Understanding-dictionary About Encryption Systems !

[Trinity]

RSA (PART 2)


Background, Part I: How to Calculate with Exponents

Here's a quick refresher on how to combine exponents. Recall that:

N^2 = N * N,
N^3 = N * N * N,
N^4 = N * N * N * N,

and so on. For example:

2^7 = 2 * 2 * 2 * 2 * 2 * 2 * 2 = 128.
If we fiddle with exponents for a bit, we will quickly realize that:

N^E * N = N^(E + 1).
So, for example:

2^7 * 2 = 128 * 2 = 256 = 2^8.
Building upon this, we can also see that:

N^E * N * N = N^(E + 2).
But N * N can also be written as N^2:

N^E * N^2 = N^(E + 2).
We can extrapolate from this, and derive a more general rule -- namely:

N^A * N^B = N^(A + B).
And, if we repeated this process on the next level up, we would find that:

(N^A)^B = N^(A * B).
These two simple facts are very useful when handling exponent-laden
formulas.

Background, Part II: Modulus Arithmetic

Most computer programmers are familiar with modulus as a "remainder"
operator, usually denoted by "%", which gives the remainder of an integer
division instead of the quotient. For example:

27 % 12 = 3.
Though the idea is the same, the mechanics here are slightly different from
what mathematicians refer to as modulus arithmetic. In essence, modulus
arithmetic consists of taking the infinitely long number-line and coiling
it around a finite circle. All the numbers that land on the same point
along the circle's edge are considered interchangeable, or congruent. Thus,
the analogue to the above example in modulus arithmetic would be expressed
as:

27 = 3 (mod 12),
or, in words:

27 is congruent to 3, modulo 12.
(Though note that mathematicians actually use a three-barred version of the
equal sign to indicate congruency.) In this case, 12 is the modulus that we
are working under, and the equation simply tells us that, under a modulus
of 12, 27 and 3 are considered to be the same number. Likewise:

11 + 16 = 3 (mod 12)
reads as:

11 plus 16 is congruent to 3, modulo 12.
Modulus arithmetic is sometimes called clockface arithmetic -- if it's
currently 11 o'clock, then 16 hours later it will be 3 o'clock. (Of course,
the analogy is less perfect when the modulus is something other than 12.)

An important feature of modulus arithmetic is that you can replace the
terms of an addition operation with congruent values, and still get the
right answer:

16 = 4 (mod 12), therefore
11 + 16 = 11 + 4 = 3 (mod 12).
Another example:

9835 = 7 (mod 12), and
1176 = 0 (mod 12), therefore
9835 + 1176 = 7 + 0 = 7 (mod 12).
Even better, this trick also works with multiplication:

9835 * 1176 = 7 * 0 = 0 (mod 12)
(and, if we check, we will see that, yes, 9835 * 1176 is 11565960, and
11565960 = 0 (mod 12)).

If our modulus was 10, then modulus arithmetic would be equivalent to
ignoring all but the last digit in our numbers:

37 = 7 (mod 10),
287 + 482 = 9 (mod 10), and
895 * 9836 = 0 (mod 10).
And, in a sense, a C program does all of its calculations in modulus
arithmetic. Since integer calculations in C are permitted to overflow, the
high bits silently falling off into the bit bucket, a C program using
32-bit integers is really doing all of its arithmetic modulo 2^32.

As you might imagine, some calculations that are time-consuming and produce
huge numbers become trivial in modulus arithmetic. The ability to reduce
values to their remainders before doing the actual calculation keeps the
calculations from getting out of hand.

Background, Part III: The Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra states that for every number, there is
exactly one way to factor that number into primes -- and vice versa: every
selection of primes multiplies into a different number. For example:

1176 = 2 * 2 * 2 * 3 * 7 * 7, or
1176 = 2^3 * 3^1 * 7^2.
It is guaranteed that there is no other way to break 1176 into prime
factors. And, certainly, any time you take three 2s, two 7s, and a three,
you're always going to get 1176 when you multiply them together. The
Fundamental Theorem of Algebra assures us that both these things are true
for every number.

(By the way, this is one of the reasons that 1 is not considered to be a
prime number: if it were, then each number would have an infinite number of
prime factorizations, all differing by how many 1s were included. Instead,
1 is considered to have no prime factors at all, and we say that a number
is prime if it has exactly one prime factor -- namely itself.)

