Department of Software The University of Babylon LECTURE NOTES ON Classical Cryptographic Techniques ( Substitution Ciphers System) By College of Information Technology, University of Babylon, Iraq Samaher@itnet.uobabylon.edu.iq
Homophonic System In the substitution system the relation between the text and key is one to one. In this type the relation is one to many (i.e each letter can enciphered in multi images) Example:- Plain text letter Cipher text homophonic letters A 17,19,34,41,56,60,67,83 I 08,22,53,65,88,90 L 03,44,76 N 02,09,15,27,32,40,59 O 01,11,23,28,42,54,70 P 33,91 T 05,10,20,29,45,58,64,78,99 Example: Plain Text: PILOT Cipher Text: 91 88 76 70 99 Plain Text : ALL Cipher Text :19 76 03
Polyalphabetic substitution ciphers Polyalphabetic substitution ciphers have multiple one-letter keys, each of which is used to encrypt one letter of the plaintext. The first key encrypts the first letter of the plaintext, the second key encrypts the second letter of the plaintext, and so on. After all the keys are used, the keys are recycled. If there were 20 one-letter keys, then every twentieth letter would be encrypted with the same key. This is called the period of the cipher. In classical cryptography, ciphers with longer periods were significantly harder to break than ciphers with short periods. There are computer techniques that can easily break substitution ciphers with very long periods.
Polyalphabetic substitution ciphers There might be five different simple substitution ciphers used: 1. Vigenère cipher 2. Beaufort cipher 3. Beale cipher also called (book cipher) where the key string is an agreed text in a book. 4. Hill cipher 5. Permutation cipher 6 October 2013
1. The Vigenere cipher Note : When key length is less than the length of text to be ciphered we repeat the key letters to be equal to the plain text letters. Plaintext : Computer Key : sun We can draw only a partial table containing the key letters as follows Cipher : The cipher text letters can be found by taking the letters apposing the key letters from the above table as follows c o m p u t e r s u n s u n s u u i z h o g w l (the cipher text) Decipher: The key letters are spreaded over the cipher text. When two letters crossected on the same row,then the letter above the crossection in the outside row is the plaintext letter.
The Vigenere cipher Example Plaintext : Good Student Key : exam Solution : g o o d s t u d e n t e x a m e x a m e x a Cipher Text : k l o p w q u p I k t Decipher As a Result, The Vigenere cipher In the Vigenere cipher the key consists of a string of k letters. These are written repeatedly below the message (from which all spaces have been removed). The message is then encrypted a letter at a time by adding the message and key letters together, working mod 26 with the letters taking values A = 0 to Z = 25. For example if the key is the three letter sequence key then the message 1
The Vigenere cipher M = THISISTHEMESSAGE is encrypted using K = KEYKEYKEYKEYKEY To give the cryptogram C=DLGCMQDLCWIQCEEO. The vigeneer cipher is slightly stronger than simple substitution. To attack it using frequency analysis is more difficult since the encryption of a particular letter is not always the same.however it is still trivial to break given a reasonable amount of ciphertext. Then, C=((P+Ki)-1) Mod n P= (C-Ki)+1) Mod n Let C= Cipher Text P= Plain Text Ki= Key; n =26, of English alphabetic
2. Set of keys Method The monoalphabetic is easily to analysed using statistical features of the letters of the language.here we assume that we have k distinct bijenctions f 1,f 2,,f k from the normal alphabet to the set of cipher to represent an encryption of a word.it uses k different alphabets, the index set {1,2,,k} is called the set of keys. Example: Let us consider a simple example using three keys with three values (a,b)=(1,3); (3,5);(2,5) in turn to give the encryption of a short text. If we want to encrypt the following message The international islamic university provides a scientific basis for the mutual interaction of the natural science and Islam Using 3 keys of mixed alphabets with the encryptions f 1,f 2,f 3. f 1 = a 1 X + b 1 (i,e., a 1 = 1, b 1 = 3) f 2 = a 2 X + b 2 (i,e., a 2 = 3, b 2 = 5) f 3 = a 3 X + b 3 (i,e., a 3 = 2, b 3 = 5) then we have to do as follows f 1 (T)=W, f 2 (H)=A, f 3 (E)=W, f 1 (I)=L, f 2 (N)=S, f 3 (T)=T f 1 (E)=H, f 2 (R)=E, f 3 (N)=P, f 1 (A)=D, f 2 (T)=K, f 3 (I)=Q The total encryption of text will be Thein terna tiona lisla micun ivers itypr ovide sasci entif icbas isfor themu tuali ntera ction ofthe natur alsci ences andis lamxx comes to be the string of ciphers WAWLS THEPD KQRSC ODOOF KLLYQ DDHEO LKSSE UYDRH HCVLQ HSTLU QFICV DOIVJ WAWPN TXFFL
3. Digraph Cipher Method Consider the rule a group of k subsequent letters changes into another group of k subsequent letters this is called Digraph cipher. In this case the plain text alphabet consists of all two letters combinations of the ordinary alphabet that s giving 26 2 =676 symbols P=(aa,bb,ac,,ba,bb,bc, zx,zy,zz) then we get the all ordered pairs that is all vectors ( x y) Є (Z/26) 2. The American mathematician Laster Hill suggested a procedure for encryption decryption such vectors by using matrices as follows p q A= r s with p,q,r,s Є Z/26. then weobtain the linear transformation X1=px + qy Y1=rx + sy which may be regard as an encryption function.this transformation is bijective iff gcd(ps-qr,26)=1 in this case the decryption function corresponds to the inverse matrix A -1 =1/d. Where d=ps qr p r q s Example : 1 2 Let, A 3 7 Of determinant 1.if we want to encrypt the word MALAYSIA we have to form the corresponding numerical string (12,0,11,0,24,18,8,0) and break it up into 4-pairs and then compute the image vectors under the transformation A.Hence we get A( 12 o)=( 12 10),A( 11 0)=( 11 7), A( 24 18)=( 8 6), A( 8 0)=( 8 24) yielding the encryption MKLHIGIY
3. Digraph Cipher Method As a result, the rule is change a group of k subsequent lettersinto another group of k subsequent letters Procedure of Encryption: 1. Convert the Plain text into pairs of Digital number x y 2. Using the following matrix A= 3. cipher text = x1 y1 Procedure of Decryption: x1 y1 = p q r s..condition (ps-qr=1) x y = px + qy rx + sy Note: we cab be perform the decryption iff (GCD(ps-qr,26)=1) 1. A -1 =1/d. 2. p r x y = A 1 q s s q r p x1 =Plain Text y1 Return to example: (Encryption) x1 y1 = ( 1 2 3 7 12 0 ) Mod 26 = 12 36 mod 26= 12 10
3. Digraph Cipher Method Return to example: (Decryption) GCD( ps-rq,26)= GCD( 7-6,26)=GCD(1,26)=1, Then, we can perform the decryption s q 1. A -1 =1/d. r p d=ps-qr=(7-6)=1 A 1 = 7 3 2 1 x y = A 1 x1 y1 = ( 7 2 * 12 10 ) mod 26 = 84 20 mod 26 36 + 10 mod 26 = 12 0 = M 3 1 A Plain Text