1. Tong Quan Ve Xu Ly Anh

7/31/2019 1. Tong Quan Ve Xu Ly Anh

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CHNG 1

TNG QUAN V XL NH S

Ths. Phm Vn TipKhoa Cng ngh thng tini hc i Nam

INTRODUCTION TO

DIGITAL IMAGE PROCESSING

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NI DUNG MN HC

Mc ch ca chng ny l: Cung cp cho ngi hc cc khi nim cbn ca mn hc

x l nh.

Gip c c ci nhn tng qut v cc k thut x l nhcng nhng dng ca n.

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HNH NH

Ti sao phi nghin cu hnh nh?

Con ngi l sinh vt c kh nng cm nhn v hnh nh rt cao. Chng ta cm nhn th gii xung quanh bng hnh nh:

Xc nh v phn loi vt th bng hnh nh

Nhn bit s khc nhau gia cc vt th da trn hnh nh

C mt ci nhn tng qut v quang cnh

Hnh nh l mt bc tranh m t s vt: Hnh nh ca con ngi, con vt, phong cnh

Hnh chp trong xt nghim y khoa

Hnh chp trong thin vn hc

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XL NH

X l nh l g? XLA l lm thay i bn cht ca hnh nh nhm :

Ci thin thng tin hnh nh cho qu trnh cm nhn ca conngi: tng sng, gim nhiu, lm ni chi titnh,Nhnbit s khc nhau gia cc vt th da trn hnh nh.

Ph hp hn cho vic cm nhn hnh nh bng my tnh:trchlc thng tin, so snh nh, nhn dng nh,

X l nh s l g?

XLA s trong ng cnh hp l vic s dng my tnh lmthay i bn cht ca nh s nh:

Lm tng cht lng hnh nh.Trch lc thng tin t hnh nh.

Ghp hnh nh.


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TNG QUAN V XL NH S (3)

Lch s v x l nh

Bt ngun t hai ng dng: nng cao cht lng thng tinhnh nhvx l s liu cho my tnh

ng dng u tin l vic truyn thng tin nh bo giaLondon v New York vo nm 1920 qua cp Bartlane.

Mha d liu nh khi phc nh

Thi gian truyn nh: T 1 tun 3 ting

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Lch s v x l nh

nh s c to ra vo nm 1921 tbng m ha ca mt my in in tn.(McFarlane)

nh s c to ra vo nm 1922 t cardc l sau 2 ln truyn qua i TyDng.Mt vi li c th nhn thy c.

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Lch s v x l nh

nh 15 cp xm c truyn t Lun nn New York, nm 1929. (McFarlane)

Trong khong thi gian ny, ngi ta ch ni n nh s, chcha cp g n x l nh s, v mt l do n gin: x lnh s th cn phi c my tnh x l, m trong giai on nykhi nim my tnh cha c.

H thng u tin c kh nng m ha hnhnh vi mc x m l 5 v tng ln 15vo nm1929

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Lch s v x l nh

Nm 1964, nh mt trng c a v tri t thng qua ccmy chp ca tu Ranger 7 ca Jet Propulsion Laboratory(Pasadena, California) cho my tnh x l: Chnh mo.

nh u tin ca mt trng c chp bi tuv tr M Ranger 7, vo 9 gi 09 pht sngngy 31/7/1964 (ngun: NASA)


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Lch sv xl nh

Song song vi cc ng dng trong khm ph khng gian,cc k thut x l nh cng bt u vo cui nhng nm1960 v u nhng nm 1970 trong y hc, theo di tinguyn tri t v thin vn hc.

n nay x l nh c mt bc tin di trong nhiungnh khoa hc, t nhng ng dng n gin n phc tp.

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Cc giai on ca x l nh

Camera

Sensor

Thu nhnnh S ha

Phn tchnh

i snhNhn dng

Hquyt nhLu tr

Lu tr

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Cc giai on ca x l nh

Thu nhn nh: nh c thu nhn t th gii thc qua mychp hnh, t tranh nh qua my qut hoc t v tinh thngqua b cm bin s hoc tng t.

S ha nh: S ha cc nh thu nhn c lu tr vomy tnh.

