Giao Trinh Xu Ly Anh

5/21/2018 Giao Trinh Xu Ly Anh

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I HC THI NGUYNKHOA CNG NGHTHNG TIN

GIO TRNH MN HC

XL NH

Ngi son : TS. NNG TON,

TS. PHM VIT BNH

Thi Nguyn, Thng 11 nm 2007


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LI NI U

Khong hn mi nm trli y, phn cng my tnh v cc thit blin quan c stin bvt bc vtc tnh ton, dung lng cha,khnng xl v.v.. v gi c gim n mc my tnh v cc thit blinquan n xl nh khng cn l thit bchuyn dng na. Khi nimnh s trnn thng dng vi hu ht mi ngi trong x hi v victhu nhn nh sbng cc thit bc nhn hay chuyn dng cng vi vica vo my tnh xl trnn n gin.

Trong hon cnh , xl nh l mt lnh vc ang c quan tm v trthnh mn hc chuyn ngnh ca sinh vin ngnh cng nghthngtin trong nhiu trng i hc trn cnc. Tuy nhin, ti liu gio trnhcn l mt iu kh khn. Hin ti chc mt st ti liu bng ting Anhhoc ting Php, ti liu bng ting Vit th rt him. Vi mong mun nggp vo snghip o to v nghin cu trong lnh vc ny, chng ti binson cun gio trnhXl nhda trn cng mn hc c duyt.Cun sch tp trung vo cc vn cbn ca xl nh nhm cung cpmt nn tng kin thc y v chn lc nhm gip ngi c c thttm hiu v xy dng cc chng trnh ng dng lin quan n xl nh.

Gio trnh c chia lm 5 chng v phn phlc: Chng 1, trnh

by Tng quan vxl nh, cc khai nim cbn, stng qut ca mththng xl nh v cc vn cbn trong xl nh. Chng 2, trnhby cc k thut nng cao cht lng nh da vo cc thao tc vi imnh, nng cao cht lng nh thng qua vic xl cc im nh trong lncn im nh ang xt. Chng ny cng trnh by cc kthut nng caocht lng nh nhvo cc php ton hnh thi. Chng 3, trnh by cc kthut cbn trong vic pht hin bin ca cc i tng nh theo chaikhuynh hng: Pht hin bin trc tip v pht hin bin gin tip. Chng4 thhin cch kthut tm xng theo khuynh hng tnh ton trc trungvv hng tip cn xp xnhcc thut ton lm mnh song song v gin

tip. V cui cng l Chng 5 vi cc kthut hu xl.Gio trnh c bin son da trn kinh nghim ging dy ca tc gi

trong nhiu nm ti cc kha i hc v cao hc ca H Cng ngh -HQG H Ni, H Khoa hc tnhin HQG H Ni, Khoa Cng nghthng tin H Thi Nguyn v.v.. Cun sch c thlm ti liu tham khocho sinh vin cc hks, cnhn v cc bn quan tm n vn nhndng v xl nh.


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Cc tc giby t lng bit n chn thnh ti cc bn ng nghiptrong Phng Nhn dng v cng nghtri thc, Vin Cng nghthng tin,Bmn Hthng thng tin, Khoa Cng nghthng tin, H Thi Nguyn,Khoa Cng nghthng tin, H Cng ngh, HQG H Ni, Khoa Ton C Tin, H Khoa hc tnhin, HQG H Ni ng vin, gp v

gip hon chnh ni dung cun sch ny. Xin cm n Lnh o KhoaCng nghthng tin, H Thi Nguyn, Ban Gim c H Thi Nguyn htrv to iu kin cho ra i gio trnh ny.

Mc d rt c gng nhng ti liu ny chc chn khng trnh khinhng sai st. Chng ti xin trn trng tip thu tt cnhng kin nggp ca bn c cng nhcc bn ng nghip c chnh l kp thi.

Thgp xin gi v: Phm Vit Bnh,Khoa Cng nghthng tin H Thi nguyn.X Quyt Thng, Tp. Thi Nguyn

in thoi: 0280.846506 Email: [email protected]

Thi Nguyn, ngy 22 thng 11 nm 2007

CC TC GI


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MC LC

LI NI U .......................................................................................................................................................................2MC LC ..................................................................................................................................................................................4

Chng 1: TNG QUAN VXL NH .....................................................................................7

1.1. XL NH, CC VN CBN TRONG XL NH ..................7

1.1.1. Xl nh l g? ............................................................................................................................................7

1.1.2. Cc vn cbn trong xl nh ........................................................................................71.1.2.1 Mt skhi nim cbn ........................................................................................................71.1.2.2 Nn chnh bin dng ....................................................................................................................8

1.1.2.3 Khnhiu.................................................................................................................................................91.1.2.4 Chnh mc xm: ...............................................................................................................................91.1.2.5 Trch chn c im ....................................................................................................................91.1.2.6 Nhn dng............................................................................................................................................ 101.1.2.7 Nn nh ................................................................................................................................................... 11

1.2. THU NHN V BIU DIN NH ........................................................................................... 11

1.2.1. Thu nhn, cc thit bthu nhn nh..................................................................................11

1.2.2. Biu din nh .............................................................................................................................................. 121.2.2.1. M hnh Raster............................................................................................................................. 12

1.2.2.2. M hnh Vector............................................................................................................................ 13

Chng 2: CC KTHUT NNG CAO CHT LNG NH...................14

2.1. CC KTHUT KHNG PHTHUC KHNG GIAN ..........................14

2.1.1. Gii thiu.........................................................................................................................................................14

2.1.2. Tng gim sng ...............................................................................................................................14

2.1.3. Tch ngng................................................................................................................................................ 15

2.1.4. B cm ............................................................................................................................................................... 152.1.5. Cn bng histogram ............................................................................................................................ 16

2.1.6. Kthut tch ngng tng ................................................................................................ 17

2.1.7. Bin i cp xm tng th ...........................................................................................................18

2.2. CC KTHUT PHTHUC KHNG GIAN .....................................................20

2.2.1. Php cun v mu .................................................................................................................................20


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2.2.2. Mt smu thng dng.................................................................................................................. 21

2.2.3. Lc trung v .................................................................................................................................................. 22

2.2.4. Lc trung bnh ........................................................................................................................................... 24

2.2.5. Lc trung bnh theo k gi trgn nht............................................................................ 25

2.3. CC PHP TON HNH THI HC ....................................................................................26

2.3.1. Cc php ton hnh thi cbn.............................................................................................. 26

2.3.2. Mt stnh cht ca php ton hnh thi.................................................................... 27

Chng 3: BIN V CC PHNG PHP PHT HIN BIN .....................32

3.1. GII THIU ............................................................................................................................................................ 32

3.2. CC PHNG PHP PHT HIN BIN TRC TIP .................................32

3.2.1. Kthut pht hin bin Gradient......................................................................................... 32

3.2.1.1. Kthut Prewitt .......................................................................................................................... 343.2.1.2. Kthut Sobel...............................................................................................................................353.2.1.3. Kthut la bn..............................................................................................................................35

3.2.2. Kthut pht hin bin Laplace ........................................................................................... 36

3.3. PHT HIN BIN GIN TIP....................................................................................................... 37

3.3.1 Mt skhi nim cbn ................................................................................................................. 37

3.3.2. Chu tuyn ca mt i tng nh....................................................................................... 38

3.3.3. Thut ton d bin tng qut .................................................................................................... 40

Chng 4: XNG V CC KTHUT TM XNG ........................................44

4.1. GII THIU ............................................................................................................................................................ 44

4.2. TM XNG DA TRN LM MNH ........................................................................... 44

4.2.1. Slc vthut ton lm mnh ........................................................................................... 44

4.2.2. Mt sthut ton lm mnh ...................................................................................................... 46

4.3. TM XNG KHNG DA TRN LM MNH................................................46

4.3.1. Khi qut vlc Voronoi................................................................................................. 47

4.3.2. Trc trung vVoronoi ri rc................................................................................................... 474.3.3. Xng Voronoi ri rc .................................................................................................................... 48

4.3.4. Thut ton tm xng ........................................................................................................................ 49

Chng 5: CC KTHUT HU XL..................................................................................52

5.1. RT GN SLNG IM BIU DIN..................................................................... 52

5.1.1. Gii thiu.........................................................................................................................................................52


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5.1.2. Thut ton Douglas Peucker..................................................................................................... 525.1.2.1. tng ................................................................................................................................................. 525.1.2.2. Chng trnh...................................................................................................................................53

5.1.3. Thut ton Band width .................................................................................................................... 54

5.1.3.1. tng ................................................................................................................................................. 545.1.3.2. Chng trnh...................................................................................................................................56

5.1.4. Thut ton Angles .................................................................................................................................575.1.4.1. tng ................................................................................................................................................. 575.1.4.2. Chng trnh...................................................................................................................................57

5.2. XP XA GIC BI CC HNH CS.................................................................... 58

5.2.1 Xp xa gic theo bt bin ng dng ........................................................................ 59

5.2.2 Xp xa gic theo bt bin aphin ......................................................................................62

5.3. BIN I HOUGH........................................................................................................................................ 63

5.3.1. Bin i Hongh cho ng thng....................................................................................... 63

5.3.2. Bin i Hough cho ng thng trong ta cc .......................................645.3.2.1. ng thng Hough trong ta cc ............................................................... 645.3.2.2. p dng bin i Hough trong pht hin gc nghing vn bn

..................................................................................................................... 65

PHLC ................................................................................................................................................................................ 68

TI LIU THAM KHO ....................................................................................................................................76


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Chng 1:

TNG QUAN VXL NH

1.1. XL NH, CC VN CBN TRONG XL NH

1.1.1. Xl nh l g?

Con ngi thu nhn thng tin qua cc gic quan, trong th gicng vai tr quan trng nht. Nhng nm trli y vi spht trin caphn cng my tnh, xl nh v ho pht trin mt cch mnh mv c nhiu ng dng trong cuc sng. Xl nh v hong mt vaitr quan trng trong tng tc ngi my.

Qu trnh xl nh c xem nhl qu trnh thao tc nh u vonhm cho ra kt qumong mun. Kt quu ra ca mt qu trnh xlnh c thl mt nh tt hn hoc mt kt lun.

Hnh 1.1.Qu trnh xl nh

nh c thxem l tp hp cc im nh v mi im nh c xemnhl c trng cng sng hay mt du hiu no ti mt vtr no ca i tng trong khng gian v n c thxem nhmt hm n binP(c1, c2,..., cn). Do , nh trong xl nh c thxem nhnh n chiu.

Stng qut ca mt hthng xl nh:

Hnh 1.2.Cc bc cbn trong mt hthng xl nh

1.1.2. Cc vn cbn trong xl nh

1.1.2.1 Mt skhi nim cbn

*nh v im nh:

Lu tr

Thu nhn nh(Scanner,

Camera,Sensor)Tin xl

Trch chnc im

Hquyt nh

i snh rtra kt lun

Huxl

XL NHnh

nhTt hn

Kt lu n


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im nh c xem nh l du hiu hay cng sng ti 1 totrong khng gian ca i tng v nh c xem nh l 1 tp hp ccim nh.

* Mc xm, mu

L scc gi trc thc ca cc im nh ca nh1.1.2.2 Nn chnh bin dng

nh thu nhn thng bbin dng do cc thit bquang hc v int.

nh thu nhn nh mong mun

Hnh 1.3. nh thu nhn v nh mong mun

khc phc ngi ta sdng cc php chiu, cc php chiu thngc xy dng trn tp cc im iu khin.

Gis(Pi, Pi) i = n,1 c n cc tp iu khin

Tm hm f: Pia f (Pi) sao cho

min)(2'

1

=ii

n

i

PPf

Gisnh bbin i chbao gm: Tnh tin, quay, tl, bin dngbc nht tuyn tnh. Khi hm f c dng:

f (x, y) = (a1x + b1y + c1, a2x + b2y + c2)

Ta c:

( ) ( )[ ]==

+++++==n

iiiiiii

n

i

ycybxaxcybxaPiPif1

2'222

2'111

2'

1

))((

cho min

Pi Pif(Pi)


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=++

=++

=++

=

=

=

= = =

= = ==

= = ==

n

i

n

i

n

iiii

n

i

n

i

n

iii

n

iiiii

n

i

n

i

n

iii

n

iiiii

xncybxa

xyycybyxa

xxxcyxbxa

c

b

a

1 1 1

'111

1 1 1

'

11

211

1 1 1

'

111

21

1

1

1

0

0

0

Gii hphng trnh tuyn tnh tm c a1, b1, c1

Tng ttm c a2, b2, c2

Xc nh c hm f

1.1.2.3 Khnhiu

C 2 loi nhiu cbn trong qu trnh thu nhn nh Nhiu h thng: l nhiu c quy lut c th kh bng cc php

bin i

Nhiu ngu nhin: vt bn khng r nguyn nhn khc phcbng cc php lc

1.1.2.4 Chnh mc xm:

Nhm khc phc tnh khng ng u ca h thng gy ra. Thngthng c 2 hng tip cn:

Gim smc xm: Thc hin bng cch nhm cc mc xm gnnhau thnh mt b. Trng hp ch c 2 mc xm th chnh lchuyn v nh en trng. ng dng: In nh mu ra my inen trng.

