xkcdentropy

7 ms·

1. 1.▲
   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
2. 2.▲
   by xkcdentropy 15y ago

   I am merely suggesting that the entropy can be less than what is estimated by looking only at the dictionary. Re 1: words still consist of characters Re 2: Certainly correct, but to ignore the possibility of English words having less entr
3. 3.▲
   by xkcdentropy 15y ago

   You deny the fact that English text can be attacked separately from your dictionary. English text is very predictable, for example e is much more common and q is almost certainly followed by u. I'm not making this up on my own either. Pleas
4. 4.▲
   by xkcdentropy 15y ago

   This is simply incorrect. If you assume you really do have 100 000 "characters" in your alphabet this is correct. However, your alphabet follows a certain pattern: It's English text. At that point its easier to brute force the individual ch
5. 5.▲
   by xkcdentropy 15y ago

   The XKCD comic is only partially correct. Depending on what source you believe English text has about 0.6 to 2.3 bits of entropy per character. This means you need somewhere between 4.7 and 18.3 characters in each word to reach 11 bits of e