- 1. I HC THI NGUYNKHOA CNG NGH THNG TINGIO TRNH MN HCX L NH Ngi
son : TS. NNG TON,TS. PHM VIT BNHThi Nguyn, Thng 11 nm 2007 1
2. LI NI U Khong hn mi nm tr li y, phn cng my tnh v cc thit blin
quan c s tin b vt bc v tc tnh ton, dung lng cha,kh nng x l v.v.. v
gi c gim n mc my tnh v cc thit b linquan n x l nh khng cn l thit b
chuyn dng na. Khi nimnh s tr nn thng dng vi hu ht mi ngi trong x hi
v victhu nhn nh s bng cc thit b c nhn hay chuyn dng cng vi vica vo
my tnh x l tr nn n gin.Trong hon cnh , x l nh l mt lnh vc ang c
quan tm v tr thnh mn hc chuyn ngnh ca sinh vin ngnh cng ngh thngtin
trong nhiu trng i hc trn c nc. Tuy nhin, ti liu gio trnhcn l mt iu
kh khn. Hin ti ch c mt s t ti liu bng ting Anhhoc ting Php, ti liu
bng ting Vit th rt him. Vi mong mun nggp vo s nghip o to v nghin cu
trong lnh vc ny, chng ti binson cun gio trnh X l nh da trn cng mn
hc c duyt.Cun sch tp trung vo cc vn c bn ca x l nh nhm cung cpmt nn
tng kin thc y v chn lc nhm gip ngi c c th ttm hiu v xy dng cc chng
trnh ng dng lin quan n x l nh.Gio trnh c chia lm 5 chng v phn ph
lc: Chng 1, trnhby Tng quan v x l nh, cc khai nim c bn, s tng qut
ca mth thng x l nh v cc vn c bn trong x l 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 x l cc im nh trong lncn im nh ang xt. Chng ny cng
trnh by cc k thut nng caocht lng nh nh vo cc php ton hnh thi. Chng
3, trnh by cc kthut c bn trong vic pht hin bin ca cc i tng nh theo
c haikhuynh hng: Pht hin bin trc tip v pht hin bin gin tip. Chng4
th hin cch k thut tm xng theo khuynh hng tnh ton trc trungv v hng
tip cn xp x nh cc thut ton lm mnh song song v gintip. V cui cng l
Chng 5 vi cc k thut hu x l. Gio trnh c bin son da trn kinh nghim
ging dy ca tc gitrong nhiu nm ti cc kha i hc v cao hc ca H Cng ngh
-HQG H Ni, H Khoa hc t nhin HQG H Ni, Khoa Cng nghthng tin H Thi
Nguyn v.v.. Cun sch c th lm ti liu tham khocho sinh vin cc h k s, c
nhn v cc bn quan tm n vn nhndng v x l nh.2 3. Cc tc gi by t lng bit
n chn thnh ti cc bn ng nghiptrong Phng Nhn dng v cng ngh tri thc,
Vin Cng ngh thng tin,B mn H thng thng tin, Khoa Cng ngh thng tin, H
Thi Nguyn,Khoa Cng ngh thng tin, H Cng ngh, HQG H Ni, Khoa Ton C
Tin, H Khoa hc t nhin, HQG H Ni ng vin, gp vgip hon chnh ni dung
cun sch ny. Xin cm n Lnh o KhoaCng ngh thng tin, H Thi Nguyn, Ban
Gim c H Thi Nguyn h tr v 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 c nhng kin nggp ca bn c cng nh cc bn ng nghip c
chnh l kp thi.Th gp xin gi v: Phm Vit Bnh,Khoa Cng ngh thng tin H
Thi nguyn.X Quyt Thng, Tp. Thi Nguynin thoi: 0280.846506 Email:
[email protected] Nguyn, ngy 22 thng 11 nm 2007 CC TC GI 3 4.
MC LCLI NI U
.......................................................................................................................................................................
2MC LC
..................................................................................................................................................................................
4Chng 1: TNG QUAN V X L NH
.....................................................................................
71.1. X L NH, CC VN C BN TRONG X L NH .................. 71.1.1. X
l nh l g?
............................................................................................................................................
71.1.2. Cc vn c bn trong x l nh
........................................................................................
7 1.1.2.1 Mt s khi nim c bn
........................................................................................................
7 1.1.2.2 Nn chnh bin dng
....................................................................................................................
8 1.1.2.3 Kh nhiu
.................................................................................................................................................
9 1.1.2.4 Chnh mc xm:
...............................................................................................................................
9 1.1.2.5 Trch chn c im
....................................................................................................................
9 1.1.2.6 Nhn dng
............................................................................................................................................
10 1.1.2.7 Nn nh
...................................................................................................................................................
111.2. THU NHN V BIU DIN NH
...........................................................................................
111.2.1. Thu nhn, cc thit b thu nhn
nh..................................................................................
111.2.2. Biu din nh
..............................................................................................................................................
12 1.2.2.1. M hnh Raster
.............................................................................................................................
12 1.2.2.2. M hnh Vector
............................................................................................................................
13Chng 2: CC K THUT NNG CAO CHT LNG NH ................... 142.1.
CC K THUT KHNG PH THUC KHNG GIAN ..........................
142.1.1. Gii
thiu.........................................................................................................................................................
142.1.2. Tng gim sng
...............................................................................................................................
142.1.3. Tch ngng
................................................................................................................................................
152.1.4. B cm
...............................................................................................................................................................
152.1.5. Cn bng histogram
............................................................................................................................
162.1.6. K thut tch ngng t ng
................................................................................................
172.1.7. Bin i cp xm tng th
...........................................................................................................
182.2. CC K THUT PH THUC KHNG GIAN
..................................................... 202.2.1. Php
cun v mu
.................................................................................................................................
204 5. 2.2.2. Mt s mu 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 tr gn nht
............................................................................
252.3. CC PHP TON HNH THI HC
....................................................................................
26 2.3.1. Cc php ton hnh thi c
bn..............................................................................................
26 2.3.2. Mt s tnh cht ca php ton hnh
thi....................................................................
27Chng 3: BIN V CC PHNG PHP PHT HIN BIN .....................
323.1. GII THIU
............................................................................................................................................................
323.2. CC PHNG PHP PHT HIN BIN TRC TIP
................................. 32 3.2.1. K thut pht hin bin
Gradient.........................................................................................
323.2.1.1. K thut Prewitt
..........................................................................................................................
343.2.1.2. K thut
Sobel...............................................................................................................................
353.2.1.3. K thut la
bn..............................................................................................................................
35 3.2.2. K thut pht hin bin Laplace
...........................................................................................
363.3. PHT HIN BIN GIN
TIP.......................................................................................................
37 3.3.1 Mt s khi nim c bn
.................................................................................................................
37 3.3.2. Chu tuyn ca mt i tng
nh.......................................................................................
38 3.3.3. Thut ton d bin tng qut
....................................................................................................
40Chng 4: XNG V CC K THUT TM XNG
........................................ 444.1. GII THIU
............................................................................................................................................................
444.2. TM XNG DA TRN LM MNH
...........................................................................
44 4.2.1. S lc v thut ton lm mnh
...........................................................................................
44 4.2.2. Mt s thut ton lm mnh
......................................................................................................
464.3. TM XNG KHNG DA TRN LM MNH
................................................ 46 4.3.1. Khi qut
v lc
Voronoi.................................................................................................
47 4.3.2. Trc trung v Voronoi ri
rc...................................................................................................
47 4.3.3. Xng Voronoi ri rc
....................................................................................................................
48 4.3.4. Thut ton tm xng
........................................................................................................................
49Chng 5: CC K THUT HU X L
..................................................................................
525.1. RT GN S LNG IM BIU
DIN.....................................................................
52 5.1.1. Gii
thiu.........................................................................................................................................................
52 5 6. 5.1.2. Thut ton Douglas Peucker
.....................................................................................................
52 5.1.2.1. tng
.................................................................................................................................................
52 5.1.2.2. Chng trnh
...................................................................................................................................
535.1.3. Thut ton Band width
....................................................................................................................
54 5.1.3.1. tng
.................................................................................................................................................
54 5.1.3.2. Chng trnh
...................................................................................................................................
565.1.4. Thut ton Angles
.................................................................................................................................
57 5.1.4.1. tng
.................................................................................................................................................
57 5.1.4.2. Chng trnh
...................................................................................................................................
575.2. XP X A GIC BI CC HNH C
S....................................................................
585.2.1 Xp x a gic theo bt bin ng dng
........................................................................
595.2.2 Xp x a gic theo bt bin aphin
......................................................................................
625.3. BIN I HOUGH
........................................................................................................................................
635.3.1. Bin i Hongh cho ng thng
.......................................................................................
635.3.2. 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.....................................................................................................................
65PH LC
................................................................................................................................................................................
68TI LIU THAM KHO
....................................................................................................................................
76 6 7. Chng 1:TNG QUAN V X L NH1.1. X L NH, CC VN C BN TRONG X L
NH1.1.1. X l nh l g? Con ngi thu nhn thng tin qua cc gic quan,
trong th gicng vai tr quan trng nht. Nhng nm tr li y vi s pht trin
caphn cng my tnh, x l nh v ho pht trin mt cch mnh mv c nhiu ng dng
trong cuc sng. X l nh v ho ng mt vaitr quan trng trong tng tc ngi
my. Qu trnh x l nh c xem nh l qu trnh thao tc nh u vonhm cho ra kt
qu mong mun. Kt qu u ra ca mt qu trnh x lnh c th l mt nh tt hn hoc
mt kt lun.nhTt hn nhX L NHKt lunHnh 1.1. Qu trnh x l nhnh c th xem
l tp hp cc im nh v mi im nh c xemnh l c trng cng sng hay mt du hiu
no ti mt v tr no ca i tng trong khng gian v n c th xem nh mt hm n
binP(c1, c2,..., cn). Do , nh trong x l nh c th xem nh nh n chiu.S
tng qut ca mt h thng x l nh:H quyt nhThu nhn nh Trch chn Tin x l Hu
i snh rt(Scanner, c im x l ra kt lunCamera,Sensor) Lu tr Hnh 1.2.
Cc bc c bn trong mt h thng x l nh1.1.2. Cc vn c bn trong x l
nh1.1.2.1 Mt s khi nim c bn* nh v im nh:7 8. im nh c xem nh l du
hiu hay cng sng ti 1 to trong khng gian ca i tng v nh c xem nh l 1
tp hp ccim nh.* Mc xm, muL s cc gi tr c th c ca cc im nh ca
nh1.1.2.2 Nn chnh bin dngnh thu nhn thng b bin dng do cc thit b
quang hc v int.Pi Pi f(Pi)nh thu nhnnh mong mun Hnh 1.3. nh thu nhn
v nh mong mun khc phc ngi ta s dng cc php chiu, cc php chiu thngc
xy dng trn tp cc im iu khin.Gi s (Pi, Pi) i = 1, n c n cc tp iu
khinTm hm f: Pi a f (Pi) sao cho n2f ( Pi ) Pi mini =1 Gi s nh b
bin i ch bao gm: Tnh tin, quay, t l, bin dngbc nht tuyn tnh. Khi hm
f c dng: f (x, y) = (a1x + b1y + c1, a2x + b2y + c2)Ta c: i =1 ni
=1 [ = ( f ( Pi ) Pi) 2 = (a1 xi + b1 yi + c1 xi ) + (a 2 xi + b2
yi + c2 yi )n22 ] cho min 8 9. n n n n a1=0 a1 x i2 + b1 x i y i +
c1 x i = x i x i i =1i =1i =1i =1 n n nn =0 a1 x i y i + b1 y i2 +
c1 y i = y i x i b1 i =1i =1i =1 i =1 n nn=0 a1 x i + b1 y i + nc1
= x i c1 i =1i =1 i =1Gii h phng trnh tuyn tnh tm c a1, b1, c1Tng t
tm c a2, b2, c2 Xc nh c hm f1.1.2.3 Kh nhiuC 2 loi nhiu c bn trong
qu trnh thu nhn nh Nhiu h thng: l nhiu c quy lut c th kh bng cc
phpbin i Nhiu ngu nhin: vt bn khng r nguyn nhn khc phcbng cc php
lc1.1.2.4 Chnh mc xm:Nhm khc phc tnh khng ng u ca h thng gy ra.
Thngthng c 2 hng tip cn: Gim s mc 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 s mc xm: Thc hin ni suy ra cc
mc xm trung gianbng k thut ni suy. K thut ny nhm tng cng mncho
nh1.1.2.5 Trch chn c im Cc c im ca i tng c trch chn tu theo mc ch
nhndng trong qu trnh x l nh. C th nu ra mt s c im ca nhsau y:c im
khng gian: Phn b mc xm, phn b xc 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 b vng c gi l mt n c9 10. 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 th c trchchn
nh ton t gradient, ton t la bn, ton t Laplace, ton t chokhng (zero
crossing) v.v.. Vic trch chn hiu qu cc c im gip cho vic nhn dng cci
tng nh chnh xc, vi tc tnh ton cao v dung lng nh lutr gim
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 th gic 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 th gn cho n mt tn gi no . V d mu c
th l nh ca vn tay,nh ca mt vt no c chp, mt ch vit, 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 danhnh mt thnh phn
ca mt lp xc nh.Hoc phn loi khng c mu (unsupervised classification
hayclustering) 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 t ng bao gm ba khu tng ng vi ba giaion ch yu
sau y: 1o. Thu nhn d liu v tin x l. 2o. Biu din d liu. 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.10 11. 4o. Phn loi da trn mng n-ron nhn to.Trong cc ng dng r
rng l khng th ch dng c mt cch tipcn n l phn loi ti u do vy cn s dng
cng mt lc nhiuphng php v cch tip cn khc nhau. Do vy, cc phng thc
phnloi t hp hay c s dng khi nhn dng v nay c nhng kt qu ctrin vng da
trn thit k cc h thng 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 th do chuyn gia xut, m phi c trch chn da trn cc
thtc phn tch d liu.1.1.2.7 Nn nh Nhm gim thiu khng gian lu tr. Thng
c tin hnh theo chai cch khuynh hng l nn c bo ton v khng bo ton thng
tin.Nn khng bo ton th thng c kh nng nn cao hn nhng kh nngphc hi th
km hn. Trn c s hai khuynh hng, c 4 cch tip cn cbn trong nn nh: Nn
nh thng k: K thut nn ny da vo vic thng k tn xut xut hin ca gi tr cc
im nh, trn c s m c chin lc m ha thch hp. Mt v d in 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. K thut li dng s ging nhau ca cc im nh trong cc
vng gn nhau. V d cho k thut ny l m nn *.PCX Nn nh s dng php bin i:
y l k thut tip cn theo hng nn khng bo ton v do vy, k thut thng nn
hiu qu hn. *.JPG chnh l tip cn theo k thut nn ny. Nn nh Fractal: S
dng tnh cht Fractal ca cc i tng nh, th hin s lp li ca cc chi tit. K
thut nn s tnh ton ch cn lu tr phn gc nh v quy lut sinh ra nh theo
nguyn l Fractal1.2. THU NHN V BIU DIN NH1.2.1. Thu nhn, cc thit b
thu nhn nh11 12. Cc thit b thu nhn nh bao gm camera, scanner cc
thit b thunhn ny c th cho nh en trngCc 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 t nh Raster. Nhn chung
cc h thng 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 nh1.2.2. Biu din nhnh trn
my tnh l kt qu thu nhn theo cc phng php s hoc nhng trong cc thit b
k thut khc nhau. Qu trnh lu tr nhnhm 2 mc ch: Tit kim b nh Gim thi
gian x lVic lu tr thng tin trong b nh c nh hng rt ln n vic hinth,
in n v x l nh c xem nh l 1 tp hp cc im vi cng kchthc nu s dng 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 tu thuc vo
nhu cu s dngv c trng ca mi nh c th, trn c s cc nh thng c biudin
theo 2 m hnh c bn1.2.2.1. M hnh Rastery 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 b nh camera, scanner. Tu theo yu cu thc th m mi im nhc
biu din qua 1 hay nhiu btM hnh Raster thun li cho hin th v in n.
Ngy nay cng nghphn cng cung cp nhng thit b thu nhn nh Raster ph hp
vi tc nhanh v cht lng cao cho c u 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 th hnh quytrnh chung hin
th nh Raster thng qua DIB. 12 13. Mt trong nhng hng nghin cu c bn
trn m hnh biu din nyl k thut nn nh cc k thut nn nh li chia ra theo
2 khuynh hng lnn bo ton v khng bo ton thng tin nn bo ton c kh nng
phchi hon ton d liu ban u cn nu khng bo ton ch c kh nngphc hi sai s
cho php no . Theo cch tip cn ny ngi ta ra nhiu quy cch khc nhau nh
BMP, TIF, GIF, PCX Hin nay trn th gii c trn 50 khun dng nh thng dng
bao gmc trong cc k thut nn c kh nng phc hi d liu 100% v nn ckh nng
phc hi vi sai s nhn c.BMP PaintPCC . . DIBCa s . Thay iHnh 1.4. Qu
trnh hin th v chnh sa, lu tr nh thng qua DIB1.2.2.2. M hnh Vector
Biu din nh ngoi mc ch tit kim khng gian lu tr d dngcho hin th v in
n cn m bo d dng trong la chn sao chp dichuyn tm kim Theo nhng yu cu
ny k thut biu din vector t rau 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 thunhn trc tip t cc thit b s ho nh Digital hoc c chuyn i
tnh Raster thng qua cc chng trnh s ho Cng ngh phn cng cung cp nhng
thit b x l vi tc nhanhv cht lng cho c u vo v ra nhng li ch h tr cho
nh Raster. Do vy, nhng nghin cu v biu din vect u tp trung t chuyni
t nh Raster.VecterRasterRASTER ha VECTORha RASTERHnh 1.5. S chuyn i
gia cc m hnh biu din nh 13 14. Chng 2:CC K THUT NNG CAO CHT LNG
NH2.1. CC K THUT KHNG PH THUC KHNG GIAN2.1.1. Gii thiuCc php ton
khng ph thuc khng gian l cc php ton khngphc thuc v tr 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 x l nh l biu tnsut (Histogram) Biu tn
sut ca mc xm g ca nh I l s im nh c gi tr gca nh I. K hiu l h(g)V d:
1 2 0 4 1 0 0 7I= 2 2 1 0 4 1 2 1 2 0 1 1 g 0 1 24 7 h(g)5 7 52
12.1.2. Tng gim sngGi s ta c I ~ kch thc m n v s nguyn cKhi , k
thut tng, gim c sng c th hin 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 i14 15. 2.1.3. Tch ngng Gi s ta c nh I ~ kch thc
m n, hai s Min, Max v ngng khi : K thut tch ngng c th hinfor (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 k thut chuyn nh
thnh nh en trng cng dng khi qut v nhn dng vn bn c th xy ra sai st
nn thnh nhhoc nh thnh nn dn n nh b t nt hoc dnh.2.1.4. B cmK thut
nhm gim bt s mc xm ca nh bng cch nhm li smc xm gn nhau thnh 1 nhm
Nu ch c 2 nhm th chnh l k thut tch ngng. Thng thngc nhiu nhm vi kch
thc khc nhau. tng qut khi bin i ngi ta s ly cng 1 kch