Tng smc xm: Thc hin ni suy ra cc mc xm trung gianbng k thut ni suy. K thut ny nhm tng cng mncho nh

1.1.2.5 Trch chn c im

Cc c im ca i tng c trch chn tutheo mc ch nhn

dng trong qu trnh x l nh. C th nu ra mt sc im ca nhsau y:

c im khng gian: Phn bmc xm, phn bxc sut, bin ,im un v.v..

c im bin i: Cc c im loi ny c trch chn bng victhc hin lc vng (zonal filtering). Cc bvng c gi l mt nc


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im (feature mask) thng l cc khe hp vi hnh dng khc nhau (chnht, tam gic, cung trn v.v..)

c im bin v ng bin:c trng cho ng bin ca itng v do vy rt hu ch trong vic trch trn cc thuc tnh bt binc dng khi nhn dng i tng. Cc c im ny c thc trchchn nh ton tgradient, ton t la bn, ton tLaplace, ton t chokhng (zero crossing) v.v..

Vic trch chn hiu qucc c im gip cho vic nhn dng cci tng nh chnh xc, vi tc tnh ton cao v dung lng nhlutrgim xung.

1.1.2.6 Nhn dng

Nhn dng t ng (automatic recognition), m t i tng, phnloi v phn nhm cc mu l nhng vn quan trng trong thgic my,c ng dng trong nhiu ngnh khoa hc khc nhau. Tuy nhin, mt cuhi t ra l: mu (pattern) l g? Watanabe, mt trong nhng ngi i utrong lnh vc ny nh ngha: Ngc li vi hn lon (chaos), mu lmt thc th(entity), c xc nh mt cch ang ng (vaguely defined) vc thgn cho n mt tn gi no . V dmu c thl nh ca vn tay,nh ca mt vt no c chp, mt chvit, khun mt ngi hocmt k tn hiu ting ni. Khi bit mt mu no , nhn dng hocphn loi mu c th:

Hoc phn loi c mu (supervised classification), chng hn phntch phn bit (discriminant analyis), trong mu u vo c nh danh

nhmt thnh phn ca mt lp xc nh.Hoc phn loi khng c mu (unsupervised classification hay

clustering) trong cc mu c gn vo cc lp khc nhau da trn mttiu chun ng dng no . Cc lp ny cho n thi im phn loi vncha bit hay cha c nh danh.

H thng nhn dng tng bao gm ba khu tng ng vi ba giaion chyu sau y:

1o. Thu nhn dliu v tin xl.

2

o

. Biu din dliu.3o. Nhn dng, ra quyt nh.

Bn cch tip cn khc nhau trong l thuyt nhn dng l:

1o. i snh mu da trn cc c trng c trch chn.

2o. Phn loi thng k.

3o. i snh cu trc.


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4o. Phn loi da trn mng n-ron nhn to.

Trong cc ng dng r rng l khng thchdng c mt cch tipcn n lphn loi ti u do vy cn sdng cng mt lc nhiuphng php v cch tip cn khc nhau. Do vy, cc phng thc phnloi thp hay c sdng khi nhn dng v nay c nhng kt quctrin vng da trn thit kcc hthng lai (hybrid system) bao gm nhium hnhkt hp.

Vic gii quyt bi ton nhn dng trong nhng ng dng mi, nysinh trong cuc sng khng ch to ra nhng thch thc v thut gii, mcn t ra nhng yu cu v tc tnh ton. c im chung ca tt cnhng ng dng l nhng c im c trng cn thit thng l nhiu,khng thdo chuyn gia xut, m phi c trch chn da trn cc thtc phn tch dliu.

1.1.2.7 Nn nhNhm gim thiu khng gian lu tr. Thng c tin hnh theo c

hai cch khuynh hng l nn c bo ton v khng bo ton thng tin.Nn khng bo ton th thng c khnng nn cao hn nhng khnngphc hi th km hn. Trn cshai khuynh hng, c 4 cch tip cn cbn trong nn nh:

Nn nh thng k: Kthut nn ny da vo vic thng k tn xutxut hin ca gi trcc im nh, trn cs m c chin lcm ha thch hp. Mt v din hnh cho k thut m ha ny

l *.TIF Nn nh khng gian: K thut ny da vo v tr khng gian ca

cc im nh tin hnh m ha. Kthut li dng sging nhauca cc im nh trong cc vng gn nhau. V dcho kthut nyl m nn *.PCX

Nn nh s dng php bin i: y l k thut tip cn theohng nn khng bo ton v do vy, kthut thng nn hiu quhn. *.JPG chnh l tip cn theo kthut nn ny.

Nn nh Fractal: Sdng tnh cht Fractal ca cc i tng nh,thhin slp li ca cc chi tit. Kthut nn stnh ton chcn lu trphn gc nh v quy lut sinh ra nh theo nguyn lFractal

1.2. THU NHN V BIU DIN NH

1.2.1. Thu nhn, cc thit bthu nhn nh


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Cc thit b thu nhn nh bao gm camera, scanner cc thit b thunhn ny c thcho nh en trng

Cc thit b thu nhn nh c 2 loi chnh ng vi 2 loi nh thngdng Raster, Vector.

Cc thit b thu nhn nh thng thng Raster l camera cc thit bthu nhn nh thng thng Vector l sensor hoc bn s ho Digitalizerhoc c chuyn i tnh Raster.

Nhn chung cc hthng thu nhn nh thc hin 1 qu trnh

Cm bin: bin i nng lng quang hc thnh nng lng in

Tng hp nng lng in thnh nh

1.2.2. Biu din nh

nh trn my tnh l kt quthu nhn theo cc phng php sho

c nhng trong cc thit b k thut khc nhau. Qu trnh lu trnhnhm 2 mc ch:

Tit kim bnh

Gim thi gian xl

Vic lu trthng tin trong bnhc nh hng rt ln n vic hinth, in n v xl nh c xem nhl 1 tp hp cc im vi cng kchthc nu sdng cng nhiu im nh th bc nh cng p, cng mn vcng th hin r hn chi tit ca nh ngi ta gi c im ny l phn gii.

Vic la chn phn gii thch hp tuthuc vo nhu cu sdngv c trng ca mi nh cth, trn cs cc nh thng c biudin theo 2 m hnh cbn

1.2.2.1. M hnh Raster

y l cch biu din nh thng dng nht hin nay, nh c biudin di dng ma trn cc im (im nh). Thng thu nhn qua ccthit bnhcamera, scanner. Tutheo yu cu thc thm mi im nhc biu din qua 1 hay nhiu bt

M hnh Raster thun li cho hin thv in n. Ngy nay cng nghphn cng cung cp nhng thit bthu nhn nh Raster ph hp vi tc nhanh v cht lng cao cho cu vo v u ra. Mt thun li cho vichin th trong mi trng Windows l Microsoft a ra khun dng nhDIB (Device Independent Bitmap) lm trung gian. Hnh 1.4 thhnh quytrnh chung hin thnh Raster thng qua DIB.


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Mt trong nhng hng nghin cu cbn trn m hnh biu din nyl kthut nn nh cc kthut nn nh li chia ra theo 2 khuynh hng lnn bo ton v khng bo ton thng tin nn bo ton c khnng phchi hon ton d liu ban u cn nu khng bo ton ch c kh nngphc hi sai scho php no . Theo cch tip cn ny ngi ta

ra nhiu quy cch khc nhau nhBMP, TIF, GIF, PCXHin nay trn thgii c trn 50 khun dng nh thng dng bao gm

ctrong cc kthut nn c khnng phc hi dliu 100% v nn ckhnng phc hi vi sai snhn c.

Hnh 1.4. Qu trnh hin thv chnh sa, lu trnh thng qua DIB

1.2.2.2. M hnh Vector

Biu din nh ngoi mc ch tit kim khng gian lu tr ddngcho hin th v in n cn m bo d dng trong la chn sao chp dichuyn tm kim Theo nhng yu cu ny kthut biu din vector trau vit hn.

Trong m hnh vector ngi ta s dng hng gia cc vector caim nh ln cn m ho v ti to hnh nh ban u nh vector c thu

nhn trc tip tcc thit bsho nhDigital hoc c chuyn i tnh Raster thng qua cc chng trnh sho

Cng nghphn cng cung cp nhng thit bxl vi tc nhanhv cht lng cho cu vo v ra nhng li chhtrcho nh Raster.

Do vy, nhng nghin cu vbiu din vectu tp trung tchuyni tnh Raster.

Hnh 1.5. Schuyn i gia cc m hnh biu din nh

BMP

PCC...

DIB Ca s

Thay i

Paint

RASTER VECTOR RASTERVecter

haRaster

ha


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Chng 2:

CC KTHUT NNG CAO CHT LNG NH

2.1. CC KTHUT KHNG PHTHUC KHNG GIAN

2.1.1. Gii thiu

Cc php ton khng ph thuc khng gian l cc php ton khngphc thuc vtr ca im nh.

V d: Php tng gim sng , php thng k tn sut, bin itn sut v.v..

Mt trong nhng khi nim quan trng trong xl nh l biu tn

sut (Histogram)Biu tn sut ca mc xm g ca nh I l sim nh c gi trg

ca nh I. K hiu l h(g)

V d:

1 2 0 4

1 0 0 7

I = 2 2 1 0

4 1 2 1

2 0 1 1

g 0 1 2 4 7

h(g) 5 7 5 2 1

2.1.2. Tng gim sng

Gista c I ~ kch thc m n v snguyn c

Khi , kthut tng, gim c sng c thhin

for (i = 0; i < m; i + +)for (j = 0; j < n; j + +)

I [i, j]= I [i, j]+ c;

Nu c > 0: nh sng ln

Nu c < 0: nh ti i


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2.1.3. Tch ngng

Gista c nh I ~ kch thc m n, hai sMin, Max v ngng khi : Kthut tch ngng c thhin

for (i = 0; i < m; i + +)

for (j = 0; j < n; j + +)I [i, j]= I [i, j]> = ? Max : Min;

* ng dng:

Nu Min = 0, Max = 1 kthut chuyn nh thnh nh en trng cng dng khi qut v nhn dng vn bn c thxy ra sai st nn thnh nhhoc nh thnh nn dn n nh bt nt hoc dnh.

2.1.4. B cm

Kthut nhm gim bt smc xm ca nh bng cch nhm li smc xm gn nhau thnh 1 nhm

Nu chc 2 nhm th chnh l kthut tch ngng. Thng thngc nhiu nhm vi kch thc khc nhau.

tng qut khi bin i ngi ta s ly cng 1 kch thcbunch_size

I [i,j]= I [i,j]/ bunch - size * bunch_size (i,j)

V d: B cm nh sau vi bunch_size= 3

1 2 4 6 7

2 1 3 4 5

I = 7 2 6 9 1

4 1 2 1 2

g

h(g)

0


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0 0 3 6 6

0 0 3 3 3

Ikq= 6 0 6 9 0

3 0 0 0 0

2.1.5. Cn bng histogram

nh I c gi l cn bng "l tng" nu vi mi mc xm g, g tac h(g) = h(g)

Gis, ta c nh I ~ kch thc m n

new_level ~ smc xm ca nh cn bng

levelnew

nmTB

_

= ~ sim nh trung bnh ca mi mc xm

ca nh cn bng

=

=g

i

ihgt0

)()(~ sim nh c mc xm g

Xc nh hm f: g a f(g)

Sao cho:

= 1

)(,0max)(

TB

gtroundgf

V d: Cn bng nh sau vi new_level= 4

1 2 4 6 7

2 1 3 4 5

I = 7 2 6 9 1

4 1 2 1 2

g h(g) t(g) f(g)

1 5 5 0

2 5 10 1

3 1 11 1

4 3 14 25 1 15 2

6 2 17 2

7 2 19 3

9 1 20 3


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0 1 2 2 3

1 0 1 2 2

Ikq= 3 1 2 3 0

2 0 1 0 1

Ch :nh sau khi thc hin cn bng cha chc l cn bng "l tng"

2.1.6. Kthut tch ngng tng

Ngng trong kthut tch ngng thng c cho bi ngi sdng. K thut tch ngng tng nhm tm ra ngng mt cch tng da vo histogram theo nguyn l trong vt l l vt thtch lm 2phn nu tng lnh trong tng phn l ti thiu.

Gis, ta c nh I ~ kch thc m n

G ~ l smc xm ca nh kckhuyt thiut(g) ~ sim nh c mc xm g

=

=g

i

ihigt

gm0

)(.)(

1)(

~ mmen qun tnh TB c mc xm g

Hm f: )(gfga

[ ]2)1()()(

)()(

= Gmgm

gtmxn

gtgf

Tm sao cho:( ) { })(max

10

gffGg


18/76

18

2 4 24 8 13 0,54 1.54

3 3 27 9 22 0,81 1.10

4 2 29 8 30 1,03 0.49

5 1 30 5 35 1,16

Ngng cn tch = 1 ng vi f()= 1.66

2.1.7. Bin i cp xm tng th

Nu bit nh v hm bin i th ta c thtnh c nh kt quv do ta sc c histogram ca nh bin i. Nhng thc tnhiu khi ta chbit histogram ca nh gc v hm bin i, cu hi t ra l liu ta c thc c histogram ca nh bin i. Nu c nhvy ta c thhiu chnhhm bin i thu c nh kt qu c phn b histogram nhmong mun.

