# Find sorrunding elements and element from an array

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Matlab on 26 Oct 2020 at 16:19
Commented: Matlab on 19 Nov 2020 at 10:15
I have an array
y = [
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0]
where Index 7,41,75 are the locations where 1 is found .
My requirement is
• create a block around true(1) with a size of 5
• get the indices like 5,6,7,8,9 and data 0 0 1 0 0

Matlab on 28 Oct 2020 at 3:20
You have A_idx = [5 6 7 8 9] but expectation is A'_idx = [7 8 9 6 5]
You have A_idx = [0 0 1 0 1] but expectation is A'_bits = [1 0 1 0 0]
Rules:
• The first / every row must start with error bit location followed by rest of bit. i.e. leading coefficeint of a the polynomial must be 1
• Tail bits of a polynomial can be any 0/1 in a block size( we get in the array). Bitreversal is done for 5,6 to 6,5
• Once a error bit is processed in a block , we don't need to processs the bit again. Block window should slide further in search of next index / error bit to make a polynomial of fixed size
• Minimum polynomial size i.e. block of 5 is reasonable .
• polynomial size should not change during the execution.
Thank you!
Matlab on 8 Nov 2020 at 16:34
Matlab on 19 Nov 2020 at 10:15

Image Analyst on 26 Oct 2020 at 17:25
Try this:
fprintf('Beginning to run %s.m ...\n', mfilename);
y = [
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0, ...
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0, ...
0 0 0 0 0 0 0 0 0]
yMax = movmax(y, 5)
props = regionprops(yMax > 0, 'PixelIdxList') % Requires the Image Processing Toolbox.
for k = 1 : length(props)
fprintf('\nFor block #%d, indexes = ', k);
indexes{k} = props(k).PixelIdxList;
fprintf('%d ', indexes{k});
end
fprintf('\nDone running %s.m ...\n', mfilename);
You'll see:
For block #1, indexes = 5 6 7 8 9
For block #2, indexes = 39 40 41 42 43
For block #3, indexes = 73 74 75 76 77

Matlab on 29 Oct 2020 at 4:07
Image Analyst - Any help here ?
Image Analyst on 29 Oct 2020 at 13:30
You said "Also please keep "bwareaopen" function c code translation, coder support is needed for c/c++ code generation, otherwise it is pretty difficult to implement on a 32 bit processor." so I was assuming that you needed the whole program to be converted to C code with the Coder Toolbox because you were going to embed this algorithm in a custom chip on some device. If you don't need Coder support and are not going to generate C code, then I don't know why you said that. Please elaboarate. Also, I don't know that Coder can generate code for 32 bit processors. I know MATLAB stopped supporting 32 bit processors with version 2016b.
If you want
For block #1, indexes = 7 8 9 6 5
For block #2, indexes = 41 42 43 40 39
For block #3, indexes = 75 76 77 74 73
then you can just tack on the first two indexes to the end, like this:
fprintf('Beginning to run %s.m ...\n', mfilename);
y = [
0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0, ...
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0, ...
0 0 0 0 0 0 0 0 0]
windowWidth = 5;
halfWindowWidth = floor(windowWidth/2);
yMax = movmax(y, windowWidth) > 0; % A logical vector.
yMax = bwareaopen(yMax, windowWidth); % Keep runs of 5 or longer ONLY.
props = regionprops(yMax, 'PixelIdxList') % Requires the Image Processing Toolbox.
for k = 1 : length(props)
fprintf('\nFor block #%d, indexes = ', k);
theseIndexes = props(k).PixelIdxList';
blockIndexes{k} = [theseIndexes(end-halfWindowWidth : end), fliplr(theseIndexes(1:halfWindowWidth))];
fprintf('%d ', blockIndexes{k});
end
fprintf('\nDone running %s.m ...\n', mfilename);
It prints out:
For block #1, indexes = 7 8 9 6 5
For block #2, indexes = 41 42 43 40 39
For block #3, indexes = 75 76 77 74 73
Matlab on 8 Nov 2020 at 16:33
Thank you! I'II update further with my comments on your queries with additional information. For the time being this is sufficient