1 I HC THI NGUYN KHOA CNG NGH THNG TIN GIO TRNH MN HC X L NH Ngi son: PGS. TS. NNG TON,TS. PHM VIT BNH Thi Nguyn, Thng 11 nm 20072 LI NI U Khong hn mi nm tr li y, phn cng my tnh v cc thit b lin 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 lin quan n x l nh khng cn l thit b chuyn dng na. Khi nim nh s tr nn thng dng vi hu ht mi ngi trong x hi v vic thu nhn nh s bng cc thit b c nhn hay chuyn dng cng vi vic a 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 thng tin trong nhiu trng i hc trn c nc. Tuy nhin, ti liu gio trnh cn l mt iu kh khn. Hin ti ch c mt s t ti liu bng ting Anh hoctingPhp,tiliubngtingVitthrthim.Vimongmun ng gp vo s nghip o to v nghin cu trong lnh vc ny, chng tibinsoncungiotrnhXlnhdatrncngmnhc c duyt. Cun sch tp trung vo cc vn c bn ca x l nh nhm cung cp mt nn tng kin thc y v chn lc nhm gip ngi c c th t tm 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, trnh byTngquanvxlnh,cckhainimcbn,stngqutca mt h thng x l nh v cc vn c bn trong x l nh. Chng 2, trnhbycckthutnngcaochtlngnhdavoccthaotcvi imnh,nngcaochtlngnhthngquavicxlccimnh trong ln cn im nh ang xt. Chng ny cng trnh by cc k thut nng cao cht lng nh nh vo cc php ton hnh thi. Chng 3, trnh bycckthutcbntrongvicphthinbincaccitngnh theo c hai khuynh hng: Pht hin bin trc tip v pht hin bin gin tip. Chng 4 th hin cch k thut tm xng theo khuynh hng tnh tontrctrungvvhngtipcnxpxnhccthuttonlmmnh song song v gin tip. 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 gi trongnhiunmticckhaihcvcaohccaHCngngh- HQG H Ni, H Khoa hc t nhin HQG H Ni, Khoa Cng ngh thng tin H Thi Nguyn v.v.. Cun sch c th lm ti liu tham kho 3 cho sinh vin cc h k s, c nhn v cc bn quan tm n vn nhn dng v x l nh. Cctcgibytlngbitnchnthnhticcbnngnghip trong 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 v gip hon chnh ni dung cun sch ny. Xin cm n Lnh o Khoa Cngnghthngtin,HThiNguyn,BanGimcHThiNguyn h tr v to iu kin cho ra i gio trnh ny. Mcdrtcgngnhngtiliunychcchnkhngtrnhkhi nhngsaist.Chngtixintrntrngtipthu tt c nhng kin ng gp 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 Nguyn in thoi: 0280.846506Email: pvbinh@ictu.edu.vn Thi Nguyn, ngy 22 thng 11 nm 2007 CC TC GI 4 MC LC LI NI U.................................................................................................................................................................. 2 MC LC............................................................................................................................................................................. 4 Chng 1: TNG QUAN V X L NH.................................................................................. 9 1.1. X L NH, CC VN C BN TRONG X L NH................. 9 1.1.1. X l nh l g?....................................................................................................................................... 9 1.1.2. Cc vn c bn trong x l nh..................................................................................10 1.1.2.1. Mt s khi nim c bn................................................................................................10 1.1.2.2. Nn chnh bin dng...........................................................................................................10 1.1.2.3. Kh nhiu.......................................................................................................................................11 1.1.2.4. Chnh mc xm.........................................................................................................................11 1.1.2.5. Phn tch nh...............................................................................................................................11 1.1.2.6. Nhn dng.......................................................................................................................................12 1.1.2.7. Nn nh.............................................................................................................................................13 1.2. THU NHN V BIU DIN NH.........................................................................................14 1.2.1. Mu sc..........................................................................................................................................................14 1.2.1.1. M hnh mu RGB (Red, Green, Bule)..........................................................14 1.2.1.2. M hnh mu CMY (Cyan, Magenta, Yellow)......................................15 1.2.1.3. M hnh mu HSV (Hue, Saturation, Value)...........................................16 1.2.1.4. M hnh mu HLS.................................................................................................................19 1.2.2. Thu nhn, cc thit b thu nhn nh...............................................................................22 1.2.2.1. Giai on ly mu..................................................................................................................23 1.2.2.2. Lng t ha...............................................................................................................................24 1.2.3. Biu din nh..........................................................................................................................................24 1.2.3.1. M hnh Raster..........................................................................................................................24 1.2.3.2. M hnh Vector.........................................................................................................................25 Chng 2: CC K THUT NNG CAO CHT LNG NH...................26 2.1. CC K THUT KHNG PH THUC KHNG GIAN..........................26 2.1.1. Gii thiu.....................................................................................................................................................26 5 2.1.2. Tng gim sng............................................................................................................................26 2.1.3. Tch ngng............................................................................................................................................27 2.1.4. B cm...........................................................................................................................................................27 2.1.5. Cn bng histogram.........................................................................................................................28 