thcbunch_sizeh(g) g0I [i,j] = I [i,j]/ bunch - size * bunch_size
(i,j)V d: B cm nh sau vi bunch_size= 31 2 4672 1 345I=7 2 6914 1
212 15 16. 00 36 600 33 3Ikq = 60 69 030 00 02.1.5. Cn bng
histogram nh I c gi l cn bng "l tng" nu vi mi mc xm g, g tac h(g) =
h(g)Gi s, ta c nhI ~ kch thc m n new_level ~ s mc xm ca nh cn bng
mnTB = ~ s im nh trung bnh ca mi mc xm new _ level ca nh cn bng gt
( g ) = h(i )i =0 ~ s im nh c mc xm gXc nh hm f: g a f(g) t(g) Sao
cho: f ( g ) = max 0, round 1 TB V d: Cn bng nh sau vi new_level= 4
1 246 7 2 134 5 I= 7269 1 4 121 2g h(g) t(g) f(g)15 5 025 10131
11143 14251 15262 17272 19391 20316 17. 0 12 231 01 22Ikq = 3 12
302 01 01Ch : nh sau khi thc hin cn bng cha chc l cn bng "l
tng"2.1.6. K thut tch ngng t ngNgng trong k thut tch ngng thng c
cho bi ngi sdng. K thut tch ngng t ng nhm tm ra ngng mt cch tng da
vo histogram theo nguyn l trong vt l l vt th tch lm 2phn nu tng lnh
trong tng phn l ti thiu.Gi s, ta c nhI ~ kch thc m n G ~ l s mc xm
ca nh k c khuyt thiu t(g) ~ s im nh c mc xm g1 gm( g ) = i.h(i) t (
g ) i =0 ~ mmen qun tnh TB c mc xm gHm f: g a f (g ) t(g ) f (g) =
[m( g ) m(G 1)]2 mxn t ( g )Tm sao cho: f ( ) = max { f ( g )}0 g 3
Bc 1: V Histogram ca nh c f(g)g018 19. Bc 2: V th hm f(g)h(g) 0 g
Bc 3: V Histogram 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))})= h(i)h(g) f(g)i
f 1 ( q ) g 0 Histogram ca nh mi thua c bng cch chng hnh v tnh gi
trtheo cc q (= f(g)) theo cng thc tnh trn. Kt qu cui thu c sau
phpquay gc 90 thun chiu kim ng h. 19 20. 2.2. CC K THUT PH THUC
KHNG GIAN2.2.1. Php cun v mu Gi s ta 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. m 1 n 1 I T (
x, y ) = i =0 I (x + i, y + j ) * T (i, j )j =0 (2.1) m 1 n 1Hoc I
T ( x, y ) = i =0 I (x i, y j )* T (i, j )j =0 (2.2)VD: 124587
211422 I=455882 121144 722152 T=10 0111 I T ( x, y ) = i =0 I (x +
i, y + j )*T (i, j ) = I (x, y )*T (0,0) + I (x + 1, y + 1)*T
(1,1)j =0 = I ( x, y ) + I ( x + 1, y + 1) 238 7 10 * 76912 4*Tnh
theo (2.1)IT= 66612 12 * 342 66 * *** ** *Tnh theo cng thc 2.2
****** *2387 10IT= *76912 4 *6661212 *342 6 6 20 21. * Nhn xt: -
Trong qu trnh thc hin php cun c mt s thao tc ra ngoi nh,nh khng c
xc nh ti nhng v tr dn n nh thu c c kchthc nh hn. - nh thc hin theo
cng thc 2.1 v 2.2 ch sai khc nhau 1 phpdch chuyn n gin ta s hiu php
cun l theo cng thc 2.12.2.2. Mt s mu thng dng - Mu:111 T1 = 111111
~ Dng kh nhiu Cc im c tn s cao VD1:1 2 4587231 1422 I= 4 5 58821 2
11447 2 215255 65 45 46 **52 58 34 35 ** I T1 = 29 27 35 35 ** ****
** **** ** p dng k thut cng hng s vi c = -27, ta c:28 38 18 19 **25
31 7 8**Ikq =2088** ****** ****** - Mu:0-1 0 T2 = -14 -10-1 021 22.
~ Dng pht hin cc im c tn s cao 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...; xn n iu tng (gim). Khi trung v ca
dy khiu l Med({xn}), c nh ngha: n + Nu n l x + 1 2 n n + Nu n chn:
x hoc x + 122 * Mnh 2.1n xx i min ti Med ({xn }) i =1Chng minh + Xt
trng hp n chn n t M = 2 Ta c:n M Mi =1x xi = x xi + x x M +i i =1 i
=1 M M = ( x xi + x M + i x ) x M + i xi i =1 i =1 M = [(x M +1 x M
) + ( x M xi )] i =1 MM = x M + i Med ({xi }) + xi Med ({xi }) i =1
i =122 23. n= xi Med ({xi })i =1 + Nu n l: B sung thm phn t Med
({xi }) vo dy. Theo trng hp n chnta c:n xxi + Med ({xi }) Med ({xi
}) min ti Med({xn}) i =1n xx i =1 i min ti Med({xn})* K thut lc
trung v Gi s ta c nh I ngng ca s W(P) v im nh P Khi k thut lc trung
v ph thuc khng gian bao gm cc bcc bn sau: + Bc 1: Tm trung v{I(q)|
q W(P)} Med (P) + Bc 2: Gn gi tr I ( P) I ( P) Med ( P ) I ( P) =
Med ( P ) Nguoclai V d: 1 2 3 2 416 2 1 I=4 2 1 1 2 1 2 1W(3 3); =
212324221 Ikq = 4 2112121Gi tr 16, sau php lc c gi tr 2, cc gi tr
cn li khng thay igi tr. 23 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:1 n AV ({xn }) = round xi n i =1 * Mnh 2.2n 2 ( x xi ) i =1
min ti AV ({xn })Chng minh: nt: ( x) = (x x )2ii =1Ta c: n ( x ) =
2 ( x x i ) i =1 ( x) = 0 n ( x xi ) = 0i =1 1 n x= xi = AV ({xi })
n i =1Mt khc, ( x) = 2n > 0 min ti x = AV ({xi })K thut lc trung
bnh Gi s ta c nh I, im nh P, ca s W(P) v ngng . Khi kthut lc trung
bnh ph thuc khng gian bao gm cc bc c bn sau:+ Bc 1: Tm trung bnh
{I(q)| q W(P)} AV(P)24 25. + Bc 2: Gn gi tr I ( P)I ( P) AV ( P) I
( P) = AV ( P)NguoclaiV d:123 24 162 1 I= 421 1212 1W(3 3); = 212
3243 21Ikq = 42 1121 21Gi tr 16 sau php lc trung bnh c gi tr 3, cc
gi tr cn li ginguyn sau php lc.2.2.5. Lc trung bnh theo k gi tr gn
nht Gi s ta c nh I, im nh P, ca s W(P), ngng v s k. Khi, lc trung
bnh theo k gi tr gn nht bao gm cc bc sau:+ Bc 1: Tm K gi tr gn nht
{I(q) q W(p)} {k gi tr gn I(P) nht}+ Bc 2: Tnh trung bnh {k gi tr
gn I(P) nht} AVk(P)+ Bc 3: Gn gi tr I ( P)I ( P ) AV k ( P ) I ( P)
= AV k ( P ) NguoclaiV d:123 24 162 1 I= 421 1212 1W(3 3); = 2; k =
3 25 26. 123 2482 1 Ikq =421 1212 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 qu khng
thay i Cht lng ca k thut ph thuc vo s phn t la chn k.2.3. CC PHP
TON HNH THI HC2.3.1. Cc php ton hnh thi c bn Hnh thi l thut ng ch s
nghin cu v cu trc hay hnh hc topoca i tng trong nh. Phn ln cc php
ton ca "Hnh thi" c nhngha t hai php ton c bn l php "gin n"
(Dilation) v php "co"(Erosion). Cc php ton ny c nh ngha nh sau: Gi
thit ta c i tngX v phn t cu trc (mu) B trong khng gian Euclide hai
chiu. K hiuBx l dch chuyn ca B ti v tr x.nh ngha 2.3 (DILATION)Php
"gin n" ca X theo mu B l hp ca tt c cc Bx vi x thucX. Ta c: X B = U
Bx xXnh ngha 2.4 (EROSION) Php "co" ca X theo B l tp hp tt c cc im
x sao cho Bx nmtrong X. Ta c: X B = {x : Bx X}0 x 0 x x x 0 x x 0 V
d: Ta c tp X nh sau:X = 0 x x 0 0 B =x0 x 0 x 00 x x x 0 26 27. 0x
x x x00 0 x 0 xx x x x00 x 0 0 000XB= x x x v X B = x 0 0 0 0x x x
x00 0 0 0 0x0 x x x x x 0 0nh 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 n lin lip theo
B. Ta c:OPEN(X,B) = (X B) BV d: Vi tp X v B trong v d trn ta c 0 00
x x 0 0x x 0 OPEN(X,B) = (XB) B = 0xx 0 0 0 00 0 0 0 xx x 0nh ngha
2.6 (CLOSE) Php ton ng (CLOSE) ca X theo cu trc B l tp hp cc imca
nh X sau khi gin n v co lin tip theo B. Ta c:CLOSE(X,B) = (X B)
BTheo v d trn ta c: 0 xx xx x xx xx CLOSE(X,B) = (X B) B = 0 xx 00
0 xx x0 0 xx x0 2.3.2. Mt s tnh cht ca php ton hnh thi* Mnh 2.3
[Tnh gia tng]:(i) X X X B X B B X B X B B (ii) B B X B XB X X B X
BX27 28. Chng minh:(i) X B = UBx UBx = X BxX xX X B = {x / Bx X }
{x / Bx X } = XB(ii) X B =UBx U Bx = X BxX xXTheo nh ngha:X B = {x
/ B x X } {x / Bx X } = X B.*Mnh 2.4 [Tnh phn phi vi php ]:(i) X (B
B) = (X B) (X B)(ii) X(B B) = (X B) (XB)Chng minh:(i) X (B B) = ( X
B) (X B)Ta c: B B BX (B B) X B (tnh gia tng)Tng t:X ( B B) X BX (B
B) (X B) (X B)(2.3)Mt khc, y X (B B) x X sao cho y (B B)x y Bx yXBy
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) = (X B) (XB)Ta c: B B B X (B B)
X B (tnh gia tng)Tng t : X(B B) XBX(B B) (X B) ( XB)(2.5) 28 29. Mt
khc,x (XB) (XB) Suy ra, x XB Bx X xXB Bx X ( B B)x X xX (B B) X(B
B) (XB) (XB) (2.6) T (2.5) v (2.6) ta c: X(B B) = (X B) (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 YX (X Y)BX BTng t:(X Y) BYB (X Y)B (X B) (YB)(2.7)Mt
khc,x (XB) (Y B)Suy ra x X B Bx XxYB Bx Y Bx X Y x ( X Y) B (X Y) B
(X B) (YB) (2.8)T (2.7) v (2.8) ta c: (X Y)B = (X B) (Y B).* Mnh
2.6 [Tnh kt hp](i) (X B) B = X (B B)(ii) (X B) B = X(B B) 29 30.
