Bai 1 Analysis

7/23/2019 Bai 1 Analysis

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PHN TCH GII THUT

Bi ging Cu trc dliu v Gii thut

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Gii thiu mn hc

Phn tch gii thut

Cu trc dliu v Gii thut


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Gii thiu mn hc Trng tm ca mn hc:Cc cu trc dliu v gii thut c lin quan Tnh ng n v phc tp (thi gian v khng gian)

Yu cu:Nhp mn v Kthut lp trnh: hm, mng, cu trc,

qui, con tr

Ngn nglp trnh: C/C++

Song song: Lp trnh hng i tng

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Tchc mn hc Tnh im (thang im 10):n: 40% Thi vit cui k(m): 60%

im cng (trnh by n): 30%

Trao i thng tin: Trang web mn hc

Email: [email protected]

Gio vin hng dn

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Gii thiu CTDL & GT Pht biu ni ting:

Chng trnh = Gii thut + Cu trc dliu. Tt cchng ta u lp trnh; do c lm quen

vi gii thut v cu trc dliu.

C lta khng thy nhng vn ny mt cch ringbit; M ta thy nhng cu trc dliu di dng nhng

thnh phn lp trnh n gin (c cung cp bi

STLthvin mu chun). Tuy nhin, cu trc dliu khc bit so vi gii thut,

v c vai tr rt quan trng.

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Mc tiu Mc tiu chnh ca mn hc ny l cung cp mt

suy nghc hthng vgii thut v cu trc dliu. Hai nguyn l nh hng vmn hc: tnh tru

tng v phn tch hnh thc. Tnh tru tng: Tp trung trn cc chc

ng dng rng ri trn nhiu bi ton khc nhau. Phn tch: Sdng mt cch hnh thc so snh

hai i tng (cu trc dliu hay gii thut). Cth, ta squa tm vtnh ng n, v cc

gii hn vthi gian v khng gian (bnh) trongtrng hp xu nht.

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Khi qut mn hc Phn tch gii thut

Kiu dliu tru tng (ADT) Danh sch lin kt

Stack/Queue Bm/Bng bm

Cu trc cy

Cc gii thut sp xp Nn/gii nn

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Gii thut v Cu trc dliu

Phn tch gii thut Cc k hiu tim cn BigO

Cc quy tc phn tch gii thut

Mt sv d

Phn tch gii thut


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Phn tch gii thut Cc nn tng ca phn tch gii thut v cu trc d

liu. Phn tch: Lm sao don hiu quca mt gii thut

Mt gii thut thc hin tt n mc no Lm sao so snh cc gii thut khc nhau cho 1 bi ton

Cu trc dliu Lm sao lu tr, truy xut v qun l dliu hiu qu

Cc cu trc dliu nh hng n hiu quca gii thut

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V d

cho gi

i thut

int factorial (int n) {

if (n


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V dcc gii thut ni ting Tnh bi schung ca Euclid

Tm cn bc hai ca Newton Bin i Fourier nhanh

Nn (Huffman, Lempel-Ziv, GIF, MPEG)

M ho DES, RSA Gii thut n gin cho lp trnh tuyn tnh

Gii thut tm ng i ngn nht(Dijkstra, Bellman-Ford)

M ho sa li(CDs, DVDs) Kim sot tc nghn TCP, nh tuyn IP

So khp mu (Hnh hc)

ng ctm kim (Search Engines)

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Vai tr ca gii thut trong thgii

hin i Lng dliu khng l Thng mi in t(Amazon, Ebay) Lu thng trn mng (theo di, tnh tin dch vvin

thng)

Giao dch csdliu (Mua bn, kim k) Cc o trong khoa hc (khng gian, a cht)

Mng sensor, cc thRFID

Sinh tin hc (genome, ngn hng protein)

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Bi ton Chui con Ln nht Cho mt chui snguyn A1, A2, , An, tm gi trln nht

c thca chui con Ai, , Aj. Sc thm.

Ta cn tm mt on lin tc vi tng ln nht.

V d: -2, 11, -4, 13, -5, -2 Kt qul 20 (chui con tA2 n A4).

Chng ta c 4 gii thut khc nhau, vi phc tp vmtthi gian l O(n3), O(n2), O(n log n), and O(n).

Vi n = 106, gii thut 1 cn > 10 nm; gii thut 4 chcnmt phn trm giy!

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int maxSum = 0;

for( int i=0; i


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Gii thut 2 tng: Vi tng ti n j-1, ta c thtnh tng ti n j

trong khong thi gian hng s. Gii thut ny bqua vng lp trong, v gim thi gian chy

xung O(n2).

into maxSum = 0;

for( int i = 0; i < n; i++ )

int thisSum = 0;

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

{ thisSum += a[ j ];

if( thisSum > maxSum )

maxSum = thisSum;

}

return maxSum;

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Gii thut 3 Gii thut ny dng cch chia tr.

Gista chia i chui ban u ti im gia. Chui con ln nht hoc nm hon ton na bn tri, hon

ton na bn phi, hoc ct ngang im gia.

