Gauss-Seidel method convergence algorithm












3












$begingroup$


From Wikipedia:




The convergence properties of the Gauss–Seidel method
are dependent on the matrix A. Namely, the procedure
is known to converge if either:



1. A is symmetric positive-definite, or

2. A is strictly or irreducibly diagonally dominant.



Is there algorithm for transforming matrix to meet the convergence criteria or must I do it manually?



I am implementing Gauss-Seidel method in C++:



#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;



int n;

double eps = 0.001;



//yygnanma sherti

bool converge(double *xk, double *xkp)

{

    double norm = 0;

    for (int i = 0; i < n; i++) 

    {

      norm += (xk[i] - xkp[i])*(xk[i] - xkp[i]);

    }

    if(sqrt(norm) >= eps)

      return false;

    return true;

}





int main()

{

    //olcheg

    printf("Olchegi giriz: ");

    scanf("%d", &n);



    //koeffisientler uchin

    double a[n][n];

    for(int i=0; i<n; i++)

    {

        for(int j=0; j<n; j++)

        {

            double tmp;

            printf("a[%d][%d] = ", i+1, j+1);

            scanf("%lf", &tmp);

            a[i][j] = tmp;

        }

    }



    //azat agzalar uchin

    double b[n];

    for(int i=0; i<n; i++)

    {

        double tmp;

        printf("b[%d] = ", i+1);

        scanf("%lf", &tmp);

        b[i] = tmp;

    }



    double x[n];

    double p[n];



    for(int i=0; i<n; i++)

    {

        x[i] = 0;

        p[i] = 0;

    }



    do

    {

        for (int i = 0; i < n; i++)

            p[i] = x[i];



        for (int i = 0; i < n; i++)

        {

            double var = 0;

            for (int j = 0; j < i; j++)

                var += (a[i][j] * x[j]);

            for (int j = i; j < n; j++)

                var += (a[i][j] * p[j]);

            x[i] = (b[i] - var) / a[i][i];

        }

    }while (!converge(x, p));



    return 0;

}


Now I want to implement matrix transformation algorithm (if there is one) to meet convergence criteria.






linear-algebra numerical-methods numerical-linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Where does your matrix A come from? Is it symmetric? Clearly arbitrarily "transforming" the matrix to meet some criteria may change the problem you are solving -- so if you tell me more about the problem I can make some suggestions.
    $endgroup$
    – RRL
    May 16 '14 at 19:21












  • $begingroup$
    A is not symmetric matrix. I am going to implement for abitrary matrix
    $endgroup$
    – torayeff
    May 16 '14 at 19:38








  • 1




    $begingroup$
    @torayeff you left out the fact that it will often converge without those conditions. Usually, if the system was small I would manually work it till there was as much diagonal dominance as possible.
    $endgroup$
    – bobbym
    May 16 '14 at 19:51










  • $begingroup$
    Why are you implementing this algorithm for a dense matrix?
    $endgroup$
    – Pawel Kowal
    Jul 14 '16 at 10:01
















3












$begingroup$


From Wikipedia:




The convergence properties of the Gauss–Seidel method
are dependent on the matrix A. Namely, the procedure
is known to converge if either:



1. A is symmetric positive-definite, or

2. A is strictly or irreducibly diagonally dominant.



Is there algorithm for transforming matrix to meet the convergence criteria or must I do it manually?



I am implementing Gauss-Seidel method in C++:



#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;



int n;

double eps = 0.001;



//yygnanma sherti

bool converge(double *xk, double *xkp)

{

    double norm = 0;

    for (int i = 0; i < n; i++) 

    {

      norm += (xk[i] - xkp[i])*(xk[i] - xkp[i]);

    }

    if(sqrt(norm) >= eps)

      return false;

    return true;

}





int main()

{

    //olcheg

    printf("Olchegi giriz: ");

    scanf("%d", &n);



    //koeffisientler uchin

    double a[n][n];

    for(int i=0; i<n; i++)

    {

        for(int j=0; j<n; j++)

        {

            double tmp;

            printf("a[%d][%d] = ", i+1, j+1);

            scanf("%lf", &tmp);

            a[i][j] = tmp;

        }

    }



    //azat agzalar uchin

    double b[n];

    for(int i=0; i<n; i++)

    {

        double tmp;

        printf("b[%d] = ", i+1);

        scanf("%lf", &tmp);

        b[i] = tmp;

