Phan Tich Cac Thuat Toan Sap Xep Va Tim Kiem

7/31/2019 Phan Tich Cac Thuat Toan Sap Xep Va Tim Kiem

1/95

1

Chng 2:Phn tch cc thut ton sp xp v tm kim

NGUYN THIN AN

SV Khoa KT CN MT

i Hc An Giang


2/95

2

Mc ch

p dng k php O ln phn tch nh gi ccphng php sp xp:

Sp xp bng phng php chn (selection sort) Sp xp bng phng php chn (insertion sort) Sp xp bng phng php i ch (bubble sort) Sp xp bng phng php Shell (Shell Sort) Sp xp bng phng php trn (merge sort)

Sp xp bng phng php vun ng (heap sort) Sp xp nhanh (quick sort) Sp xp bng phng php th (bucket sort) Sp xp bng phng php c s (radix sort)


3/95

3

Sp xp bng phng php chn

tng: Tm phn t nh nht a v u dy hin ti

Tip tc thc hin phn cn li ca dy Thut ton: AlgorithmselectSort(A)

Input: Mt mng n phn t s AOutput: Mng A c sp xp tng dn.Fori 1 ton-1 do

min iForj i+1 tondo

if A[j] < A[min] thenmin j

swap(A, i, min)

Returnarray A


4/95

4

Phn tch SX bng pp chn

Vng lp ngoi (bin i) c thi hnh n-1 ln: O(n)

Tng i: n-1 ln

Kim tra i: n ln

Gn i vo min: n-1 ln

i ch: ti a n-1 ln

Vi mi gi tr ca i, vng lp trong (bin j) c thihnh n-1-i lntng cng (n-1) + (n-2) + + 1 =(n-1)n/2 ln: O(n2)

So snh: (n-1)n/2 ln

Gn: ti a (n-1)n/2 ln


5/95

5

Phn tch SX bng pp chn (tt)

Thi gian thc thi:

T(n) = O(n) + O(n2

) = O(n2

+n) = O(n2

)


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6

Sp xp bng phng php chn

tng: Chn tng phn t mt vo dy c sp xp n bc hin ti,

vo ng v tr ca n bo m sau khi chn dy vn c th t Thut ton: AlgorithminsertSort(A)

Input: Mt mng n phn t s AOutput: Mng A c sp xp tng dn.Fori 2tondo

temp A[i]j i- 1whiletemp 0 do

A[j+1] A[j]j j- 1

A[j+1] tempReturnarray A


7/957

Phn tch thut ton SX bng pp chn

Vng lp for (bin i) c thc hin n-1 ln Tng i: n-1 ln So snh i vi n: n ln Gn gi tr vo cc bin temp, j, A[j+1]: n ln

Vi mi gi tr i, thn vng lp while (bin j) ti

thiu c thc hin 0 ln v ti a c thchin i ln Tmin(n) = n-1

Tmax(n) = 1++(n-1) = (n-1)n/2 = O(n2)

Ttb(n) = Tmax(n)


8/95


9/959

Phn tch SX bng pp i ch

Vng lp for ngoi (bin i) c thi hnh n-1 ln

Tng i: n-1 ln So snh i: n ln

Vi mi gi tr i, vng lp for trong (bin j) cthi hnh (n-1-i) ln Tng j: n(n-1)/2 ln So snh j: n(n+1)/2 ln

Php so snh: n(n-1)/2 ln

Php i ch: ti a n(n-1)/2 ln


10/9510

Phn tch SX bng pp i ch

Thi gian thc thi: T(n) = O(n) + O(n2) =

O(n2

)


11/9511

Bi tp

Ci t 3 thut ton sp xp selection sort,insertion sort,v bubble sort bng ngn ng C/C++.

Kho st thi gian thc thi 3 thut ton ln lt vi ccgi tr n khc nhau vi cng mt dy s

Thi gian thc thi ca 3 thut ton vi cng mt gi tr n

(rt ln, >10000) vi cng mt dy s c khc nhau haykhng? Nu c gii thch v sao c. Nu khng gii thchv sao khng.

V th th hin thi gian thc thi ca mi thut ton

ph thuc vo n.


12/9512

Sp xp bng phng php Shell

tng:

L mt m rng ca insertion Sort cho php dchchuyn cc phn t xa nhau.

AlgorithmShellSort(A)

Input: Mt mng n phn t s A

Output: Mng A c sp xp tng dn.


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13

h 1repeat

h 3 * h + 1untilh > nrepeat

h h div 3fori h+1 tondo

v A[i]

j iwhilea[j-h] > vandj>hdoa[j] a[j-h]

j j-hA[j] v

untilh=1

Returnarray A


14/95

14

Phng php Chia v Tr

Mt m hnh thit k thut ton c 3 bc:

Chia: Nu kch thc d liu u vo nh hn mt ngng no

th gii trc tip.

Ngc li chia nh d liu u vo thnh hai hoc nhiu tpd liu ri nhau.

qui: Gii mt cch qui cc bi ton con ly cc li gii

Tr: Kt hp cc li gii ca cc bi ton con thnh li gii ca

bi ton ban u.


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15

Sp xp bng phng php trn

p dng m hnh chia tr thit k thut ton spxp bng phng php trn.

Chia:

Nu mng A rng hoc ch c mt phn t th tr v chnh A (c th t).

Ngc li A c chia thnh 2 mng con A1 v A2, mi mng

cha n/2 phn t qui:

Sp xp mt cch qui hai mng con A1 v A2

Tr:

To mng A bng cch trn hai mng c sp xp A1 v A2.


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Cy sp xp trn

PP sp xp trn

c th biu dinbng mt cynh phn.

Chiu cao ca

cy: [log2n]+1

A

A1 A2

1. Chia i d liu

2. Gii qui 2. Gii qui

3. Trn


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Trn hai mng c th tAlgorithmmerge (A1,A2,A)

Input: Mng A1, A2 c th t tng dn.Output: Mng A c hnh thnh t A1, A2v c th t tng dn.

while not(A1.isEmpty and A2.isEmpty)if A1[0]


19/95

19

Phn tch SX bng pp trn

Hm merge c phc tp O(n1+n2) vi n1, n2 l

kch thc ca A1, A2. Gi s mng A ban u c kch thc n=2m.

Ti mc th i trong cy sp xp trn: 2int

Mi nt cha bi ton vi mng c n/2iphn t.

Thi gian thc thi: 2i*O(n/2i) = O(n)

Cy c log2n mc (chiu cao ca cy)

phc tp O(logn*n)


20/95

20

Phn tch SX bng pp trn (2)

n

n/2 n/2

n/2 n/2 n/2 n/2

.

.

.

O(n)Chiu cao

O(n)

O(n)

O(logn)


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21


Gi t(n) l thi gian thc thi ca merge-sort

, 1( )

( / 2 ) ( / 2 )

b nt n

t n t n cn


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2 2 2

2 3 2 3 3

, 1

( ) 2 ( / 2)

( ) 2 (2 ( / 2 ) / 2)) 2 ( / 2 ) 2

( ) 2 (2 ( / 2 ) / 2 )) 2 2 ( / 2 ) 3

...

( ) 2 ( / 2 )

Thay i=m:

( ) 2 ( / 2 ) (1) log ( log )

i i

m m

b n

t n t n cn

t n t n cn cn t n cn

t n t n cn cn t n cn

t n t n icn

t n t n mcn nt c nn O n n

Gi s n=2m:


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23

Sp xp nhanh (Quick Sort)

Chia:

Nu mng A rng hoc ch c mt phn t th tr v chnh A (

c th t). Ngc li chn mt phn t x bt k ca A, chia A thnh 3 mng:

L: cha cc phn t ca A nh hn x

E: cha cc phn t ca A bng x

G: cha cc phn t ca A ln hn x

qui: Sp xp mt cch qui hai mng con L v G

Tr: To mng A bng cch lin tip 3 mng L, E, G theo th t.


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24

Cy sp xp nhanh

Cy nh phn

Chiu cao khngxc nh c,ph thuc vo x.

Trong trng

hp xu nht,chiu cao cacy l n-1 (mng sp xp)

E(=x)

L(x)

1. Chia d liu theo x

2. Gii qui 2. Gii qui

3. Ghp


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Sp xp nhanhAlgorithmquickSort(A, left, right)

Input: A: mng s, left v tr cc tri, right v tr cc phiOutput: Mng A c sp xp tng dn.

if r>l thenj leftk right + 1repeat

repeatj j+1

until a[j]>=a[left]

repeat k k-1until a[k]


26/95

26

Phn tch SX nhanh

Trng hp xu nht

Dy cn sp xp c th t Cy sp xp nhanh c chiu cao l O(n)

Mi ln gi qui gim mt phn t (x)

T(n) = n + (n-1) + + 1 = O(n2)

Trng hp tt nht Mi ln chia, chia i c dy

Cy sp xp nhanh c chiu cao l O(logn)

T(n) = 2T(n/2)+cn = O(nlogn) (xem mergesort)


27/95

27

Phn tch SX nhanh (2)

im mu cht l chn phn t dng sosnh (x) chia mng.

