C = E(p) = (p + k) mod (n) p = D(C) = (C k) mod (n)

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Substitutions ciphers (monoalphabetic) A substitution technique is one in which the characters of plaintext are replaced by other characters or by numbers or symbols.

1 Substitutions ciphers (monoalphabetic) A substitution technique is one in which the characters of plaintext are replaced by other characters or by numbers or symbols. If the plaintext is viewed as a sequence of bits, then substitution involves replacing plaintext bit patterns with ciphertext bit patterns. Caesar cipher (Shifted alphabet) A simplest type of substitution cipher shifts each letter in the English alphabet forward by K positions cyclically (shifts past Z cycle back to A). K is the key to the cipher. This type of cipher is often called a Caesar cipher because Julius Caesar used it with k=3. Caesar cipher as: C = E(p) = (p + k) mod (n) p = D(C) = (C k) mod (n) Where k is the number of positions to be shifted, P is a single character of the plaintext, C is a single character of the ciphertext, and n is the size of the alphabet (for English alphabet n=26). If k = 3 then we can encrypt the following message as: Plaintext: this is a secret message Ciphertext: wklv lv d vhfuhw phvvdjh The Caesar code program: 1

2 namespace seclab2 public partial class Form1 : Form public Form1() InitializeComponent(); private void Form1_Load(object sender, EventArgs e) public string Encryption(string plian,int key ) char[] temp = new char[plian.length]; for (int i = 0; i < plian.length; i++) temp[i] = (char)(plian[i]+key); if (temp[i] > 'z') temp[i] = (char)(temp[i] - 26); string cipher = new string(temp); return cipher; public string Decryption(string cipher, int key) char[] temp = new char[cipher.length]; for (int i = 0; i < cipher.length; i++) temp[i] = (char)(cipher[i] - key); if (temp[i] < 'a') temp[i] = (char)(temp[i] + 26); string plian = new string(temp); return plian; private void button1_click(object sender, EventArgs e) int key =(int) numericupdown1.value; string output = Encryption(textBox1.Text,key); textbox2.text = output; private void button2_click(object sender, EventArgs e) int key = (int)numericupdown1.value; string output = Decryption(textBox2.Text,key); textbox3.text = output; 2

3 Multiplicative cipher Multiplicative cipher is encrypted by multiply each character in plaintext by multiplicative key k modulo 26. The key and 26 are relatively prime (GCD (key, 26)=1). To decrypt, the modular inverse of the key is the number that is multiplied. The modular inverse of key is a number X such that (key *X) mod 26= 1. Multiplicative cipher as: C = E(p) = (p * k) mod (n) p = D(C) = (C *invk) mod (n) Where k and n are relatively prime in order to produce a complete set of residues and invk is multiplicative inverse number to key. If k and n are not prime, several letters will encipher to the same ciphertext letter, and not all letters will appear in the ciphertext. Notice that only 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, and 25 have multiplicative inverses modulo 26. Here are the possible multipliers and their inverses: Number Multiplicative inverse *1= 1mod 26, 3* 9=1 mod 26, 5*21=1 mod 26 and so on, but, 2, for example, does not have an inverse, there is no number mod 26 that 2 can be multiplied by that will result in 1. Example: Plaintext: this is a secret message Ciphertext: fvyc yc a cmgzmf kmccasm 3

The Multiplicative code program: namespace seclab2multiplicative public partial class Form1 : Form public Form1() InitializeComponent(); private void Form1_Load(object sender, EventArgs e)

4 The Multiplicative code program: namespace seclab2multiplicative public partial class Form1 : Form public Form1() InitializeComponent(); private void Form1_Load(object sender, EventArgs e) public static int MultiplicativeInverse(int key) int x = 0; while (((key * x) % 26!= 1)) x++; return x; public static int GCD(int p, int q) if (q == 0) return p; int r = p % q; return GCD(q, r); 4

5 public string Encryption(string plian, int key) char[] temp = new char[plian.length]; for (int i = 0; i < plian.length; i++) temp[i] = (char)((((plian[i]-97) * key)% 26)+97); string cipher = new string(temp); return cipher; public string Decryption(string cipher, int invkey) char[] temp = new char[cipher.length]; for (int i = 0; i < cipher.length; i++) temp[i] = (char)((((cipher[i] - 97) * invkey) % 26) + 97); string plian = new string(temp); return plian; private void button1_click(object sender, EventArgs e) int key = (int)numericupdown1.value; if (GCD(key, 26)!= 1) MessageBox.Show("invalid cipher key. please try again "); string output = Encryption(textBox1.Text, key); textbox2.text = output; private void button2_click(object sender, EventArgs e) int invkey = MultiplicativeInverse((int)numericUpDown1.Value); string output = Decryption(textBox2.Text, invkey); textbox3.text = output; 5

