Ahead of this year’s World Password Day, 1Password – maker of password management software – announced a password cracking challenge. The company ostensibly wanted to find out how hard it would be to crack a three-word passphrase master password on one of their vaults, assuming that the attacker had the derived hash of the passphrase.

(As of the evening of May 31st, none of the passwords have been cracked.)

Wait, What?

First of all, the 1Password blog post announcing the challenge does a pretty good job at explaining what the challenge is all about in a pretty non-technical way. But we can step back for a minute and ask “how are passwords stored?” For example, how does Amazon (or 1Password) store my password?

They are NOT just stored in plain text. This video does a good job of explaining why, but basically it’s because if someone gets their hands on that database, they’ll have every user’s username and password right there.

Vaguely, generally, from a non-expert, the best way to store passwords (as a website) is to hash and salt the passwords. To hand-wave over that, let’s stick to “hashing”. Hashing, I’ve only recently learned, basically means you run the password through a one-way mathematical function, and then only store the result. Wikipedia defines one-way mathematical function thus:

In computer science, a one-way function is a function that is easy to compute on every input, but hard to invert given the image of a random input. Here, “easy” and “hard” are to be understood in the sense of computational complexity theory, specifically the theory of polynomial time problems. Not being one-to-one is not considered sufficient of a function for it to be called one-way.

An example of a (apparently useful) one-way function is multiplying two very large prime numbers. Multiplying two numbers isn’t hard to do computationally (even if they’re huge numbers), but if I gave you a huge number and said “guess the two prime numbers that I multiplied together to make this number” you’d be forced to do a lot of guessing and checking, by multiplication, to get the answer.

A Non-Mathematical Metaphor For My Fellow Liberal Arts Grads

One non-mathematical example that helped me get a handle on this is mixing paint colors: it’s easy to mix red and blue paint and get purple, but it’s very hard to take purple and somehow get the exact shades of red and blue back. Even if you had the purple and the blue it’d be hard to get the red back without a lot of trial and error. But if you had the blue and the purple and someone asked you to check a given shade of red, it’d be easy to check it (just mix the red with your blue and see if it’s the same as the purple you have).

Specifics aside, the general idea is that we’re going to pass some parameters (information) into a function and we’re going to get an output that we can reliably compare to a stored output. Amazon, Facebook, and 1Password all store your exact shade of purple (though you’re hopefully using different passwords for each service). You type in your password (a specific shade of red), the website or program does the mixing (the “easy” direction of the one-way function) with a standardized shade of blue and then compares the produced (or “derived”) shade of purple with the shade it has on file. If they match, you’re in; if they don’t, you’re rejected.

It’s basically a game of Go Fish – exactly what you’d want to protect a password.

To translate some of the terms I’ll be using going forward: the shade of red– the password – is the “secret”; the method by which we’ll mix the colors is called the digest algorithm; the shade of purple (correct or incorrect) is called the “derived hash”. We’ll have some other variables as well that don’t fit the metaphor too well – for example, the “salt” could be defined as the shade of blue we’re using to mix (it can be known to the attack and indeed is in the 1Password challenge), and the number of iterations could be thought of as how much paint we’re dealing with in total: the more paint the longer it takes a given computer to “mix” the colors, slowing down a brute force attack.

The one-way function that 1Password actually uses to hash its master passwords is Password-Based Key Derivation Function 2 (or PDKF2). More on this below, but let’s look at the text of the actual challenge for a second.

How 1Password’s Challenge Works

Right when the challenge started, a 1Password employee pushed the actual challenges to this Github repo. Here’s the json file, which has three samples and then seven actual challenges. Let me just be super clear here: neither the samples nor that challenges are actual passwords of users! This is all for research purposes. Don’t use my program to crack actual hashes of real user passwords!

Here’s the first sample (think of it as for practice):

  "id": "3UOKUEBO",
  "hint": "3 words",
  "sample": true,
  "prf": "HMAC-SHA256",
  "rounds": 100000,
  "salt": "e65814e4382759f85550029e723dc7e7",
  "derived": "5f37a3bd08ac1c7d163294a3cb192ed1407b62bbc6a6259fee55f6e53f754273",
  "pwd": "governor washout beak"

Since this is a sample, they give us the password here (in this case, “governor washout beak”). We also get the algorithm name, the number of rounds, the salt, and the derived hash. This is useful for making sure our cracker is working correctly.

