CSE 20 Lecture 12: Analysis of
Homogeneous Linear Recursion
CK Cheng
May 5, 2011
1
3. Analysis
•
•
•
•
3.1 Introduction
3.2 Homogeneous Linear Recursion
3.3 Pigeonhole Principle
3.4 Inclusion-Exclusion Principle
2
3.1 Introduction
Derive the bound of functions or recursions
Estimate CPU time and memory allocation
Eg.
PageRank calculation
Allocation of memory, CPU time,
Resource optimization
MRI imaging
Real time?
VLSI design
Design automation flow to meet the deadline
for tape out?
Further Study
Algorithm, Complexity
3
3.1 Introduction
• Derive the bound of functions or recursions
• Estimate CPU time and memory allocation
• Example on Fibonacci Sequence: Estimate fn.
– Index: 0 1 2 3 4 5 6 7 8 9
– fn:
0 1 1 2 3 4 5 8 13 21 34
1 1 5 n 1 1 5 n
fn 
(
) 
(
)
2
2
5
5
1 1 5 0 1 1 5 0
f0 
(
) 
(
) 0
2
2
5
5
1 1 5
1 1 5
f1 
(
)
(
) 1
2
2
5
5
4
Example: Fibonacci Sequence
0
1
2
3
4
5
6
7
8
9
0
1
1
2
3
5
8
13
21
34
1 1 5 n 1 1 5 n
fn 
(
) 
(
)
2
2
5
5
1

(1.618n  0.618n )
2.236
1
9
1
n
f9 
 1.618  33.98

 1.618
2.236
2.236
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3.2 Homogeneous Linear Recursion
• (1) Arithmetic Recursion
a, a+d, a+2d, …, a+kd
• (2) Geometric Recursion
a, ar, ar2, …, ark
• (3) Linear Recursion
an= e1an-1+e2an-2+…+ekan-k+ f(n)
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Linear Recursion and Homogeneous
Linear Recursion
• Linear Recursion: There are no powers or
products of
• Homogenous Linear Recursion: A linear
recursion with f(n)=0.
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Solving Linear Recursion
Input: Formula an  e1an1  e2 an2  ...  ek ank
and k initial values a0 , a1 ,..., ak 1
(1) characteristic
k
k 1
k 2
x

e
x

e
x
 ...  ek
polynomial:
1
2
(2)Find the root of the
characteristic polynomial (r1 , r2 ,..., rk )
(assuming ri are distinct)
an  c r  c2 r2  ...  ck rk
(3)Set
(4)Determine ci
from k initial values
n
1 1
n
n
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Solving Linear Recursion
Input: Formula an  e1an1  e2 an2  ...  ek ank
and k initial values a0 , a1 ,..., ak 1
(1) characteristic
polynomial:
Rewrite the formula with
n=k
ak  e1ak 1  e2 ak 2  ...  ek a0
Replace ai with xi
k
k 1
k 2
x  e1 x  e2 x  ...  ek
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Solving Linear Recursion
Input: Formula an  e1an1  e2 an2  ...  ek ank
and k initial values a0 , a1 ,..., ak 1
2. Find the root of the
k
k 1
k 2
x

e
x

e
x
 ...  ek
polynomial
1
2
Or,
x k  e1 x k 1  e2 x k  2  ...  ek  0
(r1 , r2 ,..., rk )
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Solving Linear Recursion
Input: Formula an  e1an1  e2 an2  ...  ek ank
and k initial values a0 , a1 ,..., ak 1
3. Set (assuming that the
roots are distinct.)
an  c r  c2 r2  ...  ck rk
n
1 1
n
4. Determine ci
from k initial values a0  c1  c2  ...  ck
a1  c1r1  c2 r2  ...  ck rk
...  ...
k 1
1 1
ak 1  c r
 c2 r2
k 1
 ...  ck rk
k 1
11
n
Solving Linear Recursion
Input: Formula an  e1an1  e2 an2  ...  ek ank
and k initial values a0 , a1 ,..., ak 1
3. Set (when the roots are
(r1 , r2 ,..., ru )
not distinct.)
u
an   g i (ri )
i 1
wi
gi (ri )   cij n r
j 1
j 1 n
i
where ri is a root of multiplicity wi
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Example on Fibonacci sequence
Input: initial values a0=0 and a1=1; and recursion
formula an=an-1+an-2.
Rewrite recursion:
an-an-1-an-2=0.
1. Characteristic polynomial: x2-x-1=0.
2. Roots of the polynomial:
1 5
r1 
,
2
3. Set:
1 5
r2 
2
an=c1r1n+c2r2n.
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Example on Fibonacci sequence
Input: initial values a0=0 and a1=1; and recursion
formula an=an-1+an-2.
4. Determine ci from k initial values
a0=c1r10+c2r20: c1+c2=0
a1=c1r11+c2r21: c1r1+c2r2=1, where
1 5
1 5
r1 
, r2 
2
2
Thus, we have
1
1
c1 
, c2  
5
5
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Example 2
Given initial values a0=1 and a1=1;
and recursion formula: an=an-1+2an-2
Rewrite recursion:
an-an-1-2an-2=0
1. Characteristic polynomial:
x2-x-2=0
2. Characteristic roots: r1=2 and r2= -1
3. We have an=c1r1n+c2r2n=c12n+c2(-1)n
4. We use two initial values for n=0 and n=1:
a0=c1+c2
a1=c12+c2(-1)
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Example 2 (cont)
Two initial values
a0=c1+c2:
c1+c2=1
a1=2c1+(-1)c2: 2c1-c2=1
Thus, we have
c1=2/3, c2=1/3.
Since an=c1r1n+c2r2n,
the formula is an=2/3*2n+1/3*(-1)n,
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Example 3 (identical roots)
Given initial values a0=1 and a1=1; and recursion
an=-2an-1-an-2
Rewrite the recursion: an+2an-1+an-2=0
1.Characteristic polynomial: x2+2x+1=0
2.Characteristic roots: r1=r2=-1
3.Formula for roots of multiplicity 2
an=c1r1n+c2nr1n=c1(-1)n+c2n(-1)n
Note the formula is different for roots of
multiplicity.
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Example 3 (identical roots)
Given initial values a0=1 and a1=1; and recursion
an=-2an-1-an-2
4. Two initial conditions:
a0=c1(-1)0+c20(-1)0=c1
a1=c1(-1)1+c21(-1)1=-c1-c2 with a0=1 and a1=1
Thus, c1= 1 and c2= -2.
Therefore, an= (-1)n-2n(-1)n
Exercise: verify the sequence a2, a3 and a4.
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