在非优化伪代码中:
def isBetween(a,b,c ):
crossproduct =(cy - ay)*(bx - ax) - (cx - ax)*(by - ay)
#与浮点数val的epsilon比较ues或!= 0如果使用整数
if abs(crossproduct)> epsilon:
return False
dotproduct =(c.x - a.x)*(b.x - a.x)+(c.y - a.y)*(b.y - a.y)
if dotproduct< 0:
return False
squaredlengthba =(b.x - a.x)*(b.x - a.x)+(b.y - a.y)*(b.y - a.y)
if dotproduct> squaredlengthba:
返回False
返回True
Let's say you have a two dimensional plane with 2 points (called a and b) on it represented by an x integer and a y integer for each point.
How can you determine if another point c is on the line segment defined by a and b?
I use python most, but examples in any language would be helpful.
解决方案
Check if the cross product of (b-a) and (c-a) is 0, as tells Darius Bacon, tells you if the points a, b and c are aligned.
But, as you want to know if c is between a and b, you also have to check that the dot product of (b-a) and (c-a) is positive and is less than the square of the distance between a and b.
In non-optimized pseudocode:
def isBetween(a, b, c):
crossproduct = (c.y - a.y) * (b.x - a.x) - (c.x - a.x) * (b.y - a.y)
# compare versus epsilon for floating point values, or != 0 if using integers
if abs(crossproduct) > epsilon:
return False
dotproduct = (c.x - a.x) * (b.x - a.x) + (c.y - a.y)*(b.y - a.y)
if dotproduct < 0:
return False
squaredlengthba = (b.x - a.x)*(b.x - a.x) + (b.y - a.y)*(b.y - a.y)
if dotproduct > squaredlengthba:
return False
return True
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