Cryptonomicon (27 page)

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Authors: Neal Stephenson

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where
i
= (1, 2, 3, . . . (∞))

more or less, depending on how close to infinitely long Turing wants to keep riding his bicycle. After a while, it seems infinitely long to Waterhouse.

Turing’s chain will fall off when his bicycle reaches the state (
theta
= 0,
C
= 0) and in light of what is written above,
this will happen when
i
(which is just a counter telling how many times the rear wheel has revolved) reaches some hypothetical value such that
in
mod
l
= 0, or, to put it in plain language, it will happen if there is some multiple of
n
(such as, oh, 2
n,
3
n,
395
n
or 109,948,368,443
n
) that just happens to be an exact multiple of
l
too. Actually there might be several of these so-called common multiples, but from a practical standpoint the only one that matters is the first one—the least common multiple, or LCM—because that’s the one that will be reached first and that will cause the chain to fall off.

If, say, the sprocket has twenty teeth (
n
= 20) and the chain has a hundred teeth (
l
= 100) then after one turn of the wheel we’ll have
C
= 20, after two turns
C
= 40, then 60, then 80, then 100. But since we are doing the arithmetic modulo 100, that value has to be changed to zero. So after five revolutions of the rear wheel, we have reached the state (
θ
= 0, C = 0) and Turing’s chain falls off. Five revolutions of the rear wheel only gets him ten meters down the road, and so with these values of
l
and
n
the bicycle is very nearly worthless. Of course, this is only true if Turing is stupid enough to begin pedaling with his bicycle in the chain-falling-off state. If, at the time he begins pedaling, it is in the state (
θ
= 0,
C
= 1) instead, then the successive values will be
C
= 21, 41, 61, 81, 1, 21, . . . and so on forever—the chain will never fall off. But this is a degenerate case, where “degenerate,” to a mathematician, means “annoyingly boring.” In theory, as long as Turing put his bicycle into the right state before parking it outside a building, no one would be able to steal it—the chain would fall off after they had ridden for no more than ten meters.

But if Turing’s chain has a hundred and one links (
l
= 101) then after five revolutions we have
C
= 100, and after six we have
C
= 19, then

 

C = 39, 59, 79, 99, 18, 38, 58, 78, 98, 17, 37, 57, 77, 97, 16, 36, 56, 76, 96, 15, 35, 55, 75, 95, 14, 34, 54, 74, 94, 13, 33, 53, 73, 93, 12, 32, 52, 72, 92, 11, 31, 51, 71, 91, 10, 30, 50, 70, 90, 9, 29, 49, 69, 89, 8, 28, 48, 68, 88, 7, 27, 47, 67, 87, 6, 26, 46, 66, 86, 5, 25, 45, 65, 85, 4, 24, 44, 64, 84, 3, 23, 43, 63, 83, 2, 22, 42, 62, 82, 1, 21, 41, 61, 81, 0

 

So not until the 101st revolution of the rear wheel does the bicycle return to the state (
θ
= 0,
C
= 0) where the chain falls off. During these hundred and one revolutions, Turing’s bicycle has proceeded for a distance of a fifth of a kilometer down the road, which is not too bad. So the bicycle is usable. However, unlike in the degenerate case, it is
not
possible for this bicycle to be placed in a state where the chain never falls off at all. This can be proved by going through the above list of values of
C,
and noticing that every possible value of
C
—every single number from 0 to 100—is on the list. What this means is that no matter what value
C
has when Turing begins to pedal, sooner or later it will work its way round to the fatal
C
= 0 and the chain will fall off. So Turing can leave his bicycle anywhere and be confident that, if stolen, it won’t go more than a fifth of a kilometer before the chain falls off.

The difference between the degenerate and nondegenerate cases has to do with the properties of the numbers involved. The combination of (
n
= 20,
l
= 100) has radically different properties from (
n
= 20,
l
= 101). The key difference is that 20 and 101 are “relatively prime” meaning that they have no factors in common. This means that their least common multiple, their LCM, is a large number—it is, in fact, equal to
l
×
n
= 20 × 101 = 2020. Whereas the LCM of 20 and 100 is only 100. The
l
= 101 bicycle has a long
period
—it passes through many different states before returning back to the beginning—whereas the
l
= 100 bicycle has a period of only a few states.

Suppose that Turing’s bicycle were a cipher machine that worked by alphabetic substitution, which is to say that it would replace each of the 26 letters of the alphabet with some other letter. An A in the plaintext might become a T in the ciphertext, B might become F, C might become M, and so on all the way through to Z. In and of itself this would be an absurdly easy cipher to break—kids-in-treehouses stuff. But suppose that the substitution scheme
changed
from one letter to the next. That is, suppose that after the first letter of the
plaintext was enciphered using one particular substitution alphabet, the second letter of plaintext was enciphered using a completely different substitution alphabet, and the third letter a different one yet, and so on. This is called a polyalphabetic cipher.

