Computer Security (Cryptography) Ch03
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The document discusses various polyalphabetic ciphers including the Vigenere cipher, one-time pad cipher, Playfair cipher, and Hill cipher. It provides examples of encrypting and decrypting messages with the Vigenere cipher and Playfair cipher. The Vigenere cipher uses a keyword to shift the letters of a plaintext by various amounts, while the Playfair cipher encrypts pairs of letters based on their positions in a 5x5 grid generated from a keyword. The document also describes how the Hill cipher works by converting the plaintext and keyword into matrices and performing matrix multiplication to encrypt messages in blocks of letters.
![Computer Security & Networks
Dr. Saif Kassim Alfraije Page 1
Polyalphabetic Cipher
Leon Battista invented the Polyalphabetic Substitution Cipher in year 1568.
“Poly means Many in Greek language”. This method uses a mixed alphabet to
encrypt a message. The Enigma machine is more complex but still fundamentally a
polyalphabetic substitution cipher.
In polyalphabetic substitution, each occurrence of a character may have a different
substitute. The Polyalphabetic Cipher was adapted as a twist on the standard
Caesar cipher to reduce the effectiveness of frequency analysis on the ciphertext.
The polyalphabetic cipher uses a text string (for example, a word) as a key, which
is then used for doing a number of shifts on the plaintext.
Polyalphabetic Cipher Types
A. Vigenère Cipher
B. One Time Pad Cipher
C. Playfair Cipher
D. Hill Cipher
Vigenere cipher
One such cipher is the famous Vigenere cipher, the Vigenere cipher uses
the power of 26 possible shift ciphers. A tool that was used for the Vigenère
Cipher was a (code wheel) or (Keyword). Vigenere cipher shifts the characters in a
string by some places. Now those some places are determined by the characters
that are present in the key string.
P=P1P2P3…… C=C1C2C3…….. k=[(K1,K2,….. , KM), (K1,K2,….. , KM),..]
Encryption : Ci = Pi + Ki Decryption : Pi = Ci - Ki](https://image.slidesharecdn.com/chapter03-151121204704-lva1-app6891/85/Computer-Security-Cryptography-Ch03-1-320.jpg)
Computer Security (Cryptography) Ch03
- 1. Computer Security & Networks Dr. Saif Kassim Alfraije Page 1 Polyalphabetic Cipher Leon Battista invented the Polyalphabetic Substitution Cipher in year 1568. “Poly means Many in Greek language”. This method uses a mixed alphabet to encrypt a message. The Enigma machine is more complex but still fundamentally a polyalphabetic substitution cipher. In polyalphabetic substitution, each occurrence of a character may have a different substitute. The Polyalphabetic Cipher was adapted as a twist on the standard Caesar cipher to reduce the effectiveness of frequency analysis on the ciphertext. The polyalphabetic cipher uses a text string (for example, a word) as a key, which is then used for doing a number of shifts on the plaintext. Polyalphabetic Cipher Types A. Vigenère Cipher B. One Time Pad Cipher C. Playfair Cipher D. Hill Cipher Vigenere cipher One such cipher is the famous Vigenere cipher, the Vigenere cipher uses the power of 26 possible shift ciphers. A tool that was used for the Vigenère Cipher was a (code wheel) or (Keyword). Vigenere cipher shifts the characters in a string by some places. Now those some places are determined by the characters that are present in the key string. P=P1P2P3…… C=C1C2C3…….. k=[(K1,K2,….. , KM), (K1,K2,….. , KM),..] Encryption : Ci = Pi + Ki Decryption : Pi = Ci - Ki
- 2. Computer Security & Networks Dr. Saif Kassim Alfraije Page 2 Example Plain Text: “Computer Science”. Key = Security, Cipher Text? Encode plain text. Write key under coded plaintext, repeating as many times as necessary. Encode key. Add key character code to corresponding plain text code (mod 26). Decoding of resulting integer is Ciphertext. Note: If (Pencoding + Kencoding) >= 26 then - 26 Answer (Encryption) Plain text C O M P U T E R S C I E N C E Pencoding 2 14 12 15 20 19 4 17 18 2 8 4 13 2 4 Key S E C U R I T Y S E C U R I T Kencoding 18 4 2 20 17 8 19 24 18 4 2 20 17 8 19 Cipher Code 20 18 14 9 11 1 23 15 10 6 10 24 4 10 23 Cipher text U S O J L B X P K G K Y E K X +
