// Daniel Knüttel Aufgabe2
/*
* Copyright(c) 2018 Daniel Knüttel
*/
/* This program is free software: you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation, either version 3 of the License, or
* (at your option) any later version.
*
* This program is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with this program. If not, see .
*
* Dieses Programm ist Freie Software: Sie können es unter den Bedingungen
* der GNU General Public License, wie von der Free Software Foundation,
* Version 3 der Lizenz oder (nach Ihrer Wahl) jeder neueren
* veröffentlichten Version, weiterverbreiten und/oder modifizieren.
*
* Dieses Programm wird in der Hoffnung, dass es nützlich sein wird, aber
* OHNE JEDE GEWÄHRLEISTUNG, bereitgestellt; sogar ohne die implizite
* Gewährleistung der MARKTFÄHIGKEIT oder EIGNUNG FÜR EINEN BESTIMMTEN ZWECK.
* Siehe die GNU General Public License für weitere Details.
*
* Sie sollten eine Kopie der GNU General Public License zusammen mit diesem
* Programm erhalten haben. Wenn nicht, siehe .
*/
#include
#include
#include
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define square(a) ((a) * (a))
#define kronecker_delta(i, j) ((i) == (j) ? 1 : 0)
/*
* \brief Computes the householder partition of A.
* Stores the generating vectors of Q in the memory passed as A,
* the diagonal elements in alpha and the rest of R in the memory
* passed as A.
*
* \param A Matrix to partition. After the partition this memory contains Q and R.
* \param alpha Will be filled with diagonal elements of R.
* \param n Length of A[i].
* \param m Length of A.
* \return Status.
* */
int householder(double ** A, double * alpha, int m, int n);
/*
* \brief Computes the solution of and n times n system using a QR partition.
* Overwrites b.
*
* \param A Matrix that has been modified by the function householder.
* \param alpha Diagonal elements of R, produced by the function householder.
* \param n Length of A and A[i].
* \param b Vector b for Ax=b. Will be overwritten with elements of x.
* \return Status.
* */
int solve_QR(double ** A, double * alpha, int n, double * b);
/*
* \brief Computes Q^Tb and stores the result in b.
*
* \param A Matrix that has been modified by the function householder.
* \param alpha Diagonal elements of R, produced by the function householder.
* \param n Length of A and A[i].
* \param b Vector b for Ax=b. Will be overwritten with elements of the result.
*
* \return Status.
* */
int qtb(double ** A, double * alpha, int n, double * b);
/*
*
* \brief Uses backwards substitution to solve R = Q^Tb. b := Q^Tb has already been calculated.
*
* \param A Matrix that has been modified by the function householder.
* \param alpha Diagonal elements of R, produced by the function householder.
* \param n Length of A and A[i].
* \param b Vector b for Ax=b. Will be overwritten with elements of the result.
*
* \return Status.
*
* */
int rw_subs(double ** A, double * alpha, int n, double * b);
/*
* \brief Print the matrix in a nicely formatted way to stream.
*
* \param stream The output stream.
* \param matrix The matrix to print.
* \param n Length of A[i].
* \param m Length of A.
*
* */
int fprintm(FILE * stream, double ** matrix, int m, int n);
/*
* \brief Print the vector in a nicely formatted way to stream.
*
* \param stream The output stream.
* \param vector The vector to print.
* \param n Length of vector.
*
* */
int fprintv(FILE * stream, double * vector, int n);
/*
* \brief Extract R from A and alpha, allocates R dynamically.
*
* \param A Marix computed by householder.
* \param alpha Vector computed by householder.
* \param m Length of A.
* \param n Length of A[i].
* \param R Pointer to the result (unallocated).
* */
int unwind_R(double ** A, double * alpha, int m, int n, double *** R);
/*
* \brief Extract v from A and alpha, allocates R dynamically.
*
* \param A Marix computed by householder.
* \param alpha Vector computed by householder.
* \param m Length of A.
* \param n Length of A[i].
* \param v Pointer to the results (unallocated).
