algorithm analysis. algorithm def an algorithm is a step-by-step procedure

AlgoritAlgorithmhm AAnalysnalysisisAlgoritAlgorithmhm AAnalysnalysisis

AlgorithmDef

An algorithm is a step-by-step procedure

AlgorithmExample - Compute the sum 10+11+12+13+…+30

public int sum (int nStart, int nEnd) {int sum = 0;for (int i = nStart; i nEnd; i++) {

sum += i;}return sum;

}

Take the first number and continuing add next number until the last number

sum = 0FOR i = nStart TO nEnd

sum = sum + iENDFORRETURN sum

Algorithm AnalysisRequirements

SimplicityReadabilityVerifiabilityValidityRobustnessModifiabilityReuseabilityEfficiency

public int sum (int nStart, int nEnd) {int sum = 0;for (int i = nStart; i nEnd; i++) {

sum += i;}return sum;

}

sum = 0FOR i = nStart TO nEnd

sum = sum + iENDFORRETURN sum

AlgorithmExample - Finding the largest value

5

8

2

7

AlgorithmExample - selectionSort

5

8

2

7

AlgorithmExample - Character permutations

A

B

C

B

A

C

A

C

B

C

A

B

B

C

A

C

B

Ann 321!

Asymtotic NotationNumber of statements - Mathematical relations

1 )1(

)1( -

1

1 1

1

6

)12)(1(

2

)1(

2

1

1

2

1

1

1

0

1

2

1

1

aa

aa

a

naai

aa

aa

nnni

nnn

ni

n

nnn

i

i

nn

i

i

n

i

n

i

n

i

Proof

n

1...111

naa

s nn 2

1 n

n ...321

n

n2222 ...321

1

210 ...

n

naaaa

n

nnaaaa ...321 321

Asymtotic NotationNumber of statements - Mathematical relationsProof by Induction

6

)672)(1(

6

)1(6

6

)12)(1()1(

6

)12)(1()1(

6

)672)(1(

6

)1)1(2)(1)1)((1( dvs,1for stemmer (*)at viseSkal

6

)12)(1( dvs ,for stemmer (*)at Anta

1for stemmer dvs 16

321

6

)112)(11(1 11:1

(*) 6

)12)(1(

2

222

1

21

1

2

2

1

1

2

1

2

21

1

2

1

2

kkk

kkkkk

kkkkii

kkk

kkkikn

kkkikn

nin

nnni

k

i

k

i

k

i

k

i

i

n

i

AlgorithmBasic Timing Analysis - Consecutive statements

count algorithm step

1 a = 4;1 b = 8;1 c = a;

3111)( nT Run time

AlgorithmBasic Timing Analysis - For loop


1 for (i = 1;n+1 i <= n;n i++)n x = x + 1;

23)1(1)( nnnnnT

for (i = 1; i <= n; i++)x = x + 1;

Run time

AlgorithmBasic Timing Analysis - For loop, if statement


1 max = array[0];1 for ( i = 1;n i < n;n-1 i++)n-1 if (array[i] > max)p max = array[i]; 0 p n-11 return max;

nnT

pnpnnnnT

4)(

131)1()1(11)(

max

Run time

AlgorithmBasic Timing Analysis


1 sum = 0;1 + (n+1) + n for ( i = 1; i <= n; i++)1*n + (n+1)*n + n*n for ( j = 1; j <= n; j++)n*n sum = sum + 1;1 return sum;

342

*1111

**)1(*1)1(11)(

2

22

nn

nnnnnn

nnnnnnnnT

Run time

AlgorithmBasic Timing Analysis - Simple For loop


1+(n+1)+n for ( i = 1; i <= n; i++)kn Sk

)()(

2)2()1(1)(

nnT

nkknnnnT

Run time

ComplexityAsymptotic Notation

AlgorithmBasic Timing Analysis - Simple For loop


for ( i = 1; i < =n; i++)kn Sk

)()(

)(

nnT

knnT

Run time

ComplexityAsymptotic Notation

AlgorithmBasic Timing Analysis - Double For loop

for (i = 1; i <= n; i++)for (j = 1; j <= n; j++)

