Simplified Data Encryption Standard | Set 2
Prerequisite – Simplified Data Encryption Standard | Set 1
Simplified Data Encryption Standard is a simple version of Data Encryption Standard having a 10-bit key and 8-bit plain text. It is much smaller than the DES algorithm as it takes only 8-bit plain text whereas DES takes 64-bit plain text. It was developed for educational purpose so that understanding DES can become easy. It is a block cipher algorithm and uses a symmetric key for its algorithm i.e. they use the same key for both encryption and decryption. It has 2 rounds for encryption which use two different keys.
First, we need to generate 2 keys before encryption. After generating keys we pass them to each individual round for s-des encryption. The below diagram shows the steps involved in the s-des algorithm.
Components :
S-DES encryption involves four functions –
1. Initial permutation(IP) –
2. Complex function (fk) –
It is the combination of permutation and substitution functions. The below image represents a round of encryption and decryption. This round is repeated twice in each encryption and decryption.
Components in fk are –
a. Expanded Permutation (EP) –
It takes a 4-bit input and converts it into an 8-bit output.
b. S-boxes (S0 and S1) –
It is a basic component of a symmetric key algorithm that performs substitution.
c. Permutation P4 –
3. Switch (SW) –
4. Inverse of Initial Permutation (IP-1) –
First, we need to generate 2 keys before encryption.
Consider, the entered 10-bit key is - 1 0 1 0 0 0 0 0 1 0
Therefore,
Key-1 is - 1 0 1 0 0 1 0 0 Key-2 is - 0 1 0 0 0 0 1 1
Encryption –
Entered 8-bit plaintext is - 1 0 0 1 0 1 1 1
Step-1:
We perform initial permutation on our 8-bit plain text using the IP table. The initial permutation is defined as –
IP(k1, k2, k3, k4, k5, k6, k7, k8) = (k2, k6, k3, k1, k4, k8, k5, k7) After ip = 0 1 0 1 1 1 0 1
Step-2:
After the initial permutation, we get an 8-bit block of text which we divide into 2 halves of 4 bit each.
l = 0 1 0 1 and r = 1 1 0 1
On the right half, we perform expanded permutation using EP table which converts 4 bits into 8 bits. Expand permutation is defined as –
EP(k1, k2, k3, k4) = (k4, k1, k2, k3, k2, k3, k4, k1)
After ep = 1 1 1 0 1 0 1 1
We perform XOR operation using the first key K1 with the output of expanded permutation.
Key-1 is - 1 0 1 0 0 1 0 0 (1 0 1 0 0 1 0 0) XOR (1 1 1 0 1 0 1 1) = 0 1 0 0 1 1 1 1 After XOR operation with 1st Key = 0 1 0 0 1 1 1 1
Again we divide the output of XOR into 2 halves of 4 bit each.
l = 0 1 0 0 and r = 1 1 1 1
We take the first and fourth bit as row and the second and third bit as a column for our S boxes.
S0 = [1,0,3,2
3,2,1,0
0,2,1,3
3,1,3,2]
S1= [0,1,2,3
2,0,1,3
3,0,1,0
2,1,0,3]
For l = 0 1 0 0
row = 00 = 0, column = 10 = 2
S0 = 3 = 11
For r = 1 1 1 1
row = 11 = 3, column = 11 = 3
S1 = 3 = 11
After first S-Boxes combining S0 and S1 = 1 1 1 1
S boxes gives a 2-bit output which we combine to get 4 bits and then perform permutation using the P4 table. P4 is defined as –
P4(k1, k2, k3, k4) = (k2, k4, k3, k1) After P4 = 1 1 1 1
We XOR the output of the P4 table with the left half of the initial permutation table i.e. IP table.
(0 1 0 1) XOR (1 1 1 1) = 1 0 1 0 After XOR operation with left nibble of after ip = 1 0 1 0
We combine both halves i.e. right half of initial permutation and output of ip.
Combine 1 1 0 1 and 1 0 1 0 After combine = 1 0 1 0 1 1 0 1
Step-3:
Now, divide the output into two halves of 4 bit each. Combine them again, but now the left part should become right and the right part should become left.
After step 3 = 1 1 0 1 1 0 1 0
Step-4:
Again perform step 2, but this time while doing XOR operation after expanded permutation use key 2 instead of key 1.
Expand permutation is defined as - 4 1 2 3 2 3 4 1 After second ep = 0 1 0 1 0 1 0 1 After XOR operation with 2nd Key = 0 0 0 1 0 1 1 0 After second S-Boxes = 1 1 1 1 P4 is defined as - 2 4 3 1 After P4 = 1 1 1 1 After XOR operation with left nibble of after first part = 0 0 1 0 After second part = 0 0 1 0 1 0 1 0 l = 1 1 0 1 and r = 1 0 1 0
On the right half, we perform expanded permutation using EP table which converts 4 bits into 8 bits. Expand permutation is defined as –
EP(k1, k2, k3, k4) = (k4, k1, k2, k3, k2, k3, k4, k1) After second ep = 0 1 0 1 0 1 0 1
We perform XOR operation using second key K2 with the output of expanded permutation.
