AppendixC. A Worked Example.  Scalable Key Cryptography.
 From: adacrypt <austin.obyrne@xxxxxxxxxxx>
 Date: Thu, 24 Dec 2009 00:36:12 0800 (PST)
Alice and Bob are the industry pseudonyms for the sending entity and
the receiving entity respectively in some crypto circles although a
lot of people do not use the expressions and I suspect it is because
it is often disconcerting to talk with such familiarity about complete
strangers. It does save the repeated verbalising of large mouthfuls
of the English language and I shall use them here for that reason.
Alice is the administrator of an office secure network.
She has set the scope of her infrastructure encryption scheme at 14250
characters of message length capability. => she wants to be able to
send messages this long if needs be =>
X = 14250 + 63 = 14313
Lower Bound of N = 14313 + 127 = 14440
Upper Bound of N = 2(14313 + 32)= 2 x 14245 = 28690
So, the range of N is 14440 … 28690 inclusive (14440<= N <=28690)
The encryption algorithm is,
[(PlainText +X) + (Key+ X)(Mod N) = a residue (modulo N)
Cipher text = residue – N
Let us say that Alice wants to encrypt the letter Q (Capital Q) ( 81
in ASCII)
Let us say also that the instantaneous key she is using is the
‘$’(dollar sign) (36 in ASCII)
Let us say also that the instantaneous modulus N is 17359.
Then,
[(81 +14313) +(36 +14313)] (MOD 17359) = (14394 + 14349)(MOD 17359) =
11384
Cipher text = 11384 – 17359 = 5975 (in practice this is made positive
by multiplying it by (1)
ð Cipher text = 5975
The decryption algorithm is,
Plaintext = Cipher text + 2N – 2X  Key
Cipher text = 5975 x (1) = 5975
Then,
Plaintext (as message text) = 5975 + (2 x 17359) – (2 x 14313) – 36 =
81
Decoded back in ASCII 81 => ‘Q’ as expected.
Comment. There are 95x95x14250 possible permutations of (Plaintext,
Key, Modulus N) as encryption transformations in Alice’s scheme i.e. a
possibility space of 128606250.
(large possibility spaces mean nothing in cryptography of course).
The reader might like to try encrypting a few of your own characters.

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