02a-RunningTime

8/13/2019 02a-RunningTime

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Dr. Ahmad R. Hadaegh

A.R. Hadaegh National University Page 1

Algorithm Analysis:

Running TimeBig Oand omega (


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Introduction

An algorithm analysis of a program is a step by step procedurefor accomplishing that program

In order to learn about an algorithm, we need to analyze it

This means we need to study the specification of the algorithmand draw conclusion about the implementation of that algorithm(the program) will perform in general


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The issues that should be considered in analyzing an algorithmare:

The running time of a program as a function of its inputs

The total or maximum memory space needed for programdata

The total size of the program code

hether the program correctly computes the desired result

The complexity of the program! "or example, how easy it is

to read, understand, and modify the program

The robustness of the program! "or example, how well does itdeal with unexpected or erroneous inputs


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In this course, we consider the running time of the algorithm!

The main factors that effect the running time are the algorithm

itself, input data, the computer system, etc!

The performance of a computer is determined by The hardware The programming language used and

The operating system

To calculate the running time of a general #$$ program, we firstneed to define a few rules!

In our rules , we are going to assume that the effect of hardwareand software systems used in the machines are independent ofthe running time of our #$$ program


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Rule 1:

The time re%uired to fetch an integer from memory is aconstant t(fetch),

The time re%uired to store an integer in memory is also aconstant t(store)

"or example the running time of

x y

is:

t(!etch" # t(store"

because we need to fetch y from memory store it into x

&imilarly the running time ofx 1

is also t(!etch" # t(store"because typically any constant is

stored in the memory before it is fetched!


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7/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page %

Rule 3: The time re%uired to call a function is a constant, t(call)

And the time re%uired to return a function is a constant, t(return)

Rule 4:

The time re%uired to pass an integer argument to a function orprocedure is the same as the time re%uired to store an integer inmemory, t(store)


8/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page &


y !(x"

is:

t(!etch" # 2t(store" # t(call" # t(!(x""

.ecause you need To fetch the alue of x: t (fetch) /ass x to the function and store it into parameter: t (store)

#all the function f(x): t (call) 0un the function: t (f(x)) &tore the returned result into y: t (store)


9/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page '

Rule 5:

The time re%uired for the address calculation implied by an arraysubscripting operation li1e a2i3 is a constant, t(2 3)!

This time does not include the time to compute the subscriptexpression, nor does it include the time to access (fetch or store)the array element

"or example, the running time of

y ai)

is:

3t(!etch" # t( )" # t(store"

.ecause you need

To fetch the alue of i: t(fetch)To fetch the alue of a: t(fetch)

To find the address of a2i3: t(2 3)

To fetch the alue of a2i3: t(fetch)

To store the alue of a2i3 into y: t(store)


10/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 1*

Rule $: The time re%uired to calculate a fixed amount of storage from the

heap using operator newis a constant, t(new) This time does not include any time re%uired for initialization ofthe storage (calling a constructor)!

&imilarly, the time re%uired to return a fixed amount of storage

to the heap using operator deleteis a constant, t(delete)! This time does not include any time spent cleaning up the

storage before it is returned to the heap (calling destructor)


int+ ,tr ne- int.

is:

t(ne-" # t(store"

.ecause you needTo allocate a memory: t(new)And to store its address into ptr: t(store)


delete ,tr.

is:

t(!etch " # t(delete"

.ecause you needTo fetch the address from ptr : t(fetch)And delete specifies location: t(delete)


11/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 11

! int &um (int n)-! 45! int result 6789! for (int i68 i6n8 i6i$);! result 6 result $ i8

/tatement Time 0ode

5 t(fetch)$ t(store) result 6 7

9a t(fetch) $ t(store) i 6

9b (-t(fetch)$t(cmp)) (n$) i6n

9c (-t(fetch)$t($) $t(store)) n $$i

; (-t(fetch)$t($) $t(store)) n 0esult $6i

< t(fetch)$ t(return) 0eturn result

Total 2



! int func (int a2 3, int n, int x)-! 45! int result 6 a2n389! for (int i6n'8 i?678 i6i');! result 6result x $ a2i38

