1. Fast Modular Exponentiation
Viny Christanti M., M.Kom

2. Modular Exponentiation
Modular Exponentiation is a type of exponentiation performed over a modulus.
A = xn mod p
Solusi:
Modular multiply
Modular square and multiply
Modular reduction

3. Kegunaan
Implementasi perhitungan RSA sangat membutuhkan modulo exponentiation.
Operasi ini harus dapat dilakukan dengan cepat dan tepat.
Bagaimana menghitung mod 4559 ?
memiliki 3397·4 digits!

4. Modular Addition 7 + 9 = 16 (7 + 9) mod 10 = 6 7 + 4 = 11?
7 + 9 = 16
(7 + 9) mod 10 = 6
7 + 4 = 11
(7 + 4) mod 10 = 1
5 + 5 = 10
(5 + 5) mod 10 = 0 ? ? ? ? ?

5. Modular Subtraction 7 - 9 = -2 (7 - 9) mod 10 = 8 7 - 4 = 3?
7 - 9 = -2
(7 - 9) mod 10 = 8
7 - 4 = 3
(7 - 4) mod 10 = 3
5 - 5 = 0
(5 - 5) mod 10 = 0 ? ? ? ? ?

6. Modular Multiplication?
7 × 9 = 63
(7 × 9) mod 10 = 3
7 × 4 = 28
(7 × 4) mod 10 = 8
5 × 5 = 25
(5 × 5) mod 10 = 5 ? ? ? ? ?

7. Modular Division 7 ÷ 9 = 7/9 (7 ÷ 9) mod 10 = 3?
7 ÷ 9 = 7/9
(7 ÷ 9) mod 10 = 3
proof: 3 × 9 mod 10 = 7
7 ÷ 4 = 7/4
(7 ÷ 4) mod 10 = ??
undefined: no x such that 4 × x mod 10 = 7
5 ÷ 5 = 1
(5 ÷ 5) mod 10 = ??
undefined: more than one x such that 5 × x mod 10 = 5 ? ? ? ? ?

8. Modular Exponentiation
79 = 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 = 40,353,607
79 mod 10 = 7
79 takes four multiplications, and the intermediate values keep getting larger:
72 = 7 × 7 = 49
74 = 72 × 72 = 49 × 49 = 2401
78 = 74 × 74 = 2401 × 2401 = 5,764,801
79 = 78 × 71 = 5,764,801 × 7 = 40,353,607
79 mod 10 takes also four (mod) multiplications, but the values stay small

9. Contoh 1113 mod 53? Jawaban: How ‘bout 2325 mod 30 ?
13 = maka,
1113 = = 118 * 114 * 111 sehingga,
11 mod 53 = 11
112 = 121, 121 mod 53 = 121 – 2*53 = 15
114 = (112 mod 53 )2 = 152 mod 53 = 225 mod 53 = 225 – 4*53 = 13
118 = (114 mod 53 )2 = 132 mod 53 = 169 mod 53 = 169 – 3*53 = 10
1113 mod 53 = 11 * 13 * 10 = mod 53 = 1430 – 26*53 = 52
1113 mod 53 = 52
How ‘bout 2325 mod 30 ?

10. Algoritma Initialise y=1; u=x mod p; Repeat
If n is odd then y:=(y*u) mod p;
n:=n div 2;
u:=(u*u) mod p;
Until n==0;
Output y;

11. Contoh 2325 mod 30
Solusi: The previous method of repeated squaring works for any exponent that’s a power of isn’t. However, we can break 25 down as a sum of such powers: 25 = Apply repeated squaring to each part, and multiply the results together. Previous calculation: 238 mod 30 = 2316 mod 30 = mod 30  (mod 30)

12. Contoh 2325 mod 30 … 2325 mod 30  2316+8+1 (mod 30)
Jawaban: mod 30 = 23

13. Contoh 2325 mod 30 … How could we have figured out the decomposition
25 = from the binary (unsigned) representation of 25?
25 = (11001)2 maka
25 = 1·16+1·8+0·4+0·2+1·1 =

14. Selamat Berputar