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[DigiBunny]

Right.

So I'm trying to multiply one matrix by another matrix, then display the output using for() loops. Spoilered for length and ease of access.

SPOILER

The Matrices are: SPOILER float P[5][4] = {{0,10,0,1},{10,0,0,1},{0,0,10,1},{-10,0,0,1},{0,0,-10,1}}; float Rx[4][4] = {{1,0,0,0},{0,0,1,0},{0,-1,0,0},{0,0,0,1}}; For context, this is a box in 3D space, which I am rotating along the X axis by 90 degrees. The Loop logic is: SPOILER for(int i = 0; i <5; i++) { for(int j = 0; j < 4; j++) { for(int k = 0; k < 4; k++) { Compiler += P[i][k] * Rx[k][i]; FinalMatrix[i][j] = Compiler; } } Compiler is a temporary storage variable, used to store the sum of the dot products the loop does. After a computing one of the numbers in the final matrix, it shunts that value to the FinalMatrix, and gets refreshed so that it can get back to storing more values. The FinalMatrix is supposed to be the "answer" matrix. However, all I seem to get is a 10 for the top right corner of the matrix, and everything else is a zero. Computing the matrix by hand tells me that this is wrong. The Display logic is: SPOILER int a,b; for(a = 0; a < 4; a++) { for(b = 0; b < 3; b++) { printf("%f \t",FinalMatrix[a][b]); } printf("\n"); }

I've had more than a few brain freeze-ups because I couldn't follow the numbers. Even when I tried to trace it on paper, I kept losing track of what would happen if I modified one of the loop counters.

I'm not expecting an instant solution, but if someone sees any logical errors with the code, please point them out.

[Perry]

What you want to do is try something simpler, like 2x3 or 2x2 matrices. Then do all the iterations on paper. You must know exactly what happen in each for loop before you can continue..

[Kiroshino]

Haven't touched programming in a while, but from what I see, Compiler is reset to zero with every iteration of the inner-most for loop. The entry in FinalMatrix is given the new value of Compiler with every iteration of the inner-most for loop. Hence the reason why most of your FinalMatrix entries are zero.

Didn't double check, but I think this should work:

for(int i = 0; i <5; i++)
{
for(int j = 0; j < 4; j++)
{
float Compiler = 0;

for(int k = 0; k < 4; k++)
{
Compiler += P[i][k] * Rx[k][i];
}
FinalMatrix[i][j] = Compiler;
}
}

This post has been edited by Kiroshino on Aug 7 2011, 04:33 PM

[DigiBunny]

The simpler matrix was a grand idea Perry. That let me scale down the problem.

Kiro; I did something similar just now before I saw your code. We're essentially identical except for where we reset Compiler's value. Mine is at the end of the j loop instead of at the beginning; but the answer remains the same.

On the bright side, I've isolated the problem. I used a new matrix set;
M = {{1,0},{0,1}} As an Identity Matrix
N = {{2,3},{5,4}} As my example.

At this point, I should only get the value for N back since multiplying by M gives me the same thing. Thus, eliminating the variable of error in human calculations.

[Y x] [x x]
[x x] and [x Y]

are the only ones being hit correctly. Everything else takes on the same values for that row. In other words, I'm missing the top right and bottom left parts of the resulting matrix, probably because I need another loop to stay on one row, while evaluating all the columns on that row, and vice versa.

[Kiroshino]

Don't have something to test it with, but I think this is the cause:

for(int i = 0; i <5; i++)
{
for(int j = 0; j < 4; j++)
{
float Compiler = 0;

for(int k = 0; k < 4; k++)
{
//Compiler += P[i][k] * Rx[k][i];
Compiler += P[i][k] * Rx[k][j];
}
FinalMatrix[i][j] = Compiler;
}
}

Note the change from i to j.

Slightly easier to follow version:

SPOILER

float product[maxrow][maxcol2]; for(int row = 0; row < maxrow; row++) { for(int col2 = 0; col2 < maxcol2; col2++) { float sum = 0; for(int col = 0; col < maxcol; col++) { sum += mat1[row][col] * mat2[col][col2]; } product[row][col2] = sum; } }

This post has been edited by Kiroshino on Aug 7 2011, 07:32 PM

[DigiBunny]

Done!

It's refreshing to finally get this thing right. I've included the source code below, should anyone, for any remote reason, find that they need this. Development notes are included.

Now all I need to do is make this work on a much larger scale. Since the logic is the same, I'm under the impression it should be as simple as replacing the Matrices, then changing the maximum values of the for() loops to suit them.

SPOILER

#include <stdio.h> #include <iostream> #include <conio.h> /* As per Perry's suggestion; this is a proof of concept to show I can multiply a matrix on a simple level. */ float M[2][2] = {{1,0},{0,1}}; //Identity Matrix. float N[2][2] = {{2,3},{5,4}}; //Random Line. Doesn't matter in the end. float P[2][2]; //The Answer matrix. Currently empty. float Compiler = 0; // The same compiling variable from last time. void main() { int a,b,c; /* Right. So here's how this is going to work. The Dot Product is contained within loops a,b,c. Actually computing the matrix is done within b and c loops. Displaying the matrix is done in the a and b loops. Development note: M[b][c] * N[c][b] Works for the two diagonal ends. This is because the way the loop works does not allow the row to evaluate two columns, and vice versa for the other matrix. | v ->[xx] [yy] [xx] [yy] This means I need ANOTHER variable to control whether the loop proceeds to the next row. (Nyeh Nyeh REDSTONE Nyeh.) What if... We use a, b and c this time? It would then go something like this: if(a = 0) { We would stay at the first row of the Matrix being multiplied. And (if b = 0) { We would stay at the first column of the second matrix. And (if c = 0) { We would stay at the first row of the second matrix, and go down from there. } } } To put that into human readable steps: 1.)Cycle through ROWS of Current column on second matrix. 2.)Move one column. Do step 1 again. 3.)Once all columns are completed, allow the next row of the first matrix to move. Development Note: It now works. Resetting your logic circuits works wonders. */ for(a = 0; a < 2; a++)//Controls flow of First Matrix Rows. (Column control here is unecessary. We're only cross producting. { for(b = 0; b < 2; b++)//Controls flow of Second Matrix Columns. { for(c = 0; c < 2; c++)//Controls flow of Second Matrix Rows. Will also control the column movement on the First Matrix, because that's how the dot product goes. { Compiler += M[a][c] * N[c][b]; // First matrix[a][c] * Second Matrix[c][b] } P[a][b] = Compiler; //P[CurrentRow][CurrentColumn] We will leave out C because it's the computing loop. a and b are the placement loops. Compiler = 0; //Reset Compiler so it's ready for the next round. } } printf("%f \t",P[0][0]); printf("%f \n",P[0][1]); printf("%f \t",P[1][0]); printf("%f \n",P[1][1]); getch(); }

Though, I'm a little puzzled as to how I'll multiply matrixes more than once every time I run the code.

[Kiroshino]

What do you mean? Matrix1 x Matrix2 x MatrixN?