* */
int unwind_v(double ** A, int m, int n, double *** v);
#define printm(matrix, n, m) fprintm(stdout, matrix, n, m)
#define printv(vector, n) fprintv(stdout, vector, n)
int main(void)
{
int status = 0;
int i, j;
int m, n;
// PART a: QR-Partition.
double A_stack_1[15] = {2, 2.23606797749979, 0
, 2, 3, -1
, 2, 3, -1
, 1, 0, -1
, 0, 2.6457513110645907, 5.669467095138408
};
m = 5;
n = 3;
double ** A_1 = malloc(sizeof(double *) * m);
for(i = 0; i < m; i++)
{
A_1[i] = malloc(sizeof(double) * n);
for(j = 0; j < n; j++)
{
A_1[i][j] = A_stack_1[n*i + j];
}
}
printf("A:\n");
printm(A_1, m, n);
double * alpha_1 = malloc(sizeof(double) * m);
status = householder(A_1, alpha_1, m, n);
if(status)
{
fprintf(stderr, "failed to compute QR decomposition\n");
goto exit;
}
double ** R_1, ** vs_1;
status = unwind_R(A_1, alpha_1, m, n, &R_1);
if(status)
{
fprintf(stderr, "failed to unwind R\n");
goto exit;
}
status = unwind_v(A_1, m, n, &vs_1);
if(status)
{
fprintf(stderr, "failed to unwind v\n");
for(i = 0; i < m; i++)
{
free(R_1[i]);
}
free(R_1);
goto exit;
}
printf("R:\n");
printm(R_1, m, n);
printf("Generating vectors of Q:\n");
for(i = 0; i < n; i++)
{
printv(vs_1[i], m);
free(vs_1[i]);
}
free(vs_1);
for(i = 0; i < m; i++)
{
free(R_1[i]);
}
free(R_1);
free(alpha_1);
for(i = 0; i < m; i++)
{
free(A_1[i]);
}
free(A_1);
double A_stack_2[16] = {2, -1, 3, 0
, 7, 0, 1.4142135623730951, 1
, 1, 1, 1, 1
, 0, 10, 3, 2
};
m = 4;
n = 4;
double ** A_2 = malloc(sizeof(double *) * m);
for(i = 0; i < m; i++)
{
A_2[i] = malloc(sizeof(double) * n);
for(j = 0; j < n; j++)
{
A_2[i][j] = A_stack_2[n*i + j];
}
}
double * b = malloc(sizeof(double) * m);
double b_stack[4] = {1, 0, 0, 0};
for(i = 0; i < m; i++)
{
b[i] = b_stack[i];
}
printf("System: A, b\n");
printm(A_2, m, m);
printv(b, m);
double * alpha_2 = malloc(sizeof(double) * m);
status = householder(A_2, alpha_2, m, n);
if(status)
{
fprintf(stderr, "failed to compute QR decomposition\n");
goto exit;
}
status = solve_QR(A_2, alpha_2, m, b);
printf("Solution for the System:\n");
printv(b, m);
free(alpha_2);
for(i = 0; i < m; i++)
{
free(A_2[i]);
}
free(A_2);
free(b);
exit:
if(status)
{
fprintf(stderr, "something went wrong\n");
}
return status;
}
int householder(double ** A, double * alpha, int m, int n)
{
if(n > m)
{
return 1;
}
int k, i, j;
double beta, gamma, delta;
/*
* This is just the sample implementation from the
* lecture script.
