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This article is posted every time the comic is mentioned but I can't understand the argument. The calculation of password complexity in the comic is made with t

This article is posted every time the comic is mentioned but I can't understand the argument. The calculation of password complexity in the comic is made with that in mind. It's assumed that the cracker knows the method used to generate the password, including the dictionary. The strength of this method does not rely on the fact that the password has many characters but that words are randomly chosen from a large dictionary. The attacker would need to do the same at minimum.

thatdude 10y ago
But the words themselves follow an identifiable pattern (their spelling). As such, a 4 letter word in your password is cracked much quicker than a portion of your password being 4 characters of random info.
- creshal 10y ago
  That only matters if, for some reason, your password is length limited. If your password must not be more than four letters long, then yes, choosing your tokens from an ascii table has the highest possible entropy. (Example: WPA2 PSKs, shitty websites.) If your password can have arbitrary length (or arbitrary enough, about ~120 letters), you can generate a 128 bit password with dictionary words as tokens. Sure, the password will be much longer (factor ~6), but also much easier to memorize.
bwindels 10y ago
This is my understanding, could be missing something though: The comic proposes that correcthorsebatterystaple would be a secure password. It is 26 characters long and contains 4 basic english words that would be in even the smallest of dictionaries. Let's say a dictionary with 100.000 entries contains these words. Combining 4 words in the dictionary gives you 100,000^4 = 1.e+20 possibilities. Not bad, but let's say 15 random uppercase and lowercase (like the first letters of every word in a phrase like Schneier proposes) letters give you (26+26)^15 = 5.49e+25, still 5 orders of magnitude better. Edit: a phrase with 26 words in to take each letter from indeed isn't doable to remember, changed it to 15.
- germanier 10y ago
  100k words might be a bit to many for a dictionary of words that are easy to remember. The comic proposes using not enough entropy (i.e. you should use more than four random words). The Schneier method is basically equivalent but with important caveats: 1) Such a phrase is not randomly chosen but follows at minimum basic English grammar and at worst is well-known and thus part of the attacker's dictionary and 2) even if that is taken account for the first letters are not distributed uniformly across the alphabet, you just need to take a look at any (printed) dictionary. That greatly reduces entropy and makes it hard to reliably estimate entropy.
  - creshal 10y ago
    > 100k words might be a bit to many for a dictionary of words that are good to remember. I've been generating all my (memorized) passwords from a 120k word dictionary for years now, can't complain. And estimating the entropy is easy.
    - germanier 10y ago
      Easy entropy calculation is the beauty of that method. But for "take the first letters in an English phrase" you can't just do (2*26)^{length} or you are way overestimating.
- hollander 10y ago
  Why limit yourself to one language? You could even use dialects. Five words, five languages, and not too common words.

thatdude 10y ago

But the words themselves follow an identifiable pattern (their spelling). As such, a 4 letter word in your password is cracked much quicker than a portion of your password being 4 characters of random info.

creshal 10y ago
That only matters if, for some reason, your password is length limited. If your password must not be more than four letters long, then yes, choosing your tokens from an ascii table has the highest possible entropy. (Example: WPA2 PSKs, shitty websites.) If your password can have arbitrary length (or arbitrary enough, about ~120 letters), you can generate a 128 bit password with dictionary words as tokens. Sure, the password will be much longer (factor ~6), but also much easier to memorize.

bwindels 10y ago

This is my understanding, could be missing something though: The comic proposes that correcthorsebatterystaple would be a secure password. It is 26 characters long and contains 4 basic english words that would be in even the smallest of dictionaries. Let's say a dictionary with 100.000 entries contains these words. Combining 4 words in the dictionary gives you 100,000^4 = 1.e+20 possibilities. Not bad, but let's say 15 random uppercase and lowercase (like the first letters of every word in a phrase like Schneier proposes) letters give you (26+26)^15 = 5.49e+25, still 5 orders of magnitude better. Edit: a phrase with 26 words in to take each letter from indeed isn't doable to remember, changed it to 15.

germanier 10y ago
100k words might be a bit to many for a dictionary of words that are easy to remember. The comic proposes using not enough entropy (i.e. you should use more than four random words). The Schneier method is basically equivalent but with important caveats: 1) Such a phrase is not randomly chosen but follows at minimum basic English grammar and at worst is well-known and thus part of the attacker's dictionary and 2) even if that is taken account for the first letters are not distributed uniformly across the alphabet, you just need to take a look at any (printed) dictionary. That greatly reduces entropy and makes it hard to reliably estimate entropy.
- creshal 10y ago
  > 100k words might be a bit to many for a dictionary of words that are good to remember. I've been generating all my (memorized) passwords from a 120k word dictionary for years now, can't complain. And estimating the entropy is easy.
  - germanier 10y ago
    Easy entropy calculation is the beauty of that method. But for "take the first letters in an English phrase" you can't just do (2*26)^{length} or you are way overestimating.
hollander 10y ago
Why limit yourself to one language? You could even use dialects. Five words, five languages, and not too common words.