03 Divide and Conquer

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DIVIDE ANDCONQUER

Design and Analysis Algorithm

Mochammad Kautsar SophanGenap 2015


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Review Divide &Conquer Divide the problem into a number of

subproblems that are smaller instances of thesame problem

!on"uer the subproblems by solving themrecursively#f the subproblem si$es are small enough% ho&ever% 'ust solve the subproblems in a straightfor&ard

manner

!ombine the solutions to the subproblems intothe solution for the original problem

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,hen the subproblems are large enough tosolve recursively% &e call that the recursivecase

-nce the subproblems become small enoughthat &e no longer recurse% &e say that therecursion .bottoms out/ and that &e have

gotten do&n to the base case

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Objective

see more algorithms based on divide)and)con"uer+he rst one solves the maimum)subarray

problem it ta(es as input an array of numbers% andit determines the contiguous subarray &hosevalues have the greatest sum



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Recurrences

is an e"uation or ine"uality that describes afunction of recursive running time

4or eample% &orst)case running time +n6 of

the M78G7)S-8+ procedure by therecurrence &hose solution &e claimed to be+ n6 9 :n lg n6



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8ecurrences can ta(e many forms 4or eample% a recursive algorithm might

divide subproblems into une"ual si$es% such

as a 2;)to)1; split #f the divide and combine steps ta(e linear

time% such an algorithm &ould give rise to the

recurrence + n6 9 + 2n;6 < + n;6


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three methods for solving recurrences>that is% forobtaining asymptotic .:/or.-/ bounds on thesolution

#n the substitution method% &e guess a bound and thenuse mathematical induction to prove our guess correct

+he recursion)tree method converts the recurrence intoa tree &hose nodes represent the costs incurred at

various levels of the recursion ,e use techni"ues forbounding summations to solve the recurrence

+he master method provides bounds for recurrences ofthe form

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Master et!od

A recurrence of the form in master method characteri$es a divide)and)con"uer algorithmthat creates a subproblems% each of &hich is 1;b

the si$e of the original problem% and in &hich thedivide and combine steps together ta(e fn6 time

+o use the master method% you &ill need tomemori$e three cases% but once you do that%you &ill easily be able to determine asymptoticbounds for many simple recurrences

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-ccasionally% &e shall see recurrences thatare not e"ualities but rather ine"ualities% suchas

ecause such a recurrence states only anupper bound on +n6% &e &ill couch itssolution using -)notation rather than :

notation

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Similarly% if the ine"uality &ere reversed to

then because the recurrence gives only a

lo&er bound on +n6% &e &ould use C)notation in its solution



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"!e a#iu$subarra% rob'e

Suppose that you been offered the opportunity toinvest in the olatile !hemical !orporation

the stoc( price of the olatile !hemical !orporation

is rather volatile Eou are allo&ed to buy one unit of stoc( only one

time and then sell it at a later date% buying and sellingafter the close of trading for the day

+o compensate for this restriction% you are allo&ed tolearn &hat the price of the stoc( &ill be in the future

Eour goal is to maimi$e your prot



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Eou may buy the stoc( at any one time%starting after day 0% &hen the price is F100per share

-f course% you &ould &ant to .buy lo&% sellhigh/>buy at the lo&est possible price andlater on sell at the highest possible price>to

maimi$e your prot



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&e &ould maimi$e prot by buying at thelo&est price% after day ?

#f this strategy al&ays &or(ed% then it &ould

be easy to determine ho& to maimi$e protnd the highest and lo&est prices% and then &or(

left from the highest price to nd the lo&est priorprice%

&or( right from the lo&est price to nd the highestlater price% and ta(e the pair &ith the greaterdifference



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aroac!

