SolveNormalizedCubic fix to return proper real root #224
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Signed-off-by: Piotr Barejko <[email protected]>
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src/Imath/ImathRoots.h
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| if (D > 0) | ||
| { | ||
| auto real_root = [] (T a, T x) -> T { | ||
| T sign = a < T (0) ? T (-1) : T (1); |
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Suggested change
| T sign = a < T (0) ? T (-1) : T (1); | |
| T sign = std::copysign(T(1), a); |
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Thanks for the suggestion @meshula.
Also, I have a question, I didn't notice but in the declaration of the function here:
https://github.com/AcademySoftwareFoundation/Imath/blob/master/src/Imath/ImathRoots.h#L67
IMATH_HOSTDEVICE is used. Does it mean we can't use std:: features? I guess code has to compile as cuda code too?
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Apologies for the delay, and thank you for the fix! I'm not entirely sure about the complete details of compatibility between cuda and std, but I'm pretty sure simple inline std:: functions are acceptable.
use copy sign Co-authored-by: Nick Porcino <[email protected]> Signed-off-by: Piotr Barejko <[email protected]>
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…oundation#224) * SolveNormalizedCubic fix to return proper real root Signed-off-by: Piotr Barejko <[email protected]> * Update src/Imath/ImathRoots.h use copy sign Co-authored-by: Nick Porcino <[email protected]> Signed-off-by: Piotr Barejko <[email protected]> Co-authored-by: Nick Porcino <[email protected]>
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* SolveNormalizedCubic fix to return proper real root Signed-off-by: Piotr Barejko <[email protected]> * Update src/Imath/ImathRoots.h use copy sign Co-authored-by: Nick Porcino <[email protected]> Signed-off-by: Piotr Barejko <[email protected]> Co-authored-by: Nick Porcino <[email protected]>
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In some cases
solveNormalizedCubiccan return a real value from principled root that is a complex root. This happens in #86 for example.In case of discriminant
D > 0there must be one real and two complex roots. Given formula:-q/2 + sqrt(D)might be a negative number. Instead of searching thoughy0,y1,y2to find the real root, we can solve it as follows: