Gauss Jordan Method in Basic C Multi

Question

I am have a multi-dimensional array that needs to be resolved, to solve for the unknown values (x1,x2,x3,..). In this multi-dimensional array, my array size in the i and j coordinate are different; B[n][n+1]. I know this Basic C, Gauss Jordan Method to solve for the unknown, is incorrect and would like someone to point in how to modify it..

This code is accurate for an Array with the same n values only; eg. A[n][n].

//Computation to solve
for(j=0; j<=n; j++)
{
    for(i=0; i<=n; i++)
    {
        if(i!=j)
        {
            c=B[i][j]/B[j][j];                    
            for(k=0;k<=n+1;k++)
            {
                B[i][k] = B[i][k] - c*B[j][k];
            }
        }
    }
}

//Print Solution
printf("\nThe solution is:\n");
for(i=0; i<=n; i++)
{
    x[i]=B[i][n+1]/B[i][i];
    printf("\n x%d=%.3f\n",i,x[i]);
}

An example would be if my n=2. The array that i would want to solve is B[2][3].

0  -20  0  -1
0  30  -10  0
0  -10  10  1

The output of this code is

x1= -inf 
x2=0.050
x3=0.000

The correct output should be

x1=0.00
x2=0.05
x3=0.15

array index start from 0 till n-1. What's wrong with above code and what output you expect is not clear. — SMA, Jan 24 '15 at 14:01
I added what's my supposed output to be. Please advise me accordingly. — Cherry_thia, Jan 24 '15 at 15:58
1st column of sample {0,0,0} => x1 = Indefinite. (x2,x3) compute by `{{-20,0,-1},{30,-10,0}}` — BLUEPIXY, Jan 25 '15 at 16:40
@BLUEPIXY I don't fully understand. Can you copy and edit to answer the solution? — Cherry_thia, Jan 25 '15 at 23:15
Solution of the your example x1 is indeterminate. x1 column of can not be obtained a unique solution to be selected how pivot because all zero. — BLUEPIXY, Jan 25 '15 at 23:38
@BLUEPIXY yes, i am understand that the first column cannot be determined because of the values are all zero. What about the code itself, since my number of columns and rows are different? — Cherry_thia, Jan 26 '15 at 15:35

score 0 · Answer 1 · answered Aug 07 '15 at 11:00

This code is accurate for an Array with the same n values only; eg. A[n][n].

No, the code already accounts for column n+1 (which holds the right-hand constant terms) in the loop

            for(k=0;k<=n+1;k++)
            {
                B[i][k] = B[i][k] - c*B[j][k];
            }

(since k runs up to n+1). So, it is correct in this regard.

An example would be if my n=2. The array that i would want to solve is B[2][3].

C arrays are declared by specifying the number of elements in the array, not the highest index, so you want B[3][4].

The main fault of the shown code is its lack of the row swapping operation, which may be necessary if B[j][j] is zero, and the related test. We can implement that by inserting

        if (!B[j][j])
        {
            for (i=j+1; i<=n; i++)
                if (B[i][j]) { swap(B[j], B[i]); break; }
            if (!B[j][j]) continue; // leave all-zero column as is
        }

at the beginning of the for(j=0; j<=n; j++) loop body (implementing swap() left to the reader).

The correct output should be
x1=0.00

Whether this is correct is debatable, but if you want a variable where every number is a solution to take the value zero, you may insert

        if (B[i][i] == 0 && B[i][n+1] == 0) x[i] = 0;

before the printf("\n x%d=%.3f\n",i,x[i]).

Gauss Jordan Method in Basic C Multi

1 Answers1