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[Docs] Licensing (#1269)
* add 3rd party * work
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‎LICENSE-3RD-PARTY‎

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Pytorch3D license ( https://github.com/facebookresearch/pytorch3d/ ):
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BSD License
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For PyTorch3D software
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Copyright (c) Meta Platforms, Inc. and affiliates. All rights reserved.
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Redistribution and use in source and binary forms, with or without modification,
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are permitted provided that the following conditions are met:
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* Redistributions of source code must retain the above copyright notice, this
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list of conditions and the following disclaimer.
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* Redistributions in binary form must reproduce the above copyright notice,
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this list of conditions and the following disclaimer in the documentation
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and/or other materials provided with the distribution.
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* Neither the name Meta nor the names of its contributors may be used to
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endorse or promote products derived from this software without specific
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prior written permission.
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THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" AND
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ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED
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WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE
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DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT HOLDER OR CONTRIBUTORS BE LIABLE FOR
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ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES
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(INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES;
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LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON
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ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
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(INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS
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SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.

‎mani_skill/utils/geometry/bounding_cylinder.py‎

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#
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# Smallest enclosing circle - Library (Python)
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#
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# Copyright (c) 2020 Project Nayuki
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# https://www.nayuki.io/page/smallest-enclosing-circle
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#
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# This program is free software: you can redistribute it and/or modify
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# it under the terms of the GNU Lesser General Public License as published by
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# the Free Software Foundation, either version 3 of the License, or
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# (at your option) any later version.
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#
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# This program is distributed in the hope that it will be useful,
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# but WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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# GNU Lesser General Public License for more details.
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#
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# You should have received a copy of the GNU Lesser General Public License
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# along with this program (see COPYING.txt and COPYING.LESSER.txt).
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# If not, see <http://www.gnu.org/licenses/>.
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#
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# fmt: off
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# isort: skip_file
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import math, random
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"""
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Smallest enclosing cylinder computation.
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Based on the algorithm from:
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https://www.nayuki.io/page/smallest-enclosing-circle
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"""
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import math
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from typing import List, Optional, Tuple
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import numpy as np
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28-
# Data conventions: A point is a pair of floats (x, y). A circle is a triple of floats (center x, center y, radius).
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# Returns the smallest circle that encloses all the given points. Runs in expected O(n) time, randomized.
