4 ms·
> "This is an answer to the batteryhorsestaple thing." Steube misunderstands the xkcd comic [1]. There's a really good comment which explains it: "It could be
by salmonellaeater 13y ago
> "This is an answer to the batteryhorsestaple thing."
Steube misunderstands the xkcd comic [1]. There's a really good comment which explains it: "It could be argued that Randall's example of 4 words is too short -- and indeed, for some applications, it is. However for a typical dictionary size, and genuinely random selection, it is massively stronger than "typical" passwords and in fact easily adequte to defeat the above-mentioned attacks." [2]
Emphasis on "genuinely random selection."
[1] https://xkcd.com/936/ https://xkcd.com/936/
[2] http://www.schneier.com/blog/archives/2013/06/a_really_good_a.html#c1483848 http://www.schneier.com/blog/archives/2013/06/a_really_good_...
- derefr 13y agoWhat password length would you need to get away with a plain-old grammatical english sentence (i.e. very much non-random selection)? For example: "and in the swept plains of winter's vale, our hero did beseech the emperor to send for his forces" -- what would be the difficulty in cracking that, given that this isn't a quote from a book or anything, but just a sentence that popped into my mind and seems easy enough to remember?
- salmonellaeater 13y agoShannon did an experiment that found the entropy of English text is about 1.6 bits per character. This is probably a high estimate, since the kinds of sentences you might think up for a password probably have lower entropy than if you used a source of random bits to generate valid sentences.
- enscr 13y agoMy God, are you going to type all of that or will you need a script to do it for you. Watch out for those touch-screen thingies people are touting around.
- danielweber 13y agoAlmost 20 years ago I saw a great password-picking article that still holds today. http://world.std.com/~reinhold/diceware.html http://world.std.com/~reinhold/diceware.html Take a list of 6^5 words. Roll 5 dice. Take that word from the list. Do this 4 more times. You now have a five-word passphrase like "moire fraud 80 row bernet". Even if someone knew the exact method and list you did to get that passphrase, there are 28430288029929701376 combinations, giving you over 64 bits of entropy. Someone has probably tried to rainbow table all those results for MD5. If a core can do 1 billion hashes per second, it would take 900 core-years to build a complete list of all those combinations, which is probably feasible for a small group to put together, but messing with the list just a little bit or adding a 6th word would likely put you past that even for a crappy MD5 hashing.
- dasil003 13y agoWhat makes you think he misunderstands it? For the cracker it's not about entropy per se, it's a game to come up with algorithms that crack more passwords for less compute power. The XKCD comic got a lot of mindshare so it makes sense to target algorithms towards that type of password. I think Schneier's suggestion of reducing it to the first letter of each word is vastly preferable because it packs the majority of entropy from random word selection into the least amount of typing.
- idProQuo 13y agoThe entropy Randall calculated for "correcthorsestaplebattery" was a lower bound, meaning that if the attacker knows that you made your password out of 4 dictionary words, it still has tons of entropy. If the attacker doesn't know how you came up with your password, it'll take them even longer.
- belorn 13y agoThe algorithm is not targeted against the type of password which the XKCD comic suggests. The algorithm is designed to exploit common human behavior, which is similar to the XKCD method but not identical. The significant difference is that human behavior in picking words is not random, while the XKCD method requires the word selection process to be truly random. The "iloveyousomuch" example by Steube is unlikely to be picked randomly. salmonellaeater is right, Steube misunderstands the comic. The idea of the comic is to pick a small random selection of the 250,000 distinct words in a oxford dictionary, rather than 8 of the 95 letters from all ASCII printable characters. A selection of 3 words has then higher entropy than 8 random characters, because 250,000^3 is a bigger number than 95^8. The question then is, will 3 random words really be easier to remember than 8 ASCII printable characters? The downside to the Schneier scheme, is that each is a common sentence (low entropy), with a chosen transformation algorithm added. Thus the quality of the password will depend on the number of transformation algorithms, and the quality of each one. If we are to use the one first described to create "tlpWENT2m", we get a password strength like: Using strictly the first letter, would only do 2x linear increase in entropy over just searching for common sentences. Change any occurrence of common numbers substitutes for words adds (0-2x) entropy increase. Writing one of the words in all caps means 6x increase in entropy. Combined, tlpWENT2m is slightly less secure than "This little piggy went to market" + two [random number below 10] or a single letter at the end.