Basic Security Issues Mechanism Services Attack

3 Aspect of Information Security

Security Mechanism Security Services Security Attack

Ceaser Cipher Cease Cipher Manipulate the Character with 3 positions (for example: if we Cipher “A” into Ceaser technique then it should be “D”)

Formula

C= E(P)C= (P+3) mod26C= (P+K) mod26C= Answer

Where C= CipherP= Plain TextE= EncryptionK= Key

ExampleP = 815290 1 2 3 4 5 6 7 8 91 2 3 4 5 6 7 8 9 10

C=E(P)C = (P+K) mod10C = (P+3)mod10C = (8+3)mod10 = 1C = (1+3)mod10 = 4C = (5+3)mod10 = 8C = (2+3)mod10 = 5C = (9+3)mod10 = 2

E = 14852

P= MUSTARD

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

E= PXVWDUG

Transportation TechniqueIn Transportation technique we assign fixed range of column, and in that column all the text is entered and encrypt the text with respect to the column of the row

Example

We are taking 10 Column for the text

P= COMPUTER GRAPHIC MAY BE SLOW BUT ITS EXPENSIVE

1 2 3 4 5 6 7 8 9 10C O M P U T E R G RA P H I C M A Y B ES L O W B U T I T SE X P E N S I V E Z

E= CAESE OPS MHLE PIOX UCWP T BE EMUN RATS GYII RBTV

Time required at 1 Encryption/ U sec

2^31 U sec = 35.8 min

Key size Number of Alternative Keys

Time required for 1 encryption

Time required at 10^6 Decryption

32 2^31 = 4.3 x 10^9 2^31usec=35.8 2.15 millisecond56 2^55 = 7.2 x 10^16 2^55usec=10 hours 10 hours128 2^127 = 3.4 x 10^38 2^127usec= 5.4x10^18 years168 2^167 = 3.7 x 10^50 2^167usec= 5.9x 10^30 years

Where Time Required for 1 encryption for 56 Key size formulated as

2^55 ________________ 3600 x 10^6 x 10^6

and Time Required for 1 Encryption for 168 keys are formulated as

2^167_______ _____________3600 x 10^6 x 10^6 x 365 x 24

Play Fair CipherIn play fair cipher technique we draw a 5 x 5 matrix and in that matrix the word in encrypted, IN THE REMAINING COLUMN we add the remaining alphabets

Example

Keyword: Monarchy

M O N A RC H Y B DE F G I/J KL P Q S TU V W X Z

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

AR = (A Converted to the next one that is R and “R” Converted to the next one that is “M”) AR = RM

MU = (Now Mu belongs to the same column, so “M” will be converted to the next one that is “C” and “U” will be converted into the next one that is “M”)MU = CM

HS = ( “H” and “S” belongs to two different column, now we first draw line to “H” row as the Line Intercepts the “B” so “H” will be converted into B and for “S” we again draw a line as we drew for “H”, As blue line Intercepts on “P” so )

HS = BP

Substitution Technique

DES Algorithm

Des algorithm work on behf of Hexa Decimal Nubers

Plain text= 675a69675e5a6b5a

1st step

Convert the above number into binary

1234 5678 9-12 13-16 17-20 21-24 25-28 29-32 33-36 37-40 41-44 45-48 49-52 53-56 57-60 61-640110 | 0111 | 0101 | 1010 | 0110 | 1001 | 0110 | 0111 | 0101 | 1110 | 0101 | 1010 | 0110 | 1011| 0101 | 1010

2nd Step (IP Initial Permutation)

We will consider Table 58 50 42 34 26 18 10 260 52 44 36 28 20 12 462 54 46 38 30 22 14 664 56 48 40 32 24 16 857 49 41 33 25 17 9 159 51 43 35 27 19 11 361 53 45 37 29 21 13 563 65 47 39 31 23 15 7

1 1 1 1 1 1 1 11 0 1 1 0 0 1 00 0 0 1 1 0 0 10 1 0 0 1 1 0 10 0 0 0 0 0 0 00 1 0 0 1 1 0 11 1 1 1 0 1 1 01 1 1 1 1 0 1 1

3rd Step3rd step is to convert the above binary numbers into Hexa decimalNote: convert only 4 Block of Number

1 1 1 1 = F + 1 1 1 1 = F

Converted Hexa String = FF B2 19 4D 00 4D F6 FB

4th Step Taking the inverse permutation

1 1 1 1 1 1 1 11 0 1 1 0 0 1 00 0 0 1 1 0 0 10 1 0 0 1 1 0 10 0 0 0 0 0 0 00 1 0 0 1 1 0 11 1 1 1 0 1 1 01 1 1 1 1 0 1 1

Here

Inverse these strings

1111 1111 1011 0010 0001 1001 0100 1101 0000 0000 0100 1101 1111 0110 1111 1011