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# Input: A sequence of pairs of floats or ints, e.g. [(0,5), (3.1,-2.7)].
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# Output: A triple of floats representing a circle.
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# Note: If 0 points are given, None is returned. If 1 point is given, a circle of radius 0 is returned.
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#
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# Initially: No boundary points known
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def make_circle(points):
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# Convert to float and randomize order
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shuffled = [(float(x), float(y)) for (x, y) in points]
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random.shuffle(shuffled)
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# Progressively add points to circle or recompute circle
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c = None
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for (i, p) in enumerate(shuffled):
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if c is None or not is_in_circle(c, p):
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c = _make_circle_one_point(shuffled[ : i + 1], p)
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return c
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# One boundary point known
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def _make_circle_one_point(points, p):
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c = (p[0], p[1], 0.0)
52-
for (i, q) in enumerate(points):
53-
if not is_in_circle(c, q):
54-
if c[2] == 0.0:
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c = make_diameter(p, q)
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def _compute_smallest_circle(
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points: List[Tuple[float, float]]
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) -> Optional[Tuple[float, float, float]]:
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points = [(float(x), float(y)) for x, y in points]
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np.random.shuffle(points)
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circle = None
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for i, point in enumerate(points):
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if circle is None or not _point_in_circle(circle, point):
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circle = _compute_circle_with_point(points[: i + 1], point)
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return circle
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def _compute_circle_with_point(
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points: List[Tuple[float, float]], p: Tuple[float, float]
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) -> Tuple[float, float, float]:
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circle = (p[0], p[1], 0.0)
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for i, q in enumerate(points):
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if not _point_in_circle(circle, q):
33+
if circle[2] == 0.0:
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circle = _get_circle_from_diameter(p, q)
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else:
57-
c = _make_circle_two_points(points[ : i + 1], p, q)
58-
return c
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circle = _compute_circle_with_two_points(points[: i + 1], p, q)
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return circle
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61-
# Two boundary points known
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def _make_circle_two_points(points, p, q):
63-
circ = make_diameter(p, q)
64-
left = None
65-
right = None
66-
px, py = p
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qx, qy = q
40+
def _compute_circle_with_two_points(
41+
points: List[Tuple[float, float]], p: Tuple[float, float], q: Tuple[float, float]
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) -> Tuple[float, float, float]:
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circle = _get_circle_from_diameter(p, q)
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left = right = None
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69-
# For each point not in the two-point circle
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for r in points:
71-
if is_in_circle(circ, r):
47+
if _point_in_circle(circle, r):
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continue
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74-
# Form a circumcircle and classify it on left or right side
75-
cross = _cross_product(px, py, qx, qy, r[0], r[1])
76-
c = make_circumcircle(p, q, r)
77-
if c is None:
50+
cross = _compute_cross_product(p[0], p[1], q[0], q[1], r[0], r[1])
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candidate = _compute_circumcircle(p, q, r)
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if candidate is None:
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continue
79-
elif cross > 0.0 and (left is None or _cross_product(px, py, qx, qy, c[0], c[1]) > _cross_product(px, py, qx, qy, left[0], left[1])):
80-
left = c
81-
elif cross < 0.0 and (right is None or _cross_product(px, py, qx, qy, c[0], c[1]) < _cross_product(px, py, qx, qy, right[0], right[1])):
82-
right = c
55+
elif cross > 0 and (
56+
left is None
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or _compute_cross_product(
