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RC4 Encryption Explained

This document summarizes the RC4 symmetric cryptosystem. It describes generating a random permutation of bytes using a key, then using that permutation to encrypt and decrypt messages by XORing bytes of the message with bytes of the permutation. A sample message "Sysadmin" is encrypted to "eOEWR[_X" using this method and then decrypted back to the original text.

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Anton Andrian
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68 views3 pages

RC4 Encryption Explained

This document summarizes the RC4 symmetric cryptosystem. It describes generating a random permutation of bytes using a key, then using that permutation to encrypt and decrypt messages by XORing bytes of the message with bytes of the permutation. A sample message "Sysadmin" is encrypted to "eOEWR[_X" using this method and then decrypted back to the original text.

Uploaded by

Anton Andrian
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Criptosisteme simetrice fluide: RC4

n=3
23 =8 – flux de biţi
Formăm vectrul S: 01234567
Alegem cheia: 26, K – vectorul cheii
Codul ASCII al cheii este: 2 → 50; 6 → 54;
Vectorul cheii este:
50 54 50 54 50 54 50 54

Depunem valorile codului ASCII al cheii:


P 0 1 2 3 4 5 6 7
S 0 1 2 3 4 5 6 7
K 50 54 50 54 50 54 50 54

Efectuăm permutări asupra biţilor vectorului S:

Iniţial i=0, j=0.


j=(j+K[i]+S[j]) mod 8

1) i=0, j=(0+K[0]+S[0]) mod 8 = (0+50+0) mod 8 = 50 mod 8 = 2;


SWAP(S[i],S[j])=SWAP(S[0],S[2])=21034567= 50 54 50 54 50 54 50 54
P 0 1 2 3 4 5 6 7
S 2 1 0 3 4 5 6 7
K 50 54 50 54 50 54 50 54

2) i=1, j=(2+K[1]+S[1]) mod 8 = (2+54+1) mod 8 = 57 mod 8 = 1;


SWAP(S[i],S[j])=SWAP(S[1],S[1])= 21034567= 50 54 50 54 50 54 50 54
P 0 1 2 3 4 5 6 7
S 2 1 0 3 4 5 6 7
K 50 54 50 54 50 54 50 54

3) i=2, j=(1+K[2]+S[2]) mod 8 = (1+50+0) mod 8 = 51 mod 8 = 3;


SWAP(S[i],S[j])=SWAP(S[2],S[3])= 21304567= 50 54 54 50 50 54 50 54
P 0 1 2 3 4 5 6 7
S 2 1 3 0 4 5 6 7
K 50 54 54 50 50 54 50 54

4) i=3, j=(3+K[3]+S[3]) mod 8 = (3+50+0) mod 8 = 53 mod 8 = 5;


SWAP(S[i],S[j])=SWAP(S[3],S[5])= 21354067= 50 54 54 54 50 50 50 54
P 0 1 2 3 4 5 6 7
S 2 1 3 5 4 0 6 7
K 50 54 54 54 50 50 50 54

1
5) i=4, j=(5+K[4]+S[4]) mod 8 = (0+50+4) mod 8 = 54 mod 8 = 6;
SWAP(S[i],S[j])=SWAP(S[4],S[6])= 21356047= 50 54 54 54 50 50 50 54
P 0 1 2 3 4 5 6 7
S 2 1 3 5 6 0 4 7
K 50 54 54 54 50 50 50 54

6) i=5, j=(6+K[5]+S[5]) mod 8 = (5+50+0) mod 8 = 55 mod 8 = 7;


SWAP(S[i],S[j])=SWAP(S[5],S[7])= 21356740= 50 54 54 54 50 54 50 50
P 0 1 2 3 4 5 6 7
S 2 1 3 5 6 7 4 0
K 50 54 54 54 50 54 50 50

7) i=6, j=(7+K[6]+S[6]) mod 8 = (6+50+4) mod 8 = 60 mod 8 = 4;


SWAP(S[i],S[j])=SWAP(S[6],S[4])= 21354760= 50 54 54 54 50 54 50 50
P 0 1 2 3 4 5 6 7
S 2 1 3 5 4 7 6 0
K 50 54 54 54 50 54 50 50