Sk

)()(

11)(

2

2

111 11 1

nnT

knknnknnkkknTn

i

n

i

n

i

n

j

n

i

n

j

AlgorithmBasic Timing Analysis - Single For loop + Double For loop

for (i = 1; i <= n; i++)Sk

for (i = 1; i <= n; i++)for (j = 1; j <= n; j++)

Sk

)()(

11)(

2

2

111 11 1

nnT

knknknnknknknnkknkknkknnTn

i

n

i

n

i

n

j

n

i

n

j

AlgorithmBasic Timing Analysis - Double For loop

for (i = 1; i <= n; i++)for (j = i; j <= n; j++)

Sk

)()(

222

1)1(1)(

2

2

111

nnT

nk

nk

nn

kinkkknTn

i

n

i

n

ij

n

i

n

ij

AlgorithmBasic Timing Analysis - Triple For loop

for (i = 1; i <= n; i++)for (j = i; j <= n; j++)

for (k=i; k <=j; k++) Sm

)()(

6

23

2

23

2

)1(

2

3

6

)12)(1(

2

1

1232

1

2

3

2

1

2

)2)(1(

)1(2

)1(1)1(...321

)1(1)(

3

23

2

1

2

11

2

1

11

111

nnT

mnnn

nnnnn

nnnn

m

nninim

ininm

inin

minm

ijmmmnT

n

i

n

i

n

i

n

i

n

i

n

i

n

i

n

ij

n

i

n

ij

j

ik

n

i

n

ij

j

ik

Different functions and their graphs

Expression Name--------------------------------------------1 constantlog n logarithmiclog2n log squaredn linearn log n n log nn2 quadraticn3 cubic2n exponential--------------------------------------------

AlgorithmGrowth of Common Functions

AlgorithmGeneral Rules

LoopsThe running time of a loop is at most the running time of the statementsinside the loop (including tests) times the number of iterations.

Nested LoopsAnalyze these inside out. The total running time of at statement insidea group of nested loops is the running time of the statement multiplied by the products of the sizes of all the loops.

Consecutive statementsThese just add.

if-then-elseThe running time of an if-then-else statement is never more thanthe running time of the test plus the larger of the running times of each part.

Basic AxiomsA 01

A 01

The time required to fetch an operand from memory is a constant:

The time required to store an operand in memory is a constant:

fetch

store

Memoryfetchstore

Basic AxiomsC 01

C 01

The time required to perform a simple assignment statementis the time required to fetch an operand plus the time required to store an operand: storefetchassignment

xfetch

store

xy

y

1y …

Basic AxiomsA 02

A 02

The time required to perform elementary operations,such as addition, subtraction, multiplication, division, and comparison,are all contants.

/

*

Basic AxiomsA 01 - A 02 - Example

Time required to perform the following statement:

storefetch 2

y

y

y

yy

1

1

Basic AxiomsA 03

A 03

The time required to call a method is a contant,and the time required to return from a method is a constant.

return

call

Basic AxiomsA 04

A 04

The time required to pass an argument to a methodis the same as the time required to store a value in memory:

storepass

Basic AxiomsA 04 - Example


)(

)(

2

xfcallstorefetch

storexfstorecallfetch

T

T

)(xfy

fetch xcall f

passing x

running time of f

assignment to y

Basic AxiomsA 01 - A 02 - A 03 - Example

n

i

i1

Running timeTime required to computethe following simple arithmetic series summation:

1 public class Example {2 public static int sum (int n) {3 int result = 0;4 for (int i = 0; i <= n; i++)5 result += i;6 return result;7 }8 }

resultreturn

iresultn

inc

ninb

ia

result

returnfetch

storefetch

storefetch

fetch

storefetch

storefetch

6

)2( 5

)2( 4

)1()2( 4

1 4

0 3

ττττ t

ττττt

nttT(n)

storefetch

returnstorefetch

226

25

2

1

21

Basic AxiomsA 05

A 05

The time required for the address calculationimplied by an array subscripting operation, a[i], is a constant.This time does not include the time to compute the subscript expression,nor does it include the time to access (fetch or store) the array element.

