Delaunay三角测量让我失去了对称性

Question

我正在使用Delaunay三角剖分将多边形分割成三角形。我使用大型代码处理FEM，并且我的一个“检查点”是对称的（如果数据是对称的，则输出也必须是对称的）。但是，由于我无法控制Delaunay三角剖分，所以它让我失去了对称性。

我写了一个小代码来说明我的问题：我们考虑两个不相交的三角形和一个与它们相交的大矩形。我们想用矩形对这些三角形的减法进行三角测量：

clear all
close all
warning off % the warning is about duplicate points, not important here

figure
hold on

p =[.3 .3
.4 .3
.3 .4
.7 .6
.6 .7
.7 .7]; % coordinates of the points for the triangles

px = 1/3;
py = 1/3;
lx = 1/3;
ly = 1/3; % size and position of the rectangle

% rearrange the polygon with clockwise-ordered vertices
[x1,y1]=poly2cw([px; px+lx; px+lx; px],[py; py; py+ly; py+ly]); % rectangle

patch(x1,y1, 1, 'EdgeColor', 'k');

for i=1:2

    pc = p(3*i-2:3*i,:); % current triangle
    % rearrange the polygon with clockwise-ordered vertices
    [x0,y0]=poly2cw(pc(:,1),pc(:,2)); % triangle

    [x2,y2] = polybool('intersection',x1,y1,x0,y0); % intersection
    [x3,y3] = polybool('subtraction',x0,y0,x2,y2); % subtraction

    DT = delaunayTriangulation(x3,y3);

    triplot(DT,'Marker','o')

end
XL = xlim; xlim(XL+[-1 +1]*diff(XL)/10);
YL = ylim; ylim(YL+[-1 +1]*diff(YL)/10);
axis equal;
box on;

如您所见，Delaunay三角剖分在两个三角形中都没有相同的行为，因此失去了对称性。

有没有简单的方法来恢复对称性？

我使用Matlab R2013a。

Answer 1

好像你正在使用MatLab R2013，因为在我的R2011b中没有delaunayTriangulation功能。为了能够运行你的代码我稍微改了一下：

clear all
close all
warning off % the warning is about duplicate points, not important here

figure
hold on

p =[.3 .3
    .4 .3
    .3 .4
    .7 .6
    .6 .7
    .7 .7]; % coordinates of the points for the triangles

px = 1/3;
py = 1/3;
lx = 1/3;
ly = 1/3; % size and position of the rectangle

% rearrange the polygon with clockwise-ordered vertices
[x1,y1]=poly2cw([px; px+lx; px+lx; px],[py; py; py+ly; py+ly]); % rectangle

patch(x1,y1, 1, 'EdgeColor', 'k');

for i=1:2

    pc = p(3*i-2:3*i,:); % current triangle
    % rearrange the polygon with clockwise-ordered vertices
    [x0,y0]=poly2cw(pc(:,1),pc(:,2)); % triangle

    [x2,y2] = polybool('intersection',x1,y1,x0,y0); % intersection
    [x3,y3] = polybool('subtraction',x0,y0,x2,y2); % subtraction

    %DT = delaunayTriangulation(x3,y3);
    %triplot(DT)

    % This is triangulation of subtraction
    DT = delaunay(x3,y3);
    triplot(DT,x3,y3, 'Marker','.', 'Color','r')

    % This is triangulation of intersection
    DT = delaunay(x2,y2);
    triplot(DT,x2,y2, 'Marker','o', 'Color','b', 'LineWidth',1)

end
axis equal;
axis tight;
box on;

XL = xlim; xlim(XL+[-1 +1]*diff(XL)/10);
YL = ylim; ylim(YL+[-1 +1]*diff(YL)/10);

text(0.5,0.55,'triangulation of subtraction',  'HorizontalAlignment','center', 'VerticalAlignment','bottom', 'Color','r');
text(0.5,0.45,'triangulation of intersection', 'HorizontalAlignment','center', 'VerticalAlignment','top',    'Color','b');

