delete elements in array bash - arrays

Question

im trying to delete two elements at same time in my array in bash script
my code is
elegidos = (1 2 3 4 5 6 7)
i=0
j=${#elegidos[#]}
delete=($i $j)
while [ $i -le $j ]; do
#elegidosmenosdos=${elegidos[#]/$i:$j}
echo ${elegidos[#]/$delete}
delete=($i $j)
let "i++"
let "j--"
done
the output that i have is
1 2 3 4 5 6 7
1 2 3 4 5 6 7
2 3 4 5 6 7
1 3 4 5 6 7
and i need 21 different combinations with five elements using seven numbers
output example
1 2 3 4 5
2 3 4 5 6
7 6 5 4 3
.
.
.
.
.(21 MORE)

Answer 1

Well, it took a minute to suss out that you were simply wanting to do an end-to-middle delete of two-elements at a time working from the ends of the array, deleting the end nodes at the same time, increment/decrement your counters and repeat until you reached the middle. Why you want to do it this way is a bit of a mystery. You can, of course, simply unset elegidos and unset all values at once. However, if you want to work in from both ends -- that fine too if you have a purpose for it.
You have several problems in your script. In bash, all arrays are zero indexed. Your array has 7-elements, so the valid indexes are 0-6. Therefore, you j was wrong to begin with. You needed to subtract 1 from the number of elements to get the index for the end-element, e.g.
i=0
j=$((${#elegidos[#]} - 1))
bash provides a C-style loop that can greatly simplify your task. While you are free to use a while a C-style loop can handle the index increment and decrement seamlessly, e.g.
for ((; i <= j; i++, j--)); do
note the i <= j. If you have an odd-number of elements in the array, on your last iteration, you will simply be deleting one-value instead of two. To handle that condition you need a simple test within the loop to check whether [ "$i" = "$j" ] (or using the arithmetic comparison (( i == j ))).
Putting that altogether, you could refine your element removal to empty the array two-elements at a time to something similar to the following:
#!/bin/bash
elegidos=(1 2 3 4 5 6 7)
i=0
j=$((${#elegidos[#]} - 1))
delete=($i $j)
for ((; i <= j; i++, j--)); do
declare -p elegidos
unset elegidos[$i]
[ "$i" != "$j" ] && unset elegidos[$j]
delete=($i $j)
done
Example Use/Output
$ bash array_del.sh
declare -a elegidos='([0]="1" [1]="2" [2]="3" [3]="4" [4]="5" [5]="6" [6]="7")'
declare -a elegidos='([1]="2" [2]="3" [3]="4" [4]="5" [5]="6")'
declare -a elegidos='([2]="3" [3]="4" [4]="5")'
declare -a elegidos='([3]="4")'
You can see above, that on the first 3-iterations, both end-elements are removed. However, notice on the last removal (there originally being and odd number of elements, only the single-last value is removed on the final iteration.
Look things over and let me know if I captured what you were attempting and whether you have any further questions. If your intent was something else, drop a comment and I'm happy to help further.