BASH Palindrome Checker

Question

This is my first time posting on here so bear with me please.

I received a bash assignment but my professor is completely unhelpful and so are his notes.

Our assig

Answer 1

The multiple greps are wasteful. You can simply do

grep -E '^([a-z])[a-z]\1$' /usr/share/dict/words

in one fell swoop, and similarly, put the expressions on grep's standard input like this:

echo '^([a-z])[a-z]\1$
^([a-z])([a-z])\2\1$
^([a-z])([a-z])[a-z]\2\1$' | grep -E -f - /usr/share/dict/words

However, regular grep does not permit backreferences beyond \9. With grep -P you can use double-digit backreferences, too.

The following script constructs the entire expression in a loop. Unfortunately, grep -P does not allow for the -f option, so we build a big thumpin' variable to hold the pattern. Then we can actually also simplify to a single pattern of the form ^(.)(?:.|(.)(?:.|(.)....\3)?\2?\1$, except we use [a-z] instead of . to restrict to just lowercase.

head=''
tail=''
for i in $(seq 1 22); do
    head="$head([a-z])(?:[a-z]|"
    tail="\\$i${tail:+)?}$tail"
done
grep -P "^${head%|})?$tail$" /usr/share/dict/words

The single grep should be a lot faster than individually invoking grep 22 or 43 times on the large input file. If you want to sort by length, just add that as a filter at the end of the pipeline; it should still be way faster than multiple passes over the entire dictionary.

The expression ${tail+:)?} evaluates to a closing parenthesis and question mark only when tail is non-empty, which is a convenient way to force the \1 back-reference to be non-optional. Somewhat similarly, ${head%|} trims the final alternation operator from the ultimate value of $head.