Suppose that Turing’s bicycle were capable of generating a different alphabet for each one of its different states. So the state (
θ
= 0,
C
= 0) would correspond to, say, this substitution alphabet:

 

 

A  Q

B  G

C  U

D  W

E  B

F  I

G  Y

H  T

I  F

J  K

K  V

L  N

M  D

N  O

O  H

P  E

Q  P

R  X

S  L

T  Z

U  R

V  C

W  A

X  S

Y  J

Z  M

 

 

but the state (
θ
= 180,
C
= 15) would correspond to this (different) one:

 

 

A  B

B  O

C  R

D  I

E  X

F  V

G  G

H  Y

I  P

J  F

K  J

L  M

M  T

N  C

O  Q

P  N

Q  H

R  A

S  Z

T  U

U  K

V  L

W  D

X  S

Y  E

Z  W

 

 

No two letters would be enciphered using the same substitution alphabet—
until,
that is, the bicycle worked its way back around to the initial state (
θ
= 0,
C
= 0) and began to repeat the cycle. This means that it is a
periodic
polyalphabetic system. Now, if this machine had a short period, it would repeat itself frequently, and would therefore be useful, as an encryption system, only against kids in treehouses. The longer its period (the more relative primeness is built into it) the less frequently it cycles back to the same substitution alphabet, and the more secure it is.

The three-wheel Enigma is just that type of system (i.e., periodic polyalphabetic). Its wheels, like the drive train of Turing’s bicycle, embody cycles within cycles. Its period is 17,576, which means that the substitution alphabet that enciphers the first letter of a message will not be used again until the 17,577th letter is reached. But with Shark the Germans have added a fourth wheel, bumping the period up to 456,976. The wheels are set in a different, randomly chosen starting position at the beginning of each message. Since the Germans’ messages are never as long as 450,000 characters, the Enigma never reuses the same substitution alphabet in the course of a given message, which is why the Germans think it’s so good.

A flight of transport planes goes over them, probably headed for the aerodrome at Bedford. The planes make a weirdly musical diatonic hum, like bagpipes playing two drones at once. This reminds Lawrence of yet another phenomenon related to the bicycle wheel and the Enigma machine. “Do you know why airplanes sound the way they do?” he says.

“No, come to think of it.” Turing pulls his gas mask off again. His jaw has gone a bit slack and his eyes are darting from side to side. Lawrence has caught him out.

“I noticed it at Pearl. Airplane engines are rotary,” Lawrence says. “Consequently they must have an odd number of cylinders.”

“How does that follow?”

“If the number were even, the cylinders would be directly opposed, a hundred and eighty degrees apart, and it wouldn’t work out mechanically.”

“Why not?”

“I forgot. It just wouldn’t work out.”

Alan raises his eyebrows, clearly not convinced.

“Something to do with cranks,” Waterhouse ventures, feeling a little defensive.

“I don’t know that I agree,” Alan says.

“Just stipulate it—think of it as a boundary condition,” Waterhouse says. But Alan is already hard at work, he suspects, mentally designing a rotary aircraft engine with an even number of cylinders.

“Anyway, if you look at them, they all have an odd number of cylinders,” Lawrence continues. “So the exhaust noise combines with the propeller noise to produce that two-tone sound.”

Alan climbs back onto his bicycle and they ride into the woods for some distance without any more talking. Actually, they have not been talking so much as mentioning certain ideas and then leaving the other to work through the implications. This is a highly efficient way to communicate; it eliminates much of the redundancy that Alan was complaining about in the case of FDR and Churchill.

Waterhouse is thinking about cycles within cycles. He’s already made up his mind that human society is one of
these cycles-within-cycles things
*
and now he’s trying to figure out whether it is like Turing’s bicycle (works fine for a while, then suddenly the chain falls off; hence the occasional world war) or like an Enigma machine (grinds away incomprehensibly for a long time, then suddenly the wheels line up like a slot machine and everything is made plain in some sort of global epiphany or, if you prefer, apocalypse) or just like a rotary airplane engine (runs and runs and runs; nothing special happens; it just makes a lot of noise).

“It’s somewhere around… here!” Alan says, and violently brakes to a stop, just to chaff Lawrence, who has to turn his bicycle around, a chancy trick on such a narrow lane, and loop back.

They lean their bicycles against trees and remove pieces of equipment from the baskets: dry cells, electronic breadboards, poles, a trenching tool, loops of wire. Alan looks about somewhat uncertainly and then strikes off into the woods.

“I’m off to America soon, to work on this voice encryption problem at Bell Labs,” Alan says.

Lawrence laughs ruefully. “We’re ships passing in the night, you and I.”

“We are
passengers
on ships passing in the night,” Alan corrects him. “It is no accident. They need you precisely because I am leaving. I’ve been doing all of the 2701 work to this point.”

“It’s Detachment 2702 now,” Lawrence says.

“Oh,” Alan says, crestfallen. “You noticed.”

“It was reckless of you, Alan.”

“On the contrary!” Alan says. “What will Rudy think if he notices that, of all the units and divisions and detachments in the Allied order of battle, there is not a single one whose number happens to be the product of two primes?”

“Well, that depends upon how common such numbers are compared to all of the other numbers, and on how many other numbers in the range are going unused…” Lawrence says, and begins to work out the first half of the problem. “Riemann Zeta function again. That thing pops up everywhere.”

“That’s the spirit!” Alan says. “Simply take a rational and common-sense approach.
They
are really quite pathetic.”

“Who?”

“Here,” Alan says, slowing to a stop and looking around at the trees, which to Lawrence look like all the other trees. “This looks familiar.” He sits down on the bole of a windfall and begins to unpack electrical gear from his bag. Lawrence squats nearby and does the same. Lawrence does not know how the device works—it is Alan’s invention—and so he acts in the role of surgical assistant, handing tools and supplies to the doctor as he puts the device together. The doctor is talking the entire time, and so he requests tools by staring at them fixedly and furrowing his brow.


They
are—well, who do you suppose? The fools who use all of the information that comes from Bletchley Park!”

“Alan!”

“Well, it is foolish! Like this Midway thing. That’s a perfect example, isn’t it?”

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