Calculate how long a function takes to run

Question

I would like to know how to find out how long the function will take to run
public int[][] MatrixMultiplication(int[][] MatrixA, boolean[][] MatrixB) {
        int result[][] = new int[number_positions + 2][number_positions + 2];
        int temp = 0;
        for (int j = 0; j < number_positions + 2; j++) {
            for (int i = 0; i < number_positions + 2; i++) {
                for (int k = 0; k < number_positions + 2; k++) {
                    temp = (MatrixB[j][k])? 1 : 0; 
                    //result[i][k] += MatrixA[i][j] * MatrixB[j][k];
                    result[i][k] += MatrixA[i][j] * temp;
        return result;
    }

Answer 1

gamewell wrote:
I would like to know how to find out how long the function will take to runNote that you need to repeat the multiplication N times and clock that. Select N so the total time is at least 1 second. Then you divide the total time with N to get the average time per call.
public int[][] MatrixMultiplication(int[][] MatrixA, boolean[][] MatrixB) {
int result[][] = new int[number_positions + 2][number_positions + 2];
int temp = 0;
for (int j = 0; j < number_positions + 2; j++) {
for (int i = 0; i < number_positions + 2; i++) {
for (int k = 0; k < number_positions + 2; k++) {
temp = (MatrixB[j][k])? 1 : 0; 
//result[i][k] += MatrixA[i][j] * MatrixB[j][k];
result[i][k] += MatrixA[i][j] * temp;
return result;
}If you care about efficiency you should remove invariants from the innermost loop. As you can see i and j don't change and because array accesses are expensive they should be removed. Also why perform the multiplication with 0 when you know it will have no effect? It's better to avoid that case.
// innermost loop
boolean[] mb = MatrixB[j];
int[] res = result;
int m = MatrixA[i][j];
for (int k = 0; k < number_positions + 2; k++) {
if (mb[k])
res[k] += m;
Now everything that's constant has been removed and only what depends on k is left. With very few iterations in the innermost loop this may not pay off but it's worth a try.