Re: Noob dumb question (extending emacs)

[Emanuel Berg]

See below, if we remain at 48 bits of entropy, but instead of
60 have an alphabet of only 3 chars, the password length rises
from 8.13 chars to 30.28 ...

Does this tell us that it is equally difficult to randomly
guess the both passwords from the two different configuration
at these lengths, or what does it tell us?

What does "48 bits of entropy" mean BTW?

;;; 48 bits, alphabet length n, password length l
;;; 2^48     = n^l            <=>
;;; 48*ln(2) = l*ln(n)        <=>
;;; l        = 48*ln(2)/ln(n)

(defun epwgen-password-length (bits abc)
  (let ((abc-len (if (numberp abc)
                     abc
                   (when (listp abc)
                     (length abc) ))))
    (when abc-len
      (/ (* bits (log 2)) (log abc-len)) )))

;; (epwgen-password-length 48 60)       ;  8.13
;; (epwgen-password-length 48 '(a b c)) ; 30.28

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