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Make solveCubic() more robust

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Also, return duplicate roots multiple times.

Part of https://github.com/behdad/fonttools/issues/617
This commit is contained in:
Behdad Esfahbod 2016-06-08 14:54:23 -07:00
parent b1474a3993
commit b2bd15d580
1 changed files with 21 additions and 12 deletions
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33
Lib/fontTools/misc/bezierTools.py
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@ -293,10 +293,14 @@ def solveCubic(a, b, c, d):
This function returns a list of roots. Note that the returned list
is neither guaranteed to be sorted nor to contain unique values!
>>> solveCubic(1, 1, -6, 0)
[-3.0, -0.0, 2.0]
>>> solveCubic(-10.0, -9.0, 48.0, -29.0)
[-2.9, 1.0]
[-2.9, 1.0, 1.0]
>>> solveCubic(-9.875, -9.0, 47.625, -28.75)
[-2.911392405063291, 1.0]
[-2.911392405063, 1.0, 1.0]
>>> solveCubic(1.0, -4.5, 6.75, -3.375)
[1.5, 1.5, 1.5]
"""
#
# adapted from:
@ -317,7 +321,10 @@ def solveCubic(a, b, c, d):
R = (2.0*a1*a1*a1 - 9.0*a1*a2 + 27.0*a3)/54.0
R2_Q3 = R*R - Q*Q*Q
if R2_Q3 < epsilon:
if abs(R) < epsilon and abs(Q) < epsilon:
x = -a1/3.0
return [x, x, x]
elif R2_Q3 < epsilon:
theta = acos(min(R/sqrt(Q*Q*Q), 1.0))
rQ2 = -2.0*sqrt(Q)
a1_3 = a1/3.0
@ -327,19 +334,21 @@ def solveCubic(a, b, c, d):
x0, x1, x2 = sorted([x0, x1, x2])
# Merge roots that are close-enough
if x1 - x0 < epsilon and x2 - x1 < epsilon:
return [round((x0 + x1 + x2) / 3., epsilonDigits)]
x0 = x1 = x2 = round((x0 + x1 + x2) / 3., epsilonDigits)
elif x1 - x0 < epsilon:
return [round((x0 + x1) / 2., epsilonDigits), x2]
x0 = x1 = round((x0 + x1) / 2., epsilonDigits)
x2 = round(x2, epsilonDigits)
elif x2 - x1 < epsilon:
return [x0, round((x1 + x2) / 2., epsilonDigits)]
x0 = round(x0, epsilonDigits)
x1 = x2 = round((x1 + x2) / 2., epsilonDigits)
else:
return [x0, x1, x2]
x0 = round(x0, epsilonDigits)
x1 = round(x1, epsilonDigits)
x2 = round(x2, epsilonDigits)
return [x0, x1, x2]
else:
if Q == 0 and R == 0:
x = 0
else:
x = pow(sqrt(R2_Q3)+abs(R), 1/3.0)
x = x + Q/x
x = pow(sqrt(R2_Q3)+abs(R), 1/3.0)
x = x + Q/x
if R >= 0.0:
x = -x
x = x - a1/3.0