    }



    double x[n];

    double p[n];



    for(int i=0; i<n; i++)

    {

        x[i] = 0;

        p[i] = 0;

    }



    do

    {

        for (int i = 0; i < n; i++)

            p[i] = x[i];



        for (int i = 0; i < n; i++)

        {

            double var = 0;

            for (int j = 0; j < i; j++)

                var += (a[i][j] * x[j]);

            for (int j = i; j < n; j++)

                var += (a[i][j] * p[j]);

            x[i] = (b[i] - var) / a[i][i];

        }

    }while (!converge(x, p));



    return 0;

}


Now I want to implement matrix transformation algorithm (if there is one) to meet convergence criteria.






linear-algebra numerical-methods numerical-linear-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Where does your matrix A come from? Is it symmetric? Clearly arbitrarily "transforming" the matrix to meet some criteria may change the problem you are solving -- so if you tell me more about the problem I can make some suggestions.
    $endgroup$
    – RRL
    May 16 '14 at 19:21












  • $begingroup$
    A is not symmetric matrix. I am going to implement for abitrary matrix
    $endgroup$
    – torayeff
    May 16 '14 at 19:38








  • 1




    $begingroup$
    @torayeff you left out the fact that it will often converge without those conditions. Usually, if the system was small I would manually work it till there was as much diagonal dominance as possible.
    $endgroup$
    – bobbym
    May 16 '14 at 19:51










  • $begingroup$
    Why are you implementing this algorithm for a dense matrix?
    $endgroup$
    – Pawel Kowal
    Jul 14 '16 at 10:01














3












3








3





$begingroup$


From Wikipedia:




The convergence properties of the Gauss–Seidel method
are dependent on the matrix A. Namely, the procedure
is known to converge if either:



1. A is symmetric positive-definite, or

2. A is strictly or irreducibly diagonally dominant.



Is there algorithm for transforming matrix to meet the convergence criteria or must I do it manually?



I am implementing Gauss-Seidel method in C++:



#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;



int n;

double eps = 0.001;



//yygnanma sherti

bool converge(double *xk, double *xkp)

{

    double norm = 0;

    for (int i = 0; i < n; i++) 

    {

      norm += (xk[i] - xkp[i])*(xk[i] - xkp[i]);

    }

    if(sqrt(norm) >= eps)

      return false;

    return true;

}





int main()

{

    //olcheg

    printf("Olchegi giriz: ");

    scanf("%d", &n);



    //koeffisientler uchin

    double a[n][n];

    for(int i=0; i<n; i++)

    {

        for(int j=0; j<n; j++)

        {

            double tmp;

            printf("a[%d][%d] = ", i+1, j+1);

            scanf("%lf", &tmp);

            a[i][j] = tmp;

        }

    }



    //azat agzalar uchin

    double b[n];

    for(int i=0; i<n; i++)

    {

        double tmp;

        printf("b[%d] = ", i+1);

        scanf("%lf", &tmp);

        b[i] = tmp;

    }



    double x[n];

    double p[n];



    for(int i=0; i<n; i++)

    {

        x[i] = 0;

        p[i] = 0;

    }



    do

    {

        for (int i = 0; i < n; i++)

            p[i] = x[i];



        for (int i = 0; i < n; i++)

        {

            double var = 0;

            for (int j = 0; j < i; j++)

                var += (a[i][j] * x[j]);

            for (int j = i; j < n; j++)

                var += (a[i][j] * p[j]);

            x[i] = (b[i] - var) / a[i][i];

        }

    }while (!converge(x, p));



    return 0;

}


Now I want to implement matrix transformation algorithm (if there is one) to meet convergence criteria.






linear-algebra numerical-methods numerical-linear-algebra







share|cite|improve this question









$endgroup$




From Wikipedia:




The convergence properties of the Gauss–Seidel method
are dependent on the matrix A. Namely, the procedure
is known to converge if either:



1. A is symmetric positive-definite, or

2. A is strictly or irreducibly diagonally dominant.



Is there algorithm for transforming matrix to meet the convergence criteria or must I do it manually?



