Boundary function fails, R2016b

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Jukka Koskinen
Jukka Koskinen 2016 年 12 月 1 日
コメント済み: Jukka Koskinen 2016 年 12 月 12 日
I try to make a concave hull type boundary around several overlapping shapes (four circles in this case) by the boundary function. The following code illustrates the problem. The boundary is drawn in black color (better seen in attached file). It doesn't look good... It seems that there is a bug in the function?
if true
clear all;
close all;
r_1 = 5;
r_2 = 3;
r_3 = 2;
r_4 = 2;
center_1 = [2 2];
center_2 = [2 -2];
center_3 = [5 5];
center_4 = [-3 5];
alpha = 0:(2*pi)/120:2*pi;
circle_1 = [center_1(1)+r_1*sin(alpha)' center_1(2)+r_1*cos(alpha)'];
circle_2 = [center_2(1)+r_2*sin(alpha)' center_2(2)+r_2*cos(alpha)'];
circle_3 = [center_3(1)+r_3*sin(alpha)' center_3(2)+r_3*cos(alpha)'];
circle_4 = [center_4(1)+r_4*sin(alpha)' center_4(2)+r_4*cos(alpha)'];
shape = [circle_1
circle_2
circle_3
circle_4];
k = boundary(shape(:,2), shape(:,1), 1);
figure(1)
plot(circle_1(:,1), circle_1(:,2), 'r')
hold on;
grid on;
plot(circle_2(:,1), circle_2(:,2), 'b')
plot(circle_3(:,1), circle_3(:,2), 'g')
plot(circle_4(:,1), circle_4(:,2), 'm')
plot(shape(k,1), shape(k,2), 'k')
axis([-10 10 -10 10])
end
  3 件のコメント
Jukka Koskinen
Jukka Koskinen 2016 年 12 月 2 日
編集済み: Jukka Koskinen 2016 年 12 月 2 日
The function works fine, when I move circle_4 center one step to the right (coordinates [-2,5]). But when coordinates are [-3,5] the function doesn't work properly, as seen above.
It should look like this:
Adam
Adam 2016 年 12 月 2 日
Ah, ok, I kind of just ignored the inner black lines thinking they wee just part of the things being bounded.

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採用された回答

Massimo Zanetti
Massimo Zanetti 2016 年 12 月 7 日
Hi Jukka, if you look at the plot of the boundary without superimposing it to the circles you can see (see first image below) that the shrinking is extreme. You set it to the maximum value of 1 with k = boundary(shape(:,2), shape(:,1), 1);. Thus, the algorithm finds points also in the interior of the major circle.
To solve for this, just relax the shrinking factor by setting it to .5 (the default value). So, invoke k = boundary(shape(:,2), shape(:,1), .5);. The result is shown in the second image. I think that is the one you need. :)
  1 件のコメント
Jukka Koskinen
Jukka Koskinen 2016 年 12 月 12 日
Actually I contacted support and the answer was pretty same: relax the shrinking factory.

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