Bash random number generator where number is not in array

Question

I am attempting to create a random number generator that generates a number between 1 and 99 but not any number that has already been generated.

In the script array1 contains the numbers already generated. To make it easier to test I have reduced the random number range to 0 - 14 and manually created an array.

I am quite new to bash scripting and am picking it up with a couple of books and the internet.

I have tried a mixture of ideas, the one that seems to make most sense is

array1=( 1 2 3 6 7 8 9 10 11 12 13 )
func1() {
    for var in "${array1[@]}"
    do
      echo $var
    done
}
rnd=$[ $RANDOM % 14 ]
until [ $rnd != func1 ]
    do
        rnd=$[ $RANDOM % 14 ]
    done
echo $rnd

however I know the problem is on line 9 the shell sees the following code:

until [ $rnd != 1 2 3 6 7 8 9 10 11 12 13 ]

I know that the solution is that line 9 needs to be:

until [ $rnd != 1 ] && [ $rnd != 2 ] && [ $rnd != 3 ] && ...

I just don't know how to make this happen automatically from the array. The array does vary in length depending on how many numbers have been generated.

Any help will be greatly appreciated!

The form rnd=$[ $RANDOM % 14 ] is deprecated. Use ((rnd = $RANDOM % 14)) instead. If you have the shuf utility, you can do shuf -i 0-14. — Dennis Williamson
– Dennis Williamson, Commented Jan 4, 2012 at 22:10

jordanm · Accepted Answer · 2012-01-03 14:34:41Z

1

This is something that I found difficulty doing in bash. The approach I came up with is to have func1() return true or false and modify the array to remove the number that has been picked.

array=( {1..15} )
func1() {
   local pick="$1"
   found=1
   total=${#array[@]}
   for ((i=0;i<total;i++)); do 
      if (( pick == ${array[i]} )); then
         echo $pick
         array=( ${array[@]:0:i} ${array[@]:((i + 1)):$total})
         found=0
         break
      fi
   done
   return $found
}
numbers=3
for ((x=0;x<numbers;x++)); do 
   until func1 $(( $RANDOM % ( ${#array[@]} ) )); do 
      continue 
   done 
done

edited Jan 3, 2012 at 14:34

answered Jan 3, 2012 at 2:55

jordanm

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4 Comments

Aleksandra Zalcman Over a year ago

have you tested it? I get a syntax error near unexpected token (': until func1 (( $RANDOM % ${array[-1]} )); do in bash 3

jordanm Over a year ago

Looks like I left out a "$" when I copied it over. I wrote all that in one line before converting it for readability.

Dennis Williamson Over a year ago

You can omit the (()) and dollar sign inside the array slice: array=( ${array[@]:0:i} ${array[@]:i + 1:total}). You can also unset an array element, but that creates a sparse array and you have to iterate over its indices using ${!array[@]}.

jordanm Over a year ago

@DennisWilliamson Thanks for the info on ${!array[@]}. I didn't use unset because I was unaware of ${!array[@]}.

SiegeX · Accepted Answer · 2012-01-05 02:29:38Z

As noted in one of the comments, using the Knuth Shuffle is an excellent way to do this

#!/bin/bash

shuffle() {
   local i tmp size max rand
   # Code from http://mywiki.wooledge.org/BashFAQ/026
   # $RANDOM % (i+1) is biased because of the limited range of $RANDOM
   # Compensate by using a range which is a multiple of the array size.
   size=${#array[*]}
   max=$(( 32768 / size * size ))

   for ((i=size-1; i>0; i--)); do
      while (( (rand=$RANDOM) >= max )); do :; done
      rand=$(( rand % (i+1) ))
      tmp=${array[i]} array[i]=${array[rand]} array[rand]=$tmp
   done
}

# Fill an array with values 1 to 99
array=({1..99});

# Shuffle the array at random
shuffle

# Echo shuffled array
echo ${array[@]}

Output

$ ./knuth
58 78 6 37 84 79 81 43 50 25 49 56 99 41 26 15 86 11 96 90 76 46 92 70 87 27 33 91 1 2 73 97 65 69 42 32 39 67 72 52 36 64 24 88 60 35 83 89 66 30 4 53 57 28 75 48 40 74 18 23 45 61 20 31 21 16 68 80 62 8 98 14 7 19 47 55 22 85 59 17 77 10 63 93 51 54 95 82 94 9 44 38 13 71 34 29 5 3 12

choroba · Accepted Answer · 2012-01-03 10:10:14Z

0

You can also use the -R switch to sort, if your version of sort supports it:

for x in {1..99} ; do echo $x ; done | sort -R

answered Jan 3, 2012 at 10:10

choroba

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1 Comment

Dennis Williamson Over a year ago

printf '%s\n' {1..99} | sort -R

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