Random number from a range in a Bash Script

Question

I need to generate a random port number between 2000-65000 from a shell script  The problem is  RANDOM is a 15-bit number  so I m stuck   PORT     RANDOM 63000 2001   would work nicely if it wasn t for the size limitation   Does anyone have an example of how I can do this  maybe by extracting something from  dev urandom and getting it within a range

User · Answer

This is how I usually generate random numbers  Then I use  NUM 1  as the variable for the port number I use  Here is a short example script      bin bash  clear echo  Choose how many digits you want for port   1-5   read PORT  NUM 1    tr -dc  0-9   lt  dev urandom   head -c  PORT    echo   NUM 1   if     PORT  -gt  5    then clear echo -e   x1b 31m Choose a number between 1 and 5   x1b 0m  sleep 3 clear exit 0 fi

User · Answer

You can do this  cat  dev urandom od -N2 -An -i awk -v f 2000 -v r 65000   printf   i n   f   r    1   65536     If you need more details see Shell Script Random Number Generator

User · Answer

Here s another one   I thought it would work on just about anything  but sort s random option isn t available on my centos box at work    seq 2000 65000   sort -R   head -n 1

User · Answer

PORT     RANDOM 63000 2001   is close to what you want I think   PORT     RANDOM RANDOM RANDOM 63000 2001   gets around the size limitation that troubles you   Since bash makes no distinctions between a number variable and a string variable  this works perfectly well  The  number   RANDOM can be concatenated like a string  and then used as a number in a calculation   Amazing

User · Answer

You can get the random number through urandom  head -200  dev urandom   cksum  Output   3310670062 52870  To retrieve the one part of the above number   head -200  dev urandom   cksum   cut -f1 -d      Then the output is  3310670062  To meet your requirement   head -200  dev urandom  cksum   cut -f1 -d       awk   print  1 63000 2001

User · Answer

Same with ruby   echo   ruby -e  puts rand 20  65       gt  65  inclusive ending  echo   ruby -e  puts rand 20   65       gt  37  exclusive ending

User · Answer

Or on OS-X the following works for me     gsort --random-sort

User · Answer

shuf -i 2000-65000 -n 1   Enjoy   Edit  The range is inclusive

User · Answer

RANDOM is a number between 0 and 32767  You want a port between 2000 and 65000  These are 63001 possible ports  If we stick to values of  RANDOM   2000 between 2000 and 33500  we cover a range of 31501 ports  If we flip a coin and then conditionally add 31501 to the result  we can get more ports  from 33501 to 65001  Then if we just drop 65001  we get the exact coverage needed  with a uniform probability distribution for all ports  it seems   random-port         while    not    found     do           2000  33500         port    RANDOM   2000           while     port -gt 33500     do             port    RANDOM   2000           done            2000  65001               RANDOM   2     0     amp  amp  port    port   31501               2000  65000             port   65001     amp  amp  continue         echo  port         break     done     Testing  i 0 while true  do     i    i   1       printf   rIteration  i         printf   05d n    random-port   gt  gt  ports txt done    Then later we check the distribution sort ports txt   uniq -c   sort -r

User · Answer

According to the bash man page   RANDOM is distributed between 0 and 32767  that is  it is an unsigned 15-bit value   Assuming  RANDOM is uniformly distributed  you can create a uniformly-distributed unsigned 30-bit integer as follows       RANDOM lt  lt 15  RANDOM     Since your range is not a power of 2  a simple modulo operation will only almost give you a uniform distribution  but with a 30-bit input range and a less-than-16-bit output range  as you have in your case  this should really be close enough   PORT       RANDOM lt  lt 15  RANDOM    63001   2000

User · Answer

On Mac OS X and FreeBSD you may also use jot   jot -r 1  2000 65000

User · Answer

The simplest general way that comes to mind is a perl one-liner   perl -e  print int rand 65000-2000     2000    You could always just use two numbers   PORT     RANDOM     RANDOM   2    32768     You still have to clip to your range  It s not a general n-bit random number method  but it ll work for your case  and it s all inside bash   If you want to be really cute and read from  dev urandom  you could do this   od -A n -N 2 -t u2  dev urandom   That ll read two bytes and print them as an unsigned int  you still have to do your clipping

User · Answer

If you re not a bash expert and were looking to get this into a variable in a Linux-based bash script  try this   VAR   shuf -i 200-700 -n 1   That gets you the range of 200 to 700 into  VAR  inclusive

User · Answer

and here s one with Python  randport   python -S -c  import random  print random randrange 2000 63000      and one with awk  awk  BEGIN srand   print int rand    63000-2000   2000

User · Answer

Bash documentation says that every time  RANDOM is referenced  a random number between 0 and 32767 is returned   If we sum two consecutive references  we get values from 0 to 65534  which covers the desired range of 63001 possibilities for a random number between 2000 and 65000   To adjust it to the exact range  we use the sum modulo 63001  which will give us a value from 0 to 63000  This in turn just needs an increment by 2000 to provide the desired random number  between 2000 and 65000  This can be summarized as follows   port      RANDOM   RANDOM    63001    2000     Testing    Generate random numbers and print the lowest and greatest found test-random-max-min         max 2000     min 65000     for i in  1  10000   do         port      RANDOM   RANDOM    63001    2000           echo -en   r port               port  -gt   max      amp  amp  max   port               port  -lt   min      amp  amp  min   port      done     echo -e   rMax   max  min   min       Sample output   Max  64990  min  2002   Max  65000  min  2004   Max  64970  min  2000   Correctness of the calculation  Here is a full  brute-force test for the correctness of the calculation  This program just tries to generate all 63001 different possibilities randomly  using the calculation under test  The --jobs parameter should make it run faster  but it s not deterministic  total of possibilities generated may be lower than 63001    test-all         start   date   s      find start   date   s      total 0  ports     i 0     rm -f ports ports   ports       mkdir -p ports     while      total  -lt   2   amp  amp    all found      yes      do         port      RANDOM   RANDOM    63001    2000    i    i 1           if    -z    ports port        then             ports   port     port              total    total   1               if       total   1000      0     then                 echo -en  Elapsed time       date   s  - find start  s  t                  echo -e  Found   port  t t Total   total tIteration   i                  find start   date   s              fi         fi     done     all found  yes      echo  Job  1 finished after  i iterations in      date   s  - start  s       out  ports  1 txt           1      0      amp  amp  out  ports  out      echo    ports       gt    out     say-total         generated ports   cat        tr       n     sed -E s     0-9  4      0 1        echo  Total generated    echo   generated ports    sort   uniq   wc -l      total-single     say-total  ports 0 txt     total-jobs     say-total  ports       all found  no       1      --jobs      amp  amp  test-all 0 63001  amp  amp  total-single  amp  amp  exit for i in  1  1000   do test-all   i  40000  amp  sleep 1  done  amp  amp  wait  amp  amp  total-jobs   For determining how many iterations are needed to get a given probability p q of all 63001 possibilities having been generated  I believe we can use the expression below  For example  here is the calculation for a probability greater than 1 2  and here for greater than 9 10

[bash] Random number from a range in a Bash Script

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