Thanks to Judd for excellent examples.
In addition, putting an extra layer of pixels around the image is a good
approach to simplify the coding, too. I used to put one layer of object (i.e.
1s), including corner pixels. This ensures that there will be no erosion from
edges. This has the same effect to duplicate the edge rows/columns as in
Judd's example. New corner pixels need to be 1s also, to avoid erosion from
corners in case of using a "ones(3)" structure element.
For gray image operation, I think that putting one layer of pixels with highest
intensity would do the same.
Thanks
Li-Ping (Li) Yuan
________________________________
From: Judd Storrs <jsto...@gmail.com>
To: Søren Hauberg <so...@hauberg.org>
Cc: octave-dev@lists.sourceforge.net; lipingy...@yahoo.com
Sent: Thu, June 24, 2010 9:25:10 AM
Subject: Re: [OctDev] imerode erodes from edges in Octave but not Matlab.
Here's another example:
Evil:
>> A = ones(5) ;
>> A(2,1) = 0 ;
>> A(5,5) = 0
A =
1 1 1 1 1
0 1 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 1 0
>> imerode(A,[0,0,0;1,1,1;0,0,0])
ans =
1 1 1 1 1
0 0 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 0 0
In octave we get:
> imerode(A,[0,0,0;1,1,1;0,0,0])
ans =
0 1 1 1 0
0 0 1 1 0
0 1 1 1 0
0 1 1 1 0
0 1 1 0 0
I think Evil just replicates the boundary values but I'm not entirely
sure what happens to the corners. The correct result can be had in
octave by:
B = ones(7) ;
B(2:6,2:6) = A ;
B(1,2:6) = A(1,:) ;
B(7,2:6) = A(5,:) ;
B(2:6,1) = A(:,1) ;
B(2:6,7) = A(:,5)
B =
0 1 1 1 1 1 0
1 1 1 1 1 1 1
0 0 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 1 1
1 1 1 1 1 0 0
0 1 1 1 1 0 0
imerode(B,[0,0,0;1,1,1;0,0,0])(2:6,2:6)
ans =
1 1 1 1 1
0 0 1 1 1
1 1 1 1 1
1 1 1 1 1
1 1 1 0 0
But I'm not sure what values goes in B(1,1), B(1,7), B(7,1), B(7,7).
--judd
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