8) i=7, j=(4+K[7]+S[7]) mod 8 = (7+50+0) mod 8 = 57 mod 8 = 1;


SWAP(S[i],S[j])=SWAP(S[7],S[1])= 20354761= 50 50 54 54 50 54 50 54
P 0 1 2 3 4 5 6 7
S 2 0 3 5 4 7 6 1
K 50 50 54 54 50 54 50 54

CRIPTARE

Alegem un mesaj pentru criptare m = Sysadmin


Iniţial i=0, j=0.
i=(i+1) mod 8 = 1 mod 8 = 1.
j=(j+S[i]) mod 8 = (j+S[1]) mod 8 = (0+0) mod 8 = 0

Calculam t:
t=(S[i]+S[j]) mod 8 = (S[1]+S[0]) mod 8 = (0+2) mod 8 = 2

Criptul r = chr(ord(m)) XOR K[t]

, K[t] = K[2] = (54)2 = (00110110)2


m[0] = S → (83)10 = (01010011)2
m[1] = y → (121)10 = (01111001)2
m[2] = s → (115)10 = (01110011)2
m[3] = a → (97)10 = (01100001)2
m[4] = d → (100)10 = (01100100)2
m[5] = m → (109)10 = (01101101)2
m[6] = i → (105)10 = (01101001)2
m[7] = n → (110)10 = (01101110)2

1) r[0] = 01010011 XOR 00110110 = (01100101)2 = (101)10 → e


2
2) r[1] = 01111001 XOR 00110110 = (01001111)2 = (79)10 → O
3) r[2] = 01110011 XOR 00110110 = (01000101)2 = (69)10 → E
4) r[3] = 01100001 XOR 00110110 = (01010111)2 = (87)10 → W
5) r[4] = 01100100 XOR 00110110 = (01010010)2 = (82)10 → R
6) r[5] = 01101101 XOR 00110110 = (01011011)2 = (91)10 → [
7) r[6] = 01101001 XOR 00110110 = (01011111)2 = (95)10 → _
8) r[7] = 01101110 XOR 00110110 = (01011000)2 = (88)10 → X

Deci, mesajul m = „Sysadmin” criptat este „eOEWR[_X”

DECRIPTARE

Cript r = eOEWR[_X;
K=26;
S= 20354761
P 0 1 2 3 4 5 6 7
S 2 0 3 5 4 7 6 1
K 50 50 54 54 50 54 50 54

Iniţial i=0, j=0.


i = (i+1) mod 8 = 1 mod 8 = 1.
j = (j+S[i]) mod 8 = (j+S[1]) mod 8 = (0+0) mod 8 = 0

Calculam t:
t=(S[i]+S[j]) mod 8 = (S[1]+S[0]) mod 8 = (0+2) mod 8 = 2

i = chr(ord(r)) XOR K[t]


r[0] = e → (101)10 = (01100101)2
r[1] = O → (79)10 = (01001111)2
r[2] = E → (69)10 = (01000101)2
r[3] = W → (87)10 = (01010111)2
r[4] = R → (82)10 = (01010010)2
r[5] = [ → (91)10 = (01011011)2
r[6] = _ → (95)10 = (01011111)2
r[7] = X → (88)10 = (01011000)2

1) m[0] = 01100101 XOR 00110110 = (01010011)2 = (83)10 → S


2) m[1] = 01001111 XOR 00110110 = (01111001)2 = (121)10 → y
3) m[2] = 01000101 XOR 00110110 = (01110011)2 = (115)10 → s
4) m[3] = 01010111 XOR 00110110 = (01100001)2 = (97)10 → a
5) m[4] = 01010010 XOR 00110110 = (01100100)2 = (100)10 → d
6) m[5] = 01011011 XOR 00110110 = (01101101)2 = (109)10 → m
7) m[6] = 01011111 XOR 00110110 = (01101001)2 = (105)10 → i
8) m[7] = 01011000 XOR 00110110 = (01101110)2 = (110)10 → n

Mesajul criptat r = „eOEWR[_X” decriptat va fi: „Sysadmin”.

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