How can I determine is (x,y) pair is inside an orthogonal area?

Hi everyone,

Suppose we have an orthogonal area where the 4 x,y pairs are (1,5) , (3,5) , (1,7) and (3,7). I would like to test if a random pair, such as (0,6), is inside the orthogonal area. I did the following code but I don't get any answer:

x = [1, 3];
y = [5, 7];

x0 = 0;
y0 = 6;
if x0 < x(2)
    if   x0 > x(1)
        if   y0 < y(2)
            if   y0 > y(1)
                answer0 == 1
            else
                answer0 == 0
            end
        end
    end
end

Answer 1

[Rik]

Your else statement is only reached in the inner part. If you want to use this structure, you should do the following:

x = [1, 3];
y = [5, 7];
x0 = 0;
y0 = 6;
answer0 =false;
if x0 < x(2)
    if   x0 > x(1)
        if   y0 < y(2)
            if   y0 > y(1)
                answer0 = true;
            end
        end
    end
end

However, a more compact/clearer way is this:

x = [1, 3];
y = [5, 7];
x0 = 0;
y0 = 6;
if x0>min(x) && x0<max(x) && ...
        y0>min(y) && y0<max(y)
    answer0=true;
else
    answer0=false;
end

[Stephen23]

The if - else is not required:

answer0 = x0>min(x) && x0<max(x) && y0>min(y) && y0<max(y)

Answer 2

[John D'Errico]

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.