/tatement Time5 5t(fetch)$ t(2 3) $ t(store)

9a -t(fetch) $ t(') $ t(store)

9b (-t(fetch)$t(cmp)) (n$)

9c (-t(fetch)$t(') $t(store)) n

; (;t(fetch)$t(2 3)$t($)$t()$t(store)) n

< t(fetch)$ t(return)

Total 2(@t(fetch) $-t(store) $ t(cmp) $t(23) $ t() $ t(')3n$ 2(t(fetch) $ -t(store) $ t(23) $ t(') $t(cmp) $ t (return) )3



Bsing constant times such as t(fetch), t(store), t(delete), t(new), t($),C, ect ma1es our running time accurate

Doweer, in order to ma1e life simple, we can consider theapproximate running time of any constant to be the same time ast()!

"or example, the running time of

y x # 1

is 5 because it includes two EfetchesF and one EstoreF in which allare constants

"or a loop there are two cases: If we 1now the exact number of iterations, the running timebecomes constant t()

If we do not 1now the exact number of iterations, the runningtime becomes t(n) where nis the number of iterations



! int Geometric (int x, int n)

-! 45! int sum 6 789! for (int i678 i6n8 $$i);! 4

=! for (int H678 Hi8 $$H)! prod 6 prod x8@! sum 6 sum $ prod87! >! return result

-! >

/tatement Time

5 -

9a -

9b 5(n$-)

9c 9(n$)

< -(n$)

=a -(n$)

=b

=c

@ 9(n$)

-

Total (*-)n-$ (9=*-)n $ -=

i67n

9 i

i67n

9 i

i67n5 i$



! int /ower (int x, int n)-! 45! if (n6 67)9! return 8

;! else if (n- 6 6 7) ** n is een

/tatement

n* n* (n is een" n* (n is odd"

5 5 5 5

9 - ' '

; ' ; ;< ' 7 $ T( n*- ) '

' ' - $ T( n*- )

Total ; $ T( n*- ) -7 $ T( n*- )


16/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 1$

5 !or n*T(n" 1T( n2 " !or n* and n is een

2*#T( n2 " !or n * and n is odd

&uppose n 6 -1for some 1?7!

Jbiously -1is an een number, we get

T(2" 1& # T(21" 1& # (1& # T(22"" 1& # 1& # (1& # T(23"" 677 677 1& # T(2" 1& # T(2*" 1& # T(1"


17/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 1%

&ince T() is add number, the running time of T() is:

T(1" 2* # T(*" 2* # 5 25

Therefore,T(2" 1& # 25

If n 6 -1, then log n 6 log-1indicating that 1 6 log n

Therefore,T(n" 1&log n # 25


18/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page1&

Asym,totic 8otation

&uppose the running time of two algorithms A and . are TA(n) and

T.(n), respectiely wherenis the size of the problem

Dow can we determine TA(n) is better than T.(n)K

Jne way to do that is if we 1now the size nahead of time for some

n=no! Then we may say that algorithm A is performing better thanalgorithm . for n= n

o

.ut this is a special case for n=no! Dow about n = n

1, or n=n

2K Is A

better than . for other cases tooK

Bnfortunately, this is not an easy answer! e cannot expect that thesize of nto be 1nown ahead of time! .ut we may be able to say thatunder certain situations TA(n) is better than T.(n) for all n ?6 n


19/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 1'

To understand the running times of the algorithms we need toma1e some definitions:

9e!inition: #onsider a function f(n) which is non'negatie for all

integers n?67! e say that f(n) is big oh of g(n)whichwe write (f(n) is !(g(n)) if there exists an integer noand a

constant c ? 7 such that for all integers n ?6no,

f(n) 6c g(n)

xam,le: &how that f(n) 6 n $ - is J(n-)

n $ - 6 c n- (lets set c 6 )7 6 cn-'n '-7 6 (n'


20/30Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 2*

g5(n)6n-g-(n)6-n

-g(n)69n-

f(n)6n$-

; 7 ; -7

-77

977

n

f(n)



