Matlab: given 4 points, fit the closest rhombus/square - matlab

Question

I need a script that "checks" if 4 given points form a square or a rhombus.
I am working in a QR code segmentation script in which I try to locate the vertex by looking for the non-negative values of a binary image traversing it by rows and columns.
There are some cases in which checking is not neccessary, like in this image:
It is a bit hard to see, but the vertex are labelled as 4 points in green, magenta, cyan and yellow. In this case the script should return the same input points, since no modification is needed.
On the other hand, there are cases in which the vertex are labeled as so:
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

Answer 1

From basic geometry about squares:
TopLeft distance to BotLeft = x
TopLeft distance to TopRight= x
TopLeft distance to BotRight= sqrt(2)*x
Use the same logic for BotLeft to other points, etc.
Allow yourself something like 10-20% margin of error to declare an incorrect point. That is, if TopLeft distance to 2 points outside of the range (80%;120%)*x , and its distance to the third point is outside of the range (80%;120%)*sqrt(2)*x, you can declare the point as placed incorrectly.
In your case, the TopLeft point fails on all distance tests:
0 instead of x to TopRight (about 100% error)
sqrt(2)*x vs x to BotLeft (about 44% error)
x vs sqrt(2)*x to BotRight) (about 31% error)
As long as the rhombus is very similar to a square, a 20% margin of error while treating it as a square should still work.