6-4 Compare Methods of Jacobi with Gauss-Seidel (50分)

Use Jacobi and Gauss-Seidel methods to solve a given n×n linear system A
​x
​⃗
​​ =
​b
​⃗
​​ with an initial approximation
​x
​⃗
​​
​(0)
​​ .

Note: When checking each a
​ii
​​ , first scan downward for the entry with maximum absolute value (a
​ii
​​ included). If that entry is non-zero, swap it to the diagonal. Otherwise if that entry is zero, scan upward for the entry with maximum absolute value. If that entry is non-zero, then add that row to the i-th row.

Format of functions:
int Jacobi( int n, double a[][MAX_SIZE], double b[], double x[], double TOL, int MAXN );

int Gauss_Seidel( int n, double a[][MAX_SIZE], double b[], double x[], double TOL, int MAXN );
where int n is the size of the matrix of which the entries are in the array double a[][MAX_SIZE] and MAX_SIZE is a constant defined by the judge; double b[] is for
​b
​⃗
​​ , double x[] passes in the initial approximation
​x
​⃗
​​
​(0)
​​ and return the solution
​x
​⃗
​​ ; double TOL is the tolerance for ∥
​x
​⃗
​​
​(k+1)
​​ −
​x
​⃗
​​
​(k)
​​ ∥
​∞
​​ ; and finally int MAXN is the maximum number of iterations.

The function must return:

k if there is a solution found after k iterations;
0 if maximum number of iterations exceeded;
1 if the matrix has a zero column and hence no unique solution exists;
2 if there is no convergence, that is, there is an entry of
​x
​⃗
​​
​(K)
​​ that is out of the range [-bound, bound] where bound is a constant defined by the judge.
Sample program of judge:
#include <stdio.h>
#include <math.h>

#define MAX_SIZE 10
#define bound pow(2, 127)
#define ZERO 1e-9 /* X is considered to be 0 if |X|<ZERO */

enum bool { false = 0, true = 1 };
#define bool enum bool

int Jacobi( int n, double a[][MAX_SIZE], double b[], double x[], double TOL, int MAXN );

int Gauss_Seidel( int n, double a[][MAX_SIZE], double b[], double x[], double TOL, int MAXN );

int main()
{
int n, MAXN, i, j, k;
double a[MAX_SIZE][MAX_SIZE], b[MAX_SIZE], x[MAX_SIZE];
double TOL;

scanf("%d", &n);
for (i=0; i<n; i++) {
    for (j=0; j<n; j++)
        scanf("%lf", &a[i][j]);
    scanf("%lf", &b[i]);
}
scanf("%lf %d", &TOL, &MAXN);

printf("Result of Jacobi method:\n");
for ( i=0; i<n; i++ )
    x[i] = 0.0;
k = Jacobi( n, a, b, x, TOL, MAXN );
switch ( k ) {
    case -2:
        printf("No convergence.\n");
        break;
    case -1: 
        printf("Matrix has a zero column.  No unique solution exists.\n");
        break;
    case 0:
        printf("Maximum number of iterations exceeded.\n");
        break;
    default:
        printf("no_iteration = %d\n", k);
        for ( j=0; j<n; j++ )
            printf("%.8f\n", x[j]);
        break;
}
printf("Result of Gauss-Seidel method:\n");
for ( i=0; i<n; i++ )
    x[i] = 0.0;
k = Gauss_Seidel( n, a, b, x, TOL, MAXN );
switch ( k ) {
    case -2:
        printf("No convergence.\n");
        break;
    case -1: 
        printf("Matrix has a zero column.  No unique solution exists.\n");
        break;
    case 0:
        printf("Maximum number of iterations exceeded.\n");
        break;
    default:
        printf("no_iteration = %d\n", k);
        for ( j=0; j<n; j++ )
            printf("%.8f\n", x[j]);
        break;
}

return 0;

}

/* Your function will be put here */
Sample Input 1:
3
2 -1 1 -1
2 2 2 4
-1 -1 2 -5
0.000000001 1000
Sample Output 1:
Result of Jacobi method:
No convergence.
Result of Gauss-Seidel method:
no_iteration = 37
1.00000000
2.00000000
-1.00000000
Sample Input 2:
3
1 2 -2 7
1 1 1 2
2 2 1 5
0.000000001 1000
Sample Output 2:
Result of Jacobi method:
no_iteration = 232
1.00000000
2.00000000
-1.00000000
Result of Gauss-Seidel method:
No convergence.
Sample Input 3:
5
2 1 0 0 0 1
1 2 1 0 0 1
0 1 2 1 0 1
0 0 1 2 1 1
0 0 0 1 2 1
0.000000001 100
Sample Output 3:
Result of Jacobi method:
Maximum number of iterations exceeded.
Result of Gauss-Seidel method:
no_iteration = 65
0.50000000
0.00000000
0.50000000
0.00000000
0.50000000

看起來很簡單的一道題，但是卡了我挺久，就是找不出錯誤。
後來發現應該是等號的問題。
測試點：
倒數第二個返回0 0
最後一個返回-1 -1（n=15）

注意別忘了換b，其他的按書上寫。
另外注意書上是最後再更新x，這樣會導致達到maxerror<TOL條件時返回的是上一次的值，所以要在返回前更新。

真是粗心要了我老命。