Command Line Passphrase Generation

I figured out another way to calculate passphrases, this time using the command line.

To cut to the chase, here is the command:

shuf -n4 $WORD_LIST | tr [:cntrl:] ' ' ; echo

I found out about a command called “shuf” which according to its man page can “generate random permutations”. You can run this against a word list. Many Unix-based systems have a word list.

On my system, the word list is at /usr/share/dict/words, which is a link to /etc/dictionaries-common/words, which itself is a link to /usr/share/dict/american-english. You can also find one online here.

After you run the list against “shuf”, you need to put them all on one line. You can do this with the “tr” command (see man page here). You can use it to “translate or delete characters”. I want to replace the carriage return-new line combination with a space. In the command, we can represent that combination with either ‘\r\n’ or [:cntrl:]. I prefer [:cntrl:] because the only way to represent the space is with a space between single quotes, and I think using one less pair cuts down on ambiguity (“tr [:cntrl:] ‘ ‘” as opposed to “tr ‘\r\n’ ‘ ‘”).

Even though the word list is “american-english”, it still has some characters that we do not use, like vowels with umlauts or the letter “A” with a circle over it (I guess that is a “super-umlaut”). There are also a lot of words in there with an apostrophe followed by an “s”, words with upper-case letters and words with less than 4 characters. Taking out all of that takes us from a file with 100K words to one with about 26K words.

Here is the full grep sequence I used:

grep -v ''s american-english |\
grep -v Å |\
grep -v é |\
grep -v ö |\
grep -v ä |\
grep -v [A-Z] |\
grep '\w\{4\}' | cat > word.list.002

You’re welcome.