Simulation for a triangle
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i'm trying to creat a simple script to find the area of a triangle using monte carlo simulation. however, i'm basically stuck. could someone please help me.
any help is appreciated,
thannks
1 个评论
Aswin Farzana Mohamed Ansar
2017-10-1
Thufz can you share the script that you tried with!
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Roger Stafford
2014-2-10
0 个投票
The theory of the Monte Carlo simulation which you describe is, I presume, that if you enclose a triangle completely in a square of known area, the area of the triangle can be estimated by multiplying the area of the square by the fraction of the random points within the square which happen to land inside the triangle. Of course this requires a very large number of random points to obtain a reasonably accurate estimate.
With the triangle you have specified with vertices (0,0), (0,1), (1,0), it is very easy to do this, and in fact you have done just that in your code. Writing x = rand and y = rand will fill the unit square with vertices at (0,0), (1,0), (1,1), and (0,1) uniformly. Since x and y are non-negative, the only test you need for being inside your triangle is for the hypotenuse, whether or not x+y <=1, and that is what your 'count' is recording. That means your code as you have shown it is almost done. Multiply the known area of the square by the fraction count/tries and you have your estimate for the area. You were almost there!
The simulation is a little more challenging with an arbitrary triangle. You have to define a square which is guaranteed to completely enclose it and then generate a large number of random points that are distributed uniformly in the square. Each of these points is tested to see if it does or does not lie within the triangle. The fraction that do lie within it is to be multiplied by the area of the square to give an estimate of the triangle's area.
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Image Analyst
2014-2-9
0 个投票
You have not specified the 3 vertex coordinates of the triangle. Once you do that you can use inpolygon() to determine if a test point is inside the triangle.
4 个评论
Thufz
2014-2-9
Image Analyst
2014-2-9
Tell me what the vertices are. Tell me what x1, y1, and x2,y2, and x3,y3 are. Then make xv and yv like this
xv = [x1, x2, x3];
yv = [y1, y2, y3];
Then see if x,y is in or out
itsInside = inpolygon(x,y,xv,yv);
If you read the help for inpolygon, you should know how to do it.
Image Analyst
2014-2-9
编辑：Image Analyst
2014-2-9
x = [0, 0, 1]
y = [0, 1, 0]
area = polyarea(x, y)
However I don't know how a Monte Carlo could be worked into this. The only way I can see it is if you run it a million times and get random coordinates, but I don't know if that is what you are after.
By the way, don't use max as a variable name or else you will destroy a very important build in function. Same for min.
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