Bi ton t ra l bit histogram ca nh, bit hm bin i hy vhistogram ca nh mi.

V d:

g 1 2 3 4

h(g) 4 2 1 2

g + 1 nu g 2

f(g)= g nu g = 3

g 1 nu g > 3

Bc 1:VHistogram ca nh c

g

f(g)

0


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19

Bc 2:Vthhm f(g)

Bc 3:VHistogram ca nh mi

t q = f(g)

h(q) = card ({P|I(P) = q})

= card ({P|I(P) = f(g)})

= card ({P|g = f-1(I(P))})

= )(1

)(qfi

ih

Histogram ca nh mi thua c bng cch chng hnh v tnh gi trtheo cc q (= f(g)) theo cng thc tnh trn. Kt qucui thu c sau phpquay gc 90 thun chiu kim ng h.

h(g) f(g)

0

g

h(g)

0


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20

2.2. CC KTHUT PHTHUC KHNG GIAN

2.2.1. Php cun v mu

Gista c nh I kch thc M N, mu T c kch thc m n khi, nh I cun theo mu T c xc nh bi cng thc.

( ) ( )jiTjyixIyxTIn

j

m

i

,*,),(1

0

1

0

=

=

++= (2.1)

Hoc ( ) ( )jiTjyixIyxTIn

j

m

i

,*,),(1

0

1

0

=

=

= (2.2)

VD:

1 2 4 5 8 72 1 1 4 2 2

I = 4 5 5 8 8 2

1 2 1 1 4 4

7 2 2 1 5 2

T = 1 0

0 1

( ) ( ) ( ) ( ) ( ) (1*1,10,0*,,*,),(1

0

1

0

TyxITyxIjiTjyixIyxTIji

+++=++=

==

( ) ( )1,1, +++= yxIyxI

2 3 8 7 10 *

7 6 9 12 4 * Tnh theo (2.1)

I T = 6 6 6 12 12 *3 4 2 6 6 *

* * * * * *

Tnh theo cng thc 2.2

* * * * * *

* 2 3 8 7 10

I T = * 7 6 9 12 4* 6 6 6 12 12

* 3 4 2 6 6


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21

* Nhn xt:

- Trong qu trnh thc hin php cun c mt sthao tc ra ngoi nh,nh khng c xc nh ti nhng vtr dn n nh thu c c kchthc nh hn.

- nh thc hin theo cng thc 2.1 v 2.2 chsai khc nhau 1 phpdch chuyn n gin ta shiu php cun l theo cng thc 2.1

2.2.2. Mt smu thng dng

- Mu:

1 1 1

T1= 1 1 1

1 1 1

~ Dng khnhiu Cc im c tn scao

VD1:

1 2 4 5 8 7

2 31 1 4 2 2

I = 4 5 5 8 8 2

1 2 1 1 4 4

7 2 2 1 5 2

55 65 45 46 * *

52 58 34 35 * *

I T1= 29 27 35 35 * ** * * * * *

* * * * * *

p dng kthut cng hng svi c = -27, ta c:

28 38 18 19 * *

25 31 7 8 * *

Ikq= 2 0 8 8 * *

* * * * * ** * * * * *

- Mu:

0 -1 0

T2= -1 4 -1

0 -1 0


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22

~ Dng pht hin cc im c tn scao

VD2:

114 -40 0 -14 * *

-22 5 14 16 * *

I T2 =-1 -6 -10 -2 * ** * * * * *

* * * * * *

2.2.3. Lc trung v

* nh ngha 2.1 (Trung v)

Cho dy x1; x2...; xnn iu tng (gim). Khi trung vca dy khiu l Med({xn}), c nh ngha:

+ Nu n l +12nx

+ Nu n chn:

2

nx hoc

+ 1

2

nx

* Mnh 2.1

min1

=

n

iixx ti { }( )nxMed

Chng minh

+ Xt trng hp n chn

t2

nM=

Ta c:

=

+==

+=M

iiM

M

ii

n

ii xxxxxx

111

( ) =

+

=

+ +=M

i

iiM

M

i

iMi xxxxxx11

( ) ( )[ ]=

+ +=M

iiMMM xxxx

11

{ }( ) { }( ) = =

+ +=M

i

M

iiiiiM xMedxxMedx

1 1


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23

{ }( )=

=n

iii xMedx

1

+ Nu n l:

B sung thm phn t { }( )ixMed vo dy. Theo trng hp n chn

ta c:

{ }( ) { }( )iin

ii xMedxMedxx +

=1min ti Med({xn})

=

n

iixx

1

min ti Med({xn})

* Kthut lc trung v

Gista c nh I ngng ca sW(P) v im nh P

Khi kthut lc trung vphthuc khng gian bao gm cc bccbn sau:

+ Bc 1:Tm trung v

{I(q)|q W(P)}Med (P)

+ Bc 2:Gn gi tr

=)(

)()(

PMed

PIPI

Nguoclai

PMedPI )()(

V d:

1 2 3 2

4 16 2 1

I = 4 2 1 1

2 1 2 1

W(3 3); = 21 2 3 2

4 2 2 1

Ikq

= 4 2 1 1

2 1 2 1

Gi tr16, sau php lc c gi tr2, cc gi trcn li khng thay igi tr.


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24

2.2.4. Lc trung bnh

* nh ngha 2.2 (Trung bnh)

Cho dy x1, x2, xn khi trung bnh ca dy k hiu AV({xn})

ddc nh ngha:{ }( )

=

=

n

iin xn

roundxAV1

1

* Mnh 2.2

( ) min2

1

=

n

iixx ti { }( )nxAV

Chng minh:

t: ( )= =n

iixxx

1

2)(

Ta c:

( )=

=n

iixxx

1

2)(

0)(' =x

( )= =n

i ixx1 0

{ }( )=

==n

iii xAVxn

x1

1

Mt khc, 02)('' >= nx

min ti { }( )ixAVx=

Kthut lc trung bnh

Gista c nh I, im nh P, ca sW(P) v ngng . Khi kthut lc trung bnh phthuc khng gian bao gm cc bc cbn sau:

+ Bc 1:Tm trung bnh

{I(q)|q W(P)}AV(P)


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25

+ Bc 2:Gn gi tr

=)(

)()(

PAV

PIPI

Nguoclai

PAVPI )()(

V d:

1 2 3 2

4 16 2 1

I = 4 2 1 1

2 1 2 1

W(3 3); = 21 2 3 2

4 3 2 1

Ikq= 4 2 1 12 1 2 1

Gi tr16 sau php lc trung bnh c gi tr3, cc gi trcn li ginguyn sau php lc.

2.2.5. Lc trung bnh theo k gi trgn nht

Gista c nh I, im nh P, ca sW(P), ngng v sk. Khi, lc trung bnh theo k gi trgn nht bao gm cc bc sau:

+ Bc 1: Tm K gi trgn nht

{I(q) q W(p)}{k gi trgn I(P) nht}+ Bc 2:Tnh trung bnh

{k gi trgn I(P) nht}AVk(P)

+ Bc 3:Gn gi tr

=)(

)()(

PAV

PIPI

k

Nguoclai

PAVPI k )()(

V d:

1 2 3 24 16 2 1

I = 4 2 1 1

2 1 2 1

W(3 3); = 2; k = 3


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26

1 2 3 2

4 8 2 1

Ikq= 4 2 1 1

2 1 2 1

* Nhn xt:

- Nu k ln hn kch thc ca s th k thut chnh l k thut lctrung bnh

- Nu k= 1 th nh kt qukhng thay i

Cht lng ca kthut phthuc vo sphn tla chn k.

2.3. CC PHP TON HNH THI HC

2.3.1. Cc php ton hnh thi cbn

Hnh thi l thut ngchsnghin cu vcu trc hay hnh hc topoca i tng trong nh. Phn ln cc php ton ca "Hnh thi" c nhngha thai php ton cbn l php "gin n" (Dilation) v php "co"(Erosion).

Cc php ton ny c nh ngha nhsau: Githit ta c i tngX v phn tcu trc (mu) B trong khng gian Euclide hai chiu. K hiuBxl dch chuyn ca B ti vtr x.

nh ngha 2.3 (DILATION)

Php "gin n" ca X theo mu B l hp ca tt ccc Bxvi x thucX. Ta c:

X B=UXx

xB

nh ngha 2.4 (EROSION)

Php "co" ca X theo B l tp hp tt ccc im x sao cho Bxnmtrong X. Ta c:

XB = {x : BxX}

V d: Ta c tp X nhsau: X =

00

000

000

00

00

xxx

xx

xx

xxx

xxx

B = x


27/76

27

X B =

xxxx

xxxx

xxx

xxxxx

xxxx

0

0

00

0

v XB =

000

00000

0000

0000

0000

xx

x

x

x

nh ngha 2.5 (OPEN)

Php ton m(OPEN) ca X theo cu trc B l tp hp cc im canh X sau khi co v gin nlin lip theo B. Ta c:

OPEN(X,B) = (XB) B

V d: Vi tp X v B trong v dtrn ta c

OPEN(X,B) =(XB) B =

0xxx0

00000

00xx00xx00

xx000

nh ngha 2.6 (CLOSE)

Php ton ng (CLOSE) ca X theo cu trc B l tp hp cc imca nh X sau khi gin nv co lin tip theo B. Ta c:

CLOSE(X,B) = (X B)B

Theo v dtrn ta c:

CLOSE(X,B) = (X B)B =

0xxx0

0xxx0

00xx0

xxxxx

xxxx0

2.3.2. Mt stnh cht ca php ton hnh thi

* Mnh 2.3[Tnh gia tng]:

(i) X X XB XB B

X B X B B

(ii) B B' XB XB' X

X B X B X


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28

Chng minh:

(i) X B = BXBBXx

xXx

x =

''

UU

XB = { } { }'// XBxXBx xx = XB

(ii) X B = '' BXBBXx

xXx

x =UU

Theo nh ngha:

XB = { } { }XBxXBx xx /'/ = XB .

*Mnh 2.4 [Tnh phn phi vi php ]:

(i) X (B B') = (X B) (X B')

(ii) X(B B') = (XB) (XB')

Chng minh:

(i) X (B B) = ( X B) (X B)

Ta c: B B B

X (B B) X B (tnh gia tng)

Tng t:

X ( B B) X B

X (B B) (X B) (X B) (2.3)

Mt khc,

y X (B B) x X sao cho y (B B)x

y Bx y X B

y Bx y X B

y (X B) (X B)

X (B B) (X B ) (X B) (2.4)

T(2.3) v (2.4) ta c: X (B B) = (X B) (X B)

(ii) X(B B) = (XB) (XB)

Ta c: B B B

X (B B) X B (tnh gia tng)

Tng t: X (B B) XB

X(B B) (XB) ( XB) (2.5)


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29

Mt khc,

x (XB) (XB)

Suy ra, x XB Bx X

x XB BxX

( B B)xX

x X(B B)

X(B B) (XB) (XB) (2.6)

T(2.5) v (2.6) ta c: X(B B) = (XB) (XB).

* ngha:

Ta c th phn tch cc mu phc tp tr thnh cc mu n ginthun tin cho vic ci t.

* Mnh 2.5 [Tnh phn phi vi php ]:

(X Y)B = (XB) (YB)

Chng minh:

Ta c, X Y X

(X Y)B XB

Tng t: (X Y)B YB

(X Y)B (XB) (YB) (2.7)

Mt khc,x (XB) (YB)

Suy ra x XB BxX

x YB BxY

Bx X Y

x ( X Y)B

(X Y)B (XB) (YB) (2.8)

T(2.7) v (2.8) ta c: (X Y)B = (XB) (YB).* Mnh 2.6[Tnh kt hp]

(i) (X B) B' = X (B B')

(ii) (XB)B' = X(B B')


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30

Chng minh:

(i) (X B) B' = X (B' B)

Ta c, (X B) B' = (UXx

xB

) B'

= UXx

x BB

)( ' = UXx

xBB

)( '

= X (B' B)

(i) (XB)B' = X(B B')

Trc ht ta i chng minh: 'xBXB xBB )( ' X

Tht vy, do 'xBXB nn y 'xB yXB

ByX

XBxBy

y U'

xBB )( ' X

Mt khc, xBB )(' X ( 'xBB) X

U'

xByyB

X

y 'xB ta c ByX

hay y 'x

B ta c y XB

Do , 'xBXB

Ta c, (XB)B' = { }XBx x / B'

= {x/ 'xB XB}

= {x/ xBB )(' X} (do chng minh trn)

= X(B B') .

