How can I check if two segments intersect?
Question:
How can I check if 2 segments intersect?
I’ve the following data:
Segment1 [ {x1,y1}, {x2,y2} ]
Segment2 [ {x1,y1}, {x2,y2} ]
I need to write a small algorithm in Python to detect if the 2 lines are intersecting.
Answers:
The equation of a line is:
f(x) = A*x + b = y
For a segment, it is exactly the same, except that x is included on an interval I.
If you have two segments, defined as follow:
Segment1 = {(X1, Y1), (X2, Y2)}
Segment2 = {(X3, Y3), (X4, Y4)}
The abcisse Xa of the potential point of intersection (Xa,Ya) must be contained in both interval I1 and I2, defined as follow :
I1 = [min(X1,X2), max(X1,X2)]
I2 = [min(X3,X4), max(X3,X4)]
And we could say that Xa is included into :
Ia = [max( min(X1,X2), min(X3,X4) ),
min( max(X1,X2), max(X3,X4) )]
Now, we need to check that this interval Ia exists :
if (max(X1,X2) < min(X3,X4)):
return False # There is no mutual abcisses
So, we have two line formula, and a mutual interval. Your line formulas are:
f1(x) = A1*x + b1 = y
f2(x) = A2*x + b2 = y
As we got two points by segment, we are able to determine A1, A2, b1 and b2:
A1 = (Y1-Y2)/(X1-X2) # Pay attention to not dividing by zero
A2 = (Y3-Y4)/(X3-X4) # Pay attention to not dividing by zero
b1 = Y1-A1*X1 = Y2-A1*X2
b2 = Y3-A2*X3 = Y4-A2*X4
If the segments are parallel, then A1 == A2 :
if (A1 == A2):
return False # Parallel segments
A point (Xa,Ya) standing on both line must verify both formulas f1 and f2:
Ya = A1 * Xa + b1
Ya = A2 * Xa + b2
A1 * Xa + b1 = A2 * Xa + b2
Xa = (b2 - b1) / (A1 - A2) # Once again, pay attention to not dividing by zero
The last thing to do is check that Xa is included into Ia:
if ( (Xa < max( min(X1,X2), min(X3,X4) )) or
(Xa > min( max(X1,X2), max(X3,X4) )) ):
return False # intersection is out of bound
else:
return True
In addition to this, you may check at startup that two of the four provided points are not equals to avoid all that testing.
if your data define line you just have to prove that they are not parallel. To do this you can compute
alpha = float(y2 - y1) / (x2 - x1).
If this coefficient is equal for both Line1 and Line2, it means the line are parallel. If not, it means they will intersect.
If they are parallel you then have to prove that they are not the same. For that, you compute
beta = y1 - alpha*x1
If beta is the same for Line1 and Line2,it means you line intersect as they are equal
If they are segment, you still have to compute alpha and beta as described above for each Line. Then you have to check that (beta1 – beta2) / (alpha1 – alpha2) is greater than Min(x1_line1, x2_line1) and less than Max(x1_line1, x2_line1)
Calculate the intersection point of the lines laying on your segments (it means basically to solve a linear equation system), then check whether is it between the starting and ending points of your segments.
You have two line segments. Define one segment by endpoints A & B and the second segment by endpoints C & D. There is a nice trick to show that they must intersect, WITHIN the bounds of the segments. (Note that the lines themselves may intersect beyond the bounds of the segments, so you must be careful. Good code will also watch for parallel lines.)
The trick is to test that points A and B must line on opposite sides of line CD, AND that points C and D must lie on opposite sides of line AB.
Since this is homework, I won’t give you an explicit solution. But a simple test to see which side of a line a point falls on, is to use a dot product. Thus, for a given line CD, compute the normal vector to that line (I’ll call it N_C.) Now, simply test the signs of these two results:
dot(A-C,N_C)
and
dot(B-C,N_C)
If those results have opposite signs, then A and B are opposite sides of line CD. Now do the same test for the other line, AB. It has normal vector N_A. Compare the signs of
dot(C-A,N_A)
and
dot(D-A,N_A)
I’ll leave it to you to figure out how to compute a normal vector. (In 2-d, that is trivial, but will your code worry about whether A and B are distinct points? Likewise, are C and D distinct?)
You still need to worry about line segments that lie along the same infinite line, or if one point actually falls on the other line segment itself. Good code will cater to every possible problem.
Suppose the two segments have endpoints A,B and C,D. The numerically robust way to determine intersection is to check the sign of the four determinants:
| Ax-Cx Bx-Cx | | Ax-Dx Bx-Dx |
| Ay-Cy By-Cy | | Ay-Dy By-Dy |
| Cx-Ax Dx-Ax | | Cx-Bx Dx-Bx |
| Cy-Ay Dy-Ay | | Cy-By Dy-By |
For intersection, each determinant on the left must have the opposite sign of the one to the right, but there need not be any relationship between the two lines. You are basically checking each point of a segment against the other segment to make sure they lie on opposite sides of the line defined by the other segment.
See here: http://www.cs.cmu.edu/~quake/robust.html
You don’t have to compute exactly where does the segments intersect, but only understand whether they intersect at all. This will simplify the solution.
The idea is to treat one segment as the “anchor” and separate the second segment into 2 points.
Now, you will have to find the relative position of each point to the “anchored” segment (OnLeft, OnRight or Collinear).
After doing so for both points, check that one of the points is OnLeft and the other is OnRight (or perhaps include Collinear position, if you wish to include improper intersections as well).
You must then repeat the process with the roles of anchor and separated segments.
An intersection exists if, and only if, one of the points is OnLeft and the other is OnRight. See this link for a more detailed explanation with example images for each possible case.
Implementing such method will be much easier than actually implementing a method that finds the intersection point (given the many corner cases which you will have to handle as well).
Update
The following functions should illustrate the idea (source: Computational Geometry in C).
Remark: This sample assumes the usage of integers. If you’re using some floating-point representation instead (which could obviously complicate things), then you should determine some epsilon value to indicate “equality” (mostly for the IsCollinear evaluation).
// points "a" and "b" forms the anchored segment.
// point "c" is the evaluated point
bool IsOnLeft(Point a, Point b, Point c)
{
return Area2(a, b, c) > 0;
}
bool IsOnRight(Point a, Point b, Point c)
{
return Area2(a, b, c) < 0;
}
bool IsCollinear(Point a, Point b, Point c)
{
return Area2(a, b, c) == 0;
}
// calculates the triangle's size (formed by the "anchor" segment and additional point)
int Area2(Point a, Point b, Point c)
{
return (b.X - a.X) * (c.Y - a.Y) -
(c.X - a.X) * (b.Y - a.Y);
}
Of course, when using these functions, one must remember to check that each segment lies “between” the other segment (since these are finite segments, and not infinite lines).
Also, using these functions you can understand whether you’ve got a proper or improper intersection.
- Proper: There are no collinear points. The segments crosses each
other “from side to side”.
- Improper: One segment only “touches” the other (at least one of
the points is collinear to the
anchored segment).
This is what I’ve got for AS3, don’t know much about python but the concept is there
public function getIntersectingPointF($A:Point, $B:Point, $C:Point, $D:Point):Number {
var A:Point = $A.clone();
var B:Point = $B.clone();
var C:Point = $C.clone();
var D:Point = $D.clone();
var f_ab:Number = (D.x - C.x) * (A.y - C.y) - (D.y - C.y) * (A.x - C.x);
// are lines parallel
if (f_ab == 0) { return Infinity };
var f_cd:Number = (B.x - A.x) * (A.y - C.y) - (B.y - A.y) * (A.x - C.x);
var f_d:Number = (D.y - C.y) * (B.x - A.x) - (D.x - C.x) * (B.y - A.y);
var f1:Number = f_ab/f_d
var f2:Number = f_cd / f_d
if (f1 == Infinity || f1 <= 0 || f1 >= 1) { return Infinity };
if (f2 == Infinity || f2 <= 0 || f2 >= 1) { return Infinity };
return f1;
}
public function getIntersectingPoint($A:Point, $B:Point, $C:Point, $D:Point):Point
{
var f:Number = getIntersectingPointF($A, $B, $C, $D);
if (f == Infinity || f <= 0 || f >= 1) { return null };
var retPoint:Point = Point.interpolate($A, $B, 1 - f);
return retPoint.clone();
}
for segments AB and CD, find the slope of CD
slope=(Dy-Cy)/(Dx-Cx)
extend CD over A and B, and take the distance to CD going straight up
dist1=slope*(Cx-Ax)+Ay-Cy
dist2=slope*(Dx-Ax)+Ay-Dy
check if they are on opposite sides
return dist1*dist2<0
User @i_4_got points to this page with a very efficent solution in Python. I reproduce it here for convenience (since it would have made me happy to have it here):
def ccw(A,B,C):
return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)
# Return true if line segments AB and CD intersect
def intersect(A,B,C,D):
return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)
Resolved but still why not with python… 🙂
def islineintersect(line1, line2):
i1 = [min(line1[0][0], line1[1][0]), max(line1[0][0], line1[1][0])]
i2 = [min(line2[0][0], line2[1][0]), max(line2[0][0], line2[1][0])]
ia = [max(i1[0], i2[0]), min(i1[1], i2[1])]
if max(line1[0][0], line1[1][0]) < min(line2[0][0], line2[1][0]):
return False
m1 = (line1[1][1] - line1[0][1]) * 1. / (line1[1][0] - line1[0][0]) * 1.
m2 = (line2[1][1] - line2[0][1]) * 1. / (line2[1][0] - line2[0][0]) * 1.
