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How to generate a a restricted binary number

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Mohsen
Mohsen am 5 Dez. 2018
Kommentiert: Mohsen am 6 Dez. 2018
Dear friends
I would like to generate a random number include 24 bits 4 bits of 1 and 20 bits of 0. Can you help me?
Best Regards
Mohsen

Akzeptierte Antwort

Jan
Jan am 5 Dez. 2018
Bearbeitet: Jan am 5 Dez. 2018
What exactly is a "number include 24 bits"?
This sets 4 elements to 1 and 20 to 0:
v = zeros(1, 24);
v(randperm(24, 4)) = 1;
What is the wanted output?
result = bin2dec('0' + v)
% or
result = v * power(2, 0:23).'
Instead of creating the sum using the dot product, you can use the indices replied by randperm directly:
result = sum(2 .^ (randperm(24, 4) - 1));
I hope this was not a homework question. Otherwise it will be hard for you to submit your own solution now. But you participate in this forum for 6 years now, such that I assume that you are not in the learning phase anymore.

Weitere Antworten (1)

Mohsen
Mohsen am 5 Dez. 2018
Bearbeitet: Mohsen am 5 Dez. 2018
Exuce me,
There is a problem. In fact, i have a optimization algorithm and the input dimention and max and min. must be determined.
a posible state can be:
[0 1 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]
or
[0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0]
or...
there are four swing "1"s in this vector and total states is equal 10626(=). the vector length is 24. How do i define number dimention, max and min to derive this array?
  2 Kommentare
Jan
Jan am 6 Dez. 2018
Bearbeitet: Jan am 6 Dez. 2018
I do not understand what "number dimension, max and min" means. Maximum and minimum of what? My answer includes a method to create random vectors already:
v = zeros(1, 24);
v(randperm(24, 4)) = 1;
Do you now want to create all these vectors? Then:
M = nchoosek(1:24, 4);
for k = 1:size(M, 1)
v = zeros(1, 24);
v(M(k, :)) = 1;
...
end
Mohsen
Mohsen am 6 Dez. 2018
I get a new idea by your post.
Thank you very much

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