5 ms·
A single password represents a distribution of possible bit values for each byte within it. The password itself is a distribution of characters used within the
by BeefWellington 5y ago
A single password represents a distribution of possible bit values for each byte within it. The password itself is a distribution of characters used within the password.
In fact, the author's article makes this very point, which is why I pointed out the logical flaw in the thinking.
I'll reduce N to 6 for simplifying the author's absurd example but it can expand to any N.
If we take the argument to hold that you roll a random die of N length (6 in our case) and the upperbound represents one strong password, while all other values equate to the word "password", the flaw is in how this logic is applied.
Imagine this is our set of possible values:
password, password, password, password, password, hj5^@l2jl9GGk;Clkm(0]
It makes little difference if you look at this as either the bytes involved in the entire set, or the average of all passwords within the set, it's going to come out looking like you are secure.
This means what they're attacking is all permutations of the following set of characters:
a, C, d, h, j, k, l, m, o, p, r, s, w, G, 0, 2, 5, 9, ;, ], @, ^, (
What an attacker must know though, is the character set used within, as well as the length.
This is the logical flaw the author made in their analysis. For an attacker, the entropy of an individual string is taken as possible character permutations required to discover the true password and NOT permutations of the entire strings themselves.
If you look at the values for each string presented in our set, what an attacker has to attack is:
a, d, o, p, r, s, w
C, h, j, k, l, m, G, 0, 2, 5, 9, ;, ], @, ^, (
But in order to attack these, they need to try the full set:
a-z
a-zA-Z0-9;:[]!@#$%^&*(){}
One of these will be VASTLY easier to break.
- less_less 5y agoI don't understand your argument at all. Why does an attacker need to try a full set of characters? Real attackers try from dictionaries or password generation methods (eg dictionary + numbers, dictionary + dictionary + number + symbol, etc), and "password" is one of the first passwords they'll try. They do this because they don't know exactly how you generated the password, but due to password leaks, they do have a pretty good idea of how most people generate passwords. In principle, you could estimate a password's strength by the order in which a cracker would be expected to guess it. But that's a pain, depends on the password cracker being used, and can change at any time. Also, it's not "entropy", which is a well-defined mathematical concept and is what the linked article is about. Entropy is supposed to be a bound that even if the attacker knows your generation method, they won't be able to do better than brute-force search. For this, the author is correct that min-entropy or a similarly conservative measure is the right one; though for the most common (uniform) generation methods this is the same as Shannon entropy. Entropy of the set of characters used in your password (well, sets don't have entropy, but let's say of the uniform distribution on that set) isn't the same as entropy of password generation mechanism, because the attacker might have more information. For example, if he knows (or correctly guesses) that your password is a dictionary word, then this is super helpful information that isn't captured in the entropy of the bytes.
- BeefWellington 5y ago> I don't understand your argument at all. Why does an attacker need to try a full set of characters? Real attackers try from dictionaries or password generation methods (eg dictionary + numbers, dictionary + dictionary + number + symbol, etc), and "password" is one of the first passwords they'll try. They do this because they don't know exactly how you generated the password, but due to password leaks, they do have a pretty good idea of how most people generate passwords. I'm well aware. How does this help the attacker attacking the higher-entropy string I outlined? How difficult is it for an attacker to attack a password consisting of four lower case english dictionary words? If you run some of these permutations through John, you'll see how long it takes just to generate even quick broken hashes like MD5 versus using something that is a long string of essentially type-able byte data. > Entropy is supposed to be a bound that even if the attacker knows your generation method, they won't be able to do better than brute-force search. For this, the author is correct that min-entropy or a similarly conservative measure is the right one; though for the most common (uniform) generation methods this is the same as Shannon entropy. I'm not sure who has dictated that this is supposed to be how entropy is used for password management. Do you have any references here? Because otherwise it looks like it's still the author and yourself assigning a set of rules to something that doesn't actually apply in the real world and doesn't represent how things are used in practice. My entire point is that the author has taken an incredibly narrow definition of what entropy must be applied to (only to the distribution of the overall set of characters used in the example) and how it must be used in this circumstance, and argued against that. Where it falls down is this: The entire purpose of using entropy as a measure of difficulty of cracking a password is precisely the character set approach. If you were to type "password" into any system employing a Shannon entropy analysis on the set of characters required to generate that password, you would at worst have to generate 26^8 combinations. Dictionary attacks are good because they reduce that from around 208 billion to about half a million. 208 billion is not a high enough number, and these systems will tell you it's weak. Smarter ones will probably alert you that it's a dictionary word as well. If the issue is that people are "misusing" the term entropy for passwords here, that's fine but that's a different article (and I'd still disagree).
- less_less 5y ago