- 3. Computer Security & Networks Dr. Saif Kassim Alfraije Page 3 Decryption Ciphertext = USOJLBXPKGKYEKX , Key = Security . Plaintext ?? Cipher text U S O J L B X P K G K Y E K X Cipher Code 20 18 14 9 11 1 23 15 10 6 10 24 4 10 23 Key S E C U R I T Y S E C U R I T Kencoding 18 4 2 20 17 8 19 24 18 4 2 20 17 8 19 Pencoding 2 14 12 15 20 19 4 17 18 2 8 4 13 2 4 Plain text C O M P U T E R S C I E N C E - 9 – 20 = – 11 : 26 – 11 = 15
- 4. Computer Security & Networks Dr. Saif Kassim Alfraije Page 4 Build Program 1. Add a three label control on the form. Set the first Text property to Plain text , second label set to Key and third to Cipher text . 2. Add three text box controls as shown. o Name the first text box control PTtxt o Name the second text box control Keytxt o Name the third text box control CTtxt 3. Add two button controls as shown. o Name the first button control Encryption o Name the second button control Decryption
- 5. Computer Security & Networks Dr. Saif Kassim Alfraije Page 5 Program Coding Double-click the Encryption control. And write follow code. Double-click the Decryption control. And write follow code. Now you can write class VigenereCipher that contain two function Encrypt For return Cipher text and Decrypt for return Result Plain to the end user Depending on the key type string , this function will receive 2 parameters when you click button Enc or Dec : Plain text Key Secret Private Sub Encbtn_Click(sender As Object, e As EventArgs) Handles Encbtn.Click CTtxt.Text = VigenereCipher.Encrypt(PTtxt.Text, Keytxt.Text) End Sub Private Sub Decbtn_Click(sender As Object, e As EventArgs) Handles Decbtn.Click CTtxt.Text = VigenereCipher.Decrypt(PTtxt.Text, Keytxt.Text) End Sub
- 6. Computer Security & Networks Dr. Saif Kassim Alfraije Page 6 Public Class VigenereCipher Public Shared Function Encrypt(ByVal plainTxt As String, ByVal key As String) Dim encryptedText As String = "" For i As Integer = 1 To cipherTxt.Length Dim temp As Integer = AscW(GetChar(cipherTxt, i)) + AscW(GetChar(key, i Mod key.Length + 1)) encryptedText += ChrW(temp) Next Return encryptedText End Function Public Shared Function Decrypt(ByVal cipherTxt As String, ByVal key As String) Dim decryptedText As String = "" For i As Integer = 1 To cipherTxt.Length Dim temp As Integer = AscW(GetChar(cipherTxt, i)) - AscW(GetChar(key, i Mod key.Length + 1)) decryptedText += ChrW(temp) Next Return decryptedText End Function End Class
- 7. Computer Security & Networks Dr. Saif Kassim Alfraije Page 7 One Time Pad Cipher known as Vernam Cipher is implemented using a random set of non repeating characters as the Key. The rest of process is same as the Vigenère cipher. The sender uses each key letter on the pad to encrypt exactly one plaintext character. Encryption is the addition modulo 26 of the plaintext character and the one-time pad key character (Pencoding + Kencoding) >= 26 then - 26 Example: ( Encryption ) Plaintext: “ HOW ARE YOU ’’ Key = NCBTZQARX Ciphertext: ? Plain text H O W A R E Y O U Pencoding 7 14 22 0 17 4 24 14 20 Key N C B T Z Q A R X Kencoding 13 2 1 19 25 16 0 17 23 Cipher Code 20 16 23 19 16 20 24 5 17 Cipher text U Q X T Q U Y F R +
- 8. Computer Security & Networks Dr. Saif Kassim Alfraije Page 8 (Decryption) Cipher text U Q X T Q U Y F R Cipher Code 20 16 23 19 16 20 24 5 17 Key N C B T Z Q A R X Kencoding 13 2 1 19 25 16 0 17 23 Pencoding 7 14 22 0 17 4 24 14 20 Plain text H O W A R E Y O U - 16 – 25 = – 9 26 – 9 = 17