Phn tch nh: Gm nhiu thao tc: Tng cng nh, khnhiu,chnh mo,trch chn c trng, ...

i snh nhn dng: Phn lp phc v cho cc mc chkhc nhau

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Cc thnh phn chnh ca h thng x l nh

Mn hnh ha

B xlTng t

Camera B nhnh

My chB xl

nh s

Bn phm Con chutMy in

B nh

ngoi


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Cc thit b cbn trong x l nh

Camera: cng ging nh con mt ca h thng. Camera chai loi: loi CCIR ng vi chun CCIR qut nh vi tn s1/25, mi nh gm 625 dng; loi CCD gm cc photo ittng ng mt cng sng ti mt im nh ng vi mtphn t nh (pixel).

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B xl tng t: B x l tng t thc hin cc chcnng sau:

Chn camera thch hp nu h thng c nhiu camera.

Chn mn hnh hin th tn hiu.

Thc hin ly mu v m ho.

Tin x lnh trong khi thu nhn.

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B xlnh s: Gm nhiu b x l chuyn dng: x l lc,trch chn ng bao, nh phn ho nh.

My ch: ng vai tr iu khin cc thnh phn nu trn.

B nhngoi: Lu tr d liu nh cng nh cc kiu dliu khc, c th chuyn giao cho qu trnh khc.

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NG DNG CA XL NH (1)

Bc x phin t ca nh sng

Di ca phc nhm theo nng lng trn photonnh sau:

Tia gama, Tia X, Tia cc tm, Tia nhn thy, Tia hng ngoi,vi sng, sng radio.

Nng lng ca photo


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NG DNG CA XL NH (2)

Bc x phin t ca nh sng

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NG DNG CA XL NH

1. ng dng trong Y khoa

- Xem xt v gii thch hnh nh X quang

- Phn tch hnh nh t bo

2. ng dng trong nng nghip

- X l hnh nh thu c t v tinh ca cc khu t -> phnloi t ->tmloi cy trng ph hp.

- Phn tch tnh hnh su, bnh thng qua x l mu l cy.

3. ng dng trong cng nghip

- Kim tra sn phm trong dy truyn sn xut tng

- Kim tra mu giy

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NG DNG CA XL NH (3)

nh Gamma

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NG DNG CA XL NH (4)

nh Gammaa b

c d

nh phng x(a) Qut b xng(b) Chp PET (Positron Emission

Tomography)nh thin vn(c) Chm sao thin nganh phn ng ht nhn(d) S bc x tia Gamma t l phn

ng


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NG DNG CA XL NH (5)

nh tia X (nh X-Quang)

H thng my chp nh X-Quang

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NG DNG CA XL NH (6)


nh X-Quang chp lng ngc nh X-Quang chp hm mt

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NG DNG CA XL NH (7)


H thng my chpnh ct lp CT

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NG DNG CAXL NH (8)

nh tia X

nh chp ct lp CT


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NG DNG CA XL NH (9)

nh trong di cc tma c

d

(a) Trng bnh thng(b) Trng bnh than(c) Chm sao thin nga

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NG DNG CA XL NH (10)

nh hng ngoi

nh hng ngoichp u con mo nh hng ngoichp u con ch

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nh hng ngoi

nh hngngoi chp bmt tri t.Nhng ni cnh sng mnhl nhng ni cngun nhitln.

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nh hng ngoi

nh hng ngoichp khng giantrn b mt tri t.nh ny cho bitlng hi nc tcht trong khng gian,phc v cho vic dbo thi tit.


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To nh trn my tnh

(a) v (b) nh Fractal cto ra bi my tnh

(c) v (d) nh 3 chiu cm hnh ha bi my tnh

a b

c d

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D tm mt ngi

D tm v tr camt ngi, lmbc tin x lcho vic nhndng mt ngi

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D tm mt ngi

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D tm mt ngi



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Trong vng nh sng nhn thy

Ci thin nh:


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Gim nhiu:


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Nhn dng ch vit:


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Nhn dng vn tay:


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Cc ng dng nhn dng khc Nhn dng du vn tay, vn mi ca con ngi.

Nhn dng gng mt con ngi.

Nhn dng ngn ng k hiu.

Nhn dng ch k.

Nhn dng mu da ngi.

Nhn dng nh v phim khiu dm.

Phn tch ch vit tay.

Chm im trc nghim bng my.

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CC KHI NIM C BN (1)

Pixcel(im nh picture element): im nh (Pixel) l mt phn tca nh s ti to(x, y) vi xm hoc mu nhtnh.

