bash palindrome grep loop if then else missing ' - linux

Question

My Syst admin prof just started teaching us bash and he wanted us to write a bash script using grep to find all 3-45 letter palindromes in the linux dictionary without using reverse. And im getting an error on my if statement saying im missing a '
UPDATED CODE:
front='\([a-z]\)'
front_s='\([a-z]\)'
numcheck=1
back='\1'
middle='[a-z]'
count=3
while [ $count -ne "45" ]; do
if [[ $(($count % 2)) == 0 ]]
then
front=$front$front_s
back=+"\\$numcheck$back"
grep "^$front$back$" /usr/share/dict/words
count=$((count+1))
else
grep "^$front$middle$back$" /usr/share/dict/words
numcheck=$((numcheck+1))
count=$((count+1))
fi
done

Answer 1

You have four obvious problems here:
First about a misplaced and unescaped backslash:
back="\\$numcheck$back" # and not back="$numcheck\$back"
Second is that you only want to increment numcheck if count is odd.
Third: in the line
front=$front$front
you're doubling the number of patterns in front! hey, that yields an exponential growth, hence the explosion Argument list too long. To fix this: add a variable, say, front_step:
front_step='\([a-z]\)'
front=$front_step
and when you increment front:
front=$front$front_step
With these fixed, you should be good!
The fourth flaw is that grep's back-references may only have one digit: from man grep:
Back References and Subexpressions
The back-reference \n, where n is a single digit, matches the substring
previously matched by the nth parenthesized subexpression of the
regular expression.
In your approach, we'll need up to 22 back-references. That's too much for grep. I doubt there are any such long palindromes, though.
Also, you're grepping the file 43 times… that's a bit too much.

Answer 2

Try this:
#!/bin/bash
for w in `grep -E "^[[:alnum:]]{3,45}$" /usr/share/dict/words`; do if [[ "$w" == "`echo $w|sed "s/\(.\)/\1\n/g"|tac|tr -d '\012'`" ]]; then echo "$w == is a palindrome"; fi; done
OR
#!/bin/bash
front='\([a-z]\)'
numcheck=1
back='\1'
middle='[a-z]'
count=3
while [ $count -ne "45" ]; do
if [[ $(($count % 2)) == 0 ]]
then
front=$front$front
back="\\$numcheck$back"
grep "^$front$back$" /usr/share/dict/words
else
grep "^$front$middle$back$" /usr/share/dict/words
## Thanks to gniourf for catching this.
numcheck=$((numcheck+1))
fi
count=$((count+1))
## Uncomment the following if you want to see one by one and run script using bash -x filename.sh
#echo Press any key to continue: ; read toratora;
done