bash palindrome grep loop if then else missing '

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