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Great Theoretical Ideas In Computer Science
V. Adamchik
D. Sleator
Lecture 8
CS 15-251
Feb 04, 2010
Spring 2010
Carnegie Mellon University
Recurrences and Continued Fractions
Solve in integers
x1 + x2 +…+ x5 = 40
xkrk;
yk=xk – k
y1 + y2 +…+ y5 = 25
yk ≥ 0;
 29 
 
4
Solve in integers
x + y + z = 11
xr0; 0y3; zr0
[X11]
1xx2 x3
(1 -x) 2
Partitions
Find the number of ways to
partition the integer n
3 = 1+1+1 = 1+2
4 = 1+1+1+1 = 1+1+2 = 1+3 = 2+2
1x1  2x2  3x3  ...  nxn  n
1
(1 -x)(1 -x2)(1 -x3)...(1 -xn )
Plan
•Review: Sat, 6-8pm in 2315 DH
•Exam: Mon in recitations
•Characteristic Equations
•Golden Ratio
•Continued Fractions
Applied Combinatorics, by Alan Tucker
The Divine Proportion, by H. E. Huntley
Leonardo Fibonacci
In 1202, Fibonacci has become
interested in rabbits and what they do
The rabbit reproduction model
•A rabbit lives forever
•The population starts as a single newborn
pair
•Every month, each productive pair begets a
new pair which will become productive after 2
months old
Fn= # of rabbit pairs at the beginning of the
nth month
month
1
2
3
4
5
6
7
rabbits
1
1
2
3
5
8
13
F0=0, F1=1,
Fn=Fn-1+Fn-2 for n≥2
What is a closed form
formula for Fn?
We won’t be using GFs!
WARNING!!!!
This lecture has
explicit mathematical
content that can be
shocking to some
students.
Characteristic Equation
Fn = Fn-1 + Fn-2
Consider solutions of the form:
Fn= n
for some (unknown) constant  ≠ 0
 must satisfy:
n - n-1 - n-2 = 0
n - n-1 - n-2 = 0
iff
n-2(2
-  - 1) = 0
Characteristic
equation
iff 2 -  - 1 = 0
 = , or  = -1/
 (“phi”) is the golden ratio
1 5

2
a,b a n + b (-1/)n
satisfies the inductive condition
So for all these values of  the
inductive condition is satisfied:
2 -  - 1= 0
Do any of them happen to satisfy the base
condition as well?
F0=0, F1=1
a,b a n + b (-1/)n
satisfies the inductive condition
Adjust a and b to fit the base
conditions.
n=0: a + b = 0
n=1: a 1 + b (-1/ )1 = 1
1
a
,
5
1
b
5
Leonhard Euler (1765)
Fn 
  ( 1  )
n
n
5
1 5

2
Fibonacci Power Series

F x
k 0
k
k
 ??
x
2
1 x  x
Fibonacci Bamboozlement
8
8
13
5
Cassini’s Identity
Fn+1Fn-1 - Fn2 = (-1)n
We dissect FnxFn square and rearrange
pieces into Fn+1xFn-1 square
Heads-on
How to convert
kilometers into miles?
Magic conversion
50 km = 34 + 13 + 3
50 = F9 + F7 + F4
F8 + F6 + F3 = 31 miles
From the previous lecture
dn  dn 1  2dn 2
d0  0, d1  1
Characteristic Equation
dn  dn 1  2dn 2
d0  0, d1  1
1  1, 2  2
   2  0
2
dn  a ( 1)  b 2
n
n
a, b
dn  a ( 1)  b 2
n
n
d0  0  a (1)  b 2  a  b
0
0
d1  1  a (1)  b 2  a  2 b
1
1
1
1
b  ,a  
3
3
Characteristic Equation
xn  2xn 1  xn 2
x0  1, x1  2
  2  1  0
2
1  2  1
Characteristic Equation
Theorem. Let  be a root of multiplicity p
of the characteristic equation. Then
λ , n λ , n λ ,..., n λ
n
n
are all solutions.
2 n
p-1 n
xn  2xn 1  xn2
The theorem says that xn=n is a solution.
This can be easily verified: n = 2 n - n
From the previous lecture:
Rogue Recurrence
an  5an 1  8an 2  4an 3
a0  0, a1  1, a2  4
Characteristic equation:
  5  8  4  0
3
2
1  1, 2  3  2
an  5an 1  8an 2  4an 3
a0  0, a1  1, a2  4
General solution:
an  a  b 2  c n 2
a0  0  a  b
a1  1  a  2b  2c
a2  4  a  4b  8c
n
n
a  b  0,
1
c
2
an  n 2
n -1
The Golden Ratio
“Some other majors have their
mystery numbers, like  and e.
In Computer Science the mystery
number is , the Golden Ratio.” –
S.Rudich
Golden Ratio -Divine Proportion
Ratio obtained when you divide a line segment
into two unequal parts such that the ratio of
the whole to the larger part is the same as
the ratio of the larger to the smaller.
AC AB