I am implementing Gauss-Seidel method in C++:



#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;



int n;

double eps = 0.001;



//yygnanma sherti

bool converge(double *xk, double *xkp)

{

    double norm = 0;

    for (int i = 0; i < n; i++) 

    {

      norm += (xk[i] - xkp[i])*(xk[i] - xkp[i]);

    }

    if(sqrt(norm) >= eps)

      return false;

    return true;

}





int main()

{

    //olcheg

    printf("Olchegi giriz: ");

    scanf("%d", &n);



    //koeffisientler uchin

    double a[n][n];

    for(int i=0; i<n; i++)

    {

        for(int j=0; j<n; j++)

        {

            double tmp;

            printf("a[%d][%d] = ", i+1, j+1);

            scanf("%lf", &tmp);

            a[i][j] = tmp;

        }

    }



    //azat agzalar uchin

    double b[n];

    for(int i=0; i<n; i++)

    {

        double tmp;

        printf("b[%d] = ", i+1);

        scanf("%lf", &tmp);

        b[i] = tmp;

    }



    double x[n];

    double p[n];



    for(int i=0; i<n; i++)

    {

        x[i] = 0;

        p[i] = 0;

    }



    do

    {

        for (int i = 0; i < n; i++)

            p[i] = x[i];



        for (int i = 0; i < n; i++)

        {

            double var = 0;

            for (int j = 0; j < i; j++)

                var += (a[i][j] * x[j]);

            for (int j = i; j < n; j++)

                var += (a[i][j] * p[j]);

            x[i] = (b[i] - var) / a[i][i];

        }

    }while (!converge(x, p));



    return 0;

}


Now I want to implement matrix transformation algorithm (if there is one) to meet convergence criteria.






linear-algebra numerical-methods numerical-linear-algebra




linear-algebra numerical-methods numerical-linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked May 16 '14 at 18:56









torayefftorayeff

1164




1164












  • $begingroup$
    Where does your matrix A come from? Is it symmetric? Clearly arbitrarily "transforming" the matrix to meet some criteria may change the problem you are solving -- so if you tell me more about the problem I can make some suggestions.
    $endgroup$
    – RRL
    May 16 '14 at 19:21












  • $begingroup$
    A is not symmetric matrix. I am going to implement for abitrary matrix
    $endgroup$
    – torayeff
    May 16 '14 at 19:38








  • 1




    $begingroup$
    @torayeff you left out the fact that it will often converge without those conditions. Usually, if the system was small I would manually work it till there was as much diagonal dominance as possible.
    $endgroup$
    – bobbym
    May 16 '14 at 19:51










  • $begingroup$
    Why are you implementing this algorithm for a dense matrix?
    $endgroup$
    – Pawel Kowal
    Jul 14 '16 at 10:01


















  • $begingroup$
    Where does your matrix A come from? Is it symmetric? Clearly arbitrarily "transforming" the matrix to meet some criteria may change the problem you are solving -- so if you tell me more about the problem I can make some suggestions.
    $endgroup$
    – RRL
    May 16 '14 at 19:21












  • $begingroup$
    A is not symmetric matrix. I am going to implement for abitrary matrix
    $endgroup$
    – torayeff
    May 16 '14 at 19:38








  • 1




    $begingroup$
    @torayeff you left out the fact that it will often converge without those conditions. Usually, if the system was small I would manually work it till there was as much diagonal dominance as possible.
    $endgroup$
    – bobbym
    May 16 '14 at 19:51










  • $begingroup$
    Why are you implementing this algorithm for a dense matrix?
    $endgroup$
    – Pawel Kowal
    Jul 14 '16 at 10:01
















$begingroup$
Where does your matrix A come from? Is it symmetric? Clearly arbitrarily "transforming" the matrix to meet some criteria may change the problem you are solving -- so if you tell me more about the problem I can make some suggestions.
$endgroup$
– RRL
May 16 '14 at 19:21






$begingroup$
Where does your matrix A come from? Is it symmetric? Clearly arbitrarily "transforming" the matrix to meet some criteria may change the problem you are solving -- so if you tell me more about the problem I can make some suggestions.
$endgroup$
– RRL
May 16 '14 at 19:21














$begingroup$
A is not symmetric matrix. I am going to implement for abitrary matrix
$endgroup$
– torayeff
May 16 '14 at 19:38






$begingroup$
A is not symmetric matrix. I am going to implement for abitrary matrix
$endgroup$
– torayeff
May 16 '14 at 19:38






1




1




$begingroup$
@torayeff you left out the fact that it will often converge without those conditions. Usually, if the system was small I would manually work it till there was as much diagonal dominance as possible.
$endgroup$
– bobbym
May 16 '14 at 19:51