以下是我看到的结果

它与你的相似吗？ 你可以在你的问题中添加一个图像来描述你得到的结果有什么问题吗？

Answer 2

这似乎不是一个错误。 实际上，您的结果来自您的数据。

我用你的代码玩了一点

clear all
close all
warning off % the warning is about duplicate points, not important here

figure
hold on

p =[.3 .3
.4 .3
.3 .4
.7 .6
.6 .7
.7 .7]; % coordinates of the points for the triangles

px = 1/3;
py = 1/3;
lx = 1/3;
ly = 1/3; % size and position of the rectangle

% rearrange the polygon with clockwise-ordered vertices
[x1,y1]=poly2cw([px; px+lx; px+lx; px],[py; py; py+ly; py+ly]); % rectangle

patch(x1,y1, 1, 'EdgeColor', 'k');

for i=1:2

    pc = p(3*i-2:3*i,:); % current triangle
    % rearrange the polygon with clockwise-ordered vertices
    [x0,y0]=poly2cw(pc(:,1),pc(:,2)); % triangle

    [x2,y2] = polybool('intersection',x1,y1,x0,y0); % intersection
    [x3,y3] = polybool('subtraction',x0,y0,x2,y2); % subtraction

    % This is for R2013a
    %DT = delaunayTriangulation(x3,y3);
    %triplot(DT,'Marker','o');

    % This is for R2011b
    %DT = DelaunayTri(x3,y3);
    %triplot(DT,'Marker','o');

    % This is plain delaunay version
    DT = delaunay(x3,y3);
    triplot(DT,x3,y3,'Marker','o')

    % we break here to analyze the first triangulation
    break

end
XL = xlim; xlim(XL+[-1 +1]*diff(XL)/10);
YL = ylim; ylim(YL+[-1 +1]*diff(YL)/10);
axis equal;
box on;

% % % % % % % % % % % % % % % % % %
% Checking the triangulation
% % % % % % % % % % % % % % % % % %

% Wrong triangulation for i=2 is hard-coded
DT2     = [
    2 1 6
    6 5 2
    5 3 2
    5 4 3
    2 3 1 ];

figure;
hold all;
triplot(DT2,x3,y3,'Marker','o', 'Color','r', 'LineWidth',1)
axis equal;
axis tight;
box on;
XL = xlim; xlim(XL+[-1 +1]*diff(XL)*0.5);
YL = ylim; ylim(YL+[-1 +1]*diff(YL)*0.5);

% circumcircle: http://www.mathworks.com/matlabcentral/fileexchange/17300
ca = linspace(0,2*pi);
cx = cos(ca);
cy = sin(ca);

hl = [];
for k=1:size(DT2,1)
    tx  = x3(DT(k,:));
    ty  = y3(DT(k,:));
    [r,cn]=circumcircle([tx,ty]',0);
    if ~isempty(hl)
        %delete(hl);
    end
    fprintf('Circle %d: Center at (%.23f,%.23f); R=%.23f\n',k,cn,r);
    text( cn(1),cn(2), sprintf('c%d',k) );
    hl = plot( cx*r+cn(1), r*cy+cn(2), 'm-' );
    drawnow;
    %pause(3); %if you prefere to go slowly 
end

这是我看到的输出：

  

圈1：居中于（0.28333333333333333000000,0.35000000000000003000000）; R = 0.05270462766947306400000   圆圈2：中心位于（0.34999999999999998000000,0.34999999999999998000000）; R = 0.02357022603955168100000   第3圈：中心位于（0.28333333333333338000000,0.34999999999999992000000）; R = 0.05270462766947289800000   第4圈：中心位于（0.35000000000000003000000,0.28333333333333355000000）; R = 0.05270462766947290500000   第5圈：中心位于（0.35000000000000003000000,0.28333333333333333000000）; R = 0.05270462766947312000000

这个数字：

因此圆圈1和圆圈3以及圆圈4和圆圈5几乎相同。 因此，您和我的结果之间的差异可能来自舍入误差，因为相应的四个点位于浮点数学精度内的同一个圆上。你必须重新设计你的积分才能获得不依赖于这些东西的可靠结果。

玩得开心; o）