Internet Security - Dang Thanh Binh's Page

ASYMMETRIC PUBLIC-KEY CRYPTOSYSTEMS 169
To decrypt a message, perform the same exponentiation process using the decryption key
d = 1019 such that:
m ≡ c d (mod n) ≡ 1570 1019 (mod 3337)
m = (1570) 512 × (1570) 256 × (1570) 128 × (1570) 64 × (1570) 32
× (1570) 16 × (1570) 8 × (1570) 2 × (1570)
= 3925000 (mod 3337) ≡ 688
Thus, the message is recovered.
To encrypt the message m, break it into a series of mi-digit blocks, 1 � i � n − 1.
Suppose each character in the message is represented by a two-digit number as shown in
Table 5.4.
Example 5.6 Encode the message ‘INFORMATION SECURITY’ using Table 5.4.
m = (0914061518130120091514001905032118092025)
Choose p = 47 and q = 71. Then
n = pq = 47 × 71 = 3337
φ(n) = (p − 1)(q − 1) = 46 × 70 = 3220
Break the message m into blocks of four digits each:
0914 0615 1813 0120 0915
1400 1905 0321 1809 2025
Choose the encryption key e = 79. Then the decryption key d becomes:
d ≡ e −1 (mod φ(n)) ≡ 79 −1 (mod 3220) ≡ 1019
The first block, m1 = 914, is encrypted by raising it to the power e = 79 and dividing by
n = 3337 and taking the remainder c1 = 3223 as the first block of ciphertext:
c1 ≡ m e 1
(mod n)
≡ 914 79 (mod 3337)
≡ 3223
Table 5.4 Two-digit number representing each character
Blank 00 E 05 J 10 O 15 T 20 Y 25
A 01 F 06 K 11 P 16 U 21 Z 26
B 02 G 07 L 12 Q 17 V 22
C 03 H 08 M 13 R 18 W 23
D 04 I 09 N 14 S 19 X 24