{"id":52,"date":"2006-04-10T07:00:15","date_gmt":"2006-04-10T11:00:15","guid":{"rendered":"https:\/\/blogs.mathworks.com\/steve\/?p=52"},"modified":"2019-10-22T09:14:04","modified_gmt":"2019-10-22T13:14:04","slug":"quick-tip-determining-uniqueness-of-neighborhood-maximum","status":"publish","type":"post","link":"https:\/\/blogs.mathworks.com\/steve\/2006\/04\/10\/quick-tip-determining-uniqueness-of-neighborhood-maximum\/","title":{"rendered":"Quick tip: Determining uniqueness of local maximum"},"content":{"rendered":"
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A MATLAB user asked me how to determine which image pixels had local maxima that weren't unique. My answer involves a creative\r\n use of ordfilt2<\/tt>.\r\n <\/p>\r\n

Let's review first the concept of a local maximum filter. This image operator replaces every pixel value with the maximum\r\n pixel value found in a neighborhood surrounding the pixel. For example:\r\n <\/p>

a = magic(5)<\/pre>
\r\na =\r\n\r\n    17    24     1     8    15\r\n    23     5     7    14    16\r\n     4     6    13    20    22\r\n    10    12    19    21     3\r\n    11    18    25     2     9\r\n\r\n<\/pre>
What should the output be for a local maximum filter using 3-by-3 neighborhoods?  The output in the 3rd row, 2nd column would\r\n      be the maximum value found in this submatrix:\r\n   <\/p>
neighborhood = a(2:4, 1:3)<\/pre>
\r\nneighborhood =\r\n\r\n    23     5     7\r\n     4     6    13\r\n    10    12    19\r\n\r\n<\/pre>
The (3,2) output is 23:<\/p>
max(neighborhood(:))<\/pre>
\r\nans =\r\n\r\n    23\r\n\r\n<\/pre>
Local maximum filtering is the same as morphological dilation, so I usually use imdilate<\/tt> to compute the result for the entire image:\r\n   <\/p>
b = imdilate(a, ones(3,3))<\/pre>
\r\nb =\r\n\r\n    24    24    24    16    16\r\n    24    24    24    22    22\r\n    23    23    21    22    22\r\n    18    25    25    25    22\r\n    18    25    25    25    21\r\n\r\n<\/pre>
That's the background; now back to the original question.  How do you determine which pixels have multiple maxima in their\r\n      neighborhoods?\r\n   <\/p>\r\n   
I suggest using ordfilt2<\/tt> twice.  This function is for ''two-dimensional order-statistic filtering.''  This operation replaces every pixel value with\r\n      the n-th sorted pixel value in its neighborhood.  It uses an ascending sort order, so if n is the number of pixels in the\r\n      neighborhood, the operation is exactly equivalent to the local maximum filter.\r\n   <\/p>\r\n   
The specific procedure is:<\/p>\r\n   
1. Call ordfilt2<\/tt> to get the highest pixel values in each neighborhood.\r\n   <\/p>\r\n   
2. Call ordfilt2<\/tt> again to get the second-highest pixel values in each neighborhood.\r\n   <\/p>\r\n   
3. Determine where the outputs of steps 1 and 2 are equal.  Wherever they are equal, the local maximum is not unique.<\/p>\r\n   
Let's try an example.  First, modify our original matrix a bit:<\/p>
ap = a;\r\nap(3,4) = 22<\/pre>
\r\nap =\r\n\r\n    17    24     1     8    15\r\n    23     5     7    14    16\r\n     4     6    13    22    22\r\n    10    12    19    21     3\r\n    11    18    25     2     9\r\n\r\n<\/pre>
Get the highest pixel values in each 3-by-3 neighborhood:<\/p>
highest = ordfilt2(ap, 9, ones(3,3))<\/pre>
\r\nhighest =\r\n\r\n    24    24    24    16    16\r\n    24    24    24    22    22\r\n    23    23    22    22    22\r\n    18    25    25    25    22\r\n    18    25    25    25    21\r\n\r\n<\/pre>
Get the second-highest pixel values:<\/p>
second_highest = ordfilt2(ap, 8, ones(3,3))<\/pre>
\r\nsecond_highest =\r\n\r\n    23    23    14    15    15\r\n    23    23    22    22    22\r\n    12    19    21    22    22\r\n    12    19    22    22    22\r\n    12    19    21    21     9\r\n\r\n<\/pre>
Determine where the highest and second highest pixel values are equal:<\/p>
non_unique_maxima_mask = (highest == second_highest)<\/pre>
\r\nnon_unique_maxima_mask =\r\n\r\n     0     0     0     0     0\r\n     0     0     0     1     1\r\n     0     0     0     1     1\r\n     0     0     0     0     1\r\n     0     0     0     0     0\r\n\r\n<\/pre>