CODEWITHSUNDEEP
optimizationmatrix
Michael
Michael

Reputation: 644

Finding position in a matrix

I have a matrix of 3x3 which looks like

1 2 3

4 5 6

7 8 9

Behind each number is an object. So I wrote this code which will return which position in the matrix I'm currently at. But it feels very repetitive, Im curious if it would be possible to optimize that piece of code. If so, how?

public int findPos(int i, int k) {
    if (i == 0 && k == 0) {
        return 1;
    }
    if (i == 0 && k == 1) {
        return 2;
    }
    if (i == 0 && k == 2) {
        return 3;
    }
    if (i == 1 && k == 0) {
        return 4;
    }
    if (i == 1 && k == 1) {
        return 5;
    }
    if (i == 1 && k == 2) {
        return 6;
    }
    if (i == 2 && k == 0) {
        return 7;
    }
    if (i == 2 && k == 1) {
        return 8;
    }
    if (i == 2 && k == 2) {
        return 9;
    }

    return 0;
}

The method is later applied in this type of fashion

if (myMatrix[i][k] == 2) {
    position = findPos(i, k);
    drawAtPos(position);
}

Upvotes: 0

Views: 54

Answers (3)

Mike Dunlavey
Mike Dunlavey

Reputation: 40669

What's the matter with this?

(i*3 + k) + 1

Why does it work? Well, subtract 1 from your matrix, so it looks like this:

0 1 2
3 4 5
6 7 8

so it should be obvious.

Upvotes: 1

ergonaut
ergonaut

Reputation: 7057

Can you just covert it to:

public int findPos(int i, int k) {
    if (0 <= i && i < COLUMN_COUNT && 0 <= k && k < row_COUNT){ // out of bounds
      return 0;
    }
    return (i * 3) + (k + 1);
}

All this is doing is reducing your conditions into math as each column returns an offset of 1 each position to the right, and each row needs to be incremented by 3.

For a more generic solution, it would be:

public int findPos(int i, int k) {
    if (0 <= i && i < COLUMN_COUNT && 0 <= k && k < row_COUNT){ // out of bounds
      return 0;
    }
    return (i * COLUMN_COUNT) + (k + 1);
}

where COLUMN_COUNT is the number of columns you have. So it would work for things like:

1 2 3  4  5  6
7 8 9 10 11 12

where COLUMN_COUNT = 6

Upvotes: 1

Henrique Costa
Henrique Costa

Reputation: 11

You can calculate your position with:

if (0 <= i && i < 3 &&
      0 <= k && k < 3) {    
   // number of rows passed * size of each row + current column + 1 for 0-to-1-based-conversion
   return (i * 3) + (k + 1);
} else {
   return 0;
}

Upvotes: 1

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