Chng minh:(i) (X B) B = X (B B)Ta c, (X B) B = ( U B x ) B xX= U
(Bx B ) = U (B B ) xx X x X= X (B B)(i) (XB)B = X (B B)Trc ht ta i
chng minh: B x XB ( B B) x XTht vy, do B x XB nn y B x yX B By XUB
y X yBx ( B B) x XMt khc, ( B B) x X ( B x B) X UB y X yBx y B x ta
c By X hay y B x ta c y XBDo , B x XBTa c, (XB)B = {x / B x X }B=
{x/ B x X B}= {x/ ( B B) x X} (do chng minh trn)=X(B B) .* nh l 2.1
[X b chn bi cc cn OPEN v CLOSE] Gi s, X l mt i tng nh, B l mu, khi
, X s b chn trnbi tp CLOSE ca X theo B v b chn di bi tp OPEN ca X
theo B.Tc l:(X B)B X (XB) B 30 31. Chng minh:Ta c: x X Bx X B (V X
B = UB x )x X x (X B)B (theo nh ngha php co) (X B)BX(2.9)Mt khc, y
(X B) B, suy ra: xXB sao cho y Bx(V (X B) B =UB x ) xXB Bx X y XSuy
ra: X (X B) B(2.10)T (2.9) v (2.10) Ta c: (X B) B X (X B) B .*H qu
2.1 [Tnh bt bin] :(i) ((X B)B) B = X B(ii) ((X B) B)B=X BChng
minh:(i) Tht vy, t nh 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 t nh 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, t nh l 2.1 ta c (XB) B X ((X B) B)B X B (do
tnh cht gia tng) (2.13)Mt khc, cng t nh l 2.1 ta c X (X B) B XDo ,
thay X bi XB ta c, X B ((X B) B) B (2.14)T (2.13) v (2.14) Ta c:
((X B) B)B=XB (pcm). 31 32. Chng 3: BIN V CC PHNG PHP PHT HIN
BIN3.1. GII THIU Bin l vn quan trng trong trch chn c im nhm tin ti
hiunh. Cho n nay cha c nh ngha chnh xc v bin, trong mi ngdng ngi ta
a ra cc o khc nhau v bin, mt trong cc o l o v s thay i t ngt v cp
xm. V d: i vi nh en trng,mt im c gi l im bin nu n l im en c t nht
mt imtrng bn cnh. Tp hp cc im bin to nn bin hay ng bao cai tng. Xut
pht t c s ny ngi ta thng s dng hai phngphp pht hin bin c bn: Pht
hin bin trc tip: Phng php ny lm ni bin da vo sbin thin mc xm ca nh.
K thut ch yu dng pht hin bin y l da vo s bin i cp xm theo hng. Cch
tip cn theo ohm bc nht ca nh da trn k thut Gradient, nu ly o hm bc
haica nh da trn bin i gia ta c k thut 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. K thut d binv phn vng nh l hai bi ton i ngu nhau v d bin thc
hin phnlp i tng m khi phn lp xong ngha l phn vng c nh vngc li, khi
phn vng nh c phn lp thnh cc i tng, do c th pht hin c bin.Phng php
pht hin bin trc tip t ra kh hiu qu v t chu nhhng ca nhiu, song nu s
bin thin sng khng t ngt, phngphp t ra 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 TIP3.2.1. K thut pht hin bin Gradient
32 33. Theo nh ngha, gradient l mt vct c cc thnh phn biu th tc thay
i gi tr ca im nh, ta c:f ( x, y )f ( x + dx, y ) f ( x, y )= fx
xdxf ( x, y )f ( x, y + dy ) f ( x, y )= fy ydy Trong , dx, dy l
khong cch (tnh bng s im) theo hng x vy.* Nhn xt:Tuy ta ni l ly o hm
nhng thc cht ch l m phng v xp xo hm bng cc k thut nhn chp (cun theo
mu) v nh s l tn hiuri rc nn o hm khng tn ti. V d: Vi dx = dy = 1,
ta c: f x f (x + 1, y ) f (x, y ) f f (x, y + 1) f (x, y ) y Do ,
mt n nhn chp theo hng x l A= ( 1 1) 1 v hng y l B= 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 * 033 * IA=3 0 0 * ; I B= 0
00 * 3 0 0 *0 00 * * * * ** ** * 0 0 0 * I A + I B= 30 0 * 3 0 0 *
* * * * 33 34. 3.2.1.1. K thut PrewittK thut s dng 2 mt n nhp chp
xp x o hm theo 2 hng xv y l:-1 01Hx =-1 01-1 01-1 -1 -1Hy = 00 0 11
1Cc bc tnh ton ca k thut Prewitt+ Bc 1: Tnh I Hx v I Hy+ Bc 2: Tnh
I Hx + I HyV d:00000 055550 055550 0 I= 55550 000000 000000 000-10
-10 * *00-15 -15 * *I Hx =00-10 -10 * *00 -5 -5* ****** ****** * 15
15 10 5 * *0 0 00* * -15 -15 -10 -5 * *I Hy = -15 -15 -10 -5 * ** *
*** ** * *** *34 35. 15 15 0 -5**00 -15 -15 ** I Hx + I Hy = -15
-15 -20 -15 ** -15 -15 -15 -10 *** * * **** * * ***3.2.1.2. K thut
Sobel Tng t nh k thut Prewitt k thut Sobel s dng 2 mt n nhnchp theo
2 hng x, y l:-1 01 Hx = -2 02-1 01-1 -2 -1 Hy =00 0 12 1 Cc bc tnh
ton tng t Prewitt + Bc 1: Tnh I Hx v I Hy + Bc 2: Tnh I Hx + I
Hy3.2.1.3. K thut la bn K thut s dng 8 mt n nhn chp theo 8 hng 00,
450, 900, 1350,180 , 2250, 2700, 3150 0 55 -3 5 5 5 H1 = 5 0 -3H2 =
-3 0 -3-3 -3 -3-3 -3 -3-3 5 5-3 -3 5 H3 = -3 0 5H4 = -3 05-3 -3
-3-3 -3 5-3 -3 -3-3 -3 -3 H5 = -3 0 5H6 = -3 0 -3-3 5 555 5-3 -3 -3
5 -3 -3 H7 = 5 0 -3H8 = 5 0 -3 55 -3 5 -3 -335 36. Cc bc tnh ton
thut ton La bn+ Bc 1: Tnh I Hi ; i = 1,88+ Bc 2:I H i =1i3.2.2. K
thut 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 qu hn l phng php s dng o hmbc hai
Laplace.Ton t Laplace c nh ngha nh sau:Ta c:2 f 2 f f = 2 + 2 2x y
2 f f = ( f ( x + 1, y ) f ( x, y ) ) x 2 x x x [ f ( x + 1, y ) f
( x, y )] [ f ( x, y ) f ( x 1, y )] f ( x + 1, y ) 2 f ( x, y ) +
f ( x 1, y )Tng t, 2 f f = ( f ( x, y + 1) f ( x, y ) ) y 2 y y y [
f ( x, y + 1) f ( x, y )] [ f ( x, y ) f ( x, y 1)] f ( x, y + 1) 2
f ( x, y ) + f ( x, y 1) 2Vy: f= f(x+1,y) + f(x,y+1) - 4f(x,y) +
f(x-1,y) + f(x,y-1)Dn ti: 0 1 0 H = 1 4 1 0 1 0 Trong thc t, ngi ta
thng dng nhiu kiu mt n khc nhau xp x ri rc o hm bc hai Laplace. Di
y l ba kiu mt nthng dng:36 37. 0 1 0 1 1 1 1 2 1 H1 = 1 4 1 H 2 = 1
8 1 H 3 = 2 4 2 0 1 0 1 1 1 1 2 1 VD:0 0 00 0 05 5 55 0 0 I= 5 5 55
0 05 5 55 0 00 0 00 0 00 0 00 0 03.3. PHT HIN BIN GIN TIP3.3.1 Mt s
khi nim c bn*nh v im nhnh s l mt mng s thc 2 chiu (Iij) c kch thc
(MN), trong mi phn t Iij(i = 1,...,M; j = 1,...,N) biu th mc xm ca
nh ti (i,j)tng ng. nh c gi l nh nh phn nu cc gi tr Iij ch nhn gi tr
0hoc 1. y ta ch xt ti nh nh phn v nh bt k c th a v dngnh phn bng k
thut 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 gingGi 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 k nhau, cc im P0,P2,
P4, P6 l cc 4-lng ging ca im P, cn cc im P0, P1, P2, P3, P4, P5,P6,
P7 l cc 8-lng ging ca P.37 38. P3 P2P1P4PP0P5 P6P7Hnh 1.3. Ma trn
8-lng ging k nhau*i tng nh Hai im Ps, Pe E, 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,...,nNhn xt: Quan h k-lin thng trong E (k=4,8) l mt quan h phn
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. V sau ta sgi mi thnh phn k-lin thng ca
nh l mt i tng nh.3.3.2. Chu tuyn ca mt i tng nhnh ngha 3.1: [Chu
tuyn] Chu tuyn ca mt i tng nh l dy cc im ca i tng nhP1,,Pn sao cho
Pi v Pi+1 l cc 8-lng ging ca nhau (i=1,...,n-1) v P1 l8-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 k tip ca chu
tuyn l di cachu tuyn v k hiu Len(C) v hng PiPi+1 l 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. P Hnh 3.1. V
d v chu tuyn ca i tng nh 38 39. nh ngha 3.2 [Chu tuyn i ngu] Hai
chu tuyn C= v C= c gi l i nguca nhau nu v ch nu i j sao cho:(i) Pi