AB BC
A
AC AB AC
 


AB BC BC
2
AC AB AC  AB
  


1
BC BC
BC
2
B
C
  1 0
2
Golden Ratio - the divine
proportion
  1 0
2
1 5

2
 = 1.6180339887498948482045…
“Phi” is named after the Greek sculptor
Phidias
Aesthetics
 plays a central role in renaissance art and
architecture.
After measuring the dimensions of pictures,
cards, books, snuff boxes, writing paper,
windows, and such, psychologist Gustav
Fechner claimed that the preferred
rectangle had sides in the golden ratio
(1871).
Which is the most attractive
rectangle?
Which is the most attractive rectangle?

1
Golden
Rectangle
The Golden Ratio
Divina Proportione
Luca Pacioli (1509)
Pacioli devoted an entire book to the
marvelous properties of . The book
was illustrated by a friend of his
named:
Leonardo Da Vinci
Table of contents
•The first considerable effect
•The second essential effect
•The third singular effect
•The fourth ineffable effect
•The fifth admirable effect
•The sixth inexpressible effect
•The seventh inestimable effect
•The ninth most excellent effect
•The twelfth incomparable effect
•The thirteenth most distinguished effect
Table of contents
“For the sake of salvation, the list
must end here” – Luca Pacioli
Divina Proportione
Luca Pacioli (1509)
"Ninth Most Excellent Effect"
two diagonals of a regular pentagon
divide each other in the Divine
Proportion.
C
B
A
AC AB


AB BC
Expanding Recursively
  1 0
2
  1
1

  1
1
1
1

Expanding Recursively
1
  1
1
1
1
1
1
1

A (Simple) Continued Fraction Is
Any Expression Of The Form:
1
a
1
b
1
c
1
d
1
e
1
f
1
g
h
1
1
i
j  ....
where a, b, c, … are whole numbers.
A Continued Fraction can have a
finite or infinite number of terms.
1
a
1
b
1
c
1
d
1
e
1
f
1
g
h
1
1
i
j  ....
We also denote this fraction by [a,b,c,d,e,f,…]
Continued Fraction Representation
8
1
 1
1
5
1
1
1
1
1
1
= [1,1,1,1,1,0,0,0,…]
Recursively Defined Form For CF
CF  whole number
1
 whole number 
CF
Proposition: Any finite
continued fraction
evaluates to a rational.
Converse: Any rational has
a finite continued fraction
representation.
Euclid’s GCD = Continued Fraction
Euclid(A,B) = Euclid(B, A mod B)
Stop when B=0
a  b q1  r1
b  r1 q2  r2
r1  r2 q3  r3
....
rk 1  rk qk 1  0
Euclid’s GCD = Continued Fraction
a  b q1  r1
b  r1 q2  r2
r1  r2 q3  r3
....
rk 1  rk qk 1  0
a
r1
1
 q1   q1  b
b
b
r1
b
r2
1
 q2   q2  r1
r1
r1
r2
r3
r1
1
 q3   q3  r2
r2
r2
r3
Euclid’s GCD = Continued Fraction
a  b q1  r1
b  r1 q2  r2
r1  r2 q3  r3
....
rk 1  rk qk 1  0
a
1
 q1  b
b
r1
1
 q1 
q2 
1
r1
r2
1
 q1 
q2  q3 1 1
r2
r3
A Pattern for 
Let r1 = [1,0,0,0,…] = 1
r2 = [1,1,0,0,0,…] = 2/1
r3 = [1,1,1,0,0,0…] = 3/2
r4 = [1,1,1,1,0,0,0…] = 5/3
and so on.
  1
Theorem:
 = Fn/Fn-1 when n ->
1
1
1