$begingroup$
@torayeff you left out the fact that it will often converge without those conditions. Usually, if the system was small I would manually work it till there was as much diagonal dominance as possible.
$endgroup$
– bobbym
May 16 '14 at 19:51












$begingroup$
Why are you implementing this algorithm for a dense matrix?
$endgroup$
– Pawel Kowal
Jul 14 '16 at 10:01




$begingroup$
Why are you implementing this algorithm for a dense matrix?
$endgroup$
– Pawel Kowal
Jul 14 '16 at 10:01










1 Answer
1






active

oldest

votes


















0












$begingroup$

I have found solution for my question and implemented it in C++:



/*

 * www.twitter.com/torayeff

 * 

 * Gauss-Seidel method:

 * http://math.semestr.ru/optim/methodzeidel.php

 * http://en.wikipedia.org/wiki/Gauss–Seidel_method

 * http://ru.wikipedia.org/wiki/Метод_Гаусса_—_Зейделя

 * 

 * To solve A*x = b transform to 

 * A(t)*A*x = A(t)*b <---> C*x = d 

 * (A(t) transpose of matrix A)

 * (in this state, Gauss-Seidel method always congerges)

*/

#include <iostream>

#include <cstdio>

#include <cmath>

using namespace std;



#define MAXN 100



int N; //matrix dimension

double eps = 0.001;

double err = eps;

double A[MAXN][MAXN];

double b[MAXN][MAXN];

double At[MAXN][MAXN];

double C[MAXN][MAXN];

double d[MAXN][MAXN];



//convergence check

bool converge(double *xk, double *xkp)

{

    double norm = 0;

    for(int i=0; i<N; i++) 

    {

      norm += (xk[i] - xkp[i])*(xk[i] - xkp[i]);

    }

    if(sqrt(norm)>=eps)

      return false;

    err = eps;

    return true;

}





int main()

{

    printf("Input dimension N: ");

    scanf("%d", &N);



    for(int i=0; i<N; i++)

    {

        for(int j=0; j<N; j++)

        {

            double tmp;

            printf("A[%d][%d] = ", i+1, j+1);

            scanf("%lf", &tmp);

            A[i][j] = tmp;

        }

    }



    for(int i=0; i<N; i++)

    {

        double tmp;

        printf("b[%d] = ", i+1);

        scanf("%lf", &tmp);

        b[i][0] = tmp;

    }





    //transpose A

    for(int i=0; i<N; i++)

    {

        for(int j=0; j<N; j++)

        {

            At[i][j] = A[j][i];

        }

    }



    //multiply At*A = C

    for(int i=0; i<N; i++)

    {

        for(int j=0; j<N; j++)

        {

            C[i][j] = 0;

            for(int k=0; k<N; k++)

            {

                C[i][j] = C[i][j] + At[i][k]*A[k][j];

            }

        }

    }



    //multiply At*b = d

    for(int i=0; i<N; i++)

    {

        for(int j=0; j<1; j++)

        {

            d[i][j] = 0;

            for(int k=0; k<N; k++)

            {

                d[i][j] = d[i][j] + At[i][k]*b[k][j];

            }

        }

    }



    //Solve Gauss-Seidel method for C*x = d

    double x[N]; //current values

    double p[N]; //previous values



    for(int i=0; i<N; i++)

    {

        x[i] = 0;

        p[i] = 0;

    }



    do

    {

        for(int i = 0; i<N; i++)

            p[i] = x[i];



        for(int i=0; i<N; i++)

        {

            double v = 0.0;

            for(int j=0; j<i; j++)

            {

                double cij;

                if(i==j) cij = 0;

                else cij = -1.0*C[i][j]/C[i][i];

                v += cij*x[j];

            }



            for(int j=i; j<N; j++)

            {

                double cij;

                if(i==j) cij = 0;

                else cij = -1.0*C[i][j]/C[i][i];

                v += cij*p[j];

            }

            v +=  1.0*d[i][0]/C[i][i];

            x[i] = v;

        }



    }while (!converge(x, p));



    //print solution

    printf("Err. val.: %lfn", err);

    for(int i=0; i<N; i++)

    {

        printf("%lf ", x[i]);

    }



    return 0;

}





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    $begingroup$

    I have found solution for my question and implemented it in C++:



    /*
    
 * www.twitter.com/torayeff
    
 * 
    
 * Gauss-Seidel method:
    