*nh l 2.1 [X bchn bi cc cn OPEN v CLOSE]

Gis, X l mt i tng nh, B l mu, khi , X sbchn trnbi tp CLOSE ca X theo B v bchn di bi tp OPEN ca X theo B.Tc l:

(X B)B X (XB) B


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31

Chng minh:

Ta c: x X Bx X B (V X B = UXx

xB

)

x (X B)B (theo nh ngha php co)

(X B)B X (2.9)

Mt khc,

y (X B) B, suy ra:

x XB sao cho y Bx (V (XB) B = UBXx

xB )

BxX y X

Suy ra: X (XB) B (2.10)

T(2.9) v (2.10) Ta c: (X B)B X (XB) B .*Hqu2.1 [Tnh bt bin] :

(i) ((X B)B) B = X B

(ii) ((XB) B)B = XB

Chng minh:

(i) Tht vy, tnh l 2.1 ta c X (X B) B

X B ((X B)B) B (do tnh cht gia tng) (2.11)

Mt khc, cng tnh l 2.1 ta c (X

B) B X XDo , thay X bi X B ta c, ((X B)B) B X B (2.12)

T(2.11) v (2.12) Ta c: ((X B)B) B = X B

(ii) Tht vy, tnh l 2.1 ta c (XB) B X

((XB) B)B XB (do tnh cht gia tng) (2.13)

Mt khc, cng tnh l 2.1 ta c X (X B) B X

Do , thay X bi XB ta c, XB ((XB) B)B (2.14)

T(2.13) v (2.14) Ta c: ((X

B) B)

B = X

B (pcm).


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Chng 3:

BIN V CC PHNG PHP PHT HIN BIN

3.1. GII THIU

Bin l vn quan trng trong trch chn c im nhm tin ti hiu

nh. Cho n nay cha c nh ngha chnh xc v bin, trong mi ng

dng ngi ta a ra cc o khc nhau vbin, mt trong cc o

l o vsthay i t ngt vcp xm. V d: i vi nh en trng,

mt im c gi l im bin nu n l im en c t nht mt im

trng bn cnh. Tp hp cc im bin to nn bin hay ng bao ca

i tng. Xut pht t c s ny ngi ta thng s dng hai phngphp pht hin bin cbn:

Pht hin bin trc tip:Phng php ny lm ni bin da vo s

bin thin mc xm ca nh. K thut chyu dng pht hin bin

y l da vo s bin i cp xm theo hng. Cch tip cn theo o

hm bc nht ca nh da trn kthut Gradient, nu ly o hm bc hai

ca nh da trn bin i gia ta c kthut Laplace.

Pht hin bin gin tip: Nu bng cch no ta phn c nhthnh cc vng th ranh gii gia cc vng gi l bin. Kthut d bin

v phn vng nh l hai bi ton i ngu nhau v d bin thc hin phn

lp i tng m khi phn lp xong ngha l phn vng c nh v

ngc li, khi phn vng nh c phn lp thnh cc i tng, do c thpht hin c bin.

Phng php pht hin bin trc tip tra kh hiu quv t chu nh

hng ca nhiu, song nu sbin thin sng khng t ngt, phng

php tra km hiu qu, phng php pht hin bin gin tip tuy kh cit, song li p dng kh tt trong trng hp ny.

3.2. CC PHNG PHP PHT HIN BIN TRC TIP

3.2.1. Kthut pht hin bin Gradient


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33

Theo nh ngha, gradient l mt vctc cc thnh phn biu thtcthay i gi trca im nh, ta c:

Trong , dx, dy l khong cch (tnh bng sim) theo hng x vy.

* Nhn xt:

Tuy ta ni l ly o hm nhng thc cht chl m phng v xp xo hm bng cc kthut nhn chp (cun theo mu) v nh sl tn hiuri rc nn o hm khng tn ti.

V d: Vi dx = dy = 1, ta c:

( ) ( )

( ) ( )

+

+

yxfyxfy

f

yxfyxfx

f

,1,

,,1

Do , mt nnhn chp theo hng x l A= ( )11

v hng y l B=

1

1

Chng hn:0 0 0 0

0 3 3 3

I = 0 3 3 3

0 3 3 3

Ta c,

0 0 0 * 0 3 3 *

I A = 3 0 0 * ; I B= 0 0 0 *3 0 0 * 0 0 0 *

* * * * * * * *

0 0 0 *

I A + I B= 3 0 0 *

3 0 0 *

* * * *

dy yxfdyyxffyyyxf

dx

yxfydxxffx

x

yxf

),(),(),(

),(),(),(

+=

+=


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34

3.2.1.1. Kthut Prewitt

K thut sdng 2 mt nnhp chp xp xo hm theo 2 hng xv y l:

-1 0 1

Hx= -1 0 1-1 0 1

-1 -1 -1

Hy= 0 0 0

1 1 1

Cc bc tnh ton ca kthut Prewitt

+ Bc 1:Tnh I Hxv I Hy

+ Bc 2: Tnh I Hx+ I Hy

V d:

0 0 0 0 0 0

5 5 5 5 0 0

5 5 5 5 0 0

I = 5 5 5 5 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 -10 -10 * *

0 0 -15 -15 * *

I Hx= 0 0 -10 -10 * *0 0 -5 -5 * *

* * * * * *

* * * * * *

15 15 10 5 * *

0 0 0 0 * *

-15 -15 -10 -5 * *

I Hy= -15 -15 -10 -5 * ** * * * * *

* * * * * *


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35

15 15 0 -5 * *

0 0 -15 -15 * *

I Hx+ I Hy= -15 -15 -20 -15 * *-15 -15 -15 -10 * *

* * * * * ** * * * * *

3.2.1.2. Kthut Sobel

Tng tnhkthut Prewitt kthut Sobel sdng 2 mt nnhnchp theo 2 hng x, y l:

-1 0 1

Hx= -2 0 2

-1 0 1

-1 -2 -1Hy= 0 0 0

1 2 1

Cc bc tnh ton tng tPrewitt

+ Bc 1:Tnh I Hxv I Hy

+ Bc 2: Tnh I Hx+ I Hy

3.2.1.3. Kthut la bn

Kthut sdng 8 mt nnhn chp theo 8 hng 00

, 450

, 900

, 1350

,1800, 2250, 2700, 3150

5 5 -3 5 5 5

H1= 5 0 -3 H2= -3 0 -3

-3 -3 -3 -3 -3 -3

-3 5 5 -3 -3 5

H3= -3 0 5 H4= -3 0 5

-3 -3 -3 -3 -3 5

-3 -3 -3 -3 -3 -3H5= -3 0 5 H6= -3 0 -3

-3 5 5 5 5 5

-3 -3 -3 5 -3 -3

H7= 5 0 -3 H8= 5 0 -3

5 5 -3 5 -3 -3


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36

Cc bc tnh ton thut ton La bn

+ Bc 1:Tnh I Hi; i = 1,8

+ Bc 2: =

8

1iiHI

3.2.2. Kthut pht hin bin Laplace

Cc phng php nh gi gradient trn lm vic kh tt khi m sng thay i r nt. Khi mc xm thay i chm, min chuyn tip trirng, phng php cho hiu quhn l phng php sdng o hmbc hai Laplace.

Ton tLaplace c nh ngha nhsau:

Ta c:

( )),(),1(2

2

yxfyxfxx

f

xx

f+

=

[ ] [ ]),1(),(2),1(

),1(),(),(),1(

yxfyxfyxf

yxfyxfyxfyxf

+++

Tng t,

( )),()1,(22

yxfyxfyy

f

yy

f

+

=

[ ] [ ])1,(),(2)1,(

)1,(),(),()1,(

+++

yxfyxfyxf

yxfyxfyxfyxf

Vy: 2 f= f(x+1,y) + f(x,y+1) - 4f(x,y) + f(x-1,y) + f(x,y-1)Dn ti:

010

141010

H

=

Trong thc t, ngi ta thng dng nhiu kiu mt nkhc nhau xp x ri rc o hm bc hai Laplace. Di y l ba kiu mt nthng dng:

2

2

2

22

y

f

x

ff

+

=


37/76

37

VD: 0 0 0 0 0 0

5 5 5 5 0 0

I = 5 5 5 5 0 0

5 5 5 5 0 0

0 0 0 0 0 0

0 0 0 0 0 0

3.3. PHT HIN BIN GIN TIP

3.3.1 Mt skhi nim cbn

*nh v im nh

nh sl mt mng sthc 2 chiu (Iij) c kch thc (MN), trong mi phn tI ij(i = 1,...,M; j = 1,...,N) biu thmc xm ca nh ti (i,j)tng ng.

nh c gi l nh nh phn nu cc gi tr Iij ch nhn gi tr 0

hoc 1.y ta chxt ti nh nhphn v nh bt kc tha vdngnhphn bng kthut phn ngng. Ta k hiu l tp cc im vng(im en) v l tp cc im nn (im trng).

*Cc im 4 v 8-lng ging

Gi s (i,j) l mt im nh, cc im 4-lng ging l cc im ktrn, di, tri, phi ca (i,j):

N4(i,j) = {(i,j) : |i-i|+|j-j| = 1},

v nhng im 8-lng ging gm:N8(i,j) = {(i,j) : max(|i-i|,|j-j|) =1}.

Trong Hnh 1.2 biu din ma trn 8 lng ging knhau, cc im P0,P2, P4, P6l cc 4-lng ging ca im P, cn cc im P0, P1, P2, P3, P4, P5,P6, P7l cc 8-lng ging ca P.

=

=

=

121

242

121

H

111

181

111

H

010

141

010

H 321


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Hnh 1.3. Ma trn 8-lng ging knhau

*i tng nh

Hai im Ps, PeE, E hoc c gi l 8-lin thng (hoc 4-lin thng) trong E nu tn ti tp cc im c gi l ng i(io,jo)...(in,jn) sao cho (io,jo)= Ps, (in,jn)= Pe, (ir,jr) E v (ir,jr) l 8-lng ging(hoc 4-lng ging tng ng) ca (ir-1,jr-1) vi r = 1,2,...,n

Nhn xt: Quan hk-lin thng trong E (k=4,8) l mt quan hphn x,i xng v bc cu. Bi vy l mt quan h tng ng. Mi lptng ng c gi l mt thnh phn k-lin thng ca nh. Vsau ta sgi mi thnh phn k-lin thng ca nh l mt i tng nh.

3.3.2. Chu tuyn ca mt i tng nh

nh ngha 3.1: [Chu tuyn]

Chu tuyn ca mt i tng nh l dy cc im ca i tng nhP1,,Pn sao cho Piv Pi+1 l cc 8-lng ging ca nhau (i=1,...,n-1) v P1 l

8-lng ging ca Pn, i Q khng thuc i tng nh v Q l 4-lng gingca Pi (hay ni cch khc i th Pi l bin 4). K hiu .

Tng cc khong cch gia hai im ktip ca chu tuyn l di cachu tuyn v k hiu Len(C) v hng PiPi+1l hng chn nu Pi v Pi+1 lcc 4 lng ging (trng hp cn li th PiPi+1 l hng l).

Hnh 3.1 di y biu din chu tuyn ca nh, trong , P l im khiu chu tuyn.

Hnh 3.1. V dvchu tuyn ca i tng nh

P5 P7P6

P4 P0P

P3 P1P2

P


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nh ngha 3.2 [Chu tuyn i ngu]

Hai chu tuyn C= v C= c gi l i ngu

ca nhau nu v chnu i j sao cho:

(i)Piv Qjl 4-lng ging ca nhau.

(ii)

Cc im Pil vng th Qjl nn v ngc li.nh ngha 3.3 [Chu tuyn ngoi]

Chu tuyn C c gi l chu tuyn ngoi (Hnh 3.2a) nu v chnu

(i)Chu tuyn i ngu Cl chu tuyn ca cc im nn

(ii)di ca C nhhn di C

nh ngha 3.4 [Chu tuyn trong]

Chu tuyn C c gi l chu tuyn trong (Hnh 3.2b) nu v chnu:

(i)

Chu tuyn i ngu Cl chu tuyn ca cc im nn(ii)di ca C ln hn di CChu tuyn C

Chu tuyn C

Chu tuyn C

Chu tuyn C

a) Chu tuyn ngoi b) Chu tuyn trong

Hnh 3.2. Chu tuyn trong, chu tuyn ngoi

nh ngha 3.5 [im trong v im ngoi chu tuyn]

GisC= l chu tuyn ca mt i tng nh v P l mtim nh. Khi :

(i)Nu na ng thng xut pht tP sct chu tuyn C ti slln,th P c gi l im trong chu tuyn C v k hiu in(P,C)

(ii)

Nu PC v P khng phi l im trong ca C, th P c gi lim ngoi chu tuyn C v k hiu out(P,C).

B3.1 [Chu tuyn i ngu]

GisE l mt i tng nh v C= < P1P2..Pn> l chu tuyn caE, C= l chu tuyn i ngu tng ng. Khi :

(i)Nu C l chu tuyn trong th in(Qi,C) i (i=1,....,m)


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(ii)Nu C l chu tuyn ngoi th in(Pi,C) i (i=1,...,n)

B3.2 [Phn trong/ngoi ca chu tuyn]

GisE l mt i tng nh v C l chu tuyn ca E. Khi :

(i)Nu C l chu tuyn ngoi th x E sao cho xC, ta c in(x,C)

(ii)Nu C l chu tuyn trong th x E sao cho xC, ta c out(x,C)

nh l 3.1 [Tnh duy nht ca chu tuyn ngoi]

GisE l mt i tng nh v CEl chu tuyn ngoi ca E.Khi CEl duy nht.