][

Basic AxiomsA 05 - Example


storefetch

storefetchfetchfetch

][

][

3

][iay

fetch afetch i

address subscripting calcutaion

fetch a[i]

assignment to y

Basic AxiomsA 01 - A 02 - A 03 - A 04 - Example - Horner’s Rule

n

i

ii xa

0

1 public class Example {2 public static int horner (int[] a, int n, int x) {3 int result = a[n];4 for (int i = n-1; i >= 0; i--)5 result = result * x + a[i];6 return result;7 }8 }

resultreturn

iresultn

inc

inb

nia

naresult

returnfetch

storexfetch

storefetch

fetch

storefetch

storefetch

6

)5( 5

)2( 4

0 )1()2( 4

1 2 4

][ 3 3

][

][

xstorefetch

returnstorefetch

ττττ t

ττττt

nttT(n)

][2

][1

21

29

28

Basic AxiomsA 01 - A 02 - A 03 - A 04 - Recursive Methods - Factorial [1/2]

0 1

0 1

0

0 1

!

1

n)!(n-n

n

ni

n

n n

i

1 public class Example {2 public static int factorial (int n) {3 if (n == 0)4 return 1;5 else6 return n * factorial(n-1);7 }8 }

)T(n-returncall

storefetch

returnfetch

fetchfetch

1

3 6

4

2 2 3

*

returncallxstorefetch

returnfetch

ττττ t

τττt

ntnT

ntnT

5

3

0 )1(

0 )(

2

1

2

1

Basic AxiomsA 01 - A 02 - A 03 - A 04 - Recursive Methods - Factorial [2/2]

0 1

0 1!

n)!(n-n

nn

returncallxstorefetch

returnfetch

ττττ t

τττt

ntnT

ntnT

5

3

0 )1(

0 )(

2

1

2

1

21

2

2

222

222

2

)0(

)(

...

3)3(2)3(

2)2()2(

)1()(

ntt

ntT

ktknT

tnTttnT

tnTttnT

tnTnT

Basic AxiomsA 01 … A 05 - Finding the Largest Element of an Array [1/2]

iani

max0

1 public class Example {2 public static int max (int[] a) {3 int result = a[0];4 for (int i = 1; i < a.length; i++)5 if (a[i] > result)6 result = a[i];7 return result;8 }9 }

storefetch

][storefetch

fetchstore

aa

n

in

τττt

τττττ t

ττττt

tnttaaanT

jiji

][3

2

1

)(max

1

1321110

3

28

2

),...,,,(

0

resultreturn

resultn

inc

lengthainb

ia

aresult

storefetch

storefetch

fetch

storefetch

fetch

storefetch

storefetch

3 7

a[i]result ?)(3 6

a[i] )1()4( 5

)1()2( 4

. )2( 4

1 4

]0[ 3 3

][

][

][

Basic AxiomsA 01 … A 05 - Finding the Largest Element of an Array [2/2]

iani

max0

)(max

1

1321110

0

),...,,,(

jiji aa

n

in tnttaaanT

ntntttnT

ni

HtHntt

ti

ntt

ti

ntt

iaaPptpnttnT

avg

n

inn

n

i

n

i

jij

ii

n

iiavg

ln)1()(

ln1

)1(

11

1

1

1

1)max( )(

3231

1321

31

21

3

1

121

0

1

1321

pi probabilitythat line 6 is executed

)ln()ln(11

1

1

1

11 nxdx

xiH n

nn

in

Asymtotic NotationDefinition 01 - An Asymtotic Upper Bound - Big Oh

00 )()(0 :0, |)())(( nnncgnfncnfngO

D 01

Consider a function f(n).We say that f(n) is big oh g(n), which we write f(n) O(g(n)),if there exists an integer n0 and a constant c > 0 such that for all integers n n0, 0 f(n) cg(n)

cg(n)

f(n)

n0

Asymtotic NotationDefinition 01 - An Asymtotic Upper Bound - Big OhAlternative definition

