5 ms·

You don't compare the hashes. When the user changes the password to "myOldPassword22", you hash "myOldPassword21" and "myOldPassword20" etc and compare those to

You don't compare the hashes. When the user changes the password to "myOldPassword22", you hash "myOldPassword21" and "myOldPassword20" etc and compare those to the old hashes.

gregmac 10y ago
Do you know how many variations there could possibly be?? For example, I could use: myoldpasswordA1 / myoldpasswordB2 / myoldpasswordC3. Probably obvious to a human my next one will be D4, but it is a comparable problem to "play every possible game of chess" for a computer to figure that out. Combined with using a secure computationally-expensive password storage mechanism (eg: bcrypt or PBKDF2) and I'm not exaggerating: it now takes billions of years to verify a password change is allowed. And this is trying to solve a compromised password problem. If there's no compromise, rotating passwords is pointless. If the password IS compromised, rotating the password doesn't really fix it, it merely limits access -- maybe. It doesn't address the original means of the compromise (so you have to assume that's still there). It doesn't address backdoors installed. It doesn't address information stolen. So this is a hell of a lot of work to accomplish .. basically nothing.
- lmm 10y ago
  > For example, I could use: myoldpasswordA1 / myoldpasswordB2 / myoldpasswordC3. Probably obvious to a human my next one will be D4, but it is a comparable problem to "play every possible game of chess" for a computer to figure that out. Two possible changes (forward/backwards) for two characters out of 15. 420 combinations. Not exactly billions.
  - JadeNB 10y ago
    > Two possible changes (forward/backwards) for two characters out of 15. Yes, but you can only cut down the sample space in this way because you can look at "myoldpasswordA1" and know that the 'A1' characters are the iterators. A computer doesn't know that.
    - deleted 10y ago
      [deleted]
    - lmm 10y ago
      My figure is for testing all possible ways of changing two characters by one each (and the ways of changing only one character are a tiny subset of that). This doesn't catch people who change 3 or more characters, or who change characters by more than 1, but at that point is it really still so easy to guess?
      - gregmac 10y ago
        You're focused on the wrong problem! Why does someone have an old password in the first place? You should be asking how it got compromised, who knew the credentials, if it was compromised while still active, and if so, how many backdoors were installed, what data/systems were compromised, and how do you prevent this from happening again. None of the problems could have been avoided if only there was a better password iteration detection algorithm.
        lmm 10y ago
        > Why does someone have an old password in the first place? You should be asking how it got compromised, who knew the credentials, if it was compromised while still active, and if so, how many backdoors were installed, what data/systems were compromised, and how do you prevent this from happening again. All true. Nevertheless, there are cases where someone gets a one-off copy of a password - perhaps overheard. If you force them to actively install a backdoor then you increase the costs of attacking greatly. It's not about perfect security, it's about the cost/benefit - a lot of crime is crime of opportunity, especially internal fraud.
        JadeNB 10y ago
        > If you force them to actively install a backdoor then you increase the costs of attacking greatly. I think that it's my fault, but I can't understand this sentence. Doesn't forcing someone to install a backdoor decrease security? The only reading that makes sense to me is "If you make it so that it's user action, rather than inaction, that results in the installation of a backdoor …"? (For what it's worth, although I disagree with you, I respect your dedication to arguing this determinedly but civilly.)
- blacksmythe 10y ago
  >> comparable problem to "play every possible game of chess" for a computer to figure that out. my_password + iterator is a very common password convention. It would be a poorly programmed computer that didn't consider this possibility.
deleted 10y ago
[deleted]

You don't compare the hashes. When the user changes the password to "myOldPassword22", you hash "myOldPassword21" and "myOldPassword20" etc and compare those to the old hashes.

gregmac 10y ago

Do you know how many variations there could possibly be?? For example, I could use: myoldpasswordA1 / myoldpasswordB2 / myoldpasswordC3. Probably obvious to a human my next one will be D4, but it is a comparable problem to "play every possible game of chess" for a computer to figure that out. Combined with using a secure computationally-expensive password storage mechanism (eg: bcrypt or PBKDF2) and I'm not exaggerating: it now takes billions of years to verify a password change is allowed. And this is trying to solve a compromised password problem. If there's no compromise, rotating passwords is pointless. If the password IS compromised, rotating the password doesn't really fix it, it merely limits access -- maybe. It doesn't address the original means of the compromise (so you have to assume that's still there). It doesn't address backdoors installed. It doesn't address information stolen. So this is a hell of a lot of work to accomplish .. basically nothing.

lmm 10y ago
> For example, I could use: myoldpasswordA1 / myoldpasswordB2 / myoldpasswordC3. Probably obvious to a human my next one will be D4, but it is a comparable problem to "play every possible game of chess" for a computer to figure that out. Two possible changes (forward/backwards) for two characters out of 15. 420 combinations. Not exactly billions.
- JadeNB 10y ago
  > Two possible changes (forward/backwards) for two characters out of 15. Yes, but you can only cut down the sample space in this way because you can look at "myoldpasswordA1" and know that the 'A1' characters are the iterators. A computer doesn't know that.
  - deleted 10y ago
    [deleted]
  - lmm 10y ago
    My figure is for testing all possible ways of changing two characters by one each (and the ways of changing only one character are a tiny subset of that). This doesn't catch people who change 3 or more characters, or who change characters by more than 1, but at that point is it really still so easy to guess?
    - gregmac 10y ago
      You're focused on the wrong problem! Why does someone have an old password in the first place? You should be asking how it got compromised, who knew the credentials, if it was compromised while still active, and if so, how many backdoors were installed, what data/systems were compromised, and how do you prevent this from happening again. None of the problems could have been avoided if only there was a better password iteration detection algorithm.
      - lmm 10y ago
        > Why does someone have an old password in the first place? You should be asking how it got compromised, who knew the credentials, if it was compromised while still active, and if so, how many backdoors were installed, what data/systems were compromised, and how do you prevent this from happening again. All true. Nevertheless, there are cases where someone gets a one-off copy of a password - perhaps overheard. If you force them to actively install a backdoor then you increase the costs of attacking greatly. It's not about perfect security, it's about the cost/benefit - a lot of crime is crime of opportunity, especially internal fraud.
        JadeNB 10y ago
        > If you force them to actively install a backdoor then you increase the costs of attacking greatly. I think that it's my fault, but I can't understand this sentence. Doesn't forcing someone to install a backdoor decrease security? The only reading that makes sense to me is "If you make it so that it's user action, rather than inaction, that results in the installation of a backdoor …"? (For what it's worth, although I disagree with you, I respect your dedication to arguing this determinedly but civilly.)
blacksmythe 10y ago
>> comparable problem to "play every possible game of chess" for a computer to figure that out. my_password + iterator is a very common password convention. It would be a poorly programmed computer that didn't consider this possibility.

deleted 10y ago

[deleted]