Cryptography Essay

1052 Words3 Pages

Cryptography the method of applying protocols to information to make it seem like nonsense to hurt those who do not know the protocols. Cryptology is used to encrypt messages and create secure codes; the earliest known use of Cryptographs was in Egyptian times but over time several different types of cryptographs were made today we use it to keep information secure and to create registration numbers.
Focus:
This IA will focus on the math involved in certain ciphers, these ciphers are.
The classic cipher
Substitution cipher
ISBN codes
Public Key
Credit Card codes
Aim
My aim is to understand the ciphers and see if I can make some of my own
Investigation

Transition/Classic Cipher
This was a simple cipher and one of the earliest established. It was said that Julius Caesar used this cipher to protect messages of military importance, it is also known as the Caesar shift. The reason it is simple is because during the time not many people could read.
# 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
Plain a b c d e f g h i j k l m n o p q r s t u v w x y z cipher C D E F G H I J K L M N O P Q R S T U V W X Y Z A B

In this example each letter is given a corresponding numerical value, add 2 to that number then turn it back into a letter . It can be represented by the mathematical formula f(x)=x+2 and to change it back it would be the inverse of the function (x)^(-1)=x-2 . Using this formula a phrase like “Hello my name is” would become “jgkkqoaqcogku”. It seems like complete nonsense there are also no spaces in between words to make it harder to decrypt, to anyone who saw it and didn’t know that it was would just leave that information alone but knowing the cipher it is easily deciphered so it isn’t used ...

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Choose two distinct prim numbers p and q. p=2 , q =13
Work out n, n= pq n = 26
Work out S, S= (p-1)(p-2) S = (2-1)(13-1) = 1 x 12 = 12
Choose a number A which is smaller than S and has no common factors. A = 5
Find H, HA-1 is a multiple of S 5H-1 is a multiple of 12
H = 5
To encrypt Find N by working out V=NA (mod n) Calculated below
To decrypt Find V by working out N= VS (mod n)

V N
1 15 (mod 26) =1
2 25 (mod 26) = 32 (mod 26) = 6
3 35 (mod 26) = 243 (mod 26) = 9
4 45 (mod 26) = 1024 (mod 26) = 10
Once again the numbers refer to letters in the alphabet. The only number that needs to remain secret is S because only it can decipher the code.
As it can be seen the number for S is 12 which is a very high number to use as an exponent and can be difficult to calculate but using math we can figure it out. Using the formula ab(mod n) = ((a mod n)(b mod n)) mod n

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