3.3.3. Thut ton d bin tng qut

Biu din i tng nh theo chu tuyn thng da trn cc kthutd bin. C hai kthut d bin cbn. Kthut thnht xt nh bin thu

c tnh vng sau mt ln duyt nhmt th, sau p dng cc thutton duyt cnh th. Kthut thhai da trn nh vng, kt hp ng thiqu trnh d bin v tch bin. y ta quan tm cch tip cn thhai.

Trc ht, gisnh c xt chbao gm mt vng nh 8-lin thng, c bao bc bi mt vnh ai cc im nn. Dthy l mt vng 4-lin thng chl mt trng ring ca trng hp trn.

V c bn, cc thut ton d bin trn mt vng u bao gm ccbc sau:

Xc nh im bin xut pht

Dbo v xc nh im bin tip theo

Lp bc 2 cho n khi gp im xut pht

Do xut pht t nhng tiu chun v nh ngha khc nhau vimbin, v quan hlin thng, cc thut ton d bin cho ta cc ng binmang cc sc thi rt khc nhau.

Kt qutc ng ca ton td bin ln mt im bin ril im binri+1 (8-lng ging ca ri). Thng thng cc ton tny c xy dng nhmt hm i sBoolean trn cc 8-lng ging ca ri. Mi cch xy dng

cc ton tu phthuc vo nh ngha quan hlin thng v im bin.Do s gy kh khn cho vic kho st cc tnh cht ca ng bin.Ngoi ra, v mi bc d bin u phi kim tra tt ccc 8-lng ging cami im nn thut ton thng km hiu qu. khc phc cc hn chtrn, thay v sdng mt im bin ta sdng cp im bin (mt thuc ,mt thuc ), cc cp im ny to nn tp nn vng, k hiu l NV vphn tch ton td bin thnh 2 bc:


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Xc nh cp im nn vng tip theo.

La chn im bin

Trong bc thnht thc hin chc nng ca mt nh xtrn tpNV ln NV v bc thhai thc hin chc nng chn im bin.

Thut ton d bin tng qutBc 1: Xc nh cp nn-vng xut pht

Bc 2: Xc nh cp nn-vng tip theo

Bc 3: La chn im bin vng

Bc 4: Nu gp li cp xut pht th dng, nu khng quay libc 2.

Vic xc nh cp nn-vng xut pht c thc hin bng cch duytnh ln lt ttrn xung di v ttri qua phi ri kim tra iu kin la

chn cp nn-vng. Do vic chn im bin chmang tnh cht quy c, nnta gi nh xxc nh cp nn-vng tip theo l ton td bin.

nh ngha 3.6 [Ton td bin]

GisT l mt nh xnhsau: T: NV NV

(b,r) a (b,r)

Gi T l mt ton td bin csnu n thomn iu kin: b,r lcc 8-lng ging ca r.

Gi s (b,r) NV; gi K(b,r) l hm chn im bin. Bin ca mtdng c thnh ngha theo mt trong ba cch:

Tp nhng im thuc c mt trn NV, tc l K(b,r)= r

Tp nhng im thuc c trn NV, tc l K(b,r)= b

Tp nhng im o nm gia cp nn-vng, tc l K(b,r) l nhngim nm gia hai im b v r.

Cch nh ngha thba tng ng mi cp nn-vng vi mt imbin. Cn i vi cch nh ngha th nht v th hai mt s cp nn-vng c thc chung mt im bin. Bi vy, qu trnh chn im bin

c thc hin nhsau:i:= 1; (bi,ri):= (bo,ro);

WhileK(bi,ri)K(bn,rn) and i8 do

Begin (bi+1,ri+1)= T(bi,ri); i:= i+1; End;

iu kin dng

Cp nn-vng thn trng vi cp nn vng xut pht: (bn,rn)= (bo,ro)


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* Xc nh cp nn vng xut pht

Cp nn vng xut pht c xc nh bng cch duyt nh ln lt ttrn xung di v ttri sang phi im em u tin gp c cng viim trng trc (theo hng 4) to nn cp nn vng xut pht.

* Xc nh cp nn vng tip theou vo: pt, dir

V d: (3, 2) 4

Point orient []= {(1,0);(1;-1);(0;-1);(-1;-1);(-1;0);(-1,1);(0,1);(1,1)};

//Hm tm hng c im en gn nht

BYTE GextNextDir(POINT pt, BYTE dir)

{

BYTE pdir= (dir + 7)%8;

do{

if(getpixel(pt. x+orient [pdir]. x,pt.y+orient [pdir]. y))==BLACK)

return pdir;

pdir = (pdir + 7) %8;

}while(pdir ! = dir);

return. ERR; //im c lp

}

//Gn gi trcho bc tip theo

pdir = GetNextDir(pt, dir);

if(pdir==ERR) //Kim tra c l im c lp khng?

return. ERR; //im c lp

pt. x = pt. x + orient [pdir]. x;

pt. y = pt. y + orient [pdir]. y ;

tnh gi trcho hng tip theo ta lp bng da trn gi trpdir

tnh c trc theo cc khnng c thxy ra:


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pdir im trng trc Trng so vi en mi

0 1 2

1 2 4

2 3 4

3 4 6

4 5 6

5 6 0

6 7 0

7 0 2

Do cng thc tnh hng tip theo sl :

dir= ((pdir+3)/ 2 * 2)%8 ;


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Chng 4:

XNG V CC KTHUT TM XNG

4.1. GII THIU

Xng c coi nhhnh dng cbn ca mt i tng, vi stcc im nh cbn. Ta c thly c cc thng tin vhnh dng nguynbn ca mt i tng thng qua xng.

Mt nh ngha xc tch vxng da trn tnh continuum (tng tnhhin tng chy ng c) c a ra bi Blum (1976) nhsau: Githit rng i tng l ng nht c phbi ckh v sau dng lnmt vng bin la. Xng c nh ngha nhni gp ca cc vt la v

ti chng c dp tt.

a) nh gc b) nh xng

Hnh 4.1.V dvnh v xng

K thut tm xng lun l ch nghin cu trong x l nhnhng nm gn y. Mc d c nhng n lc cho vic pht trin ccthut ton tm xng, nhng cc phng php c a ra u b mtmt thng tin. C thchia thnh hai loi thut ton tm xng cbn:

Cc thut ton tm xng da trn lm mnh

Cc thut ton tm xng khng da trn lm mnh

4.2. TM XNG DA TRN LM MNH

4.2.1. Slc vthut ton lm mnh

Thut ton lm mnh nh snhphn l mt trong cc thut ton quantrng trong xl nh v nhn dng. Xng cha nhng thng tin bt binvcu trc ca nh, gip cho qu trnh nhn dng hoc vectho sau ny.


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Thut ton lm mnh l qu trnh lp duyt v kim tra tt ccc imthuc i tng. Trong mi ln lp tt ccc im ca i tng sckim tra: nu nhchng thomn iu kin xo no tuthuc vo mithut ton th n sbxo i. Qu trnh clp li cho n khi khng cnim bin no c xo. i tng c bc dn lp bin cho n khi no

bthu mnh li chcn cc im bin.Cc thut ton lm mnh c phn loi da trn phng php xl

cc im l thut ton lm mnh song song v thut ton lm mnh tun t.

Thut ton lm mnh song song, l thut ton m trong cc imc xl theo phng php song song, tc l c xl cng mt lc.Gi trca mi im sau mt ln lp chphthuc vo gi trca cc lngging bn cnh (thng l 8-lng ging) m gi tr ca cc im ny c xc nh trong ln lp trc . Trong my c nhiu bvi xl mivi xl sxl mt vng ca i tng, n c quyn c tcc im

vng khc nhng chc ghi trn vng ca n xl.Trong thut ton lm mnh tun tcc im thuc i tng sc

kim tra theo mt thtno (chng hn cc im c xt ttri quaphi, t trn xung di). Gi tr ca im sau mi ln lp khng nhngph thuc vo gi trca cc lng ging bn cnh m cn ph thuc vocc im c xt trc trong chnh ln lp ang xt.

Cht lng ca thut ton lm mnh c nh gi theo cc tiuchun c lit k di y nhng khng nht thit phi thomn ngthi tt ccc tiu chun.

Bo ton tnh lin thng ca i tng v phn b ca i tng Stng hp gia xng v cu trc ca nh i tng

Bo ton cc thnh phn lin thng

Bo ton cc im ct

Xng chgm cc im bin, cng mnh cng tt

Bn vng i vi nhiu

Xng cho php khi phc nh ban u ca i tng

Xng thu c chnh gia ng nt ca i tng clm mnh

Xng nhn c bt bin vi php quay.


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4.2.2. Mt sthut ton lm mnh

Trong phn ny im qua mt sc im, u v khuyt im ca ccthut ton c nghin cu.

1o. Thut ton lm mnh cin l thut ton song song, to ra xng8 lin thng, tuy nhin n rt chm, gy t nt, xo hon tonmt scu hnh nh.

2o. Thut ton lm mnh ca Toumazet bo ton tt ccc im ctkhng gy t nt i tng. Tuy nhin, thut ton c nhc iml rt chm, rt nhy cm vi nhiu, xng chl 4-lin thng vkhng lm mnh c vi mt scu hnh phc tp

3o. Thut ton lm mnh ca Y.Xia da trn ng bin ca itng, c thci t theo cphng php song song v tun t.Tc ca thut ton rt nhanh. N c nhc im l gy t nt,xng to ra l xng gi(c dy l 2 phn tnh).

4o. Thut ton lm mnh ca N.J.Naccache v R.Shinghal. Thut tonc u im l nhanh, xng to ra c khnng khi phc nh banu ca i tng. Nhc im chnh ca thut ton l rt nhyvi nhiu, xng nhn c phn nh cu trc ca i tng thp.

5o. Thut ton lm mnh ca H.E.Lu P.S.P Wang tng i nhanh,gi c tnh lin thng ca nh, nhng li c nhc im lxng to ra l xng 4-lin thng v xo mt mt s cuhnh nh.

6o. Thut ton lm mnh ca P.S.P Wang v Y.Y.Zhang da trnng bin ca i tng, c thci t theo phng php songsong hoc tun t, xng l 8-lin thng, t chu nh hng canhiu. Nhc im chnh ca thut ton l tc chm.

7o. Thut ton lm mnh song song thun tu nhanh nht trong ccthut ton trn, bo ton tnh lin thng, t chu nh hng canhiu. Nhc im l xo hon ton mt scu hnh nh, xngto ra l xng 4-lin thng.

4.3. TM XNG KHNG DA TRN LM MNHtch c xng ca i tng c thsdng ng bin ca i

tng. Vi im p bt k trn i tng, ta bao n bi mt ng bin.Nu nhc nhiu im bin c cng khong cch ngn nht ti p th p nmtrn trc trung v. Tp tt ccc im nhvy lp thnh trc trung vhayxng ca i tng. Vic xc nh xng c tin hnh thng quahai bc:


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Bc thnht, tnh khong cch tmi im nh ca i tng nim bin gn nht. Nhvy cn phi tnh ton khong cch ti tt ccc im bin ca nh.

Bc thhai, khong cch nh c tnh ton v cc im nh cgi trln nht c xem l nm trn xng ca i tng.

4.3.1. Khi qut vlc Voronoi

Lc Voronoi l mt cng chiu qutrong hnh hc tnh ton.Cho hai im Pi, Pjl hai phn tca tp gm n im trong mt phng.Tp cc im trong mt phng gn Pihn Pj l na mt phng H(Pi, Pj)cha im Piv bgii hn bi ng trung trc ca on thng P iPj. Do, tp cc im gn Pihn bt kim Pjno c ththu c bng cchgiao n-1 cc na mt phng H(Pi, Pj):

V(Pi) = H(Pi, Pj) ij (i= 1,...,n) (4.1)

nh ngha 4.1 [a gic/SVoronoi]

SVoronoi ca l hp ca tt ccc V(Pi)

Vor() = V(Pi) Pi(l mt a gic) (4.2)

nh ngha 4.2 [a gic Voronoi tng qut]

Cho tp cc im , a gic Voronoi ca tp con U ca c nhngha nhsau:

V(U) = {P|v U, w \ U : d(P,v) < d(P,w)}

= V(Pi) PiU (4.3)

4.3.2. Trc trung vVoronoi ri rc

nh ngha 4.3 [Bn khong cch - Distance Map]

Cho i tng S, i vi mi (x, y)S, ta tnh gi tr khong cchmap(x, y) vi hm khong cch d(.,.) nhsau:

(x, y)S: map(x, y) = min d[(x, y), (xi, yi)] (4.4)

trong (xi, yi) B(S) - tp cc im bin ca S

Tp tt ccc map(x, y), k hiu l DM(S), c gi l bn khongcch ca S.