A brute)force solution A transformation

divide)and)con"uer

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A brute$(orce so'ution

'ust try every possible pair of buy and selldates in &hich the buy date precedes the selldate

A period of n days has n;2 such pairs ofdates

Since n;2 is :n26 and the best &e can hope

for is to evaluate each pair of dates inconstant time% this approach &ould ta(e Cn26time

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A trans(oration

&e &ill loo( at the input in a slightly different&ay

,e &ant to nd a se"uence of days over &hich

the net change from the rst day to the last ismaximum

#nstead of loo(ing at the daily prices% let us

instead consider the daily change in price%&here the change on day i is the differencebet&een the prices after day i ) 1 and after day i

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+he table sho&s daily changes in the bottomro&

#f &e treat this ro& as an array A% &e no&

&ant to nd the nonempty% contiguoussubarray of A &hose values have the largestsum

,e call this contiguous subarray themaximum subarray

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4or eample% in the array the maimum subarray of AH1 1=I is AH@ 11I% &ith the

sum 3+hus% you &ould &ant to buy the stoc( 'ust before

day @ that is% after day ?6 and sell it after day 11%earning a prot of F3 per share



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A so'ution usin) divide$and$conquer

Suppose &e &ant to nd a maimumsubarray of the subarray AHlo& highI

Divide)and)con"uer suggests that &e divide

the subarray into t&o subarrays of as e"ualsi$e as possible

+hat is% &e nd the midpoint% say mid% of the

subarray% and consider the subarrays AHlo& midI and AHmid < 1 highI



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any contiguous subarray AHi ' I of AHlo& highI must lie in eactly one of the follo&ingplaces

entirely in the subarray AHlo& midI% so thatlo& J i J ' J mid%

entirely in the subarray AHmid < 1 highI% so that mid i J ' J high

crossing the midpoint% so thatlo& J i J mid ' J high

a maimum subarray of AHlo& LighI must lie ineactly one of these places



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Ana'%*in) t!e divide$and$conquer a')orit!

&e ma(e the simplifying assumption that theoriginal problem si$e is a po&er of 2% so thatall subproblem si$es are integers

,e denote by +n6 the running time of 4#D)MAN#M*M)S*A88AE on a subarray of n elements

4or starters% line 1 ta(es constant time +he base case% &hen n 9 1% is easyO line 2

ta(es constant time% and so +169:16

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+he recursive case occurs &hen nP1

Qines 1 and ta(e constant time

7ach of the subproblems solved in lines 3 and 5 is on a

subarray of n;2 elements our assumption that theoriginal problem si$e is a po&er of 2 ensures that n;2 isan integer6%

and so &e spend +n;26 time solving each of them

ecause &e have to solve t&o subproblems>for the leftsubarray and for the right subarray>the contribution tothe running time from lines 3 and 5 comes to 2+n;26

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the call to 4#D)MAN)!8-SS#G)S*A88AE in line = ta(es :n6 time

Qines ?R11 ta(e only :16 time 4or the recursive case% therefore% &e have

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!ombining e"uations 356 and 3=6 gives usa recurrence for the running time +n6 of4#D)MAN#M*M)S*A88AEO

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+hus% &e see that the divide)and)con"uermethod yields an algorithm that isasymptotically faster than the brute)force

method ,ith merge sort and no& the maimum)

subarray problem% &e begin to get an idea of

ho& po&erful the divide)and)con"uer methodcan be



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Assignment (1)

1 #mplement 4ind)Maimum)Subarray

2 ,hat does 4#D)MAN#M*M)S*A88AEreturn &hen all elements of A are negative

,rite pseudocode for the brute)force methodof solving the maimum)subarray problemState your looping count



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atri# u'ti'ication

#f A 9 ai'6 and 9 bi'6 are s"uare n n

matrices% then in the product! 9 A % &e dene the entry ci' %

for i %' 9 1%2Tn% by



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ecause each of the triply)nested for loopsruns eactly n iterations% and each eecutionof line ? ta(es constant time% the SU*A87)

MA+8#N)M*Q+#VQE procedure ta(es :n6 time



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Assignment (2)

1 #mplement Matri multiplication

2 State your looping count for n n matri



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The master method forsolving recurrences

+he master method provides a .coo(boo(/method for solving recurrences of the form

&here a W 1 and b P1 are constants and fn6 isan asymptotically positive function

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+he recurrence describes the running time ofan algorithm that divides a problem of si$e n into a subproblems% each of si$e n;b%&here a

and b are positive constants +he a subproblems are solved recursively%

each in time + n;b6

+he function fn6 encompasses the cost ofdividing the problem and combining theresults of the subproblems

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03 Divide and Conquer

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