Matlab：二进制图像打开到最小矩形大小

Question

I have a binary image:

and I'm trying to see if a 100x150 rectangle will fit in any blank space on the map. I tried creating a rectangular strel & then eroding & dilating the picture to get rid of any areas smaller than needed:

se = strel('rectangle',[150, 100]);
BW = imerode(BW,se);
BW = imdilate(BW,se);

Unfortunately, it finds a hole prematurely

which is only 80x150. I think the erosion is failing since it's against the wall & only needs half the width, but don't know how to fix it.

Also, if I'm headed down the wrong path, please feel free to set me straight. Ultimately, I just need to find the upper-left corner of the a blank space at least as big as 100x150.

Answer 1

The approach below works quite well and runs fairly quickly. It has a few nested loops but you can probably further optimize the performance, I mainly just wanted to get it working for you. Keep in mind though that if you comment out the fprintf() and the plotting command, that will speed things up.

I downloaded your image from your Stack post but I believe that my downloaded version has a different size (398x398) than the raw data you are working with, so keep that in mind when viewing my results below.

As indicated in the code, you supply the width (w) and height (h), the algorithm then returns all of the (col, row) positions where the rectangle can fit.

Side Note: I believe this provides a solution to a 2D version of the Bin packing problem , but I'm not sure about that, you can check out the link above if thats of interest to you.

Either way, its a great example of computational problem where an exhaustive search can be a carried out rather quickly.

To verify the results, I added simple plotting of the rectangles. Keep in mind that if the rectangle fits in more than one position, a plot of the multiple rectangles begins to look rather jumbled as they are drawn repeatedly on top of one another (with offsets).

As an example case where only a single rectangle is found, I use: w = 29; h = 102; w = 29; h = 102; and then the result shows that the only position where this particular rectangle can fit, has the upper left corner = (row = 295, col = 368) (this rectangle size will likely only work for my downloaded version of your data):

In summary, I first I load the data and then convert to a binary map (0's and 1's):

% Note: '0' = black; '1' = white
data = round(im2double(rgb2gray(imread(filepath))));
figure(1);imshow(data); set(gcf,'Color',[1 1 1]);
hold on;

Input the search width and height:

w = 29;
h = 102;

sze = size(data);
numRows = sze(1);
numCols = sze(2);

Next we just do a search to see what will fit at each row and col position:

for col = 1:numCols - w - 1
    for row = 1:numRows - h - 1
        doesFit = fitshere(data, row,col, w, h);
        if (doesFit == 1)
            fprintf('row = %d; col = %d \n',row,col);
            colX = [col col+w col+w col col];
            colY = [row row row+h row+h row];
            line(colX,colY,'Color','r','linewidth',2);
        end
    end
end

hold off;

You will need the following function to check if a given rectangle can fit in the array:

function [val] = fitshere(data, row, col, w, h)
val = 1;
for i = col:col + w
    for j = row:row + h
        if (data(j,i) == 0) % if this is true, we are in the black!
            val = 0;
            return;
        end
    end
end
return;

If your interested in knowing if your rectangle will fit at all (say either width X height or height X width ), you can simply repeat the search after swapping the width and height.

Hope this helps.

Answer 2

lets do this with some matlab idioms

M=binaryImage;
sz=size(M);
nrows = 100;
ncols = 150 ;

colsum = cumsum(M,1);
cols_are_good = colsum(nrows+1:end,:)-colsum(1:end-nrows+1,:)==0; 
    %  nrows empty rows below this point. in this column
rows_are_also_good = cols_are_good(:,ncols+1:end)-cols_are_good(:,1:end+1-ncols)==0;

and Bob's your uncle, that last variable contains 1 in all places that have nrows below them clear and each of those has ncols to the side