Put another way, the Fundamental Theorem of Algebra states that the set of
all numbers and the set of all selections of prime numbers are "isomorphic"
-- there is a perfect one-to-one mapping between the two. A number is
therefore defined by its prime factorization.

Background, Part IV: Relatively Prime Numbers

The greatest common divisor (abbreviated GCD) of two numbers is the largest
number that evenly divides into both of them. For example:

GCD(15, 10) = 5,
GCD(18, 10) = 2,
GCD(21, 10) = 1, and
GCD(170, 102) = 34.
Or, another way to look at it is to say that the GCD is the intersection of
the two numbers' set of prime factors:

GCD((2^3 * 3^1 * 7^2), (2^2 * 5^1 * 7^3)) = 2^2 * 7^2, so
GCD(1176, 6860) = 196.
When two numbers have no common factors, their GCD will be 1, and the two
numbers are said to be relatively prime (or coprime). For example, we can
see in our list up above that 21 and 10 are relatively prime.

Since a prime number has no factors besides itself, clearly a prime number
is relatively prime to every other number. And the same can be said of the
number 1.

Okay. Enough background material. Let's get to the good stuff.

------------------------------------------------------------------------

Euler's Totient Function

Euler's Totient Function is denoted by the Greek letter phi, and is defined
as follows:

phi(N) = how many numbers between 1 and N - 1 which are
relatively prime to N.

Thus:

phi(4) = 2 (1 and 3 are relatively prime to 4),
phi(5) = 4 (1, 2, 3, and 4 are relatively prime to 5),
phi(6) = 2 (1 and 5 are relatively prime to 6),
phi(7) = 6 (1, 2, 3, 4, 5, and 6 are relatively prime to 7),
phi( = 4 (1, 3, 5, and 7 are relatively prime to , and
phi(9) = 6 (1, 2, 4, 5, 7, and 8 are relatively prime to 9).

Here is the same definition expressed as C code:

phi = 1;
for (i = 2 ; i < N ; ++i)
if (gcd(i, N) == 1)
++phi;

(By the way, notice that phi(1) is specially defined to be 1.)

It should be easy to see that phi(N) will be N - 1 whenever N is prime.
Somewhat less obvious is the useful fact that phi(N) is also easy to
calculate when N has exactly two different prime factors:

phi(P * Q) = (P - 1) * (Q - 1), if P and Q are prime.
(The proof of this fact is left as an exercise for the reader.) Thus, for
example:

phi(15) = 2 * 4 = 8 (1, 2, 4, 7, 8, 11, 13, and 14).

The two prime factors cannot be the same number for this to work, and in
fact you can see above that phi(9) does not equal 4.

Euler's Totient Theorem

This theorem is one of the important keys to the RSA algorithm:

If GCD(T, R) = 1 and T < R, then T^(phi(R)) = 1 (mod R).

Or, in words:

If T and R are relatively prime, with T being the smaller number,
then when we multiply T with itself phi(R) times and divide the
result by R, the remainder will always be 1.

We can test this theorem on some smaller numbers for which we have already
calculated the totient. Using 5 for T and 6 for R, we get:

phi(6) = (2 - 1) * (3 - 1) = 1 * 2 = 2, so
5^(phi(6)) = 5^2 = 25, and
25 = 24 + 1 = 6 * 4 + 1, therefore
25 = 1 (mod 6).
Using 2 for T and 15 for R, we have:

phi(15) = (3 - 1) * (5 - 1) = 2 * 4 = 8, so
2^(phi(15)) = 2^8 = 256, and
256 = 255 + 1 = 17 * 15 + 1, therefore
256 = 1 (mod 15).

[Trinity]

RSA (PART 3)

VARATIONS ON A THEME

Here again is the equation of Euler's Totient Theorem:

T^(phi(R)) = 1 (mod R)
(remembering that T < R, and T and R are relatively prime). Thanks to the
way that modulus arithmetic works on multiplication, we can easily see
that:

T^(phi(R)) * T^(phi(R)) = 1 * 1 (mod R),
which can be rewritten, using the laws of exponents, as:

T^(phi(R) + phi(R)) = 1 * 1 (mod R),
or:

T^(2 * phi(R)) = 1 (mod R).
If we ran through this sequence again, we would get:

T^(3 * phi(R)) = 1 (mod R).
Clearly, we could keep doing this as many times as we like. So, we can
expand on Euler's Totient Theorem, and state a more general corollary:

If GCD(T, R) = 1 and T < R, then T^(S * phi(R)) = 1 (mod R),
where S can be any number.