The entries a bit further down in that file are the actual challenges, and they are the same as the samples except they don’t give the password (that’s for us to find out!). Here’s the first one.

    "id": "NO4VRU4S",
    "hint": "3 words",
    "prf": "HMAC-SHA256",
    "rounds": 100000,
    "salt": "8ad1712ab5d632d8c4dac07b792ebb17",
    "derived": "a3a8b8eb8e739c86f67332d17364b149cd88f33bb11eedae066ac366711ec266"

What I Wrote

Now I’m pretty sure that the tool you’re supposed to use for something like this is hashcat or John the Ripper – here’s one user’s screenshot of hashcat or something similar. But I’ve been slowly teaching myself a programming language called Rust, and I wanted to learn more about password hashing (plus, with only access to my personal laptop, I felt I didn’t really have a chance of winning the challenge no matter the method I used). Note: during almost every step of the way outlined below I got some help from Fediverse users (if any of y’all want a shout-out here, let me know, though I assumed you wouldn’t).

For the eager, here’s my Github repo: the master branch uses Rayon to use threading, while the “no-threads” branch does not (it’s easier to read my crappy Rust without my implementation of threading).

PDKDF2 from the Ring crate

I found a Rust “crate” or library called Ring that has a pdkdf2 function already defined (it’s defined here). It takes five variables as inputs: algorithm type, number of iterations, the “salt”, the secret (aka our password guess), and the output variable you want to store the derived hash in.

To actually use Ring’s pdkdf2 function to attack the 1Password challenge format, I found that I needed to write a wrapper function that did some basic text formatting before and after actually calling the function from the Ring crate. Here is that wrapper function, plus some global variables I grabbed from the example the Ring library gave:

static DIGEST_ALG: &'static digest::Algorithm = &digest::SHA256;
const CREDENTIAL_LEN: usize = digest::SHA256_OUTPUT_LEN; // or just put 32 I think
pub type Credential = [u8; CREDENTIAL_LEN];

fn derive(iterations: u32, salt: &str, password: &str) -> String {
    // first, make salt_vec (thanks to https://stackoverflow.com/a/44532957)
    let mut salt_vec = vec![];
    for i in 0..(salt.len() / 2) {
        let mut byte = u8::from_str_radix(&salt[2 * i..2 * i + 2].to_string(), 16).unwrap();

    let mut derived_hash: Credential = [0u8; CREDENTIAL_LEN];

        &mut derived_hash,

    let mut lower = String::new();
    for &byte in derived_hash.iter() {
        write!(&mut lower, "{:02x}", byte).expect("Unable to write byte");
    return lower;

Getting the salt formatted correctly took a bit of trial and error and frustration. Likewise, as you can see, after calling the library’s derive function I had to give some zero-padding to each byte of the derived hash, as well as make it lowercase, before I could compare it to the derived hashes that 1Password presents in its challenge.

I wrote a few tests of this derive function. Here’s one of those tests:

fn derive_example1() {
    let password = "tanbark artistic callus";
    let salt = "00bb202b205f064e30f6fae101162a2e";
    let derived = "91976be95cd28e55e580ee9f69a2139202a9b65eabfbbf33c99bc42e3665564d";
    assert_eq!(derive(100000, salt, password), derived);

It basically says: given a number of rounds, a salt, and a password, will the resulting derived hash be equal to the one 1Password gives in one of their samples (here’s the sample)? In other words, derive(100000, salt, password) should equal "91976be95cd28e55e580ee9f69a2139202a9b65eabfbbf33c99bc42e3665564d". And when we run cargo test Cargo informs us that it passes this test.

Now we’re ready to look at the shorter guess function. This function takes our guess, the number of iterations, the salt, and the derived hashed that we want to compare our output to. If they match, it returns the boolean True. If they don’t match, it returns False.

fn guess(password_guess: &str, iterations: u32, salt: &str, derived: &str) -> bool {
    println!("Guessing {}", password_guess);
    derive(iterations, salt, password_guess) == derived

Obviously we’re going to be running this guess function a lot (we are going to be mixing a lot of shades of red and comparing the outputted shade of purple with the given shade of purple).