Theorem: If f(n) is J(g(n)) and f-(n) is J(g-(n)), then

f(n) $ f-(n) 6 J (max(g(n), g-(n)))

;roo!:

If f(n) is J(g(n)) then f(n) 6 cg(n) for some cand n ?6 n

If f-(n) is J(g-(n)) then f-(n) 6 c-g-(n) for some c-and n ?6 n-

Let no6 max(n, n-) and co6 -(max(c, c-)), consider the sum f(n)

$ f-(n) for some n ?6 no

f(n) $ f-(n) 6 cg(n) $ c-g-(n)

6 co(g(n) $ g-(n) )*-

6 co(max (g(n), g-(n))

Therefore, f(n) $ f-(n) is J (max(g(n), g-(n)) )



Theorem: If f(n) is J(g(n)) and f-(n) is J(g-(n)), then

!1(n" + !2(n" O(g1(n"+g2(n" "

;roo!:

If f(n) is J(g(n)) then f(n) 6 cg(n) for some cand n?6n

If f-(n) is J(g-(n)) then f-(n) 6 c-g-(n) for some c-and n?6n-

Let no6 max(n, n-) and co6 cc-, consider the product of

f(n)f-(n) for some n?6no

f(n) f-(n) 6 cg(n) c-g-(n)

6 co(g(n) g-(n) )

Therefore, f(n) f-(n) is J (g(n)g-(n) )



Theorem: If f(n) is J(g(n)) and g(n) is J(h(n)), then

!(n" is O(h(n""

;roo!:

If f(n) is J(g(n)) then f(n) 6 cg(n) for some cand n?6n If g(n) is J(h(n)) then g(n) 6 c-h(n) for some c-and n?6n-

Let no6 max(n, n-) and co6 cc-, then

f(n) 6 cg(n)

6 cc-h(n)

6 coh(n)

Therefore, f(n) is J (h(n))



The names o! common



0onentions !or -riting Big Oh x,ression

#ertain conentions hae eoled which concern how big ohexpression normally written:

"irst, it is common practice when writing big oh expression

to drop all but the most significant items! Thus instead ofJ(n-$ nlogn $ n) we simply write J(n-)

&econd, it is common practice to drop constant coefficients!

Thus, instead of J(5n-), we write J(n-)! As a special caseof this rule, if the function is a constant, instead of, sayJ(7-9), we simply write J()


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Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 2$

Asym,totic =o-er Bound ("9e!inition:

#onsider a function f(n), which is non'negatie for allintegers n?67! e say that f(n) is omega of g(n) whichwe write(f(n) is (g(n))"if there exists an integer noand a

constant c ? 7 such that for all integers n?6 no, f(n) ?6 c g(n)

xam,le: &how that f(n) 6 ;n-'


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Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 2%

f(n)6n-

f(n)6-n-

f(n)6;n-'


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Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 2&

Other de!initions9e!inition:

#onsider a function f(n) which is non'negatie for allintegers n?67! e say that f(n) is theta of g(n) which wewrite(f(n) is (g(n)) if and only if f(n) is J(g(n)) and f(n) is(g(n))

9e!inition: #onsider a function f(n) which is non'negatie for all

integers n?67! e say that f(n) is little o of g(n) which wewrite(f(n) is o(g(n))if and only if f(n) is J(g(n)) and f(n) isnot (g(n))

Oow lets consider some of the preious examples in terms ofthe big J notations:


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Dr. Ahmad R. HadaeghA.R. Hadaegh National University Page 2'

! int func (int a2 3, int n, int x)-! 45! int result 6 a2n389! for (int i6n'8 i?678 ''i)

;! result 6result x $ a2i38

/tatement /im,le

Timemodel

Big O

5 ; J()9a 9 J()9b 5n $ 5 J(n)9c 9n J(n); @n J(n)< - J()

Total


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! int /refix&ums (int a2 3, int n)-! 45! for (int H6n'8 i?678 ''H)9! 4;! int sum 6 787! return result! >

/tatement Big O

5a J()5a J()J(n)

5c J()J(n); J()J(n)

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