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Is it possible to detect a specefic form of matrices from a big one?

2 vues (au cours des 30 derniers jours)
mika
mika le 21 Nov 2013
Commenté : mika le 21 Avr 2014
Hello, I need to find those 5 forms of matrices from a binary 'big' matrice
X=[1 0 1 ;
0 1 0 ;
1 0 1 ;];
Y=[1 0 1 ;
0 1 0 ;
1 0 0 ;];
O=[0 1 0 ;
1 0 1 ;
0 1 0 ;];
-=[0 0 0 ;
1 1 1 ;
0 0 0 ;];
L=[1 0 0 ;
1 0 0 ;
1 1 1 ;];
Any idea? help is much appreciated
  3 commentaires
mika
mika le 21 Nov 2013
it's 1000*1001
talha
talha le 21 Nov 2013
how to connect blocks of two different library together ?

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Réponse acceptée

Cedric
Cedric le 21 Nov 2013
Modifié(e) : Cedric le 22 Nov 2013
EDITED: got rid of a side effect in the dash case (illustrated in the 4th picture below.. I didn't update it though) by using 2*A-1 instead of A and numel(X) instead of nnz(X).
Someone might provide you with a solution based on the Image Processing Toolbox (which I am not familiar with). Here is one potential solution and a hint in the meantime..
1. Based on CONV2
A = double( rand(10) > 0.4 ) ; % Small example.
X = [1 0 1 ; ...
0 1 0 ; ...
1 0 1 ] ;
ker = rot90(rot90( 2*X -1 )) ;
[r, c] = find( conv2(2*A-1, ker, 'same') == numel(X) ) ;
With that and a few runs (need a little luck for RAND to output one or more fillings with matches):
>> A
A =
1 1 1 1 1 1 0 1 0 1
0 0 0 1 1 0 1 1 0 1
1 0 0 0 0 1 0 1 0 1
1 0 1 0 1 0 1 1 1 1
0 1 0 1 0 1 1 1 1 0
0 1 1 0 1 0 0 1 1 0
0 1 1 0 1 1 0 0 0 1
0 1 0 0 1 0 1 1 1 1
0 1 0 1 1 1 1 0 1 1
1 1 0 1 0 1 1 1 0 0
>> [r,c]
ans =
5 4
3 6
which indicates two matches, centered in (5,4) and (3,6). This approach takes ~13ms on my laptop with a 1000x1001 A matrix.
2. By looping over elements of patterns and vectorizing tests on A rather than the opposite.
.. I leave that for later if CONV2 is not suitable.
--------------------------------------------------------------------------------------------------------------------------------
EDIT: here is a more elaborate example..
n = 50 ;
A = double( rand(n, n+1) > 0.5 ) ;
B = 2*A - 1 ;
patterns = {[1 0 1; 0 1 0; 1 0 1]; ... % X
[1 0 1; 0 1 0; 1 0 0]; ... % Y
[0 1 0; 1 0 1; 0 1 0]; ... % O
[0 0 0; 1 1 1; 0 0 0]; ... % -
[1 0 0; 1 0 0; 1 1 1]} ; % L
labels = {'X', 'Y', 'O', '-', 'L'} ;
matches = cell( size(labels )) ;
for pId = 1 : numel( patterns )
nel = numel( patterns{pId} ) ;
ker = 2*patterns{pId} - 1 ;
[r, c] = find( conv2(B, rot90(rot90(ker)), 'same') == nel ) ;
matches{pId} = [r, c] ;
figure(pId) ; clf ; hold on ;
spy(A) ;
plot( c, r, 'r+', 'MarkerSize', 20 ) ;
title( sprintf( 'Matches for "%s" pattern', labels{pId} )) ;
set( gcf, 'Units', 'normalized' ) ;
set( gcf, 'Position', [0 0 0.4 0.7] ) ;
end
which outputs the following figures..
  6 commentaires
Cedric
Cedric le 28 Nov 2013
Modifié(e) : Cedric le 28 Nov 2013
Ok, this is quite different from what the initial question suggested. I would advise you to re-post the question including images of the skeleton and zooms over regions which represent the different patterns. You should also attach the source array or image so people can perform tests.
mika
mika le 13 Jan 2014
Hello,
Can you tell me please how to make the output into the same plot. i'm trying to make the result as one plot but in vain can you help me please.
thank you.

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Plus de réponses (2)

talha
talha le 21 Nov 2013
how to connect blocks of two different library together ?
  2 commentaires
Cedric
Cedric le 21 Nov 2013
Please post a new question and provide more details about your issue.
mika
mika le 21 Avr 2014
okk sorry for late reply
i' m still looking for the solution

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Image Analyst
Image Analyst le 26 Nov 2013
Try normalized cross correlation (see attached demo below in blue text), or the hit or miss transform, done by bwhitmiss() in the Image Processing Toolbox.

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