Discrrete mathematics for computer science arithmetic

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 > L
2. T(n) = Θ(ne log n) if e = L
3. 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(n 1.58…) [Case 3]


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 2+1+2 = 5 multiplications in all!
lec 4F.8


Fast Exponentiation
compute ab using registers
X,Y,Z,R
X:= a; Y:= 1;
REPEAT:

Z:= b;

if Z=0, then return Y
R:= remdr(Z,2); Z:= quotnt(Z,2)
if R=1,then Y:= X⋅Y

X:= X2

lec 4F.9


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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III. Modular arithmetic
0
7

1

6

2

5

3
4

6 + 5 = 3 (mod 8)
February 28, 2007

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Math Quiz
2x6 = 1

mod 11

2x6x5 =5

mod 11

23 =
2300 =

1
1

mod 7
mod 7

= (23)100 = 1100 = 1
February 28, 2007

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(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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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

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FINIS