Grid detection in matlab

Question

I have a grid in a binary image (may be rotated). How can I know approximate formula for that grid using MATLAB?

Example image:

_{(source: sjtu.edu.cn)}

Sometimes these black dots are missing, so I need formula or ‘a way’ to estimate possible center of these black dots.

I have tried by using regionprops, it help me to get center of these exist black dots, but no idea if black dots a missing

clear all
im = imread('print5.jpg');
im = im2bw(im);
[sy,sx] = size(im);
im = imcomplement(im);
im(150:200,100:150) = 0; % let some dots missing!
im = imclearborder(im);
st = regionprops(im, 'Centroid');

imshow(im) hold on;
for j = 1:numel(st)
    px = round(st(j).Centroid(1,1));
    py = round(st(j).Centroid(1,2));
    plot(px,py,'b+')
end

Answer 1

here's a way using fft in 1D over the x and y projection:

First, I'll blur the image a bit to smooth the high freq noise by convolving with a gaussian:

m=double(imread('print5.jpg'));
m=abs(m-max(m(:))); % optional line if you want to look on the black square as "signal"
H=fspecial('gaussian',7,1);
m2=conv2(m,H,'same');

then I'll take the fft of a projection of each axis:

delta=1;
N=size(m,1);
df=1/(N*delta);        % the frequency resolution (df=1/max_T)
f_vector= df*((1:N)-1-N/2);     % frequency vector 

freq_vec=f_vector;
fft_vecx=fftshift(fft(sum(m2)));
fft_vecy=fftshift(fft(sum(m2')));
plot(freq_vec,abs(fft_vecx),freq_vec,abs(fft_vecy))

So we can see both axis yield a peak at 0.07422 which translate to a period of 1/0.07422 pixels or ~ 13.5 pixels.

A better way to get also the angle info is to go 2D, that is:

ml= log( abs( fftshift (fft2(m2)))+1);
imagesc(ml) 
colormap(bone)

and then apply tools such as simple geometry or regionprops if you want, you can get the angle and size of the squares. The size of the square is 1/ the size of the big rotated square on the background ( bit fuzzy because I blurred the image so try to do that without that), and the angle is atan(y/x). The distance between the squares is 1/ the distance between the strong peaks in the center part to the image center.

so if you threshold ml properly image say

 imagesc(ml>11)

you can access the center peaks for that...

yet another approach will be morphological operation on a binary image, for example I threshold the blurred image and shrink objects to points. It removes pixels so that objects without holes shrink to a point:

BW=m2>100;
BW2 = bwmorph(BW,'shrink',Inf);
figure, imshow(BW2)

Then you practically have a one pixel per lattice site grid! so you can feed it to Amro's solution using Hough transform, or analyze it with fft, or fit a block, etc...

Answer 2

You could apply Hough transform to detect the grid lines. Once we have those you can infer the grid locations and the rotation angle:

%# load image, and process it
img = imread('print5.jpg');
img = imfilter(img, fspecial('gaussian',7,1));
BW = imcomplement(im2bw(img));
BW = imclearborder(BW);
BW(150:200,100:150) = 0;    %# simulate a missing chunk!

%# detect dots centers
st = regionprops(BW, 'Centroid');
c = vertcat(st.Centroid);

%# hough transform, detect peaks, then get lines segments
[H,T,R] = hough(BW);
P  = houghpeaks(H, 25);
L = houghlines(BW, T, R, P);

%# show image with overlayed connected components, their centers + detected lines
I = imoverlay(img, BW, [0.9 0.1 0.1]);
imshow(I, 'InitialMag',200, 'Border','tight'), hold on
line(c(:,1), c(:,2), 'LineStyle','none', 'Marker','+', 'Color','b')
for k = 1:length(L)
    xy = [L(k).point1; L(k).point2];
    plot(xy(:,1), xy(:,2), 'g-', 'LineWidth',2);
end
hold off

(I'm using imoverlay function from the File Exchange)

The result:

Here is the accumulator matrix with the peaks corresponding to the lines detected highlighted:

---

Now we can recover the angle of rotation by computing the mean slope of detected lines, filtered to those in one of the two directions (horizontals or verticals):

%# filter lines to extract almost vertical ones
%# Note that theta range is (-90:89), angle = theta + 90
LL = L( abs([L.theta]) < 30 );

%# compute the mean slope of those lines
slopes = vertcat(LL.point2) - vertcat(LL.point1);
slopes = atan2(slopes(:,2),slopes(:,1));
r = mean(slopes);

%# transform image by applying the inverse of the rotation
tform = maketform('affine', [cos(r) sin(r) 0; -sin(r) cos(r) 0; 0 0 1]);
img_align = imtransform(img, fliptform(tform));
imshow(img_align)

Here is the image rotated back so that the grid is aligned with the xy-axes: