2005-05-23 Behandelte Themen

Reply to: 052456 LOG Grundlagen des Informationsmanagements am Arbeitsplatz (Angelika Bursig)
Date: 05/22/2005 - 10:00 PM
Author: Ludwig Nastansky
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052456 LOG Grundlagen des Informationsmanagements am Arbeitsplatz07.03.2005 16:14Angelika Bursig
. . 2005-05-09 Behandelte Themen09.05.2005 08:34Ludwig Nastansky
. . 2005-05-02 Behandelte Themen09.05.2005 08:34Ludwig Nastansky
. . 2005-04-11 Behandelte Themen22.05.2005 21:36Ludwig Nastansky
. . 2005-04-18 Behandelte Themen22.05.2005 21:39Ludwig Nastansky
. . 2005-04-25 Behandelte Themen22.05.2005 21:39Ludwig Nastansky
. . 2005-05-23 Behandelte Themen22.05.2005 22:00Ludwig Nastansky
. . 2005-05-30 Behandelte Themen29.05.2005 19:15Ludwig Nastansky
. . 2005-06-06 Behandelte Themen02.06.2005 15:17Bernd Hesse
. . 2005-06-13 Behandelte Themen13.06.2005 08:27Ludwig Nastansky
. . 2005-06-20 Behandelte Themen19.06.2005 22:41Ludwig Nastansky
. . 2005-07-11 Fotos27.07.2005 14:32Ludwig Nastansky
2005-06-06 Gastvortrag von Veit Florian Lier02.06.2005 15:16Bernd Hesse
 
... RSA-Prinzip
      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: 19
      5 = 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: 66
      77 = (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