Intersection method of a Line.

[akshayah3]

Gives wrong result in the following case.
a = Ray(Point(0,0),Point(0,4))
b = Ray(Point(0,1),Point(0,-1))
a.intersection(b)
Ray(Point(0,0),Point(0,4))
But the result should be
Segment(Point(0,0),Point(0,1))

[smichr]

The test that identified this should be added to the test suite.

[akshayah3]

@smichr There is no test for this currently, I found the drawback while looking at the code(For rays it checked the xdirection whereas we have to check both x and y directions respectively).I will add some tests in the test suite.

[smichr]

Does the following work? (I suspect that the 2nd test regarding source is not sufficiently general.)

z = Point(0, 0)
for o in (1,4),(-1,4),(-1,-4),(1,-4):
  o=Point(*o)
  p = o/2
  assert Ray(z,o).intersect(Ray(z, p)) == Ray(p,o)
  assert Ray(z,o),intersect(Ray(p,z)) == Segment(z, p)

[akshayah3]

@smichr It is enough because it is not the full code because it just checks for the cases when the rays are similar or when they are parallel.

a1, b1, c1 = self.coefficients
a2, b2, c2 = o.coefficients
t = simplify(a1_b2 - a2_b1)
if t.equals(0) is not False: # assume they are parallel

[smichr]

I think the lines are parallel in every case; you can assert that with Ray(z,o).is_parallel(Ray(z, p) (or something to that effect). So the code should work. Can you modify it to do so?

[akshayah3]

@smichr I think there is a bit of misunderstanding, I already modified the code and it works for the above cases you mentioned.

[smichr]

it works

It seems like you are assuming it work. Please post your results. I get

>>> z = Point(0,0)
>>> for o in (1,4),(-1,4),(-1,-4),(1,-4):
...   o=Point(*o)
...   p = o/2
...   assert Ray(z,o).intersection(Ray(z, p)) == Ray(p,o)
...   assert Ray(z,o),intersection(Ray(p,z)) == Segment(z, p)
...
Traceback (most recent call last):
  File "<stdin>", line 4, in <module>
AssertionError

[akshayah3]

@smichr Use [Segment(z, p)] because in the code it returns like that.Same for the Ray too.

[smichr]

Note: the code I gave had some errors itself. The following highlights the problem:

>>> for o in (1,4),(-1,4),(-1,-4),(1,-4):
...   o=Point(*o)
...   p = o/2
...   assert Ray(z,o).intersection(Ray(p,o)) == [Ray(p,o)]
...   assert Ray(z,o),intersection(Ray(p,z)) == [Segment(z, p)]
...
Traceback (most recent call last):
  File "<stdin>", line 4, in <module>
AssertionError
>>> o
Point(-1, 4)
>>> Ray(z,o).intersection(Ray(p,o)),[Ray(p,o)]
([Ray(Point(0, 0), Point(-1, 4))], [Ray(Point(-1/2, 2), Point(-1, 4))])

The raw code is

z=Point(0,0)
for o in (1,4),(-1,4),(-1,-4),(1,-4):
  o=Point(*o)
  p = o/2
  assert Ray(z,o).intersection(Ray(p,o)) == [Ray(p,o)]
  assert Ray(z,o),intersection(Ray(p,z)) == [Segment(z, p)]

print o

[akshayah3]

@smichr Have a look now, It's working.

[smichr]

It might be helpful to draw a picture

>>> Ray((0,0),(1,1)).intersection(Ray((-2,-2),(-1.8,-1.8)))
[Ray(Point(-2, -2), Point(-9/5, -9/5))]

That should be [Ray((0,0),(1,1))]. Perhaps you want to see if o.contains(self.source) then return self else return o?

[smichr]

btw, stating it the way I did will guarantee that self is returned in the case that both share the same source

[akshayah3]

@smichr Sorry I got a little confused,But I think this does the job?
if self.xdirection == o.xdirection and self.ydirection == o.ydirection: if self.source in o:
return [self]
return [o]
With indentation.

[smichr]

you should raise an error in this case -- we already know that the rays are coincident. In fact, I would be surprised if this ever got triggered. I think return [self] if (self.source in o) else [o] is fine. Did you find otherwise?

[akshayah3]

@smichr I have made the changes.

[smichr]

I opened modifications at #2940 -- can you take a look