v Qj l 4-lng ging ca nhau.(ii) Cc im Pi l vng th Qj l 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 ch nu(i) Chu tuyn i ngu C l chu tuyn ca cc im nn(ii) di
ca C nh hn di Cnh ngha 3.4 [Chu tuyn trong]Chu tuyn C c gi l chu
tuyn trong (Hnh 3.2b) nu v ch nu:(i) Chu tuyn i ngu C l chu tuyn ca
cc im nn(ii) di ca C ln hn di C Chu tuyn CChu tuyn CChu tuyn C Chu
tuyn C a) Chu tuyn ngoi b) Chu tuyn trongHnh 3.2. Chu tuyn trong,
chu tuyn ngoinh ngha 3.5 [im trong v im ngoi chu tuyn]Gi s C= l chu
tuyn ca mt i tng nh v P l mtim nh. Khi :(i) Nu na ng thng xut pht t
P s ct chu tuyn C ti s l ln,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 l im ngoi
chu tuyn C v k hiu out(P,C).B 3.1 [Chu tuyn i ngu] Gi s E 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)
39 40. (ii) Nu C l chu tuyn ngoi th in(Pi,C) i (i=1,...,n)B 3.2
[Phn trong/ngoi ca chu tuyn] Gi s E 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] Gi s E l mt i tng
nh v CE l chu tuyn ngoi ca E.Khi CE l duy nht.3.3.3. Thut ton d bin
tng qut Biu din i tng nh theo chu tuyn thng da trn cc k thutd bin.
C hai k thut d bin c bn. K thut th nht xt nh bin thuc t nh vng sau
mt ln duyt nh mt th, sau p dng cc thutton duyt cnh th. K thut th
hai da trn nh vng, kt hp ng thiqu trnh d bin v tch bin. y ta quan
tm cch tip cn th hai.Trc ht, gi s nh c xt ch bao gm mt vng nh 8-lin
thng, c bao bc bi mt vnh ai cc im nn. D thy l mt vng 4-lin thng ch
l 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 D bo 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 v imbin, v quan h lin thng, cc thut ton d bin cho
ta cc ng binmang cc sc thi rt khc nhau. Kt qu tc ng ca ton t d bin
ln mt im bin ri l im binri+1 (8-lng ging ca ri). Thng thng cc ton t
ny c xy dng nhmt hm i s Boolean trn cc 8-lng ging ca ri. Mi cch xy
dngcc ton t u ph thuc vo nh ngha quan h lin 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 c cc 8-lng ging cami im nn thut ton thng km hiu qu.
khc phc cc hn chtrn, thay v s dng mt im bin ta s dng cp im bin (mt
thuc ,mt thuc ), cc cp im ny to nn tp nn vng, k hiu l NV vphn tch
ton t d bin thnh 2 bc:40 41. Xc nh cp im nn vng tip theo. La chn im
binTrong bc th nht thc hin chc nng ca mt nh x trn tpNV ln NV v bc
th hai thc hin chc nng chn im bin. Thut ton d bin tng qut Bc 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 li bc 2. Vic
xc nh cp nn-vng xut pht c thc hin bng cch duytnh ln lt t trn xung
di v t tri qua phi ri kim tra iu kin lachn cp nn-vng. Do vic chn im
bin ch mang tnh cht quy c, nnta gi nh x xc nh cp nn-vng tip theo l
ton t d bin.nh ngha 3.6 [Ton t d bin] Gi s T l mt nh x nh sau: T:
NV NV(b,r) a (b,r) Gi T l mt ton t d bin c s nu n tho mn 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 th nh 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 nhng im nm gia hai im
b v r. Cch nh ngha th ba 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 th c chung mt im bin.
Bi vy, qu trnh chn im binc thc hin nh sau: i:= 1; (bi,ri):=
(bo,ro); While K(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 th n trng vi cp nn vng
xut pht: (bn,rn)= (bo,ro) 41 42. * 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 t tri 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 theo u 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 tr cho
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 tr cho hng tip theo ta
lp bng da trn gi tr pdir tnh c trc theo cc kh nng c th xy ra: 42
43. pdir im trng trc Trng so vi en mi01 212 423 434 645 656 067 070
2 Do cng thc tnh hng tip theo s l :dir= ((pdir+3)/ 2 * 2)%8 ; 43
44. Chng 4: XNG V CC K THUT TM XNG4.1. GII THIUXng c coi nh hnh dng
c bn ca mt i tng, vi s tcc im nh c bn. Ta c th ly c cc thng tin v
hnh dng nguynbn ca mt i tng thng qua xng.Mt nh ngha xc tch v xng da
trn tnh continuum (tng tnh hin tng chy ng c) c a ra bi Blum (1976)
nh sau: Githit rng i tng l ng nht c ph bi c kh v sau dng lnmt vng
bin la. Xng c nh ngha nh ni gp ca cc vt la vti chng c dp tt. a) nh
gc b) nh xngHnh 4.1. V d v nh 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 th chia
thnh hai loi thut ton tm xng c bn: Cc thut ton tm xng da trn lm mnh
Cc thut ton tm xng khng da trn lm mnh4.2. TM XNG DA TRN LM
MNH4.2.1. S lc v thut ton lm mnh Thut ton lm mnh nh s nh phn l mt
trong cc thut ton quantrng trong x l nh v nhn dng. Xng cha nhng
thng tin bt binv cu trc ca nh, gip cho qu trnh nhn dng hoc vect ho
sau ny. 44 45. Thut ton lm mnh l qu trnh lp duyt v kim tra tt c cc
imthuc i tng. Trong mi ln lp tt c cc im ca i tng s ckim tra: nu nh
chng tho mn iu kin xo no tu thuc vo mithut ton th n s b xo i. Qu
trnh c lp li cho n khi khng cnim bin no c xo. i tng c bc dn lp bin
cho n khi nob thu mnh li ch cn cc im bin. Cc thut ton lm mnh c phn
loi da trn phng php x lcc 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
x l theo phng php song song, tc l c x l cng mt lc.Gi tr ca mi im
sau mt ln lp ch ph thuc vo gi tr ca 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
b vi x l mivi x l s x l mt vng ca i tng, n c quyn c t cc im vng khc
nhng ch c ghi trn vng ca n x l. Trong thut ton lm mnh tun t cc im
thuc i tng s ckim tra theo mt th t no (chng hn cc im c xt t tri
quaphi, t trn xung di). Gi tr ca im sau mi ln lp khng nhngph thuc
vo gi tr ca 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 tho mn ngthi tt c cc
tiu chun. Bo ton tnh lin thng ca i tng v phn b ca i tng S tng hp
gia xng v cu trc ca nh i tng Bo ton cc thnh phn lin thng Bo ton cc
im ct Xng ch gm 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 c
lm mnh Xng nhn c bt bin vi php quay. 45 46. 4.2.2. Mt s thut ton lm
mnh Trong phn ny im qua mt s c im, u v khuyt im ca ccthut ton c
nghin cu. 1o. Thut ton lm mnh c in l thut ton song song, to ra xng
8 lin thng, tuy nhin n rt chm, gy t nt, xo hon ton mt s cu hnh nh.
2o. Thut ton lm mnh ca Toumazet bo ton tt c cc im ct khng gy t nt i
tng. Tuy nhin, thut ton c nhc im l rt chm, rt nhy cm vi nhiu, xng
ch l 4-lin thng v khng lm mnh c vi mt s cu hnh phc tp 3o. Thut ton
lm mnh ca Y.Xia da trn ng bin ca i tng, c th ci t theo c phng 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 t nh). 4o. Thut ton lm mnh ca
N.J.Naccache v R.Shinghal. Thut ton c u im l nhanh, xng to ra c kh
nng khi phc nh ban u ca i tng. Nhc im chnh ca thut ton l rt nhy vi
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 l xng to ra l xng 4-lin thng v xo mt mt s cu hnh nh. 6o.