Ch : Nu hm khong cch d(.,.) l khong cch Euclide, th phngtrnh (4.4) chnh l khong cch ngn nht tmt im bn trong i tngti bin. Do , bn khong cch c gi l bn khong cchEuclide EDM(S) ca S. nh ngha trn c dng cho chnh ri rc lnlin tc.

i


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nh ngha 4.4 [Tp cc im bin sinh]

Cho map(x, y) l khong cch ngn nht t(x, y) n bin (theo nhngha 4.3). Ta nh ngha: map-1(x, y) = {p| p B(S), d(p,(x, y)):=map(x, y)}

Khi tp cc im bin sinh ^B(S) c nh ngha bi:^B(S) = map-1(x, y), (x, y)S (4.5)

Do S c thcha cc ng bin ri nhau, nn ^B(S) bao gm nhiutp con, mi tp m tmt ng bin phn bit:

^B(S)={B1(S),..BN(S)} (4.6)

nh ngha 4.5[Trc trung vVoronoi ri rc (DVMA)]

Trc trung vVoronoi ri rc c nh ngha l kt qu ca sVoronoi bc nht ri rc ca tp cc im bin sinh giao vi hnh sinh S :

DVMA(^B(S)) = Vor(^B(S)) S (4.7)

4.3.3. Xng Voronoi ri rc

nh ngha 4.6 [Xng Voronoi ri rc - DiscreteVoronoi Skeleton]

Xng Voronoi ri rc theo ngng T, k hiu l SkeDVMA(^B(S),T)(hoc Ske(^B(S),T)) l mt tp con ca trc trung vVoronoi:

SkeDVMA(^B(S),T)= {(x,y)| (x,y)DVMA(^B(S)), (x,y) > T} (4.8)

: l hm hiu chnh.

D thy nu ngng T cng ln th cng th s lng im tham giatrong xng Vonoroi cng t (Hnh 4.2).

Hnh 4.2. Xng Voronoi ri rc nh hng ca cc hm hiu chnh khc nhau.(a) nh nhphn. (b) SVoronoi. (c) Hiu chnh bi hm Potential, T=9.0.

(d) Hiu chnh bi hm Potential, T=18.0

a) b)

c) d)


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4.3.4. Thut ton tm xng

Trong mc ny strnh by tng cbn ca thut ton tm xngv m tbng ngn ngta Pascal.

Tng trng: Vic tnh ton sVoronoi c bt u tmt im

sinh trong mt phng. Sau im sinh thhai c thm vo v qu trnhtnh ton tip tc vi a gic Voronoi tm c vi im va c thmvo . Cnhth, qu trnh tnh ton sVoronoi c thc hin chon khi khng cn im sinh no c thm vo. Nhc im ca chinlc ny l mi khi mt im mi c thm vo, n c thgy ra sphnvng ton bcc a gic Voronoi c tnh.

Chia tr: Tp cc im bin u tin c chia thnh hai tp imc kch cbng nhau. Sau thut ton tnh ton sVoronoi cho chaitp con im bin . Cui cng, ngi ta thc hin vic ghp chai s

Voronoi trn thu c kt qumong mun. Tuy nhin, vic chia tp ccim bin thnh hai phn khng phi c thc hin mt ln, m c lpli nhiu ln cho n khi vic tnh ton sVoronoi trnn n gin. Vth, vic tnh sVoronoi trthnh vn lm thno trn hai sVoronoi li vi nhau.

Thut ton strnh by y l skt hp ca hai tng trn. Tuynhin, n smang nhiu dng dp ca thut ton chia tr.

Hnh 4.3 minh ho tng ca thut ton ny. Mi mt im binc chia thnh hai phn (bn tri: 1- 6, bn phi: 7-11) bi ng gp

khc , v hai sVoronoi tng ng Vor(SL) v Vor(SR). thu csVornonoi Vor(SLSR), ta thc hin vic trn hai strn v xcnh li mt sa gic sbsa i do nh hng ca cc im bn cnhthuc skia. Mi phn tca sl mt bphn ca ng trung trcni hai im m mt im thuc Vor(SL) v mt thuc Vor(SR). Trc khixy dng , ta tm ra phn tu v cui ca n. Nhn vo hnh trn, tanhn thy rng cnh 1v 5l cc tia. Dnhn thy rng vic tm ra cccnh u v cui ca trthnh vic tm cnh vo tv cnh ra t.

Hnh 4.3.Minh hothut ton trn hai sVoronoi

1

2

4

3

6

5

7

9

8

11

10

CH(SR)

CH(SL)


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Sau khi tm c tv t, cc im cui ca tc sdng xydng phn tu tin ca (1trong hnh trn). Sau thut ton tm imgiao ca vi Vor(SL) v Vor(SR). Trong v d trn, u tin giao viV(3). Kty, cc im nm trn phn ko di sgn im 6 hn im3. Do , phn ttip theo 2ca sthuc vo ng trung trc ca im6 v im 7. Sau im giao tip theo ca sthuc v Vor(SL); bygisi vo V(9) v 2sc thay thbi 3. Qu trnh ny skt thckhi gp phn tcui 5.

Trn y chl minh hocho thut trn hai sVoronoi trong chinlc chia tr. Tuy nhin, trong thut ton s trnh by y th s thchin c khc mt cht. Tp cc im nh khng phi c a vo ngay tu m sc qut vo tng dng mt. Gisti bc thi, ta thu cmt sVoronoi gm i-1 hng cc im sinh Vor(Si-1). Tip theo, ta qutly mt hng Licc im nh t tp cc im bin cn li. Thc hin vic

tnh sVoronoi Vor(Li) cho hng ny, sau trn Vor(Si-1) vi Vor(Li).Kt quta sc mt smi, v li thc hin vic qut hng Li+1 ccim sinh cn li v.v.. Qu trnh ny skt thc khi khng cn im bin no thm vo sVoronoi. Do Vor(Li) sc dng rng lc (nu Lic kim th Vor(Li) sgm k-1 ng thng ng), nn vic trn Vor(Si-1) viVor(Li) c phn n gin hn.

Hnh 4.4.Minh hothut ton thm mt im bin vo sVoronoi

Gii thut trn c thc m tbng ngn ngta Pascal nhsau:

ProcedureVORONOI

(*Si: Tp cc im ca i dng qut u tin,

0


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Begin

i:=0; Si:=rng;

While(i


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Chng 5:

CC KTHUT HU XL

5.1. RT GN SLNG IM BIU DIN

5.1.1. Gii thiu

Rt gn slng im biu din l kthut thuc phn hu xl. Ktquca phn d bin hay trch xng thu c 1 dy cc im lin tip.Vn t ra l hiu c thb bt cc im thu c gim thiu khngquan lu trv thun tin cho vic i sch hay khng.

Bi ton:

Cho ng cong gm n im trong mt phng (x1, y1), (x2, y2)(xn,yn). Hy bbt 1 sim thuc ng cong sao cho ng cong minhn c l (Xi1; Yi1), (Xi2; Yi2) (Xim; Yim) gn ging vi ng congban u.

* Mt so gn ging

+ Chiu di (chiu rng) ca hnh chnht nh nht cha ng cong

+ Khong cch ln nht tng cong n on thng ni 2 u mtca ng cong

+ Tlgia chiu di v chiu rng ca hnh chnht nh nht chang con

+ Sln ng cong ct on thng ni 2 u mt

5.1.2. Thut ton Douglas Peucker

5.1.2.1. tng

Hnh 5.1. n gin ha ng cng theo thut ton Douglas Peucker

tng c bn ca thut ton Douglas-Peucker l xt xem khongcch ln nht tng cong ti on thng ni hai u mt ng cong(xem Hnh 5.1) c ln hn ngng khng. Nu iu ny ng th im xanht c gili lm im chia ng cong v thut ton c thc hin

h >


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tng tvi hai ng cong va tm c. Trong trng hp ngc li,kt quca thut ton n gin ho l hai im u mt ca ng cong.

Thut ton Douglas-Peucker:

Bc 1: Chn ngng .

Bc 2: Tm khong cch ln nht tng cong ti on thngni hai u on ng cong h.

Bc 3: Nu h th dng.

Bc 4: Nu h > th gili im t cc i ny v quay trlibc 1.

Nhn xt:Thut ton ny tra thun li i vi cc ng cong thu nhnc m gc l cc on thng, ph hp vi vic n gin ho trong qutrnh vctcc bn vkthut, sthit kmch in v.v..

5.1.2.2. Chng trnh//Hm tnh ng cao tdinh n on thng ni hai im dau, cuoi

float Tinhduongcao (POINT dau, POINT cuoi, POINT dinh)

{

floot h;

tnh ng cao

returm h ;

}//Hm quy nhm nh du loi bcc im trong ng cong

void DPSimple(POINT *pLINE,int dau,int cuoi,BOOL *chiso,float )

{

int i, index = dau;

float h, hmax = 0;

for(i = dau + 1; i < cuoi; i++)

{

h= Tinhduongcao(pLINE[dau], pLINE[cuoi]; pLINE[i]);

if(h > hmax)

{

hmax = h;

index = i;


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}

}

if(hmax )

for(i= dau + 1; i < cuoi, i++)

chiso[i]= FALSE;

else

{

DPSimple(PLINE, dau, index, chiso, );

DPSimple(PLINE, index, cuoi, chiso, ) ;

}

}

//Hm rt gn slng im DouglasPeucker

int DouglasPeucker(POINT *pLINE, int n, float )

{

int i, j;

BOOL chiso [MAX_PT];

for(i = 0; i < m; i++) //Tt ccc im c gili

chiso[i]= TRUE;

DPSimple(pLINE, 0, n 1, chiso, );for(i = j = 0; i < n; i ++)

if (chiso [i]==TRUE)

pLINE[j++]= pLINE[i];

return j;

}

5.1.3. Thut ton Band width

5.1.3.1. tng

Trong thut ton Band Width, ta hnh dung c mt di bng di chuynt u mt ng cong dc theo ng cong sao cho ng cong nmtrong di bng cho n khi c im thuc ng cong chm vo bin cadi bng, im ny sc gili. Qu trnh ny c thc hin vi phncn li ca ng cong bt u t im va tm c cho n khi htng cong. Cthnhsau:


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Hnh 5.2. n gin ha ng cong vi thut ton Band Width

Bt u bng vic xc nh im u tin trn ng cong v coi nhl mt im cht (P1). im thba (P3) c coi l im ng. imgia im cht v im ng (P2) l im trung gian. Ban u khong cchtim trung gian n on thng ni im cht v im ng c tnhton v kim tra. Nu khong cch tnh c ny nhhn mt ngng cho trc th im trung gian c thbi, tin trnh tip tc vi im chtl im cht c, im trung gian l im ng cv im ng l im k

tip sau im ng c. Trong trng hp ngc li, khong cch tnh cln hn ngng cho trc th im trung gian sc gili, tin trnhtip tc vi im cht l in trung gian, im trung gian l im ng cv im ng l im k tip sau im ng c. Tin trnh c lp chon ht ng cong (Hnh 5.2 minh ha thut ton Band-Width).

Thut ton Band-Width:

Bc 1: Xc nh im u tin trn ng cong v coi nhlmt im cht (P1). im th ba (P3) c coi l imng. im gia im cht v im ng (P2) l im

trung gian. Bc 2: Tnh khong cch tim trung gian n on thng ni

hai im cht v im ng.

Bc 3: Kim tra khong cch tm c nu nhhn mt ngngcho trc th im trung gian c thbi. Trong trnghp ngc li im cht chuyn n im trung gian.

Bc 4: Chu trnh c lp li th im trung gian c chuynn im ng v im k tip sau im ng c chnh lm im ng mi..

Nhn xt:Thut ton ny tng tc trong trng hp ng ng chanhiu im, iu c ngha l lch gia cc im trong ng thng lnh, hay dy nt ca ng c vctho l mnh.

di

P3

P2 P4

dk

P5P1


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5.1.3.2. Chng trnh

//Hm tnh ng cao tnh n on thng ni hai im dau, cuoi

float Tinhduongcao(POINT dau, POINT cuoi, POINT dinh)

{

floot h;

tnh ng cao

returm h ;

}

//Hm quy nhm nh du loi bcc im trong ng cong

void BWSimple(POINT *pLINE, int chot, int tg, BOOL *chiso,

float , int n)

{

if(Tinhduongcao(pLINE[chot], pLINE[tg+1], pLINE[tg]) )

chiso[tg]= 0;

else

chot = tg;

tg = tg + 1

if(tg < n - 1)

BWSimple (pLINE, chot, tg, chiso, , n) ;}

//Hm rt gn slng im BandWidth

int BandWidth(POINT *pLINE, int n, floot )

{

int i, j;

BOOL chiso [MAX_PT];

for (i = 0; i < n; i++)chiso[i]= TRUE; //Tt ccc im c gili

BWSimple(pLINE, 0, 1, chiso, , n);

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

if(chiso [i]== TRUE)


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pLINE [j ++1]= pLINE [i];

return j;

}

5.1.4. Thut ton Angles5.1.4.1. tng

Tng t nh thut ton Band Width nhng thay vic tnh tonkhong cch bi tnh gc. C th thut ton bt u vi im u ngcong (P1) l im cht.