6 Affine cipher The key for the Affine cipher consists of two numbers, we'll call them key1and key2. Key1 should be chosen to be relatively prime to n (26 character alphabet), and key2 is the number of positions to be shifted. The encryption and decryption function for a single letter is: C = E(p) = ((key1 * p) + key2 ) mod (n) p = D(C) = (C key2)*invkey1 mod (n) Where invkey1 is the multiplicative inverse of key1 in the group of modulo n Example: using key1=5, key2=8 (notes that the invkey will be 21) Plaintext: this is a secret message Ciphertext: zrwu wu i ucspcz qcuuimc The Affine code program: 6

7 namespace seclab2affine public partial class Form1 : Form public Form1() InitializeComponent(); public static int MultiplicativeInverse(int key1) int x = 0; while (((key1 * x) % 26!= 1)) x++; return x; public static int GCD(int p, int q) if (q == 0) return p; int r = p % q; return GCD(q, r); public string Encryption(string plian, int key1, int key2) char[] temp = new char[plian.length]; for (int i = 0; i < plian.length; i++) temp[i] =(char)(((((plian[i] - 97) * key1) + key2) % 26) + 97); string cipher = new string(temp); return cipher; public string Decryption(string cipher, int invkey,int key2) char[] temp = new char[cipher.length]; for (int i = 0; i < cipher.length; i++) temp[i] =(char)((((((cipher[i] - 97) -key2)+26)* invkey) % 26) + 97); string plian = new string(temp); return plian; 7

8 private void button1_click(object sender, EventArgs e) int key1 = (int)numericupdown1.value; if (GCD(key1, 26)!= 1) MessageBox.Show("invalid cipher key. please try again "); int key2 = (int)numericupdown2.value; string output = Encryption(textBox1.Text, key1,key2 ); textbox2.text = output; private void button2_click(object sender, EventArgs e) int invkey = MultiplicativeInverse((int)numericUpDown1.Value); int key2 = (int)numericupdown2.value; string output = Decryption(textBox2.Text, invkey,key2); textbox3.text = output; 8

9 Vigenere Cipher (polyalphabetic) In cryptography, a Vigenere cipher is a method of encrypting alphabetic text by using a series of different Caesar ciphers based on the letters of a keyword. It is a simple form of polyalphabetic substitution. This scheme of cipher uses a text string as a key, which is then used for doing a number of shifts on the plaintext. Vigenere cipher as: Example: C i = E(p i ) = (p i + k i ) mod (n) p i = D(C i ) = (C i k i ) mod (n) The sender and the receiver decide on a key. For example, let s assume the key is 'cipher'. Each alphabet of the key is converted to its respective numeric value: In this case: c 2, i 8, p 15, h 7, e 4, r 17. Thus, the numeric representation of this key is ' ' The sender wants to encrypt the message, for example 'cryptography'. He will arrange plaintext and numeric key as follows: c r y p t o g r a p h y He now shifts each plaintext alphabet by the number written below it to create ciphertext as shown below : c r y p t o g r a p h y e z n w x f i z p w l p Here, each plaintext character has been shifted by a different amount and that amount is determined by the key. The key must be less than or equal to the size of the message. For decryption, the receiver uses the same key and shifts received ciphertext in reverse order to obtain the plaintext. 9

e z n w x f i z p w l p 2 8 15 7 4 17 2 8 15 7 4 17 c r y p t o g r a p h y The Vigenere code program: namespace Vigenere public partial class Form1 : Form public Form1() InitializeComponent();

10 e z n w x f i z p w l p c r y p t o g r a p h y The Vigenere code program: namespace Vigenere public partial class Form1 : Form public Form1() InitializeComponent(); public string Encryption(string plian) char[] key = textbox1.text.tochararray(); char[] temp = new char[plian.length]; int j = 0; // counter for key for (int i = 0; i < plian.length; i++) temp[i] = (char)(plian[i] +( key[j]-97)); if (temp[i] > 'z') temp[i] = (char)(temp[i] - 26); j++; if (j == key.length) j = 0; string cipher = new string(temp); return cipher; 10

