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CSE 20 Lecture 12: Analysis of Homogeneous Linear Recursion

CK ChengMay 5, 2011

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3. Analysis

• 3.1 Introduction • 3.2 Homogeneous Linear Recursion• 3.3 Pigeonhole Principle• 3.4 Inclusion-Exclusion Principle

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3.1 Introduction Derive the bound of functions or recursions Estimate CPU time and memory allocationEg. PageRank calculation

Allocation of memory, CPU time,Resource optimization

MRI imagingReal time?

VLSI designDesign automation flow to meet the deadline for tape out?

Further StudyAlgorithm, Complexity

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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 1 1 5( ) ( )2 25 5

n nnf

0 00

1 1 5 1 1 5( ) ( ) 02 25 5

f

1

1 1 5 1 1 5( ) ( ) 12 25 5

f

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Example: Fibonacci Sequence0 1 2 3 4 5 6 7 8 9

0 1 1 2 3 5 8 13 21 34

1 1 5 1 1 5( ) ( )2 25 5

1(1.618 0.618 )

2.2361

1.6182.236

n nn

n n

n

f

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9

11.618 33.98

2.236f

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3.2 Homogeneous Linear Recursion

• (1) Arithmetic Recursiona, a+d, a+2d, …, a+kd

• (2) Geometric Recursiona, ar, ar2, …, ark

• (3) Linear Recursionan= 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: Formulaand k initial values(1) characteristic

polynomial:(2)Find the root of the

characteristic polynomial (assuming ri are distinct)

(3)Set(4)Determine ci from k initial values

1 1 2 2 ...n n n k n ka e a e a e a 0 1 1, ,..., ka a a

1 21 2 ...k k k

kx e x e x e

1 2( , ,..., )kr r r

1 1 2 2 ...n n nn k ka c r c r c r

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Solving Linear Recursion

Input: Formulaand k initial values(1) characteristic

polynomial:Rewrite the formula with

n=kReplace ai with xi

1 1 2 2 ...n n n k n ka e a e a e a 0 1 1, ,..., ka a a

1 21 2 ...k k k

kx e x e x e

1 1 2 2 0...k k k ka e a e a e a

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Solving Linear Recursion

Input: Formulaand k initial values2. Find the root of the

polynomialOr,

1 1 2 2 ...n n n k n ka e a e a e a 0 1 1, ,..., ka a a

1 21 2 ...k k k

kx e x e x e

1 2( , ,..., )kr r r

1 21 2 ... 0k k k

kx e x e x e

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Solving Linear Recursion

Input: Formulaand k initial values3. Set (assuming that the

roots are distinct.)

4. Determine ci from k initial values

1 1 2 2 ...n n n k n ka e a e a e a 0 1 1, ,..., ka a a

1 1 2 2 ...n n nn k ka c r c r c r

0 1 2 ... ka c c c

1 1 1 2 2 ... k ka c r c r c r

1 1 11 1 1 2 2 ...k k k

k k ka c r c r c r

... ...

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Solving Linear Recursion

Input: Formulaand k initial values3. Set (when the roots are

not distinct.)

1 1 2 2 ...n n n k n ka e a e a e a 0 1 1, ,..., ka a a

1

( )u

n i ii

a g r

1

1

( )iw

j ni i ij i

j

g r c n r

1 2( , ,..., )ur r r

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

1 5,

2r

2

1 5

2r

3. Set: an=c1r1n+c2r2

n.

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Example on Fibonacci sequenceInput: initial values a0=0 and a1=1; and recursion

formula an=an-1+an-2.

4. Determine ci from k initial valuesa0=c1r1

0+c2r20: c1+c2=0

a1=c1r11+c2r2

1:c1r1+c2r2=1, where

1

1 5,

2r

2

1 5

2r

Thus, we have1 2

1 1,5 5

c c

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Example 2Given 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=02. Characteristic roots: r1=2 and r2= -1

3. We have an=c1r1n+c2r2

n=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 valuesa0=c1+c2: c1+c2=1

a1=2c1+(-1)c2: 2c1-c2=1

Thus, we havec1=2/3, c2=1/3.

Since an=c1r1n+c2r2

n,

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=02.Characteristic roots: r1=r2=-1

3.Formula for roots of multiplicity 2an=c1r1

n+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.