GCC Teaching: 052445 LOG eSpace Concepts 1. LOG of the course, Groupware Competence Center, Paderborn 2002.

Log of the course eSpace Concepts 1

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YEAR: 2002
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User: Anonymous


PLACES: Paderborn
TIME: 2002
 
16.12.
Example: RSA Algorithm
.
I. RSA: Generating the encryption keys
.
The public key :
.
1. Choose an odd number E: E = 5
.
2. Choose 2 prime-numbers,
p and q with p not q, and make sure
that it is possible to divide
(p-1)(q-1)-1 with E: p = 7, q = 17
.
3. Multiply p with q and get N: N = p * q = 7 * 17 = 119
.
4. The numbers N and E are now
the Public Key: (N, E) = (119, 5)
.
.
The private key :
.
1. Subtract 1 from p,q and e,
multiply the result and add 1: (p-1) (q-1) (E-1) + 1 =
6 * 16 * 4 + 1 = 385
.
2. Divide the result with E and get D: D = 385 / 5 = 77
.
3. The numbers N and D are now
the Private Key: (N, D) = (119, 77)
.


II. How to encrypt using the keys
.
Encrypt the message using the public key:
.
1. Translate the text-strings into a
collection of numbers, like appropriate
number of the char in the alphabetical
order; by example char "s" is now 19: Original Char = 19
.
2. The algorithm :
a) get the Power of the text-
chars with E: 195 = 2476099
.
b) divide with N and get the 2476099 / 119 = 20807 + 66/119
remainder:
.
3. The remainder is the encrypted
Char "s": Encrypted Char = 66
.
Decrypt the message using the private key:
.
1. The algorithm:
a) get the Power of the
encrypted Text with D: 6677 = (1,27,.......)140
.
b) divide with N: (1,27,...)140 / 119 = (1,069,...)138
+19 / 119
.
2. The rest of this term is the decrypted
char: Decrypted Char = 19