How can I put 5 points inside of the triangular by using rand (5,2)?

I have triangular ABC, where

A=[5,60];

B=[50,90];

C=[50,30];

I need to put 5 points just inside of this ABC triangular. x and y values of 5 points should be randomly chosen, by using rand(5,2).

How can I do? Could anyone help me?

Answer 1

[Image Analyst]

Yet another solution using simple geometry and the congruence of triangles.

A=[5,60];
B=[50,90];
C=[50,30];
x = [A(1), B(1), C(1), A(1)];
y = [A(2), B(2), C(2), A(2)];
x1 = min(x);
x2 = max(x);
plot(x, y, 'b*-', 'LineWidth', 2);
grid on;
hold on;
% Get an array of 5x2 random numbers, like required.
r = rand(5,2);
% Get 5 random x
xt = x1 + (x2-x1)*r(:,1)
% Get 5 random y and plot them
yt = (xt - x1) .* (90-60) ./ (x2 - x1) .* (2*r(:, 2)-1) + 60
plot(xt, yt, 'r+', 'LineWidth', 2);

Answer 2

[the cyclist]

For each of the 5 points, do this

t = sqrt(rand());
s = rand();

P = (1-t)*A + t*((1-s)*B+s*C)

I found that method in this thread, after a very brief google keyword search.

Here is a vectorized version:

N = 5;

A=[5,60];
B=[50,90];
C=[50,30];

AN = repmat(A,N,1);
BN = repmat(B,N,1);
CN = repmat(C,N,1);

t = sqrt(rand(N,1));
s = rand(N,1);

P = AN + t.*(s.*(CN-BN)+(BN-AN));

figure
hold on
plot([A(1) B(1) C(1) A(1)],[A(2) B(2) C(2) A(2)],'r-')
h = plot(P(:,1),P(:,2),'.');
set(h,'MarkerSize',24)

[the cyclist]

I don't know if it is important to you that the points be uniformly distributed within the triangle, but this solution has that property.

Answer 3

[Image Analyst]

Not sure by what you mean by "just" inside. Does it need to be pretty close to the sides and not in the middle of the triangle? Anyway, here's a brute force but easy to understand way using a for loop to try locations until you get the required number of points inside, as determined by the inpolygon() function:

A=[5,60];
B=[50,90];
C=[50,30];
x = [A(1), B(1), C(1), A(1)];
y = [A(2), B(2), C(2), A(2)];
x1 = min(x);
x2 = max(x);
y1 = min(y);
y2 = max(y);
plot(x, y, 'b*-', 'LineWidth', 2);
grid on;
hold on;
numInside = 0;
loopCounter = 0;
maxIterations = 1000;
% Keep looping and checking with inpolygon
% until we get the required number.
while numInside < 5 && loopCounter < maxIterations
  xTrial = x1 + (x2-x1)*rand;
  yTrial = y1 + (y2-y1)*rand;
  if inpolygon(xTrial, yTrial, x, y)
    xInTriangle = xTrial;
    yInTriangle = yTrial;
    plot(xInTriangle, yInTriangle, 'r+', 'MarkerSize', 16, 'LineWidth', 2);
    numInside = numInside + 1
  end  
  loopCounter = loopCounter + 1;
end
xInTriangle % Echo to command window.
yInTriangle

Answer 4

[Star Strider]

It’s not possible to use rand alone to put them inside the triangle, because rand only produces values on the interval [0,1], and those are outside the triangle. So a bit of code is needed to place them inside the triangle and plot the points and the triangle:

A=[5,60];
B=[50,90];
C=[50,30];

Mpatch = cat(1,A,B,C)';                                         % Patch Vertices
Lims = mean([min(Mpatch,[],2)  max(Mpatch,[],2)],2);            % Mean Of Coordinates
Mult = 10;                                                      % Multiplier To Separate Points (Optional)
pts = bsxfun(@plus,Mult*rand(5,2),Lims');                       % Add Mean Of Coordinates To Random Matrix

figure(1)
patch(Mpatch(1,:), Mpatch(2,:), 'g')
hold on
scatter(pts(:,1), pts(:,2), 'bp')
hold off
axis([0  60    20  100])

I added a multiplier to separate the points, because otherwise they cluster together. The code will work without the multiplier.

A sample plot: