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Design and Analysis of Algorithms PART II-- Median & Runtime
Analysis
RecordedbyChandrasekarVijayarenuandShridharanMuthu.DateJan302008
MediansandOrderStatistics
Minimumandmaximum
Howmanycomparisonsarenecessarytodeterminetheminimumofasetofn
elements?Wecaneasilyobtainanupperboundofn1comparisons:examine
eachelementofthesetinturnandkeeptrackofthesmallestelementseenso
far.Inthefollowingprocedure,weassumethatthesetresidesinarrayA,where
length[A]=n.
MINIMUM(A)
1minA[1]
2fori2tolength[A]
3doifmin>A[i]
4thenminA[i]
5returnmin
TogettheminimumormaximumtheoperationislinearandhencetheorderforminandmaxisO(n).
Mean
Meanistheaverageofallthenumbers.
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=avg(A)
Median
Medianisexactlythemiddleoffallnumbers.
Thisfunctionismorestableandrobustsincethereisnosquarefactorcomparedtomean
Algorithmsforfindingthemedian.
1. Simplea. SortA[]
OrderforsortisO(nlogn)
b. Median=A[size/2]
2. Randamizedalgorithma1,a2.an
a. step1:Order(formgroups)b. step2:Findmedianineachgroupc.
step3:findmedianofthemedian(m1,m2,m3,..m5)=Md.
step4:DopartitionofAbyM
LR
If|L|=n/21thenk=n/2andthemiddletermisthemedian.
If|L|=n/2thenmedian(L,k)
Example:
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ConsiderA[]={2544442555321839987212647199131656241021971}
Formagroupof5numbers
25444425
55321839
987212647
199131656
241021971
Findthemedianforeachrow
2544442525
5532183918
98721264726
19913165616
24102197119
Findthemedianofthemedians
Medianof2518261619=19
ThereforeM=19
DopartitionofA[]byM
2544442555321839987212647199131656241021971
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254
24445425
245542544
245525445432
2455184454322539
2455189543225394487212647
245518919322539448721264754
2455189199253944872126475432
245518919913394487212647543225
245518919913164487212647543225395624
24551891991316108721264754322539562444
245518919913161022126475432253956244487
2455189199131610219264754322539562444872171
SinceM=19isthemiddleterm,themedianis19.
Thisexampleshowsthatiftherearekelements=pivot,thesekelementsdonotnecessarilylocatetogetherinthefinalresults.
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DateFeb4th2008
RuntimeanalysisofQuickSort:
T(n)=T(n/10)+T(9/10n)+cn=O(nlog2n)
AverageCase:
Thesplitupofsubarraysintheaveragecasewouldbe
BestCase:(LUCKY)
Forthebestcase,thearraywillbesplitexactlyinthemiddlei.e.pivotelementwillbeexactlyinthemiddleofthearray.
RuntimeAnalysisofMedianFind:
Letstakeanexample
Heretheentirearrayisbeingsplitintodatachunksofsize5i.e.n/5
0 1 2 3 4
2 54 44 4 255 5 32 18 399 87 21 26 4719 9 13 16 5624 10 2 19
71
1/10n 9/10n
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Aftersortingeachandeverycolumn,wewillget
Intheabovefigure,themiddlerowwhichiscircledisthemedianofthecorrespondingcolumns.
Nowweneedtofindthemedianofmedians.Sothemiddlerowneedstobesorted.Aftersortingthemiddlerow,themiddleelementwillbethemedianofmedians.
18isthemedianofmedians.Intheabovefigurethecolumn3andcolumn4areswapped.
Generallythepartitionwillbelike
3n/10
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T(n)=T(1/10n)+T(9/10n)+cnWeneedtoprovethisisoforderO(nlogn)
2.
ForMedianfindingT(n)=T(1/5n)+T(7/10n)+cnWeneedtoprovethisisoforderO(n)
Recurrence1(QuickSort)T(n)=T(1/10n)+T(9/10n)+cn
=T((1/10)2n)+T(9/10*1/10n)+cn/10+
T(1/10*9/10n)+T((9/10)2n)+9cn/10+cn
=T((1/10)2n)+2T(9/10*1/10n)+T((9/10)2n)+2cnGeneralizingtheaboveequationweget
T((1/10)kn)+(K1)T((1/10)
k19/10n)+(K2)T((1/10)k2(9/10)2n)+..+(K1)T((1/10)
2(9/10)k1n)+(K0)T((9/10)
kn)+kcn
Theabovecircledfactoristhelargestfactorintheaboveequationbecauseofthefollowingreasons1.
(K0)willgivethelargestnumber.2.
(9/10)kwillbethelargestfactorintheaboveequation.Thesmallestfactoris(1/10)k.
WeneedtodeducetheaboveequationtoClosedFormi.e.toT(1)Therefore
(9/10)kn=1 n=(10/9)k logn=klog(10/9)
k=(logn)/log(10/9)substitutingvalueofkandtakingthelargestfactoroutsideintheaboveequation,weget
T(n)
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=O(nlog2n)HenceT(n)=O(nlog2n)
Recurrence2(MedianFinding)T(n)=T(1/5n)+T(7/10n)+cn
=T((1/5)2n)+T(1/5*7/10n)+cn/5 +T(7/10*1/5n)+T((7/10)2n)+7cn/10 +cn
=T((1/5)2n)+2T(1/5*7/10n)+T((7/10)2n)+(9/10+1)cn
=T((1/5)kn)+(K1)T((1/5)k17/10n)+(K2)T((1/5)
k2(7/10)2n)+..+(K1)T((1/5)2(7/10)k1
n)+(K0)T((7/10)kn)+[(9/10)k1+(9/10)k2+..+1]cn
Kterms=[(1(9/10)k+1)/(1(9/10))]cnTodeducetheaboveequationtoClosedForm,weset(7/10)kn=1n=(10/7)k
Takinglogonbothsideslogn=klog(10/7)
k=logn/log(10/7)Substitutingthevalueofkintheaboveequationweget
T(n)
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Normallywepartitionthearrayby5i.e.n/5
Whatifwepartitionthearrayby3i.e.n/3(sortingwillbeeasierfor3elements)n/3
X X X
X M X
X X X
Therearetotally6halflinesintheabovediagram.2/6n>guaranteedlowerelementsthanY.Theotherhalfis12/6n=4/6n=2/3nSotheruntimeanalysisT(n)
- Theoreticallyn/7+5/7n=6/7nwhichis