V d: na tri | na phi

4 -3 5 -2 | -1 2 6 -2

Ln nht bn tri l 6 (A1 n A3); ln nht bn phi l 8(A6 n A7). Nhng ct ngang im gia ln nht l 11 (A1n A7).

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V d:

na tri | na phi4 -3 5 -2 | -1 2 6 -2 Chui con ln nht hai bn c thtm c bng qui. Lm cch no tm c chui con ln nht ct ngang im

gia? im lu : Bn tri ca chui ct ngang im gia l chui con ln

nht kt thc -2. Bn phi l chui con ln nht bt u -1.

C thtnh chui ny bng 1 dng qut tuyn tnh trong thi

gian O(n).

Gii thut 3 (tt)17


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Gii thut 3: Phn tch

Cch tt nht phn tch chia trl hi qui:

T(1) = 1

T(n) = 2T(n/2) + O(n)

Kt quca biu thc hi qui: T(n) = O(n log n).

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Gii thut 4

phc tp thi gian r rng l O(n)

Nhng liu gii thut trn c chnh xc? (Cn chng minh)

2, 3, -2, 1, -5, 4, 1, -3, 4, -1, 2int maxSum = 0, thisSum = 0;

for( int j = 0; j < n; j++ )

{

thisSum += a[j];

if ( thisSum > maxSum )

maxSum = thisSum;

else if ( thisSum < 0 )

thisSum = 0;

}

return maxSum;

}

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Gii thut 4: chng minh Chui con ln nht khng thbt u hay kt thc ti Ai m.

Tng qut hn, chui con ln nht khng thc phn u vitng m.

Vd: -2 11 -4 13 -5 -2

Do , bt ckhi no thy Ai n Aj c tng < 0, tng c thtin i n j+1

CM: Gisj l chsu tin sau i m tng < 0

Chui con ln nht khng thbt u tbt kvtr p nmgia i v j. V Ain Ap-1 dng, do bt u ti stthn.

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Gii thut 4int maxSum = 0, thisSum = 0;

for( int j = 0; j < n; j++ ){

thisSum += a[j];

if ( thisSum > maxSum )

maxSum = thisSum;

else if ( thisSum < 0 )

thisSum = 0;

}

return maxSum

Gii thutt li gi tr khi phnu < 0. Ngcli, n tnh tng mi v cp nht maxSum trongmt lt.

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Ti sao c

n Cc Gii thut Hi

u qu?

GisN = 106

Mt PC c thc/xl N mu tin trong 1 giy. Nhng nu gii thut thc hin N*N php tnh, n cn 1 triu

giy = 11 ngy!!!

Bi ton Ngi bn hng (TSP) i qua 100 thnh ph. Mt siu my tnh c thkim tra 100 tvng/giy vn cn

10100 nm!

Cc gii thut phn tch tha snhanh c thbcc m hnhm ho. Nghin cu cc gii thut gip xc nh di mho an ton. (> 100 chs)

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Lm cch no nh gi hiu

qugii thut? o no dng nh gi cc gii thut?

di ca chng trnh (s dng lnh)D lp trnh (pht hin li, qun l)

B nh

Thi gian thc thi

Thi gian thc thi l tiu chun quytnh.C th lng gic v d dng so snh

Thng lyu t c chai quan trng

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Tnh tru tng Mt gii thut c ththc thi khc nhau tuthuc vo:

nn tng phn cng (PC, Cray, Sun) ngn nglp trnh (C, Java, C++) lp trnh vin (you, me, Bill Joy)

Tuy khc nhau vchi tit, tt cm hnh phn cng vchng trnh u c im tng tnhau: chng l cc myTuring.

Chcn m cc php ton cbn l .

o n gin nhng c gi tri vi hiu quca cc thutton l mt hm ca kch thc input.

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Trng hp Trung bnh, Xu nht

v Tt nht Hiu quca gii thut c nh gi trong trng

hp input no? Trng hp trung bnh: Kh don phn phi ca thgii tht

Trng hp tt nht: C vkhng thc t

Trng hp xu nht:a ra bo m tuyt i Ta sdng o trong trng hp xu nht.

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V d

Cng vector Z = A+B

for (int i=0; i


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Nhn (ngoi) vector Z = A*BT

for (int i=0; i


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n gin ho binT(n) = ckn

k+ ck-1nk-1+ ck-2 n

k-2+ + c1n + co

qu phc tp qu nhiu shng

Rt kh so snh hai biu thc m mi biu thc c 10 hay

20 shng Ta c thc scn nhiu biu thc nhvy khng?

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Chcn gimt shng!

shng tng nhanh nht (chi phi thi gian thc thi) Khng cn gicc hshng Cc hshng chu nh hng bi phn cng, ngn

ng,

Gi trtim cn (khi n ln) c xc nh hon ton bi shng dn u. V d T(n) = 10 n3 + n2 + 40n + 800

Nu n = 1,000, th T(n) = 10,001,040,800 sai sl 0.01% nu loi bmi shng tr n3

Trong mt dy chuyn lp rp ngi thchm nht quytnh tc sn xut.

