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    <p><strong>Udate</strong></p> <p>With new information: </p> <blockquote> <p>Assuming that a coordinate diagonally is 2 away</p> </blockquote> <p>This algorithm would work. The algorithm searches outwards in a spiral kinda way testing each point in each 'ring' started from the inside.</p> <p>Note that it does not handle out of bounds situations. So you should change this to fit your needs. </p> <pre><code>int xs, ys; // Start coordinates // Check point (xs, ys) for (int d = 1; d&lt;maxDistance; d++) { for (int i = 0; i &lt; d + 1; i++) { int x1 = xs - d + i; int y1 = ys - i; // Check point (x1, y1) int x2 = xs + d - i; int y2 = ys + i; // Check point (x2, y2) } for (int i = 1; i &lt; d; i++) { int x1 = xs - i; int y1 = ys + d - i; // Check point (x1, y1) int x2 = xs + d - i; int y2 = ys - i; // Check point (x2, y2) } } </code></pre> <p><strong>Old version</strong></p> <p>Assuming that in your 2D grid the distance between (0, 0) and (1, 0) is the same as (0, 0) and (1, 1). This algorithm would work. The algorithm searches outwards in a spiral kinda way testing each point in each 'ring' started from the inside.</p> <p>Note that it does not handle out of bounds situations. So you should change this to fit your needs. </p> <pre><code>int xs, ys; // Start coordinates if (CheckPoint(xs, ys) == true) { return (xs, ys); } for (int d = 0; d&lt;maxDistance; d++) { for (int x = xs-d; x &lt; xs+d+1; x++) { // Point to check: (x, ys - d) and (x, ys + d) if (CheckPoint(x, ys - d) == true) { return (x, ys - d); } if (CheckPoint(x, ys + d) == true) { return (x, ys - d); } } for (int y = ys-d+1; y &lt; ys+d; y++) { // Point to check = (xs - d, y) and (xs + d, y) if (CheckPoint(x, ys - d) == true) { return (xs - d, y); } if (CheckPoint(x, ys + d) == true) { return (xs - d, y); } } } </code></pre>
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    1. POAlgorithm for finding nearest object on 2D grid
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    1. USMartin Ingvar Kofoed Jensen
    1. USMartin Ingvar Kofoed Jensen
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    1. POAlgorithm for finding nearest object on 2D grid
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    1. COForgot to specify that a coordinate diagonally is 2 away, not 1. Maybe this can be modified to adjust for it.
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    2. CO@random: Yes it can be modified. Ill try to put it in there in a moment
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    3. CO@random: I have added a new version of the algorithm taking into account the new information :)
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