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    <p>Here's a perfectly simple and practical solution (C++):</p> <pre><code>#include &lt;iostream&gt; #include &lt;vector&gt; struct Run { int x, w; }; enum { EAST, NORTHEAST, NORTH, NORTHWEST, WEST, SOUTHWEST, SOUTH, SOUTHEAST }; int main() { const Run data[] = { { 7, 2 }, { 5, 6 }, { 5, 7 }, { 5, 7 }, { 6, 6 }, { 0, 12 }, { 0, 12 }, { 0, 11 }, { 1, 7 }, { 3, 4 }, { 3, 4 }, { 3, 5 }, { 3, 7 }, { 3, 7 }, { 5, 5 } }; std::vector&lt;Run&gt; runs(data, data + 15); std::vector&lt;int&gt; before; std::vector&lt;int&gt; after; unsigned int i; int j; for (i = 0; i &lt; runs.size() - 1; ++i) { if (runs[i].x &lt; runs[i + 1].x) { for (j = 0; j &lt; runs[i + 1].x - runs[i].x - 1; ++j) before.push_back(WEST); before.push_back(NORTHWEST); } else if (runs[i].x &gt; runs[i + 1].x) { before.push_back(NORTHEAST); for (j = 0; j &lt; runs[i].x - runs[i + 1].x - 1; ++j) before.push_back(EAST); } else { before.push_back(NORTH); } int first_right(runs[i].x + runs[i].w); int second_right(runs[i + 1].x + runs[i + 1].w); if (first_right &lt; second_right) { after.push_back(SOUTHEAST); for (j = 0; j &lt; second_right - first_right - 1; ++j) after.push_back(EAST); } else if (first_right &gt; second_right) { for (j = 0; j &lt; first_right - second_right - 1; ++j) after.push_back(WEST); after.push_back(SOUTHWEST); } else { after.push_back(SOUTH); } } for (j = 0; j &lt; runs.back().w - 1; ++j) after.push_back(WEST); std::reverse(before.begin(), before.end()); after.insert(after.end(), before.begin(), before.end()); for (j = 0; j &lt; int(after.size()); ++j) { switch (after[j]) { case EAST: std::cout &lt;&lt; "EAST\n"; break; case NORTHEAST: std::cout &lt;&lt; "NORTHEAST\n"; break; case NORTH: std::cout &lt;&lt; "NORTH\n"; break; case NORTHWEST: std::cout &lt;&lt; "NORTHWEST\n"; break; case WEST: std::cout &lt;&lt; "WEST\n"; break; case SOUTHWEST: std::cout &lt;&lt; "SOUTHWEST\n"; break; case SOUTH: std::cout &lt;&lt; "SOUTH\n"; break; case SOUTHEAST: std::cout &lt;&lt; "SOUTHEAST\n"; break; } } } </code></pre> <p>This works by iterating over the runs, testing the left and right endpoints for the direction they're jumping to, and adding the appropriate number of chain elements to two vectors: one in forward order, for the right side, and one in reverse order, for the left. It then connects the two chains by adding the appropriate number of links for the last scanline, then reverses the left side chain and appends it to the right one to produce the final chain.</p> <p>Hope this is what you're looking for!</p>
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    1. COI'm not sure ton understand what your Run struct holds, is it x coordinate and length? I don't get how to relate that to the y coordinate
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    2. COAlso I'm not sure this is gonna work since if you get a U shape you will have to go back in your data array in some way
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    3. COYes, it's an x and a length. The y coordinate doesn't matter as long as the scan lines are contiguous; assume that runs[0] holds the run with the lowest y value, and so on up. And no, it doesn't work if there are multiple runs with the same y value; you can easily solve this by splitting the shape into two sets of runs, finding the chains, and joining them afterward.
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