That's our first variation. Now, let's tweak this further by multiplying
both sides by T:

T^(S * phi(R)) * T = 1 * T (mod R).
Simplifying leaves us with:

T^(S * phi(R) + 1) = T (mod R).
If we repeat this, we will get:

T^(S * phi(R) + 1) * T = T * T (mod R),
or:

T^(S * phi(R) + 2) = T^2 (mod R).
Doing this yet again will give us:

T^(S * phi(R) + 3) = T^3 (mod R),
and so on.

What makes this so interesting is that S can be any number. It means that,
when calculating the value of:

T^E (mod R),
if E happens to be greater than phi(R), we can subtract phi(R) from E, and
keep on subtracting it until we have a number less than phi(R), and the new
formula will still produce the same value.

Guess what? This is just another way of describing modulus arithmetic. So,
what this boils down to is the rather surprising rule:

T^E = T^F (mod R) whenever E = F (mod phi(R)).
(again, as long as T < R, and T and R are relatively prime). In other
words, when doing modulus arithmetic, exponentiation works differently than
addition and multiplication. You can reduce an exponent, but you need to
use phi(R), and not R itself.

The Plot Thickens

We just blew past something very important. Let's back up and look at this
equation more closely:

T^(S * phi(R) + 1) = T (mod R).
Notice what we have here. We take a number, T, and raise it to a power, and
when we do the calculation in modulus arithmetic, we wind up with T again.
In short, we have a recipe for a function that returns its own input
(presuming that R has been chosen ahead of time, and T is relatively prime
to R).

If you're thinking to yourself, "What's so interesting about that?", then
consider what we would have if we broke this function up into two separate
steps. Specifically, let's imagine that we can find two new numbers P and Q
such that:

P * Q = S * phi(R) + 1, for some value of S that we choose.
Then we could rewrite:

T^(S * phi(R) + 1) = T (mod R)
as:

T^(P * Q) = T (mod R).
Now, this is equivalent to:

(T^P)^Q = T (mod R),
and this is something that can be broken up into two steps:

T^P = X (mod R), and then
X^Q = T (mod R).
Now, if you don't see the value in doing this, imagine now that the two
steps are performed on separate computers. And that X is sent from the
first computer to the second over an insecure phone line....

Does This Really Work?

T stands for the plaintext, the message that is to be sent. P, Q, and R
together form the cipher's keys -- P and R make up the public key, and Q
and R make up the private key. And X becomes the encrypted message.

Here, again, is the central equation that makes it all possible:

P * Q = S * phi(R) + 1.
Note here that P and Q will both automatically be relatively prime to
phi(R). (Why? Because their product, P * Q, is one more than a multiple of
phi(R). Any factors of P or Q must also be factors of P * Q. Any number
that is a factor of S * phi(R) + 1 can't also be a factor of phi(R).) This
is important, since it helps to assure us that a P and Q can actually be
found.

Imagine a clockface, with just an hour hand, and imagine yourself placing
the hour hand on midnight and then moving it forward by jumps, over and
over, each jump covering N hours. If you pick a value for N that is
divisible by 2 or 3 (the prime factors of 12), then you will find that you
will only hit certain numbers before you return to midnight, and the
sequence will then repeat. If N is 2, then the hour hand will visit 12, 2,
4, 6, 8, 10, 12, 2, 4, 6, 8, 10, 12 ...

If, however, your N is relatively prime with 12, then you will wind up
hitting every number exactly once before finally returning to midnight 12
jumps later. For example, using 7 for your N, the itinerary would be: 12,
7, 2, 9, 4, 11, 6, 1, 8, 3, 10, 5, 12, ... In addition, the order in which
you visit the numbers is entirely dependent on what value you pick for N.

In a similar vein, it is important that both P and Q be relatively prime to
phi(R). Because of this, we know that every possible value for T, when
raised to the power P modulo R, will land on a different number. (Remember
that when doing exponents in modulus arithmetic, it is actually phi(R), and
not R itself, that determines the length of the cycles.) If this weren't
true -- if P, for example, shared a factor in common with phi(R) -- then
some values for T could get mapped to the same value for X, and it would
clearly be impossible to tell which was originally which. There could not
be one value for Q that would correctly map X back to T every time.