Kch thc v khong cch gia cc im nh c chn thch hpsao cho mt ngicm nhn s lin tc v khng gian v mc xm (hocmu) ca nh s gn nhnh tht. Miphn t trong ma trn c gi lmt phn tnh.

phn gii (Resolution) ca nh l mt im nh c nnh trn mt nh sc hin th.

Khong cch gia cc im nh phi c chn sao cho mt ngi vnthy c s lin tc ca nh. Vic la chn khong cch thch hp tonn mt mt phn b, chnh l phn gii v c phn b theo

trcx v y trong khng gian hai chiu.

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CC KHI NIM C BN (2)

im nh (pixel picture element)

a b c

d e f

(a) 1024 1024

(b) 512 512

(c) 256 256

(d) 128 128(e) 64 64

(f) 32 32


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CC KHI NIM C BN (3)

Mc xm (Gray): Mc xm ca im nh l cng sng ca n c gn bng gi tr s ti im .

Cc thang gi tr mc xm thng thng: 16, 32, 64, 128,256.

V d: ti im nh ta (20, 40) c mc xm l 60, tiim nh ta (30, 40) c mc xm l 23, ...

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CC KHI NIM C BN (4)

Mc xm (Gray)

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CC KHI NIM C BN (5) nh c thc chia lm 3 loi:

1. nh trng en (2 mu)

- Mu en: gi tr 0

- Mu trng: gi tr 1

2. nh mu (nhiu hn 2 mu)

- Mi mu l mt t hp ca 3 gi tr mu thnh phn: R(Red), G(Green), B (Blue)

- Thng thng, R, G, B c gi tr t 0 n 255

3. nh mc xm (l kt qu bin i ca nh mu)

-Gi tr mc xm c tnh t cc thnh phn R, G, B

-Thng thng, mc xm c gi tr t 0 n 255

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CC KHI NIM C BN (5) nh nh phn: cn c bit vi tn gi nh trng en.

Gi tr ca pixel l 0 (mu en), 1 (mu trng)


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CC KHI NIM C BN (5) nh mc xm: Gi tr ca pixel l gi tr mc xm t 0 (mu

en) n 255 (mu trng)

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CC KHI NIM C BN (5) nh mu: Gi tr ca pixel l mt b gi tr RGB

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CC KHI NIM C BN (5) nh ch mc: L nh mu s dng bng mu v bng ch mc

km theo.

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CC KHI NIM C BN (5)

nh (Image)

Tp hp cc im nh c biu din thng qua mng haichiuI(n, p): n dng v p ct, lc s lng im nh s ln p. Ta k hiu I(x,y) ch gi tr mc xm ca im nhti v tr to(x,y).

Mt hm hai bin f(x, y), trong (x, y) l ta trongkhng gian hai chiu v f l ln ti ta (x, y) c gil mc xm ca nh ti im .

Khi mi im (x, y) trong khng gian hai chiu biu din cpxm c ln f hu hn, xc nh v c lng ha ri rcta gi l nh s.


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CC KHI NIM C BN (6)

nh (Image)

Cc gi trim nh ti vng mu trong nh bn tri 50

CC KHI NIM C BN (7)

Biu din nh (Image Representation)

Cc phn tc trng cbn ca nh l im nh

Cc m hnh thng s dng l: m hnh ton hc biudin nh thng qua cc hm hai bin trc giao, m hnhthng k biu din thng qua cc i lng k vng,phng sai, moment.

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CC KHI NIM C BN (8)

Tng cng nh (Image Enhancement)Lm ni bt cc c trng chn.Cc k thut c chn: lc tng phn, kh nhiu, nimu, ni bin, gin tng phn.

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CC KHI NIM C BN (9)

Tng cng nh (Image Enhancement)


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CC KHI NIM C BN (10)

Khi phc nh (Image Restoration)Loi b hay ti thiu ha cc nh hng ca mi trng bnngoi hay h thng thu nhn nh gy ra.

Kt qu thu c l nh gn ging vi nh gc.

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Khi phc nh (Image Restoration)

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CC KHI NIM C BN (12) Bin i nh (Image Transform)

Bin i th hin ca nh di cc gc nhn khc nhau tincho vic x l, phn tch nh.

Cc k thut thng s dng bin i nh: Bin iFourier, Sin, Cosin, Karhumen Loeve, ...

