如何在bash中获得0到999999999之间的随机（/ dev / random）数字？

Question

For example,

echo "$(( ($(od -An -N2 -i /dev/random) )%(1000-0+1) ))"

can be used. But it doesn't scale for bigger numbers.

How can I get a random (/dev/random) number between 0 and 999999999 in bash?

Answer 1

I'd use:

shuf -i0-999999999 -n1

although there is no guarantee that shuf uses /dev/random . With GNU shuf you can specify --random-source=/dev/random if you really want to.

IIRC, FreeBSD (and probably Mac OS X) call this utility shuffle , and it takes slightly different arguments ( shuffle -n1000000000 -p1 ).

If you really want to use /dev/random directly, you can generate a four-byte number by using od -An -N4 -tu4 but remember that there is a bias generated by using %1000000000 , since 2 ³² is not divisible by 1000000000 . To correct for that, in the particular case of generating random numbers in the range 0-999999999 , you need to reject the four-byte random number if it is greater than or equal to 4000000000 .

Answer 2

Explenation

The key to doing this using /dev/random is thinking about what produces a uniform distribution. /dev/random does, but doing

 od -An -N2 -i /dev/random

Provides a 2-byte output, which is uniform over 0...2 ¹⁶ -1 or 0...65535. But obviously not uniform over 0...99999.

Taking it mod 1000, will change the output to 0...999, but the distribution still won't be uniform over it.

Solution

Let's say we have a pseudo-random generator that produces 0,1,2,3,4, but we want one that produces 0,1. We just ignore any output 2,3,4 and keep outputs 0,1 which by definition must still be pseudo-random.

So, you can do something like this.

To get a single digit you can do something like this

while :; do ran=$(echo $(od -An -N1 -i /dev/random)) && [[ $ran -lt 250 ]] && \
echo ${ran: -1} && break; done

since the distribution of the least significant digit between 0...249 is uniform.

Then to build a 9 digit random number, just do something like

#!/bin/bash

digit() {
  while :; do ran=$(echo $(od -An -N1 -i /dev/random)) && [[ $ran -lt 250 ]] && \
  echo ${ran: -1} && break; done
}

for ((i=0; i<9; i++)) {
  num+=$(digit)
}

echo $num

Answer 3

你可以使用awk ，它几乎可以在任何类似UNIX的盒子上使用：

awk 'BEGIN{print int(rand()*999999999)}'

Answer 4

 cat /dev/random | tr -cd 0-9 | dd bs=1 count=9

Cat random data from /dev/random , delete all except 0 , 1 , ... 9 ; then wait only for 9 digits with dd.