Cc bc n gin ho29


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Shng tng nhanh hn (v d2n) sdn dn ln t

cc shng tng chm hn(vd, 1000n) bt khsca chng l g!

Ni cch khc, vi cng mt sgia tng vthigian cho php, mt gii thut c bc cao hn skhng thu c li trong vic gii quyt cc bi

ton ln hn.

n gin ho30


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T(n) gimt s

hng

bhs

3n2+4n+1 3 n2 n2

101 n2+102 101 n2 n2

15 n2+6n 15 n2 n2

a n2+bn+c a n2 n2

Tt cu c tc tng nhnhau

Cc tim cn31


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T(n)

n n nlog n n2 n3 n4 n10 2n

10 .01s .03s .1s 1s 10s 10s 1s

20 .02s .09s .4s 8s 160s 2.84h 1ms

30 .03s .15s .9s 27s 810s 6.83d 1s

40 .04s .21s 1.6s 64s 2.56ms 121d 18m

50 .05s .28s 2.5s 125s 6.25ms 3.1y 13d

100 .1s .66s 10s 1ms 100ms 3171y 41013

y

103

1s 9.96s 1ms 1s 16.67m 3.171013

y 3210283

y

104

10s 130s 100ms 16.67m 115.7d 3.171023

y

10

5

100s 1.66ms 10s 11.57d 3171y 3.1710

33

y10

6 1ms 19.92ms 16.67m 31.71y 3.17107y 3.171043y

Gismy tnh thc hin1 tthao tc mi giy.

phc tp v tnh khthi32


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Khng chp nht vo cu ch

mt thut gii 100n t hn mt thut gii n2 khi n N.

c dng nhiu nht

Big-Omega, cn di: T(n) = (f(n)) Tn ti c>0 v N sao cho T(n) cf(n) khi n > N. Tng tf(n) = O(T(n)).

Big-Theta, cn trn v di: T(n) = (f(n)) Tng tT(n) = O(f(n)) v T(n) = (f(n))

Cc k hiu tim cn34


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33

25

10 0

(?)(?)

nn

nn

nn

O

23 2)( nnnT +=

V d

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)(log/

))/((!)(

)(

)(

)(

)()(c

)(

ni

ennnrr

ni

ni

ni

nnc

nf

n

i

n

nini

kkni

ni

ni

kii

ki

=

=

+

=

=

=

=

1

1

1

0

1

1

32

1

2

1

1

V d

Tim cn

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T(n) = cknk+ ck-1n

k-1+ ck-2 nk-2+ + c1n + co

Qu phc tp O(nk)

mt shng duy nht loi bcc hshng

n gin hn nhiu, cc shng v hshng khckhng c nh hngn tim cn

Cc chskhc rt kh nh lng

Tm tt (Ti sao O(n)?)37


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Phn tch Th

i gian th

c thi Cc quy tc

pht biu n gin (c, ghi, gn)O(1)

cc php tnh n gin (+ - * / == > >= <


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Hai qui tc quan trngQuy tc cng nu ta thc hin mt sthao tc theo tht, thi gian

thc thi bchi phi bi thao tc tn nhiu thi gian nht

Quy tc nhn nu ta lp li mt thao tc mt sln, tng thi gian thc

thi l thi gian thc thi ca thao tc nhn vi svng lp

Phn tch Th

i gian th

c thi (tt)39


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if (cond) then O(1)

body1 T1(n)else

body2 T2(n)

T(n) = O(max (T1(n), T2(n))

Phn tch Th

i gian th

c thi (tt)40


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Gi hm

A gi BB gi C

Mt chui thao tc gi hm ng cp:T(n) = max(TA(n), TB(n), TC(n))

Phn tch Th

i gian th

c thi (tt)41


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V d

for (i=1; i maxVal){

maxVal= A[i];

maxPos= i;

}

phc tp tim cn: O(n)

Php so snh: O(n) Php gn: Xu nht: O(n)

Trung bnh: O(n)

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V d

for (i=0; i= i+1; j--)

if (A[j-1] > A[j]){

temp = A[j-1];

A[j-1] = A[j];

A[j] = tmp;}

}

}

phc tp tim cn l O(n2)

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Th

i gian Th

c thi cho

qui T(n) cnh ngha qui theo T(k), k


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V d: Giai tha

int factorial (int n) {

if (n


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V d: Giai tha (tt)

int factorial1(int n) {

if (n


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Bi tp

Vit chng trnh v Tnh phc tp :

Tm phn t ln th nh ca 1 mngBi ton tm di chui con tng lin tip di nht

ca mt dys: vit t nht hai chng trnh vi

phc tp khc nhau.V d: 19 43 23 5 36 72 => 3

(*) Bi ton tm di chui con tngkhng lin tipdi nht ca mt dys:V d: 19 43 23 5 36 72 => 4

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Date post:	15-Feb-2018
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