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by jpeter 1587 days ago
Or use a new password every day like worle. So you have a community effort to guess it
2 comments

That sounds like Bitcoin with extra steps

Jokes aside, that would actually be fun if the password is actually reasonably guess-able, I would definitely give it a try if that existed

Oh no, now someone will make wordlecoin, where you have to work with others sharing hints to mine each block.
Proof-of-Wordle
Is any of the information (yellow/green for characters) presented getting you closer to the real answer in any meaningful way though?
According to the best current knowledge of humanity, it provides no information whatsoever.

However, proving that is difficult. It is possible that there exists an algorithm that could narrow in on the answer from hashes. Such an algorithm could run quickly, but it could also potentially take quite significant computation. We don't know what the true, optimal answer to this question is.

> According to the best current knowledge of humanity, it provides no information whatsoever.

??? My first guess has two green letters, or 8 bits of the hash are known. This excludes 255/256 of possible passwords-- so if there's a dictionary, it's way cut down. I also know for the other 30 digits a value that they are not-- this is about .1 bits apiece, for 3 more bits. And I get a few more bits from knowing the population count for each digit.

One guess has reduced the search space by a factor of 10000+. If I say, know the word is in /usr/share/dict/words, the number of possibilities has dwindled from 230,000 to something around 20.

Now, in this case, with a 14 character randomized password-- the amount of benefit is limited. The search space is still significantly shrunk by each guess, but in a way that is difficult to iterate.

This is one of those places where it's easy to conflate computer bits with information theory bits. You may have eight computer bits, but in order for you to have eight bits of information, you must have your search space cut down by a factor of 256, not just the abstract concept of a search space cut down.

Can you enumerate the remaining 1/256th of the search space? Not with anything other than a brute force search, minus the one password you tried. The exact same brute force search that you would have needed to solve the problem in the first place. Your one password attempt has yielded one password's worth of knowledge. You, a human, don't have eight bits of information. You have almost nothing.

In principle, such a guess does eliminate 8 bits of information, but we have no way of manifesting that. In principle if we had a full list of the shortest passwords that led to the given hash, we could strike off the non-matching entries, but no human can do that. In principle an easier algorithm than the brute-force search exists, but we have no idea what it is, and we don't know what it would look like, whether it would be an incremental improvement over brute force or if there's hypothetically an algorithm that could do it on your cell phone in a couple of seconds or what.

Hashing and cryptography in general hide in this space between the theoretical information leakage and the practical inability to do anything with it. You have 8 theoretical bits and just shy of 0 real, practical bits.

> Can you enumerate the remaining 1/256th of the search space? Not with anything other than a brute force search, minus the one password you tried. The exact same brute force search that you would have needed to solve the problem in the first place. Your one password attempt has yielded one password's worth of knowledge. You, a human, don't have eight bits of information. You have almost nothing.

Eh, the actual search space for reasonable online guesses is cut down by 10000x.

Yes, you still need to search an impractically large number of passwords here-- 2^92 or so.

But you only have to provide 10 guesses to the oracle. Described here: https://news.ycombinator.com/item?id=30367095

Or, if you tell me that the password is in /usr/share/dict/words, I can figure out what the password is in 2 online guesses.

I can give you the full hash so that you can be done in one guess if you have a giant rainbow table of precomputed hashes. Still, the full hash doesn’t reduce the search space at all, assuming SHA256 is secure. Sure, you can cut down on the number of oracle queries, but that’s not the limiting factor of this game.
"Eh, the actual search space for reasonable online guesses is cut down by 10000x."

Only in theory. In order to determine which 9999 out of 10000 guesses are no longer relevant, the only known method you have is to compute the hashes of all the 10000 representatives anyhow... which is, again, the exact same problem you started out with at the beginning. You have theoretical information because you've made theoretical progress, but you have no real information, because you've made no real progress.

This program uses a number of random characters each time you load it. You have no list for this program.

In principle you could look at your random number generator and possibly narrow it down beyond the sheer size of the SHA256 space, if it has fewer bits of internal state. I don't know how many bits of internal state it has or even if the answer is constant per browser, and that's really just a practical detail.

To put this in even more stark relief, suppose I bring up Passwordle and by some magic, I hand you a password at the beginning that has a hash that is identical to the hash of the answer in all but one bit. In theory, that constitutes enough information to name the answer on the next guess. In practice, you can't do that.

In fact, we can play that game right now. The SHA256 hash [1] of "mlyle" is "CAD9051E126DA9BC7CB4048C4CA28804CCFEE0E3824F4E63FC151BC5E30B96D0". Using this information, please produce a password with the hash CAD9051E126DA9BC7CB4048C4CA28804CCFEE0E3824F4E63FC151BC5E30B96D1, differing only in the last bit. Ideally the shortest password using letters, numbers, and symbols in US ASCII, but honestly I'll take any binary string.

How much help does that provide you? In theory, like I said, you should be able to do it in one guess now, if what you say is true. In practice, you don't have the lookup table to do it, you can't have the lookup table to do it in our real universe, and we have no known better algorithm for it.

(Observant people may note that providing the mlyle hash is irrelevant and this challenge is equivalent to simply directly asking for something that hashes to the target string. And that's the point. Providing you the hash of mlyle provides zero assistence. You must still enumerate everything.)

[1]: https://passwordsgenerator.net/sha256-hash-generator/ if you want to play along.

"??? My first guess has two green letters, or 8 bits of the hash are known. This excludes 255/256 of possible passwords"

Sha256 is a one-way hash. Knowing some of the sha256 doesn't tell you anything about the plaintext.

Put another way, the matching SHA characters are just a decoy. That's the joke. They could give you the SHA256 hash up front and you'd still have to search the entire password space.

Are you sure thats how evenly distributed hash algorithms work? change one letter of your string, or just make it longer or shorter - none of your green fields will stay.
Nothing about this algorithm relies on similar words producing similar hashes. If the word “foobar” has a 0 in the first digit of its hash, and you see a green 1 in the first digit in Passwordle, then you know that the answer can’t be foobar.
> Are you sure thats how evenly distributed hash algorithms work? change one letter of your string, or just make it longer or shorter - none of your green fields will stay.

True. But still, I know the vast majority of words in my dictionary don't match those two green fields after hashing, and can be eliminated from further consideration as the password.

The password is not a dictionary word, it’s randomly generated though?
Inherently with a (proper) hashing algorithm, the value and placement of characters in the hash means next-to-nothing in terms of the actual original text. For example:

password = 5e884898da28047151d0e56f8dc6292773603d0d6aabbdd62a11ef721d1542d8

passwurd = 1966e583daff0fce5630d5de44f303f0e77f77940f02c7d648defadc31059c7b

Notice they're very different results, even though the original text only has 1 character difference.

The Avalanche effect for anyone interested in reading more.

https://en.wikipedia.org/wiki/Avalanche_effect

No, that's the joke of the site.
It reduces the search space to find the hash, but not the search space of what hashes to that value.
Sort off, if you already have a lookup table of possible solutions.
... which you won't, because the space is too large (around 90 bits of entropy if I'm not mistaken, bit less, so 10^27-ish possible solutions).