00 )()( :0,

))(()(

nnncgnfcn

ngOnf

D 01

Consider a function f(n) that is non-negative for all integers n >= 0.We say that f(n) is big oh g(n), which we write f(n) = O(g(n)),if there exists an integer n0 and a constant c > 0 such that for all integers n n0, f(n) cg(n)

cg(n)

f(n)

n0

Asymtotic NotationDefinition 01 - An Asymtotic Upper Bound - Big Oh

))(()(1 ngOnf

00 )()( :0,

))(()(

nnncgnfcn

ngOnf

00 )()(0 :, |)())(( nnncgnfncnfngO

cg(n)

f(n)

n0

))(()(1 ngOnf

Asymtotic NotationAn Asymtotic Upper Bound - Big Oh

)()(

50

501110

5050

5050

505)(

00

nOnf

nncn

cn

cn

cnn

nnf

f(n) = 5n + 50 We wish to show that f(n) O(n)


Asymtotic NotationAn Asymtotic Upper Bound - Big Oh 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

2

02

0

2

22

)(

)(10 ,1

0)5)(10(

0505

505)(

1cVelger

505)(

nOnf

nncnnfnc

nn

nn

nncnnf

nnf

f(n) = 5n + 50 We wish to show that f(n) = O(n2)

2

2

2

4)(3

2)(2

)(1

505 )(

nng

nng

nng

nnf

Asymtotic NotationT 01 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

T 01

If f1(n) = O(g1(n)) and f2(n) = O(g2(n)), then f1(n) + f2(n) = O(max(g1(n),g2(n))).

Proof

)))(),((max()()(

))()(max(

))()((2

)()()()(),max(

),max(2

2121

210

210

221121

0

210

210

ngngOnfnf

ngngc

ngngc

ngcngcnfnf

nn

nnn

ccc

Asymtotic NotationExamples 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

)( !

)!( 2

)2(

))((log

)( )(log

)( 3001004

)( 3001004

)1( 350

)( 201004

)( 20

10

210

32

22

323

22

n

n

nk

nOn

nO

On

nOn

nOnn

nOnn

nOnn

O

nOnn

nOn

Asymtotic NotationExample - 1+2+3+…+n 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

Outer BodyLevel Loop

1 12

1

10 or 10 or 10 or 1

)()( nOnT

Asymtotic NotationExample - Selection sort 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

Outer Body BodyLevel Outer Inner

Loop Loop

1+1 1+111+1 1+1

10 or 1

10 or 10 or 10 or 1

Asymtotic NotationNumber of statements 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

)1(41)1()1(441031

1

1

1

ininn

ij

n

ij

2

0

2

0

loopinner for count 9301 loopinner for count 5n

i

n

i

Outer loop

2

0

loopinner for count 92 loopouter for count 2n

i

Total number

Inner loop for a certain i

Asymtotic NotationNumber of statements 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

772

442244552

2

)1)(2(4)1()45(2

41)45(2

4452

1)-i-(n4 92

loopinner for count 92)(

2

22

2

0

2

0

2

0

2

0

2

0

nn

nnnnnn

nnnn

in

in

nTS

n

i

n

i

n

i

n

i

n

issort

)()( 2nOnTSssort


))(()(

nnncgnfcn

ngOnf

T 02

If f(n) = f1(n) + f2(n) where f1(n) and f2(n) both are non-negative for all intergers n >= 0 such that limn-> f2(n)/f1(n) = L for some limit L >= 0,then f(n) = O(f1(n)).

