Create binary matrix in Matlab reporting increasing number of ones

Question

I would like your advise to write a Matlab code that creates a binary matrix A of size 31x5 such that

the first row of A is [1 1 1 1 1]

from the 2nd to the 6th of A we have 1 only once per row

from the 7th to the 16th row we have 1 twice per row
```
[1 1 0 0 0
 1 0 1 0 0
 1 0 0 1 0
 ...]
```
from the 17th to the 26th row we have 1 three times per row
from the 26th to the 31th row we have 1 four times per row

I could that manually, but I would like to know if there is a faster way to proceed.

Luis Mendo · Accepted Answer · 2020-07-27 10:28:15Z

3

Here's an approach:

Generate all possible rows containing zeros and ones, except all zeros or all ones;
Sort rows (atomically) based on row sums, then on negated row values, to produce the desired order;
Prepend a row of ones to build the result.

N = 5;
A = dec2bin(1:2^N-2)-'0'; % step 1
[~, ind] = sortrows([sum(A,2) -A]); % step 2
result = [ones(1,N); A(ind,:)]; % step 3

answered Jul 27, 2020 at 10:28

Luis Mendo

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Comments

Zaragesh · Accepted Answer · 2020-07-27 10:30:05Z

0

I recommend you the command bin2dec()/dec2bin() with a for-loop. bin2dec('10000') writes 16 and your next writes bin2dec('01000') 8 so I guess you are following some sort of pattern?

put all of your wanted numbers in an array like that:

clear all;
nums = [16 8 4 2 0];
mat = [];
for(a=1:1:size(nums,2))
    mBinChar = dec2bin(nums(a));
    
    for(b=1:1:length(mBinChar))
        mat(a,b) = str2double(mBinChar(1));
    end
end

answered Jul 27, 2020 at 10:30

Zaragesh

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2 Comments

Luis Mendo Over a year ago

This doesn't seem to produce the desired result

Zaragesh Over a year ago

of course only under the pretense that his numbers follow a certain pattern

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