if m1 == m2:
return False
b1 = line1[0][1] - m1 * line1[0][0]
b2 = line2[0][1] - m2 * line2[0][0]
x1 = (b2 - b1) / (m1 - m2)
if (x1 < max(i1[0], i2[0])) or (x1 > min(i1[1], i2[1])):
return False
return True
This:
print islineintersect([(15, 20), (100, 200)], [(210, 5), (23, 119)])
Output:
True
And this:
print islineintersect([(15, 20), (100, 200)], [(-1, -5), (-5, -5)])
Output:
False
Implemented in JAVA. However It seems that it does not work for co-linear lines (aka line segments that exist within each other L1(0,0)(10,10) L2(1,1)(2,2)
public class TestCode
{
public class Point
{
public double x = 0;
public double y = 0;
public Point(){}
}
public class Line
{
public Point p1, p2;
public Line( double x1, double y1, double x2, double y2)
{
p1 = new Point();
p2 = new Point();
p1.x = x1;
p1.y = y1;
p2.x = x2;
p2.y = y2;
}
}
//line segments
private static Line s1;
private static Line s2;
public TestCode()
{
s1 = new Line(0,0,0,10);
s2 = new Line(-1,0,0,10);
}
public TestCode(double x1, double y1,
double x2, double y2,
double x3, double y3,
double x4, double y4)
{
s1 = new Line(x1,y1, x2,y2);
s2 = new Line(x3,y3, x4,y4);
}
public static void main(String args[])
{
TestCode code = null;
////////////////////////////
code = new TestCode(0,0,0,10,
0,1,0,5);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,0,10,
0,1,0,10);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,10,0,
5,0,15,0);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,10,0,
0,0,15,0);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,10,10,
1,1,5,5);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,0,10,
-1,-1,0,10);
if( intersect(code) )
{ System.out.println( "OK SLOPE END: INTERSECTS" ); }
else
{ System.out.println( "ERROR SLOPE END: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(-10,-10,10,10,
-10,10,10,-10);
if( intersect(code) )
{ System.out.println( "OK SLOPE Intersect(0,0): INTERSECTS" ); }
else
{ System.out.println( "ERROR SLOPE Intersect(0,0): DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(-10,-10,10,10,
-3,-2,50,-2);
if( intersect(code) )
{ System.out.println( "OK SLOPE Line2 VERTIAL: INTERSECTS" ); }
else
{ System.out.println( "ERROR SLOPE Line2 VERTICAL: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(-10,-10,10,10,
50,-2,-3,-2);
if( intersect(code) )
{ System.out.println( "OK SLOPE Line2 (reversed) VERTIAL: INTERSECTS" ); }
else
{ System.out.println( "ERROR SLOPE Line2 (reversed) VERTICAL: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,0,10,
1,0,1,10);
if( intersect(code) )
{ System.out.println( "ERROR PARALLEL VERTICAL: INTERSECTS" ); }
else
{ System.out.println( "OK PARALLEL VERTICAL: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,2,10,2,
0,10,10,10);
if( intersect(code) )
{ System.out.println( "ERROR PARALLEL HORIZONTAL: INTERSECTS" ); }
else
{ System.out.println( "OK PARALLEL HORIZONTAL: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,10,5,13.75,
0,18.75,10,15);
if( intersect(code) )
{ System.out.println( "ERROR PARALLEL SLOPE=.75: INTERSECTS" ); }
else
{ System.out.println( "OK PARALLEL SLOPE=.75: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,1,1,
2,-1,2,10);
if( intersect(code) )
{ System.out.println( "ERROR SEPERATE SEGMENTS: INTERSECTS" ); }
else
{ System.out.println( "OK SEPERATE SEGMENTS: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,1,1,
-1,-10,-5,10);
if( intersect(code) )
{ System.out.println( "ERROR SEPERATE SEGMENTS 2: INTERSECTS" ); }
else
{ System.out.println( "OK SEPERATE SEGMENTS 2: DO NOT INTERSECT" ); }
}
public static boolean intersect( TestCode code )
{
return intersect( code.s1, code.s2);
}
public static boolean intersect( Line line1, Line line2 )
{
double i1min = Math.min(line1.p1.x, line1.p2.x);
double i1max = Math.max(line1.p1.x, line1.p2.x);
double i2min = Math.min(line2.p1.x, line2.p2.x);
double i2max = Math.max(line2.p1.x, line2.p2.x);
double iamax = Math.max(i1min, i2min);
double iamin = Math.min(i1max, i2max);
if( Math.max(line1.p1.x, line1.p2.x) < Math.min(line2.p1.x, line2.p2.x) )
return false;
double m1 = (line1.p2.y - line1.p1.y) / (line1.p2.x - line1.p1.x );
double m2 = (line2.p2.y - line2.p1.y) / (line2.p2.x - line2.p1.x );
if( m1 == m2 )
return false;
//b1 = line1[0][1] - m1 * line1[0][0]
//b2 = line2[0][1] - m2 * line2[0][0]
double b1 = line1.p1.y - m1 * line1.p1.x;
double b2 = line2.p1.y - m2 * line2.p1.x;
double x1 = (b2 - b1) / (m1 - m2);
if( (x1 < Math.max(i1min, i2min)) || (x1 > Math.min(i1max, i2max)) )
return false;
return true;
}
}
Output thus far is
ERROR COLINEAR: DO NOT INTERSECT
ERROR COLINEAR: DO NOT INTERSECT
ERROR COLINEAR: DO NOT INTERSECT
ERROR COLINEAR: DO NOT INTERSECT
ERROR COLINEAR: DO NOT INTERSECT
OK SLOPE END: INTERSECTS
OK SLOPE Intersect(0,0): INTERSECTS
OK SLOPE Line2 VERTIAL: INTERSECTS
OK SLOPE Line2 (reversed) VERTIAL: INTERSECTS
OK PARALLEL VERTICAL: DO NOT INTERSECT
OK PARALLEL HORIZONTAL: DO NOT INTERSECT
OK PARALLEL SLOPE=.75: DO NOT INTERSECT
OK SEPERATE SEGMENTS: DO NOT INTERSECT
OK SEPERATE SEGMENTS 2: DO NOT INTERSECT
Here is C code to check if two points are on the opposite sides of the line segment. Using this code you can check if two segments intersect as well.
// true if points p1, p2 lie on the opposite sides of segment s1--s2
bool oppositeSide (Point2f s1, Point2f s2, Point2f p1, Point2f p2) {
//calculate normal to the segment
Point2f vec = s1-s2;
Point2f normal(vec.y, -vec.x); // no need to normalize
// vectors to the points
Point2f v1 = p1-s1;
Point2f v2 = p2-s1;
// compare signs of the projections of v1, v2 onto the normal
float proj1 = v1.dot(normal);
float proj2 = v2.dot(normal);
if (proj1==0 || proj2==0)
cout<<"collinear points"<<endl;
return(SIGN(proj1) != SIGN(proj2));
}
Based on Liran’s and Grumdrig’s excellent answers here is a complete Python code to verify if closed segments do intersect. Works for collinear segments, segments parallel to axis Y, degenerate segments (devil is in details). Assumes integer coordinates. Floating point coordinates require a modification to points equality test.
def side(a,b,c):
""" Returns a position of the point c relative to the line going through a and b
Points a, b are expected to be different
"""
d = (c[1]-a[1])*(b[0]-a[0]) - (b[1]-a[1])*(c[0]-a[0])
return 1 if d > 0 else (-1 if d < 0 else 0)
def is_point_in_closed_segment(a, b, c):
""" Returns True if c is inside closed segment, False otherwise.
a, b, c are expected to be collinear
"""
if a[0] < b[0]:
return a[0] <= c[0] and c[0] <= b[0]
if b[0] < a[0]:
return b[0] <= c[0] and c[0] <= a[0]
if a[1] < b[1]:
return a[1] <= c[1] and c[1] <= b[1]
if b[1] < a[1]:
return b[1] <= c[1] and c[1] <= a[1]
return a[0] == c[0] and a[1] == c[1]
#
def closed_segment_intersect(a,b,c,d):
""" Verifies if closed segments a, b, c, d do intersect.
"""
if a == b:
return a == c or a == d
if c == d:
return c == a or c == b
s1 = side(a,b,c)
s2 = side(a,b,d)
# All points are collinear
if s1 == 0 and s2 == 0:
return
is_point_in_closed_segment(a, b, c) or is_point_in_closed_segment(a, b, d) or
is_point_in_closed_segment(c, d, a) or is_point_in_closed_segment(c, d, b)
# No touching and on the same side
if s1 and s1 == s2:
return False
s1 = side(c,d,a)
s2 = side(c,d,b)
# No touching and on the same side
if s1 and s1 == s2:
return False
return True
Since you do not mention that you want to find the intersection point of the line, the problem becomes simpler to solve. If you need the intersection point, then the answer by OMG_peanuts is a faster approach. However, if you just want to find whether the lines intersect or not, you can do so by using the line equation (ax + by + c = 0). The approach is as follows:
-
Let’s start with two line segments: segment 1 and segment 2.
segment1 = [[x1,y1], [x2,y2]]
segment2 = [[x3,y3], [x4,y4]]
-
Check if the two line segments are non zero length line and distinct segments.
-
From hereon, I assume that the two segments are non-zero length and distinct. For each line segment, compute the slope of the line and then obtain the equation of a line in the form of ax + by + c = 0. Now, compute the value of f = ax + by + c for the two points of the other line segment (repeat this for the other line segment as well).
a2 = (y3-y4)/(x3-x4);
b1 = -1;
b2 = -1;
c1 = y1 - a1*x1;
c2 = y3 - a2*x3;
// using the sign function from numpy
f1_1 = sign(a1*x3 + b1*y3 + c1);
f1_2 = sign(a1*x4 + b1*y4 + c1);
f2_1 = sign(a2*x1 + b2*y1 + c2);
f2_2 = sign(a2*x2 + b2*y2 + c2);
-
Now all that is left is the different cases. If f = 0 for any point, then the two lines touch at a point. If f1_1 and f1_2 are equal or f2_1 and f2_2 are equal, then the lines do not intersect. If f1_1 and f1_2 are unequal and f2_1 and f2_2 are unequal, then the line segments intersect. Depending on whether you want to consider the lines which touch as “intersecting” or not, you can adapt your conditions.
Here is another python code to check whether closed segments intersect. It is the rewritten version of the C++ code in http://www.cdn.geeksforgeeks.org/check-if-two-given-line-segments-intersect/. This implementation covers all special cases (e.g. all points colinear).
def on_segment(p, q, r):
'''Given three colinear points p, q, r, the function checks if
point q lies on line segment "pr"
'''
if (q[0] <= max(p[0], r[0]) and q[0] >= min(p[0], r[0]) and
q[1] <= max(p[1], r[1]) and q[1] >= min(p[1], r[1])):
return True
return False
def orientation(p, q, r):
'''Find orientation of ordered triplet (p, q, r).
The function returns following values
0 --> p, q and r are colinear
1 --> Clockwise
2 --> Counterclockwise
'''
val = ((q[1] - p[1]) * (r[0] - q[0]) -
(q[0] - p[0]) * (r[1] - q[1]))
if val == 0:
return 0 # colinear
elif val > 0:
return 1 # clockwise
else:
return 2 # counter-clockwise
def do_intersect(p1, q1, p2, q2):
'''Main function to check whether the closed line segments p1 - q1 and p2
- q2 intersect'''
o1 = orientation(p1, q1, p2)
o2 = orientation(p1, q1, q2)
o3 = orientation(p2, q2, p1)
o4 = orientation(p2, q2, q1)
# General case
if (o1 != o2 and o3 != o4):
return True
# Special Cases
# p1, q1 and p2 are colinear and p2 lies on segment p1q1
if (o1 == 0 and on_segment(p1, p2, q1)):
return True
# p1, q1 and p2 are colinear and q2 lies on segment p1q1
if (o2 == 0 and on_segment(p1, q2, q1)):
return True
# p2, q2 and p1 are colinear and p1 lies on segment p2q2
if (o3 == 0 and on_segment(p2, p1, q2)):
return True
# p2, q2 and q1 are colinear and q1 lies on segment p2q2
if (o4 == 0 and on_segment(p2, q1, q2)):
return True
return False # Doesn't fall in any of the above cases
Below is a test function to verify that it works.
import matplotlib.pyplot as plt
def test_intersect_func():
p1 = (1, 1)
q1 = (10, 1)
p2 = (1, 2)
q2 = (10, 2)
fig, ax = plt.subplots()
ax.plot([p1[0], q1[0]], [p1[1], q1[1]], 'x-')
ax.plot([p2[0], q2[0]], [p2[1], q2[1]], 'x-')
print(do_intersect(p1, q1, p2, q2))
p1 = (10, 0)
q1 = (0, 10)
p2 = (0, 0)
q2 = (10, 10)
fig, ax = plt.subplots()
ax.plot([p1[0], q1[0]], [p1[1], q1[1]], 'x-')
ax.plot([p2[0], q2[0]], [p2[1], q2[1]], 'x-')
print(do_intersect(p1, q1, p2, q2))
p1 = (-5, -5)
q1 = (0, 0)
p2 = (1, 1)
q2 = (10, 10)
fig, ax = plt.subplots()
ax.plot([p1[0], q1[0]], [p1[1], q1[1]], 'x-')
ax.plot([p2[0], q2[0]], [p2[1], q2[1]], 'x-')
print(do_intersect(p1, q1, p2, q2))
p1 = (0, 0)
q1 = (1, 1)
p2 = (1, 1)
q2 = (10, 10)
fig, ax = plt.subplots()
ax.plot([p1[0], q1[0]], [p1[1], q1[1]], 'x-')
ax.plot([p2[0], q2[0]], [p2[1], q2[1]], 'x-')
print(do_intersect(p1, q1, p2, q2))
I thought I’d contribute a nice Swift solution:
struct Pt {
var x: Double
var y: Double
}
struct LineSegment {
var p1: Pt
var p2: Pt
}
func doLineSegmentsIntersect(ls1: LineSegment, ls2: LineSegment) -> Bool {
if (ls1.p2.x-ls1.p1.x == 0) { //handle vertical segment1
if (ls2.p2.x-ls2.p1.x == 0) {
//both lines are vertical and parallel
return false
}
let x = ls1.p1.x
let slope2 = (ls2.p2.y-ls2.p1.y)/(ls2.p2.x-ls2.p1.x)
let c2 = ls2.p1.y-slope2*ls2.p1.x
let y = x*slope2+c2 // y intersection point
return (y > ls1.p1.y && x < ls1.p2.y) || (y > ls1.p2.y && y < ls1.p1.y) // check if y is between y1,y2 in segment1
}
if (ls2.p2.x-ls2.p1.x == 0) { //handle vertical segment2
let x = ls2.p1.x
let slope1 = (ls1.p2.y-ls1.p1.y)/(ls1.p2.x-ls1.p1.x)
let c1 = ls1.p1.y-slope1*ls1.p1.x
let y = x*slope1+c1 // y intersection point
return (y > ls2.p1.y && x < ls2.p2.y) || (y > ls2.p2.y && y < ls2.p1.y) // validate that y is between y1,y2 in segment2
}
let slope1 = (ls1.p2.y-ls1.p1.y)/(ls1.p2.x-ls1.p1.x)
let slope2 = (ls2.p2.y-ls2.p1.y)/(ls2.p2.x-ls2.p1.x)
if (slope1 == slope2) { //segments are parallel
return false
}
let c1 = ls1.p1.y-slope1*ls1.p1.x
let c2 = ls2.p1.y-slope2*ls2.p1.x
let x = (c2-c1)/(slope1-slope2)
return (((x > ls1.p1.x && x < ls1.p2.x) || (x > ls1.p2.x && x < ls1.p1.x)) &&
((x > ls2.p1.x && x < ls2.p2.x) || (x > ls2.p2.x && x < ls2.p1.x)))
//validate that x is between x1,x2 in both segments
}
We can also solve this utilizing vectors.
Let’s define the segments as [start, end]. Given two such segments [A, B] and [C, D] that both have non-zero length, we can choose one of the endpoints to be used as a reference point so that we get three vectors:
x = 0
y = 1
p = A-C = [C[x]-A[x], C[y]-A[y]]
q = B-A = [B[x]-A[x], B[y]-A[y]]
r = D-C = [D[x]-C[x], D[y]-C[y]]
From there, we can look for an intersection by calculating t and u in p + t*r = u*q. After playing around with the equation a little, we get:
t = (q[y]*p[x] - q[x]*p[y])/(q[x]*r[y] - q[y]*r[x])
u = (p[x] + t*r[x])/q[x]
Thus, the function is:
def intersects(a, b):
p = [b[0][0]-a[0][0], b[0][1]-a[0][1]]
q = [a[1][0]-a[0][0], a[1][1]-a[0][1]]
r = [b[1][0]-b[0][0], b[1][1]-b[0][1]]
t = (q[1]*p[0] - q[0]*p[1])/(q[0]*r[1] - q[1]*r[0])
if (q[0]*r[1] - q[1]*r[0]) != 0
else (q[1]*p[0] - q[0]*p[1])
u = (p[0] + t*r[0])/q[0]
if q[0] != 0
else (p[1] + t*r[1])/q[1]
return t >= 0 and t <= 1 and u >= 0 and u <= 1
This is my way of checking for line crossing and where the intersection occurs. Lets use x1 through x4 and y1 through y4
Segment1 = ((X1, Y1), (X2, Y2))
Segment2 = ((X3, Y3), (X4, Y4))
Then we need some vectors to represent them
dx1 = X2 - X1
dx2 = X4 - X3
dy1 = Y2 - Y1
dy2 = Y4 - Y3
Now we look at the determinant
det = dx1 * dy2 - dx2 * dy1
If the determinant is 0.0, then the line segments are parallel. This could mean they overlap. If they overlap just at endpoints, then there is one intersection solution. Otherwise there will be infinite solutions. With infinitely many solutions, what do say is your point of intersection? So it’s an interesting special case. If you know ahead of time that the lines can’t overlap then you can just check if det == 0.0 and if so just say they don’t intersect and be done. Otherwise, lets continue on
dx3 = X1 - X3
dy3 = Y1 - Y3
det1 = dx1 * dy3 - dx3 * dy1
det2 = dx2 * dy3 - dx3 * dy2
Now, if det, det1 and det2 are all zero, then your lines are co-linear and could overlap. If det is zero but either det1 or det2 are not, then they are not co-linear, but are parallel, so there is no intersection. So what’s left now if det is zero is a 1D problem instead of 2D. We will need to check one of two ways, depending if dx1 is zero or not (so we can avoid division by zero). If dx1 is zero then just do the same logic with y values rather than x below.
s = X3 / dx1
t = X4 / dx1
This computes two scalers, such that if we scale the vector (dx1, dy1) by s we get point (x3, y3), and by t we get (x4, y4). So if either s or t is between 0.0 and 1.0, then point 3 or 4 lies on our first line. Negative would mean the point is behind the start of our vector, while > 1.0 means it is further ahead of the end of our vector. 0.0 means it is at (x1, y1) and 1.0 means it is at (x2, y2). If both s and t are < 0.0 or both are > 1.0, then they don’t intersect. And that handles the parallel lines special case.
Now, if det != 0.0 then
s = det1 / det
t = det2 / det
if s < 0.0 or s > 1.0 or t < 0.0 or t > 1.0:
return false # no intersect
This is similar to what we were doing above really. Now if we pass the above test, then our line segments intersect, and we can calculate the intersection quite easily like so:
Ix = X1 + t * dx1
Iy = Y1 + t * dy1
If you want to dig deeper into what the math is doing, look into Cramer’s Rule.
Edit: I’ve fixed two errors that were present so it should be correct now. I’ve learned enough python now to write actual code for this now. It works except for some special cases, though it does handle some special cases correctly. Special cases get really hard to deal with and I’ve spend enough time on it and want to move on. If someone requires better then they have a good starting point at least to try and improve it.
import math
def line_intersection(line1, line2):
x1, x2, x3, x4 = line1[0][0], line1[1][0], line2[0][0], line2[1][0]
y1, y2, y3, y4 = line1[0][1], line1[1][1], line2[0][1], line2[1][1]
dx1 = x2 - x1
dx2 = x4 - x3
dy1 = y2 - y1
dy2 = y4 - y3
dx3 = x1 - x3
dy3 = y1 - y3
det = dx1 * dy2 - dx2 * dy1
det1 = dx1 * dy3 - dx3 * dy1
det2 = dx2 * dy3 - dx3 * dy2
if det == 0.0: # lines are parallel
if det1 != 0.0 or det2 != 0.0: # lines are not co-linear
return None # so no solution
if dx1:
if x1 < x3 < x2 or x1 > x3 > x2:
return math.inf # infinitely many solutions
else:
if y1 < y3 < y2 or y1 > y3 > y2:
return math.inf # infinitely many solutions
if line1[0] == line2[0] or line1[1] == line2[0]:
return line2[0]
elif line1[0] == line2[1] or line1[1] == line2[1]:
return line2[1]
return None # no intersection
s = det1 / det
t = det2 / det
if 0.0 < s < 1.0 and 0.0 < t < 1.0:
return x1 + t * dx1, y1 + t * dy1
print("one intersection")
print(line_intersection(((0.0,0.0), (6.0,6.0)),((0.0,9.0), (9.0,0.0))))
print(line_intersection(((-2, -2), (2, 2)), ((2, -2), (-2, 2))))
print(line_intersection(((0.5, 0.5), (1.5, 0.5)), ((1.0, 0.0), (1.0, 2.0))))
print(line_intersection(((0, -1), (0, 0)), ((0, 0), (0, 1))))
print(line_intersection(((-1, 0), (0, 0)), ((0, 0), (1, 0))))
print()
print("no intersection")
print(line_intersection(((-1, -1), (0, 0)), ((2, -4), (2, 4))))
print(line_intersection(((0.0,0.0), (0.0,9.0)),((9.0,0.0), (9.0,99.0))))
print(line_intersection(((0, 0), (1, 1)), ((1, 0), (2, 1))))
print(line_intersection(((-1, 1), (0, 1)), ((0, 0), (1, 0))))
print(line_intersection(((1, -1), (1, 0)), ((0, 0), (0, -1))))
print()
print("infinite intersection")
print(line_intersection(((-1, -1), (1, 1)), ((0, 0), (2, 2))))
print(line_intersection(((-1, 0), (1, 0)), ((0, 0), (2, 0))))
print(line_intersection(((0, -1), (0, 1)), ((0, 0), (0, 2))))
print(line_intersection(((-1, 0), (0, 0)), ((0, 0), (-1, 0))))
print(line_intersection(((1, 0), (0, 0)), ((0, 0), (1, 0))))
Checking if line segments intersect is very easy with Shapely library using intersects method:
from shapely.geometry import LineString
line = LineString([(0, 0), (1, 1)])
other = LineString([(0, 1), (1, 0)])
print(line.intersects(other))
# True
line = LineString([(0, 0), (1, 1)])
other = LineString([(0, 1), (1, 2)])
print(line.intersects(other))
# False
One of the solutions above worked so well I decided to write a complete demonstration program using wxPython. You should be able to run this program like this: python “your file name“
# Click on the window to draw a line.
# The program will tell you if this and the other line intersect.
import wx
class Point:
def __init__(self, newX, newY):
self.x = newX
self.y = newY
app = wx.App()
frame = wx.Frame(None, wx.ID_ANY, "Main")
p1 = Point(90,200)
p2 = Point(150,80)
mp = Point(0,0) # mouse point
highestX = 0
def ccw(A,B,C):
return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)
# Return true if line segments AB and CD intersect
def intersect(A,B,C,D):
return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)
def is_intersection(p1, p2, p3, p4):
return intersect(p1, p2, p3, p4)
def drawIntersection(pc):
mp2 = Point(highestX, mp.y)
if is_intersection(p1, p2, mp, mp2):
pc.DrawText("intersection", 10, 10)
else:
pc.DrawText("no intersection", 10, 10)
def do_paint(evt):
pc = wx.PaintDC(frame)
pc.DrawLine(p1.x, p1.y, p2.x, p2.y)
pc.DrawLine(mp.x, mp.y, highestX, mp.y)
drawIntersection(pc)
def do_left_mouse(evt):
global mp, highestX
point = evt.GetPosition()
mp = Point(point[0], point[1])
highestX = frame.Size[0]
frame.Refresh()
frame.Bind(wx.EVT_PAINT, do_paint)
frame.Bind(wx.EVT_LEFT_DOWN, do_left_mouse)
frame.Show()
app.MainLoop()
The answer by Georgy is the cleanest to implement, by far. Had to chase this down, since the brycboe example, while simple as well, had issues with colinearity.
Code for testing:
#!/usr/bin/python
#
# Notes on intersection:
#
# https://bryceboe.com/2006/10/23/line-segment-intersection-algorithm/
#
# https://stackoverflow.com/questions/3838329/how-can-i-check-if-two-segments-intersect
from shapely.geometry import LineString
class Point:
def __init__(self,x,y):
self.x = x
self.y = y
def ccw(A,B,C):
return (C.y-A.y)*(B.x-A.x) > (B.y-A.y)*(C.x-A.x)
def intersect(A,B,C,D):
return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)
def ShapelyIntersect(A,B,C,D):
return LineString([(A.x,A.y),(B.x,B.y)]).intersects(LineString([(C.x,C.y),(D.x,D.y)]))
a = Point(0,0)
b = Point(0,1)
c = Point(1,1)
d = Point(1,0)
'''
Test points:
b(0,1) c(1,1)
a(0,0) d(1,0)
'''
# F
print(intersect(a,b,c,d))
# T
print(intersect(a,c,b,d))
print(intersect(b,d,a,c))
print(intersect(d,b,a,c))
# F
print(intersect(a,d,b,c))
# same end point cases:
print("same end points")
# F - not intersected
print(intersect(a,b,a,d))
# T - This shows as intersected
print(intersect(b,a,a,d))
# F - this does not
print(intersect(b,a,d,a))
# F - this does not
print(intersect(a,b,d,a))
print("same end points, using shapely")
# T
print(ShapelyIntersect(a,b,a,d))
# T
print(ShapelyIntersect(b,a,a,d))
# T
print(ShapelyIntersect(b,a,d,a))
# T
print(ShapelyIntersect(a,b,d,a))
Here’s a solution using dot products:
# assumes line segments are stored in the format [(x0,y0),(x1,y1)]
def intersects(s0,s1):
dx0 = s0[1][0]-s0[0][0]
dx1 = s1[1][0]-s1[0][0]
dy0 = s0[1][1]-s0[0][1]
dy1 = s1[1][1]-s1[0][1]
p0 = dy1*(s1[1][0]-s0[0][0]) - dx1*(s1[1][1]-s0[0][1])
p1 = dy1*(s1[1][0]-s0[1][0]) - dx1*(s1[1][1]-s0[1][1])
p2 = dy0*(s0[1][0]-s1[0][0]) - dx0*(s0[1][1]-s1[0][1])
p3 = dy0*(s0[1][0]-s1[1][0]) - dx0*(s0[1][1]-s1[1][1])
return (p0*p1<=0) & (p2*p3<=0)
Here’s a visualization in Desmos: Line Segment Intersection
Using OMG_Peanuts solution, I translated to SQL.
(HANA Scalar Function)
Thanks OMG_Peanuts, it works great.
I am using round earth, but distances are small, so I figure its okay.
FUNCTION GA_INTERSECT" ( IN LAT_A1 DOUBLE,
IN LONG_A1 DOUBLE,
IN LAT_A2 DOUBLE,
IN LONG_A2 DOUBLE,
IN LAT_B1 DOUBLE,
IN LONG_B1 DOUBLE,
IN LAT_B2 DOUBLE,
IN LONG_B2 DOUBLE)
RETURNS RET_DOESINTERSECT DOUBLE
LANGUAGE SQLSCRIPT
SQL SECURITY INVOKER AS
BEGIN
DECLARE MA DOUBLE;
DECLARE MB DOUBLE;
DECLARE BA DOUBLE;
DECLARE BB DOUBLE;
DECLARE XA DOUBLE;
DECLARE MAX_MIN_X DOUBLE;
DECLARE MIN_MAX_X DOUBLE;
DECLARE DOESINTERSECT INTEGER;
SELECT 1 INTO DOESINTERSECT FROM DUMMY;
IF LAT_A2-LAT_A1 != 0 AND LAT_B2-LAT_B1 != 0 THEN
SELECT (LONG_A2 - LONG_A1)/(LAT_A2 - LAT_A1) INTO MA FROM DUMMY;
SELECT (LONG_B2 - LONG_B1)/(LAT_B2 - LAT_B1) INTO MB FROM DUMMY;
IF MA = MB THEN
SELECT 0 INTO DOESINTERSECT FROM DUMMY;
END IF;
END IF;
SELECT LONG_A1-MA*LAT_A1 INTO BA FROM DUMMY;
SELECT LONG_B1-MB*LAT_B1 INTO BB FROM DUMMY;
SELECT (BB - BA) / (MA - MB) INTO XA FROM DUMMY;
-- Max of Mins
IF LAT_A1 < LAT_A2 THEN -- MIN(LAT_A1, LAT_A2) = LAT_A1
IF LAT_B1 < LAT_B2 THEN -- MIN(LAT_B1, LAT_B2) = LAT_B1
IF LAT_A1 > LAT_B1 THEN -- MAX(LAT_A1, LAT_B1) = LAT_A1
SELECT LAT_A1 INTO MAX_MIN_X FROM DUMMY;
ELSE -- MAX(LAT_A1, LAT_B1) = LAT_B1
SELECT LAT_B1 INTO MAX_MIN_X FROM DUMMY;
END IF;
ELSEIF LAT_B2 < LAT_B1 THEN -- MIN(LAT_B1, LAT_B2) = LAT_B2
IF LAT_A1 > LAT_B2 THEN -- MAX(LAT_A1, LAT_B2) = LAT_A1
SELECT LAT_A1 INTO MAX_MIN_X FROM DUMMY;
ELSE -- MAX(LAT_A1, LAT_B2) = LAT_B2
SELECT LAT_B2 INTO MAX_MIN_X FROM DUMMY;
END IF;
END IF;
ELSEIF LAT_A2 < LAT_A1 THEN -- MIN(LAT_A1, LAT_A2) = LAT_A2
IF LAT_B1 < LAT_B2 THEN -- MIN(LAT_B1, LAT_B2) = LAT_B1
IF LAT_A2 > LAT_B1 THEN -- MAX(LAT_A2, LAT_B1) = LAT_A2
SELECT LAT_A2 INTO MAX_MIN_X FROM DUMMY;
ELSE -- MAX(LAT_A2, LAT_B1) = LAT_B1
SELECT LAT_B1 INTO MAX_MIN_X FROM DUMMY;
END IF;
ELSEIF LAT_B2 < LAT_B1 THEN -- MIN(LAT_B1, LAT_B2) = LAT_B2
IF LAT_A2 > LAT_B2 THEN -- MAX(LAT_A2, LAT_B2) = LAT_A2
SELECT LAT_A2 INTO MAX_MIN_X FROM DUMMY;
ELSE -- MAX(LAT_A2, LAT_B2) = LAT_B2
SELECT LAT_B2 INTO MAX_MIN_X FROM DUMMY;
END IF;
END IF;
END IF;
-- Min of Max
IF LAT_A1 > LAT_A2 THEN -- MAX(LAT_A1, LAT_A2) = LAT_A1
IF LAT_B1 > LAT_B2 THEN -- MAX(LAT_B1, LAT_B2) = LAT_B1
IF LAT_A1 < LAT_B1 THEN -- MIN(LAT_A1, LAT_B1) = LAT_A1
SELECT LAT_A1 INTO MIN_MAX_X FROM DUMMY;
ELSE -- MIN(LAT_A1, LAT_B1) = LAT_B1
SELECT LAT_B1 INTO MIN_MAX_X FROM DUMMY;
END IF;
ELSEIF LAT_B2 > LAT_B1 THEN -- MAX(LAT_B1, LAT_B2) = LAT_B2
IF LAT_A1 < LAT_B2 THEN -- MIN(LAT_A1, LAT_B2) = LAT_A1
SELECT LAT_A1 INTO MIN_MAX_X FROM DUMMY;
ELSE -- MIN(LAT_A1, LAT_B2) = LAT_B2
SELECT LAT_B2 INTO MIN_MAX_X FROM DUMMY;
END IF;
END IF;
ELSEIF LAT_A2 > LAT_A1 THEN -- MAX(LAT_A1, LAT_A2) = LAT_A2
IF LAT_B1 > LAT_B2 THEN -- MAX(LAT_B1, LAT_B2) = LAT_B1
IF LAT_A2 < LAT_B1 THEN -- MIN(LAT_A2, LAT_B1) = LAT_A2
SELECT LAT_A2 INTO MIN_MAX_X FROM DUMMY;
ELSE -- MIN(LAT_A2, LAT_B1) = LAT_B1
SELECT LAT_B1 INTO MIN_MAX_X FROM DUMMY;
END IF;
ELSEIF LAT_B2 > LAT_B1 THEN -- MAX(LAT_B1, LAT_B2) = LAT_B2
IF LAT_A2 < LAT_B2 THEN -- MIN(LAT_A2, LAT_B2) = LAT_A2
SELECT LAT_A2 INTO MIN_MAX_X FROM DUMMY;
ELSE -- MIN(LAT_A2, LAT_B2) = LAT_B2
SELECT LAT_B2 INTO MIN_MAX_X FROM DUMMY;
END IF;
END IF;
END IF;
IF XA < MAX_MIN_X OR
XA > MIN_MAX_X THEN
SELECT 0 INTO DOESINTERSECT FROM DUMMY;
END IF;
RET_DOESINTERSECT := :DOESINTERSECT;
END;
Digging up an old thread and modifying Grumdrig’s code …
template <typename T>
struct Point {
T x;
T y;
Point(T x_, T y_) : x(x_), y(y_) {};
};
template <typename T>
inline T CrossProduct(const Point<T>& pt1, const Point<T>& pt2, const Point<T>& pt3)
{
// nb: watch out for overflow
return ((pt2.x - pt1.x) * (pt3.y - pt2.y) - (pt2.y - pt1.y) * (pt3.x - pt2.x));
}
template <typename T>
bool SegmentsIntersect(const Point<T>& a, const Point<T>& b,
const Point<T>& c, const Point<T>& d)
{
return
CrossProduct(a, c, d) * CrossProduct(b, c, d) < 0 &&
CrossProduct(c, a, b) * CrossProduct(d, a, b) < 0;
}
if (SegmentsIntersect(Point<int>(50, 0), Point<int>(50, 100),
Point<int>(0, 50), Point<int>(100, 50)))
std::cout << "it works!" << std::endl;
if anybody wants to do a C-like speed look up for multiple lines intersections you could use this code done in numba and python
Note
the epsilon argument should be proportional to your line distance. Also,
the imports are just import numba and import numpy as np
@numba.njit('float64[:,::1], float64[::1], float64', fastmath=True)
def nb_isBetween(line_ab, c, epsilon):
"""
:param line_ab: like --> np.array([[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]])
:param c: point to check like --> np.array([731362.47087528, 9746708.78767337])
:param epsilon:
:return: check if points is on line or not netween point a and b
"""
a, b = line_ab
a_x, a_y = a
b_x, b_y = b
c_x, c_y = 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
@numba.njit('float64[:,::1], float64[:,::1]', fastmath=True)
def nb_get_line_intersect(line_ab, line_cd):
"""
:param line_ab: like --> np.array([[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]])
:param line_cd: like --> np.array([[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]])
:return: get point of intersection, if the points in on line ab or cd returns the point if not retunrs 0
"""
A, B = line_ab
C, D = line_cd
# a1x + b1y = c1
a1 = B[1] - A[1]
b1 = A[0] - B[0]
c1 = a1 * (A[0]) + b1 * (A[1])
# a2x + b2y = c2
a2 = D[1] - C[1]
b2 = C[0] - D[0]
c2 = a2 * (C[0]) + b2 * (C[1])
# determinant
det = a1 * b2 - a2 * b1
# parallel line
if det == 0:
return np.array([np.nan, np.nan])
# intersect point(x,y)
x = ((b2 * c1) - (b1 * c2)) / det
y = ((a1 * c2) - (a2 * c1)) / det
#check if x and y area in the line segment interval
if nb_isBetween(line_ab, np.array([x, y]), epsilon=0.001) and nb_isBetween(line_cd, np.array([x, y]), epsilon=0.001):
return np.array([x, y])
else:
return np.array([np.nan, np.nan])
@numba.njit('float64[:, :, ::1], float64[:, :, ::1]', parallel=True, fastmath=True)
def nb_get_line_intersect(m_ramales_lines, interference_lines):
"""
:param m_ramales_lines: like --> np.array([[[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]] , [[731297.282, 9746727.286], [ 731290.048, 9746724.403]]])
:param interference_lines: like --> np.array([[[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]] , [[731297.282, 9746727.286], [ 731290.048, 9746724.403]]])
:return: m_ramales_lines x interference_lines x 2
"""
#empty matrix to fill
m_ramales_interference = np.empty(shape=(len(m_ramales_lines), len(interference_lines), 2))
for pos1 in range(len(m_ramales_lines)):
line_ab = m_ramales_lines[pos1]
for pos2 in numba.prange(len(interference_lines)):
# interference line
line_cd = interference_lines[pos2].T
# get crossing point
cross_point = nb_get_line_intersect(line_ab.copy(), line_cd.copy())
#fill 2D array
m_ramales_interference[pos1, pos2] = cross_point
return m_ramales_interference
How can I check if 2 segments intersect?
I’ve the following data:
Segment1 [ {x1,y1}, {x2,y2} ]
Segment2 [ {x1,y1}, {x2,y2} ]
I need to write a small algorithm in Python to detect if the 2 lines are intersecting.
The equation of a line is:
f(x) = A*x + b = y
For a segment, it is exactly the same, except that x is included on an interval I.
If you have two segments, defined as follow:
Segment1 = {(X1, Y1), (X2, Y2)}
Segment2 = {(X3, Y3), (X4, Y4)}
The abcisse Xa of the potential point of intersection (Xa,Ya) must be contained in both interval I1 and I2, defined as follow :
I1 = [min(X1,X2), max(X1,X2)]
I2 = [min(X3,X4), max(X3,X4)]
And we could say that Xa is included into :
Ia = [max( min(X1,X2), min(X3,X4) ),
min( max(X1,X2), max(X3,X4) )]
Now, we need to check that this interval Ia exists :
if (max(X1,X2) < min(X3,X4)):
return False # There is no mutual abcisses
So, we have two line formula, and a mutual interval. Your line formulas are:
f1(x) = A1*x + b1 = y
f2(x) = A2*x + b2 = y
As we got two points by segment, we are able to determine A1, A2, b1 and b2:
A1 = (Y1-Y2)/(X1-X2) # Pay attention to not dividing by zero
A2 = (Y3-Y4)/(X3-X4) # Pay attention to not dividing by zero
b1 = Y1-A1*X1 = Y2-A1*X2
b2 = Y3-A2*X3 = Y4-A2*X4
If the segments are parallel, then A1 == A2 :
if (A1 == A2):
return False # Parallel segments
A point (Xa,Ya) standing on both line must verify both formulas f1 and f2:
Ya = A1 * Xa + b1
Ya = A2 * Xa + b2
A1 * Xa + b1 = A2 * Xa + b2
Xa = (b2 - b1) / (A1 - A2) # Once again, pay attention to not dividing by zero
The last thing to do is check that Xa is included into Ia:
if ( (Xa < max( min(X1,X2), min(X3,X4) )) or
(Xa > min( max(X1,X2), max(X3,X4) )) ):
return False # intersection is out of bound
else:
return True
In addition to this, you may check at startup that two of the four provided points are not equals to avoid all that testing.
if your data define line you just have to prove that they are not parallel. To do this you can compute
alpha = float(y2 - y1) / (x2 - x1).
If this coefficient is equal for both Line1 and Line2, it means the line are parallel. If not, it means they will intersect.
If they are parallel you then have to prove that they are not the same. For that, you compute
beta = y1 - alpha*x1
If beta is the same for Line1 and Line2,it means you line intersect as they are equal
If they are segment, you still have to compute alpha and beta as described above for each Line. Then you have to check that (beta1 – beta2) / (alpha1 – alpha2) is greater than Min(x1_line1, x2_line1) and less than Max(x1_line1, x2_line1)
Calculate the intersection point of the lines laying on your segments (it means basically to solve a linear equation system), then check whether is it between the starting and ending points of your segments.
You have two line segments. Define one segment by endpoints A & B and the second segment by endpoints C & D. There is a nice trick to show that they must intersect, WITHIN the bounds of the segments. (Note that the lines themselves may intersect beyond the bounds of the segments, so you must be careful. Good code will also watch for parallel lines.)
The trick is to test that points A and B must line on opposite sides of line CD, AND that points C and D must lie on opposite sides of line AB.
Since this is homework, I won’t give you an explicit solution. But a simple test to see which side of a line a point falls on, is to use a dot product. Thus, for a given line CD, compute the normal vector to that line (I’ll call it N_C.) Now, simply test the signs of these two results:
dot(A-C,N_C)
and
dot(B-C,N_C)
If those results have opposite signs, then A and B are opposite sides of line CD. Now do the same test for the other line, AB. It has normal vector N_A. Compare the signs of
dot(C-A,N_A)
and
dot(D-A,N_A)
I’ll leave it to you to figure out how to compute a normal vector. (In 2-d, that is trivial, but will your code worry about whether A and B are distinct points? Likewise, are C and D distinct?)
You still need to worry about line segments that lie along the same infinite line, or if one point actually falls on the other line segment itself. Good code will cater to every possible problem.
Suppose the two segments have endpoints A,B and C,D. The numerically robust way to determine intersection is to check the sign of the four determinants:
| Ax-Cx Bx-Cx | | Ax-Dx Bx-Dx |
| Ay-Cy By-Cy | | Ay-Dy By-Dy |
| Cx-Ax Dx-Ax | | Cx-Bx Dx-Bx |
| Cy-Ay Dy-Ay | | Cy-By Dy-By |
For intersection, each determinant on the left must have the opposite sign of the one to the right, but there need not be any relationship between the two lines. You are basically checking each point of a segment against the other segment to make sure they lie on opposite sides of the line defined by the other segment.
See here: http://www.cs.cmu.edu/~quake/robust.html
You don’t have to compute exactly where does the segments intersect, but only understand whether they intersect at all. This will simplify the solution.
The idea is to treat one segment as the “anchor” and separate the second segment into 2 points.
Now, you will have to find the relative position of each point to the “anchored” segment (OnLeft, OnRight or Collinear).
After doing so for both points, check that one of the points is OnLeft and the other is OnRight (or perhaps include Collinear position, if you wish to include improper intersections as well).
You must then repeat the process with the roles of anchor and separated segments.
An intersection exists if, and only if, one of the points is OnLeft and the other is OnRight. See this link for a more detailed explanation with example images for each possible case.
Implementing such method will be much easier than actually implementing a method that finds the intersection point (given the many corner cases which you will have to handle as well).
Update
The following functions should illustrate the idea (source: Computational Geometry in C).
Remark: This sample assumes the usage of integers. If you’re using some floating-point representation instead (which could obviously complicate things), then you should determine some epsilon value to indicate “equality” (mostly for the IsCollinear evaluation).
// points "a" and "b" forms the anchored segment.
// point "c" is the evaluated point
bool IsOnLeft(Point a, Point b, Point c)
{
return Area2(a, b, c) > 0;
}
bool IsOnRight(Point a, Point b, Point c)
{
return Area2(a, b, c) < 0;
}
bool IsCollinear(Point a, Point b, Point c)
{
return Area2(a, b, c) == 0;
}
// calculates the triangle's size (formed by the "anchor" segment and additional point)
int Area2(Point a, Point b, Point c)
{
return (b.X - a.X) * (c.Y - a.Y) -
(c.X - a.X) * (b.Y - a.Y);
}
Of course, when using these functions, one must remember to check that each segment lies “between” the other segment (since these are finite segments, and not infinite lines).
Also, using these functions you can understand whether you’ve got a proper or improper intersection.
- Proper: There are no collinear points. The segments crosses each
other “from side to side”. - Improper: One segment only “touches” the other (at least one of
the points is collinear to the
anchored segment).
This is what I’ve got for AS3, don’t know much about python but the concept is there
public function getIntersectingPointF($A:Point, $B:Point, $C:Point, $D:Point):Number {
var A:Point = $A.clone();
var B:Point = $B.clone();
var C:Point = $C.clone();
var D:Point = $D.clone();
var f_ab:Number = (D.x - C.x) * (A.y - C.y) - (D.y - C.y) * (A.x - C.x);
// are lines parallel
if (f_ab == 0) { return Infinity };
var f_cd:Number = (B.x - A.x) * (A.y - C.y) - (B.y - A.y) * (A.x - C.x);
var f_d:Number = (D.y - C.y) * (B.x - A.x) - (D.x - C.x) * (B.y - A.y);
var f1:Number = f_ab/f_d
var f2:Number = f_cd / f_d
if (f1 == Infinity || f1 <= 0 || f1 >= 1) { return Infinity };
if (f2 == Infinity || f2 <= 0 || f2 >= 1) { return Infinity };
return f1;
}
public function getIntersectingPoint($A:Point, $B:Point, $C:Point, $D:Point):Point
{
var f:Number = getIntersectingPointF($A, $B, $C, $D);
if (f == Infinity || f <= 0 || f >= 1) { return null };
var retPoint:Point = Point.interpolate($A, $B, 1 - f);
return retPoint.clone();
}
for segments AB and CD, find the slope of CD
slope=(Dy-Cy)/(Dx-Cx)
extend CD over A and B, and take the distance to CD going straight up
dist1=slope*(Cx-Ax)+Ay-Cy
dist2=slope*(Dx-Ax)+Ay-Dy
check if they are on opposite sides
return dist1*dist2<0
User @i_4_got points to this page with a very efficent solution in Python. I reproduce it here for convenience (since it would have made me happy to have it here):
def ccw(A,B,C):
return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)
# Return true if line segments AB and CD intersect
def intersect(A,B,C,D):
return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)
Resolved but still why not with python… 🙂
def islineintersect(line1, line2):
i1 = [min(line1[0][0], line1[1][0]), max(line1[0][0], line1[1][0])]
i2 = [min(line2[0][0], line2[1][0]), max(line2[0][0], line2[1][0])]
ia = [max(i1[0], i2[0]), min(i1[1], i2[1])]
if max(line1[0][0], line1[1][0]) < min(line2[0][0], line2[1][0]):
return False
m1 = (line1[1][1] - line1[0][1]) * 1. / (line1[1][0] - line1[0][0]) * 1.
m2 = (line2[1][1] - line2[0][1]) * 1. / (line2[1][0] - line2[0][0]) * 1.
if m1 == m2:
return False
b1 = line1[0][1] - m1 * line1[0][0]
b2 = line2[0][1] - m2 * line2[0][0]
x1 = (b2 - b1) / (m1 - m2)
if (x1 < max(i1[0], i2[0])) or (x1 > min(i1[1], i2[1])):
return False
return True
This:
print islineintersect([(15, 20), (100, 200)], [(210, 5), (23, 119)])
Output:
True
And this:
print islineintersect([(15, 20), (100, 200)], [(-1, -5), (-5, -5)])
Output:
False
Implemented in JAVA. However It seems that it does not work for co-linear lines (aka line segments that exist within each other L1(0,0)(10,10) L2(1,1)(2,2)
public class TestCode
{
public class Point
{
public double x = 0;
public double y = 0;
public Point(){}
}
public class Line
{
public Point p1, p2;
public Line( double x1, double y1, double x2, double y2)
{
p1 = new Point();
p2 = new Point();
p1.x = x1;
p1.y = y1;
p2.x = x2;
p2.y = y2;
}
}
//line segments
private static Line s1;
private static Line s2;
public TestCode()
{
s1 = new Line(0,0,0,10);
s2 = new Line(-1,0,0,10);
}
public TestCode(double x1, double y1,
double x2, double y2,
double x3, double y3,
double x4, double y4)
{
s1 = new Line(x1,y1, x2,y2);
s2 = new Line(x3,y3, x4,y4);
}
public static void main(String args[])
{
TestCode code = null;
////////////////////////////
code = new TestCode(0,0,0,10,
0,1,0,5);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,0,10,
0,1,0,10);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,10,0,
5,0,15,0);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,10,0,
0,0,15,0);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,10,10,
1,1,5,5);
if( intersect(code) )
{ System.out.println( "OK COLINEAR: INTERSECTS" ); }
else
{ System.out.println( "ERROR COLINEAR: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,0,10,
-1,-1,0,10);
if( intersect(code) )
{ System.out.println( "OK SLOPE END: INTERSECTS" ); }
else
{ System.out.println( "ERROR SLOPE END: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(-10,-10,10,10,
-10,10,10,-10);
if( intersect(code) )
{ System.out.println( "OK SLOPE Intersect(0,0): INTERSECTS" ); }
else
{ System.out.println( "ERROR SLOPE Intersect(0,0): DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(-10,-10,10,10,
-3,-2,50,-2);
if( intersect(code) )
{ System.out.println( "OK SLOPE Line2 VERTIAL: INTERSECTS" ); }
else
{ System.out.println( "ERROR SLOPE Line2 VERTICAL: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(-10,-10,10,10,
50,-2,-3,-2);
if( intersect(code) )
{ System.out.println( "OK SLOPE Line2 (reversed) VERTIAL: INTERSECTS" ); }
else
{ System.out.println( "ERROR SLOPE Line2 (reversed) VERTICAL: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,0,10,
1,0,1,10);
if( intersect(code) )
{ System.out.println( "ERROR PARALLEL VERTICAL: INTERSECTS" ); }
else
{ System.out.println( "OK PARALLEL VERTICAL: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,2,10,2,
0,10,10,10);
if( intersect(code) )
{ System.out.println( "ERROR PARALLEL HORIZONTAL: INTERSECTS" ); }
else
{ System.out.println( "OK PARALLEL HORIZONTAL: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,10,5,13.75,
0,18.75,10,15);
if( intersect(code) )
{ System.out.println( "ERROR PARALLEL SLOPE=.75: INTERSECTS" ); }
else
{ System.out.println( "OK PARALLEL SLOPE=.75: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,1,1,
2,-1,2,10);
if( intersect(code) )
{ System.out.println( "ERROR SEPERATE SEGMENTS: INTERSECTS" ); }
else
{ System.out.println( "OK SEPERATE SEGMENTS: DO NOT INTERSECT" ); }
////////////////////////////
code = new TestCode(0,0,1,1,
-1,-10,-5,10);
if( intersect(code) )
{ System.out.println( "ERROR SEPERATE SEGMENTS 2: INTERSECTS" ); }
else
{ System.out.println( "OK SEPERATE SEGMENTS 2: DO NOT INTERSECT" ); }
}
public static boolean intersect( TestCode code )
{
return intersect( code.s1, code.s2);
}
public static boolean intersect( Line line1, Line line2 )
{
double i1min = Math.min(line1.p1.x, line1.p2.x);
double i1max = Math.max(line1.p1.x, line1.p2.x);
double i2min = Math.min(line2.p1.x, line2.p2.x);
double i2max = Math.max(line2.p1.x, line2.p2.x);
double iamax = Math.max(i1min, i2min);
double iamin = Math.min(i1max, i2max);
if( Math.max(line1.p1.x, line1.p2.x) < Math.min(line2.p1.x, line2.p2.x) )
return false;
double m1 = (line1.p2.y - line1.p1.y) / (line1.p2.x - line1.p1.x );
double m2 = (line2.p2.y - line2.p1.y) / (line2.p2.x - line2.p1.x );
if( m1 == m2 )
return false;
//b1 = line1[0][1] - m1 * line1[0][0]
//b2 = line2[0][1] - m2 * line2[0][0]
double b1 = line1.p1.y - m1 * line1.p1.x;
double b2 = line2.p1.y - m2 * line2.p1.x;
double x1 = (b2 - b1) / (m1 - m2);
if( (x1 < Math.max(i1min, i2min)) || (x1 > Math.min(i1max, i2max)) )
return false;
return true;
}
}
Output thus far is
ERROR COLINEAR: DO NOT INTERSECT
ERROR COLINEAR: DO NOT INTERSECT
ERROR COLINEAR: DO NOT INTERSECT
ERROR COLINEAR: DO NOT INTERSECT
ERROR COLINEAR: DO NOT INTERSECT
OK SLOPE END: INTERSECTS
OK SLOPE Intersect(0,0): INTERSECTS
OK SLOPE Line2 VERTIAL: INTERSECTS
OK SLOPE Line2 (reversed) VERTIAL: INTERSECTS
OK PARALLEL VERTICAL: DO NOT INTERSECT
OK PARALLEL HORIZONTAL: DO NOT INTERSECT
OK PARALLEL SLOPE=.75: DO NOT INTERSECT
OK SEPERATE SEGMENTS: DO NOT INTERSECT
OK SEPERATE SEGMENTS 2: DO NOT INTERSECT
Here is C code to check if two points are on the opposite sides of the line segment. Using this code you can check if two segments intersect as well.
// true if points p1, p2 lie on the opposite sides of segment s1--s2
bool oppositeSide (Point2f s1, Point2f s2, Point2f p1, Point2f p2) {
//calculate normal to the segment
Point2f vec = s1-s2;
Point2f normal(vec.y, -vec.x); // no need to normalize
// vectors to the points
Point2f v1 = p1-s1;
Point2f v2 = p2-s1;
// compare signs of the projections of v1, v2 onto the normal
float proj1 = v1.dot(normal);
float proj2 = v2.dot(normal);
if (proj1==0 || proj2==0)
cout<<"collinear points"<<endl;
return(SIGN(proj1) != SIGN(proj2));
}
Based on Liran’s and Grumdrig’s excellent answers here is a complete Python code to verify if closed segments do intersect. Works for collinear segments, segments parallel to axis Y, degenerate segments (devil is in details). Assumes integer coordinates. Floating point coordinates require a modification to points equality test.
def side(a,b,c):
""" Returns a position of the point c relative to the line going through a and b
Points a, b are expected to be different
"""
d = (c[1]-a[1])*(b[0]-a[0]) - (b[1]-a[1])*(c[0]-a[0])
return 1 if d > 0 else (-1 if d < 0 else 0)
def is_point_in_closed_segment(a, b, c):
""" Returns True if c is inside closed segment, False otherwise.
a, b, c are expected to be collinear
"""
if a[0] < b[0]:
return a[0] <= c[0] and c[0] <= b[0]
if b[0] < a[0]:
return b[0] <= c[0] and c[0] <= a[0]
if a[1] < b[1]:
return a[1] <= c[1] and c[1] <= b[1]
if b[1] < a[1]:
return b[1] <= c[1] and c[1] <= a[1]
return a[0] == c[0] and a[1] == c[1]
#
def closed_segment_intersect(a,b,c,d):
""" Verifies if closed segments a, b, c, d do intersect.
"""
if a == b:
return a == c or a == d
if c == d:
return c == a or c == b
s1 = side(a,b,c)
s2 = side(a,b,d)
# All points are collinear
if s1 == 0 and s2 == 0:
return
is_point_in_closed_segment(a, b, c) or is_point_in_closed_segment(a, b, d) or
is_point_in_closed_segment(c, d, a) or is_point_in_closed_segment(c, d, b)
# No touching and on the same side
if s1 and s1 == s2:
return False
s1 = side(c,d,a)
s2 = side(c,d,b)
# No touching and on the same side
if s1 and s1 == s2:
return False
return True
Since you do not mention that you want to find the intersection point of the line, the problem becomes simpler to solve. If you need the intersection point, then the answer by OMG_peanuts is a faster approach. However, if you just want to find whether the lines intersect or not, you can do so by using the line equation (ax + by + c = 0). The approach is as follows:
-
Let’s start with two line segments: segment 1 and segment 2.
segment1 = [[x1,y1], [x2,y2]] segment2 = [[x3,y3], [x4,y4]] -
Check if the two line segments are non zero length line and distinct segments.
-
From hereon, I assume that the two segments are non-zero length and distinct. For each line segment, compute the slope of the line and then obtain the equation of a line in the form of ax + by + c = 0. Now, compute the value of f = ax + by + c for the two points of the other line segment (repeat this for the other line segment as well).
a2 = (y3-y4)/(x3-x4); b1 = -1; b2 = -1; c1 = y1 - a1*x1; c2 = y3 - a2*x3; // using the sign function from numpy f1_1 = sign(a1*x3 + b1*y3 + c1); f1_2 = sign(a1*x4 + b1*y4 + c1); f2_1 = sign(a2*x1 + b2*y1 + c2); f2_2 = sign(a2*x2 + b2*y2 + c2); -
Now all that is left is the different cases. If f = 0 for any point, then the two lines touch at a point. If f1_1 and f1_2 are equal or f2_1 and f2_2 are equal, then the lines do not intersect. If f1_1 and f1_2 are unequal and f2_1 and f2_2 are unequal, then the line segments intersect. Depending on whether you want to consider the lines which touch as “intersecting” or not, you can adapt your conditions.
Here is another python code to check whether closed segments intersect. It is the rewritten version of the C++ code in http://www.cdn.geeksforgeeks.org/check-if-two-given-line-segments-intersect/. This implementation covers all special cases (e.g. all points colinear).
def on_segment(p, q, r):
'''Given three colinear points p, q, r, the function checks if
point q lies on line segment "pr"
'''
if (q[0] <= max(p[0], r[0]) and q[0] >= min(p[0], r[0]) and
q[1] <= max(p[1], r[1]) and q[1] >= min(p[1], r[1])):
return True
return False
def orientation(p, q, r):
'''Find orientation of ordered triplet (p, q, r).
The function returns following values
0 --> p, q and r are colinear
1 --> Clockwise
2 --> Counterclockwise
'''
val = ((q[1] - p[1]) * (r[0] - q[0]) -
(q[0] - p[0]) * (r[1] - q[1]))
if val == 0:
return 0 # colinear
elif val > 0:
return 1 # clockwise
else:
return 2 # counter-clockwise
def do_intersect(p1, q1, p2, q2):
'''Main function to check whether the closed line segments p1 - q1 and p2
- q2 intersect'''
o1 = orientation(p1, q1, p2)
o2 = orientation(p1, q1, q2)
o3 = orientation(p2, q2, p1)
o4 = orientation(p2, q2, q1)
# General case
if (o1 != o2 and o3 != o4):
return True
# Special Cases
# p1, q1 and p2 are colinear and p2 lies on segment p1q1
if (o1 == 0 and on_segment(p1, p2, q1)):
return True
# p1, q1 and p2 are colinear and q2 lies on segment p1q1
if (o2 == 0 and on_segment(p1, q2, q1)):
return True
# p2, q2 and p1 are colinear and p1 lies on segment p2q2
if (o3 == 0 and on_segment(p2, p1, q2)):
return True
# p2, q2 and q1 are colinear and q1 lies on segment p2q2
if (o4 == 0 and on_segment(p2, q1, q2)):
return True
return False # Doesn't fall in any of the above cases
Below is a test function to verify that it works.
import matplotlib.pyplot as plt
def test_intersect_func():
p1 = (1, 1)
q1 = (10, 1)
p2 = (1, 2)
q2 = (10, 2)
fig, ax = plt.subplots()
ax.plot([p1[0], q1[0]], [p1[1], q1[1]], 'x-')
ax.plot([p2[0], q2[0]], [p2[1], q2[1]], 'x-')
print(do_intersect(p1, q1, p2, q2))
p1 = (10, 0)
q1 = (0, 10)
p2 = (0, 0)
q2 = (10, 10)
fig, ax = plt.subplots()
ax.plot([p1[0], q1[0]], [p1[1], q1[1]], 'x-')
ax.plot([p2[0], q2[0]], [p2[1], q2[1]], 'x-')
print(do_intersect(p1, q1, p2, q2))
p1 = (-5, -5)
q1 = (0, 0)
p2 = (1, 1)
q2 = (10, 10)
fig, ax = plt.subplots()
ax.plot([p1[0], q1[0]], [p1[1], q1[1]], 'x-')
ax.plot([p2[0], q2[0]], [p2[1], q2[1]], 'x-')
print(do_intersect(p1, q1, p2, q2))
p1 = (0, 0)
q1 = (1, 1)
p2 = (1, 1)
q2 = (10, 10)
fig, ax = plt.subplots()
ax.plot([p1[0], q1[0]], [p1[1], q1[1]], 'x-')
ax.plot([p2[0], q2[0]], [p2[1], q2[1]], 'x-')
print(do_intersect(p1, q1, p2, q2))
I thought I’d contribute a nice Swift solution:
struct Pt {
var x: Double
var y: Double
}
struct LineSegment {
var p1: Pt
var p2: Pt
}
func doLineSegmentsIntersect(ls1: LineSegment, ls2: LineSegment) -> Bool {
if (ls1.p2.x-ls1.p1.x == 0) { //handle vertical segment1
if (ls2.p2.x-ls2.p1.x == 0) {
//both lines are vertical and parallel
return false
}
let x = ls1.p1.x
let slope2 = (ls2.p2.y-ls2.p1.y)/(ls2.p2.x-ls2.p1.x)
let c2 = ls2.p1.y-slope2*ls2.p1.x
let y = x*slope2+c2 // y intersection point
return (y > ls1.p1.y && x < ls1.p2.y) || (y > ls1.p2.y && y < ls1.p1.y) // check if y is between y1,y2 in segment1
}
if (ls2.p2.x-ls2.p1.x == 0) { //handle vertical segment2
let x = ls2.p1.x
let slope1 = (ls1.p2.y-ls1.p1.y)/(ls1.p2.x-ls1.p1.x)
let c1 = ls1.p1.y-slope1*ls1.p1.x
let y = x*slope1+c1 // y intersection point
return (y > ls2.p1.y && x < ls2.p2.y) || (y > ls2.p2.y && y < ls2.p1.y) // validate that y is between y1,y2 in segment2
}
let slope1 = (ls1.p2.y-ls1.p1.y)/(ls1.p2.x-ls1.p1.x)
let slope2 = (ls2.p2.y-ls2.p1.y)/(ls2.p2.x-ls2.p1.x)
if (slope1 == slope2) { //segments are parallel
return false
}
let c1 = ls1.p1.y-slope1*ls1.p1.x
let c2 = ls2.p1.y-slope2*ls2.p1.x
let x = (c2-c1)/(slope1-slope2)
return (((x > ls1.p1.x && x < ls1.p2.x) || (x > ls1.p2.x && x < ls1.p1.x)) &&
((x > ls2.p1.x && x < ls2.p2.x) || (x > ls2.p2.x && x < ls2.p1.x)))
//validate that x is between x1,x2 in both segments
}
We can also solve this utilizing vectors.
Let’s define the segments as [start, end]. Given two such segments [A, B] and [C, D] that both have non-zero length, we can choose one of the endpoints to be used as a reference point so that we get three vectors:
x = 0
y = 1
p = A-C = [C[x]-A[x], C[y]-A[y]]
q = B-A = [B[x]-A[x], B[y]-A[y]]
r = D-C = [D[x]-C[x], D[y]-C[y]]
From there, we can look for an intersection by calculating t and u in p + t*r = u*q. After playing around with the equation a little, we get:
t = (q[y]*p[x] - q[x]*p[y])/(q[x]*r[y] - q[y]*r[x])
u = (p[x] + t*r[x])/q[x]
Thus, the function is:
def intersects(a, b):
p = [b[0][0]-a[0][0], b[0][1]-a[0][1]]
q = [a[1][0]-a[0][0], a[1][1]-a[0][1]]
r = [b[1][0]-b[0][0], b[1][1]-b[0][1]]
t = (q[1]*p[0] - q[0]*p[1])/(q[0]*r[1] - q[1]*r[0])
if (q[0]*r[1] - q[1]*r[0]) != 0
else (q[1]*p[0] - q[0]*p[1])
u = (p[0] + t*r[0])/q[0]
if q[0] != 0
else (p[1] + t*r[1])/q[1]
return t >= 0 and t <= 1 and u >= 0 and u <= 1
This is my way of checking for line crossing and where the intersection occurs. Lets use x1 through x4 and y1 through y4
Segment1 = ((X1, Y1), (X2, Y2))
Segment2 = ((X3, Y3), (X4, Y4))
Then we need some vectors to represent them
dx1 = X2 - X1
dx2 = X4 - X3
dy1 = Y2 - Y1
dy2 = Y4 - Y3
Now we look at the determinant
det = dx1 * dy2 - dx2 * dy1
If the determinant is 0.0, then the line segments are parallel. This could mean they overlap. If they overlap just at endpoints, then there is one intersection solution. Otherwise there will be infinite solutions. With infinitely many solutions, what do say is your point of intersection? So it’s an interesting special case. If you know ahead of time that the lines can’t overlap then you can just check if det == 0.0 and if so just say they don’t intersect and be done. Otherwise, lets continue on
dx3 = X1 - X3
dy3 = Y1 - Y3
det1 = dx1 * dy3 - dx3 * dy1
det2 = dx2 * dy3 - dx3 * dy2
Now, if det, det1 and det2 are all zero, then your lines are co-linear and could overlap. If det is zero but either det1 or det2 are not, then they are not co-linear, but are parallel, so there is no intersection. So what’s left now if det is zero is a 1D problem instead of 2D. We will need to check one of two ways, depending if dx1 is zero or not (so we can avoid division by zero). If dx1 is zero then just do the same logic with y values rather than x below.
s = X3 / dx1
t = X4 / dx1
This computes two scalers, such that if we scale the vector (dx1, dy1) by s we get point (x3, y3), and by t we get (x4, y4). So if either s or t is between 0.0 and 1.0, then point 3 or 4 lies on our first line. Negative would mean the point is behind the start of our vector, while > 1.0 means it is further ahead of the end of our vector. 0.0 means it is at (x1, y1) and 1.0 means it is at (x2, y2). If both s and t are < 0.0 or both are > 1.0, then they don’t intersect. And that handles the parallel lines special case.
Now, if det != 0.0 then
s = det1 / det
t = det2 / det
if s < 0.0 or s > 1.0 or t < 0.0 or t > 1.0:
return false # no intersect
This is similar to what we were doing above really. Now if we pass the above test, then our line segments intersect, and we can calculate the intersection quite easily like so:
Ix = X1 + t * dx1
Iy = Y1 + t * dy1
If you want to dig deeper into what the math is doing, look into Cramer’s Rule.
Edit: I’ve fixed two errors that were present so it should be correct now. I’ve learned enough python now to write actual code for this now. It works except for some special cases, though it does handle some special cases correctly. Special cases get really hard to deal with and I’ve spend enough time on it and want to move on. If someone requires better then they have a good starting point at least to try and improve it.
import math
def line_intersection(line1, line2):
x1, x2, x3, x4 = line1[0][0], line1[1][0], line2[0][0], line2[1][0]
y1, y2, y3, y4 = line1[0][1], line1[1][1], line2[0][1], line2[1][1]
dx1 = x2 - x1
dx2 = x4 - x3
dy1 = y2 - y1
dy2 = y4 - y3
dx3 = x1 - x3
dy3 = y1 - y3
det = dx1 * dy2 - dx2 * dy1
det1 = dx1 * dy3 - dx3 * dy1
det2 = dx2 * dy3 - dx3 * dy2
if det == 0.0: # lines are parallel
if det1 != 0.0 or det2 != 0.0: # lines are not co-linear
return None # so no solution
if dx1:
if x1 < x3 < x2 or x1 > x3 > x2:
return math.inf # infinitely many solutions
else:
if y1 < y3 < y2 or y1 > y3 > y2:
return math.inf # infinitely many solutions
if line1[0] == line2[0] or line1[1] == line2[0]:
return line2[0]
elif line1[0] == line2[1] or line1[1] == line2[1]:
return line2[1]
return None # no intersection
s = det1 / det
t = det2 / det
if 0.0 < s < 1.0 and 0.0 < t < 1.0:
return x1 + t * dx1, y1 + t * dy1
print("one intersection")
print(line_intersection(((0.0,0.0), (6.0,6.0)),((0.0,9.0), (9.0,0.0))))
print(line_intersection(((-2, -2), (2, 2)), ((2, -2), (-2, 2))))
print(line_intersection(((0.5, 0.5), (1.5, 0.5)), ((1.0, 0.0), (1.0, 2.0))))
print(line_intersection(((0, -1), (0, 0)), ((0, 0), (0, 1))))
print(line_intersection(((-1, 0), (0, 0)), ((0, 0), (1, 0))))
print()
print("no intersection")
print(line_intersection(((-1, -1), (0, 0)), ((2, -4), (2, 4))))
print(line_intersection(((0.0,0.0), (0.0,9.0)),((9.0,0.0), (9.0,99.0))))
print(line_intersection(((0, 0), (1, 1)), ((1, 0), (2, 1))))
print(line_intersection(((-1, 1), (0, 1)), ((0, 0), (1, 0))))
print(line_intersection(((1, -1), (1, 0)), ((0, 0), (0, -1))))
print()
print("infinite intersection")
print(line_intersection(((-1, -1), (1, 1)), ((0, 0), (2, 2))))
print(line_intersection(((-1, 0), (1, 0)), ((0, 0), (2, 0))))
print(line_intersection(((0, -1), (0, 1)), ((0, 0), (0, 2))))
print(line_intersection(((-1, 0), (0, 0)), ((0, 0), (-1, 0))))
print(line_intersection(((1, 0), (0, 0)), ((0, 0), (1, 0))))
Checking if line segments intersect is very easy with Shapely library using intersects method:
from shapely.geometry import LineString
line = LineString([(0, 0), (1, 1)])
other = LineString([(0, 1), (1, 0)])
print(line.intersects(other))
# True
line = LineString([(0, 0), (1, 1)])
other = LineString([(0, 1), (1, 2)])
print(line.intersects(other))
# False
One of the solutions above worked so well I decided to write a complete demonstration program using wxPython. You should be able to run this program like this: python “your file name“
# Click on the window to draw a line.
# The program will tell you if this and the other line intersect.
import wx
class Point:
def __init__(self, newX, newY):
self.x = newX
self.y = newY
app = wx.App()
frame = wx.Frame(None, wx.ID_ANY, "Main")
p1 = Point(90,200)
p2 = Point(150,80)
mp = Point(0,0) # mouse point
highestX = 0
def ccw(A,B,C):
return (C.y-A.y) * (B.x-A.x) > (B.y-A.y) * (C.x-A.x)
# Return true if line segments AB and CD intersect
def intersect(A,B,C,D):
return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)
def is_intersection(p1, p2, p3, p4):
return intersect(p1, p2, p3, p4)
def drawIntersection(pc):
mp2 = Point(highestX, mp.y)
if is_intersection(p1, p2, mp, mp2):
pc.DrawText("intersection", 10, 10)
else:
pc.DrawText("no intersection", 10, 10)
def do_paint(evt):
pc = wx.PaintDC(frame)
pc.DrawLine(p1.x, p1.y, p2.x, p2.y)
pc.DrawLine(mp.x, mp.y, highestX, mp.y)
drawIntersection(pc)
def do_left_mouse(evt):
global mp, highestX
point = evt.GetPosition()
mp = Point(point[0], point[1])
highestX = frame.Size[0]
frame.Refresh()
frame.Bind(wx.EVT_PAINT, do_paint)
frame.Bind(wx.EVT_LEFT_DOWN, do_left_mouse)
frame.Show()
app.MainLoop()
The answer by Georgy is the cleanest to implement, by far. Had to chase this down, since the brycboe example, while simple as well, had issues with colinearity.
Code for testing:
#!/usr/bin/python
#
# Notes on intersection:
#
# https://bryceboe.com/2006/10/23/line-segment-intersection-algorithm/
#
# https://stackoverflow.com/questions/3838329/how-can-i-check-if-two-segments-intersect
from shapely.geometry import LineString
class Point:
def __init__(self,x,y):
self.x = x
self.y = y
def ccw(A,B,C):
return (C.y-A.y)*(B.x-A.x) > (B.y-A.y)*(C.x-A.x)
def intersect(A,B,C,D):
return ccw(A,C,D) != ccw(B,C,D) and ccw(A,B,C) != ccw(A,B,D)
def ShapelyIntersect(A,B,C,D):
return LineString([(A.x,A.y),(B.x,B.y)]).intersects(LineString([(C.x,C.y),(D.x,D.y)]))
a = Point(0,0)
b = Point(0,1)
c = Point(1,1)
d = Point(1,0)
'''
Test points:
b(0,1) c(1,1)
a(0,0) d(1,0)
'''
# F
print(intersect(a,b,c,d))
# T
print(intersect(a,c,b,d))
print(intersect(b,d,a,c))
print(intersect(d,b,a,c))
# F
print(intersect(a,d,b,c))
# same end point cases:
print("same end points")
# F - not intersected
print(intersect(a,b,a,d))
# T - This shows as intersected
print(intersect(b,a,a,d))
# F - this does not
print(intersect(b,a,d,a))
# F - this does not
print(intersect(a,b,d,a))
print("same end points, using shapely")
# T
print(ShapelyIntersect(a,b,a,d))
# T
print(ShapelyIntersect(b,a,a,d))
# T
print(ShapelyIntersect(b,a,d,a))
# T
print(ShapelyIntersect(a,b,d,a))
Here’s a solution using dot products:
# assumes line segments are stored in the format [(x0,y0),(x1,y1)]
def intersects(s0,s1):
dx0 = s0[1][0]-s0[0][0]
dx1 = s1[1][0]-s1[0][0]
dy0 = s0[1][1]-s0[0][1]
dy1 = s1[1][1]-s1[0][1]
p0 = dy1*(s1[1][0]-s0[0][0]) - dx1*(s1[1][1]-s0[0][1])
p1 = dy1*(s1[1][0]-s0[1][0]) - dx1*(s1[1][1]-s0[1][1])
p2 = dy0*(s0[1][0]-s1[0][0]) - dx0*(s0[1][1]-s1[0][1])
p3 = dy0*(s0[1][0]-s1[1][0]) - dx0*(s0[1][1]-s1[1][1])
return (p0*p1<=0) & (p2*p3<=0)
Here’s a visualization in Desmos: Line Segment Intersection
Using OMG_Peanuts solution, I translated to SQL.
(HANA Scalar Function)
Thanks OMG_Peanuts, it works great.
I am using round earth, but distances are small, so I figure its okay.
FUNCTION GA_INTERSECT" ( IN LAT_A1 DOUBLE,
IN LONG_A1 DOUBLE,
IN LAT_A2 DOUBLE,
IN LONG_A2 DOUBLE,
IN LAT_B1 DOUBLE,
IN LONG_B1 DOUBLE,
IN LAT_B2 DOUBLE,
IN LONG_B2 DOUBLE)
RETURNS RET_DOESINTERSECT DOUBLE
LANGUAGE SQLSCRIPT
SQL SECURITY INVOKER AS
BEGIN
DECLARE MA DOUBLE;
DECLARE MB DOUBLE;
DECLARE BA DOUBLE;
DECLARE BB DOUBLE;
DECLARE XA DOUBLE;
DECLARE MAX_MIN_X DOUBLE;
DECLARE MIN_MAX_X DOUBLE;
DECLARE DOESINTERSECT INTEGER;
SELECT 1 INTO DOESINTERSECT FROM DUMMY;
IF LAT_A2-LAT_A1 != 0 AND LAT_B2-LAT_B1 != 0 THEN
SELECT (LONG_A2 - LONG_A1)/(LAT_A2 - LAT_A1) INTO MA FROM DUMMY;
SELECT (LONG_B2 - LONG_B1)/(LAT_B2 - LAT_B1) INTO MB FROM DUMMY;
IF MA = MB THEN
SELECT 0 INTO DOESINTERSECT FROM DUMMY;
END IF;
END IF;
SELECT LONG_A1-MA*LAT_A1 INTO BA FROM DUMMY;
SELECT LONG_B1-MB*LAT_B1 INTO BB FROM DUMMY;
SELECT (BB - BA) / (MA - MB) INTO XA FROM DUMMY;
-- Max of Mins
IF LAT_A1 < LAT_A2 THEN -- MIN(LAT_A1, LAT_A2) = LAT_A1
IF LAT_B1 < LAT_B2 THEN -- MIN(LAT_B1, LAT_B2) = LAT_B1
IF LAT_A1 > LAT_B1 THEN -- MAX(LAT_A1, LAT_B1) = LAT_A1
SELECT LAT_A1 INTO MAX_MIN_X FROM DUMMY;
ELSE -- MAX(LAT_A1, LAT_B1) = LAT_B1
SELECT LAT_B1 INTO MAX_MIN_X FROM DUMMY;
END IF;
ELSEIF LAT_B2 < LAT_B1 THEN -- MIN(LAT_B1, LAT_B2) = LAT_B2
IF LAT_A1 > LAT_B2 THEN -- MAX(LAT_A1, LAT_B2) = LAT_A1
SELECT LAT_A1 INTO MAX_MIN_X FROM DUMMY;
ELSE -- MAX(LAT_A1, LAT_B2) = LAT_B2
SELECT LAT_B2 INTO MAX_MIN_X FROM DUMMY;
END IF;
END IF;
ELSEIF LAT_A2 < LAT_A1 THEN -- MIN(LAT_A1, LAT_A2) = LAT_A2
IF LAT_B1 < LAT_B2 THEN -- MIN(LAT_B1, LAT_B2) = LAT_B1
IF LAT_A2 > LAT_B1 THEN -- MAX(LAT_A2, LAT_B1) = LAT_A2
SELECT LAT_A2 INTO MAX_MIN_X FROM DUMMY;
ELSE -- MAX(LAT_A2, LAT_B1) = LAT_B1
SELECT LAT_B1 INTO MAX_MIN_X FROM DUMMY;
END IF;
ELSEIF LAT_B2 < LAT_B1 THEN -- MIN(LAT_B1, LAT_B2) = LAT_B2
IF LAT_A2 > LAT_B2 THEN -- MAX(LAT_A2, LAT_B2) = LAT_A2
SELECT LAT_A2 INTO MAX_MIN_X FROM DUMMY;
ELSE -- MAX(LAT_A2, LAT_B2) = LAT_B2
SELECT LAT_B2 INTO MAX_MIN_X FROM DUMMY;
END IF;
END IF;
END IF;
-- Min of Max
IF LAT_A1 > LAT_A2 THEN -- MAX(LAT_A1, LAT_A2) = LAT_A1
IF LAT_B1 > LAT_B2 THEN -- MAX(LAT_B1, LAT_B2) = LAT_B1
IF LAT_A1 < LAT_B1 THEN -- MIN(LAT_A1, LAT_B1) = LAT_A1
SELECT LAT_A1 INTO MIN_MAX_X FROM DUMMY;
ELSE -- MIN(LAT_A1, LAT_B1) = LAT_B1
SELECT LAT_B1 INTO MIN_MAX_X FROM DUMMY;
END IF;
ELSEIF LAT_B2 > LAT_B1 THEN -- MAX(LAT_B1, LAT_B2) = LAT_B2
IF LAT_A1 < LAT_B2 THEN -- MIN(LAT_A1, LAT_B2) = LAT_A1
SELECT LAT_A1 INTO MIN_MAX_X FROM DUMMY;
ELSE -- MIN(LAT_A1, LAT_B2) = LAT_B2
SELECT LAT_B2 INTO MIN_MAX_X FROM DUMMY;
END IF;
END IF;
ELSEIF LAT_A2 > LAT_A1 THEN -- MAX(LAT_A1, LAT_A2) = LAT_A2
IF LAT_B1 > LAT_B2 THEN -- MAX(LAT_B1, LAT_B2) = LAT_B1
IF LAT_A2 < LAT_B1 THEN -- MIN(LAT_A2, LAT_B1) = LAT_A2
SELECT LAT_A2 INTO MIN_MAX_X FROM DUMMY;
ELSE -- MIN(LAT_A2, LAT_B1) = LAT_B1
SELECT LAT_B1 INTO MIN_MAX_X FROM DUMMY;
END IF;
ELSEIF LAT_B2 > LAT_B1 THEN -- MAX(LAT_B1, LAT_B2) = LAT_B2
IF LAT_A2 < LAT_B2 THEN -- MIN(LAT_A2, LAT_B2) = LAT_A2
SELECT LAT_A2 INTO MIN_MAX_X FROM DUMMY;
ELSE -- MIN(LAT_A2, LAT_B2) = LAT_B2
SELECT LAT_B2 INTO MIN_MAX_X FROM DUMMY;
END IF;
END IF;
END IF;
IF XA < MAX_MIN_X OR
XA > MIN_MAX_X THEN
SELECT 0 INTO DOESINTERSECT FROM DUMMY;
END IF;
RET_DOESINTERSECT := :DOESINTERSECT;
END;
Digging up an old thread and modifying Grumdrig’s code …
template <typename T>
struct Point {
T x;
T y;
Point(T x_, T y_) : x(x_), y(y_) {};
};
template <typename T>
inline T CrossProduct(const Point<T>& pt1, const Point<T>& pt2, const Point<T>& pt3)
{
// nb: watch out for overflow
return ((pt2.x - pt1.x) * (pt3.y - pt2.y) - (pt2.y - pt1.y) * (pt3.x - pt2.x));
}
template <typename T>
bool SegmentsIntersect(const Point<T>& a, const Point<T>& b,
const Point<T>& c, const Point<T>& d)
{
return
CrossProduct(a, c, d) * CrossProduct(b, c, d) < 0 &&
CrossProduct(c, a, b) * CrossProduct(d, a, b) < 0;
}
if (SegmentsIntersect(Point<int>(50, 0), Point<int>(50, 100),
Point<int>(0, 50), Point<int>(100, 50)))
std::cout << "it works!" << std::endl;
if anybody wants to do a C-like speed look up for multiple lines intersections you could use this code done in numba and python
Note
the epsilon argument should be proportional to your line distance. Also,
the imports are just import numba and import numpy as np
@numba.njit('float64[:,::1], float64[::1], float64', fastmath=True)
def nb_isBetween(line_ab, c, epsilon):
"""
:param line_ab: like --> np.array([[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]])
:param c: point to check like --> np.array([731362.47087528, 9746708.78767337])
:param epsilon:
:return: check if points is on line or not netween point a and b
"""
a, b = line_ab
a_x, a_y = a
b_x, b_y = b
c_x, c_y = 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
@numba.njit('float64[:,::1], float64[:,::1]', fastmath=True)
def nb_get_line_intersect(line_ab, line_cd):
"""
:param line_ab: like --> np.array([[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]])
:param line_cd: like --> np.array([[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]])
:return: get point of intersection, if the points in on line ab or cd returns the point if not retunrs 0
"""
A, B = line_ab
C, D = line_cd
# a1x + b1y = c1
a1 = B[1] - A[1]
b1 = A[0] - B[0]
c1 = a1 * (A[0]) + b1 * (A[1])
# a2x + b2y = c2
a2 = D[1] - C[1]
b2 = C[0] - D[0]
c2 = a2 * (C[0]) + b2 * (C[1])
# determinant
det = a1 * b2 - a2 * b1
# parallel line
if det == 0:
return np.array([np.nan, np.nan])
# intersect point(x,y)
x = ((b2 * c1) - (b1 * c2)) / det
y = ((a1 * c2) - (a2 * c1)) / det
#check if x and y area in the line segment interval
if nb_isBetween(line_ab, np.array([x, y]), epsilon=0.001) and nb_isBetween(line_cd, np.array([x, y]), epsilon=0.001):
return np.array([x, y])
else:
return np.array([np.nan, np.nan])
@numba.njit('float64[:, :, ::1], float64[:, :, ::1]', parallel=True, fastmath=True)
def nb_get_line_intersect(m_ramales_lines, interference_lines):
"""
:param m_ramales_lines: like --> np.array([[[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]] , [[731297.282, 9746727.286], [ 731290.048, 9746724.403]]])
:param interference_lines: like --> np.array([[[731362.47087528, 9746708.78767337], [731297.282, 9746727.286]] , [[731297.282, 9746727.286], [ 731290.048, 9746724.403]]])
:return: m_ramales_lines x interference_lines x 2
"""
#empty matrix to fill
m_ramales_interference = np.empty(shape=(len(m_ramales_lines), len(interference_lines), 2))
for pos1 in range(len(m_ramales_lines)):
line_ab = m_ramales_lines[pos1]
for pos2 in numba.prange(len(interference_lines)):
# interference line
line_cd = interference_lines[pos2].T
# get crossing point
cross_point = nb_get_line_intersect(line_ab.copy(), line_cd.copy())
#fill 2D array
m_ramales_interference[pos1, pos2] = cross_point
return m_ramales_interference