Proof

))(()(

)( )(

)()()(

)()(

)()()(

)()()( )(

)(

)(

)(lim

1

021010

10211

2211

21

1020 01

2

1

2

nfOnf

Lcccnfc

nfLcnfc

nfcnfc

nfnfnf

nfLnfnnLnf

nfL

nf

nfn

Asymtotic NotationT 01 - Example 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

)(

)()()(

01

limlim)(

)(lim

)(

)(

)()()(

3

23

21

3

2

1

2

22

31

21

nO

nn

nfnfnf

nn

n

nf

nf

nnf

nnf

nfnfnf

nnn

f(n) = f1(n) + f2(n) = n3 + n2


))(()(

nnncgnfcn

ngOnf

T 03

If f1(n) = O(g1(n)) and f2(n) = O(g2(n)) then f1(n) x f2(n) = O(g1(n) x g2(n)).

Proof

))()(()()(

))()(()()()()(

),,(

0 0)( 0 0)(

)()( )()(:,,,

2121

0210221121

210210

11

222211112121

ngngOnfnf

nnngngcngcngcnfnf

cccnnmaxn

nnfnnf

nnngcnfnnngcnfccnn


))(()(

nnncgnfcn

ngOnf

f1(n) = n3 + n2 + n + 1 = O(n3)

f2(n) = n2 + n + 1 = O(n2)

f1(n) x f2(n) = O(n3 x n2) = O(n5)


))(()(

nnncgnfcn

ngOnf

T 04

If f1(n) = O(g1(n)) and g2(n) is a function whose value is non-negativefor integers n >= 0, then f1(n) x g2(n) = O(g1(n) x g2(n)).

Proof

))()(()()(

)()()()(

0 0)(

)()(:,

2121

021021

2

010100

ngngOngnf

nnngngcngnf

nng

nnngcnfcn

Asymtotic NotationT 05 - Transitive Property 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

T 05

If f(n) = O(g(n)) and g(n) = O(h(n)) then f(n) = O(h(n)).

Proof

)(

)(

)()(

),,(

)()( )()(:,,,

0

021

11

210210

22112121

nhc

nnnhcc

nnngcnf

cccnnmaxn

nnnhcngnnngcnfccnn


))(()(

nnncgnfcn

ngOnf

f1(n) = 7n3 = O(n3)

f2(n) = 2n2 = O(n2)

f1(n) + f2(n) = O(f1(n)) because lim f2(n)/ f1(n) = 0f1(n) + f2(n) = O(f1(n)) = O(n3)

Asymtotic NotationT 06 - Polynomials 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

T 06

Proof

m

mm

mm

m

m

i

ii

nOnf

aanananana

nanf

)(

0 ...

)(

012

21

1

0

m

n

c

m

ii

ng

mm

iimi

m

m

i

mii

mm

i

ii

m

i

ii

nOnf

nann

an

nnannananf

)(

1 1

1 )(

00)(0

000

Asymtotic NotationLogarithms 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

logn < n => logn = O(n)

Asymtotic NotationT 07 - Logarithms 00 )()( :0,

))(()(

nnncgnfcn

ngOnf

T 07

For every integer k >= 1, logkn = O(n).

Proof

0log

limln

1

ln1

log)1(lim

loglim :1

0log

lim:m1,2,3,...,kfor holdsit Assume

01

limln

1

1ln1

limlog

lim :1

:Induction

0log

lim :Hypothesis

''1

''

n

n

a

m

nan

nm

n

nmk

n

n

naan

n

nk

n

n

ma

n

ma

n

rulesHopitalLma

n

ma

n

nn

rulesHopitalLa

n

ka

n

Names of Common Big Oh Expressions

Expression Name--------------------------------------------O(1) constantO(log n) logarithmicO(log2n) log squaredO(n) linearO(n log n) n log nO(n2) quadraticO(n3) cubicO(2n) exponential--------------------------------------------

Asymtotic Lower Bound - Omega

00 )()( :0,

))(()(

nnncgnfcn

ngnf

D 01

Consider a function f(n) that is non-negative for all integers n >= 0.We say that f(n) is omega g(n), which we write f(n) = (g(n)),if there exists an integer n0 and a constant c > 0 such that for all integers n >= n0, f(n) >= cg(n)

cg(n)

f(n)

n0

Asymtotic Upper and Lower Bound - Theta

))(()( ))(()(

))(()(

ngnfngOnf

ngnf

D 01

Consider a function f(n) that is non-negative for all integers n >= 0.We say that f(n) is theta g(n), which we write f(n) = (g(n)),if and only if f(n) = O(g(n)) and f(n) = (g(n)).