And we want to make sure that when it does test the correct password, that it really returns true (otherwise we’d blow right by the correct password). So here’s how I tested this guess function:

fn guess_example1() {
    let incorrect_password = "smith artistic callus";
    let correct_password = "tanbark artistic callus";

    let salt = "00bb202b205f064e30f6fae101162a2e";
    let derived = "91976be95cd28e55e580ee9f69a2139202a9b65eabfbbf33c99bc42e3665564d";

    assert_eq!(guess(incorrect_password, 100000, salt, derived), false);
    assert_eq!(guess(correct_password, 100000, salt, derived), true);

The guessing loops

1Password tells us that the passwords will be three words RANDOMLY chosen from this word list, with a space in between each word. (To understand why it’s important to have a character in between each word, you can read an earlier post of mine.)

We want to run through every possible password (since their made up of three words we could also call them passphrases), performing what’s called a brute force attack. For example, our first guess is going to be aardvark aardvark aardvark, and then our second guess will be aardvark aardvark abaci, and so on, until our produced derived hash matches the given derived hash.

Here is the run_crack function (as seen in the “no-threads” branch of the Github repo). It sports three nested for loops, each of which work through the same array of words created from the word list text file. If the guess function (shown above) returns true, it prints the correct guess and then returns it; else it prints the incorrect guess and moves on to the next guess.

fn run_crack(given_iterations: u32, given_salt: &str, given_derived: &str) -> Option<String> {
    let words = make_word_list("agile_words.txt");

    for word1 in &words {
        for word2 in &words {
            for word3 in &words {
                let password_guess = format!("{} {} {}", word1, word2, word3);
                if guess(&password_guess, given_iterations, given_salt, given_derived) {
                    println!("Found it! {}", password_guess);
                    return Some(password_guess);
                } else {
                    println!("Tried {} unsuccessfully", password_guess);

At this point you may be wondering how many different passphrases are possible given 1Password’s stated rules. The number of words on the word list is 18,328, so the number of possible passphrases is 18,328 * 18,328 * 18,328 – roughly 6,156,000,000,000, or over 6 trillion.

So… How long?

So how long would it take my System76 Oryx Pro with a not-too-shabby Intel i7-7700HQ? As a test I set the mystery password as aardvark aardvark accolade, the 100th passphrase that my program would guess. My janky benchmark function tells me that my laptop does those 100 guesses in about 4,167 milliseconds, which works out to about 24 guess per second. Considering there are literally trillions of possible passphrases, this is… comically much, much too slow. If we extrapolate, it would take 8,193 years to check all of the passphrases (using one core).

We can speed up the cracking by telling Rust to use all eight threads that my laptop has in parallel. I did this using the Rayon crate – you can see that version of the run crack function here, but it is a bit messier than the above version, and I couldn’t figure out how to get it to stop running when it found the correct password.

If I were to use all eight threads, and assume they’d all run as fast as when I running on just one core, that’s still over a thousand years. Even if the password we were looking for was square in the middle of the list, we’re still at five centuries!

I’m hoping that the code works slowly because my CPUs just aren’t powerful enough to run the 100,000 rounds of the PDKDF2 function that quickly (I think hashcat uses GPUs rather than CPUs), but it’s possible that my code could be more efficient. My goal was to get it working though.

Unfazed, I ran the finished script (using cargo run --release) for about 36 hours for the lulz, figuring I might get lucky. I didn’t trust my benchmark – maybe it was guessing faster–, and as I fell asleep to the sounds of my laptop’s fan working at what was presumably its top speed, I fell asleep daring to dream about miraculously winning the $4,096 top prize, and the fame it would gain me, a mere social media producer who just started writing Rust. But of course, I was just mixing the wrong shades of paint very quickly, casting for fish in a very large sea.

How I Might Go About Solving This

Without a single powerful computer, I think the only way to make progress on the trillions of guesses is to use multiple computers.