How do you create a random point generation with the values of 1 or 2?

Question

Ian Bouchard 2017-12-4

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https://ww2.mathworks.cn/matlabcentral/answers/370867-how-do-you-create-a-random-point-generation-with-the-values-of-1-or-2

回答： Roger Stafford 2017-12-4

Hello, So I have a 2 point question. The first question I have is how do you generate random points with the value of either 1 or 2? I am currently creating a code where I create random points inside a polygon and then use delaunay triangulation on the points. After creating the triangulation I then find the incenter point of each triangle which leads me to my next question. How do you average the three points of each triangulation and then average them into a single value which is the incenter? Thank you so much for your help!!

P.S. I have received much help on this code and could use a small explanation behind the steps used to get there. Thank you!

states = shaperead('usastatehi.shp');
st = states(47); %creates a polgon in the shape of Washington State
stBB = st.BoundingBox;
st_minlat = min(stBB(:,2 )); 
st_maxlat = max(stBB(:,2 )); 
st_latspan = st_maxlat - st_minlat;
st_minlong = min(stBB(:,1 )); 
st_maxlong = max(stBB(:,1 )); 
st_longspan = st_maxlong - st_minlong;
stX = st.X ; 
stY = st.Y;
numPointsIn = 42;
for i = 1:numPointsIn 
flagIsIn = 0;
while ~flagIsIn 
x(i) = st_minlong + rand(1) * st_longspan ; 
y(i) = st_minlat + rand(1) * st_latspan ; 
flagIsIn = inpolygon(x(i), y(i), stX, stY ); 
end 
end
mapshow(st, 'edgecolor', 'r', 'facecolor', 'none ') 
hold on 
scatter(x, y , '.') 
dt=delaunayTriangulation(x',y')
IC=incenter(dt)
dt1=delaunayTriangulation(IC)
hold on
triplot(dt, '--g')
triplot(dt1, '--r')
[XV, YV] = voronoi(IC(:,1),IC(:,2 )); 
plot(XV,YV,'k') 
axis([min(stX) max(stX) min(stY) max(stY)])

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Answer 1

Roger Stafford 2017-12-4

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/370867-how-do-you-create-a-random-point-generation-with-the-values-of-1-or-2#answer_294653

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As to your question of how to find the location of the incenter of a triangle, given its three vertices, here is how. Let (x1,y1), (x2,y2), and (x3,y3) be the 2D cartesian coordinates of the three vertices of a triangle. Let (xc,yc) be the desired coordinates of its incenter - that is, the point which is an equal orthogonal distance from each of its three sides.

% The three side lengths
a = sqrt((x2-x3)^2+(y2-y3)^2);
b = sqrt((x3-x1)^2+(y3-y1)^2);
c = sqrt((x1-x2)^2+(y1-y2)^2);
% The desired cartesian coordinates of the incenter
xc = (a*x1+b*x2+c*x3)/(a+b+c);
yc = (a*y1+b*y2+c*y3)/(a+b+c);

In other words, the incenter is the "average" of the three vertices, weighted respectively by the triangle's opposite side lengths.