Matlab：给出4点，拟合最接近的菱形/平方

我需要一个脚本来“检查”给定的4个点是形成正方形还是菱形。

我正在使用QR码分段脚本，在其中尝试通过查找按行和列遍历二进制图像的非负值来定位顶点。

在某些情况下，不需要进行检查，如下图所示：

很难看到它，但是顶点被标记为4点，分别是绿色，品红色，青色和黄色。在这种情况下，脚本应该返回相同的输入点，因为不需要修改。

另一方面，在某些情况下，顶点被标记为：

It can be seen that the magenta and cyan labels rely on the top right corner of the image. This is obviously not correct, but it fullfills the specified condition: traverse each row of the image until you find a row satisfying sum(row)>1 (greater than 1 to avoid single, noisy pixels).

How can I locate the misplaced vertex and place it using the remaining vertex coordinates?

EDIT

Solved the problem. I'm posting the code of the function in case someone needs it:

function correctedCorners = square(corners)
    correctedCorners = corners;
    X = corners(:,1);
    Y = corners(:,2);
    sortedX = sort(corners(:,1));
    sortedY = sort(corners(:,2));
    %% DISTANCES BW POINTS
    for i=1:4
        for j=1:4
            distances(i,j) = sqrt((corners(i,1)-corners(j,1))^2+        (corners(i,2)-corners(j,2))^2);
        end
    end
    %% relationship bw distances
    % check corner 1
    d11 = distances(1,1);%0
    d12 = distances(1,2);%x
    d13 = distances(1,3);%sqrt(2)*x
    d14 = distances(1,4);%x
    bool1 = [(d12*0.8<=d14)&(d12*1.2>=d14) (d12*0.8*sqrt(2)<=d13)&                (d12*1.2*sqrt(2)>=d13) (d14*0.8<=d12)&(d14*1.2>=d12) (d14*0.8*sqrt(2)<=d13)&(d14*1.2*sqrt(2)>=d13)];
    % check corner 2
    d21 = distances(2,1);%x
    d22 = distances(2,2);%0
    d23 = distances(2,3);%x
    d24 = distances(2,4);%sqrt(2)*x
    bool2 = [(d21*0.8<=d23)&(d21*1.2>=d23) (d21*0.8*sqrt(2)<=d24)&(d21*1.2*sqrt(2)>=d24) (d23*0.8<=d21)&(d23*1.2>=d21) (d23*0.8*sqrt(2)<=d24)&(d23*1.2*sqrt(2)>=d24)];
    % check corner 3
    d31 = distances(3,1);%sqrt(2)*x
    d32 = distances(3,2);%x
    d33 = distances(3,3);%0
    d34 = distances(3,4);%x
    bool3 = [(d32*0.8<=d34)&(d32*1.2>=d34) (d32*0.8*sqrt(2)<=d31)&(d32*1.2*sqrt(2)>=d31) (d34*0.8<=d32)&(d34*1.2>=d32) (d34*0.8*sqrt(2)<=d31)&(d34*1.2*sqrt(2)>=d31)];
    % check corner 4
    d41 = distances(4,1);%x
    d42 = distances(4,2);%sqrt(2)*x
    d43 = distances(4,3);%x
    d44 = distances(4,4);%0
    bool4 = [(d41*0.8<=d43)&(d41*1.2>=d43) (d41*0.8*sqrt(2)<=d42)&(d41*1.2*sqrt(2)>=d42) (d43*0.8<=d41)&(d43*1.2>=d41) (d43*0.8*sqrt(2)<=d42)&(d43*1.2*sqrt(2)>=d42)];
    bool = [bool1; bool2;bool3;bool4];
    idx = 0;
    for i=1:4
        if (sum(bool(i,:))==0)
            idx = [idx i];
        end
    end
    if (length(idx)>=2)
        for i=2:length(idx)
            switch idx(i)
                case 1
                    correctedCorners(1,:) =         abs(corners(4,:)-(corners(3,:)-corners(2,:)));
                case 2
                    correctedCorners(2,:) = abs(corners(3,:)-(corners(4,:)-corners(1,:)));
                case 3
                    correctedCorners(3,:) = abs(corners(2,:)+(corners(1,:)-corners(1,:)));
                case 4
                    correctedCorners(4,:) = abs(corners(1,:)+(corners(3,:)-corners(2,:)));
            end
        end
    end