11 public string Decryption(string cipher) char[] key = textbox1.text.tochararray(); char[] temp = new char[cipher.length]; int j = 0; for (int i = 0; i < cipher.length; i++) temp[i] = (char)(cipher[i] - (key[j]-97)); if (temp[i] < 'a') temp[i] = (char)(temp[i] + 26); j++; if (j == key.length) j = 0; string plian = new string(temp); return plian; private void button1_click(object sender, EventArgs e) string output = Encryption(textBox2.Text); textbox3.text = output; private void button2_click(object sender, EventArgs e) string output = Decryption(textBox3.Text); textbox4.text = output; Playfair Cipher (polygraphic) In this scheme, pairs of letters are encrypted, instead of single letters as in the case of simple substitution cipher. The playfair cipher starts with creating a key table. The key table is a 5 5 grid of letters that will act as the key for encrypting the plaintext. Each of the 25 letters must be unique and one letter of the alphabet is omitted from the table (as there are 25 spots and 26 letters in the alphabet). The algorithm first, split a plaintext message into pairs of two letters, and if the second letter of a pair is the same as the first letter, 'X' is inserted.. If there is an odd number of letters, 'X' is added to the last letter. 11

The sender and the receiver decide on a particular key, for example hello world as the key.

The rest of the table will be filled with the remaining letters of the alphabet, in order.

digraph are in relation to each other. The rules are: If both letters are in the same column, take the letter below each one (going back to the top if at the bottom).

12 The sender and the receiver decide on a particular key, for example hello world as the key. The first characters (going left to right) in the table will be the phrase, with duplicate letters removed. The rest of the table will be filled with the remaining letters of the alphabet, in order. The key table would look like this: h e l o w r d a b C f g i k m n p q s t u v x y z The Playfair cipher uses a few simple rules relating to where the letters of each digraph are in relation to each other. The rules are: If both letters are in the same column, take the letter below each one (going back to the top if at the bottom). If both letters are in the same row, take the letter to the right of each one (going back to the left if at the farthest right). If neither of the preceding two rules are true, form a rectangle with the two letters and take the letters on the horizontal opposite corner of the rectangle. The Playfair code program: namespace playfair public partial class Form1 : Form public Form1() InitializeComponent(); 12

13 public char[,] KeyGenerator(string input) char[,] key = new char[5, 5]; int count = 0; char[] inputkey = input.tochararray(); char[] alphabet = "abcdefghiklmnopqrstuvwxyz".tochararray(); char[] tempkey = inputkey.union(alphabet).toarray(); for (int i = 0; i < 5; i++) for (int j = 0; j < 5; j++) key[i,j] = tempkey[count]; count++; return key; private int[] GetPosition(char[,] key, char ch) int[] position = new int[2]; for (int i = 0; i < 5; ++i) for (int j = 0; j < 5; ++j) if (key[i, j] == ch) position[0] = i; position[1] = j; return position; public string Encryption(string plian,char[,] key) int[] p1 = new int[2]; int[] p2 = new int[2]; //If the second letter of pair is the same as the first letter, 'x' is inserted. for (int i = 0; i < plian.length - 1; i++) if (plian[i] == plian[i + 1]) plian = plian.insert(i + 1, "x"); // If the length of the string is odd, x' is appended to make it an even length. if (plian.length % 2!= 0) plian += "x"; char[] temp = new char[plian.length]; for (int i = 0; i <= plian.length-2; i+=2) p1 = GetPosition(key,plian[i]); //get position of the first char p2 = GetPosition(key,plian[i + 1]);//get position of the second char 13

14 // check if the pair of char at the same row. if (p1[0] == p2[0]) temp[i] = key[p1[0], (p1[1]+1) % 5]; temp[i + 1] = key[p2[0], (p2[1] + 2) % 5]; // check if the pair of char at the same col. else if (p1[1] == p2[1]) temp[i] = key[ (p1[0] + 1) % 5, p1[1]]; temp[i + 1] = key[ (p2[0] + 2) % 5, p2[1]]; // check if the pair of char at rectangle shape. else temp[i] = key[p1[0],p2[1]]; temp[i + 1] = key[p2[0], p1[1]]; string cipher = new string(temp); return cipher; public string Decryption(string cipher, char[,] key) int[] p1 = new int[2]; int[] p2 = new int[2]; char[] temp = new char[cipher.length]; for (int i = 0; i <= cipher.length - 2; i += 2) p1= GetPosition(key,cipher[i]); //get position of the first char p2= GetPosition(key,cipher[i+1]); //get position of the second char // check if the pair of char at the same row. if (p1[0] == p2[0]) temp[i] = key[p1[0], ((p1[1] - 1)+5) % 5]; temp[i + 1] = key[p2[0], ((p2[1] - 2)+5) % 5]; // check if the pair of char at the same col. else if (p1[1] == p2[1]) temp[i] = key[((p1[0] - 1)+5) % 5, p1[1]]; temp[i + 1] = key[((p2[0] - 2)+5) % 5, p2[1]]; 14

