#!/usr/bin/python
# -*- coding: utf-8 -*-
"""Generate a random, memorizable password: http://xkcd.com/936/
Example run:
kragen@inexorable:~/devel/inexorable-misc$ ./mkpass.py 5 12
Your password is "learned damage saved residential stages".
That's equivalent to a 60-bit key.
That password would take 2.5e+03 CPU-years to crack
on my inexpensive Celeron E1200 from 2008,
assuming an offline attack on a MS-Cache hash,
which is the worst password hashing algorithm in common use,
slightly worse than even simple MD5.
The most common password-hashing algorithm these days is FreeBSD’s
iterated MD5; cracking such a hash would take 5.2e+06 CPU-years.
But a modern GPU can crack about 250 times as fast,
so that same iterated MD5 would fall in 2e+04 GPU-years.
That GPU costs about US$1.45 per day to run in 2011,
so cracking the password would cost about US$3e+09.
I've started using a password generated this way in place of a 9-printable-
ASCII-character random password, which is equally strong. Munroe's assertion
that these passwords are much easier to memorize is correct. However, there is
still a problem: because there are many fewer bits of entropy per character
(about 1.7 instead of 6.6) there is a lot of redundancy in the password, and so
attacks such as the ssh timing-channel attack (the Song, Wagner, and Tian
Herbivore attack, which I learned about from Bram Cohen in the Bagdad Café in
the wee hours one morning, years ago) and keyboard audio recording attacks have
a much better chance of capturing enough information to make the password
attackable.
My countermeasure to the Herbivore attack, which works well with 9-character
password but is extremely annoying with my new password, is to type the
password with a half-second delay between characters, so that the timing
channel does not carry much information about the actual characters used.
Additionally, the lower length of the 9-character password inherently gives the
Herbivore approach much less information to chew on.
Other possible countermeasures include using Emacs shell-mode, which prompts
you locally for the password when it recognizes a password prompt and then
sends the whole password at once, and copying and pasting the password from
somewhere else.
As you'd expect, this password also takes a little while longer to type: about
6 seconds instead of about 3 seconds.
"""
import random, itertools, os, sys
def main(argv):
try:
nwords = int(argv[1])
except IndexError:
return usage(argv[0])
try:
nbits = int(argv[2])
except IndexError:
nbits = 11
filename = os.path.join(os.environ['HOME'], 'devel', 'wordlist')
wordlist = read_file(filename, nbits)
if len(wordlist) != 2**nbits:
sys.stderr.write("%r contains only %d words, not %d.\n" %
(filename, len(wordlist), 2**nbits))
return 2
display_password(generate_password(nwords, wordlist), nwords, nbits)
return 0
def usage(argv0):
p = sys.stderr.write
p("Usage: %s nwords [nbits]\n" % argv0)
p("Generates a password of nwords words, each with nbits bits\n")
p("of entropy, choosing words from the first entries in\n")
p("$HOME/devel/wordlist, which should be in the same format as\n")
p("<http://canonical.org/~kragen/sw/wordlist>, which is a text file\n")
p("with one word per line, preceded by its frequency, most frequent\n")
p("words first.\n")
p("\nRecommended:\n")
p(" %s 5 12\n" % argv0)
p(" %s 6\n" % argv0)
return 1
def read_file(filename, nbits):
return [line.split()[1] for line in
itertools.islice(open(filename), 2**nbits)]
def generate_password(nwords, wordlist):
choice = random.SystemRandom().choice
return ' '.join(choice(wordlist) for ii in range(nwords))
def display_password(password, nwords, nbits):
print 'Your password is "%s".' % password
entropy = nwords * nbits
print "That's equivalent to a %d-bit key." % entropy
print
# My Celeron E1200
#
(<http://ark.intel.com/products/34440/Intel-Celeron-Processor-E1200-(512K-Cache-1_60-GHz-800-MHz-FSB)>)
# was released on January 20, 2008. Running it in 32-bit mode,
# john --test (<http://www.openwall.com/john/>) reports that it
# can do 7303000 MD5 operations per second, but I’m pretty sure
# that’s a single-core number (I don’t think John is
# multithreaded) on a dual-core processor.
t = years(entropy, 7303000 * 2)
print "That password would take %.2g CPU-years to crack" % t
print "on my inexpensive Celeron E1200 from 2008,"
print "assuming an offline attack on a MS-Cache hash,"
print "which is the worst password hashing algorithm in common use,"
print "slightly worse than even simple MD5."
print
t = years(entropy, 3539 * 2)
print "The most common password-hashing algorithm these days is FreeBSD’s"
print "iterated MD5; cracking such a hash would take %.2g CPU-years." % t
print
# (As it happens, my own machines use Drepper’s SHA-2-based
# hashing algorithm that was developed to replace the one
# mentioned above; I am assuming that it’s at least as slow as the
# MD5-crypt.)
# <https://en.bitcoin.it/wiki/Mining_hardware_comparison> says a
# Core 2 Duo U7600 can do 1.1 Mhash/s (of Bitcoin) at a 1.2GHz
# clock with one thread. The Celeron in my machine that I
# benchmarked is basically a Core 2 Duo with a smaller cache, so
# I’m going to assume that it could probably do about 1.5Mhash/s.
# All common password-hashing algorithms (the ones mentioned
# above, the others implemented in John, and bcrypt, but not
# scrypt) use very little memory and, I believe, should scale on
# GPUs comparably to the SHA-256 used in Bitcoin.
# The same mining-hardware comparison says a Radeon 5870 card can
# do 393.46 Mhash/s for US$350.
print "But a modern GPU can crack about 250 times as fast,"
print "so that same iterated MD5 would fall in %.1g GPU-years." % (t / 250)
print
# Suppose we depreciate the video card by Moore’s law,
# i.e. halving in value every 18 months. That's a loss of about
# 0.13% in value every day; at US$350, that’s about 44¢ per day,
# or US$160 per GPU-year. If someone wanted your password as
# quickly as possible, they could distribute the cracking job
# across a network of millions of these cards. The cards
# additionally use about 200 watts of power, which at 16¢/kWh
# works out to 77¢ per day. If we assume an additional 20%
# overhead, that’s US$1.45/day or US$529/GPU-year.
cost_per_day = 1.45
cost_per_crack = cost_per_day * 365 * t
print "That GPU costs about US$%.2f per day to run in 2011," % cost_per_day
print "so cracking the password would cost about US$%.1g." % cost_per_crack
def years(entropy, crypts_per_second):
return float(2**entropy) / crypts_per_second / 86400 / 365.2422
if __name__ == '__main__':
sys.exit(main(sys.argv))
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