120 lines
3.5 KiB
Python
120 lines
3.5 KiB
Python
# -*- coding: utf-8 -*-
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"""
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Created on Sat Nov 7 13:26:09 2020
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Höhere Mathematik 1, Serie 8, Gerüst für Aufgabe 2
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Description: calculates the QR factorization of A so that A = QR
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Input Parameters: A: array, n*n matrix
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Output Parameters: Q : n*n orthogonal matrix
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R : n*n upper right triangular matrix
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Remarks: none
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Example: A = np.array([[1,2,-1],[4,-2,6],[3,1,0]])
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[Q,R]=Serie8_Aufg2(A)
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@author: knaa
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"""
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import numpy as np
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import timeit
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# Aufgabe 2a
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def Serie8_Aufg2(A):
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A = np.copy(A) #necessary to prevent changes in the original matrix A_in
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A = A.astype('float64') #change to float
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n = np.shape(A)[0]
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if n != np.shape(A)[1]:
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raise Exception('Matrix is not square')
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Q = np.eye(n)
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R = A
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for j in np.arange(0,n-1):
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# erzeuge Nullen in R in der j-ten Spalte unterhalb der Diagonalen:
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#a = (Q @ A)[j:,j:][:,0]
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a = np.copy((Q @ A)[j:,j:][:,0]).reshape(n-j,1)
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e = np.eye(n-j)[:,0].reshape(n-j,1)
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length_a = np.linalg.norm(a)
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if a[0] >= 0: sig = 1
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else: sig = -1
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v = a + sig * length_a * e # vj := aj + sign(a1j) · |aj| · ej
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u = 1 / np.linalg.norm(v) * v # uj := 1/|vj|*vj
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ut = u.T
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ua = u @ ut
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ub = 2 * ua
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x = np.eye(n)
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H = np.eye(n-j) - (2 * (u @ u.T)) # Hj := In − 2u1u1T bestimme die (n − j + 1) × (n − j + 1) Householder-Matrix Hj
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Qj = np.eye(n)
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Qj[j:,j:] = H # erweitere Hi durch einen Ii−1 Block links oben zur n × n Matrix Qi
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R = Qj @ R # R := Qj · R
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Q = Q @ Qj.T # Q := Q · QjT
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return(Q,R)
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if __name__ == '__main__':
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# Beispiel aus Skript
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# A = np.array([[1, 2, -1], [4, -2, 6], [3, 1, 0]])
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# b = np.array([
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# [9],
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# [-4],
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# [9]
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# ])
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# Beispiel aus Aufgabe 1
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A = np.array([
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[1, -2, 3],
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[-5, 4, 1],
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[2, -1, 3]
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])
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b = np.array([
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[1],
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[9],
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[5]
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])
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[Q,R]=Serie8_Aufg2(A)
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# Aufgabe 2b
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n = len(b) - 1
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QTb = Q.T @ b
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result = [0 for i in range(n+1)]
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row = n
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while row >= 0:
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value = QTb[row][0]
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column = n
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while column > row:
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value -= R[row][column] * result[column]
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column -= 1
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value = value / R[row,row]
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result[row] = value
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row -= 1
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print("\nQ:\n", Q)
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print("\nR:\n", R)
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print("\Result:\n", result)
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# Aufgabe 2c
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t1 = timeit.repeat("Serie8_Aufg2(A)", "from __main__ import Serie8_Aufg2, A", number=100)
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t2 = timeit.repeat("np.linalg.qr(A)", "from __main__ import np, A", number=100)
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avg_t1 = np.average(t1) / 100
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avg_t2 = np.average(t2) / 100
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print("Geschwindigkeit mit 3x3 Matrix:")
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print("Benötigte Zeit mit eigener Funktion:", avg_t1)
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print("Benötigte Zeit mit Numpy:", avg_t2)
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# Aufgabe 2d
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Test = np.random.rand(100,100)
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t1 = timeit.repeat("Serie8_Aufg2(Test)", "from __main__ import Serie8_Aufg2, Test", number=100)
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t2 = timeit.repeat("np.linalg.qr(Test)", "from __main__ import np, Test", number=100)
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avg_t1 = np.average(t1) / 100
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avg_t2 = np.average(t2) / 100
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print("Geschwindigkeit mit 100x100 Matrix:")
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print("Benötigte Zeit mit eigener Funktion:", avg_t1)
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print("Benötigte Zeit mit Numpy:", avg_t2)
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# Die von Numpy zur Verfügung gestellt Funktion ist wesentlich effizienter. |