Thut ton lm mnh ca P.S.P Wang v Y.Y.Zhang da trn ng bin ca i tng, c
th ci t theo phng php song song hoc tun t, xng l 8-lin thng, t chu
nh hng ca nhiu. Nhc im chnh ca thut ton l tc chm. 7o. Thut ton lm
mnh song song thun tu nhanh nht trong cc thut ton trn, bo ton tnh
lin thng, t chu nh hng ca nhiu. Nhc im l xo hon ton mt s cu hnh nh,
xng to ra l xng 4-lin thng.4.3. TM XNG KHNG DA TRN LM MNH tch c xng
ca i tng c th s dng ng bin ca itng. Vi im p bt k trn i tng, ta bao
n bi mt ng bin.Nu nh c nhiu im bin c cng khong cch ngn nht ti p th
p nmtrn trc trung v. Tp tt c cc im nh vy lp thnh trc trung v hayxng
ca i tng. Vic xc nh xng c tin hnh thng quahai bc:46 47. Bc th nht,
tnh khong cch t mi im nh ca i tng n im bin gn nht. Nh vy cn phi tnh
ton khong cch ti tt c cc im bin ca nh.Bc th hai, khong cch nh c tnh
ton v cc im nh c gi tr ln nht c xem l nm trn xng ca i tng.4.3.1.
Khi qut v lc Voronoi Lc Voronoi l mt cng c hiu qu trong hnh hc tnh
ton.Cho hai im Pi, Pj l hai phn t ca tp gm n im trong mt phng.Tp cc
im trong mt phng gn Pi hn Pj l na mt phng H(Pi, Pj)cha im Pi v b
gii hn bi ng trung trc ca on thng PiPj. Do, tp cc im gn Pi hn bt k
im Pj no c th thu 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/S Voronoi] S
Voronoi ca l hp ca tt c cc 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 nh sau: V(U) = {P| v U, w U :
d(P,v) < d(P,w)} = V(Pi) Pi U (4.3)4.3.2. Trc trung v Voronoi ri
rcnh 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(.,.) nh sau:
(x, y)S: map(x, y) = min d[(x, y), (xi, yi)] (4.4) itrong (x , y )
B(S) - tp cc im bin ca S i i Tp tt c cc 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 t mt 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 c hnh ri rc lnlin tc.47 48. 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 th cha cc ng bin ri
nhau, nn ^B(S) bao gm nhiutp con, mi tp m t mt ng bin phn
bit:^B(S)={B1(S),..BN(S)}(4.6)nh ngha 4.5 [Trc trung v Voronoi ri
rc (DVMA)]Trc trung v Voronoi ri rc c nh ngha l kt qu ca s Voronoi
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 rcnh 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 v
Voronoi: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).a) b) c)d) Hnh 4.2. Xng
Voronoi ri rc nh hng ca cc hm hiu chnh khc nhau. (a) nh nh phn. (b)
S Voronoi. (c) Hiu chnh bi hm Potential, T=9.0. (d) Hiu chnh bi hm
Potential, T=18.0 48 49. 4.3.4. Thut ton tm xngTrong mc ny s trnh
by tng c bn ca thut ton tm xngv m t bng ngn ng ta Pascal.Tng trng:
Vic tnh ton s Voronoi c bt u t mt imsinh trong mt phng. Sau im sinh
th hai c thm vo v qu trnhtnh ton tip tc vi a gic Voronoi tm c vi im
va c thmvo . C nh th, qu trnh tnh ton s Voronoi 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 th gy ra s phnvng ton b cc a gic Voronoi c tnh.Chia
tr: Tp cc im bin u tin c chia thnh hai tp imc kch c bng nhau. Sau
thut ton tnh ton s Voronoi cho c haitp con im bin . Cui cng, ngi ta
thc hin vic ghp c hai s Voronoi trn thu c kt qu mong 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 s Voronoi tr nn n gin. Vth, vic
tnh s Voronoi tr thnh vn lm th no trn hai s Voronoi li vi nhau.
Thut ton s trnh by y l s kt hp ca hai tng trn. Tuynhin, n s mang
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 gpkhc , v hai s Voronoi tng ng Vor(SL) v Vor(SR). thu cs
Vornonoi Vor(SL SR), ta thc hin vic trn hai s trn v xcnh li mt s a
gic s b sa i do nh hng ca cc im bn cnhthuc s kia. Mi phn t ca s l
mt b phn ca ng trung trcni hai im m mt im thuc Vor(SL) v mt thuc
Vor(SR). Trc khixy dng , ta tm ra phn t u v cui ca n. Nhn vo hnh
trn, tanhn thy rng cnh 1 v 5 l cc tia. D nhn thy rng vic tm ra
cccnh u v cui ca tr thnh vic tm cnh vo t v cnh ra t.3 1 t 7 CH(SL)1
11 CH(SR)6492105t58 Hnh 4.3. Minh ho thut ton trn hai s Voronoi 49
50. Sau khi tm c t v t, cc im cui ca t c s dng xydng phn t u tin ca
(1 trong hnh trn). Sau thut ton tm imgiao ca vi Vor(SL) v Vor(SR).
Trong v d trn, u tin giao viV(3). K t y, cc im nm trn phn ko di s
gn im 6 hn im3. Do , phn t tip theo 2 ca s thuc vo ng trung trc ca
im6 v im 7. Sau im giao tip theo ca s thuc v Vor(SL); bygi s i vo
V(9) v 2 s c thay th bi 3. Qu trnh ny s kt thckhi gp phn t cui 5.
Trn y ch l minh ho cho thut trn hai s Voronoi 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 s c qut vo tng dng mt. Gi s ti
bc th i, ta thu cmt s Voronoi gm i-1 hng cc im sinh Vor(Si-1). Tip
theo, ta qutly mt hng Li cc im nh t tp cc im bin cn li. Thc hin
victnh s Voronoi Vor(Li) cho hng ny, sau trn Vor(Si-1) vi
Vor(Li).Kt qu ta s c mt s mi, v li thc hin vic qut hng Li+1 ccim
sinh cn li v.v.. Qu trnh ny s kt thc khi khng cn im bin no thm vo s
Voronoi. Do Vor(Li) s c dng rng lc (nu Li c kim th Vor(Li) s gm k-1
ng thng ng), nn vic trn Vor(Si-1) viVor(Li) c phn n gin hn. v5
p8p9v2 p10t p6 v4 p7v1v3 tp5v6p4 Cc im thucp1 p2 p3 Si-1Hnh 4.4.
Minh ho thut ton thm mt im bin vo s Voronoi Gii thut trn c th c m t
bng ngn ng ta Pascal nh sau:Procedure VORONOI (*Si: Tp cc im ca i
dng qut u tin, 0 hmax){hmax = h;index = i; 53 54. }}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 s lng im DouglasPeuckerint DouglasPeucker(POINT *pLINE, int
n, float ){int i, j;BOOL chiso [MAX_PT];for(i = 0; i < m; i++)
//Tt c cc im c gi lichiso[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 width5.1.3.1. 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 s c gi li. Qu trnh ny c thc hin
vi phncn li ca ng cong bt u t im va tm c cho n khi htng cong. C th
nh sau:54 55. P3P2 diP4 dk P1 P5 Hnh 5.2. n gin ha ng cong vi thut
ton Band WidthBt u bng vic xc nh im u tin trn ng cong v coi nh l mt
im cht (P1). im th ba (P3) c coi l im ng. imgia im cht v im ng (P2)
l im trung gian. Ban u khong ccht im trung gian n on thng ni im cht
v im ng c tnhton v kim tra. Nu khong cch tnh c ny nh hn mt ngng cho
trc th im trung gian c th b i, tin trnh tip tc vi im chtl im cht c,
im trung gian l im ng c v im ng l im ktip sau im ng c. Trong trng
hp ngc li, khong cch tnh cln hn ngng cho trc th im trung gian s c
gi li, 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 nh l mt im cht (P1). im th ba (P3)
c coi l im ng. im gia im cht v im ng (P2) l im trung gian. Bc 2:
Tnh khong cch t im trung gian n on thng ni hai im cht v im ng. Bc
3: Kim tra khong cch tm c nu nh hn mt ngng cho trc th im trung gian
c th b i. Trong trng hp ngc li im cht chuyn n im trung gian. Bc 4:
Chu trnh c lp li th im trung gian c chuyn n im ng v im k tip sau im
ng c ch nh 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 vct ho l mnh.55 56. 5.1.3.2. Chng trnh//Hm tnh ng cao t
nh n on thng ni hai im dau, cuoifloat Tinhduongcao(POINT dau, POINT
cuoi, POINT dinh){floot h;tnh ng caoreturm h ;}//Hm quy nhm nh du
loi b cc im trong ng congvoid BWSimple(POINT *pLINE, int chot, int
tg, BOOL *chiso, float , int n){if(Tinhduongcao(pLINE[chot],
pLINE[tg+1], pLINE[tg]) )chiso[tg] = 0;elsechot = tg;tg = tg +
1if(tg < n - 1)BWSimple (pLINE, chot, tg, chiso, , n) ;}//Hm rt
gn s lng im BandWidthint BandWidth(POINT *pLINE, int n, floot
){inti, j;BOOL chiso [MAX_PT];for (i = 0; i < n; i++)chiso[i]=
TRUE; //Tt c cc im c gi liBWSimple(pLINE, 0, 1, chiso, , n);for(i=
j= 0; i < n; i++)if(chiso [i]== TRUE) 56 57. pLINE [j ++1] =
pLINE [i];return j;}5.1.4. Thut ton Angles5.1.4.1. tngTng 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.P3 kP2 i P4 P5P1Hnh 5.3. n
gin ha ng cong vi thut ton Anglesim th 3 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 th b i trong trng hp ngc li imcht s l im trung gian c v qu
trnh lp vi im trung gian l imng c, im ng mi l im k tip sau im ng c.
