[静态后端W21]Review Swift's dispatch implementation# MobileDevelopment - 移动开发
f*2
1 楼
建议在香港报纸做广告呼吁抵制DISNEYLAND
i*D
2 楼
我想判断3d空间中一个线段和在一个点的关系. 这个点在这个线段的直线上,我想判断
它在线段的两个端点之间还是在之外. 这个问题本身是一个明确的判断,不能容忍error
和tolerance.
比如线段是AB, 另外一个点是C. 我的方法是判断if AB-(AC+BC)<0. 但是在实际运算中
,我发现即便是C在AB之间,也有很多时候AB-(AC+BC)<0. 应该是因为round off error.
不知道有什么好的trick或者策略来解决这个问题.
它在线段的两个端点之间还是在之外. 这个问题本身是一个明确的判断,不能容忍error
和tolerance.
比如线段是AB, 另外一个点是C. 我的方法是判断if AB-(AC+BC)<0. 但是在实际运算中
,我发现即便是C在AB之间,也有很多时候AB-(AC+BC)<0. 应该是因为round off error.
不知道有什么好的trick或者策略来解决这个问题.
i*e
3 楼
Fucking Eric Schmidt is a fucking pussy.
-Steve Ballmer
-Steve Ballmer
x*x
4 楼
这个意见可以考虑。如果需要我捐钱
k*g
5 楼
If it is given beyond doubt that C is a point on the line extension of AB,
then C is outside the finite segment AB iff
C is outside the axis-aligned bounding box
the cartesian product with
( all x in (xmin, xmax) ) and
( all y in (ymin, ymax) )
where xmin = min ( A_x , B_x )
xmax = max ( A_x , B_x )
etc etc.
for 3D, 4D etc just apply cartesian product over all dimensions.
Upon thinking about that, it seems that even if you apply a unitary
coordinate transform (unitary rotation and/or mirroring) the "iff" still
holds. Not very sure, but seems interesting.
Not a proof.. just a sketch of idea. No guarantee. If there is a mistake
please leave a comment. Thanks.
then C is outside the finite segment AB iff
C is outside the axis-aligned bounding box
the cartesian product with
( all x in (xmin, xmax) ) and
( all y in (ymin, ymax) )
where xmin = min ( A_x , B_x )
xmax = max ( A_x , B_x )
etc etc.
for 3D, 4D etc just apply cartesian product over all dimensions.
Upon thinking about that, it seems that even if you apply a unitary
coordinate transform (unitary rotation and/or mirroring) the "iff" still
holds. Not very sure, but seems interesting.
Not a proof.. just a sketch of idea. No guarantee. If there is a mistake
please leave a comment. Thanks.
i*D
6 楼
我想判断3d空间中一个线段和在一个点的关系. 这个点在这个线段的直线上,我想判断
它在线段的两个端点之间还是在之外. 这个问题本身是一个明确的判断,不能容忍error
和tolerance.
比如线段是AB, 另外一个点是C. 我的方法是判断if AB-(AC+BC)<0. 但是在实际运算中
,我发现即便是C在AB之间,也有很多时候AB-(AC+BC)<0. 应该是因为round off error.
不知道有什么好的trick或者策略来解决这个问题.
它在线段的两个端点之间还是在之外. 这个问题本身是一个明确的判断,不能容忍error
和tolerance.
比如线段是AB, 另外一个点是C. 我的方法是判断if AB-(AC+BC)<0. 但是在实际运算中
,我发现即便是C在AB之间,也有很多时候AB-(AC+BC)<0. 应该是因为round off error.
不知道有什么好的trick或者策略来解决这个问题.
k*g
7 楼
If it is given beyond doubt that C is a point on the line extension of AB,
then C is outside the finite segment AB iff
C is outside the axis-aligned bounding box
the cartesian product with
( all x in (xmin, xmax) ) and
( all y in (ymin, ymax) )
where xmin = min ( A_x , B_x )
xmax = max ( A_x , B_x )
etc etc.
for 3D, 4D etc just apply cartesian product over all dimensions.
Upon thinking about that, it seems that even if you apply a unitary
coordinate transform (unitary rotation and/or mirroring) the "iff" still
holds. Not very sure, but seems interesting.
Not a proof.. just a sketch of idea. No guarantee. If there is a mistake
please leave a comment. Thanks.
then C is outside the finite segment AB iff
C is outside the axis-aligned bounding box
the cartesian product with
( all x in (xmin, xmax) ) and
( all y in (ymin, ymax) )
where xmin = min ( A_x , B_x )
xmax = max ( A_x , B_x )
etc etc.
for 3D, 4D etc just apply cartesian product over all dimensions.
Upon thinking about that, it seems that even if you apply a unitary
coordinate transform (unitary rotation and/or mirroring) the "iff" still
holds. Not very sure, but seems interesting.
Not a proof.. just a sketch of idea. No guarantee. If there is a mistake
please leave a comment. Thanks.
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