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Looping in a spiral

Asked 2008-12-29T18:40:29.320
182

A friend was in need of an algorithm that would let him loop through the elements of an NxM matrix (N and M are odd). I came up with a solution, but I wanted to see if my fellow SO'ers could come up with a better solution.

I'm posting my solution as an answer to this question.

Example Output:

For a 3x3 matrix, the output should be:

(0, 0) (1, 0) (1, 1) (0, 1) (-1, 1) (-1, 0) (-1, -1) (0, -1) (1, -1)

3x3 matrix

Furthermore, the algorithm should support non-square matrices, so for example for a 5x3 matrix, the output should be:

(0, 0) (1, 0) (1, 1) (0, 1) (-1, 1) (-1, 0) (-1, -1) (0, -1) (1, -1) (2, -1) (2, 0) (2, 1) (-2, 1) (-2, 0) (-2, -1)

5x3 matrix

algorithm matrix loops spiral
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1 Answer

8

Java spiral "Code golf" attempt, based on the C++ variant.

public static void Spiral(int X, int Y) {
    int x=0, y=0, dx = 0, dy = -1;
    int t = Math.max(X,Y);
    int maxI = t*t;

    for (int i=0; i < maxI; i++){
        if ((-X/2 <= x) && (x <= X/2) && (-Y/2 <= y) && (y <= Y/2)) {
            System.out.println(x+","+y);
            //DO STUFF
        }

        if( (x == y) || ((x < 0) && (x == -y)) || ((x > 0) && (x == 1-y))) {
            t=dx; dx=-dy; dy=t;
        }   
        x+=dx; y+=dy;
    }
}
answered 2012-05-15T18:56:45.663

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