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How to generate a random number of n bits length?

Asked by Mohsin Shah on 6 Sep 2018
Latest activity Commented on by Mohsin Shah on 6 Sep 2018
Let's say a function takes input n as the bit length and outputs a random number. For example, for n = 4, the output range is [0 - 15] which also includes the range for n = 3 i.e. [0 - 7]. The function should generate the number in the range [8 - 15] and not in the range [0 - 15]. How to do this and how to generalize it for any n?

  3 Comments

The function should generate the number in the range [8 - 15] and not in the range [0 - 15]. How to do this and how to generalize it for any n?
Can you clarify this statement? if n=3 then ??
What I am getting you want three-bit representation(when n=3) in the range [8 - 15]??
For n = 3 then the output range is [0-7] which includes the range [0-3] which is for n = 2
@Mohsin Check my answer, as per your requirement? If any modification required, let me know.

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2 Answers

Answer by Guillaume
on 6 Sep 2018
 Accepted Answer

For n = 4 it should generate a number in the range [8-15]
If I understand correctly, you want a 4 bit random number with the MSB always set to 1. That sounds like a strange requirement, probably not thought properly, but this is simply equivalent to generating a 3 bit random number and tacking 1 as the MSB. The generic version of that (generating a n-1 bit number + MSB of 1) is:
n = 4
number = randi([1, 2^(n-1)]) + 2^(n-1) - 1

  1 Comment

It worked for me. Thank you.

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Answer by KALYAN ACHARJYA on 6 Sep 2018
Edited by KALYAN ACHARJYA on 6 Sep 2018

function bit_result=rand_bits(n)
result=randi([2^(n-1), 2^n]);
bit_result=de2bi(result,n)
end

  3 Comments

For n = 4, it generates a number in the range [0-7] which is not required. For n = 4 it should generate a number in the range [8-15].
@Mohsin I have edited the answer, can you verify it?
I appreciate your efforts but the edited function still doesn't work. Check the answer of Guillaume which worked for me.

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