Divine Proportion
n
 1
    
n

Fn


 lim
 lim n 1  
lim
n 1
n  Fn 1
n 
n  


1
n 1
    
 
n
Quadratic Equations
X2 – 3x – 1 = 0
X2 = 3X + 1
3  13
X
2
X = 3 + 1/X
X = 3 + 1/X = 3 + 1/[3 + 1/X] = …
A Periodic CF
3  13
 3
2
3
1
1
1
3
1
3
1
3
1
3
1
3
3
1
1
3
3  ....
A period-2 CF
1 2

3
1
1
1
1
4
1
8
4
1
1
8
4  ...
Proposition: Any quadratic
solution has a periodic
continued fraction.
Converse: Any periodic
continued fraction is the
solution of a quadratic
equation
What about those
non-periodic
continued fractions?
What is the
continued fraction
expansion for ?
Euclid’s GCD = Continued Fraction
a  b q1  r1
b  r1 q2  r2
r1  r2 q3  r3
....
rk 1  rk qk 1  0
π  3  1  0.1415
1  7  0.1415  0.00885
0.1415  15  0.00885  0.00882
0.00885  1  0.00882  0.000078
??
What is the pattern?
1
  3
1
7
1
15 
1
1
1
292 
1
1
1
1
1
1
1
2
1  ....
What is the pattern?
1
e 1  1
1
1
1
2
1
1
1
1
1
4
1
1
1
1
1
6
1  ....
Every irrational number
greater than 1 is the
limit of a
unique infinite
continued fraction.
Every irrational number greater than 1 is
the limit of a unique infinite continued
fraction.
Suppose that y>1 is irrational and y = [b0;b1,b2, . . .].
1
1
y  b0 
 b0 
b1  ...
y1
where y1 > 1.
Thus,
1
b0  y   y 
y1
Similarly, for all bk.
To calculate the continued
fraction for a real number
y, set y0 =y and then
bk  yk 
yk 1
1

yk  bk
What a cool
representation!
Finite CF: Rationals
Periodic CF: Quadratic
roots
And some numbers reveal
hidden regularity.
Let us embark now on
approximations
1
  3
1
7
1
15 
1
1
1
292 
1
1
1
1
1
1
1
2
1  ....
CF for Approximations
π 3
1
π  3   3.1428
7
1
π  3
 3.14151
1
7
15
We say that an positive irrational number y
is approximable by rationals to order n
if there exist a positive constant c and
infinitely many rationals p/q with q > 0
such that
p c
y  n
q q
We will see that algebraic numbers are
not approximable to arbitrarily high order.
Positive rational numbers are
approximable to order 1
Let y = a/b be a rational
number, with gcd(a,b) = 1. By
Euclid’s algorithm we can find
p0,q0 such that p0 b − q0 a = 1.
Then
a p0
1


b q0 q0
Positive rational numbers are approximable
to order 1
p0 b  q0 a  1
a 1
p0  q0 
b b
p0 a
1
 
q0 b bq0
a p0
1
1



b q0 bq0 q0
Positive rational numbers are approximable
to order 1
a p0
1
1



b q0 bq0 q0
Similarly,
a pm
1


b qm qm
Liouville’s number

y   10
k!
k 1
Transcendental number
y  0.1100010000000000000000010. . .
Liouville’s number
 and e are
transcendental number
What about e ?
Apery’s constant
irrational number
Riemann Zeta function
1
ζ(3)   3
k 1 k

transcendental number - ???
• Review GCD algorithm
• Recurrences, Phi and CF
• Solving Recurrences
Study Bee