15 // check if the pair of char at rectangle shape. else temp[i] = key[p1[0], p2[1]]; temp[i + 1] = key[p2[0], p1[1]]; string plian = new string(temp); return plian; private void button1_click(object sender, EventArgs e) char[,] key = KeyGenerator(textBox1.Text); string output = Encryption(textBox2.Text,key); textbox3.text = output; private void button2_click(object sender, EventArgs e) char[,] key = KeyGenerator(textBox1.Text); string output = Decryption(textBox3.Text, key); textbox4.text = output; 15

Linear Feedback Shift Register (LFSR) A linear feedback shift register is composed of a shift register R which contains a sequence of bits and a feedback function f which is the bit sum (Exclusive OR

Each time a bit is needed the LFSR performs step operations that: Entry in cell R n-1 is output. Entry in cell R i is passed to cell R i+1 (Shift the bits one position).

16 Linear Feedback Shift Register (LFSR) A linear feedback shift register is composed of a shift register R which contains a sequence of bits and a feedback function f which is the bit sum (Exclusive OR (XOR) of a subset of the entries of the shift register. The shift register contains n cells, labeled R 0, R 1,.....,R n 1 each holding one bit. Each time a bit is needed the LFSR performs step operations that: Entry in cell R n-1 is output. Entry in cell R i is passed to cell R i+1 (Shift the bits one position). The cell R 0 is updated with the value f(r) (Exclusive OR (XOR) of the bit shifted off and the bit at a given tap position in the register). An LFSR has three parameters that characterize the sequence of bits it produces: the number of bits N, the initial Seed (the sequence of bits that initializes the register), and the tap position. The figure below illustrates one step of an 11-bit LFSR with initial seed and tap position 8 and 10. Encryption and decryption applications The unusual sequence of values generated by an LFSR can be employed in the encryption and decryption of data. A stream of data bits can be encrypted by XOR them with the output from an LFSR. 16

The LFSR code program: namespace LFSR public partial class Form1 : Form public Form1() InitializeComponent(); private void Form1_Load(object sender, EventArgs e) public int [] LfsrGgenerator( int

17 The LFSR code program: namespace LFSR public partial class Form1 : Form public Form1() InitializeComponent(); private void Form1_Load(object sender, EventArgs e) public int [] LfsrGgenerator( int len, string seed) int[] lfsr = new int[seed.length]; int[] key = new int[len]; int tap1 = lfsr[4]; int tap2 = lfsr[2]; // lfsr initialization vector for (int i = 0; i < lfsr.length; i++) if (seed[i] == '1') lfsr[i] = 1; else lfsr[i] = 0; 17

18 for (int a = 0; a < len; a++) int[] temp = new int[seed.length]; temp[0] = tap1 ^ tap2; // shift lfsr for (int i = 0; i < lfsr.length - 1; i++) temp[i + 1] = lfsr[i]; key[a] = lfsr[lfsr.length - 1]; lfsr = temp; return key; public string Encryption(string plian, int [] key) int[] temp = new int[plian.length]; char[] cipher = new char[plian.length]; for (int i = 0; i < plian.length; i++) if (plian[i] == '1') temp[i] = 1; else temp[i] = 0; for (int i = 0; i < temp.length; i++) temp[i] = temp[i]^key[i]; for (int i = 0; i < temp.length; i++) if (temp[i] == 1) cipher[i] = '1'; else cipher[i] = '0'; return new string(cipher); public string Decryption(string cipher, int[] key) int[] temp = new int[cipher.length]; char[] plian = new char[cipher.length]; for (int i = 0; i < cipher.length; i++) if (cipher[i] == '1') temp[i] = 1; else temp[i] = 0; 18