c1g(n)

f(n)

n0

c2g(n)

Asymtotic Lower Bound - Little Oh

))(()( ))(()(

))(()(

ngnfngOnf

ngonf

D 01

Consider a function f(n) that is non-negative for all integers n >= 0.We say that f(n) is little oh g(n), which we write f(n) = o(g(n)),if and only if f(n) = O(g(n)) and f(n) (g(n)).

cg(n)

f(n)

n0

Asymtotic NotationBig Oh - Omega - Theta - Little Oh

021021 )()()(0 :0,, |)())(( nnngcnfngcnccnfng

00 )()(0 :0, |)())(( nnnfncgncnfng

00 )()(0 :0,0 |)())(( nnncgnfncnfngO

0101

0202

)()(0 :0,

)()(0 :0, |)())((

nnnfngcnc

nnngcnfncnfngo

Asymtotic NotationExample An Asymtotic Upper and Lower Bound - Theta

)()(

)()()()(

3)(

nnf

nnfnOnf

nnf

f(n) = n + 3

Asymtotic NotationExample An Asymtotic Upper and Lower Bound - Little Oh


)()(

)()()()(

3)(

2

22

nonf

nnfnOnf

nnf

f(n) = n + 3

Asymtotic NotationT 06 - Polynomials 00 )()( :0,

))(()(

nnncgnfcn

ngnf

T 08

Proof

m

mm

mm

m

m

i

ii

nnf

aanananana

nanf

)(

0 ...

)(

012

21

1

0

1

0

1

0

1

1

0)1(

1

1

0

)1(1

1

0

1

00

1

1

1 1

1

)(

m

iim

m

m

ii

mmm

m

iimi

mmm

m

i

mii

mmm

m

i

ii

mm

m

i

ii

mm

m

i

ii

an

an

nanna

nn

anna

nnanna

nana

nanananf

m

c

m

iim

ng

m

m

ii

m

m

ii

m

m

iim

m

iim

m

iim

nnf

nnan

annf

aa

n

aa

n

an

a

an

aan

a

)(

1

)(

11

1

01

11

0

1

00)(

1

00

1

00

1

00

1

00

1

0

Asymtotic NotationExample An Asymtotic Upper and Lower Bound - Theta


)()(

)()()()(

...)( 011

1

m

mm

mm

mm

nnf

nnfnOnf

ananananf

f(n) = amnm + am-1nm-1 + … + a1n + a0

Asymtotic NotationRules for Big Oh Analysis


Sequence S1;S2;…Sm;

Iteration for (S1; S2; S3)S4;

Conditional if (S1)S2;

elseS3;

O(max ( T1(n), T2(n),…,Tm(n))

O(max ( T1(n), T2(n) x (I(n)+1),T3(n) x I(n),T4(n) x I(n)

)

O(max ( T1(n), T2(n),T3(n)

)

Actual Lower BoundsAssuming a 100 MHz Clock, c = 1 Cycle, n0 = 0

yearsyearssns

yearsssnsn

hourssnsnsn

mssnsnsnn

mssnsnsn

nsnsnsnsn

nsnsnsns

KnKnnn

n 510293

3

2

101056.210)2(

3657.1012.510)(

05.32.1064010)(

2102.1024010)log(

5.1002.18010)(

2001003010)(log

10101010)1(

1024181

RecursionOne recursive call with half size - Binary Search

11

2

11)( n

nT

nnT

1)(log)( 2 nnT

))(log()( nnT

RecursionOne recursive call with half size - Binary Search - Proof I

1)(log )(

...