Php bin i Fourier

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Phn tch nh (Image Analyze)

Tm ra cc c trng ca nh, xy dng quan h gia chngda vo cc c trng cc b.

Cc k thut thng s dng: k thut lc, k thut tch, kthut hp da trn cc tiu chun nh gi v mu sc,cng , kt cu, ... V cc k thut phn lp da trn cutrc khc.

Xl nh Phn tchnh

Th gic my


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Nn nh (Image Compression)

Tm cch loi b thng tin d tha trong nh gc lmgim dung lng lu tr, thun li cho vic truyn d liu.

Cc 2 phng php nn nh:

Nn nh khng mt mt thng tin - nh sau khi nn c khi phc

ging ht nh gc.

Nn nh c mt mt thng tin - nh sau khi nn c khi phc li

gn ging vi nh gc.

Vy, thng tin mt mt y l g?

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Nn nh (Image Compression)

nh gcnh nn

(1/10)

nh nn

(1/20)nh nn

(1/30)

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Nhn dng nh (Image Recognition)

L qu trnh phn loi i tng c biu din theo mtm hnh no v gn chng vo mt lp da theo nhngquy lut v cc mu chun.

Nhn dng p dng trong vic bo mt, an ninh, nhn dngch vit, ...

Cc phng php nhn dng: Nhn dng da vo phnhoch khng gian, nhn dng theo cu trc, nhn dng datheo mng nron, m hnh Markov n.

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Nhn dng vn tay

Kha nhn dng vn tayMy tnh c bo mt bi cng

ngh nhn dng vn tay


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Tra cu nh (Image Retrieval)

Tm cc nh tha mn cc yu cu cho trc trong mt csd liu ln.

Tra cu nh c p dng trong th vin, y hc, h thng anninh, bo mt,...

C hai k thut tra cu nh:

Tra cu nh da trn t kha

Tra cu nh da trn ni dung

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Tra cunh logo

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Tra cunh logo

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BA CP XL BNG MY TNH

Cp thp: u vo, u ra l nhS dng cc ton t v php bin i cbn nh gimnhiu, tng cng, gin tng phn

Cp trung bnh: u vo l nh, u ra l cc thuctnh c trch chn tnh

Phn on, biu din i tng, phn lp cc i tng.

Cp cao: u vo l nh, u ra l nh gi vnhPhn tch nh.


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CC MI QUAN H C BNGIA CC IM NH (1)

Quy c chung

Mt nh sc k hiu biI(n, p) hocI

Mt im nh c ta l (x, y) v cp xm lf(x, y)

x k hiu cho ta theo trc honh (ngang, rng)

y k hiu cho ta theo trc tung (ng, cao)

Cc ch ci thngp, q, ... k hiu cho cc im nh.x

y

00

f(x, y)66

CC MI QUAN H C BNGIA CC IM NH (2)

Ln cn ca im nh

Tnh lin k

ng i

Tnh lin thng

Vng & bin ca vng

Gn nhn cho cc thnh phn lin thng

o khong cch

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LN CN CA IM NH

Mt im nh p ti ta (x, y) c

4 ln cn ngang - dc ca p: K hiu l N4(p)(x+1,y), (x-1,y), (x,y+1), (x,y-1)

4 ln cn cho ca p: K hiu l ND(p)(x+1,y+1), (x+1,y-1), (x-1,y+1), (x-1,y-1)

8 ln cn ca p: K hiu N8(p)l s kt hp ca N4(p) v ND(p)(x+1,y), (x-1,y), (x,y+1), (x,y-1),(x+1,y+1), (x+1,y-1), (x-1,y+1), (x-1,y-1)

x

x p x

x

x x

p

x x

x x xx p x

x x x

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TNH LIN K (1)

Cho V l tp cc gi tr cp xm, chng ta c4 lin k: Hai im nh p v q vi cc gi tr t V l 4 link nu q thuc tp N4(p).

V d: Cho tp d liu sau y v V = {1}, tm 4 lin k caim p = (2,2)

0 1 2 3 4

0 1 0 1 1 0

1 0 1 0 1 0

2 0 1 1 1 0

3 1 1 0 0 1

4 1 1 0 0 1

4 ln cn ca im (2, 2) - N4(p) l:(1,2), (2,1), (3,2), (2,3).

