(My apologies to those who are seeing this twice. I sent it
yesterday but it does not seem to have made it through.)
A number of us have been concerned about how PGP generates entropy.
Striking the keyboard beats using time as a source of random numbers,
but the degree of entropy is not well understood. Are there machines
where - for some reason - the keyboard strikes fall into some sort
of pattern?
And that's just when you are generating your public/private key pair.
What happens when you are just generating 128-bit keys for individual
messages? Where is the entropy coming from? I don't understand
completely, but somehow PGP collects entropy from the system and then
runs it into IDEA and then uses the numbers from the output. When the
program is not running, a pool of this data is kept in randseed.bin.
We already know it's a problem because it is hard to understand what
it is even doing, much less determine if it is consistent with sound
engineering practice.
This problem is certainly not confined to PGP. A lot of people have
worked hard to make computers deterministic. What do you do when you
need entropy?
Yes, you can buy a chip which generates random numbers at nearly any
bandwidth you like. But, right there you have a problem. How do you
know the output is really random? The answer is that you do not.
Playing cards are a nice source of randomness because they are widely
available and their behavior has been under study for a long time by
people with strong financial reasons for finding flaws. I slightly
prefer cards to dice because dice may be slightly predictable or even
loaded.
It would be nice if cryptography software would allow you to enter
randomly selected playing cards from time to time to increase the
entropy of keys. Careful people (Black Unicorn?) would enter the
cards prior to sending each message.
A well shuffled deck of 54 cards has about 237 bits of entropy. This
is easy to use: the program asks the order of the cards, converts this
to a string, and runs it through a one-way hash. (Entering the cards
is a bit of a nuisance. Is there an easy way to have them read
automatically?)
The Economist reports that seven riffle shuffles bring a deck "very
close to being random". ("Science and Technology: How to win at poker
and other science lessons" The Economist, October 12, 1996, pp:88-89.)
Cards have other uses. You can assign a (large) number to each
configuration of the deck. This allows you to use a deck of cards to
represent numbers. That is certainly convenient for generating large
random numbers. I have included some Lisp code below which converts
both ways between shuffled decks of cards and unique numbers.
This could be quite useful for strong steganography. Let's say you
have 230 bits you want to deliver across a border. It is unlikely
that anybody will study a deck of cards very carefully. Of course,
you need not limit yourself to cards. Any set of objects which can be
ordered will do the trick. For instance, a shelf with 100 books may
be used to store 524 bits. One thousand books may store 8529 bits.
How many bits can you store in that database which is in apparently
random order?
You can hide your data in two decks. There is a natural order to
cards: pick your favorite ordering of the suits and then do the rest
numerically. However, if you shuffle one deck of cards and lay them
in a row, you could use that ordering to define the mapping of the
second deck of cards to numbers. The Enemy may seize one deck or the
other, but without both, the number will not be revealed. While I
hesitate to use the term, this is a one-time pad. The first deck can
be thought of as the key and the second deck is the message.
How the Lisp Code Works
-----------------------
Let's use as an example a deck of five cards numbered from 0 to 4.
There are 5! = 120 combinations of these cards. We can think
of each card as a "digit" in a slightly odd numbering system.
In decimal arithmetic, we have columns whose value increments by 10^0,
10^1, 10^2, etc. In our numbering system of five cards, columns
increment by 0, 1, 2!, 3!, and 4!.
Consider the card sequence of (4 3 2 1 0). We wish to compute the
value stored in the leftmost column. Since there are five choices of
cards, this divides the number of possible combinations into five
pieces of 4! combinations each. So, in this example, the value in the
first column will be 4*4! = 96. This leaves us with cards (3 2 1 0).
We can look at this as a smaller number represented by a deck of four
cards. This smaller number is added later to the larger number to get
the final total number for our original deck. The first column of our
smaller deck is 3*3! = 18. The next one is 2*2! = 4, the one after
that is 1*1! = 1, and the last card never matters because there is
only one card to choose (0*0!). The total is 96 + 18 + 4 + 1 = 119.
This is the highest number we can represent with a deck of five
cards.
Consider the card sequence (0 1 2 3 4). This is the smallest value we
can represent, 0. The value of the first column is 0*4! = 0. This
leaves us with (1 2 3 4). But what has happened here? We can't treat
this as a normal smaller deck because it is missing a card. Besides,
1*1! = 1, which is not the zero we expected. We must renumber the
cards so that it is a reasonable deck again. The sequence of cards we
should really be looking at is (0 1 2 3). This means the value of the
next column should be 0*3!. Then we renumber the deck and continue
until we run out of cards. The answer is 0.
For an exercise, what is the value of (4 2 1 3 0)? (Answer at end.)
