1. Arithmetic

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

3. 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

4. Grade School Algorithm is Θ(n2)
But answer is only O(n) bits: Can we do better?

5. 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!!!

6. How Fast?
T(1)=1
T(n) = 3T(n/2) + cn
But how to solve this?

7. Master Theorem on D+C recurrences
T(n) = aT(n/b) + cne
Let L = logba
Recurrence has the solution:
T(n) = Θ(ne) if e > L
T(n) = Θ(ne log n) if e = L
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]

8. II: Fast Exponentiation
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 = 5 multiplications in all!

9. Fast Exponentiation 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

10. 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)
February 28, 2007
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11. III. Modular arithmetic1 2 3 4 5 6 7
6 + 5 = 3 (mod 8)
February 28, 2007

12. Math Quiz
2 x 6 =
mod 11
2 x 6 x 5 23
mod 7
2300 1 5 1 1
= (23)100 = 1100 = 1
February 28, 2007

13. (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)
February 28, 2007
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14. 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.
February 28, 2007
Harvard Bits

15. FINIS