Passphrase generation using awk

Given a requirement of generating a temporary passphrase that can be communicated over the phone to another person, I thought of XKCD #936 which suggests using four random words together as a passphrase. Then there’s just the question of how to generate that easily.

On each system there’s a file containing a list of words somewhere, on OS X it’s located at /usr/share/dict/words. This contains a good ~236,000 words on my machine so that seems like a good enough corpus to pull our four words from. We could take the easy route and lean on gnu sort to sort the file randomly & then take the top four words using head:

$ gsort -R /usr/share/dict/words | head -n4 | xargs
fountained irretraceable unoil barylalia

This works, but isn’t that performant (takes nearly 5 seconds on my machine.) There should be a more performant way to do this, we’re just reading a file in as a list & picking four random elements from said list. How about we lean on awk instead:

$ awk '
BEGIN { srand() }
{ words[NR] = $0 }
END {
  for (i = 1; i <= 4; i++) {
    f = "%s "
    if (i == 4) { f = "%s" }
    printf f, words[int(rand() * NR)]
  }
}' /usr/share/dict/words
swording memorability sneakingness readily

This runs in around 350ms on the same machine, which is a nice speedup and quick enough it feels near-instant.

Next time you need to generate a random passphrase, remember our old friend awk!

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