* */
for(k = 0; k < min(n, m - 1); k++)
{
alpha[k] = square(A[k][k]);
for(j = k + 1; j < m; j++)
{
alpha[k] += square(A[j][k]);
}
alpha[k] = sqrt(alpha[k]);
if(A[k][k] < 0)
{
alpha[k] *= -1;
}
beta = alpha[k] * (A[k][k] - alpha[k]);
A[k][k] -= alpha[k];
for(i = k + 1; i < n ; i++)
{
gamma = A[k][k] * A[k][i];
for(j = k + 1; j < m; j++)
{
gamma += A[j][k] * A[j][i];
}
delta = gamma / beta;
for(j = k; j < m; j++)
{
A[j][i] += delta * A[j][k];
}
}
}
return 0;
}
int fprintm(FILE * stream, double ** matrix, int m, int n)
{
int i, j, status = 0;
for(i = 0; i < m; i++)
{
if(i == 0)
{
fprintf(stream, "T ");
}
else if(i == m - 1)
{
fprintf(stream, "L ");
}
else
{
fprintf(stream, "| ");
}
for(j = 0; j < n; j++)
{
fprintf(stream, "%6.2f ", matrix[i][j]);
}
if(i == 0)
{
fprintf(stream, "T\n");
}
else if(i == m - 1)
{
fprintf(stream, "J\n");
}
else
{
fprintf(stream, "|\n");
}
}
fflush(stream);
return status;
}
int fprintv(FILE * stream, double * vector, int n)
{
int i;
for(i = 0; i < n; i++)
{
if(i == 0)
{
fprintf(stream, "T ");
}
else if(i == n - 1)
{
fprintf(stream, "L ");
}
else
{
fprintf(stream, "| ");
}
fprintf(stream, "%6.2f ", vector[i]);
if(i == 0)
{
fprintf(stream, "T\n");
}
else if(i == n - 1)
{
fprintf(stream, "J\n");
}
else
{
fprintf(stream, "|\n");
}
}
fflush(stream);
return 0;
}
int qtb(double ** A, double * alpha, int n, double * b)
{
/*
* Some short computing gives that:
*
* Q[i][j] = kronecker_delta(i, j) - 2*v[i]v[j]
* and for
* Q^T b = x
* x[i] = sum(over j)(Q[j][i]*b[j])
* */
int i, j;
double x;
for(i = 0; i < n; i++)
{
for(j = 0; j < n; j++)
{
x += b[j] * (kronecker_delta(i, j)
- 2 * A[i][i] * A[i][j]);
}
b[i] = x;
}
return 0;
}
int rw_subs(double ** A, double * alpha, int n, double * b)
{
/*
* In the lecture section 1 about the gaussian algorithm
* we have given the following formula:
*
* x[j] = 1/A[j][j] * (b[j] - sum(from j + 1 over k to n)(a[j][k]*x[k]))
*
* So loop from k = n to 1 to avoid recursion and
* substitute the content of b with the solution.
* */
int j, k;
double tmp;
// Note: indices start at 0.
for(j = n - 1; j >= 0; j--)
{
// Note: we start from k + 1, so there are no
// diagonal elements of A contained.
for(k = j + 1; k < n; k++)
{
tmp += b[k] * A[j][k];
}
// A[j][j] = alpha[j].
b[j] = (b[j] - tmp) / alpha[j];
}
return 0;
}
int solve_QR(double ** A, double * alpha, int n, double * b)
{
if(qtb(A, alpha, n, b))
{
return 1;
}
if(rw_subs(A, alpha, n, b))
{
return 1;
}
return 0;
}
int unwind_R(double ** A, double * alpha, int m, int n, double *** R)
{
double ** result = malloc(sizeof(double *) * m);
int i, j;
int have_to_free_all = 0;
if(!result)
{
return 1;
}
// Allocate the memory.
for(i = 0; i < m; i++)
{
result[i] = malloc(sizeof(double) * n);
if(!result[i])
{
have_to_free_all = 1;
break;
}
}
// Out of Memory.
if(have_to_free_all)
{
for(j = 0; j < i; j++)
{
free(result[i]);
}
free(result);
return 1;
}
// Alloc is OK.
// Calculate the result.
for(i = 0; i < m; i++)
{
for(j = 0; j < n; j++)
{
if(i > j)
{
result[i][j] = 0;
}
if(i == j)
{
result[i][j] = alpha[j];
}
if(i < j)
{
result[i][j] = A[i][j];
}
}
}
*R = result;
return 0;
}
int unwind_v(double ** A, int m, int n, double *** v)
{
// We have n vectors.
double ** result = malloc(sizeof(double *) * n);
int i, j;
int have_to_free_all = 0;
if(!result)
{
return 1;
}
// Allocate the memory.
for(i = 0; i < n; i++)
{
// They are m long.
result[i] = malloc(sizeof(double) * m);
if(!result[i])
{
have_to_free_all = 1;
break;
}
}
// Out of Memory.
if(have_to_free_all)
{
for(j = 0; j < i; j++)
{
free(result[i]);
}
free(result);
return 1;
}
// Alloc is OK.
// Calculate the result.
for(i = 0; i < m; i++)
{
for(j = 0; j < n; j++)
{
if(i > j)
{
result[j][i] = 0;
}
else
{
result[j][i] = A[j][i];
}
}
}
*v = result;
return 0;
}