4 lin k ca im (2,2) l:(1,2) v (3,2).


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TNH LIN K (2)

Bi tp:Cho tp d liu sau y v V = {1, 2, 5}, tm 4 lin k caim p = (2,4)

0 1 2 3 4 5 6 7

0 1 0 2 2 4 5 6 6

1 1 1 3 2 7 4 5 6

2 6 2 4 4 6 7 6 4

3 5 7 3 3 7 3 0 6

4 1 4 2 2 2 5 0 1

5 3 5 5 3 1 0 2 3

6 3 3 5 7 1 0 2 3

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TNH LIN K (3)


0 1 2 3 4 5 6 7

0 1 0 2 2 4 5 6 6

1 1 1 3 2 7 4 5 6

2 6 2 4 4 6 7 6 4

3 5 7 3 3 7 3 0 6

4 1 4 2 2 2 5 0 1

5 3 5 5 3 1 0 2 3

6 3 3 5 7 1 0 2 3

4 ln cn ca im (2, 4) -N4(p) l: (1,4), (2,3), (3,4),(2,5).

4 lin k ca im (2,4) l:(2,5) v (3,4).

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TNH LIN K (4)

Cho V l tp cc gi tr cp xm chng ta c8 lin k: Hai im nh p v q vi cc gi tr t V l 8 link nu q thuc tp N8(p).

V d: Cho tp d liu sau y v V = {1}, tm 8 lin k caim p = (2,2)

0 1 2 3 4

0 1 0 1 1 0

1 0 1 0 1 0

2 0 1 1 1 0

3 1 1 0 0 1

4 1 1 0 0 1

8 ln cn ca im (2, 2) N8(p) l:(1,1), (1,2), (1,3), (2,1), (2,3), (3,1),(3,2), (3,3)

8 lin k ca im (2,2) l:(1,1), (1,2), (1,3), (3,1), (3,2)

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TNH LIN K (5)


0 1 2 3 4 5 6 7

0 1 0 2 2 4 5 6 6

1 1 1 3 2 7 4 5 6

2 6 2 4 4 6 7 6 4

3 5 7 3 3 7 3 0 6

4 1 4 2 2 2 5 0 1

5 3 5 5 3 1 0 2 3

6 3 3 5 7 1 0 2 3


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TNH LIN K (6)


0 1 2 3 4 5 6 7

0 1 0 2 2 4 5 6 6

1 1 1 3 2 7 4 5 6

2 6 2 4 4 6 7 6 4

3 5 7 3 3 7 3 0 6

4 1 4 2 2 2 5 0 1

5 3 5 5 3 1 0 2 3

6 3 3 5 7 1 0 2 3

8 ln cn ca im (2, 4) - N8(p)l: (1,3), (1,4), (1,5), (2,3), (2,5),(3,3), (3,4), (3,5)

8 lin k ca im (2,4) l:(1,5), (2,5), (3,4)

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TNH LIN K (7)

Cho V l tp cc gi tr cp xm chng ta cm lin k (lin k hn hp): Hai im nh p v q vi cc gitr t V l m lin k nu:

q thuc N4(p) hoc

q thuc ND(p) v tp N4(p) N4(q) rng - khng c im nh no cgi tr t V. (Tp cc im nh l 4 ln cn ca p v q giao nhaukhng c gi tr nao t V)

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TNH LIN K (8)

Cho V l tp cc gi tr cp xm chng ta cV d: Cho tp d liu sau y v V = {1}, tm m lin k caim p = (2,2)

0 1 2 3 4

0 1 0 1 1 0

1 0 1 0 1 0

2 0 0 1 1 0

3 1 1 0 0 1

4 1 1 0 0 1

4 ln cn ca im (2, 2) - N4(p) l:(1,2), (2,1), (3,2), (2,3).

4 ln cn cho ca (2,2) ND(p) l:(1,1), (1,3), (3,1), (3,3).

m lin k ca (2,2) l:

(1,1), (1,3), (3,2).

