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Projekt1_Knuettel_Daniel.c
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Projekt1_Knuettel_Daniel.c
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// Daniel Knüttel Aufgabe2
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/*
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* Copyright(c) 2018 Daniel Knüttel
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*/
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/* This program is free software: you can redistribute it and/or modify
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* it under the terms of the GNU General Public License as published by
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* the Free Software Foundation, either version 3 of the License, or
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* (at your option) any later version.
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*
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* This program is distributed in the hope that it will be useful,
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* but WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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* GNU General Public License for more details.
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*
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* You should have received a copy of the GNU General Public License
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* along with this program. If not, see <http://www.gnu.org/licenses/>.
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*
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* Dieses Programm ist Freie Software: Sie können es unter den Bedingungen
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* der GNU General Public License, wie von der Free Software Foundation,
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* Version 3 der Lizenz oder (nach Ihrer Wahl) jeder neueren
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* veröffentlichten Version, weiterverbreiten und/oder modifizieren.
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*
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* Dieses Programm wird in der Hoffnung, dass es nützlich sein wird, aber
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* OHNE JEDE GEWÄHRLEISTUNG, bereitgestellt; sogar ohne die implizite
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* Gewährleistung der MARKTFÄHIGKEIT oder EIGNUNG FÜR EINEN BESTIMMTEN ZWECK.
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* Siehe die GNU General Public License für weitere Details.
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*
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* Sie sollten eine Kopie der GNU General Public License zusammen mit diesem
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* Programm erhalten haben. Wenn nicht, siehe <http://www.gnu.org/licenses/>.
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*/
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#include <stdio.h>
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#include <stdlib.h>
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/*
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* \brief Computes the householder partition of A.
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* Stores the generating vectors of Q in the memory passed as A,
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* the diagonal elements in alpha and the rest of R in the memory
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* passed as A.
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*
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* \param A Matrix to partition. After the partition this memory contains Q and R.
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* \param alpha Will be filled with diagonal elements of R.
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* \param n Length of A[i].
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* \param m Length of A.
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* \return Status.
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* */
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int householder(double ** A, double * alpha, int n, int m);
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/*
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* \brief Computes the solution of and n times n system using a QR partition.
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* Overwrites b.
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*
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* \param A Matrix that has been modified by the function householder.
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* \param alpha Diagonal elements of R, produced by the function householder.
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* \param n Length of A and A[i].
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* \param b Vector b for Ax=b. Will be overwritten with elements of x.
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* \return Status.
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* */
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int solve_QR(double ** A, double * alpha, int n, double * b);
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/*
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* \brief Computes Q^Tb and stores the result in b.
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*
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* \param A Matrix that has been modified by the function householder.
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* \param alpha Diagonal elements of R, produced by the function householder.
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* \param n Length of A and A[i].
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* \param b Vector b for Ax=b. Will be overwritten with elements of the result.
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*
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* \return Status.
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* */
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int qtb(double ** A, double * alpha, int n, double * b);
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/*
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*
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* \brief Uses backwards substitution to solve R = Q^Tb.
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*
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* \param A Matrix that has been modified by the function householder.
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* \param alpha Diagonal elements of R, produced by the function householder.
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* \param n Length of A and A[i].
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* \param b Vector b for Ax=b. Will be overwritten with elements of the result.
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*
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* \return Status.
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*
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* */
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int rw_subs(double ** A, double * alpha, int n, double * b);
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/*
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* \brief Print the matrix in a nicely formatted way to stream.
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*
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* \param stream The output stream.
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* \param matrix The matrix to print.
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* \param n Length of matrix[i].
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* \param m Length of matrix.
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*
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* */
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int fprintm(FILE * stream, double ** matrix, int n, int m);
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/*
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* \brief Print the vector in a nicely formatted way to stream.
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*
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* \param stream The output stream.
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* \param vector The vector to print.
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* \param n Length of vector.
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*
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* */
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int fprintv(FILE * stream, double * vector, int n);
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#define printm(matrix, n, m) fprintm(stdout, matrix, n, m)
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#define printv(vector, n) fprintv(stdout, vector, n)
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int main(void)
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{
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int status = 0;
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return status;
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}
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int householder(double ** A, double * alpha, int n, int m)
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{
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int i, j;
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for(i = 0; i < n; i++)
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{
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// Bring this column in triangular shape.
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}
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}
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75
gaussian.py
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gaussian.py
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import copy
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import pprint
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import numpy
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def gaussian_with_pivot(A, b, round = lambda x: x):
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A = numpy.array(A, dtype = numpy.float64)
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b = numpy.array(b, dtype = numpy.float64)
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L = numpy.identity(len(A), dtype = numpy.float64)
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s = numpy.zeros(len(A), dtype = numpy.float64)
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for k in range(len(A)):
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max_a = numpy.absolute(A[k:,k]).argmax()
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A[[k, k + max_a]] = A[[k + max_a, k]]
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b[[k, k + max_a]] = b[[k + max_a, k]]
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s[k] = k + max_a
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for i in range(k + 1, len(A)):
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L[i][k] = round(A[i][k]/A[k][k])
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b[i] = round(b[i] - round(L[i][k] * b[k]))
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for j in range(k + 1, len(A)):
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A[i][j] = round(A[i][j] - round(L[i][k] * A[k][j]))
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A[i][k] = 0
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return A, b, L, s
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def gaussian_no_pivot(A, b, round = lambda x: x):
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A = numpy.array(A, dtype = numpy.float64)
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b = numpy.array(b, dtype = numpy.float64)
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L = numpy.identity(len(A), dtype = numpy.float64)
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for k in range(len(A)):
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for i in range(k + 1, len(A)):
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L[i][k] = round(A[i][k]/A[k][k])
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b[i] = round(b[i] - round(L[i][k] * b[k]))
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for j in range(k + 1, len(A)):
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A[i][j] = round(A[i][j] - round(L[i][k] * A[k][j]))
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A[i][k] = 0
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return A, b, L
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if( __name__ == "__main__"):
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# For Exercise 1
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A = [[10, 1, 0], [15, 2, 1], [5, 0, 1]]
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b = [12, 18, 4]
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pprint.pprint(gaussian_with_pivot(A, b))
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# For Exercise 2
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A = [[7, 1, 1]
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, [10, 1, 1]
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, [1000, 0, 1]]
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b = [10, 13, 1001]
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pprint.pprint(A)
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def round4(x):
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return float("%.4g" % x)
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pprint.pprint(gaussian_with_pivot(A, b, round4))
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pprint.pprint(gaussian_no_pivot(A, b, round4))
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A = numpy.zeros((10, 10))
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for i in range(10):
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if(i - 1 > 0):
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A[i][i - 1] = -1
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A[i][i] = 2
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if(i + 1 < 10):
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A[i][i + 1] = -1
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pprint.pprint(gaussian_no_pivot(A, numpy.ones(10)))
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