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This tutorial covers the basics of the science of cryptography. It explains how programmers and network professionals can use cryptography to maintain the privacy of computer data. Starting with the origins of cryptography, it moves on to explain cryptosystems, various traditional and modern ciphers, public key encryption, data integration, message authentication, and digital signatures.

Friday 13 September 2019

MANUAL RSA ENCRYPTION AND DECRYPTION WITH JAVA

                                    RSA Algorithm

For Example :

p value is 47
q value is 71
public key value is 79
Message : "HARI INI"

Question :
How Is Encryption And Decryption Process ?

Answer :

MANUAL RSA ENCRYPTION PROCESS

Prime 1 (p) value :47
Prima 2 (q) value :71
public (e) key :79
N (p x q) value :3337
phi (N) --> (p-1) x (q-1) :3220
Private Key (d) : (1+k*3220)/79; k=1,2,3,....
find d with rounded results by trying k values
obtained private key value (d) :1019
So :
Public Key : (79,3220)
Private Key: (1019,3220)

Message (M) = HARI INI

===============
ENCRYPTION PROCESS
===============

Conversion Message to Decimal Format
H = 72
A = 65
R = 82
I = 73
  = 32
I = 73
N = 78
I = 73

For this encryption process, I break m into smaller blocks, for example m is broken into six blocks that require 3 digits

M0=726
M1=582
M2=733
M3=273
M4=787
M5=003

CipherText (C) = Plaintext (M) ^ e mod N

C0 = 726 ^ 79 mod 3337 = 215
C1 = 582 ^ 79 mod 3337 = 776
C2 = 733 ^ 79 mod 3337 = 1743
C3 = 273 ^ 79 mod 3337 = 933
C4 = 787 ^ 79 mod 3337 = 1731
C5 = 3 ^ 79 mod 3337 = 158

CipherText: 215.776.1743.933.1731.158

Plaintext (M) =Ciphertext (C) ^ d mod N

P0 = 215 ^ 1019 mod 3337 = 726
P1 = 776 ^ 1019 mod 3337 = 582
P2 = 1743 ^ 1019 mod 3337 = 733
P3 = 933 ^ 1019 mod 3337 = 273
P4 = 1731 ^ 1019 mod 3337 = 787
P5 = 158 ^ 1019 mod 3337 = 3

Convert Desimal to Ascii
Return Decryption
72=H
65=A
82=R
73=I
32=
73=I
78=N
73=I

---- FINISH----


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