Now, how do we construct the deck given only the number defining its
combination? The code below is recursive. It computes the first
digit and then calls itself to recursively find the remaining cards.
The cards returned are always an internally consistent deck of cards;
that is, if sorted they will be consecutively numbered starting from
0. When a new card is inserted, the cards have to be renumbered to
accomodate the new one by incrementing every card which has the same
or greater face value as the card being inserted. (Insertion is
slightly misleading here because you place the card at the head of the
list. Insertion refers to the conceptual process of inserting the
card into the pre-existing order and adjusting the other values to
allow it.)
In other languages, this might be a little more complicated to
implement. Lisp provides a nice bignum package and also (I believe) a
compiler which knows how to convert the recursive routines below into
iterative routines. This code performs quite well even on a small
slow machine.
;;;-*- Mode: Lisp; Package: COMMON-LISP-USER -*-
(defun compute-combination-number (cards)
"Converts list of numbered cards into a unique number representing
their order."
(cond ((> (length cards) 1)
(+ (* (car cards)
(factorial (1- (length cards))))
(compute-combination-number (renumber-cards cards))))
(t 0)))
(defun renumber-cards (cards)
"Removes lead card from deck, decrements higher numbered cards
by one so there are no gaps."
(let ((renumbered-cards-reversed)
(lead-card (car cards)))
(dolist (card (cdr cards))
(cond ((> card lead-card)
(push (1- card) renumbered-cards-reversed))
(t
(push card renumbered-cards-reversed))))
(reverse renumbered-cards-reversed)))
(defun reconstruct-deck (combination-number deck-size)
"Converts unique number representing order of a deck of cards
and returns a list of numbers representing the deck."
(cond ((not (<= deck-size 1))
(multiple-value-bind (digit remaining-combination-number)
(floor combination-number
(factorial (1- deck-size)))
(insert-card digit (reconstruct-deck
remaining-combination-number
(1- deck-size)))))
(t (list 0))))
(defun insert-card (new-card card-list)
"Inserts a card into a deck, increasing by one every card which
is of higher or equal number so there are no duplicate cards."
(let ((new-deck-reversed))
(push new-card new-deck-reversed)
(dolist (card card-list)
(cond ((>= card new-card)
(push (1+ card) new-deck-reversed))
(t
(push card new-deck-reversed))))
(reverse new-deck-reversed)))
(defun factorial (number)
(cond ((or (= number 1) (= number 0)) 1)
(t (* number (factorial (1- number))))))
;;; Testing routines
(defun shuffle-deck (deck-size)
(randomize-list (build-card-list deck-size)))
(defun randomize-list (some-list)
(do ((randomized-list nil))
((null some-list) randomized-list)
(let ((item-number (random (length some-list))))
(push (elt some-list item-number)
randomized-list)
(setf some-list
(remove (elt some-list item-number) some-list)))))
(defun build-card-list (deck-size)
"Returns a list of consecutive numbers representing a deck of
cards."
(do ((card-list nil)
(count 0 (1+ count)))
((= deck-size count) card-list)
(push count card-list)))
(defun identicalp (list-one list-two)
"Returns non-nil if the two lists are identical."
(eval (cons 'and (map 'list #'equal list-one list-two))))
(defun exhaustively-test-card-combinations (deck-size)
"Verifies that the correct deck is reconstructed from the unique
order number for every combination of a small deck of cards."
(let ((combinations (factorial deck-size)))
(do ((combo-number 0 (1+ combo-number)))
((= combinations combo-number) t)
(cond ((not (= (compute-combination-number
(reconstruct-deck combo-number deck-size))
combo-number))
(warn "Test failed for ~D" combo-number))))))
(defun test-one-deck (deck-size)
"Shuffles a deck of cards and verifies that it may be reconstructed
from the unique number representing its order."
(let* ((original-deck (shuffle-deck deck-size))
(reconstructed-deck (reconstruct-deck
(compute-combination-number original-deck)
deck-size)))
(cond ((not (identicalp original-deck reconstructed-deck))
(warn "Test failed for ~A" original-deck)))))
(defun test-many-decks (trials max-deck-size)
"Shuffles trials decks of maximum size max-deck-size and verifies
that their order may be reconstructed from the unique number we compute."
(do ((trial-number 0 (1+ trial-number)))
((= trials trial-number) t)
(test-one-deck (1+ (random max-deck-size)))))
(defun complete-combination-test ()
"Good test of all the card combination routines."
(format t "Performing exhaustive test.~%")
(exhaustively-test-card-combinations 6)
(format t "Performing random test on large decks.~%")
(test-many-decks 10 100))
;; (Exercise Answer: (4 2 1 3 0) = 111)