Tin trnh thchin cho n ht ng cong.5.1.4.2. Chng trnh//Hm tnh ng cao
t nh n on thng ni hai im dau, cuoifloat Tinhgoc(POINT dau, POINT
cuoi, POINT dinh){float ;tinhgoc (t vit)return ;}//Hm quy nhm nh du
loi b cc im trong ng congvoid ALSimple(POINT *pLINE,int chot,int
tg,BOOL *chiso,float ,int n){ 57 58. 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 s lng im Anglesint Angles(POINT *pLINE, int n, float
){int i, j, chiso [MAX];for (i = 0; i < n; i++) //Tt c cc im c
gi li 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 s b i cc imgia
thng hng. Thut ton Angles phi c = 180o b i cc im giathng hng.5.2.
XP X A GIC BI CC HNH C S Cc i tng hnh hc c pht hin thng thng qua cc
k thutd bin, kt qu tm c ny l cc ng bin xc nh i tng. l,mt dy cc im
lin tip ng knh, s dng cc thut ton n gin honh Douglas Peucker, Band
Width, Angle v.v.. ta s thu 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? Nh ta bit mt a gic c th c hnh dng ta nh mt hnh c 58
59. s, c th c nhiu cch tip cn xp x khc nhau. Cch xp x da trn ccc
trng c bn sau:c trng ton cc: Cc m men thng k, s o hnh hc nh chuvi,
din tch, tp ti u cc hnh ch nht ph hay ni tip a gic v.v.. c trng a
phng: Cc s o c trng ca ng cong nhgc, im li, lm, un, cc tr v.v..Nhn
dng i tng Bt bin Bt binng dngAphinng trn EllipseEllipse Tam gic Hnh
ch nht T gicHnh 5.4. S phn loi cc i tng theo bt bin Vic xp x t ra
rt c hiu qu i vi mt s hnh phng c bitnh tam gic, ng trn, hnh ch nht,
hnh vung, hnh ellipse, hnhtrn v mt a gic mu.5.2.1 Xp x a gic theo
bt bin ng dng Hnh 5.5. Xp x a gic bi mt a gic muMt a gic vi cc nh
V0,..,Vm-1 c xp x vi a gic muU0,..,Un-1 vi o xp x nh sau: dE (V , U
) = min,0 d m 1 n 59 60. Trong n 12 area (V0 LVm1 ) d =min r 0 2 ,
R 2 j =0kR U j + a V( j + d ) mod m , k= area (U 0 LU n 1 ) , vi R
ld php quay quanh gc to mt gc . Trong , d c tnh hiu qu bng cng thc
sau: n 1 1 n 1 n 1 n 1 d = |V( j + d ) mod m |2 | V( j + d ) mod m
|2 + k 2 |U j |2 2 k | U jV( j + d ) mod m | j =0n j =0 j =0 j =0d
y Uj, Vj c hiu l cc s phc ti cc nh tng ng. Khi m >> n th phc
tp tnh ton rt ln. Vi cc hnh c bit nh hnh trn, ellipse, hnh ch nht,
hnh xc nh duy nht bi tm v mt nh (a gic u ) ta c th vn dng cc phng
php n gin hn nh bnh phng ti thiu, cc bt bin thng k v hnh hc. nh
ngha 5.1 Cho a gic Pg c cc nh U0, U1,..., Un(U0 Un) Khi m men bc
p+q c xc nh nh sau:M pq = x p y q dxdy . PgTrong thc hnh tnh tch
phn trn ngi ta thng s dng cng thc Green hoc c th phn tch phn bn
trong a gic thnh tng i s ca cc tam gic c hng OUiUi+1 . U U2 1 f (
x, y) xPgpy q dxdy =- n 1 U0U3 sign( x yi =0i i +1 xi +1 yi )
O(0,0) - f ( x, y) xOU iU i +1py q dxdy Un- Hnh 5.6. Phn tch min a
gic thnh tng i s cc min tam gic 60 61. a. Xp x a gic bng ng trn Dng
phng php bnh phng ti thiu, ta c o xp x: 1 n 2E(Pg,Cr)= min a ,b ,cR
( xi + yi2 + axi + byi + c) 2 n i =1 Hnh 5.7. Xp x a gic bng ng
trnb. Xp x a gic bng ellipse Cng nh i vi ng trn phng trnh xp x i vi
ellipse ccho bi cng thc:1 n 2 E(Pg,El)=min a ,b ,c ,d ,eR ( x +
ayi2 + bxi yi + cxi + dyi + e) 2n i =1 i Mt bin th khc ca phng php
bnh phng ti thiu khi xp xcc ng cong bc hai c a ra trong [7].c. Xp x
a gic bi hnh ch nht S dng tnh cht din tch bt bin qua php quay, xp x
theo dintch nh sau: 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:2 11 tg2 =20 - 02
.Gi din tch ca hnh ch nht nh nht 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) y x Hnh
5.8. Xp x a gic bng hnh ch nht 61 62. d. Xp x a 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
x a gic theo bt bin aphin Trong [7] a ra m hnh chun tc v bt bin
aphin, cho php chngta c th chuyn bi ton xp x i tng bi bt bin aphin
v bi tonxp x mu trn cc dng chun tc. Nh vy c th a vic i snh cci tng
vi mu bi cc bt bin ng dng, chng hn vic xp x bitam gic, hnh bnh hnh,
ellipse tng ng vi xp x tam gic u, hnhvung, hnh trn v.v... Th tc xp
x theo bt bin aphin mt a gic vihnh c s c thc hin tun t nh sau: + Bc
0: Phn loi bt bin aphin cc dng hnh c s Dng hnh c s Dng chun tc Tam
gicTam gic u Hnh bnh hnhHnh vung Ellipse ng trn + Bc 1: Tm dng chun
tc c s Pg tho mn iu kin:m01 = m10 = 0 (php tnh tin)m02 = m20 = 1
(php co dn theo hai trc x, y) (**)m = m = 0 1331 + Bc 2:Xc nh bin i
aphin T chuyn a gic thnh a gic Pg dngchun tc (tho mn tnh cht (**)).
Xp x a gic Pg vi dng chun tc c s Pg tm c bc 1 vi o xp x E(Pg,Pg). +
Bc 3: Kt lun, a gic ban u xp x T-1(Pg) vi o xp x E(Pg,Pg). 62 63. i
vi bc 1 trong [7] a ra hai v d sau:V d 1: Tn ti duy nht tam gic u
P1P2P3 tho mn tnh cht (**) l428 3 = P1=(0,-2),P2= ( 3, ) , P3= ( 3,
) , 3 .V d 2: Tn ti hai hnh vungP1P2 P3 P4 tho mn tnh cht (**) Hnh
vung th nht c 4 nh tng ng l (-p,-p),(-p,p), (p,- 34p),(p,p), vi p=
4 Hnh vung th hai c 4 nh tng ng l (-p,0),(p,0), (0,-p),(0,p),4vi p=
3 .5.3. BIN I HOUGH5.3.1. Bin i Hongh cho ng thngBng cch no ta thu
c mt s im vn t ra l cn phikim tra xem cc im c l ng thng hay khngBi
ton: Cho n im (xi; yi) i = 1, n v ngng hy kim tra n im c tothnh ng
thng hay khng?* tngGi 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 Nh vy, mi im (xi; yi) trong mt phng
s tng ng vi mt sng thng b = - xia + yi trong mt phng tham s a, b. n
im (xi; yi) i =1, n thuc ng thng trong mt phng tng ng vi n ng
thngtrong mt phng tham s a, 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.63 64. *
Phng php:- Xy dng mng ch s [a, b] v gn gi tr 0 ban u cho tt c ccphn
t ca mng- Vi mi (xi; yi) v a, b l ch s ca phn t mng tho mnb = - xia
+ yi tng gi tr ca phn t mng tng ng ln 1 - Tm phn t mng c gi tr ln
nht nu gi tr ln nht tm c sovi s phn t ln hn hoc bng ngng cho trc th
ta c th kt luncc im nm trn cng 1 ng thng v ng thng c phng trnhy =
ax + b trong a, b tng ng l ch s ca phn t mng c gi tr lnnht 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 ch s [a, b] v gn gi tr 0+ (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 t ln nht c gi tr 44/5 = 80% - Kt
lun: 5 im ny nm trn cng 1 ng thngPhng trnh: y = 2x + 15.3.2. Bin i
Hough cho ng thng trong ta cc5.3.2.1. ng thng Hough trong ta cc 64
65. 0yr x.cos+y.sin=rH x Hnh 5.9. ng thng Hough trong to cc Mi im
(x,y) trong mt phng c biu din bi cp (r,) trong ta cc. Tng t mi ng
thng trong mt phng cng c th biu din bimt cp (r,) trong ta cc vi r l
khong cch t gc 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. Gi s
M(x,y) l m im thuc ng thng c biu din bi (r,),gi H(X,Y) l hnh chiu
ca gc to O trn ng thng ta c: X= r. cos v Y= r.sin Mt khc, ta c:
OH.HA=0 T ta c mi lin h gia (x,y) v (r,) nh sau: x*cos+y*sin= r. Xt
n im thng hng trong ta cc c phng trnhx*cos0+y*sin0= r0. Bin i Hough
nh x n im ny thnh n ng sintrong ta cc m cc ng ny u i qua (r0,0).
