如果matlab中兩個矩形的交點爲零

Question

通過x軸（水平軸）瞭解兩個矩形的中心和它們的角度，在Matlab中如何識別它們的交點是否爲零？任何包含這些信息的答案都是非常感謝的矩形的寬度和長度也是已知的

Answer 1

這是一個編程問題，如果你想解決它的數值。對於確切的解決方案，您可以使用幾何方程。

---

第一個問題：從它的寬度，高度和中心限定一矩形的角落：

C1 = [0, 0]; % Centre of rectangle 1 (x,y) 
C2 = [1, 1]; % Centre of rectangle 2 (x,y) 
W1 = 5; W2 = 3; % Widths of rectangles 1 and 2 
H1 = 2; H2 = 3; % Heights of rectangles 1 and 2 
% Define the corner points of the rectangles using the above 
R1 = [C1(1) + [W1; W1; -W1; -W1]/2, C1(2) + [H1; -H1; -H1; H1]/2]; 
R2 = [C2(1) + [W2; W2; -W2; -W2]/2, C2(2) + [H2; -H2; -H2; H2]/2]; 

接下來的問題是創造許多點代表矩形的邊緣。如果您想查看相交區域，則可以在內生成許多點。

n = 1000;  % Define some number of points to use 
% Use interp1 to interpolate around the rectangles 
R1points = interp1(1:5, [R1; R1(1,:)], linspace(1,5,n)); 
R2points = interp1(1:5, [R2; R2(1,:)], linspace(1,5,n)); 

然後旋轉矩形：

a1 = deg2rad(0); a2 = deg2rad(30); % angles of rotation for rectangle 1 and 2 respectively 
R1rotated(:,1) = (R1points(:,1)-C1(1))*cos(a1) - (R1points(:,2)-C1(2))*sin(a1) + C1(1); 
R1rotated(:,2) = (R1points(:,1)-C1(1))*sin(a1) + (R1points(:,2)-C1(2))*cos(a1) + C1(2); 
R2rotated(:,1) = (R2points(:,1)-C2(1))*cos(a2) - (R2points(:,2)-C2(2))*sin(a2) + C2(1); 
R2rotated(:,2) = (R2points(:,1)-C2(1))*sin(a2) + (R2points(:,2)-C2(2))*cos(a2) + C2(2); 

最後，檢查交叉口inpolygon：

in1 = inpolygon(R1rotated(:,1), R1rotated(:,2), R2rotated(:,1), R2rotated(:,2)); 
in2 = inpolygon(R2rotated(:,1), R2rotated(:,2), R1rotated(:,1), R1rotated(:,2)); 

如果nnz(in1)>0或nnz(in2)>0那麼你有一個交集！使用分散它想象：

hold on 
scatter(R2rotated(:,1), R2rotated(:,2), '.b') 
scatter(R2rotated(in2,1), R2rotated(in2,2), 'xc') 
scatter(R1rotated(:,1), R1rotated(:,2), '.r') 
scatter(R1rotated(in1,1), R1rotated(in1,2), 'xg') 

結果：