1)16(log 5141)8( )16( 16

1)8(log 4131)4( )8( 8

1)4(log 3121)2( )4( 4

1)2(log 2111)1( )2( 2

1)1(log 1 )1( 1

2

2

2

2

2

2

nnTn

TT

TT

TT

TT

TProof

11

2

11)( n

nT

nnT

1)(log)( 2 nnT

))(log()( nnT

RecursionOne recursive call with half size - Binary Search - Proof 2

1)(log

111)(log

11)2(log)(log

112

log

12

12

log2

Assume

1)1(log1)1(

2

2

22

2

2

2

k

k

k

k

kTkT

kkT

TProof

11

2

11)( n

nT

nnT

1)(log)( 2 nnT

))(log()( nnT

RecursionTwo recursive calls with half size - Sum

11

22

11)( n

nT

nnT

12)( nnT

)()( nnT

RecursionTwo recursive calls with half size - Sum - Proof I

12)(

...

1162 311301)8(2 )16( 16

182 151141)4(2 )8( 8

142 5161)2(2 )4( 4

122 3121)1(2 )2( 2

112 1 )1( 1

nnTn

TT

TT

TT

TT

TProof

11

22

11)( n

nT

nnT

12)( nnT

)()( nnT

RecursionTwo recursive calls with size (about) n-1 - Fibonacci number - Recursive

11)2()1(

101)(

nnTnT

nnnT

)2()( nOnT

nnT 2)(

RecursionTwo recursive calls with size (about) n-1 - Fibonacci number - Proof 1

k

k

kk

kk

kkk

kk

kk

kTkTkT

kTkT

T

T

2

22

22

222

222

122

1)2()1(

2)2( 2)1( Assume

21)1(

221)0(

1

11

21

221

21

21

1

00

Proof

11)2()1(

101)(

nnTnT

nnnT

)2()( nOnT

nnT 2)(

RecursionFibonacci number - Iterative

)()( nOnT

RecursionMore than two recursive calls in every level - Permutation

1)1(

11)(

nnnT

nnT

)!()( nnT

!)( nnT

RecursionMore than two recursive calls in every level - Permutation - Proof I

!)(

...

!5 120245)4(5 )5( 5

!4 2464)3(4 )4( 4

!3 623)2(3 )3( 3

!2 212)1(2 )2( 2

!1 1 )1( 1

nnTn

TT

TT

TT

TT

T

Proof

1)1(

11)(

nnnT

nnT

)!()( nnT

!)( nnT

Recursive calls

abnnT

abnnnT

abnnT

kbancnb

naT

nnT

kk

kk

ka

k

b

)(

log)(

)(

011 1

11)(

log

Recursive callsExamples

loop simple size half with call recursive One )( )2/()( )(

loop simple size half with calls recursive Two log)( )2/(2)(

size half with call recursive One log)( )2/()( log)(

size half with calls recursive Two )( )2/(2)( )(

1

11)(

log

nnTcnnTnTabnnT

nnnTcnnTnT

nnTcnTnTabnnnT

nnTcnTnTabnnT

ncnb

naT

nnT

kk

kk

ka

k

b

Recursive callsProof

nnmmk

km

mkmk

mm

mm

i

i

kb

kmmmm

i

mmmk

m

ii

ikmmm

m

ii

ik

m

m

m

ii

ikm

ii

im

ii

i

m

mk

m

m

m

m

mkmkmm

m

k

bb

k

k

naaanTab

nba

banTab

x

xxx

x

x

x

xxxx

nnnTncncmbmcaacaabTab

a

bcaabT

a

bc

a

bT

a

bT

a

bc

a

bT

a

bT

a

bc

a

bT

a

bT

bcbaTbcb

baTbT

kbancnb

naT

nnT

loglog

1

1

1

10

0

111

1

1

1

1

1

1)(

)(

1 )1(

1 )()1(

11

1 1

)log()()log1()1(1)(

)()(

)()()(

)()()(

)()()(

)()()()(

011 1

11)(

Complexity

Algorithm Running time------------------------------------------------------Linear search O(n)Binary search O(logn)Bubble sort O(n2)Quick sort O(nlogn)------------------------------------------------------

ENDENDENDEND

algorithm analysis. algorithm def an algorithm is a step-by-step procedure

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