Harvard University
Computer Science 20
In-Class Problems 34
Monday, April 25, 2016
Author: Crystal Chang, edited by Jack Dent
Executive Summary
1. Fast Multiplication and the Master Theorem on Divide and Conquer
• Fast Multiplication:
To compute a · b each with 2m bits (or digits), write a = a1 2m + a0 , b = b1 2m + b0 ,
where a1 , a0 , b1 , b0 are m-bit numbers — the first and last m bits of a and b. Then
a · b = a1 b1 (22m + 2m ) − (a1 − a0 )(b1 − b0 )2m + a0 b0 (1 + 2m )
so multiplication requires only 3 m-bit multiplications plus additions.
Then T (1) = 1 and T (2m) = 3T (m) + cm,
whose solution is T (m) = Θ(mlog2 3 )
• Master Theorem on D + C recurrences:
– T (1) = 1, T (n) = aT (n/b) + cne Let L = logb a
– Recurrence has the solution:
∗ T (n) = Θ(ne ) if e > L
∗ T (n) = Θ(ne log2 n) if e = L
∗ T (n) = Θ(nL ) if e < L
2. Fast Exponentiation:
We can calculate ab more efficiently by squaring and multiplying by a, e.g. a10 = ((a2 )2 a)2 .
The process takes Θ(log2 b) operations.
Algorithm:
Compute ab using registers X, Y, Z, R
X := a; Y := 1; Z := b;
REPEAT:
if Z = 0, then return Y
R:= remdr(Z,2); Z:= quotnt(Z,2)
if R = 1,then Y := X ∗ Y
X := X2
3. Modular Arithmetic:
• The notation a ≡ b (mod n) means that a and b have the same remainder when divided
by n.
• It is most common to pick a prime p for the modulus.
• For example, 4 ≡ 7 (mod 3), 1 ≡ −1 (mod 2), and 2300 ≡ 8100 ≡ 1 (mod 7).
• Successive squaring for fast exponentiation also works modp.
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PROBLEM 1
Fast Multiplication: Use the fast multiplication algorithm to compute the following products.
(A) 112 × 102
(B) 11102 × 10112
Solution.
(A) m = 1,
a = 1 ∗ 2 + 1 → a1 = 1, a0 = 1
b = 1 ∗ 2 + 0 → b1 = 1, b0 = 0
ab = a1 b1 (22m + 2m ) − (a1 − a0 )(b1 − b0 )2m + a0 b0 (1 + 2m )
= 1 ∗ 1 ∗ (22 + 2) − (1 − 1)(1 − 0) ∗ 2 + 1 ∗ 0(1 + 2) = 1 ∗ 22 + 1 ∗ 2 = 110
(B) m = 2,
a = 11 ∗ 22 + 10 → a1 = 11, a0 = 10
b = 10 ∗ 22 + 11 → b1 = 10, b0 = 11
ab = a1 b1 (22m + 2m ) − (a1 − a0 )(b1 − b0 )2m + a0 b0 (1 + 2m )
= 11 ∗ 10 ∗ (24 + 22 ) − (11 − 10)(10 − 11) ∗ 22 + 10 ∗ 11 ∗ (1 + 22 )
= 110 ∗ (24 + 22 ) − 1 ∗ (−1) ∗ 22 + 110 ∗ (1 + 22 )
= 110 ∗ 24 + 110 ∗ 22 + 1 ∗ 22 + 110 ∗ 1 + 110 ∗ 22
= 1100000 + 11000 + 100 + 110 + 11000 = 10011010
PROBLEM 2
Master Theorem: For each of the following recurrences, use the Master Theorem to give an
expression for the runtime T (n).
(A) T (n) = 8T (n/2) + 2n4
√
(B) T (n) = 5T (n/5) + n
(C) T (n) = 4T (n/2) + n2
Solution.
(A) a=8, b=2, e=4, L=log2 8 = 3, e > L → (Case 1) → T (n) = Θ(ne ) = Θ(n4 )
(B) a=5, b=5, e=1/2, L=log5 5 = 1, e < L → (Case 3) → T (n) = Θ(nL ) = Θ(n)
(C) a=4, b=2, e=2, L=log2 4 = 2, e = L → (Case 2) → T (n) = Θ(n2 log2 n)
PROBLEM 3
Modular Arithmetic: Compute the following exponentiations using the fast exponentiation algorithm.
(A) 1315 mod 23
(B) 1711 mod 50
Solution.
(A) 1315 = ((((132 ) ∗ 13)2 ) ∗ 13)2 ∗ 13 ≡ ((((169 mod 23) ∗ 13)2 ) ∗ 13)2 ∗ 13
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≡ ((((8 ∗ 13) mod 23)2 ) ∗ 13)2 ∗ 13 ≡ ((122 mod 23) ∗ 13)2 ∗ 13 ≡ ((6 ∗ 13) mod 23)2 ∗ 13
≡ (92 mod 23) ∗ 13 ≡ (12 ∗ 13) mod 23 = 18
(B) 1711 = ((((172 )2 ) ∗ 17)2 ) ∗ 17 ≡ ((((289 mod 50)2 ) ∗ 17)2 ) ∗ 17
≡ (((392 mod 50) ∗ 17)2 ) ∗ 17 ≡ (((21 ∗ 17) mod 50)2 ) ∗ 17 ≡ ((72 ) ∗ 17) mod 50 = 33
PROBLEM 4
(Bonus) Modify ab = a1 b1 (22m + 2m ) − (a1 − a0 )(b1 − b0 )2m + a0 b0 (1 + 2m ) for base 10 multiplication
and then use the resulting formula to calculate 89 × 78.
Solution.
Base 10 formula: ab = a1 b1 (102m + 10m ) − (a1 − a0 )(b1 − b0 )10m + a0 b0 (1 + 10m )
m = 1,
a = 8 ∗ 10 + 9 → a1 = 8, a0 = 9
b = 7 ∗ 10 + 8 → b1 = 7, b0 = 8
ab = a1 b1 (102m + 10m ) − (a1 − a0 )(b1 − b0 )10m + a0 b0 (1 + 10m )
= 8 ∗ 7(102 + 10) − (8 − 9)(7 − 8) ∗ 10 + 9 ∗ 8(1 + 10)
= 56 ∗ 102 + 56 ∗ 10 − 10 + 72 ∗ 1 + 72 ∗ 10 = 6942
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