = ctx.fsum( A[i,j]*ctx.conj(A[i,j]) for i in xrange(j+1, m) )
|
||
|
|
xnorm = ctx.re( ctx.sqrt(xnorm) )
|
||
|
|
else:
|
||
|
|
xnorm = rzero
|
||
|
|
|
||
|
|
if (xnorm == rzero) and (alphi == rzero):
|
||
|
|
tau[j] = zero
|
||
|
|
continue
|
||
|
|
|
||
|
|
if alphr < rzero:
|
||
|
|
beta = ctx.sqrt(alphr**2 + alphi**2 + xnorm**2)
|
||
|
|
else:
|
||
|
|
beta = -ctx.sqrt(alphr**2 + alphi**2 + xnorm**2)
|
||
|
|
|
||
|
|
tau[j] = ctx.mpc( (beta - alphr) / beta, -alphi / beta )
|
||
|
|
t = -ctx.conj(tau[j])
|
||
|
|
za = one / (alpha - beta)
|
||
|
|
|
||
|
|
for i in xrange(j+1, m):
|
||
|
|
A[i,j] *= za
|
||
|
|
|
||
|
|
A[j,j] = one
|
||
|
|
for k in xrange(j+1, n):
|
||
|
|
y = ctx.fsum(A[i,j] * ctx.conj(A[i,k]) for i in xrange(j, m))
|
||
|
|
temp = t * ctx.conj(y)
|
||
|
|
for i in xrange(j, m):
|
||
|
|
A[i,k] += A[i,j] * temp
|
||
|
|
|
||
|
|
A[j,j] = ctx.mpc(beta, '0.0')
|
||
|
|
else:
|
||
|
|
one = ctx.mpf('1.0')
|
||
|
|
zero = ctx.mpf('0.0')
|
||
|
|
|
||
|
|
# main loop to factor A (real)
|
||
|
|
for j in xrange(0, n):
|
||
|
|
alpha = A[j,j]
|
||
|
|
|
||
|
|
if (m-j) > 2:
|
||
|
|
xnorm = ctx.fsum( (A[i,j])**2 for i in xrange(j+1, m) )
|
||
|
|
xnorm = ctx.sqrt(xnorm)
|
||
|
|
elif (m-j) == 2:
|
||
|
|
xnorm = abs( A[m-1,j] )
|
||
|
|
else:
|
||
|
|
xnorm = zero
|
||
|
|
|
||
|
|
if xnorm == zero:
|
||
|
|
tau[j] = zero
|
||
|
|
continue
|
||
|
|
|
||
|
|
if alpha < zero:
|
||
|
|
beta = ctx.sqrt(alpha**2 + xnorm**2)
|
||
|
|
else:
|
||
|
|
beta = -ctx.sqrt(alpha**2 + xnorm**2)
|
||
|
|
|
||
|
|
tau[j] = (beta - alpha) / beta
|
||
|
|
t = -tau[j]
|
||
|
|
da = one / (alpha - beta)
|
||
|
|
|
||
|
|
for i in xrange(j+1, m):
|
||
|
|
A[i,j] *= da
|
||
|
|
|
||
|
|
A[j,j] = one
|
||
|
|
for k in xrange(j+1, n):
|
||
|
|
y = ctx.fsum( A[i,j] * A[i,k] for i in xrange(j, m) )
|
||
|
|
temp = t * y
|
||
|
|
for i in xrange(j,m):
|
||
|
|
A[i,k] += A[i,j] * temp
|
||
|
|
|
||
|
|
A[j,j] = beta
|
||
|
|
|
||
|
|
# return factorization in same internal format as LAPACK
|
||
|
|
if (mode == 'raw') or (mode == 'RAW'):
|
||
|
|
return A, tau
|
||
|
|
|
||
|
|
# ----------------------------------
|
||
|
|
# FORM Q USING BACKWARD ACCUMULATION
|
||
|
|
# ----------------------------------
|
||
|
|
|
||
|
|
# form R before the values are overwritten
|
||
|
|
R = A.copy()
|
||
|
|
for j in xrange(0, n):
|
||
|
|
for i in xrange(j+1, m):
|
||
|
|
R[i,j] = zero
|
||
|
|
|
||
|
|
# set the value of p (number of columns of Q to return)
|
||
|
|
p = m
|
||
|
|
if (mode == 'skinny') or (mode == 'SKINNY'):
|
||
|
|
p = n
|
||
|
|
|
||
|
|
# add columns to A if needed and initialize
|
||
|
|
A.cols += (p-n)
|
||
|
|
for j in xrange(0, p):
|
||
|
|
A[j,j] = one
|
||
|
|
for i in xrange(0, j):
|
||
|
|
A[i,j] = zero
|
||
|
|
|
||
|
|
# main loop to form Q
|
||
|
|
for j in xrange(n-1, -1, -1):
|
||
|
|
t = -tau[j]
|
||
|
|
A[j,j] += t
|
||
|
|
|
||
|
|
for k in xrange(j+1, p):
|
||
|
|
if cmplx:
|
||
|
|
y = ctx.fsum(A[i,j] * ctx.conj(A[i,k]) for i in xrange(j+1, m))
|
||
|
|
temp = t * ctx.conj(y)
|
||
|
|
else:
|
||
|
|
y = ctx.fsum(A[i,j] * A[i,k] for i in xrange(j+1, m))
|
||
|
|
temp = t * y
|
||
|
|
A[j,k] = temp
|
||
|
|
for i in xrange(j+1, m):
|
||
|
|
A[i,k] += A[i,j] * temp
|
||
|
|
|
||
|
|
for i in xrange(j+1, m):
|
||
|
|
A[i, j] *= t
|
||
|
|
|
||
|
|
return A, R[0:p,0:n]
|
||
|
|
|
||
|
|
# ------------------
|
||
|
|
# END OF FUNCTION QR
|
||
|
|
# ------------------
|