2.1.6. K thut tm tch ngng t ng...................................................................................29 2.1.7. Bin i cp xm tng th........................................................................................................30 2.2. CC K THUT PH THUC KHNG GIAN....................................................31 2.2.1. Php nhn chp v mu...............................................................................................................31 2.2.2. Mt s mu thng dng...............................................................................................................33 2.2.3. Lc trung v...............................................................................................................................................34 2.2.4. Lc trung bnh........................................................................................................................................36 2.2.5. Lc trung bnh theo k gi tr gn nht .........................................................................37 2.3. CC PHP TON HNH THI HC...................................................................................38 2.3.1. Cc php ton hnh thi c bn...........................................................................................38 2.3.2. Mt s tnh cht ca php ton hnh thi .................................................................39 Chng 3: BIN V CC PHNG PHP PHT HIN BIN.....................44 3.1. GII THIU........................................................................................................................................................44 3.2. CC PHNG PHP PHT HIN BIN TRC TIP.................................44 3.2.1. K thut pht hin bin Gradient......................................................................................44 3.2.1.1. K thut Prewitt .......................................................................................................................46 3.2.1.2. K thut Sobel ...........................................................................................................................47 3.2.1.3. K thut la bn..........................................................................................................................47 3.2.2. K thut pht hin bin Laplace........................................................................................48 3.2.3. K thut Canny.....................................................................................................................................49 3.3. PHT HIN BIN GIN TIP....................................................................................................50 3.3.1 Mt s khi nim c bn..............................................................................................................50 3.3.2. Chu tuyn ca mt i tng nh....................................................................................51 3.3.3. Thut ton d bin tng qut.................................................................................................53 3.4.PHT HIN BIN DA VO TRUNG BNH CC B.............................56 3.4.1. Bin v bin i v mc xm.......................................................................................56 6 3.4.2. Pht hin bin da vo trung bnh cc b...............................................................57 3.5. PHT HIN BIN DA VO CC PHP TON HNH THI..........60 3.5.1. Xp x trn v xp x di i tng nh................................................................60 3.5.1. Thut ton pht hin bin da vo php ton hnh thi ...........................61 Chng 4: XNG V CC K THUT TM XNG........................................63 4.1. GII THIU........................................................................................................................................................63 4.2. TM XNG DA TRN LM MNH.........................................................................63 4.2.1. S lc v thut ton lm mnh.........................................................................................63 4.2.2. Mt s thut ton lm mnh...................................................................................................65 4.3. TM XNG KHNG DA TRN LM MNH...............................................65 4.3.1. Khi qut v lc Voronoi..............................................................................................66 4.3.2. Trc trung v Voronoi ri rc................................................................................................66 4.3.3. Xng Voronoi ri rc.................................................................................................................67 4.3.4. Thut ton tm xng.....................................................................................................................68 Chng 5: CC K THUT HU X L................................................................................71 5.1. RT GN S LNG IM BIU DIN...................................................................71 5.1.1. Gii thiu.....................................................................................................................................................71 5.1.2. Thut ton Douglas Peucker ..................................................................................................71 5.1.2.1. tng..............................................................................................................................................71 5.1.2.2. Chng trnh...............................................................................................................................72 5.1.3. Thut ton Band width.................................................................................................................73 5.1.3.1. tng..............................................................................................................................................73 5.1.3.2. Chng trnh...............................................................................................................................75 5.1.4. Thut ton Angles..............................................................................................................................76 5.1.4.1. tng..............................................................................................................................................76 