Any X digit PIN as good as the next one (mostly)?
Comments
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Though none of them are very strong, most cracking attempts will start with the more commonly used PINs e.g.
000000
,000001
,123456
, and so on. As such one of those frequently used PINs is likely to be cracked sooner than a random one. That said, with modern computing power, cracking a 6-digit PIN is fairly trivial, even if it is random, taking ~24 hours as a worst case (best case?) scenario.Ben
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As you know sometimes we have no choice ie the site in question requires exactly 6 digits for its PIN. (At least it is 'better' than another site which only allows for 4 digits :( )
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This might be interesting:
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Thanks for that XIII.
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Though none of them are very strong
I have tried three 6 digit PINS for a login requiring a 6 digit PIN and each one results in 1P classifying it as a "Vulnerable Password". This would lead me to believe that all 6 digit PINS would be classified as "Vulnerable".
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That is likely correct. I wouldn't be surprised if every possible combination of 6 digits has been used as a password by someone and has been compromised at some point.
Ben
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I wouldn't be surprised if every possible combination of 6 digits has been used as a password by someone and has been compromised at some point.
Maybe. However,
948137
is not in the HIBP database...(No, I'm not using that anywhere - I used some scripting and command line tools to find one)
(I only checked this one, but if I made no mistake 1,151 out of 1,000,000 are not in there)
Unfortunately I can't check what 1Password thinks about it since I don't get
000000
(ormonkey
) listed as a vulnerable password in 1Password7 for Mac (7.4.3), while both are in the HIBP database...Bug?
(PS: I could not find any 4 digit PIN code that's not in the HIBP database)
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(I only checked this one, but if I made no mistake 1,151 out of 1,000,000 are not in there)
I need a couple of these "uncompromised" 6 digit PINS ;). Is there an "easy" way for me to generate one?
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Is there an "easy" way for me to generate one?
Depends on your skills I guess.
I might post some general instructions on how I did it (when I’m using a physical keyboard).
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Why isn't the case that 1P only generates PINS (or even passwords) which aren't "vulnerable"?
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Why isn't the case that 1P only generates PINS (or even passwords) which aren't "vulnerable"?
(Why) would it matter?
The number of 6 digit PINs that are “not vulnerable” is so small that a hacker can easily just brute force all 6 digit combinations and hit these as well...
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This is roughly what I did:
- Write a Python script that writes all possible 6 digit PIN codes and their SHA-1 hashes to a file (one PIN and its hash per line)
- Use the
cut
&sort
commands to create a new file that only contains the sorted SHA-1 hashes - Download the (almost 10 GB!) passwords from HIBP, sorted by SHA-1 hash
- Use the
comm
command to find SHA-1 hashes that are in the file from step 2, but not in the file from step 3 - Use the file from step 1 to find the PIN codes generating those hashes
Does that help?
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:) :+1:
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