Grep regex to unscramble a word

Question

I want to unscramble a word using the grep command.

I am using below code. I know there are other ways to do it, but I think I'm missing something here:

 grep "^[yxusonlia]\{9\}$" /usr/share/dict/words

should produce one output:

anxiously

but it produces:

annulosan                      
innoxious

and many more. Basically I can't find how I should specify that characters can only be matched once, so that I get only one output.

I apologise if it seems very simple but I tried a lot and can't find anything.

Answer 1

You can use grep -P (PCRE regex) with negative lookahead

grep -P '^(?:([yxusonlia])(?!.*?\1)){9}$' /usr/share/dict/words
anxiously 

Explanation:

This grep regex uses negative lookahead (?!.*?\1) for each character matched by group #1 i.e. \1. Each character is matched only and only when it is not followed by the same character again in the string till the end.

Answer 2

You can use lookaheads to make sure that each letter is matched exactly one time. It is verbose and requires a version of grep that supports lookaheads (e.g. via -P). It may be better to build the search string programmatically.

grep -P "^(?=.*y)(?=.*x)(?=.*u)(?=.*s)(?=.*o)(?=.*n)(?=.*l)(?=.*i)(?=.*a)[yxusonlia]{9}$" /usr/share/dict/words