The question of which T-values will wind up going to which X-values depends
entirely on the value used for P -- and here's the rub for the would-be
codebreaker: Just about every possible mapping of T-values to X-values does
in fact exist. Somewhere out there is a P that will make that mapping.

If this:

T^P = X
X^Q = T
was the cipher's scheme, there'd be no cipher. With P already being public
knowledge, it's tedious but no great trick to take an X and compute
backwards to T. But, instead, we have this:

T^P = X (mod R)
X^Q = T (mod R)
as the cipher's scheme, and that changes everything. The modulus arithmetic
erases too much information. There's no way to deduce how many times the
hour hand needs to spin around the clockface when it goes from X back to T.
Without knowing what Q is, a given X could wind up going to any of the
possible values for T.

But what is really maddening to our would-be codebreaker is that even when
T and P and X are all known, Q still can't be deduced! (Of course, it
actually can -- but not necessarily within anyone's lifetime. But we're
getting ahead of ourselves.)

So, let's see how to make this recipe work.

[Trinity]

RSA (PART 4)

MAKING A PAIR OF KEYS

To construct our own personal cipher keys, we need an appropriate value for
R. So, we start by randomly picking two prime numbers, U and V, and
multiplying them together:

R = U * V.
There are two good reasons for selecting a value for R that has exactly two
prime factors. Not only do we have an easy formula for calculating phi(R):

phi(R) = (U - 1) * (V - 1),
but we also want R to be hard to factor. The fewer factors a number has,
the longer it takes to find them.

We then need to figure out values for P, Q, and S, such that:

P * Q = S * phi(R) + 1.
When the numbers have been chosen, P and R together become the public key,
and Q and R make up the private key. U, V, and S are no longer needed, and
can be forgotten.

An Example

In order to see all this in action, we want to stick with numbers that we
can actually work with. So, for our example, we will select the primes 5
and 7 to be our U and V. This gives R a value of 35, and:

phi(35) = (5 - 1) * (7 - 1) = 4 * 6 = 24.
Now, we need to find numbers to fit the equation:

P * Q = S * 24 + 1.
We'd prefer P and Q to not share any factors with each other, since that
would make it easier to derive one from the other. (And we certainly don't
want P and Q to be the same number, since that would turn our trapdoor
cipher into a garden-variety symmetric cipher!) So, P and Q should be
relatively prime. We will try assigning values to S and see what S * 24 + 1
equals. If we can divide the resulting number into two relatively prime
values, then we have a worthy pair.

2 * 24 + 1 = 49 = 7 * 7 (no, factors must be different)
3 * 24 + 1 = 73 = 73 (we need two numbers)
4 * 24 + 1 = 97 = 97 (ditto)
5 * 24 + 1 = 121 = 11 * 11 (nope)
6 * 24 + 1 = 145 = 5 * 29 (jackpot)

We could continue searching -- nothing requires us to take the first value
that works -- but this is fine for our purposes. So, we have 5 for P, our
public key, and 29 for Q, our private key.

To make our cipher work, you may recall that the values we use for T must
be less than R, and also relatively prime to R. We also don't want to use 1
for T, because 1 raised to any power whatsoever is going to remain 1. That
leaves us with 23 values we can use for T. We'll use the following
character set:

2 3 4 6 8 9 11 1213 16 17 1819 22 23 2426 27 29 3132 33 34
A B D E F G H I J K L M N O P R S T U V W Y Z

(We're missing a few letters, but this can be worked around with some
creative misspellings: KS for X, KW for Q, and K or S for C. In any case,
it's not important for this example.)

The message we will encrypt is "VENIO" (Latin for "I come"):

V E N I O
31 6 19 1222

To encode it, we simply need to raise each number to the power of P modulo
R.

V: 31^5 (mod 35) = 28629151 (mod 35) = 26
E: 6^5 (mod 35) = 7776 (mod 35) = 6
N: 19^5 (mod 35) = 2476099 (mod 35) = 24
I: 12^5 (mod 35) = 248832 (mod 35) = 17
O: 22^5 (mod 35) = 5153632 (mod 35) = 22

So, our encrypted message is 26, 6, 24, 17, 22 -- or "SERLO" in our
personalized character set.

When the message "SERLO" arrives on the other end of our insecure phone
line, we can decrypt it simply by repeating the process -- this time using
Q, our private key, in place of P.