Trng hp trung bnh Mi phn t u c xc sut c chn l phn

t dng so snh l nh nhau v xc sut l

1/n


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28


1

1

1

1

1

1( ) ( 1) ( ( 1) ( ))

2( ) ( 1) ( 1)

Nhan 2 ve voi n:

( ) ( 1) 2 ( 1)

( 1) ( 1) ( 1) 2 ( 1)

( ) ( 1) ( 1) ( 1) ( 1) 2 ( 1)

( ) 2 ( 1) ( 1)Chia 2

n

k

n

k

n

k

n

k

T n n T k T n k n

T n n T k n

n T n n n T k

n T n n n T k

nT n n T n n n n n T n

nT n n n T n

2

ve cho n(n+1):

( ) ( 1) 2

1 1

( ) ( 2) 2 2

1 1 1

...

( ) (0) 21 1 1

n

k

T n T n

n n n

T n T n

n n n n

T n T

n k


29/95

29


1 1

( ) 1 1

2 2 2 ln1

ln ln 2 log 0.69 log

2 ln 1.38 log

nn

k

T n

dx nn k x

n n n

n n n n

Trng hp tt nht tt hn 38% so vitrng hp trung bnh

phc tp O(nlogn)


30/95

30

Sp xp vun ng

Mt s khi nim v cy nh ngha cy Cy nh phn Cy nh phn c tnh cht vun ng Biu din cy nh phn y bng mng

Cc thao tc trn cy nh phn c tnh cht vunng Thm mt phn t Xa mt phn t

Sp xp vun ng


31/95

31

Mt s khi nim v cy

Cy:

Rng Mt nt

Mt nt v cc cy con

Cy nh phn

Cy c s nt cy con ti mi nt ti a l 2

Cy nh phn c tnh vun ng (heap binary tree) Gi tr ti nt gc ln hn gi tr ti tt c cc nt

thuc 2 cy con ca n.


32/95

32

Biu din cy nh phn y bng mng

Xt phn t A[k] C 2 con l A[2*k] v A[2*k+1]

V d: A = (10, 3, 4, 2, 6, 7, 8)

10

2

43

6 7 8


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33

Cc thao tc trn cy NP vun ng

Thm mt phn t vo cy

Xa phn t khi cy (phn t gc)


34/95

34

Thm mt phn t vo cy

tng:

Thm phn t mi vo cui ca mng tng ngvi cy.

Phn t mi thm vo c th vi phm tnh cht heapvi nt cha ca n. Do phi iu chnh v tr ca

phn t mi thm vo. Tip tc iu chnh v tr phn t mi thm vo.


35/95

35

Thao tc upheap

Algorithm upheap(A, n, k)

Input:

A: mng tng ng vi cy heap c th b vi phmn: s phn t ca mngk: v tr phn t cn iu chnh (di ln trn)

Output:

Cy ng th t heap

v A[k]A[0] maxint

while A[k / 2]


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Thm mt phn t vo cy

Algorithm insert(A, n , v)

Input:

A: mng tng ng vi cy c n phn tv: gi tr thm vo cy

Output:

cy mi thm vo phn t gi tr v

n n + 1

A[n] v

upheap(A, n, n)


37/95

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Phn tch

upheap: S ln di chuyn nhiu nht tngng l chiu cao ca cy O(logn)

Thao tc thm mt phn t c phc tp lO(logn).


38/95

38

Thao tc xa mt phn t khi cy

tng:

Lun lun ly phn t gc Hon i phn t cui cng ca cy vi phn t gc.

Phn t gc mi c th vi phm tnh cht heap (nh

hn mt trong hai nt con) hon i v tr ca n.

Thc hin thao tc di ch phn t gc xung di

cho n khi n nm ng v tr


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39

downheapAlgorithm downheap(A, n, k)

Input:

A: mng tng ng vi cy heap c th b vi phm

n: s phn t ca mngk: v tr phn t cn iu chnh (di xung di)Output:

Cy ng th t heapv A[k]

while k = A[j] then

break

A[k] A[j]

k

jA[k] v


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Thao tc xa mt phn t khi cyAlgorithm remove(A, n)

Input:

A: mng c n phn t tng ng vi cy heapOutput:

Cy c n-1 phn t sau khi ly phn t gc rax: phn t gc b loi b

x A[1]

A[1] A[n]

n n 1downheap(A, n, 1)


41/95

41

Phn tch

Downheap: di ch ti a tng ng vichiu cao ca cy

Thao tc xa mt phn t O(logn)


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Heapsort

Algorithm heapsort(A, n)

Input:

A: mng c n phn tOutput:

Mng A c sp xpm 0

for k 1 to n do

insert(A, m, A[k])

for k n downto 1 do

A[k] = remove(A, m)


43/95

43

Phn tch

phc tp O(nlogn)


44/95

44

Sp xp da trn s so snh

Cc phng php kho st u da trnphp so snh l php ton chnh.

chng minh l chn di trong trng hpxu nht l O(nlogn)khng c phng phpsp xp no da trn s so snh c phc tp

nh hn O(nlogn) trong trng hp xu nht. p dng trong trng hp tng qut.