 * http://math.semestr.ru/optim/methodzeidel.php
    
 * http://en.wikipedia.org/wiki/Gauss–Seidel_method
    
 * http://ru.wikipedia.org/wiki/Метод_Гаусса_—_Зейделя
    
 * 
    
 * To solve A*x = b transform to 
    
 * A(t)*A*x = A(t)*b <---> C*x = d 
    
 * (A(t) transpose of matrix A)
    
 * (in this state, Gauss-Seidel method always congerges)
    
*/
    
#include <iostream>
    
#include <cstdio>
    
#include <cmath>
    
using namespace std;
    

    
#define MAXN 100
    

    
int N; //matrix dimension
    
double eps = 0.001;
    
double err = eps;
    
double A[MAXN][MAXN];
    
double b[MAXN][MAXN];
    
double At[MAXN][MAXN];
    
double C[MAXN][MAXN];
    
double d[MAXN][MAXN];
    

    
//convergence check
    
bool converge(double *xk, double *xkp)
    
{
    
    double norm = 0;
    
    for(int i=0; i<N; i++) 
    
    {
    
      norm += (xk[i] - xkp[i])*(xk[i] - xkp[i]);
    
    }
    
    if(sqrt(norm)>=eps)
    
      return false;
    
    err = eps;
    
    return true;
    
}
    

    

    
int main()
    
{
    
    printf("Input dimension N: ");
    
    scanf("%d", &N);
    

    
    for(int i=0; i<N; i++)
    
    {
    
        for(int j=0; j<N; j++)
    
        {
    
            double tmp;
    
            printf("A[%d][%d] = ", i+1, j+1);
    
            scanf("%lf", &tmp);
    
            A[i][j] = tmp;
    
        }
    
    }
    

    
    for(int i=0; i<N; i++)
    
    {
    
        double tmp;
    
        printf("b[%d] = ", i+1);
    
        scanf("%lf", &tmp);
    
        b[i][0] = tmp;
    
    }
    

    

    
    //transpose A
    
    for(int i=0; i<N; i++)
    
    {
    
        for(int j=0; j<N; j++)
    
        {
    
            At[i][j] = A[j][i];
    
        }
    
    }
    

    
    //multiply At*A = C
    
    for(int i=0; i<N; i++)
    
    {
    
        for(int j=0; j<N; j++)
    
        {
    
            C[i][j] = 0;
    
            for(int k=0; k<N; k++)
    
            {
    
                C[i][j] = C[i][j] + At[i][k]*A[k][j];
    
            }
    
        }
    
    }
    

    
    //multiply At*b = d
    
    for(int i=0; i<N; i++)
    
    {
    
        for(int j=0; j<1; j++)
    
        {
    
            d[i][j] = 0;
    
            for(int k=0; k<N; k++)
    
            {
    
                d[i][j] = d[i][j] + At[i][k]*b[k][j];
    
            }
    
        }
    
    }
    

    
    //Solve Gauss-Seidel method for C*x = d
    
    double x[N]; //current values
    
    double p[N]; //previous values
    

    
    for(int i=0; i<N; i++)
    
    {
    
        x[i] = 0;
    
        p[i] = 0;
    
    }
    

    
    do
    
    {
    
        for(int i = 0; i<N; i++)
    
            p[i] = x[i];
    

    
        for(int i=0; i<N; i++)
    
        {
    
            double v = 0.0;
    
            for(int j=0; j<i; j++)
    
            {
    
                double cij;
    
                if(i==j) cij = 0;
    
                else cij = -1.0*C[i][j]/C[i][i];
    
                v += cij*x[j];
    
            }
    

    
            for(int j=i; j<N; j++)
    
            {
    
                double cij;
    
                if(i==j) cij = 0;
    
                else cij = -1.0*C[i][j]/C[i][i];
    
                v += cij*p[j];
    
            }
    
            v +=  1.0*d[i][0]/C[i][i];
    
            x[i] = v;
    
        }
    

    
    }while (!converge(x, p));
    

    
    //print solution
    
    printf("Err. val.: %lfn", err);
    
    for(int i=0; i<N; i++)
    
    {
    
        printf("%lf ", x[i]);
    
    }
    

    
    return 0;
    
}





    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I have found solution for my question and implemented it in C++:



      /*
      
 * www.twitter.com/torayeff
      
 * 
      
 * Gauss-Seidel method:
      
