Finding the determinant of a matrix

[curiously enough]

I am having trouble making this recursive C function work for square matrices of size greater than 3 rows and columns.

#include<stdio.h>
#include<conio.h>
#include<math.h>
int M[100];
float DET(float A[][100],int dim)
{
int i=100,j,k,l;
static int det=0;
static int DIM=dim;
if(dim>2)
    for(i=0;i<DIM;i++)
    {
    if(i==0&&DIM==dim)for(j=0;j<100;j++)M[j]=100;
    for(j=DIM-1;j>=2;j--)
    if(M[j]==i)continue;
    M[dim-1]=i;
    det+=pow(-1,i+dim-1)*A[dim-1][i]*DET(A,dim-1);
    }
    if(i==DIM-1)M[dim-1]=100;
if(dim==2)
      {
    for(i=0;i<DIM;i++)
    {
    k=0;
      for(j=DIM-1;j>=2;j--)
    if(i!=M[j])k++;
    if(k==DIM-2)break;
    }
    for(j=0;j<DIM;j++)
    {
    k=0;
    for(l=DIM-1;l>=2;l--)
    if(j!=M[l])k++;
    if(k==DIM-2&&j!=i)break;
    }
    return(A[0][i]*A[1][j]-A[0][j]*A[1][i]);
      }
if(dim==1)return(A[0][0]);
return(det);
}
 
 
void main()
{
float A[100][100];
int n,k,l;
puts("Input size of square matrix:");
scanf("%d",&n);
puts("Input square matrix:");
for(k=0;k<n;k++)
{
printf("\n");
for(l=0;l<n;l++)
{
printf("Input a%d%d: ",k+1,l+1);
scanf("%f",&A[k][l]);
}
}
printf("\n\nThe determenant of the matrix is %0.2f",DET(A,n));
getch();
}
 

If anyone can tell me where I went wrong in writing the program I would be grateful.
P.S.:Here are some points to make the program clearer:
_This program works by using method of minor.
_M[100] is an array of integers used to store in decreasing order of its elements which column in the matrix does not belong to the minor associated with the element chosen on a certain row.
_In the function, dim represents wich row is being used, while i(in the first for loop) represents which column is being used.

[arnaudk]

Yup correct (using my LUP function) ;-)

Phew! I usually prefer to work in special units where 1 = -1 = 2 = 1/2. In those units I never have to hunt down minus signs or factors of 2 and my answers are always correct. :-)

[JosAH]

Phew! I usually prefer to work in special units where 1 = -1 = 2 = 1/2. In those units I never have to hunt down minus signs or factors of 2 and my answers are always correct. :-)

Well, my LUP implementation isn't the most efficient one when it comes to pivot
selection but then again it has to deal with symbolics as well and it does a fine
job when it comes to that e.g.

| a b |
| c d | = a*d-b*c
 

It can do any larger dimensions as well ;-)

kind regards,

Jos

[curiously enough]

So... were the minors displayed correct or not? For example, run your program on this invertible matrix and check the minors and cofactors are correct and have the right sign

.
   (1 2 3 4)
det(1 2 1 1) = -12
   (1 1 2 1)
   (5 6 7 8)
 

and [code]...[/code] tags are good. Please use them.

OK, now I've perfected the program, and all the minors are displayed correctly, but It still gives the wrong answer, and I'm still trying to find out what's wrong.
here's the program, test it on C++(V5.02) if you have it:

#include<stdio.h>
#include<conio.h>
#include<math.h>
int M[100];
float DET(float A[][100],int dim)
{
int i,j,k,l,m;
static float det=0;
static int DIM=dim;
if(dim>2)
    for(i=0;i<DIM;i++)
    {
    if(i==0&&DIM==dim)for(j=0;j<100;j++)M[j]=100;
    m=0;
    for(j=DIM-1;j>=2;j--)
    if(M[j]==i)m++;
    if(m!=0)continue;
       /*here printing starts*/
       for(k=0;k<dim;k++)
       {
    printf("\n");
       for(j=0;j<DIM;j++)
        {
        m=0;
        for(l=DIM-1;l>=2;l--)
        if(j!=M[l])m++;
        if(m==DIM-2)printf("%0.0f   ",A[k][j]);
        }
    }
       printf("\n\n");
    /*here printing ends*/
    M[dim-1]=i;
    det+=pow(-1,i+dim-1)*A[dim-1][i]*DET(A,dim-1);
    M[dim-1]=100;
    }
if(dim==2)
 {
    /*here printing starts*/
       for(k=0;k<dim;k++)
       {
    printf("\n");
           for(j=0;j<DIM;j++)
        {
        m=0;
        for(l=DIM-1;l>=2;l--)
        if(j!=M[l])m++;
        if(m==DIM-2)printf("%0.0f   ",A[k][j]);
        }
    }
       printf("\n\n");
    /*here printing ends*/
    for(i=0;i<DIM;i++)
    {
    k=0;
      for(j=DIM-1;j>=2;j--)
    if(i!=M[j])k++;
    if(k==DIM-2)break;
    }
    for(j=0;j<DIM;j++)
    {
    k=0;
    for(l=DIM-1;l>=2;l--)
    if(j!=M[l])k++;
    if(k==DIM-2&&j!=i)break;
    }
    return(A[0][i]*A[1][j]-A[0][j]*A[1][i]);
 }
if(dim==1)return(A[0][0]);
return(det);
}
 
 
void main()
{
float A[100][100];
int n,k,l;
puts("Input size of square matrix:");
scanf("%d",&n);
puts("Input square matrix:");
for(k=0;k<n;k++)
{
printf("\n");
for(l=0;l<n;l++)
{
printf("Input a%d%d: ",k+1,l+1);
scanf("%f",&A[k][l]);
}
}
printf("\n\nThe determenant of the matrix is %0.2f",DET(A,n));
getch();
}
 

I just have one question:
When I write
det+=pow(-1,i+dim-1)*A[dim-1][i]*DET(A,dim-1); where the return value of the DET(A,dim-1) function is det, is the value added to det just (sign)*A[dim-1][i] multiplied by the determinant of its minor?
P.S: I tried your matrix, and got -8550.

[JosAH]

When I write
det+=pow(-1,i+dim-1)*A[dim-1][i]*DET(A,dim-1); where the return value of the DET(A,dim-1) function is det, is the value added to det just (sign)*A[dim-1][i] multiplied by the determinant of its minor?

When you write X+= Y then the entire expression Y is evaluated and its value
is added to X for whatever the values of X and Y here the new value of X is
Y larger than it was before.

kind regards,

Jos

ps. you definitely should use other variable names and use a proper indentation
and spaces to make your code more readable.

[curiously enough]

When you write X+= Y then the entire expression Y is evaluated and its value
is added to X for whatever the values of X and Y here the new value of X is
Y larger than it was before.

kind regards,

Jos

ps. you definitely should use other variable names and use a proper indentation
and spaces to make your code more readable.

You didn't understand what I meant.
When I add to the value of the det static variable a return value which is that very same static variable(which is happening in the function as long as dim-1 is not equal to 2), will it add itself twice? Or will it just add the return value associated with the dim-1 part of the recursivity to the initial value of det in the part of the recursivity where I made the statement det+=pow(-1,i+dim-1)*A[dim-1][i]*DET(A,dim-1); ?

[JosAH]

You didn't understand what I meant.
When I add to the value of the det static variable a return value which is that very same static variable(which is happening in the function as long as dim-1 is not equal to 2), will it add itself twice? Or will it just add the return value associated with the dim-1 part of the recursivity to the initial value of det in the part of the recursivity where I made the statement det+=pow(-1,i+dim-1)*A[dim-1][i]*DET(A,dim-1); ?