S: 26^29 (mod 35) = 10819995774...2921981566976 (mod 35) = 31
E: 6^29 (mod 35) = 36845653286788892983296 (mod 35) = 6
R: 24^29 (mod 35) = 10620036506...3621199028224 (mod 35) = 19
L: 17^29 (mod 35) = 48196857210...1825223071697 (mod 35) = 12
O: 22^29 (mod 35) = 85164331908...9721106030592 (mod 35) = 22

The result is 31, 6, 19, 12, 22 -- or "VENIO", our original message.

How to Crack RSA

Now, let's switch hats. Imagine that we've just managed to pluck the
message "SERLO" off of our wiretap. By looking up the message's destination
in the public-key directory, we find that our message was encrypted with a
value of 35 for R and 5 for P. How do we go about decrypting it when we
don't know the value for Q?

Well, we know that that:

P * Q = S * phi(R) + 1.
We can divide both sides of the equation by P, which gives us a formula for
Q:

Q = (S * phi(R) + 1) / P.
S is also unknown, though, so we will have to try plugging in different
numbers for S, and look for values for Q that meet all the necessary
constraints.

(1 * 24 + 1) / 5 = 25 / 5 = 5 (no, same number as P)
(2 * 24 + 1) / 5 = 49 / 5 (doesn't divide evenly)
(3 * 24 + 1) / 5 = 73 / 5 (ditto)
(4 * 24 + 1) / 5 = 97 / 5 (ditto)
(5 * 24 + 1) / 5 = 121 / 5 (ditto)
(6 * 24 + 1) / 5 = 135 / 5 = 29 (this could be it!)

Each time we find a candidate for Q, we could test it out on a few
messages, and see if it works. (Note that, for example, when S = 11, Q will
have a value of 53, which also meets all the constraints, but does not
decrypt correctly with P = 5.) If 29 hadn't worked and we needed to
continue the search, it would be pretty obvious that we only needed to test
every fifth number, since those are the only numbers which will give us a
result that is evenly divisible by 5. So, even though this process involves
a brute-force search, it is very simple and very fast.

So what's the catch? Well, in order to do any of this, we first need to
know the value of phi(R). Of course, we already know that R has exactly two
prime factors, so calculating phi(R) is a snap once we know what those
factors are.

Famous last words.

[Trinity]

RSA (PART 5)

HOW TO MAKE RSA UNCKRACKABLE

Of course, in our case the factors of R can be found by consulting a times
table. So it's not much of a challenge. (For that matter, since we're
encrypting one character at a time, our coded messages would also be
vulnerable to good old-fashioned cryptanalysis).

To make it less easy to find R's factors, we need to pick larger prime
numbers for U and V to begin with. If, instead of 5 and 7, we had chosen
673 and 24971, we would have a value of 16805483 for R, and phi(R) would be
16779840. (This would also give us enough room to encrypt three bytes at a
time, which pretty much ruins any hope of cryptanalysis.) Looking for a P
and Q pair is no longer something you want to do with pencil and paper, of
course, but it took me less than three minutes to find the usable pair 397
and 211333 -- including the time it took to write and debug a Perl script.

On the other hand, it also took me less than three seconds to run "factor"
on 16805483 to obtain 673 and 24971. Armed with those numbers, it wouldn't
take long to derive 211333 from 397. So even these numbers aren't close to
being large enough. We need really big numbers.

Well, we can certainly find huge values for R that are difficult to factor.
But how far can we go before it becomes too difficult for us to use the
numbers in the first place?

Huge Exponents in Modulus Arithmetic

The problem is this: The bigger R gets, the bigger P and Q will be, and P
and Q are to be used as exponents! Even the relatively tame-looking:

9^(9^9)
produces a number over 350 million decimal digits long. How are we going to
be able to encrypt anything without needing terabytes of storage?

The trick is that we only need to calculate these exponential values modulo
R. As always, modulus arithmetic simplifies things a great deal.

Let's revisit our example, and look at how we could decrypt the first
number, 31, remembering that R = 35 and Q = 29:

31^29 (mod 35) = ?
To start with, we look at Q's binary representation. 29 in binary is 11101,
which means that:

29 = 16 + 8 + 4 + 1, or
29 = 2^4 + 2^3 + 2^2 + 2^0.
We can now break the exponential calculation apart into several smaller
ones:

31^29 = 31^(2^4 + 2^3 + 2^2 + 2^0)
= 31^(2^4) * 31^(2^3) * 31^(2^2) * 31^(2^0)
= 31^(2 * 2 * 2 * 2) * 31^(2 * 2 * 2) * 31^(2 * 2) * 31
= (((31^2)^2)^2)^2 * ((31^2)^2)^2 * (31^2)^2 * 31.

