Introduction: Efficient Algorithms
for the Problem of Computing
Fibonocci Numbers
Analysis of Algorithms
Prepared by
John Reif, Ph.D.
Readings
• Main Reading Selection:
– CLR, Chapter 1
Goal
• Devise Algorithms to solve problems
on a TI calculator
• assume each step
(a multiplication, addition, or control
branch)
takes 1 m sec = 10-6 sec
on 16 bit words
Time Efficiency
Fibonocci Sequence
• 0, 1, 1, 2, 3, 5, 8, 13…
• Recursive Definition
• Fn = if n < 1 then n
else Fn-1 + Fn-2
• Problem
• Compute Fn for n=109
fast.
Fibonacci Growth
• Can show as n  ∞
  ( )
n
Fn ~
n
5
• Golden ratio
1 5

 1.62...
2
• So Fn ~ .45 ∙ 2 .7n is .7n bit number
grows exponentially!
Obvious Method
• Fn =
n
if n < 1
Fn-1 + Fn-2 if n > 1
costs at least
n
adds of
2
.7n
 .35n bit numbers
2
n
 .35n bits 
Total Cost   adds 

2
 16 bits 
> .01 n2 steps
> 1016 m sec for n = 109
= 1010 sec ~ 317 years!
Wanted!
• An efficient algorithm for Fn
• Weapons:
• Special properties of computational
problems
= combinatorics (in this case)
Theorem:
11 


10 


n
Fn 
 Fn 1


F

F
n 1 
 n
• Proof by Induction
– Basis Step
• Holds by definition for n=1
– Inductive Step
• Assume holds for some n>0
Inductive Step
n 1
11 
1 1   Fn 1 Fn 

 
   
10 
1 0   Fn Fn 1 
 Fn 1  Fn Fn  Fn 1 

 
Fn
 Fn 1

 Fn  2 Fn 1 

 
 Fn 1 Fn 
Powering Trick
• Fix M =
1 1
1 0
• To compute Mn when n is a power of 2
For i=1 to log n do
  M 
M  M
2i
2i-1
2i-1
2
22
gives M, M , M , ... M
2
logn
General Case of Powering
• Decompose n = 2 j1 + 2
as sum of powers of 2
• Example
• Compute
j2
+ ... + 2
23  2 4  2 2  21  20
 16  4  2  1
 10111 2
M =
n
ÕM
i=1,...,k
2 ji
jk
Algorithm
1 1 
M 
10 



2
(1) compute M, M , ..., M
2
logn
by power method
(2) if n = 2 + 2 +... + 2 then M =
j1
j2
jk
n
ÕM
i=1,...,k
(3) output Fn = upper right of M n
2 ji
Bounding Mults
• = 2 log n matrix products on
symmetric matrices of size 2 x 2
• Each matrix product costs 6 integer
mults
• Total Cost = 12 (log n) integer mults
> 360 for n = 109 integer mults
Time Analysis
• Fn ~ .45 2 .7n
• So Fn is m = .7n bit integer
• New method to compute Fn requires
multiplying m bit number
• BUT
• Grammar School Method for Mult
takes m2 bool ops
= m2/16 steps
~ 1016 steps for n = 109
~ 1016 msec
~ 1010 seconds
~ 317 years!
Fast Multiplication
of m bit integers a,b by
• “Divide and Conquer”
a  a L  2k  a R
b  bL  2  bR
k
aL
bL
aR
bR
m
where k 
2
a  b  a L b L 2 2k  (a L b R  a R b L )2 k  a R b R
• seems to take 4 multiplications, but…
3 Mult Trick: Karatsubu Algorithm
(1) x  a L  b L
(2) y  a R  b R
(3) z  (a L  a R )  (b L  b R )  (x  y)
 a LbR  a R bL
Requires only 3 mults
m
on - bit integers and 6 adds on m - bits
2
a  b  a L b L 2 2k  (a L b R  a R b L )2 k  a R b R
 x 2  z2  y
2k
k
Recursive Mult Algorithm
• Cost
C m bit ops C1  1
C m  3C m  6m for m  1
2
i
3
    6m
i 0  2 
log m
3
 
2
log m
6m
 6m log2 3
~ 6m1.4 bit ops
~ 1013 bit ops if m  .7 109
13
10
~
steps
16
~.05 years
Schonhage-Strassen Integer
Multiplication Algorithm
10m log m  log log m bit ops
~ .7 1012 bit ops for m  .7 109
~ 4.3 1010 steps
~4.3 104 sec.
~3 days
• This is feasible to compute!
Improved Algorithm
• Computed Fn for n = 109
using 360 = 12 log n integer mults
each taking ~ 3 days
• Total time to compute Fn
• ~ 3 years with 1 msec/step
How Did We Get Even More
Speed-Up?
1) New trick
• a 2 log n mult algorithm
(rather than n adds)
2) Old Trick
• use known efficient algorithms for
integer multiplication
(rather than Grammar School
mult)
3) Also used careful analysis of efficiency
of algorithms to compare methods
Problem (to be pondered later…)
• How fast can 109 integers be sorted?
• What algorithms would be used?
Introduction: Efficient Algorithms
for the Problem of Computing
Fibonocci Numbers
Analysis of Algorithms
Prepared by
John Reif, Ph.D.