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7/31/2019 1. Tong Quan Ve Xu Ly Anh
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05/22/20
1
CHNG 1
TNG QUAN V XL NH S
Ths. Phm Vn TipKhoa Cng ngh thng tini hc i Nam
INTRODUCTION TO
DIGITAL IMAGE PROCESSING
2
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.
3
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
4
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
6
TNG QUAN V XL NH S (4)
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.
7
TNG QUAN V XL NH S (5)
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
8
TNG QUAN V XL NH S (6)
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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TNG QUAN V XL NH S (7)
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.
10
TNG QUAN V XL NH S (8)
Cc giai on ca x l nh
Camera
Sensor
Thu nhnnh S ha
Phn tchnh
i snhNhn dng
Hquyt nhLu tr
Lu tr
11
TNG QUAN V XL NH S (9)
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
12
TNG QUAN V XL NH S (10)
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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TNG QUAN V XL NH S (11)
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).
14
TNG QUAN V XL NH S (12)
Cc thit b cbn trong x l nh
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.
15
TNG QUAN V XL NH S (13)
Cc thit b cbn trong x l nh
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.
16
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
18
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
19
NG DNG CA XL NH (3)
nh Gamma
20
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
22
NG DNG CA XL NH (6)
nh tia X (nh X-Quang)
nh X-Quang chp lng ngc nh X-Quang chp hm mt
23
NG DNG CA XL NH (7)
nh tia X (nh X-Quang)
H thng my chpnh ct lp CT
24
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
26
NG DNG CA XL NH (10)
nh hng ngoi
nh hng ngoichp u con mo nh hng ngoichp u con ch
27
NG DNG CA XL NH (11)
nh hng ngoi
nh hngngoi chp bmt tri t.Nhng ni cnh sng mnhl nhng ni cngun nhitln.
28
NG DNG CA XL NH (12)
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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NG DNG CA XL NH (13)
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
30
NG DNG CA XL NH (14)
D tm mt ngi
D tm v tr camt ngi, lmbc tin x lcho vic nhndng mt ngi
31
NG DNG CA XL NH (15)
D tm mt ngi
32
D tm mt ngi
NG DNG CA XL NH (16)
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NG DNG CA XL NH (17)
33
Trong vng nh sng nhn thy
Ci thin nh:
NG DNG CA XL NH (18)
34
Gim nhiu:
NG DNG CA XL NH (19)
35
Nhn dng ch vit:
NG DNG CA XL NH (20)
36
Nhn dng vn tay:
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NG DNG CA XL NH (21)
37
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.
38
39
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.
40
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, ...
42
CC KHI NIM C BN (4)
Mc xm (Gray)
43
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
44
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
47
CC KHI NIM C BN (5) nh ch mc: L nh mu s dng bng mu v bng ch mc
km theo.
48
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.
51
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.
52
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.
54
CC KHI NIM C BN (11)
Khi phc nh (Image Restoration)
55
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
56
CC KHI NIM C BN (13)
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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CC KHI NIM C BN (14)
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?
58
CC KHI NIM C BN (15)
Nn nh (Image Compression)
nh gcnh nn
(1/10)
nh nn
(1/20)nh nn
(1/30)
59
CC KHI NIM C BN (16)
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.
60
CC KHI NIM C BN (17)
Nhn dng vn tay
Kha nhn dng vn tayMy tnh c bo mt bi cng
ngh nhn dng vn tay
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CC KHI NIM C BN (18)
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
62
CC KHI NIM C BN (19)
Tra cu nh (Image Retrieval)
Tra cunh logo
63
CC KHI NIM C BN (20)
Tra cu nh (Image Retrieval)
Tra cunh logo
64
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
67
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
68
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
70
TNH LIN K (3)
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
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).
71
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)
72
TNH LIN K (5)
Bi tp:Cho tp d liu sau y v V = {1, 2, 5}, tm 8 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 (6)
Bi tp:Cho tp d liu sau y v V = {1, 2, 5}, tm 8 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
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)
74
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)
75
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).
76
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)
80
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.
82
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
83
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.
Tm ng i t p = (6,0) n q = (1,5) theo quan h 8 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:
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 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:
Tm ng i t p = (6,0) n q = (1,5) theo quan h 8 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 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
GN NHN CHO CC THNH PHNLIN THNG VI 4 LIN K (2)
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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GN NHN CHO CC THNH PHNLIN THNG VI 8 LIN K (3)
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
GN NHN CHO CC THNH PHNLIN THNG (2)
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
GN NHN CHO CC THNH PHNLIN THNG (3)
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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