58+
p[0], p[1], q[0], q[1], candidate[0], candidate[1]
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)
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> _compute_cross_product(p[0], p[1], q[0], q[1], left[0], left[1])
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):
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left = candidate
63+
elif cross < 0 and (
64+
right is None
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or _compute_cross_product(
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p[0], p[1], q[0], q[1], candidate[0], candidate[1]
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)
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< _compute_cross_product(p[0], p[1], q[0], q[1], right[0], right[1])
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):
70+
right = candidate
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84-
# Select which circle to return
8572
if left is None and right is None:
86-
return circ
73+
return circle
8774
elif left is None:
8875
return right
8976
elif right is None:
9077
return left
9178
else:
92-
return left if (left[2] <= right[2]) else right
93-
94-
95-
def make_diameter(a, b):
96-
cx = (a[0] + b[0]) / 2
97-
cy = (a[1] + b[1]) / 2
98-
r0 = math.hypot(cx - a[0], cy - a[1])
99-
r1 = math.hypot(cx - b[0], cy - b[1])
100-
return (cx, cy, max(r0, r1))
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103-
def make_circumcircle(a, b, c):
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# Mathematical algorithm from Wikipedia: Circumscribed circle
105-
ox = (min(a[0], b[0], c[0]) + max(a[0], b[0], c[0])) / 2
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oy = (min(a[1], b[1], c[1]) + max(a[1], b[1], c[1])) / 2
107-
ax = a[0] - ox; ay = a[1] - oy
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bx = b[0] - ox; by = b[1] - oy
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cx = c[0] - ox; cy = c[1] - oy
110-
d = (ax * (by - cy) + bx * (cy - ay) + cx * (ay - by)) * 2.0
111-
if d == 0.0:
79+
return left if left[2] <= right[2] else right
80+
81+
82+
def _get_circle_from_diameter(
83+
a: Tuple[float, float], b: Tuple[float, float]
84+
) -> Tuple[float, float, float]:
85+
center_x = (a[0] + b[0]) / 2
86+
center_y = (a[1] + b[1]) / 2
87+
radius = max(
88+
math.hypot(center_x - a[0], center_y - a[1]),
89+
math.hypot(center_x - b[0], center_y - b[1]),
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)
91+
return (center_x, center_y, radius)
92+
93+
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def _compute_circumcircle(
95+
a: Tuple[float, float], b: Tuple[float, float], c: Tuple[float, float]
96+
) -> Optional[Tuple[float, float, float]]:
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# Center point
98+
mid_x = (min(a[0], b[0], c[0]) + max(a[0], b[0], c[0])) / 2
99+
mid_y = (min(a[1], b[1], c[1]) + max(a[1], b[1], c[1])) / 2
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101+
ax, ay = a[0] - mid_x, a[1] - mid_y
102+
bx, by = b[0] - mid_x, b[1] - mid_y
103+
cx, cy = c[0] - mid_x, c[1] - mid_y
104+
105+
det = (ax * (by - cy) + bx * (cy - ay) + cx * (ay - by)) * 2
106+
if det == 0:
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return None
113-
x = ox + ((ax*ax + ay*ay) * (by - cy) + (bx*bx + by*by) * (cy - ay) + (cx*cx + cy*cy) * (ay - by)) / d
114-
y = oy + ((ax*ax + ay*ay) * (cx - bx) + (bx*bx + by*by) * (ax - cx) + (cx*cx + cy*cy) * (bx - ax)) / d
115-
ra = math.hypot(x - a[0], y - a[1])
116-
rb = math.hypot(x - b[0], y - b[1])
117-
rc = math.hypot(x - c[0], y - c[1])
118-
return (x, y, max(ra, rb, rc))
119-
120-
121-
_MULTIPLICATIVE_EPSILON = 1 + 1e-14
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123-
def is_in_circle(c, p):
124-
return c is not None and math.hypot(p[0] - c[0], p[1] - c[1]) <= c[2] * _MULTIPLICATIVE_EPSILON
109+
center_x = (
110+
mid_x
111+
+ (
112+
(ax * ax + ay * ay) * (by - cy)
113+
+ (bx * bx + by * by) * (cy - ay)
114+
+ (cx * cx + cy * cy) * (ay - by)
115+
)
116+
/ det
117+
)
118+
center_y = (
119+
mid_y
120+
+ (
121+
(ax * ax + ay * ay) * (cx - bx)
122+
+ (bx * bx + by * by) * (ax - cx)
123+
+ (cx * cx + cy * cy) * (bx - ax)
124+
)
125+
/ det
126+
)
127+
radius = max(math.hypot(center_x - p[0], center_y - p[1]) for p in (a, b, c))
128+
129+
return (center_x, center_y, radius)
130+
131+
132+
def _point_in_circle(
133+
circle: Optional[Tuple[float, float, float]], point: Tuple[float, float]
134+
) -> bool:
135+
if circle is None:
136+
return False
137+
return math.hypot(point[0] - circle[0], point[1] - circle[1]) <= circle[2] * (
138+
1 + 1e-14
139+
)
140+
141+
142+
def _compute_cross_product(
143+
x0: float, y0: float, x1: float, y1: float, x2: float, y2: float
144+
) -> float:
145+
return (x1 - x0) * (y2 - y0) - (y1 - y0) * (x2 - x0)
125146

126147

127-
# Returns twice the signed area of the triangle defined by (x0, y0), (x1, y1), (x2, y2).
128-
def _cross_product(x0, y0, x1, y1, x2, y2):
129-
return (x1 - x0) * (y2 - y0) - (y1 - y0) * (x2 - x0)
148+
def aabc(points: np.ndarray) -> Tuple[float, float, float, float, float]:
149+
"""
150+
Compute axis-aligned bounding cylinder for 3D points.
130151
152+
Args:
153+
points: Nx3 array of points
131154
132-
def aabc(points):
155+
Returns:
156+
(center_x, center_y, radius, min_z, max_z) tuple
157+
"""
133158
points = np.asarray(points)
134-
z_min = points[:, 2].min()
135-
z_max = points[:, 2].max()
136-
x, y, r = make_circle(points[:, :2])
137-
return x, y, r, z_min, z_max
159+
z_bounds = points[:, 2].min(), points[:, 2].max()
160+
center_x, center_y, radius = _compute_smallest_circle(points[:, :2])
161+
return center_x, center_y, radius, z_bounds[0], z_bounds[1]

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