Hnh 5.3. n gin ha ng cong vi thut ton Angles

im th3 ca ng cong (P3) l im ng, im gia im cht vim ng (P2) l im trung gian

Gc to bi im cht, trung gian, ng vi im trung gian l nhvic tnh ton v kim tra

Nu th im trung gian c thbi trong trng hp ngc li im

cht sl im trung gian cv qu trnh lp vi im trung gian l imng c, im ng mi l im ktip sau im ng c. Tin trnh thchin cho n ht ng cong.

5.1.4.2. Chng trnh

//Hm tnh ng cao tnh n on thng ni hai im dau, cuoi

float Tinhgoc(POINT dau, POINT cuoi, POINT dinh)

{

float ;

tinhgoc (tvit)return ;

}

//Hm quy nhm nh du loi bcc im trong ng cong

void ALSimple(POINT *pLINE,int chot,int tg,BOOL *chiso,float ,int n)

{

P1

P3

P2 P4

k

P5


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if(Tinhgoc(pLINE[chot], pLINE[tg], pLINE[tg+1]) > )

chiso[tg]= FALSE;

else

chot = tg;

tg = tg + 1;

if(tg < n - 1)

ALSimple(pLINE, chot, tg, chiso, , n);

}

//Hm rt gn slng im Angles

int Angles(POINT *pLINE, int n, float )

{

int i, j, chiso [MAX];for (i = 0; i < n; i++) //Tt ccc im c gili

chiso[i]= TRUE;

ALSiple (PLINE, 0, 1 chiso, , n) ;

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

if (chiso ==TRUE)

pLINE[j++]= pLINE [i];

return j;}

* Ch :

Vi = 0 thut ton DouglasPeucker v BandWidth sbi cc imgia thng hng. Thut ton Angles phi c = 180obi cc im giathng hng.

5.2. XP XA GIC BI CC HNH CS

Cc i tng hnh hc c pht hin thng thng qua cc kthutd bin, kt qutm c ny l cc ng bin xc nh i tng. l,mt dy cc im lin tip ng knh, sdng cc thut ton n gin honhDouglas Peucker, Band Width, Angle v.v.. ta sthu c mt polylinehay ni khc i l thu c mt a gic xc nh i tng du. Vn lta cn phi xc nh xem i tng c phi l i tng cn tch haykhng? Nhta bit mt a gic c thc hnh dng ta nhmt hnh c


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59

s, c thc nhiu cch tip cn xp xkhc nhau. Cch xp xda trn ccc trng cbn sau:

c trng ton cc: Cc m men thng k, so hnh hc nhchuvi, din tch, tp ti u cc hnh chnht phhay ni tip a gic v.v..

c trng a phng: Cc so c trng ca ng cong nhgc, im li, lm, un, cc trv.v..

Hnh 5.4. Sphn loi cc i tng theo bt bin

Vic xp x t ra rt c hiu qui vi mt shnh phng c bitnh tam gic, ng trn, hnh ch nht, hnh vung, hnh ellipse, hnhtrn v mt a gic mu.

5.2.1 Xp xa gic theo bt bin ng dng

Hnh 5.5. Xp xa gic bi mt a gic mu

Mt a gic vi cc nh V0,..,Vm-1 c xp x vi a gic muU0,..,Un-1vi o xp xnhsau:

E V U nd md

( , ) min= 0 1

,

Nhn dng i tng

Bt binng dng

Bt binAphin

ng trnEllipse

Hnh chnht

EllipseTam gicTgic


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60

Trong

dR

j j d mj

n

d

kR U a V = +

+=

min,

( ) mod0 2 0

1 2

2

r , karea V V

area U U m

n

=

( )

( )0 1

0 1

L

L , vi R l

php quay quanh gc tomt gc .

Trong , d c tnh hiu qubng cng thc sau:

d j d m j d mj

n

j j j d mj

n

j

n

j

n

d

Vn

V k U k U V = + + +=

+=

=

=

| | | | | | | |( ) mod ( ) mod ( ) mod20

12 2 2

0

1

0

1

0

1 12

y Uj, Vjc hiu l cc s phc ti cc nh tng ng. Khim >> n th phc tp tnh ton rt ln. Vi cc hnh c bit nhhnhtrn, ellipse, hnh chnht, hnh xc nh duy nht bi tm v mt nh (agic u ) ta c th vn dng cc phng php n gin hn nh bnhphng ti thiu, cc bt bin thng k v hnh hc.

nh ngha 5.1

Cho a gic Pg c cc nh U0, U1,..., Un(U0Un) Khi m men bcp+q c xc nh nhsau:

M x y dxdypqp q=

Pg

.

Trong thc hnh tnh tch phn trn ngi ta thng sdng cngthc Green hoc c thphn tch phn bn trong a gic thnh tng i sca cc tam gic c hng OUiUi+1 .

O(0,0)

f x y x y dxdy

sign x y x y

f x y x y dxdy

p q

i i i ii

n

p q

OU Ui i

( , )

( )

( , )

Pg

=

+ +=

+

1 10

1

1

Hnh 5.6. Phn tch min a gic thnh tng i scc min tam gic

U0

U1U2

U3

Un-


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61

a. Xp xa gic bng ng trn

Dng phng php bnh phng ti thiu, ta c o xp x:

E(Pg,Cr)= min ( ), ,a b c R

i i i ii

n

nx y ax by c

=

+ + + +1 2 2 21

b. Xp xa gic bng ellipse

Cng nhi vi ng trn phng trnh xp xi vi ellipse ccho bi cng thc:

E(Pg,El)= min ( ), , , ,a b c d e

i i i i i ii

n

nx ay bx y cx dy e

=

+ + + + +R

1 2 2 21

Mt bin thkhc ca phng php bnh phng ti thiu khi xp xcc ng cong bc hai c a ra trong [7].

c. Xp xa gic bi hnh chnht

Sdng tnh cht din tch bt bin qua php quay, xp xtheo dintch nhsau: Gi 11 20 02, , l cc m men bc hai ca a gic (tnh theodin tch). Khi gc quay c tnh bi cng thc sau:

tg2

=2

- .11

20 02

Gi din tch ca hnh chnht nhnht c cc cnh song song vicc trc qun tnh v bao quanh a gic Pg l S.

K hiu E(Pg, Rect)= S area Pg ( )

Hnh 5.8. Xp xa gic bng hnh chnht

x

Hnh 5.7. Xp xa gic bng ng trn


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62

d. Xp xa gic bi a gic u n cnh

Gi M(x0,y0) l trng tm ca a gic, ly mt nh Q tu ca agic, xt a gic u n cnh Pg to bi nh Q vi tm l M.

K hiu E(Pg, Pg)= area Pg area Pg ( ) ( ' )

E(Pg, En)=min E(Pg,Pg) khi Q chy khp cc nh ca a gic.

5.2.2 Xp xa gic theo bt bin aphin

Trong [7] a ra m hnh chun tc vbt bin aphin, cho php chngta c thchuyn bi ton xp xi tng bi bt bin aphin vbi tonxp xmu trn cc dng chun tc. Nhvy c tha vic i snh cci tng vi mu bi cc bt bin ng dng, chng hn vic xp xbitam gic, hnh bnh hnh, ellipse tng ng vi xp xtam gic u, hnhvung, hnh trn v.v... Thtc xp xtheo bt bin aphin mt a gic vi

hnh csc thc hin tun tnhsau:+ Bc 0:

Phn loi bt bin aphin cc dng hnh cs

Dng hnh cs Dng chun tc

Tam gic Tam gic u

Hnh bnh hnh Hnh vung

Ellipse ng trn

+Bc 1:

Tm dng chun tc csPg' thomn iu kin:

m m

m m

m m

01 10

02 20

13 31

0

1

0

= == == =

(php tnh tin)

(php co dn theo hai trc x, y) (**)

+Bc 2:

Xc nh bin i aphin T chuyn a gic thnh a gic Pg dngchun tc (thomn tnh cht (**)).

Xp xa gic Pg vi dng chun tc csPg tm c bc 1 vio xp xE(Pg,Pg).

+Bc 3:

Kt lun, a gic ban u xp xT-1(Pg) vi o xp xE(Pg,Pg).


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i vi bc 1 trong [7] a ra hai v dsau:

V d1:

Tn ti duy nht tam gic u P1P2P3 thomn tnh cht (**) l

P1=(0,-2),P2= ),( 3 , P3= ),( 3 , 33284

= .

V d2:

Tn ti hai hnh vung P1P2 P3 P4 thomn tnh cht (**)

Hnh vung th nht c 4 nh tng ng l (-p,-p),(-p,p), (p,-

p),(p,p), vi p=

3

44

Hnh vung thhai c 4 nh tng ng l (-p,0),(p,0), (0,-p),(0,p),

vi p=34

.5.3. BIN I HOUGH

5.3.1. Bin i Hongh cho ng thng

Bng cch no ta thu c mt sim vn t ra l cn phikim tra xem cc im c l ng thng hay khng

Bi ton:

Cho n im (xi; yi) i = 1, n v ngng hy kim tra n im c to

thnh

ng thng hay khng?* tng

Gi s n im nm trn cng mt ng thng v ng thng cphng trnh

y = ax + b

V (xi, yi) i = 1, n thuc ng thng nn y1= ax1+ b, i = 1, n

b = - xia + y1; i = 1, n

Nhvy, mi im (xi; yi) trong mt phng stng ng vi mt s

ng thng b = - xia + yitrong mt phng tham sa, b. n im (xi; yi) i =1, n thuc ng thng trong mt phng tng ng vi n ng thngtrong mt phng tham sa, b giao nhau ti 1 im v im giao chnh l a,b. Chnh l h s xc nh phng trnh ca ng thng m cc imnm vo.


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64

* Phng php:

- Xy dng mng chs[a, b]v gn gi tr0 ban u cho tt cccphn tca mng

- Vi mi (xi; yi) v a, b l ch s ca phn t mng tho mnb = - xia + yitng gi trca phn tmng tng ng ln 1

- Tm phn tmng c gi trln nht nu gi trln nht tm c sovi sphn tln hn hoc bng ngng cho trc th ta c thkt luncc im nm trn cng 1 ng thng v ng thng c phng trnhy = ax + b trong a, b tng ng l chsca phn tmng c gi trlnnht tm c:

V d:

Cho 5 im (0, 1); (1, 3); (2, 5); (3, 5); (4, 9) v = 80%. Hy kimtra xem 5 im cho c nm trn cng mt ng thng hay khng? Hycho bit phng trnh ng thng nu c?

- Lp bng chs[a, b]v gn gi tr0

+ (0, 1): b = 1

+ (1, 3): b = -a + 3

+ (2, 5): b = -2a + 5

+ (3, 5): b = -3a + 5

+ (4, 9): b = -4a + 9

- Tm phn tln nht c gi tr4

4/5 = 80%- Kt lun: 5 im ny nm trn cng 1 ng thng

Phng trnh: y = 2x + 1

5.3.2. Bin i Hough cho ng thng trong ta cc

5.3.2.1. ng thng Hough trong ta cc


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OH.HA=0

Mi im (x,y) trong mt phng c biu din bi cp (r,) trong tacc.

Tng tmi ng thng trong mt phng cng c thbiu din bimt cp (r,) trong ta cc vi r l khong cch tgc ta ti ngthng v l gc to bi trc 0X vi ng thng vung gc vi n,hnh 5.9 biu din ng thng hough trong ta Decard.

Ngc li, mi mt cp (r,) trong to cc cng tng ng biudim mt ng thng trong mt phng.

GisM(x,y) l mim thuc ng thng c biu din bi (r,),gi H(X,Y) l hnh chiu ca gc toO trn ng thng ta c:

X= r. cosv Y= r.sin

Mt khc, ta c:

T ta c mi lin hgia (x,y) v (r,) nhsau: x*cos+y*sin= r.

Xt n im thng hng trong ta cc c phng trnhx*cos0+y*sin0= r0. Bin i Hough nh xn im ny thnh n ng sintrong ta cc m cc ng ny u i qua (r0,0). Giao im (r0,0) can ng sin sxc nh mt ng thng trong hta cc. Nhvy,nhng ng thng i qua im (x,y) scho duy nht mt cp (r,) v cbao nhiu ng qua (x,y) sc by nhiu cp gi tr(r,).

5.3.2.2. p dng bin i Hough trong pht hin gc nghing vn bn tng ca vic p dng bin i Hough trong pht hin gc nghing

vn bn l dng mt mng tch lu m s im nh nm trn mtng thng trong khng gian nh. Mng tch lu l mt mng hai chiuvi chshng ca mng cho bit gc lch ca mt ng thng v chsct chnh l gi trr khong cch tgc toti ng thng . Sau tnh tng sim nh nm trn nhng ng thng song song nhau theo

y

0

H

x

x.cos

y.sin

=r

Hnh 5.9. ng thng Hough trong tocc

r


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cc gc lch thay i. Gc nghing vn bn tng ng vi gc c tng gatrmng tch lucc i.

Theo bin i Hough, mi mt ng thng trong mt phng tngng c biu din bi mt cp (r,). Gista c mt im nh (x,y) trongmt phng, v qua im nh ny c v sng thng, mi ng thng licho mt cp (r,) nn vi mi im nh ta sxc nh c mt s cp(r,) thomn phng trnh Hough.

