Line segment intersection
Give the coordinates of two points of several line segments, and output the number of intersection points of the two line segments.
Condition 1: The first line segment is completely above (below) the second line segment, it can be judged that there is no intersection.
Condition 2: If AB and CD intersect, then A and B must be on both sides of CD, so (CD×CA)(CD×CB) <0.
C and D must be on both sides of AB, so (AC×AB)(AD×AB) <0.
Suppose A(x1,y1), B(x2,y2), C(x3,y3), D(x4,y4).
CA = (x1-x3, y1-y3), CB = (x2-x3, y2-y3).
CD = (x4-x3, y4-y3).
AC = (x3-x1, y3-y1), AD = (x4-x1, y4-y1).
AB = (x2-x1, y2-y1).
So, (CD×CA)(CD×CB) = ((x4-x3) (y1-y3)-(x1-x3) (y4-y3))((x4-x3) (y2-y3)-(x2 -x3) (y4-y3))> 0 holds,
or (AC×AB)(AD×AB) = ((x3-x1) (y2-y1)-(x2-x1) (y3-y1))( (x4-x1) (y2-y1)-(x2-x1) (y4-y1))> 0 is true, there
is no intersection.
All other situations have intersections.
Dot multiplication:
x=(a1,a2,…,an），y=(b1,b2,…,bn）
x·y=a1b1+……+anbn .
Cross product:
Let the coordinates of the two vectors be u (u1, u2, u3), v(v1, v2, v3).
Then u×v = (u2 v3-u3 v2)-(u1 v3-u3 v1) + (u1 v2-u2 v1).
Two-dimensional vector cross product formula: x1 y2-x2 y1.
Obtain
the clockwise and reverse relationship between the two vectors: if l1 × l2> 0, then l1 is in the clockwise direction of l2;
if l1 × l2 <0, then l1 is in the counterclockwise direction of l2;
if l1 × l2 = 0, Then l1 and l2 are collinear (may be in the same direction, or reverse);