How binary Random number generation upto n nodes ?
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i was do this
clc;
clear all;
close all;
% Link is (n-1)*node/2
node = 3;
offset = 1;
L = (node-offset)*node/2;
c =2^L;
% Generate c binary values:
r = randi([0, 1], c,L);
disp(r);
output:
0 0 1
1 0 0
1 0 0
1 0 1
1 1 0
1 0 0
0 1 0
0 0 1
but i want
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
please help me how to generate binary random numbers up to n in order wise
1 个评论
James Tursa
2021-3-15
Your current code generates random patterns. But what you want is not random at all ... it is an ordered list. Why do you write you want random output but then show a non-random ordered list as what you want?
采纳的回答
James Tursa
2021-3-15
Here is how to get your stated desired output, but this is not random at all:
r = dec2bin(0:(2^L-1)) - '0';
4 个评论
ankanna
2021-3-15
James Tursa
2021-3-15
编辑：James Tursa
2021-3-15
The memory requirements for generating the complete list of patterns can easily exceed the amount of memory that your computer has. This method is only practical for small numbers. I could suggest that you use logical variables instead of double, but even then if you have such a large list how do you intend to use this downstream in your code? Depending on how involved your downstream processing is, it might take days or weeks or months or years to process all of the individual patterns. You will need a different approach to solving your problem if you have to work with large numbers.
ankanna
2021-4-24
Jan
2021-4-24
This is a new question. Please ask it in a new thread.
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