 * http://math.semestr.ru/optim/methodzeidel.php
      
 * http://en.wikipedia.org/wiki/Gauss–Seidel_method
      
 * http://ru.wikipedia.org/wiki/Метод_Гаусса_—_Зейделя
      
 * 
      
 * To solve A*x = b transform to 
      
 * A(t)*A*x = A(t)*b <---> C*x = d 
      
 * (A(t) transpose of matrix A)
      
 * (in this state, Gauss-Seidel method always congerges)
      
*/
      
#include <iostream>
      
#include <cstdio>
      
#include <cmath>
      
using namespace std;
      

      
#define MAXN 100
      

      
int N; //matrix dimension
      
double eps = 0.001;
      
double err = eps;
      
double A[MAXN][MAXN];
      
double b[MAXN][MAXN];
      
double At[MAXN][MAXN];
      
double C[MAXN][MAXN];
      
double d[MAXN][MAXN];
      

      
//convergence check
      
bool converge(double *xk, double *xkp)
      
{
      
    double norm = 0;
      
    for(int i=0; i<N; i++) 
      
    {
      
      norm += (xk[i] - xkp[i])*(xk[i] - xkp[i]);
      
    }
      
    if(sqrt(norm)>=eps)
      
      return false;
      
    err = eps;
      
    return true;
      
}
      

      

      
int main()
      
{
      
    printf("Input dimension N: ");
      
    scanf("%d", &N);
      

      
    for(int i=0; i<N; i++)
      
    {
      
        for(int j=0; j<N; j++)
      
        {
      
            double tmp;
      
            printf("A[%d][%d] = ", i+1, j+1);
      
            scanf("%lf", &tmp);
      
            A[i][j] = tmp;
      
        }
      
    }
      

      
    for(int i=0; i<N; i++)
      
    {
      
        double tmp;
      
        printf("b[%d] = ", i+1);
      
        scanf("%lf", &tmp);
      
        b[i][0] = tmp;
      
    }
      

      

      
    //transpose A
      
    for(int i=0; i<N; i++)
      
    {
      
        for(int j=0; j<N; j++)
      
        {
      
            At[i][j] = A[j][i];
      
        }
      
    }
      

      
    //multiply At*A = C
      
    for(int i=0; i<N; i++)
      
    {
      
        for(int j=0; j<N; j++)
      
        {
      
            C[i][j] = 0;
      
            for(int k=0; k<N; k++)
      
            {
      
                C[i][j] = C[i][j] + At[i][k]*A[k][j];
      
            }
      
        }
      
    }
      

      
    //multiply At*b = d
      
    for(int i=0; i<N; i++)
      
    {
      
        for(int j=0; j<1; j++)
      
        {
      
            d[i][j] = 0;
      
            for(int k=0; k<N; k++)
      
            {
      
                d[i][j] = d[i][j] + At[i][k]*b[k][j];
      
            }
      
        }
      
    }
      

      
    //Solve Gauss-Seidel method for C*x = d
      
    double x[N]; //current values
      
    double p[N]; //previous values
      

      
    for(int i=0; i<N; i++)
      
    {
      
        x[i] = 0;
      
        p[i] = 0;
      
    }
      

      
    do
      
    {
      
        for(int i = 0; i<N; i++)
      
            p[i] = x[i];
      

      
        for(int i=0; i<N; i++)
      
        {
      
            double v = 0.0;
      
            for(int j=0; j<i; j++)
      
            {
      
                double cij;
      
                if(i==j) cij = 0;
      
                else cij = -1.0*C[i][j]/C[i][i];
      
                v += cij*x[j];
      
            }
      

      
            for(int j=i; j<N; j++)
      
            {
      
                double cij;
      
                if(i==j) cij = 0;
      
                else cij = -1.0*C[i][j]/C[i][i];
      
                v += cij*p[j];
      
            }
      
            v +=  1.0*d[i][0]/C[i][i];
      
            x[i] = v;
      
        }
      

      
    }while (!converge(x, p));
      

      
    //print solution
      
    printf("Err. val.: %lfn", err);
      
    for(int i=0; i<N; i++)
      
    {
      
        printf("%lf ", x[i]);
      
    }
      

      
    return 0;
      
}





      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I have found solution for my question and implemented it in C++:



        /*
        
 * www.twitter.com/torayeff
        
 * 
        
 * Gauss-Seidel method:
        