Giao im (r0,0) can ng sin s xc nh mt ng thng trong h ta cc. Nh
vy,nhng ng thng i qua im (x,y) s cho duy nht mt cp (r,) v cbao nhiu
ng qua (x,y) s c 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 nghingvn 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 ch s hng ca
mng cho bit gc lch ca mt ng thng v chs ct chnh l gi tr r khong cch
t gc to ti ng thng . Sau tnh tng s im nh nm trn nhng ng thng song
song nhau theo65 66. cc gc lch thay i. Gc nghing vn bn tng ng vi gc
c tng gatr mng tch lu cc i.Theo bin i Hough, mi mt ng thng trong mt
phng tngng c biu din bi mt cp (r,). Gi s ta c mt im nh (x,y)
trongmt phng, v qua im nh ny c v s ng thng, mi ng thng licho mt cp
(r,) nn vi mi im nh ta s xc nh c mt s cp(r,) tho mn phng trnh
Hough.x.cos+y.sin=r1 0 Hough[][r1]=3y x.cos+y.sin=r2Hough[][r1]=4x
Hnh 5.10. ng dng bin i Hough pht hin gc Hnh v trn minh ho cch dng
bin i Hough pht hin gcnghing vn bn. Gi s ta c mt s im 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 th nht c ba im
nh nn gi tr mng tch lu bng3, ng thng th hai c gia tr mng tch lu bng
4. Do , tng gi trmng tch lu cho cng gc trng hp ny bng 7. Gi
Hough[360][Max] l mng tch ly, gi s M v N tng ng lchiu rng v chiu
cao ca nh, ta c cc bc chnh trong qu trnh pdng bin i Hough pht hin
gc nghing vn bn nh sau: + Bc 1: Khai bo mng ch s Hough[][r] vi 0
3600 v 0 r M * M + N * N . + Bc 2: Gn gi tr khi to bng 0 cho cc phn
t ca mng. + Bc 3: Vi mi cp (x,y) l im gia y ca hnh ch nht ngoi tip
mt i tng. - Vi mi i t 0 n 360 tnh gi tr ri theo cng thc ri=
x.cosi+y.sin - Lm trn gi tr ri thnh s nguyn gn nht l r0 - Tng gi tr
ca phn t mng Hough[i][r0] ln mt n v.66 67. + Bc 4: Trong mng
Hough[][r] tnh tng gi tr cc phn t theotng dng v xc nh dng c tng gi
tr ln nht. Do s phn t ca mt phn t mng Hough[0][r0] chnh l s imnh
thuc ng thng x.cos0+y.sin0= r0 v vy tng s phn t ca mthng chnh l tng
s im 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 tr cc phn t mng ln nht. 67
68. Ph lc 1:MT S NH DNG TRONG X L NH Hin nay trn th gii c trn 50
khun dng nh thng dng.Sau y l mt s nh dng nh hay dng trong qu trnh x
l nhhin nay.1. nh dng nh IMGnh 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 k lin nhau m dy ny s c lp li mt s lnno
. S ln lp ny s c lu trong byte m. Nhiu dyging nhau c lu trong mt
byte. 4 byte tip: m t kch c pixel. 2 byte tip: s pixel trn mt dng
nh. 2 byte cui: s dng 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 nh sau: 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
nh sau: 0x00 Count. Byte th hai 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 nh sau: 0x80 Count. Byte th hai cho
bit di dy cc pixel khng ging nhau khng nn c. 68 69. Loi 4: Dy cc
Pixel ging nhau. Tu theo cc bt cao ca byte u tin c bt hay tt. Nu
btcao c bt (gi tr 1) th y l gi nn cc byte ch gm 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 ch cn 1 bt cho 1 pixel
thay v 4 hoc 8 nh ni trn. Ton b nh ch c nhng im sng v ti tng ng vi
gitr 1 hoc 0. T l nn ca kiu nh dng ny l kh cao.2. nh dng nh PCXnh
dng nh PCX l mt trong nhng nh dng nh c in.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), d liu nh (Image data) v bng
mu m rng.Header ca tp PCX c kch thc c nh gm 128 byte vc phn b nh
sau: 1 byte: ch ra kiu nh dng.Nu l PCX/PCC th n lun cgi tr l 0Ah. 1
byte: ch ra version s dng nn nh, c th c cc gitr sau:+ 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 c bng mu. 1 byte: ch ra phng php m ho. Nu l 0
th m ho theophng php BYTE PACKED, ngc li l phngphp RLE. 1 byte: S
bt cho mt im nh plane. 1 word: to gc tri ca nh. Vi kiu PCX n c gi
tr l(0,0), cn PCC th khc (0,0). 1 word: to gc phi di. 1 word: kch
thc b rng v b cao ca nh. 69 70. 1 word: s im 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 v mt thanh ghi mu. Nh vy ta c 16thanh ghi mu. 1 byte:
khng dng n v lun t l 0. 1 byte: s bt plane m nh s dng. Vi nh 16 mu,
gi trny l 4, vi nh 256 mu (1pixel/8bits) th s bt plane lil 1. 1
byte: s bytes cho mt dng qut nh. 1 word: kiu bng mu. 58 byte: khng
dng. nh dng nh PCX thng c dng lu tr nh v thaotc n gin, cho php nn v
gii nn nhanh. Tuy nhin, v cu trcca n c nh, nn trong mt s trng hp lm
tng kch thc lutr. Cng v nhc im ny m mt s ng dng s dng mt kiunh dng
khc mm do hn: nh dng TIFF (Targed Image FileFormat) s m t di y.3.
nh dng nh TIFF Kiu nh dng TIFF c thit k lm nh bt cc vn lin quan n
vic m rng tp nh c nh. 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 my Macintosh. Hai loi ny khc nhau rt ln
th t cc byte lu tr trong cc s di 2 hay 4 byte. Nu trng ny c gi tr l
4D4Dh th l nh cho my Macintosh, nu l 4949h l ca my PC. + 1 word:
version. t ny lun c gi tr l 42. y l c trng ca file TIFF v khng thay
i. + 2 word: gi tr Offset theo byte tnh t u ti cu trc IFD l cu trc
th hai ca file. Th t cc byte ny ph thuc vo du hiu trng u tin. 70
71. Phn th 2(IFD): Khng ngay sau cu trc IFH m v trc xc nh bi trng
Offset trong u tp. C th c 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 tr ti
IFD tip theo. Nu y l IFD cui cng th trng ny c gi tr 0. Phn th 3: cc
DE: cc DE c d di c nh gm 12 bytev chia lm 4 phn: + 2 byte: ch ra du
hiu m tp nh c xy dng. + 2 byte: kiu d liu ca tham s nh. C 5 kiu
tham s c bn: 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 s lng
ch mc ca kiu d liu ch ra. N khng phi l tng s byte cn thit lu tr. c
s liu ny ta cn nhn s ch mc vi kiu d liu dng. + 4 byte: l Offset ti
im bt u d liu lin quan ti du hiu, tc l lin quan vi DE khng phi lu
tr vt l cng vi n nm mt v tr no trong file.D liu cha trong tp thng c
t chc thnh cc nhmdng (ct) qut ca d liu nh. Cch t chc ny lm gim b
nhcn thit cho vic c tp. Vic gii nn c thc hin theo 4 kiukhc nhau c
lu tr trong byte du hiu nn.71 72. 4. nh dng file nh BITMAP Mi file
BITMAP gm u file cha cc thng tin chung v file,u thng tin cha cc
thng tin v nh, mt bng mu v mt mngd liu nh. Khun dng c cho nh
sau:BITMAPFILEHEADER bmfh;BITMAPINFOHEADER
bmih;RGBQUADaColors[];BYTE aBitmapBits[];Trong , cc cu trc c nh
ngha nh sau:typedef struct tagBITMAPFILEHEADER {/* bmfh */
UINTbfType; DWORD bfSize; UINTbfReserved1; UINTbfReserved2; DWORD
bfOffBits;} BITMAPFILEHEADER;typedef struct tagBITMAPINFOHEADER {/*
bmih */ DWORD biSize; LONGbiWidth; LONGbiHeight; WORDbiPlanes;
WORDbiBitCount; DWORD biCompression; DWORD biSizeImage;
LONGbiXPelsPerMeter; LONGbiYPelsPerMeter; DWORD biClrUsed; DWORD
biClrImportant;} BITMAPINFOHEADER, *LPBITMAPINFOHEADER; vibiSize
kch thc ca BITMAPINFOHEADERbiWidthChiu rng ca nh, tnh bng s im
nhbiHeight Chiu cao ca nh, tnh bng s im nh 72 73. biPlanesS plane
ca thit b, phi bng 1biBitCountS bit cho mt im nhbiCompression Kiu
nnbiSizeImage Kch thc ca nh tnh bng byte phn gii ngang ca thit b,
tnh bng im nh trnbiXPelsPerMetermet phn gii dc ca thit b, tnh bng
im nh trnbiYPelsPerMetermetbiClrUsed S lng cc mu thc s c s dngS lng
cc mu cn thit cho vic hin th, bng 0 nubiClrImportanttt c cc mu u cn
hin thNu bmih.biBitCount > 8 th mng mu rgbq[] trng, ngc li thmng
mu c 2
