CODEWITHSUNDEEP
javaarraysmatrixcombinations
Connors
Connors

Reputation: 1

How to generate combinations of 0's and 1's in a 1024X 10 Matrix

I have to generate all the possible combinations of 0's and 1's in the matrix.. Like:

0000000000, 0000000001, 0000000010, 0000000011.... etc.

Is there a better way to do it rather than using nested for loops as shown below?

class Reliability {
// Trying to create combinations of 0 and 1s
public static void main(String[] args) {
// Creating an array of size 1024X10
    int connectMat[][] = new int[1024][10];

    int count1 = 0;
// Intitially all rows are set to zero
    for (int r1 = 0; r1 <= 1; r1++) {
// fill all rows with 0 and the 1
        for (int r2 = 0; r2 <= 1; r2++) {
            for (int r3 = 0; r3 <= 1; r3++) {
                for (int r4 = 0; r4 <= 1; r4++) {
                                        // Updating the elements of each row 
                                            connectMat[count1][0] = r1;
                                            connectMat[count1][1] = r2;
                                            connectMat[count1][2] = r3;
                                            connectMat[count1][3] = r4;

                             // Incrementing count to point to the next row
                                            count1++;
                                        }
                                    }
                                }
                            }

Upvotes: 0

Views: 69

Answers (1)

John Kugelman
John Kugelman

Reputation: 361729

This problem translates directly into looking at the binary representation of each row number. We can use bit operations to pull out the necessary values.

final int BITS = 10;

int[][] connectMat = new int[1 << BITS][BITS];

for (int i = 0; i < (1 << BITS); ++i) {
    for (int j = 0; j < BITS; ++j) {
        connectMat[i][j] = (i >> (BITS-1 - j)) & 1;
    }
}

Note that 1 << 10 is equal to 210, or 1024. That explains 1 << BITS.

To understand (i >> (BITS-1 - j)) & 1, let's look at some example values. Let's say i == 673, or 1010100001 in binary. And let's say j == 2 meaning we want the third bit from the left. Replacing all the variables, we have:

connectMat[673][2] = (673 >> (10-1 - 2)) & 1;

The shift is 673 >> 10-1 - 2, or 673 >> 7. Shifting 1010100001 to the right 7 places cuts off the seven rightmost bits, giving us 1010100001. See how the bit we wanted is now the rightmost bit? The final & 1 extracts the rightmost bit, so we get 1 as the final result. The assignment ends up as:

connectMat[673][2] = 1;

Upvotes: 1

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