Conversion of a binary matrix(256*256) into another matrix which is decimal(85*85)?
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We have a 256*256 binary matrix , from this we take 3*3 matrix which keeps on increasing firstly row wise upto the end point and then 3 units column wise and then upto the end point , this process continues upto the last pixel . now in the 3*3 matrix we start from (1,1) and move clockwise untill we reach (2,1) storing all the values in an array and converting this binary string into decimal value and storing it into a different matrix.
3 Comments
Walter Roberson
on 8 Dec 2012
1-3, 4-6, 7-9, 10-12, 13-15, 16-18, 19-21, 22-24, 25-27, 28-30, 31-33, 34-36, 37-39, 40-42, 43-45, 46-48, 49-51, 52-54, 55-57, 58-60, 61-63, 64-66, 67-69, 70-72, 73-75, 76-78, 79-81, 82-84, 85-87, 88-90, 91-93, 94-96, 97-99, [....] 238-240, 241-243, 244-246, 247-249, 250-252, 253-255, 256 - ????
Answers (1)
Matt J
on 6 Dec 2012
Edited: Matt J
on 6 Dec 2012
Assuming the binary matrix is 255x255, you could do it the following way, where I use MAT2TILES from the File Exchange.
A=rand(255)>.5; %fake binary input image
C=mat2tiles(char(A+'0'),[3,3]);
idx=[1 4 7 8 9 6 3 2];
B=cellfun(@(c)c(idx),C,'uni',0);
B=bin2dec(vertcat(B{:}));
result=reshape(B,size(C));
2 Comments
Matt J
on 8 Dec 2012
Edited: Matt J
on 8 Dec 2012
I'm not sure how that's relevant to the solution I gave you, or why it's even an issue. The solution I gave you performs all steps, including saving the result to another matrix.
Also, since you know how to create matrices, (e.g. the 256x256 matrix you have now) and to pull data out of them (e.g. 3x3 blocks), it's not clear what difficulty you're having putting data back into one.
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