The answer is
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))
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
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).
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
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.
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 - X4
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 = X3 - X1
dy3 = Y3 - Y1
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 || s > 1.0 || t < 0.0 || 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.
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
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
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();
}
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.
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
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));
}
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)
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()
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;
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
}
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
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.
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'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
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))
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)
Source: Stackoverflow.com