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感谢大佬开源,我结合论文和代码看了一下有一点疑问 #36
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你少了下面 4 种情况:
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process.py中有pair_marking_points函数,这其中我枚举了一下是有以下12种情况能够返回1或者-1,进而组成slot:
但是看一些解读以及论文中的Fig.5看确实有16种组合情况。请问下是我在哪里遗漏了吗?
def pair_marking_points(point_a, point_b):
"""See whether two marking points form a slot."""
vector_ab = np.array([point_b.x - point_a.x, point_b.y - point_a.y])
vector_ab = vector_ab / np.linalg.norm(vector_ab)
point_shape_a = detemine_point_shape(point_a, vector_ab)
point_shape_b = detemine_point_shape(point_b, -vector_ab)
if point_shape_a.value == 0 or point_shape_b.value == 0:
return 0
if point_shape_a.value == 3 and point_shape_b.value == 3:
return 0
if point_shape_a.value > 3 and point_shape_b.value > 3:
return 0
if point_shape_a.value < 3 and point_shape_b.value < 3:
return 0
if point_shape_a.value != 3:
if point_shape_a.value > 3:
return 1
if point_shape_a.value < 3:
return -1
if point_shape_a.value == 3:
if point_shape_b.value < 3:
return 1
if point_shape_b.value > 3:
return -1
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