If the DET method in that last assignment expression also changes static variable
det then the entire thing causes undefined behaviour; you are changing the value
of an LHV (here: det) more than once before a sequence point is reached.
Anything can happen here.

kind regards,

Jos

[arnaudk]

curiously enough, you're really making your life difficult by insisting on having those static variables. :-) Why don't you take heed of the comments in post #6?

[curiously enough]

curiously enough, you're really making your life difficult by insisting on having those static variables. :-) Why don't you take heed of the comments in post #6?

I'm writing this so you don't find the program arcane anymore,PLEASE try to read it, cause it still gives the wrong answer for DIM greater than 3

#include<stdio.h>
#include<conio.h>
#include<math.h>
int M[100];
float DET(float A[][100],int dim,int DIM,float det)
{
int i=100,j,k,l;
if(dim>2)
    for(i=0;i<DIM;i++)
    {
    if(i==0&&DIM==dim)
    for(j=0;j<100;j++)M[j]=100;
    for(j=DIM-1;j>=2;j--)
    if(M[j]==i)continue;
    M[dim-1]=i;
    det+=pow(-1,i+dim-1)*A[dim-1][i]*DET(A,dim-1,DIM,det);
    }
    if(i==DIM-1)M[dim-1]=100;
if(dim==2)
      {
    for(i=0;i<DIM;i++)
    {
    k=0;
    for(j=DIM-1;j>=2;j--)
    if(i!=M[j])k++;
    if(k==DIM-2)break;
    }
    for(j=0;j<DIM;j++)
    {
    k=0;
    for(l=DIM-1;l>=2;l--)
    if(j!=M[l])k++;
    if(k==DIM-2&&j!=i)break;
    }
    return(A[0][i]*A[1][j]-A[0][j]*A[1][i]);
      }
if(dim==1)return(A[0][0]);
return(det);
}
 
 
void main()
{
float A[100][100];
int n,k,l;
puts("Input size of square matrix:");
scanf("%d",&n);
puts("Input square matrix:");
for(k=0;k<n;k++)
{
printf("\n");
for(l=0;l<n;l++)
{
printf("Input a%d%d: ",k+1,l+1);
scanf("%f",&A[k][l]);
}
}
printf("\n\nThe determenant of the matrix is %0.2f",DET(A,n,n,0));
getch();
}
 

[newb16]

    if(i==0&&DIM==dim)
    for(j=0;j<100;j++)M[j]=100;
    for(j=DIM-1;j>=2;j--)
    if(M[j]==i)continue;
    M[dim-1]=i;
 

This is still arcane.

    if(i==0&&DIM==dim)
       for(j=0;j<100;j++)
           M[j]=100;
    for(j=DIM-1;j>=2;j--)
        if(M[j]==i)continue;
    M[dim-1]=i;
 

This is less so.

[arnaudk]

No one here is going to debug your code for you, you'll have to do that yourself. You can get advice here on specific problems which is pretty good for free.

You don't need a special case for dim==2. That you continue to pollute your code with this non-special case seems to indicate that you don't fully understand how do do a minor expansion.
I still find your code is too compact and variable names abstruse, and the fact you can't get it to work proves you can't see what's going on either. Code should be written for human beings, not just for computers.
If your minors and cofactors are correct, then you should have gotten the correct answer. Therefore, I don't believe they are correct.
The minors expansion for the aforementioned matrix looks like this:

|1 2 3 4|   
|1 2 1 1|     |2 1 1|     |1 1 1|     |1 2 1|    |1 2 1|
|1 1 2 1| = 1 |1 2 1| - 2 |1 2 1| + 3 |1 1 1| - 4|1 1 2|
|5 6 7 8|     |6 7 8|     |5 7 8|     |5 6 8|    |5 6 7|
 

Do you get these minors, with the columns in the right order and the correct cofactors and signs? Are you sure you then get the correct minors for the 3x3 matrices? Remember that swapping columns introduces minus signs.