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TNH LIN K (9)

Bi tp:Cho tp d liu sau y v V = {1, 2, 5}, tm m lin k caim p = (2,4)

0 1 2 3 4 5 6 7

0 1 0 2 2 4 5 6 6

1 1 1 3 2 7 4 5 6

2 6 2 4 4 6 7 6 4

3 5 2 3 1 7 3 0 6

4 1 4 2 2 2 5 0 1

5 3 5 5 3 1 0 2 3

6 3 3 5 7 1 0 2 3


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TNH LIN K (9)

Bi tp:Cho tp d liu sau y v V = {1, 2, 5}, tm m lin k caim p = (2,4)

0 1 2 3 4 5 6 7

0 1 0 2 2 4 5 6 6

1 1 1 3 2 7 4 5 6

2 6 2 4 4 6 7 6 4

3 5 2 3 1 7 3 0 6

4 1 4 2 2 2 5 0 1

5 3 5 5 3 1 0 2 3

6 3 3 5 7 1 0 2 3

4 ln cn ca im (2, 4) N4(p)l: (1,4), (2,3), (2,5), (3,4).

4 ln cn cho ca (2,4) l:(1,3), (1,5), (3,3), (3,5).

m lin k ca (2,4) l:(1,3), (2,5), (3,4).

78

TNH LIN K (10)

Bi tp:Cho tp d liu sau y v V = {1, 2, 5}, tm 8 lin k caim (2,4)

0 1 2 3 4 5 6 7

0 1 0 2 2 4 5 6 6

1 1 1 3 2 7 4 5 6

2 6 2 4 4 6 7 6 4

3 5 7 3 3 7 3 0 6

4 1 4 2 2 2 5 0 1

5 3 5 5 3 1 0 2 3

6 3 3 5 7 1 0 2 3

79

TNH LIN K (10)

Bi tp:Cho tp d liu sau y v V = {1, 2, 5}, tm 8 lin k caim (2,4)

0 1 2 3 4 5 6 7

0 1 0 2 2 4 5 6 6

1 1 1 3 2 7 4 5 6

2 6 2 4 4 6 7 6 4

3 5 7 3 3 7 3 0 6

4 1 4 2 2 2 5 0 1

5 3 5 5 3 1 0 2 3

6 3 3 5 7 1 0 2 3

8 ln cn ca im (2, 4) - N8(p)l: (1,3), (1,4), (1,5), (2,3), (2,5),(3,3), (3,4), (3,5)

8 lin k ca im (2,4) l:(1,5), (2,5), (3,4)

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TNH LIN K (11)

V d: V = {1}, p l im nh trung tm.

0 1 1

0 1 0

0 0 1 100

010

110

m lin k caim nh trung tm

m lin k dng c lng tnh nhp nhng ca 8 lin k

100

010

110

8 lin k caim nh trung tm

Cc im nh

0 1 1

0 1 0

0 0 1

4 lin k caim nh trung tm


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TNH LIN K (12)

Tnh lin k ca cc tp con trong nhHai tp con S1 v S2 ca nh c gi l lin k nu tn timt im nh trong S1 lin k vi mt im nh trong S2.

Cc im lin k s dng y c th l 4 lin k, 8 lin khoc m lin k tng ng cc tp S1, S2 c gi l 4, 8, mlin k.

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TNH LIN K (4)

Bi tpXt hai tp con S1, S2 ca mt nh.

0 0 0 0 0 0 0 1 1 0

1 0 0 1 0 0 1 0 0 1

1 0 0 1 0 1 1 0 0 0

0 0 1 1 1 0 0 0 0 0

0 0 1 1 1 0 0 1 1 1Vi quan h no di y th S1 v S2 lin k vi nhau?

Quan h 4 lin k

Quan h 8 lin k

Quan h m lin k

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NG I (1)

ng i gia p v q

Mt ng i tim p c ta (x, y) n im q c ta (s, t) l mt chui tun t cc im nh phn bit nhau cta :

(x0, y0), (x1, y1), ..., (xn, yn)trong (x0, y0) = (x, y) v (xn, yn) = (s, t) v (xi, yi) lin kvi (xi-1, yi-1) vi 1 i n.

n l di ca ng i

Nu (x0, y0) = (xn, yn) th ng i c gi l mt ng ing hay chu trnh.

C th s dng cc nh ngha 4, 8, m lin k ty thuc votng mc ch c th. 84

NG I (2)

V d:

Tm ng i t p = (1,1) n q = (5,4) theo quan h 4 link vi V = {1} trong tp S sau.


