Cheating at Wordle with grep

I usually try to make a good effort at Wordle, but sometimes I get down to the last one or two chances and need some help. Instead of anything fancy-pants, I usually turn to a dictionary file and my friend, grep.

So, first thing's first, you need a dictionary file, which is just a file with a bunch of words one-per-line. These are usually found in /usr/share/dict. I happen to have cracklib-small on my machine so we'll use that.

First, you'll want to get all the five-letter words out of the file. I use the regex ^\w{5}$, which is ^ for the start of the line, \w for an alphanumeric character, {5} saying that there are 5 of them, and $ for the end of the line. The beginning and end of line markers are important, otherwise you'll get words that contain 5 or more letters.

grep -E '^\w{5}$' /usr/share/dict/cracklib-small

At this point, I've already made some guesses and have a few letters in the right spot, and some letters which are correct but in the wrong spot. We'll bucket these into two greps.

For letters in the right spot, I stick those right into a regex. For example, let's say I know the word starts with 'f' and 'a':

grep -E 'fa\w\w\w'

For the right letters in the wrong spot, I simply use the letter as the regex. If I have multiple letters, I can chain those together by piping grep like so:

grep s | grep t

So chaining them all together:

$ grep -E '^\w{5}$' /usr/share/dict/cracklib-small \
    | grep -e 'fa\w\w\w' \
    | grep s | grep t

You could eliminate words that have letters you've already eliminiated by chaining grep -v <letter> to the end, but I find that pretty cumbersome. There's only a handful of possible words at this point, so I think it's easier to just remove them in my head.