This may look like anything but an improvement, at first. But on a closer
examination, you'll see that we actually have many repeated subterms. This
simplifies matters, particularly when we take advantage of the fact that we
are calculating in modulo 35.

We compute the first square in modulus arithmetic:

31^2 = 961 = 16 (mod 35).
By substituting this value into our equation:

31^29 = (((31^2)^2)^2)^2 * ((31^2)^2)^2 * (31^2)^2 * 31 (mod 35),
we get:

31^29 = ((16^2)^2)^2 * (16^2)^2 * 16^2 * 31 (mod 35).
Now by computing that square:

16^2 = 256 = 11 (mod 35),
we will have:

31^29 = (11^2)^2 * 11^2 * 11 * 31 (mod 35).
We keep simplifying in the same fashion:

11^2 = 121 = 16 (mod 35), and
16^2 = 256 = 11 (mod 35),
and so:

31^29 = 16^2 * 16 * 11 * 31 (mod 35)
= 11 * 16 * 11 * 31 (mod 35).

We can continue to take advantage of the modulus when we do the final
multiplications:

31^29 = 11 * 16 * 11 * 31 (mod 35)
= 11 * 16 * 341 (mod 35)
= 11 * 16 * 26 (mod 35)
= 11 * 416 (mod 35)
= 11 * 31 (mod 35)
= 341 (mod 35)
= 26 (mod 35).

Lo and behold: 26, the same result as when we did it the hard way.

This binary technique is really no different than how computers normally
compute integer powers. However, the fact that we can break the process
down to successive multiplications allows us to apply the modulus at every
step of the way. This assures us that at no point will our algorithm have
to handle a number larger than (R - 1)^2.

[Trinity]

RSA (PART 6)

Safety in Numbers

Okay. So we know that the process of encryption and decryption is still
practical, even if R is immense. But all of this is moot unless we can
still generate the keys. If we want R to be so big that it can't be
factored easily, how are we going to find it in the first place?

Well, it turns out that there is an interesting little asymmetry here.
There happen to be relatively cheap ways to test a number for primality.
One of the most famous is based on what is called Fermat's Little Theorem.
Here is the version of this Theorem that we're interested in:

If P is prime, then N^(P - 1) = 1 (mod P) is true for every
number N < P.

(If this seems suspiciously reminiscent of Euler's Totient Theorem, it
should. Euler was the first person to publish a proof of this Theorem, and
his Totient Theorem is a generalization of Fermat's. You can see this for
yourself by remembering that phi(P) = P - 1 when P is prime.)

Of course, from a mathematician's point of view, this theorem is mainly
useful for proving that a given number is composite. For example, it just
so happens that:

4^14 (mod 15) = 268435456 (mod 15) = 1,
even though 15 is no prime. Nonetheless, it is also true that:

3^14 (mod 15) = 4782969 (mod 15) = 9, and
5^14 (mod 15) = 6103515625 (mod 15) = 10.
On the other hand, 17, which is prime, results in 1 every time:

3^16 (mod 17) = 43046721 (mod 17) = 1,
4^16 (mod 17) = 4294967296 (mod 17) = 1,
5^16 (mod 17) = 152587890625 (mod 17) = 1, and so on.
So, if we want to know if a number is prime, we can run it through this
test, using (say) 2 as the base. If anything besides 1 results, we know
with certainty that the number is composite. If the answer is 1, however,
we try the test again with 3, and then 4, and so on. If we keep getting
back 1 as the result, it soon becomes astronomically unlikely that the
number is anything but prime. This doesn't constitute proof, mathematically
speaking, but it certainly works for our purposes.

There are other tests for primality, some of which are faster. But they all
have at least one thing in common with this test: When they reject a
number, it tells us only that the number can be factored. The test results
give us no information at all as to what the factors might be. How
unfortunate!

Unfortunate for the mathematicians, that is. Very fortunate for us.

