Eliminate recursion with a fixed-size stack #35
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Need to spend time on it to understand why it will produce the same result 😯 |
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@delfrrr basically, it "recurses" (increases stack) only in one direction, while the other one just runs as the next iteration of the loop (while keeping |
delfrrr
reviewed
Oct 3, 2018
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| this._legalize(a); | ||
| return this._legalize(br); | ||
| if (i < EDGE_STACK.length) { |
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I guess you want here i + 1 < EDGE_STACK.length?
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Doesn't matter much I think
| return this._legalize(br); | ||
| if (i < EDGE_STACK.length) { | ||
| EDGE_STACK[i++] = br; | ||
| } |
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Should we throw exception in else statement?
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I don't think so — I'd prefer the algorithm to finish triangulation even if it has minor (e.g. floating-point-error-induced) errors in extremely degenerate cases.
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Managed to get a perf win with recursion elimination here too — but in a simpler/faster way, by only increasing the stack once on each illegal edge. cc @delfrrr