- 9. Computer Security & Networks Dr. Saif Kassim Alfraije Page 9 Playfair Cipher The Playfair cipher encrypts pairs of letters, instead of single letters. This is significantly harder to break. Each plaintext letter is replaced by a diagram in this cipher. User chooses a keyword and puts it in the cells of a 5 x 5 matrix. Letter I and J always stay in one cell . Duplicate letters appear only once. Number of diagrams is 26 x 26 = 676. To encrypt a message, one would break the message into groups of 2 letters. If there is a dangling letter at the end, we add an X. For example. "Secret Message" becomes "SE CR ET ME SS AG EX". We now take each group and find them out on the table. we apply the following rules, in order. Playfair Cipher Rules 1. If the digraph consists of the same letter twice (or there is only one letter left by itself at the end of the plaintext) then insert the letter "X" between the same letters (or at the end), and then continue with the rest of the steps. 2. If the two letters appear on the same row in the square, then replace each letter by the letter immediately to the right of it in the square (cycling round to the left hand side if necessary). 3. If the two letters appear in the same column in the square, then replace each letter by the letter immediately below it in the square (cycling round to the top of the square if necessary). 4. Otherwise, form the rectangle for which the two plaintext letters are two opposite corners. Then replace each plaintext letter with the letter that forms the other corner of the rectangle that lies on the same row as that plaintext letter (being careful to maintain the order).
- 10. Computer Security & Networks Dr. Saif Kassim Alfraije Page 10 Example Plain text = Hide the gold in the tree stump , key = playfair example , Cipher text ?? Firstly we must generate the Polybius Square that we are going to use. We do this by setting out a 5x5 grid, and filling it with the alphabet, starting with the letters of the keyphrase, and ignoring any letters we already have in the square. We are also going to combine "I" and "J" in the square. We must now split the plaintext into digraphs, and split up any double letter digraphs by inserting an "x" between them. The first image below shows the initial digraph split of the plaintext, and the second image displays how we split up the "ee" into "ex" and "es". In this case, when we insert this extra "x", we no longer need to have one at the end of the plaintext. HI DE TH EG OL DI NT HE TR EE ST UM P P L A Y F I R E X M B C D G H K N O Q S T U V W Z HI DE TH EG OL DI NT HE TR EX ES TU MP
- 11. Computer Security & Networks Dr. Saif Kassim Alfraije Page 11 We now take each digraph in turn and apply rule 2, 3 or 4 as necessary. Each step is show below with a visual representation of what is done for each digraph.
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- 15. Computer Security & Networks Dr. Saif Kassim Alfraije Page 15 We can now take each of the ciphertext digraphs that we produced and put them all together. BM OD ZB XD NA BE KU DM UI XM MO UV IF Homework Decrypt the Cipher text = BMODZBXDNABEKUDMUIXMMOUVIF . When the key = playfair example
- 16. Computer Security & Networks Dr. Saif Kassim Alfraije Page 16 Hill Cipher The Hill Cipher was invented by Lester S. Hill in 1929, and like the other Digraphic Ciphersit acts on groups of letters. Unlike the others though it is extendable to work on different sized blocks of letters. So, technically it is a polygraphic substitution cipher, as it can work on digraphs, trigraphs (3 letter blocks) or theoretically any sized blocks. The Hill Cipher uses an area of mathematics called Linear Algebra, and in particular requires the user to have an elementary understanding of matrices. It also make use of Modulo Arithmetic (like the Affine Cipher). Because of this, the cipher has a significantly more mathematical nature than some of the others. However, it is this nature that allows it to act (relatively) easily on larger blocks of letters. Encryption To encrypt a message using the Hill Cipher we must first turn our keyword into a key matrix (a 2 x 2 matrix for working with digraphs, a 3 x 3 matrix for working with trigraphs, etc). We also turn the plaintext into digraphs (or trigraphs) and each of these into a column vector. We then perform matrix multiplication modulo the length of the alphabet (i.e. 26) on each vector. These vectors are then converted back into letters to produce the ciphertext. 2 x 2 Example We shall encrypt the plaintext message "short example" using the keyword hill and a 2 x 2 matrix. The first step is to turn the keyword into a matrix. With the keyword in a matrix, we need to convert this into a key matrix.