 * http://math.semestr.ru/optim/methodzeidel.php
        
 * http://en.wikipedia.org/wiki/Gauss–Seidel_method
        
 * http://ru.wikipedia.org/wiki/Метод_Гаусса_—_Зейделя
        
 * 
        
 * To solve A*x = b transform to 
        
 * A(t)*A*x = A(t)*b <---> C*x = d 
        
 * (A(t) transpose of matrix A)
        
 * (in this state, Gauss-Seidel method always congerges)
        
*/
        
#include <iostream>
        
#include <cstdio>
        
#include <cmath>
        
using namespace std;
        

        
#define MAXN 100
        

        
int N; //matrix dimension
        
double eps = 0.001;
        
double err = eps;
        
double A[MAXN][MAXN];
        
double b[MAXN][MAXN];
        
double At[MAXN][MAXN];
        
double C[MAXN][MAXN];
        
double d[MAXN][MAXN];
        

        
//convergence check
        
bool converge(double *xk, double *xkp)
        
{
        
    double norm = 0;
        
    for(int i=0; i<N; i++) 
        
    {
        
      norm += (xk[i] - xkp[i])*(xk[i] - xkp[i]);
        
    }
        
    if(sqrt(norm)>=eps)
        
      return false;
        
    err = eps;
        
    return true;
        
}
        

        

        
int main()
        
{
        
    printf("Input dimension N: ");
        
    scanf("%d", &N);
        

        
    for(int i=0; i<N; i++)
        
    {
        
        for(int j=0; j<N; j++)
        
        {
        
            double tmp;
        
            printf("A[%d][%d] = ", i+1, j+1);
        
            scanf("%lf", &tmp);
        
            A[i][j] = tmp;
        
        }
        
    }
        

        
    for(int i=0; i<N; i++)
        
    {
        
        double tmp;
        
        printf("b[%d] = ", i+1);
        
        scanf("%lf", &tmp);
        
        b[i][0] = tmp;
        
    }
        

        

        
    //transpose A
        
    for(int i=0; i<N; i++)
        
    {
        
        for(int j=0; j<N; j++)
        
        {
        
            At[i][j] = A[j][i];
        
        }
        
    }
        

        
    //multiply At*A = C
        
    for(int i=0; i<N; i++)
        
    {
        
        for(int j=0; j<N; j++)
        
        {
        
            C[i][j] = 0;
        
            for(int k=0; k<N; k++)
        
            {
        
                C[i][j] = C[i][j] + At[i][k]*A[k][j];
        
            }
        
        }
        
    }
        

        
    //multiply At*b = d
        
    for(int i=0; i<N; i++)
        
    {
        
        for(int j=0; j<1; j++)
        
        {
        
            d[i][j] = 0;
        
            for(int k=0; k<N; k++)
        
            {
        
                d[i][j] = d[i][j] + At[i][k]*b[k][j];
        
            }
        
        }
        
    }
        

        
    //Solve Gauss-Seidel method for C*x = d
        
    double x[N]; //current values
        
    double p[N]; //previous values
        

        
    for(int i=0; i<N; i++)
        
    {
        
        x[i] = 0;
        
        p[i] = 0;
        
    }
        

        
    do
        
    {
        
        for(int i = 0; i<N; i++)
        
            p[i] = x[i];
        

        
        for(int i=0; i<N; i++)
        
        {
        
            double v = 0.0;
        
            for(int j=0; j<i; j++)
        
            {
        
                double cij;
        
                if(i==j) cij = 0;
        
                else cij = -1.0*C[i][j]/C[i][i];
        
                v += cij*x[j];
        
            }
        

        
            for(int j=i; j<N; j++)
        
            {
        
                double cij;
        
                if(i==j) cij = 0;
        
                else cij = -1.0*C[i][j]/C[i][i];
        
                v += cij*p[j];
        
            }
        
            v +=  1.0*d[i][0]/C[i][i];
        
            x[i] = v;
        
        }
        

        
    }while (!converge(x, p));
        

        
    //print solution
        
    printf("Err. val.: %lfn", err);
        
    for(int i=0; i<N; i++)
        
    {
        
        printf("%lf ", x[i]);
        
    }
        

        
    return 0;
        
}





        share|cite|improve this answer









        $endgroup$



        I have found solution for my question and implemented it in C++:



        /*
        
 * www.twitter.com/torayeff
        
 * 
        
 * Gauss-Seidel method:
        
 * http://math.semestr.ru/optim/methodzeidel.php
        
 * http://en.wikipedia.org/wiki/Gauss–Seidel_method
        
 * http://ru.wikipedia.org/wiki/Метод_Гаусса_—_Зейделя
        
 * 
        
 * To solve A*x = b transform to 
        
 * A(t)*A*x = A(t)*b <---> C*x = d 
        
 * (A(t) transpose of matrix A)
        
 * (in this state, Gauss-Seidel method always congerges)
        
*/
        
#include <iostream>
        
#include <cstdio>
        
#include <cmath>
        
using namespace std;
        

        
#define MAXN 100
        

        
int N; //matrix dimension
        
double eps = 0.001;
        
double err = eps;
        
double A[MAXN][MAXN];
        
double b[MAXN][MAXN];
        
double At[MAXN][MAXN];
        
double C[MAXN][MAXN];
        
double d[MAXN][MAXN];
        

        
//convergence check
        
bool converge(double *xk, double *xkp)
        
{
        
    double norm = 0;
        
    for(int i=0; i<N; i++) 
        
    {
        
      norm += (xk[i] - xkp[i])*(xk[i] - xkp[i]);
        
    }
        
    if(sqrt(norm)>=eps)
        
      return false;
        
    err = eps;
        
    return true;
        
}
        

        

        
int main()
        
{
        
    printf("Input dimension N: ");
        
    scanf("%d", &N);
        

        
    for(int i=0; i<N; i++)
        
    {
        
        for(int j=0; j<N; j++)
        
        {
        
            double tmp;
        
            printf("A[%d][%d] = ", i+1, j+1);
        
            scanf("%lf", &tmp);
        
            A[i][j] = tmp;
        
        }
        
    }
        

        
    for(int i=0; i<N; i++)
        
    {
        
        double tmp;
        
        printf("b[%d] = ", i+1);
        
        scanf("%lf", &tmp);
        
        b[i][0] = tmp;
        
    }
        

        

        
    //transpose A
        
    for(int i=0; i<N; i++)
        
    {
        
        for(int j=0; j<N; j++)
        
        {
        
            At[i][j] = A[j][i];
        
        }
        
    }
        

        
    //multiply At*A = C
        
    for(int i=0; i<N; i++)
        
    {
        
        for(int j=0; j<N; j++)
        
        {
        
            C[i][j] = 0;
        
            for(int k=0; k<N; k++)
        
            {
        
                C[i][j] = C[i][j] + At[i][k]*A[k][j];
        
            }
        
        }
        
    }
        

        
    //multiply At*b = d
        
    for(int i=0; i<N; i++)
        
    {
        
        for(int j=0; j<1; j++)
        
        {
        
            d[i][j] = 0;
        
            for(int k=0; k<N; k++)
        
            {
        
                d[i][j] = d[i][j] + At[i][k]*b[k][j];
        
            }
        
        }
        
    }
        

        
    //Solve Gauss-Seidel method for C*x = d
        
    double x[N]; //current values
        
    double p[N]; //previous values
        

        
    for(int i=0; i<N; i++)
        
    {
        
        x[i] = 0;
        
        p[i] = 0;
        
    }
        

        
    do
        
    {
        
        for(int i = 0; i<N; i++)
        
            p[i] = x[i];
        

        
        for(int i=0; i<N; i++)
        
        {
        
            double v = 0.0;
        
            for(int j=0; j<i; j++)
        
            {
        
                double cij;
        
                if(i==j) cij = 0;
        
                else cij = -1.0*C[i][j]/C[i][i];
        
                v += cij*x[j];
        
            }
        

        
            for(int j=i; j<N; j++)
        
            {
        
                double cij;
        
                if(i==j) cij = 0;
        
                else cij = -1.0*C[i][j]/C[i][i];
        
                v += cij*p[j];
        
            }
        
            v +=  1.0*d[i][0]/C[i][i];
        
            x[i] = v;
        
        }
        

        
    }while (!converge(x, p));
        

        
    //print solution
        
    printf("Err. val.: %lfn", err);
        
    for(int i=0; i<N; i++)
        
    {
        
        printf("%lf ", x[i]);
        
    }
        

        
    return 0;
        
}






        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered May 17 '14 at 6:35









        torayefftorayeff

        1164




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