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How to split a polygon.

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Carlos Zúñiga
Carlos Zúñiga am 31 Aug. 2020
Kommentiert: Bruno Luong am 31 Aug. 2020
Hello everyone.
If I have a polygon with the following coordinates:
x=[0 4 7 5 1]; %Polygon x-coordinates
y=[0 -2 0 10 8]; %Polygon y-coordinates
How can I split the polygon formed by the coordinates shown bellow in for example six parts which area is equal to each other?
  2 Kommentare
the cyclist
the cyclist am 31 Aug. 2020
Two questions before anyone spends time thinking about this:
  • Is this a homework assignment?
  • Is the only requirement that the six parts have equal area? I'm wary of other assumptions you may be neglecting to mention. For example, would it be ok to just make vertical slices? Or do you need to find a single point in the interior, such that lines to the vertices separate the area equally?
Carlos Zúñiga
Carlos Zúñiga am 31 Aug. 2020
Bearbeitet: Carlos Zúñiga am 31 Aug. 2020
Hello, thank you for your answer.
Actually it is not a homework. It is a problem that I couldn't achieve.
Yes, just as I said, all the areas must be equal. About the vertical slices, yes, we can use vertical slices because I'm traying to do the same but rotating the coordinets according to a slope with a rotation matrix.
Greetings and thank you so much for your time!

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Bruno Luong
Bruno Luong am 31 Aug. 2020
Bearbeitet: Bruno Luong am 31 Aug. 2020
Each slice has area of 9.5
x=[0 4 7 5 1]; %Polygon x-coordinates
y=[0 -2 0 10 8]; %Polygon y-coordinates
n = 6;
P = polyshape(x,y);
A = P.area/n;
xmin = min(x); xmax = max(x);
ymin = min(y); ymax = max(y);
x0 = xmin+0.01;
b = zeros(1,n-1);
Q = cell(1,n);
Qk = polyshape(); % empty
for k=1:n-1
x0 = fzero(@(x) areafun(P, xmin, x, ymin, ymax)-k*A, x0);
b(k) = x0;
Qp = Qk;
[s, Qk] = areafun(P, xmin, b(k) , ymin, ymax);
Q{k} = subtract(Qk, Qp);
end
Q{n} = subtract(P, Qk);
close all;
figure
hold on
for k=1:n
Q{k}.area
plot(Q{k});
end
axis equal
function [s, Q] = areafun(P, xmin, xmax, ymin, ymax)
R = polyshape([xmin xmax xmax xmin],[ymin ymin ymax ymax]);
Q = intersect(P,R);
s = Q.area;
end
  6 Kommentare
Carlos Zúñiga
Carlos Zúñiga am 31 Aug. 2020
Actually I already answer myself!
Thank you so much Mr. Bruno!
Bruno Luong
Bruno Luong am 31 Aug. 2020
Star-like partitioning
x=[0 4 7 5 1]; %Polygon x-coordinates
y=[0 -2 0 10 8]; %Polygon y-coordinates
n = 6;
P = polyshape(x,y);
A = P.area/n;
xmin = min(x); xmax = max(x);
ymin = min(y); ymax = max(y);
b = zeros(1,n-1);
Q = cell(1,n);
[xc,yc] = P.centroid;
r = sqrt(max((x-xc).^2+(y-yc).^2))*1.1;
Qk = polyshape(); % empty
x0 = 2*pi/n;
for k=1:n-1
x0 = fzero(@(tt) areafun(P, xc, yc, tt, r)-k*A, x0);
b(k) = x0;
Qp = Qk;
[s, Qk] = areafun(P, xc, yc, x0, r);
Q{k} = subtract(Qk, Qp);
end
Q{n} = subtract(P, Qk);
close all;
figure
hold on
for k=1:n
Q{k}.area
plot(Q{k});
end
axis equal
function [s, Q] = areafun(P, xc, yc, tt, r)
ntt = max(ceil(abs(tt)*128),2);
phi = linspace(0,tt,ntt);
Q = polyshape([xc xc+r*cos(phi)],[yc yc+r*sin(phi)]);
Q = intersect(P,Q);
s = sign(tt)*Q.area;
end

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