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Fix error in get_quadratic_approximation_of_cubic

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Grant Sanderson 2020-02-07 09:31:06 -08:00
parent 1a13d32452
commit db0029c32b
1 changed files with 13 additions and 4 deletions
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17
manimlib/utils/bezier.py
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@ -193,8 +193,17 @@ def get_quadratic_approximation_of_cubic(a0, h0, h1, a1):
disc = b * b - 4 * a * c disc = b * b - 4 * a * c
has_infl &= (disc > 0) has_infl &= (disc > 0)
sqrt_disc = np.sqrt(abs(disc)) sqrt_disc = np.sqrt(abs(disc))
ti_min = fdiv(-b + sqrt_disc, 2 * a, zero_over_zero_value=0) # print(a, b, c, sqrt_disc)
ti_max = fdiv(-b - sqrt_disc, 2 * a, zero_over_zero_value=0) settings = np.seterr(all='ignore')
ti_min, ti_max = [
np.true_divide(
-b + sgn * sqrt_disc, 2 * a,
out=(-c / b),
where=(a != 0),
)
for sgn in [-1, +1]
]
np.seterr(**settings)
ti_min_in_range = (0 < ti_min) & (ti_min < 1) ti_min_in_range = (0 < ti_min) & (ti_min < 1)
ti_max_in_range = (0 < ti_max) & (ti_max < 1) ti_max_in_range = (0 < ti_max) & (ti_max < 1)
@ -203,8 +212,8 @@ def get_quadratic_approximation_of_cubic(a0, h0, h1, a1):
# if they lie between 0 and 1 # if they lie between 0 and 1
t_mid = 0.5 * np.ones(len(a0)) t_mid = 0.5 * np.ones(len(a0))
t_mid[ti_min_in_range] = ti_min t_mid[ti_min_in_range] = ti_min[ti_min_in_range]
t_mid[ti_max_in_range] = ti_max t_mid[ti_max_in_range] = ti_max[ti_max_in_range]
m, n = a0.shape m, n = a0.shape
t_mid = t_mid.repeat(n).reshape((m, n)) t_mid = t_mid.repeat(n).reshape((m, n))