Key-2 is - 0 1 0 0 0 0 1 1 (0 1 0 0 0 0 1 1) XOR (0 1 0 1 0 1 0 1) = 0 0 0 1 0 1 1 0 After XOR operation with 2nd Key = 0 0 0 1 0 1 1 0
Again we divide the output of XOR into 2 halves of 4 bit each.
l = 0 0 0 1 and r = 0 1 1 0
We take the first and fourth bit as row and the second and third bit as a column for our S boxes.
S0 = [1,0,3,2
3,2,1,0
0,2,1,3
3,1,3,2]
S1 = [0,1,2,3
2,0,1,3
3,0,1,0
2,1,0,3]
For l = 0 0 0 1
row = 01 = 1 , column = 00 = 0
S0 = 3 = 11
For r = 0 1 1 0
row = 00 = 0 , column = 11 = 3
S1 = 3 = 11
After first S-Boxes combining S0 and S1 = 1 1 1 1
S boxes gives a 2-bit output which we combine to get 4 bits and then perform permutation using the P4 table. P4 is defined as –
P4(k1, k2, k3, k4) = (k2, k4, k3, k1) After P4 = 1 1 1 1
We XOR the output of the P4 table with the left half of the initial permutation table i.e. IP table.
(1 1 0 1) XOR (1 1 1 1) = 0 0 1 0 After XOR operation with left nibble of after first part = 0 0 1 0
We combine both halves i.e. right half of initial permutation and output of ip.
Combine 1 0 1 0 and 0 0 1 0 After combine = 0 0 1 0 1 0 1 0 After second part = 0 0 1 0 1 0 1 0
Step-5:
Perform inverse initial permutation. The output of this table is the cipher text of 8 bit.
Output of step 4 : 0 0 1 0 1 0 1 0
Inverse Initial permutation is defined as –
IP-1(k1, k2, k3, k4, k5, k6, k7, k8) = (k4, k1, k3, k5, k7, k2, k8, k6)
8-bit Cipher Text will be = 0 0 1 1 1 0 0 0
Java
/*package whatever //do not write package name here */import java.io.*;public class GFG { // int key[]= {0,0,1,0,0,1,0,1,1,1}; int key[] = { 1, 0, 1, 0, 0, 0, 0, 0, 1, 0 }; // extra example for checking purpose int P10[] = { 3, 5, 2, 7, 4, 10, 1, 9, 8, 6 }; int P8[] = { 6, 3, 7, 4, 8, 5, 10, 9 }; int key1[] = new int[8]; int key2[] = new int[8]; int[] IP = { 2, 6, 3, 1, 4, 8, 5, 7 }; int[] EP = { 4, 1, 2, 3, 2, 3, 4, 1 }; int[] P4 = { 2, 4, 3, 1 }; int[] IP_inv = { 4, 1, 3, 5, 7, 2, 8, 6 }; int[][] S0 = { { 1, 0, 3, 2 }, { 3, 2, 1, 0 }, { 0, 2, 1, 3 }, { 3, 1, 3, 2 } }; int[][] S1 = { { 0, 1, 2, 3 }, { 2, 0, 1, 3 }, { 3, 0, 1, 0 }, { 2, 1, 0, 3 } }; // this function basically generates the key(key1 and //key2) using P10 and P8 with (1 and 2)left shifts void key_generation() { int key_[] = new int[10]; for (int i = 0; i < 10; i++) { key_[i] = key[P10[i] - 1]; } int Ls[] = new int[5]; int Rs[] = new int[5]; for (int i = 0; i < 5; i++) { Ls[i] = key_[i]; Rs[i] = key_[i + 5]; } int[] Ls_1 = shift(Ls, 1); int[] Rs_1 = shift(Rs, 1); for (int i = 0; i < 5; i++) { key_[i] = Ls_1[i]; key_[i + 5] = Rs_1[i]; } for (int i = 0; i < 8; i++) { key1[i] = key_[P8[i] - 1]; } int[] Ls_2 = shift(Ls, 2); int[] Rs_2 = shift(Rs, 2); for (int i = 0; i < 5; i++) { key_[i] = Ls_2[i]; key_[i + 5] = Rs_2[i]; } for (int i = 0; i < 8; i++) { key2[i] = key_[P8[i] - 1]; } System.out.println("Your Key-1 :"); for (int i = 0; i < 8; i++) System.out.print(key1[i] + " "); System.out.println(); System.out.println("Your Key-2 :"); for (int i = 0; i < 8; i++) System.out.print(key2[i] + " "); } // this function is use full for shifting(circular) the //array n position towards left int[] shift(int[] ar, int n) { while (n > 0) { int temp = ar[0]; for (int i = 0; i < ar.length - 1; i++) { ar[i] = ar[i + 1]; } ar[ar.length - 1] = temp; n--; } return ar; } // this is main encryption function takes plain text as //input uses another functions and returns the array of //cipher text int[] encryption(int[] plaintext) { int[] arr = new int[8]; for (int i = 0; i < 8; i++) { arr[i] = plaintext[IP[i] - 1]; } int[] arr1 = function_(arr, key1); int[] after_swap = swap(arr1, arr1.length / 2); int[] arr2 = function_(after_swap, key2); int[] ciphertext = new int[8]; for (int i = 0; i < 8; i++) { ciphertext[i] = arr2[IP_inv[i] - 1]; } return ciphertext; } // decimal to binary string 0-3 String binary_(int val) { if (val == 0) return "00"; else if (val == 1) return "01"; else if (val == 2) return "10"; else return "11"; } // this function is doing core things like expansion // then xor with desired key then S0 and S1 //substitution P4 permutation and again xor we have used //this function 2 times(key-1 and key-2) during //encryption and 2 times(key-2 and key-1) during //decryption int[] function_(int[] ar, int[] key_) { int[] l = new int[4]; int[] r = new int[4]; for (int i = 0; i < 4; i++) { l[i] = ar[i]; r[i] = ar[i + 4]; } int[] ep = new int[8]; for (int i = 0; i < 8; i++) { ep[i] = r[EP[i] - 1]; } for (int i = 0; i < 8; i++) { ar[i] = key_[i] ^ ep[i]; } int[] l_1 = new int[4]; int[] r_1 = new int[4]; for (int i = 0; i < 4; i++) { l_1[i] = ar[i]; r_1[i] = ar[i + 4]; } int row, col, val; row = Integer.parseInt("" + l_1[0] + l_1[3], 2); col = Integer.parseInt("" + l_1[1] + l_1[2], 2); val = S0[row][col]; String str_l = binary_(val); row = Integer.parseInt("" + r_1[0] + r_1[3], 2); col = Integer.parseInt("" + r_1[1] + r_1[2], 2); val = S1[row][col]; String str_r = binary_(val); int[] r_ = new int[4]; for (int i = 0; i < 2; i++) { char c1 = str_l.charAt(i); char c2 = str_r.charAt(i); r_[i] = Character.getNumericValue(c1); r_[i + 2] = Character.getNumericValue(c2); } int[] r_p4 = new int[4]; for (int i = 0; i < 4; i++) { r_p4[i] = r_[P4[i] - 1]; } for (int i = 0; i < 4; i++) { l[i] = l[i] ^ r_p4[i]; } int[] output = new int[8]; for (int i = 0; i < 4; i++) { output[i] = l[i]; output[i + 4] = r[i]; } return output; } // this function swaps the nibble of size n(4) int[] swap(int[] array, int n) { int[] l = new int[n]; int[] r = new int[n]; for (int i = 0; i < n; i++) { l[i] = array[i]; r[i] = array[i + n]; } int[] output = new int[2 * n]; for (int i = 0; i < n; i++) { output[i] = r[i]; output[i + n] = l[i]; } return output; } // this is main decryption function // here we have used all previously defined function // it takes cipher text as input and returns the array //of decrypted text int[] decryption(int[] ar) { int[] arr = new int[8]; for (int i = 0; i < 8; i++) { arr[i] = ar[IP[i] - 1]; } int[] arr1 = function_(arr, key2); int[] after_swap = swap(arr1, arr1.length / 2); int[] arr2 = function_(after_swap, key1); int[] decrypted = new int[8]; for (int i = 0; i < 8; i++) { decrypted[i] = arr2[IP_inv[i] - 1]; } return decrypted; } public static void main(String[] args) { GFG obj = new GFG(); obj.key_generation(); // call to key generation // function // int []plaintext= {1,0,1,0,0,1,0,1}; int[] plaintext = { 1, 0, 0, 1, 0, 1, 1, 1 }; // extra example for checking purpose System.out.println(); System.out.println("Your plain Text is :"); for (int i = 0; i < 8; i++) // printing the // plaintext System.out.print(plaintext[i] + " "); int[] ciphertext = obj.encryption(plaintext); System.out.println(); System.out.println( "Your cipher Text is :"); // printing the cipher // text for (int i = 0; i < 8; i++) System.out.print(ciphertext[i] + " "); int[] decrypted = obj.decryption(ciphertext); System.out.println(); System.out.println( "Your decrypted Text is :"); // printing the // decrypted text for (int i = 0; i < 8; i++) System.out.print(decrypted[i] + " "); }}//Omkar Varhadi |
Your Key-1 : 1 0 1 0 0 1 0 0 Your Key-2 : 0 1 0 0 0 0 1 1 Your plain Text is : 1 0 0 1 0 1 1 1 Your cipher Text is : 0 0 1 1 1 0 0 0 Your decrypted Text is : 1 0 0 1 0 1 1 1
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