Dishonored 2 ][ Answer posted

I thought about creating a wordlist first but that takes way too much :D
I guess I need a new approach.

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cool! let's try it!

I appreciate you doing these good giveaways, I just don't understand how to enter any of them though 🤷🏻‍♀️
Good luck to all the nerds out there.

This thing really really doesn't like windows (no matter what I do, the "..hashes loaded" problem appears)..I don't think I can get any further with it :/

password does not contain 9

Had an idea what the password could be, no luck though. Still, could it be that

(EDIT: My question / hint has been posted: "password contains one 8")

I thought i was clever when I noticed that 2003 was ‭011111010011‬ in binary which was 12 digits long. but alas my wild guess was incorrect.

i am to stupid

Can you provide an example for that: 1 digit repeats side by side?
Thx

This is great idea:) I am gonna pass that one since brute force with mask would take a long time but cheers mate:)

Already have the game, but tempted to chase the puzzle. Alas, there is too much to DL and study. XD

tried kraken but have no idea how to use 5-7 hints in it, so it would take around 19 years to check all the possible numbers :D

I am not sure how to set additional custom rules in JTR (like only 2 numbers can be repeated in a sequence)

Even setting the min and max char to 12 and using 0-8, it will take 7 years on an i5-9300H running on 8 threads.

I think you set the password length too high :| I am using incremental=digits mode

Great challenge tho!

I have 1651p/s that's will take 5 years to crack...
I'm using Intel i7-8750H with 12 OpenMP threads.
Yes the password is too long.
With 1600p/s a 7 long pwd will took 1.7 hrs, so maybe give us 5 number in the pwd... ;)

Here's my thinking so far on how to interpret the hints:

• "password is 12 characters long": 12 spots total to fill
• "password does not contain 9" + "password contains one 8": set aside one spot for the number 8, the rest to fill with numbers from 0 to 7
• "3 digits appear more than once" + "if you are using crunch then 2@": means we have 3 digits each appearing exactly twice, so basically we do n-choose-k with n=0..7 and k=3 then repeat the chosen numbers once, that is we filled 6 more spots in addition to one spot for the number 8
• the remaining 5 spots should be distinct numbers from the set of numbers not previously picked above

Some quick Python to code the above:

from itertools import combinations
arr = [0,1,2,3,4,5,6,7]
res = [x+x+(8,)+tuple(set(arr)-set(x)) for x in combinations(arr, 3)]
for r in res:
  print r

We get 56 lines.

Obviously for each line the order of elements is still not accounted for, so we need to generate all permutations for each line (unrestricted there are 12! factorial per line). Fortunately we still have the following two hints to further restrict the valid permutations:

• "password starts with odd number": self explanatory
• "only 1 digit repeats side by side": from the 3 digits repeated, ensure only one pair of same digits is consecutive, the others are dispersed across the pattern

I haven't coded this part, but I estimate now the upper limit of number of passwords to generate is something like factorial(11)*56, or rounded down let's say 1 billion passwords. If you can check 1000 passwords/sec, that would still take over a week to crack :(

Assuming I made no mistakes, maybe someone else might improve it further, I give up 🤷

EDIT2:

More hints added which cut down the possibilities significantly.

• "password starts with 1" + "password starts with pair": so 11xxxxxxxxxx
• "third digit is smaller than 5 "

I adjusted my code and generated all possible passwords in a ~300MB file with about 25 million passwords. At a rate of 1k pass/sec, that'll take like 7 hours to crack, much better but still a bit too much brute forcing for me...

Maybe another hint or two and it'll be within reach ;)

damn, adding that 8 hint sure took possibilities down :D
difference in file size between "starts with 11" and "starts with 11 and 6th is 8" is 31GB :D

Heh, I haven't finish generating last batch and new hints are already it...

Just a question, if you can answer:

• if you are using crunch, then 2@ (123456 is good, 123345 is good, 123334 is not good)
• only one pair of digits (e.g. 99456)
• password starts with 11

Given that last hint, is it safe to assume that there are no other pairs in the password? (meaning that the first hint becomes 1@)
or is it "one pair other than the one I gave you"?

"3 digits appear more than once" is most evil of the hints... I had to make it last check as my current implementation for this one is poor :)

password contains one 8

exactly 1?

Tried crunch, never used anything like it before....the file was going to be about 4gb but it stopped after a few seconds :/
No idea how to do >5 or <5 though....also isn't 2@ wrong? ^^

This is like a Skinner box lesson on JTH. :p

Bump for solved.

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