19 for (int i = 0; i < temp.length; i++) temp[i] = temp[i] ^ key[i]; for (int i = 0; i < temp.length; i++) if (temp[i] == 1) plian[i] = '1'; else plian[i] = '0'; return new string(plian); private void button1_click(object sender, EventArgs e) int[] key = LfsrGgenerator(textBox2.Text.Length, textbox1.text); string output = Encryption(textBox2.Text, key); textbox3.text = output; private void button2_click(object sender, EventArgs e) int[] key = LfsrGgenerator(textBox2.Text.Length, textbox1.text); string output = Decryption(textBox3.Text, key); textbox4.text = output; 19

C = M e mod n M = C d mod n Both sender and receiver must know the value of n.

20 RSA Algorithm (Public-Key) The RSA scheme is a cipher in which the plaintext and ciphertext are integers between 0 and n - 1 for some n. Encryption and decryption are of the following form, for some plaintext block M and ciphertext block C. C = M e mod n M = C d mod n Both sender and receiver must know the value of n. The sender knows the value of e, and only the receiver knows the value of d. This is a public-key encryption algorithm with a public key of PU = e, n and a private key of PR = d, n. The algorithm of RSA: The example of RSA : 20

For this example, the keys were generated as follows: 1. Select two prime numbers, p = 17 and q = 11. 2. Calculate n = p*q = 17 * 11 = 187. 3. Calculate f (n) = (p - 1)(q - 1) = 16 * 10 = 160. 4.

21 For this example, the keys were generated as follows: 1. Select two prime numbers, p = 17 and q = Calculate n = p*q = 17 * 11 = Calculate f (n) = (p - 1)(q - 1) = 16 * 10 = Select e such that e is relatively prime to f (n) = 160 and less than f(n); we choose e = Determine d such that (d*e) mod 160=1 and d < 160. The correct value is d = 23, because 23 * 7 = 161 mod 160=1. The resulting keys are public key PU = 7, 187 and private key PR = 23, 187. The example shows the use of these keys for a plaintext input of M = 88. For encryption: calculate C = 88 7 mod 187 that will be 894,432 mod 187 = 11 For decryption: calculate M = mod 187 that will be 79,720,245 mod 187 = 88 The RSA code program: namespace RSA public partial class Form1 : Form public Form1() InitializeComponent(); private void Form1_Load(object sender, EventArgs e) 21

22 public static int GCD(int p, int q) if (q == 0) return p; int r = p % q; return GCD(q, r); public static bool Prime(int no) for (int i = 2; i < no - 1; i++) if (no % i == 0) return false; return true; public static int PRKey (int fn) int e = 2; while (GCD(e,fn)!= 1) e++; return e; public static int PUKey(int fn) int e = PRKey(fn); int d = 1; while (((e * d) % fn!= 1)) d++; return d; 22

23 public double Encryption(int M, int p,int q) int n = p * q; if (int.parse(textbox3.text) >= n) MessageBox.Show("invalid plian number.please try anthor one "); int fn = (p - 1) * (q - 1); // calculate private key int e = PRKey(fn); double C= ( Math.Pow(M, e)) % n; return C; public double Decryption(int C, int p, int q) int n = p * q; int fn = (p - 1) * (q - 1); // calculate public key int d = PUKey(fn); double M = (Math.Pow(C, d)) % n; return M; private void button1_click(object sender, EventArgs e) int keyp = int.parse(textbox1.text); int keyq = int.parse(textbox2.text); if ((!Prime(keyp)!Prime(keyq)) (keyp==keyq)) MessageBox.Show("invalid cipher key. please try again "); double output = Encryption(int.Parse(textBox3.Text), keyp,keyq); textbox4.text = output.tostring(); private void button2_click(object sender, EventArgs e) int keyp = int.parse(textbox1.text); int keyq = int.parse(textbox2.text); double output = Decryption(int.Parse(textBox4.Text), keyp,k eyq); textbox5.text = output.tostring(); 23

Sankalchand Patel College of Engineering, Visnagar B.E. Semester V (CE/IT) INFORMATION SECURITY Practical List

Sankalchand Patel College of Engineering, Visnagar B.E. Semester V (CE/IT) INFORMATION SECURITY Practical List 1. IMPLEMENT CAESAR CIPHER WITH VARIABLE KEY It is an encryption technique in which each plaintext letter is to be replaced with one a fixed number of places (in following implementation, key) down the