- 17. Computer Security & Networks Dr. Saif Kassim Alfraije Page 17 We now split the plaintext into digraphs, and write these as column vectors. That is, in the first column vector we write the first plaintext letter at the top, and the second letter at the bottom Now we must convert the plaintext column vectors in the same way that we converted the keyword into the key matrix. Now we must perform some matrix multiplication. We multiply the key matrix by each column vector in turn. To perform matrix multiplication we "combine" the top row of the key matrix with the column vector to get the top element of the resulting column vector. We then "combine" the bottom row of the key matrix with the column vector to get the bottom element of the resulting column vector.
- 18. Computer Security & Networks Dr. Saif Kassim Alfraije Page 18 Next we have to take each of these numbers, in our resultant column vector, modulo 26 (remember that means divide by 26 and take the remainder). Finally we have to convert these numbers back to letters, so 0 becomes "A" and 15 becomes "P", and our first two letters of the ciphertext are "AP".
- 19. Computer Security & Networks Dr. Saif Kassim Alfraije Page 19 Encryption process of the first digraph
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- 22. Computer Security & Networks Dr. Saif Kassim Alfraije Page 22 Decryption To decrypt a ciphertext encoded using the Hill Cipher, we must find the inverse matrix. Once we have the inverse matrix, the process is the same as encrypting. In general, to find the inverse of the key matrix, we perform the calculation below, where Kis the key matrix, d is the determinant of the key matrix and adj(K) is the adjugate matrix of K. 2 x 2 Example We shall decrypt the example above, so we are using the keyword hill and our ciphertext is "APADJ TFTWLFJ". Step 1 - Find the Multiplicative Inverse of the Determinant The determinant is a number that relates directly to the entries of the matrix. It is found by multiplying the top left number by the bottom right number and subtracting from this the product of the top right number and the bottom left number.
- 23. Computer Security & Networks Dr. Saif Kassim Alfraije Page 23 We now have to find the multiplicative inverse of the determinant working modulo 26. That is, the number between 1 and 25 that gives an answer of 1 when we multiply it by the determinant. So, in this case, we are looking for the number that we need to multiply 15 by to get an answer of 1 modulo 26. So the multiplicative inverse of the determinant modulo 26 is 7. We shall need this number later. Step 2 - Find the Adjugate Matrix The adjugate matrix is a matrix of the same size as the original. For a 2 x 2 matrix, this is fairly straightforward as it is just moving the elements to different positions and changing a couple of signs.
- 24. Computer Security & Networks Dr. Saif Kassim Alfraije Page 24 Again, once we have these values we will need to take each of them modulo 26 (in particular, we need to add 26 to the negative values to get a number between 0 and 25. For our example we get the matrix below. Step 3 - Multiply the Multiplicative Inverse of the Determinant by the Adjugate Matrix To get the inverse key matrix, we now multiply the inverse determinant (that was 7 in our case) from step 1 by each of the elements of the adjugate matrix from step 2. Then we take each of these answers modulo 26. Now we have the inverse key matrix, we have to convert the ciphertext into column vectors and multiply the inverse matrix by each column vector in turn, take the results modulo 26 and convert these back into letters to get the plaintext.