Using Bash to Generate Lotto Numbers - arrays

Question

I've got a simple bash script that generates the following:
These are your winning lottery numbers:
27 6 29 17 15 47
19 16 33 15 20 14
29 34 48 19 33 40
Here is the code for it:
#!/bin/bash
tickets="$1"
function get_tickets { printf "How many tickets are you going to get? "; read tickets;}
function gen_numbers { printf "\nThese are your winning lottery numbers: \n";
for ((z=1 ; z<=tickets ; z++)); do
for ((i=0; i<6; i++ )); do
x=`echo $[ 1 + $[ RANDOM % 49 ]]`;
printf "\t $x";
done;
printf "\n"
done;
printf "\n"; }
############################
if [[ -z $tickets ]] ; then
get_tickets
gen_numbers
else
gen_numbers
fi
My question is, does anyone know how to modify it to prevent duplicate numbers on each row from appearing? I am guess I'd use uniq, and an array, but am not sure how that would look. Any advice would be appreciated; thanks!
This is just a script for fun.

Answer 1

Your attempt is pretty good. However, I think it can be easier and safer to get random values by using the shuf command:
$ shuf -i 1-49 -n18 | xargs -n6
39 42 43 7 14 23
10 27 5 13 49 8
31 36 19 47 28 4
shuf -i X-Y -nZ gives Z random numbers in between X and Y. Then xargs -nT formats them in groups of T numbers per line.
Update
Now I see the comment:
Yes; to avoid duplicate numbers within a row (by ticket).
In that case, you can simply do shuf -i 1-49 -n6 to get 6 random numbers. The output is line separated, so you can use tr '\n' ' ' to make it space separated.
In case you want many rows, for example 5, you can do:
for i in {1..5}; do shuf -i 1-49 -n6; done | xargs -n6
Sample output:
$ for i in {1..5}; do shuf -i 1-49 -n6; done | xargs -n6
4 45 12 42 37 46
42 20 29 22 12 5
40 41 14 28 4 2
35 24 16 22 2 39
14 46 47 20 21 41

Answer 2

To avoid duplicates by row use an array to mark the used values. Here is a modified version of your script:
for ((z = 1; z<=tickets ; z++)); do
i=0
used=( 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 )
for (( ; i<6; )); do
x=`echo $[ 1 + $[ RANDOM % 49 ]]`;
if [ ${used[$x]} -eq 0 ]
then
printf "\t $x";
i=$((i + 1))
used[$x]=1
fi
done;
printf "\n"
done;
printf "\n"; }
EDIT: Explanation.
The idea is very simple, we use the array to mark those values already used.
Initially all the elements (50 elements that corresponds to the values [0, 49]) of the array are set to zero, i.e. all the numbers are available.
For each random number x we check if that number is available (used[x] = 0), if so we increase the counter i and mark that value (used[x] = 1) to avoid repetitions.
If the value is already taken simply try again and again until an unused value is found.

Answer 3

Check this one, it's quite good example of implementing lotto numbers generator using just JavaScript: https://lotto6aus49online.de/#zaufallszahlen; you can stop the generator by your own click for every number, which adds additional randomness into the algorithm. hope you like it!