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Diagonal heuristic computes square root of 2 every time it is used #51

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peterjwest opened this issue Apr 26, 2016 · 1 comment

peterjwest commented Apr 26, 2016

Would help performance to do this as a constant outside the function.

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peterjwest commented Apr 26, 2016

Also, the final line of heuristics.diagonal can be simplified to:
return D * Math.abs(d1 - d2) + Math.min(d1, d2) * D2;

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