0 1 2 3 4 5 6 7

0 1 0 0 0 1 0 1 0

1 1 1 1 1 1 1 0 1

2 1 1 0 0 0 1 1 03 0 1 1 1 0 1 0 1

4 1 1 0 1 1 1 0 0

5 1 1 0 0 1 0 1 0


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NG I (3)

V d:


0 1 2 3 4 5 6 7

0 1 0 0 0 1 0 1 0

1 1 1 1 1 1 1 0 1

2 1 1 0 0 0 1 1 0

3 0 1 1 1 0 1 0 1

4 1 1 0 1 1 1 0 0

5 1 1 0 0 1 0 1 0

86

NG I (4)

V d:


0 1 2 3 4 5 6 7

0 1 0 0 0 1 0 1 0

1 1 1 1 1 1 1 0 1

2 1 1 0 0 0 1 1 0

3 0 1 1 1 0 1 0 1

4 1 1 0 1 1 1 0 0

5 1 1 0 0 1 0 1 0

87

TNH LIN THNG (1)

Tnh lin thng ca cc im nhCho tp con S ca nh, hai im p v q trong S c gi llin thng nu tn ti mt ng i t p n q vi tt c ccim trung gian nm trong S.

0 1 2 3 4 5 6

0 0 0 0 1 0 0 0

1 1 1 0 1 0 0 1

2 1 0 0 1 0 1 1

3 1 0 1 1 1 0 04 1 1 1 1 1 0 0p=(3,0), q=(2,3) lin thng theo

4 lin k vi V = {1}

0 1 2 3 4 5 6

0 0 0 0 1 0 0 0

1 1 1 0 1 0 0 1

2 1 0 0 1 0 1 1

3 1 0 1 1 1 0 04 1 1 1 1 1 0 0

p=(1,1), q=(2,3) khng lin thng theo4 lin k vi V = {1} 88

TNH LIN THNG (2)

Thnh phn lin thngCho tp con S ca nh v mt im p bt k thuc S, tp ccim nh lin thng vi n trong S c gi l thnh phnlin thng ca S.

Nu S ch c mt thnh phn lin thng th S c gi lmt tp lin thng.

0 1 2 3 4 5 6

0 0 0 0 1 0 0 0

11 1 0 1 0 0 1

2 1 0 0 1 0 1 1

3 1 0 1 1 1 0 0

4 1 1 1 1 1 0 0

0 1 2 3 4 5 6

0 0 1 1 1 1 0 0

11 1 1 1 1 0 1

2 1 1 1 1 1 1 1

3 1 1 1 1 1 0 0

4 1 1 1 1 1 0 0


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VNG & BIN CA VNG (1)

VngCho R l mt tp con ca nh. R c gi l vng nu R lmt tp lin thng.

0 0 1 1 0 0 0

0 1 1 1 0 0 1

0 1 1 1 1 0 1

0 1 1 1 1 0 0

0 1 1 1 1 1 0

0 1 1 1 1 0 0

90

VNG & BIN CA VNG (2)

Bin ca vngBin ca vng R l tp hp cc im trong vng R m cmt hoc nhiu ln cn khng thuc R.

Nu R ph ton nh th bin ca n l dng u tin, ct utin, dng cui cng, ct cui cng ca nh.

0 0 1 1 0 0 0

0 1 1 1 0 0 1

0 1 1 1 1 0 1

0 1 1 1 1 0 0

0 1 1 1 1 1 0

0 1 1 1 1 0 0

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 1

91

GN NHN CHO CC THNH PHNLIN THNG VI 4 LIN K (1)

Qut nh t tri sang phi, t trn xung di.

K hiu p l im ti bc ang cn x l.

K hiu r l ln cn trn ca p.

K hiu t l ln cn tri ca p.

Khi chng ta xt p th cc im r v t c xc nh v gnnhn nu chng thuc tp V.

r

t p

0 1 0 0

1 1 1 0

1 0 1 00 1 0 1

Th t qut cc im nh

Ln cn trn v tri ca ptheo quan h 4 lin k

92


Nu gi tr ca p V, b qua v xt im tip theo.

Nu gi tr ca p V, kim tra r v t.Nu c r v t V, gn nhn mi cho p.

Nu ch c r hoc t V, gn nhn ca ln cn thuc V cho p.

Nu c hai gi tr r v t V.

Nu chng c cng nhn, gn nhn cho p

Nu chng khc nhn, gn mt trong hai nhn cho p v hai nhn ca r vt by gi l tng ng (lin thng qua p).

