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- Fixed ZeroDivisionError in solveCubic(). The solution is mathematically

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dubious (I don't think 0.0/0.0 == 0.0...) but the result seems to be
correct.
- Documented that soleCubic() and solveQuadratic() are not guaranteed to
return the roots in order, and nor that they are guaranteed to not return
duplicate roots.


git-svn-id: svn://svn.code.sf.net/p/fonttools/code/trunk@431 4cde692c-a291-49d1-8350-778aa11640f8
This commit is contained in:
jvr 2003-08-26 19:20:33 +00:00
parent d028b7be3c
commit bfadfe33db
1 changed files with 10 additions and 5 deletions
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15
Lib/fontTools/misc/bezierTools.py
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@ -159,7 +159,8 @@ def solveQuadratic(a, b, c,
sqrt=sqrt): sqrt=sqrt):
"""Solve a quadratic equation where a, b and c are real. """Solve a quadratic equation where a, b and c are real.
a*x*x + b*x + c = 0 a*x*x + b*x + c = 0
This function returns a list of roots. This function returns a list of roots. Note that the returned list
is neither guaranteed to be sorted nor to contain unique values!
""" """
if a == 0.0: if a == 0.0:
if b == 0.0: if b == 0.0:
@ -184,7 +185,8 @@ def solveCubic(a, b, c, d,
abs=abs, pow=pow, sqrt=sqrt, cos=cos, acos=acos, pi=pi): abs=abs, pow=pow, sqrt=sqrt, cos=cos, acos=acos, pi=pi):
"""Solve a cubic equation where a, b, c and d are real. """Solve a cubic equation where a, b, c and d are real.
a*x*x*x + b*x*x + c*x + d = 0 a*x*x*x + b*x*x + c*x + d = 0
This function returns a list of roots. This function returns a list of roots. Note that the returned list
is neither guaranteed to be sorted nor to contain unique values!
""" """
# #
# adapted from: # adapted from:
@ -205,7 +207,7 @@ def solveCubic(a, b, c, d,
R = (2.0*a1*a1*a1 - 9.0*a1*a2 + 27.0*a3)/54.0 R = (2.0*a1*a1*a1 - 9.0*a1*a2 + 27.0*a3)/54.0
R2_Q3 = R*R - Q*Q*Q R2_Q3 = R*R - Q*Q*Q
if R2_Q3 <= 0: if R2_Q3 < 0:
theta = acos(R/sqrt(Q*Q*Q)) theta = acos(R/sqrt(Q*Q*Q))
rQ2 = -2.0*sqrt(Q) rQ2 = -2.0*sqrt(Q)
x0 = rQ2*cos(theta/3.0) - a1/3.0 x0 = rQ2*cos(theta/3.0) - a1/3.0
@ -213,8 +215,11 @@ def solveCubic(a, b, c, d,
x2 = rQ2*cos((theta+4.0*pi)/3.0) - a1/3.0 x2 = rQ2*cos((theta+4.0*pi)/3.0) - a1/3.0
return [x0, x1, x2] return [x0, x1, x2]
else: else:
x = pow(sqrt(R2_Q3)+abs(R), 1/3.0) if Q == 0 and R == 0:
x = x + Q/x x = 0
else:
x = pow(sqrt(R2_Q3)+abs(R), 1/3.0)
x = x + Q/x
if R >= 0.0: if R >= 0.0:
x = -x x = -x
x = x - a1/3.0 x = x - a1/3.0