How to find if a pair is within an unorthogonal area with parallel sides?

Hi everyone,

Suppose I have the area within the following x,y pairs, (3,1), (3,5), (12,1), (12,7), (6,5) and (6,7). I would like to test whether a random pair, say (5,6) is inside the area of interest. Is there any way to generalize the code to test for any random pairs?

I would appreciate any help. Thanks in advance.

[John D'Errico]

Why did you feel it necessary to post the same question? See that I already answered it in your last post.

[gsourop]

It is not the same question. This is a polugon. In this questions you need to find the closest bit of this parallel and then interpolate somehow. It is completely different.

[John D'Errico]

And I stated explicitly that the answer I posted to your last question had no need for parallel sides, that it would work on ANY polygon. READ THE ANSWER. It does not even require the polygon has only 4 vertices. As long as the points form a polygon, which convhull can help you to fix, it works. And if your polygon is not convex, it still works.

Answer 1

[John D'Errico]

Was it really necessary to post this, when I just answered that same question in your last? Sigh.

A better solution, rather than such nested, specific tests, is to use a tool like inpolygon. Make sure they are sorted, so it truly represents a polygon, in that order. But the points are easily sorted in terms of angle.

px = [1 , 3 , 3 , 1];
py = [5 , 5 , 7, 7];
inpolygon(2,6,px,py)
ans =
  logical
   1

In newer releases of MATLAB, we also have the polyshape tools.

PS = polyshape(px,py)
PS = 
  polyshape with properties:

      Vertices: [4×2 double]
    NumRegions: 1
      NumHoles: 0

isinterior(PS,2,6)
ans =
  logical
   1

The nice thing about inpolygon and polyshape is these tools are not restricted to simple rectangular, 90 degree polygons.

So, if your points are not known to lie in a good order to represent a polygon, we could convert to polar coordinates, and then sort the sequence of points around the centroid. Simpler is to just use a convex hull to do the heavy thinking for you. So if I swap points 3 and 4, so the polygon turns into sort of a figure 8, we can easily recover a proper order:

px = [1 , 3 , 1, 3];
py = [5 , 5 , 7, 7];

edgelist = convhull(px,py)
edgelist =
     1
     2
     4
     3
     1

PX = px(edgelist)
PX =
     1     3     3     1     1

PY = py(edgelist)
PY =
     5     5     7     7     5

So even though the points in px and py were mi-sorted, convhull reordered them. It also connected the polygon, so the first and last point were the same, but tools like inpolygon and polyshape won't care.