{"id":211,"date":"2008-05-13T07:00:08","date_gmt":"2008-05-13T11:00:08","guid":{"rendered":"https:\/\/blogs.mathworks.com\/steve\/2008\/05\/13\/lookup-tables-makelut-applylut\/"},"modified":"2019-10-28T09:31:38","modified_gmt":"2019-10-28T13:31:38","slug":"lookup-tables-makelut-applylut","status":"publish","type":"post","link":"https:\/\/blogs.mathworks.com\/steve\/2008\/05\/13\/lookup-tables-makelut-applylut\/","title":{"rendered":"Lookup tables for binary image processing—makelut and applylut"},"content":{"rendered":"
\r\n

This is the second in a short series on using lookup tables for binary image neighborhood operations. See the first post<\/a> for the basic idea.\r\n <\/p>\r\n

The Image Processing Toolbox has two related functions:<\/p>\r\n

\r\n
    \r\n
  • applylut<\/tt> - Process binary image using lookup table\r\n <\/li>\r\n
  • makelut<\/tt> - Utility function to help you construct a lookup table.\r\n <\/li>\r\n <\/ul>\r\n <\/div>\r\n

    Since you have to have a lookup table in order to use applylut<\/tt>, let's start with makelut<\/tt>. It's syntax is:\r\n <\/p>

     lut = makelut(f, n)<\/pre>
    f<\/tt> is a function handle, and n<\/tt> is either 2 or 3, depending on whether your lookup table is to be used with 2-by-2 or 3-by-3 neighborhoods.\r\n   <\/p>\r\n   
    For 3-by-3 neighborhoods, makelut<\/tt> will call your function handle f<\/tt> 512 times.  In each call, makelut<\/tt> will pass as an argument to f<\/tt> one of the 512 possible 3-by-3 binary neighborhoods. f<\/tt> simply has to return the desired output value for a given neighborhood.\r\n   <\/p>\r\n   
    For example, suppose we want the output pixel to be 1 if and only if the corresponding input neighborhood looks like this:<\/p>
      1 0 0\r\n  0 1 0\r\n  0 0 1<\/pre>
    The you would call makelut<\/tt> as follows:\r\n   <\/p>
    pattern = [1 0 0; 0 1 0; 0 0 1];\r\nlut = makelut(@(neighborhood) isequal(neighborhood, pattern), 3);\r\nsize(lut)<\/pre>
    \r\nans =\r\n\r\n   512     1\r\n\r\n<\/pre>
    Let's try that on a small sample image.<\/p>
    bw = eye(5) | fliplr(eye(5))<\/pre>
    \r\nbw =\r\n\r\n     1     0     0     0     1\r\n     0     1     0     1     0\r\n     0     0     1     0     0\r\n     0     1     0     1     0\r\n     1     0     0     0     1\r\n\r\n<\/pre>
    bw2 = applylut(bw, lut)<\/pre>
    \r\nbw2 =\r\n\r\n     0     0     0     0     0\r\n     0     1     0     0     0\r\n     0     0     0     0     0\r\n     0     0     0     1     0\r\n     0     0     0     0     0\r\n\r\n<\/pre>
    So we see that only two input pixels, (2,2) and (4,4), have neighborhoods that exactly match the pattern.<\/p>\r\n   
    Here's another example: Find all \"endpoint\" pixels.  I'll define an endpoint pixel as a foreground pixel with exactly one\r\n      foreground neighbor.\r\n   <\/p>
    endpoint_fcn = @(nhood) (nhood(2,2) ~= 0) && (sum(nhood(:)) == 2);\r\nendpoint_lut = makelut(endpoint_fcn, 3);<\/pre>
    Here's the result on sample image from above:<\/p>
    bw3 = applylut(bw, endpoint_lut)<\/pre>
    \r\nbw3 =\r\n\r\n     1     0     0     0     1\r\n     0     0     0     0     0\r\n     0     0     0     0     0\r\n     0     0     0     0     0\r\n     1     0     0     0     1\r\n\r\n<\/pre>
    Only the four corner pixels are endpoints.<\/p>\r\n   
    In my next post on this topic, I'll use makelut<\/tt> and applylut<\/tt> to play Conway's Game of Life.  Look for it in a week or so.\r\n   <\/p>