4 ms·
I understand randomly selected words are not the same as the English text Shannon is talking about. However my point is that entropy may be lower than what it a
by xkcdentropy 15y ago
I understand randomly selected words are not the same as the English text Shannon is talking about. However my point is that entropy may be lower than what it appears to be. I'm not saying it is 0.6 bits per character (or any other number).
I'm saying that unless you words are long enough it is very likely that the entropy is lower than simply taking log(dictionary_size^word_per_passphrase)/log(2)
The references to Shannon and Applied Cryptography were only made as supporting evidence that entropy of English text is lower than what it appears. I'm not claiming their exact numbers apply in this situation.
If we can't agree on that, the we must simply agree to disagree.
- demallien 15y ago* I'm not claiming their exact numbers apply in this situation.* You see, the thing is, you kind of are making that claim. If entropy goes out to 3 bits per character, the one feeble point that you did have gets blown out of the water. The fact that you don't seem to understand that indicates that you really need to go back and reread a few books on Information Theory. Look, for your own edification, try and come up with a scheme that will reliably beat a dictionary attack in terms of the number of attempts needed before finding a password taken from the dictionary. You could even write a simple program to test the idea. Take a dictionary of 5000 words with a length of 6 characters or longer (much shorter words than what your theory suggests should be secure). Select 100 words from the dictionary at random, and then run any attack of your devising against those words. If you can reliably get a better average number of attempts than 2500, I'll concede the point. Until then, I'm here to tell you that you don't understand information theory as well as you seem to think you do.