Arithmetic

I. Fast Multiplication and the Master Theorem on Divide and Conquer

How fast can we multiply?

• Adding two n-bit numbers takes O(n) operations

• How many operations to multiply two n-bit numbers?

• Or two n-decimal-digit numbers – Difference is a factor of log210 ≈ 3.32

but the individual operations are harder

Grade School Algorithm is Θ(n2)

• But answer is only O(n) bits: Can we do better?

A Divide and Conquer Algorithm

• Suppose n is even, n = 2m• To compute a∙b• Write a = a1∙2m + a0, b = b1∙2m + b0, where

a1, a0, b1, b0 are m-bit numbers (numbers < 2m) – the first and last m bits of a and b

a∙b = a1b1∙22m + (a1b0+a0b1)∙2m + a0b0

= a1b1∙(22m+2m) + (a1-a0)(b0-b1)∙2m + a0b0∙(2m+1)

Only 3 m-bit multiplications!!!

How Fast?

• T(1)=1• T(n) = 3T(n/2) + cn• But how to solve this?

Master Theorem on D+C recurrences

• T(1) = 1• T(n) = aT(n/b) + cne

• Let L = logba

• Recurrence has the solution:1. T(n) = Θ(ne) if e > L2. T(n) = Θ(ne log n) if e = L3. T(n) = Θ(nL) if e < L

• Binary search: a=1, b=2, e=0, L=0 [Case 2]• Merge sort: a=2, b=2, e=1, L=1 [Case 2]• Ordinary mult: a=4, b=2, e=1, L=2 [Case 3]• Fast mult: a=3, b=2, e=1, L=lg 3 so Θ(n1.58…) [Case 3]

lec 4F.8

Compute 313:

313 = 3∙3∙3∙3∙3∙3∙3∙3∙3∙3∙3∙3∙3 (12 multiplications, or Θ(exponent))

313 = 36∙36∙3 (2 multiplications)36 = 33∙33 (1 multiplication)33 can be computed with 2 multiplications

So 2+1+2 = 5 multiplications in all!

II: Fast Exponentiation

lec 4F.9

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

Fast Exponentiation

Harvard Bits 10February 28, 2007

Powers by Repeated Squaring

• Problem: compute ab • Method 1: multiply a by itself n-1 times

– Requires n-1 multiplications• Method 2: use successive squaring

– How many times can you divide n by 2 before it is reduced to 1?

– Repeated squaring requires between log2n and 2∙log2n multiplications

– Huge savings! n = 1000 => at most 20 multiplications! (since log21000 < 10)

11February 28, 2007

III. Modular arithmetic

1

2

3

4

5

6

7

0

6 + 5 = 3 (mod 8)

12February 28, 2007

Math Quiz

2 x 6 = mod 11

2 x 6 x 5 = mod 11

23 = mod 7

2300 = mod 7

1

1

5

1

= (23)100 = 1100 = 1

Harvard Bits 13February 28, 2007

(mod p) notation

• Think of the (mod p) at the end of the line as referring to everything in the equation

• (23)100 = 1100 = 1 (mod 7) means“(23)100 , 1100 , and 1 are all equivalent if

you divide by 7 and keep just the remainder”

Often written a ≡ b (mod p)

Harvard Bits 14February 28, 2007

Fast Modular Exponentiation

• Problem: Given q and p and n, find y < p such that

qn = y (mod p)• Method 1: multiply q by itself n-1 times

– Requires n-1 multiplications• Method 2: use successive squaring

– Requires about log2n multiplications• Same idea works for multiplication modulo p• Example: If n is a 500-digit number, we can

compute qn (mod p) in about 1700 (= lg 10500) steps.

FINIS