??? 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.
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.
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.
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.
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.
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.
