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

closest value

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.

Best Answer

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.

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MATLAB: How to determine is (x,y) pair is inside an orthogonal area

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

MATLAB: Pretty simple question regarding convhulln. Now that I have K…

(Somebody should post a thing on the FEX explaining all the basics about convex hulls, tessellations, etc. One more thing to add to the round tuit list.)

Think in 2-d first. A convex hull of some data points is a simple polygon, composed of linear segments. Any vertex of the convex hull is also one of the data points. A convex hull does not introduce new points. So essentially a convex hull is just a set of references into the original set of points. Since these things are best done visually, lets generate an example.

>> xy = rand(8,2)
xy =
      0.80028      0.84913
      0.14189      0.93399
      0.42176      0.67874
      0.91574      0.75774
      0.79221      0.74313
      0.95949      0.39223
      0.65574      0.65548
     0.035712      0.17119
>> plot(xy(:,1),xy(:,2),'o')

See that there appear to be 5 of those points that lie on the boundary of the convex hull in this case, and 3 points lie clearly in the interior.

We can find the hull using either convhull, or convhulln.

>> convhull(xy(:,1),xy(:,2))
ans =
     1
     2
     8
     6
     4
     1
>> convhulln(xy)
ans =
     8     6
     6     4
     4     1
     1     2
     2     8

What do these numbers mean? Each index returned corresponds to ONE vertex of the hull polygon, but also to one of our original data points. Thus, the convex hull goes between the points [1 2 8 6 4 1] in the set IN THAT ORDER. The points that are inside the hull are of course then numbered [3 5 7].

See that each row of the array returned by convhulln is the same information, just in a slightly different form. It tells us that one edge of the hull is the edge [8 6], connecting point 8 and point 6. Then we connect points 6 and 4, with another edge, etc. These edges in the array returned from convhulln are given in no specific order, but they will be the same information overall as that returned by convhull.

The difference is that convhulln will work in any number of dimensions. (Ok, it can get pretty nasty to visualize in high dimensions. But 3 or 4 or 5 dimensions is entirely adequate. And we can't even draw pictures past 3-d anyway.)

So think about the 2-d case. The convex hull in 2-d is composed of "facets" composed of edges, line segments. In 3-d, the convex hull will always be composed of triangles. A 3-d convex hull will have planar facets. They will always break down into triangular facets though. Yes, one could imagine the set of points that comprise the unit 3-cube. (A cube in R^3.) Clearly its convex hull is a set of squares.

So here are the 8 vertices of the unit 3-cube.

V =
     0     0     0
     1     0     0
     0     1     0
     1     1     0
     0     0     1
     1     0     1
     0     1     1
     1     1     1

ndgrid could essentially help you to generate that list of points if you wanted.

Let us generate the convex hull using matlab. It is a set of triangles.

>> convhulln(V)
ans =
     4     3     1
     2     4     1
     6     2     1
     5     6     1
     6     4     2
     4     6     8
     3     7     1
     7     5     1
     4     7     3
     7     4     8
     7     6     5
     6     7     8

Each square has been split, into a pair of triangles. Since there are 6 square faces of a cube, there will be 12 triangles.

>> size(ans)
ans =
    12     3

If you prefer to visualize them this way, we can sort the triangles in increasing order of the vertices, and then use sortrows. This does not change the information in any serious way.

>> sortrows(sort(convhulln(V),2))
ans =
     1     2     4
     1     2     6
     1     3     4
     1     3     7
     1     5     6
     1     5     7
     2     4     6
     3     4     7
     4     6     8
     4     7     8
     5     6     7
     6     7     8

Still the same information. Each row is ONE triangle. A set of references into the original set of data points. In the case of the unit 3-cube of course, EVERY point lies one the surface, in at least a few of the triangles of that hull. In fact, depending on how those surface squares are split into triangles each point may lie in from 3 to 6 different triangles.

What you need to see is that a convex hull in 3-d is composed of triangles. If we go to higher dimensions, a convex hull in n-dimensions will be composed of (n-1)-simplexes. (A k-simplex is composed of k+1 points. So an (n-1)-simplex is defined by n vertices.) For example...

>> size(convhulln(rand(100,4)))
ans =
   290     4
>> size(convhulln(rand(100,5)))
ans =
        1068           5
>> size(convhulln(rand(100,6)))
ans =
        4047           6
>> size(convhulln(rand(100,7)))
ans =
       16094           7

See that each row of the result is a set of indexes into the original array of points, and that in n-dimensions, each row will be a facet, an (n-1) simplex. We can't plot these things on a 2-d monitor or piece of paper, but who cares?

Best Answer

Related Solutions

MATLAB: How to determine is (x,y) pair is inside an orthogonal area

MATLAB: Pretty simple question regarding convhulln. Now that I have K…

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