Hnh v trn minh ho cch dng bin i Hough pht hin gcnghing vn bn. Gi s ta c mt sim nh, y l nhng im giay cc hnh ch nht ngoi tip cc i tng c la chn t ccbc trc. y, ta thy trn mt phng c hai ng thng song songnhau. ng thng thnht c ba im nh nn gi trmng tch lubng3, ng thng thhai c gia trmng tch lubng 4. Do , tng gi tr

mng tch lucho cng gc trng hp ny bng 7.Gi Hough[360][Max] l mng tch ly, gisM v N tng ng l

chiu rng v chiu cao ca nh, ta c cc bc chnh trong qu trnh pdng bin i Hough pht hin gc nghing vn bn nhsau:

+Bc 1: Khai bo mng ch s Hough[][r] vi 0 3600v 0r NNMM ** + .

+Bc 2: Gn gi trkhi to bng 0 cho cc phn tca mng.

+Bc 3: Vi mi cp (x,y) l im gia y ca hnh ch nht

ngoi tip mt i tng.- Vi mi i t0 n 360 tnh gi tr ri theo cng thcri= x.cosi+y.sin

- Lm trn gi trrithnh snguyn gn nht l r0

- Tng gi tr ca phn t mng Hough[i][r0] ln mtn v.

y

x.cos+y.sin=r1

Hough[][r1]=3

x

0

x.cos+y.sin=r2

Hough[][r1]=4

Hnh 5.10. ng dng bin i Hough pht hin gc


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+ Bc 4: Trong mng Hough[][r] tnh tng gi trcc phn ttheotng dng v xc nh dng c tng gi trln nht.

Do sphn tca mt phn tmng Hough[0][r0] chnh l simnh thuc ng thng x.cos0+y.sin0= r0v vy tng sphn tca mthng chnh l tng sim nh thuc cc ng thng tng ng cbiu din bi gc ca hng . Do , gc nghing ca ton vn bnchnh l hng c tng gi trcc phn tmng ln nht.


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Phlc 1:

MT SNH DNG TRONG XL NH

Hin nay trn thgii c trn 50 khun dng nh thng dng.Sau y l mt snh dng nh hay dng trong qu trnh xl nhhin nay.

1. nh dng nh IMG

nh IMG l nh en trng, phn u ca nh IMG c 16 bytecha cc thng tin:

6 byte u: dng nh du nh dng nh. Gi tr ca 6byte ny vit di dng Hexa: 0x0001 0x0008 0x0001

2 byte tip theo: cha di mu tin. l di ca dycc byte klin nhau m dy ny sc lp li mt slnno . Sln lp ny sc lu trong byte m. Nhiu dyging nhau c lu trong mt byte.

4 byte tip: m tkch cpixel.

2 byte tip: spixel trn mt dng nh.

2 byte cui: sdng nh trong nh.

nh IMG c nn theo tng dng, mi dng bao gm cc gi(pack). Cc dng ging nhau cng c nn thnh mt gi. C 4 loigi sau:

Loi 1: Gi cc dng ging nhau.

Quy cch gi tin ny nhsau: 0x00 0x00 0xFF Count. Ba byteu tin cho bit s cc dy ging nhau, byte cui cho bit s ccdng ging nhau.

Loi 2: Gi cc dy ging nhau.

Quy cch gi tin ny nhsau: 0x00 Count. Byte thhai cho bits cc dy ging nhau c nn trong gi. di ca dy ghi u tp.

Loi 3: Dy cc Pixel khng ging nhau, khng lp li vkhng nn c.

Quy cch gi tin ny nhsau: 0x80 Count. Byte thhai cho bitdi dy cc pixel khng ging nhau khng nn c.


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Loi 4: Dy cc Pixel ging nhau.

Tutheo cc bt cao ca byte u tin c bt hay tt. Nu btcao c bt (gi tr1) thy l gi nn cc byte chgm bt 0, scc byte c nn c tnh bi 7 bt thp cn li. Nu bt cao tt(gi tr 0) th y l gi nn cc byte gm ton bt 1. S cc bytec nn c tnh bi 7 bt cn li.

Cc gi tin ca file IMG rt a dng do nh IMG l nh entrng, do vy chcn 1 bt cho 1 pixel thay v 4 hoc 8 nh ni trn. Ton bnh chc nhng im sng v ti tng ng vi gitr1 hoc 0. Tlnn ca kiu nh dng ny l kh cao.

2. nh dng nh PCX

nh dng nh PCX l mt trong nhng nh dng nh cin.N s dng phng php m ho lot di RLE (Run Length

Encoded) nn d liu nh. Qu trnh nn v gii nn c thchin trn tng dng nh. Thc t, phng php gii nn PCX kmhiu qu hn so vi kiu IMG. Tp PCX gm 3 phn: u tp(header), dliu nh (Image data) v bng mu mrng.

Header ca tp PCX c kch thc c nh gm 128 byte vc phn bnhsau:

1 byte: chra kiu nh dng.Nu l PCX/PCC th n lun cgi trl 0Ah.

1 byte: chra version sdng nn nh, c thc cc gitrsau:

+ 0: version 2.5.

+ 2: version 2.8 vi bng mu.

+ 3: version 2.8 hay 3.0 khng c bng mu.

+ 5: version 3.0 cbng mu.

1 byte: chra phng php m ho. Nu l 0 th m ho theophng php BYTE PACKED, ngc li l phngphp RLE.

1 byte: Sbt cho mt im nh plane.

1 word: togc tri ca nh. Vi kiu PCX n c gi trl(0,0), cn PCC th khc (0,0).

1 word: togc phi di.

1 word: kch thc brng v bcao ca nh.


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1 word: sim nh.

1 word: phn gii mn hnh.

1 word.

48 byte: chia n thnh 16 nhm, mi nhm 3 byte. Mi nhmny cha thng tin vmt thanh ghi mu. Nhvy ta c 16thanh ghi mu.

1 byte: khng dng n v lun t l 0.

1 byte: sbt plane m nh sdng. Vi nh 16 mu, gi trny l 4, vi nh 256 mu (1pixel/8bits) th sbt plane lil 1.

1 byte: sbytes cho mt dng qut nh.

1 word: kiu bng mu.

58 byte: khng dng.

nh dng nh PCX thng c dng lu trnh v thaotc n gin, cho php nn v gii nn nhanh. Tuy nhin, v cu trcca n cnh, nn trong mt strng hp lm tng kch thc lutr. Cng v nhc im ny m mt sng dng sdng mt kiunh dng khc mm do hn: nh dng TIFF (Targed Image FileFormat) sm tdi y.

3. nh dng nh TIFF

Kiu nh dng TIFF c thit k lm nhbt cc vn lin quan n vic m rng tp nh cnh. V cu trc, n cnggm 3 phn chnh:

Phn Header(IFH): c trong tt c cc tp TIFF v gm8 byte:

+ 1 word: ch ra kiu to tp trn my tnh PC hay myMacintosh. Hai loi ny khc nhau rt ln th t cc

byte lu tr trong cc s di 2 hay 4 byte. Nu trngny c gi trl 4D4Dh th l nh cho my Macintosh,

nu l 4949h l ca my PC.+ 1 word: version. tny lun c gi tr l 42. y l c

trng ca file TIFF v khng thay i.

+ 2 word: gi trOffset theo byte tnh tu ti cu trcIFD l cu trc thhai ca file. Thtcc byte ny phthuc vo du hiu trng u tin.


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Phn th2(IFD): Khng ngay sau cu trc IFH m v trc xc nh bi trng Offset trong u tp. C thc mthay nhiu IFD cng tn ti trong mt file.

Mt IFD bao gm:

+ 2 byte: cha cc DE ( Directory Entry).+ 12 byte l cc DE xp lin tip, mi DE chim 12 byte.

+ 4 byte: cha Offset trti IFD tip theo. Nu y l IFDcui cng th trng ny c gi tr0.

Phn th3: cc DE: cc DE c ddi cnh gm 12 bytev chia lm 4 phn:

+ 2 byte: chra du hiu m tp nh c xy dng.

+ 2 byte: kiu dliu ca tham snh. C 5 kiu tham s

cbn:1: BYTE (1 byte)

2: ASCII (1 byte)

3: SHORT (2 byte).

4: LONG (4 byte)

5: RATIONAL (8 byte)

+ 4 byte: trng di cha slng chmc ca kiu dliu chra. N khng phi l tng sbyte cn thit

lu tr. c sliu ny ta cn nhn schmc vi kiudliu dng.

+ 4 byte: l Offset ti im bt u dliu lin quan tidu hiu, tc l lin quan vi DE khng phi lu trvtl cng vi n nm mt vtr no trong file.

D liu cha trong tp thng c t chc thnh cc nhmdng (ct) qut ca dliu nh. Cch tchc ny lm gim bnhcn thit cho vic c tp. Vic gii nn c thc hin theo 4 kiukhc nhau c lu trtrong byte du hiu nn.


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4. nh dng file nh BITMAP

Mi file BITMAP gm u file cha cc thng tin chung vfile,u thng tin cha cc thng tin vnh, mt bng mu v mt mngdliu nh. Khun dng c cho nhsau:

BITMAPFILEHEADER bmfh;

BITMAPINFOHEADER bmih;

RGBQUAD aColors[];

BYTE aBitmapBits[];

Trong , cc cu trc c nh ngha nhsau:

typedef struct tagBITMAPFILEHEADER { /* bmfh */

UINT bfType;

DWORD bfSize;

UINT bfReserved1;

UINT bfReserved2;

DWORD bfOffBits;

} BITMAPFILEHEADER;

typedef struct tagBITMAPINFOHEADER { /* bmih */

DWORD biSize;

LONG biWidth;

LONG biHeight;

WORD biPlanes;

WORD biBitCount;

DWORD biCompression;

DWORD biSizeImage;

LONG biXPelsPerMeter;

LONG biYPelsPerMeter;

DWORD biClrUsed;

DWORD biClrImportant;

} BITMAPINFOHEADER, *LPBITMAPINFOHEADER;

vi

biSize kch thc ca BITMAPINFOHEADER

biWidth Chiu rng ca nh, tnh bng sim nh

biHeight Chiu cao ca nh, tnh bng sim nh


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biPlanes Splane ca thit b, phi bng 1

biBitCount Sbit cho mt im nh

biCompression Kiu nn

biSizeImage Kch thc ca nh tnh bng byte

biXPelsPerMeterphn gii ngang ca thit b, tnh bng im nh trnmet

biYPelsPerMeterphn gii dc ca thit b, tnh bng im nh trnmet

biClrUsed Slng cc mu thc sc sdng

biClrImportantSlng cc mu cn thit cho vic hin th, bng 0 nutt ccc mu u cn hin th

Nu bmih.biBitCount > 8 th mng mu rgbq[] trng, ngc li thmng mu c 2


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Phlc 2:

CC BC THAO TC VI FILE AVI

AVI l chun video thng c tch hp trong cc thvin cacc mi trng lp trnh. xl video, cn c cc thao tc cbnchuyn vxl nh cc khung hnh (cc frames).

1. Bc 1: Mv ng thvin

Trc mi thao tc vi file AVI, chng ta phi mthvin:

AVIFileInit( )

Hm ny khng cn tham s, c nhim vkhi ng thvincung cp cc hm thao tc vi file AVI. ( l thvin vfw32.lib,c khai bo trong file vfw.h).

Sau tt c cc thao tc bn phi nhng th vin m lcu, chbng lnh:

AVIFileExit( )

Nu thiu bt chm no, d l mhay ng thvin th trnhbin dch u sthng bo li.

2. Bc 2: Mv ng file AVI thao tc:Sau khi mthvin, bn phi mfile AVI bn nh thao tc:

AVIFileOpen(PAVIFILE* ppfile, LPCSTR fname, UINT mode,

CLSID pclsidHandler)

Thc cht, hm ny to ra mt vng m cha con tr trnfile c tn l fname cn m. V ppfile l con trtrn vng bm. Tham smode quy nh kiu mfile; chng hn OF_CREATEto mi, OF_READ c, OF_WRITE ghi . Tham scuidng

l NULL.

Trc khi ng thvin, bn phi ng file AVI m, bngcch dng hm:

AVIFileRelease(PAVIFILE pfile)

Trong , pfile l con trtrn file cn ng.


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3. Bc 3:

Mdng dliu hnh nh hay m thanh trong file AVI mrathao tc:

AVIFileGetStream(PAVIFILE pfile, PAVISTREAM * ppavi,DWORD fccType, LONG lParam)

Trong , pfile l con trn file m; ppavi trn dng dliu kt qu; fccType l loi dng d liu chn m, lstreamtypeAUDIO nu l ting v streamtypeVIDEO nu l hnh,lParam m sloi dng c m, l 0 nu chthao tc vi mt loidng dliu.

Sau cc thao tc vi dng dliu ny, bn nhphi ng n li:

AVIStreamRelease(PAVITREAM pavi).

4. Bc 4: Trng hp thao tc vi dliu hnh ca phimChun bcho thao tc vi khung hnh (frames):

AVIStreamGetFrameOpen(PAVISTRE

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Giao Trinh Xu Ly Anh