The basic truth is that, in order to find the factors of a composite
number, we're pretty much stuck with using brute force: Divide the number
by all the primes you can get your hands on until one of them goes in
evenly. There are ways to improve on this approach (the Number Field Sieve
currently being the best), but they are complicated, and all they do is
allow you to narrow the scope of the search. They don't reduce the search
enough to make this problem tractable in general.

Nor is it likely that new approaches will, either! The real issue is that
the encrypting and decrypting algorithms have a running time that is linear
with respect to the length of R. That is to say, doubling the number of
digits in R doubles the amount of time (roughly) needed to encrypt,
decrypt, and to select the two primes to make a key with. But the
algorithms for factoring R have a running time that is exponential with
respect to the length of R. That is to say, the time (roughly) doubles with
every few digits! (Because every digit added to R makes it ten times
larger, and thus multiplies the number of potential candidates for its
measly two factors.)

So if a new technique is suddenly found that makes it a trillion times
faster to factor a number, all we have to do is increase the size of R we
use by enough digits, and the situation will be right back where it started
-- and all it means to us is that it takes a little bit longer to send and
receive our messages. Already some people are using keys that, in order to
factor with the Number Field Sieve, would require more energy than exists
in the known universe.

An illustration: To the best of my knowledge, the number used as the
modulus for the RSA-140 challenge is the largest general number than has
been independently factored. (By general, I'm excluding creatures like
Mersenne numbers and Fermat numbers, which have specialized factoring
techniques that are inapplicable elsewhere.) It was completed on February
2, 1999. Now, the record previous to this was the RSA-130 number, and the
process of factoring it was estimated as taking a total of 1000 MIPS-years
of computer time. RSA-140, a number only 10 decimal digits longer, required
twice that amount.

This, finally, is the heart of what makes RSA a trapdoor function: the gap
between obtaining a number with two prime factors, and rediscovering the
factors from the number itself. And the gap just keeps expanding as the
numbers get larger.

The breakthrough that would completely destroy RSA's security would be an
algorithm that actually produced a number's factors directly, instead of
merely narrowing the search's scope. Such a thing has not been proven
impossible, and it would probably be very hard to do so. But considering
the renewed attention that has been focused on this problem in the last two
decades, the likelihood of the existence of such an algorithm is once again
starting to appear quite remote. Discovering one would change the face of
number theory as much as RSA has changed the face of cryptography.

However -- if this were to happen, there are other trapdoor functions out
there, waiting to be found. Whatever the future of RSA may be, the trapdoor
cipher has certainly changed the face of cryptography forever.

------------------------------------------------------------------------

References

1. Clawson, Calvin C.: "Mathematical Mysteries", 1996, Plenum Press.
(Clawson devotes an entire chapter to the mathematics behind RSA, and it is
this that gave me the inspiration to create this text.)

2. Gardner, Martin: "Penrose Tiles to Trapdoor Ciphers", 1989, W.H. Freeman
& Co. (This is another anthology of Gardner's wonderful columns for
"Scientific American", and includes the column which was the first widely
published description of the RSA cipher -- the one which set the NSA to
frantically running around in circles.)

3. Ribenboim, Paulo: "The Little Book of Big Primes", 1991,
Springer-Verlag. (The title should actually be "The Little Book of Big
Number Theory" -- the book is chock full of theorems and conjectures that
relate to prime numbers.)

4. Wells, David: "The Penguin Dictionary of Curious and Interesting
Numbers", 1986, Penguin Books. (I had to pull this out at the last minute
to find out how many digits were in 9^(9^9). For the curious whose
libraries lack this little gem, the exact number of digits is 369693100.)

[beer4life]

Hi all future "Einsteins",

This has got to be the best description of decryption that I have ever seen. I'm sure that it will take me the next 20 years to grasp the fundamentals, if ever.

In the meantime, I Heartily congratulate and thank "Trinity" for his insight into this intriguing facet of our hobby. I will certainly look forward to any further elucidation with bated breath.

Thanks Cobber, Kindest Regards, Bill. :cool: :smilewink

[beer4life]

Hi all, not as easy as I thought. Let me know if this is any better than the PDF file.
Open to suggestions. It's 47 Pages.
Kindest Regards, Bill.:cry:

[beer4life]

Hi all, It has been brought to my attention that some may prefer the PDF file version.

We aim to please. I think it's 69 pages. :smilewink

With special thanks to "Trinity", and thanks to "cr3pt" and "urmans" for their help.

Kindest Regards to all concerned, Bill.