// Standard ASCII Characters
add('ASCII Lowercase', /[a-z]/, 26);
add('ASCII Uppercase', /[A-Z]/, 26);
add('ASCII Numbers', /\d/, 10);
add('ASCII Top Row Symbols', /[!@£#\$%\^&\*\(\)\-_=\+]/, 15);
add('ASCII Other Symbols', /[\?\/\.>\,<`~\\|"';:\]\}\[\{\s]/, 19);
[...]
45**22/4e9/3600/24/365.25 = ~18 quintillion years.If the attacker knows you've used four dictionary words in a row. Now you need to multiply the number by the odds of that..
Not to mention that "chili dog monkey" are not exactly rare words and even a 10k word english dictionary would contain those.
What does that have to do with the output of the password in terms of brute-force guessing? Aren't you kind of assuming that an opponent won't try a broad dictionary attack? Equal distribution of hash outputs fundamentally assumes equivalent distributions of hash inputs, and that's not true for the distributions of common passwords.
Again - everyone knows the passphrase patterns everyone uses, everyone knows the password transformations everyone follows (eg l33t) and overall those patterns are significantly weak against attack because of this public knowledge. By those algorithm, with knowledge of salts/etc those passwords are very weak.
Should you use a song, or a slight modification, or a l33t transformation? Certainly not - if the input space isn't uniformly distributed across the input space, the probability of hitting pay dirt definitely isn't uniformly distributed across the output space.
Even the class of "valid transformations of sensical English sentences" is not an equal distribution because of the above. If you can compress a language you can attack it, because by the assumption of compression the data is not at its minimum entropy. Maybe not feasibly if you're lucky - but don't bet on it just because Webster's is big.
Also, you seem to imply that there is a distinguishable difference in sha1 outputs of good random inputs and English word inputs, which violates a security property of cryptographic hashes. So there is a lot of fame to be had if you can prove that.
So to give an example, any natural language phrase like:
"I took a walk in a park" is easier to crack.
I am certain there is somebody coding an infinite-monkey-type bruter to crack diceware as we speak:
Moreover, diceware can be set up with any dictionary, and nobody stops you from computing your own dictionary. If you do that, good luck to the bruter attempting to break your passphrase!
BTW there is no limit on the number of terms in your diceware passphrase, so if/when brute force makes 5-term phrases too weak, users may just add one or two more terms.
If NLG upsets you, you can always settle for a Markov chain and some atmospheric noise to seed the random values
Let's assume a word list of 7776 words. All words are lower case alpha. The attacker has pur wordlist. And the attacker knows there are seven words in the passphrase.
That's still 7776^7.
Even if we prune the wordlist ("I'm not using 'zerg' in my phrase", "I rolled 11111, that's not random so I'll roll again") it's still not an attackable space.
So, there are weaknesses if humans roll the 5 dice and get 1, 1, 1, 1, 1 and then say "That's not random, I'll roll again"[1] or they roll a number, look it up and say "I'll never remember 'zerg', I'll roll again".
My buddy Tom always generates UUIDs by hitting the button several times. It gives him pleasure to 'waste' all those random numbers.
When someone rolls 1,1,1,1,1 and rejects that word they've cut down the word list to 7775 words. That's tiny, but it is a change.
An attacker can under certain circumstances exploit this behavior. For example, if they notice you are picking shorter words or words that consist of certain characters, they will have an easier time to crack the password. Whole categories or patterns can be rejected because of that, reducing the search space.
An extreme version of your method, is picking only the patterns that relate exclusively to you, like picking your favorite music or hobbies. Then the whole scheme becomes useless.
> chili dog monkey nutso
Is definitely not 18 quintillion years at best it's approximately 250000^4 which is 2^72.
A good (though still requiring some memorizing effort approach) would be something along the lines of https://github.com/bitcoin/bips/blob/master/bip-0039.mediawi... which uses a standard dictionary of 1626 easily memorable english words with each password being a 12 word mnemonic phrase. 1626^12 ~= 12^128