The java code for the KSA algorithm is illustrated as follows for your

reference:

public class RC4KSA

{

public static void main(String[] args)

{

String key = "12345678";

int j = 0, temp = 0;

int[] S = new int[256];

int[] T = new int[256];

int[] K = new int[key.length()];

for (int a = 0; a < key.length(); a++)

{

K[a] = key.charAt(a);

}

int keyLength = key.length();

// Generation of the S-Box

for (int a = 0; a < 256; a++)

{

S[a] = a;

T[a] = Integer.parseInt(Integer.toHexString((char)K[a %

(keyLength)]), 16);

}

for (int a = 0; a < 256; a++)

{

j = (j + S[a] + T[a]) % 256;

temp = S[a];

S[a] = S[j];

S[j] = temp;

}

System.out.println("The final S-box after using KSA algorithm is...");

}

}

we have the S-box, we can do the RC4 encryption. Given a plaintext message hello and the initial key 12345678, we follow the steps below to encrypt the message hello using RC4 encryption algorithm,

Step 1. RC4 encrypt the plaintext byte by byte, so the first step is to

encrypt

h

, before encrypting h we have:

i

= j = 0

Initially, for encrypting first character

h

, we have:

i

= i+1 mod 256 = 0 + 1 mod 256 = 1

j = j + S[i] mod 256 = 0+94 mod 256 = 94

Then swap the value of S[1] and S[94] in the S-box

After swapping we have

S[1] = 177 and S[94]=94

which is a new S-box to

b

e used by the encryption next

Step 2. Then we calculate:

t = (S[1] + S[94]) (mod 256) = (177 + 94) mod 256 = 15

Step 3. Finally we calculate the exclusive OR of plaintext h in the format of

hexadecimal which is hex number 68 and S[15] in the format of

hexadecimal number which is hex number BB

And as a result, the encrypted hex number of plaintext h is:

01101000 XOR 10111011 = 11010011 = d3

which is the ciphertext for

h

S

tep 4. We repeat the steps 1, 2 and 3 to encrypt e, l, l, o respectively and

after RC4 encryption we have:

Plaintext in ASCII

Ciphertext in Hex

hd3e96l55lb8o66

The pseudo code of RC4 encryption is as follow,

Initially i = j = 0;

for each message byte Mi

i = (i + 1) (mod 256)

j = (j + S[i]) (mod 256)

swap(S[i], S[j])

t = (S[i] + S[j]) (mod 256)

Ci = Mi XOR S[t]

in which

M0 = h = 68

M1 = e = 65

M2 = l = 6c

M3 = l = 6c

M4 = o = 6f