Trong mt s trng hp c bit c th cnhng phng php sp xp tt hn: O(n).


45/95

45

Sp xp th (Bucket Sort)

Xt dy A c n kha v min gi tr ca ccphn t l [0, m-1].

tng: S dng mt mng B gi l mng th (bucket

array). Mng th c m phn t. S dng gi tr ca A chnh l ch s trong mng

B. t cc phn t ca A vo B vi v tr tng ng

vi gi tr ca n.


46/95

46

Sp xp th (bucket sort) (tt)

AlgorithmbucketSort(A, n)

Input: Mt mng n phn t s A c min gi tr [0,m-1]

Output: Mng A c sp xp tng dn.

Gi B l mng c m phn t, ban u u trng htfori 1 tondo

insert(B[A[i]], A[i])

remove(A,i)

fori 0 to m-1

while (B[i] empty)

insert(A, i)

returnarray A


47/95

47

Sp xp th (Bucket Sort)

phc tp: Vng for u tin: O(n)

Vng for th hai: O(m)

O(m+n)

Nu m t l vi n: m = cn th phc tp l

O(n + cn)= O(n) phc tp l tuyn tnh. Lu : phc tp v khng gian O(m+n).


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48

Tnh n nh trong sp xp

Th t sau khi sp xp ca cc phn t ckha bng nhau khng thay i so vi th ttrc khi sp xp.

V d:


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49

Radix Sort

M rng tng ca sp xp th.

Mi phn t khng phi l mt gi tr n lm n c to thnh t nhiu thnh phnkhc nhau.

V d: So snh Long, Loan: L = L, o = o, a < n Loan keythen

returnx

else

ifk < x->keythen

returnTK_NPTK(x->left, k)else

returnTK_NPTK(x->right, k)


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91

Thut ton tm kim trong cy NPTK:nh gi

Trng hp xu nht: phc tp thut ton t l vi ng i di nht

trong cy = chiu cao ca cy T(n) = O(h)

Trng hp trung bnh:

T(n) = O(logn)


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92

Chng minh

Trng hp tm kim thnh cng Gi S(n) l thi gian trung bnh tm kim thnh

cng

Gi I(n) l tng cc mc ca cc nt trong cy cn nt

npl s nt trong cy con phi.

nt l s nt trong cy con tri. nt = n np- 1

I(n) = I(nt) + I(np) + n-1 (do co n-1 nt con)


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93

1

0

1

0

1

0

1( ) ( ( ) ( 1) 1)

( ) ( ( ) ( 1) 1)

( ) 2 ( ) ( 1)

( ) ( 1) ( 1) 2 ( 1) ( 1) ( 1)( 2)

( ) ( 1) ( 1) 2

n

tb tb tb

i

n

tb tb tb

i

n

tb tb

i

tb tb tb

tb tb

I n I i I n i nn

nI n I i I n i n

nI n I i n n

nI n n I n I n n n n n

nI n n I n I

( 1) 2 2

( ) ( 1) ( 1) 2( 1)

Chia 2 ve cho n(n+1)

( ) ( 1) 2( 1)

1 ( 1)

tb

tb tb

tb tb

n n

nI n n I n n

I n I n n

n n n n


94/95

94

1

1

( 1) ( 2) 2( 2)Thay

1 ( 1)

( ) ( 2) 2( 2) 2( 1)1 1 ( 1) ( 1)

...

( ) (1) 12

1 2 ( 1)

( ) 1( ) (lg )

1

( ) ( lg )

( )( ) (lg )

tb tb

tb tb

n

tb tb

i

n

tb

i

tb

tb

I n I n n

n n n n

I n I n n n

n n n n n n

I n I i

n i i

I nO O n

n i

I n O n n

I nS n O n

n


95/95

Chng minh

Trng hp tm kim khng thnh cng Gi U(n) l thi gian trung bnh tm kim khng

thnh cng

Gi E(n) l tng cc mc ca cc nt trong cyc n nt v 2n nt rng

E(n) = I(n) + 2n

U(n) = O(lgn)

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Phan Tich Cac Thuat Toan Sap Xep Va Tim Kiem