Kt thc qu trnh qut th tt c cc im c gi tr thuc V uc nh du

Ln qut th hai, gn mt nhn mi cho cc thnh phn c nhntng ng nhau.


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Thc hin ging nh trong trng hp 4 lin k, nhng chng tas xt vi ln cn cho trn v ln cn tri.

Nu gi tr ca p V, b qua v xt im tip theo

Nu gi tr ca p V, kim tra bn ln cn ca nNu tt u khng thuc V, gn nhn mi cho p.

Nu ch c mt trong 4 ln cn thuc V, gn nhn ca ln cn thuc V chop.

Nu c t hai ln cn thuc V trln, gn cho chng mt nhn trong s v nh du tnh tng ng ca cc nhn ca ln cn thuc V (lin thng qua p)

Ln qut th hai, gn mt nhn mi cho cc thnh phn c nhn

tng ng nhau.

q r s

t p

94

GN NHN CHO CC THNH PHNLIN THNG (1)

Bi tpTm cc thnh phn lin thng ca tp S di y theo quanh4 lin k v 8 lin k vi V = {1}

0 0 1 0 1 0 0 1 1 1 0 0 0 1

0 0 0 1 1 1 0 1 1 1 0 1 0 1

1 0 0 1 1 0 1 0 1 0 1 1 0 0

1 1 0 0 0 0 0 0 0 0 0 0 0 1

0 0 0 0 1 1 1 1 0 0 0 0 0 1

0 0 0 1 0 0 1 1 0 1 0 0 0 0

1 0 0 1 1 0 0 0 1 0 1 0 1 0

1 0 0 1 1 0 0 0 1 0 0 0 0 1

95


Kt quCc thnh phn lin thng ca tp S theo quan h4 lin k vi V = {1} c th hin nh sau

0 0 1 0 1 0 0 1 1 1 0 0 0 1

0 0 0 1 1 1 0 1 1 1 0 1 0 1

1 0 0 1 1 0 1 0 1 0 1 1 0 0

1 1 0 0 0 0 0 0 0 0 0 0 0 1

0 0 0 0 1 1 1 1 0 0 0 0 0 1

0 0 0 1 0 0 1 1 0 1 0 0 0 0

1 0 0 1 1 0 0 0 1 0 1 0 1 0

1 0 0 1 1 0 0 0 1 0 0 0 0 1 96


Kt quCc thnh phn lin thng ca tp S theo quan h8 lin k vi V = {1} c th hin nh sau

0 0 1 0 1 0 0 1 1 1 0 0 0 1

0 0 0 1 1 1 0 1 1 1 0 1 0 1

1 0 0 1 1 0 1 0 1 0 1 1 0 0

1 1 0 0 0 0 0 0 0 0 0 0 0 1

0 0 0 0 1 1 1 1 0 0 0 0 0 1

0 0 0 1 0 0 1 1 0 1 0 0 0 0

1 0 0 1 1 0 0 0 1 0 1 0 1 0

1 0 0 1 1 0 0 0 1 0 0 0 0 1


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KHONG CCH EUCLIDE

Khong cch Euclide gia hai im p v q.22)()(),( tysxqpD

e+=

p(x, y)

q(s, t)

r

s-x

t-y

Cc im nh c khong cch De i vi im (x, y) nh hn

hoc bng r to thnh mt hnh trn tm (x, y) 102

KHONG CCH KHI (K/C D4)

Khong cch D4 gia hai im p v q.

tysxqpD +=),(4

Cc im nh c khong cch D4 i viim (x, y) nh hn hoc bng r tothnh mt hnh thoi tm (x, y)

D4 = 1 l 4 ln cn ca im (x, y)

2

2 1 2

2 1 0 1 2

2 1 2

2

Hnh trn m t D4 = 2

103

KHONG CCH BN C (K/C D8)

Khong cch D8 gia hai im p v q.

),max(),(8

tysxqpD =

Cc im nh c khong cch D8 i viim (x, y) nh hn hoc bng r tothnh mt hnh vung tm (x, y)

D8 = 1 l 8 ln cn ca im (x, y)

2 2 2 2 2

2 1 1 1 2

2 1 0 1 2

2 1 1 1 2

2 2 2 2 2Hnh trn m t D8 = 2

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1. Tong Quan Ve Xu Ly Anh