5.1.4.2. Chng trnh...............................................................................................................................76 5.2. XP X A GIC BI CC HNH C S..................................................................77 5.2.1 Xp x a gic theo bt bin ng dng......................................................................78 5.2.1.1. Xp x a gic bng ng trn.............................................................................80 5.2.1.2. Xp x a gic bng ellipse..........................................................................................80 5.2.1.3. Xp x a gic bi hnh ch nht ..........................................................................80 7 5.2.1.4. Xp x a gic bi a gic u n cnh.............................................................81 5.2.2 Xp x a gic theo bt bin aphin....................................................................................81 5.3. BIN I HOUGH.....................................................................................................................................82 5.3.1. Bin i Hongh cho ng thng....................................................................................82 5.3.2. Bin i Hough cho ng thng trong ta cc......................................84 Chng 6: NG DNG X L NH..............................................................................................85 6.1. PHT HIN GC NGHING VN BN DA VOCHU TUYN.......................................................................................................................................................85 6.1.1. Tnh ton kch thc ch o ca cc i tng nh.................................85 6.1.2. Bin i Hough v pht hin gc nghing vn bn......................................87 6.1.2.1. p dng bin i Hough trong pht hin gc nghingvn bn................................................................................................................................................87 6.1.2.2. Thut ton pht hin v hiu chnh gc nghing vn bn.........88 6.1.2.3. Thc nghim v kt qu..................................................................................................91 6.2. PHN TCH TRANG TI LIU.................................................................................................93 6.2.1. Quan h Q.................................................................................................................................................93 6.2.2. Phn tch trang vn bn nh khong cch Hausdorff bi quanh Q.................................................................................................................................................................94 6.2.3. Phn tch trang vn bn da vo mu..........................................................................96 6.2.3.1. nh gi lch cu trc vn bn theo mu...........................................96 6.2.3.2. Thut ton phn tch trang vn bn da vo mu...............................99 6.3. CT CH IN DNH DA VO CHU TUYN.................................................... 101 6.3.1. t vn ............................................................................................................................................... 101 6.3.2. Mt s khi nim c bn........................................................................................................ 103 6.3.3. Thut ton ct ch in dnh da vo chu tuyn................................................ 104 6.3.3.1. Phn tch bi ton............................................................................................................... 104 6.3.3.2. Thut ton CutCHARACTER ct ch in dnhda vo chu tuyn.................................................................................................................................... 106 6.4. NHN DNG CH VIT.............................................................................................................. 107 6.5. TCH CC I TNG HNH HC TRONG PHIU IU TRADNG DU................................................................................................................................................... 108 6.5.1. Gii thiu................................................................................................................................................. 108 8 6.5.2. Tch cc i tng nh s dng chu tuyn....................................................... 109 6.6. TCH BNG DA TRN TP CC HNH CH NHT RI RC........................................................................................................................................................... 110 6.6.1. Phn tch bi ton........................................................................................................................... 111 6.7. PHT HIN I TNG CHUYN NG......................................................... 113 6.7.1. Pht hin i tng chuyn ng da theo hng tip cn tr khung hnh lin k......................................................................................................................... 113 6.7.2. Pht hin i tng chuyn ng theo hng tip cnkt hp 117 6.7.2.1. Tr nh v nh du Iwb........................................................................................... 117 6.7.2.2. Lc nhiu v pht hin dch chuyn..................................................... 118 6.7.2.3. Pht hin bin nh a cp xm Igc.................................................................. 118 6.7.2.4. Kt hp nh Igc vi Iwb............................................................................................. 119 Ph lc 1: MT S NH DNG TRONG X L NH..................................... 121 1. nh dng nh IMG............................................................................................................................... 121 2. nh dng nh PCX............................................................................................................................... 122 3. nh dng nh TIFF............................................................................................................................... 123 4. nh dng file nh BITMAP........................................................................................................ 125 Ph lc 2: CC BC THAO TC VI FILE AVI.................................................... 127 1. Bc 1: M v ng th vin..................................................................................................... 127 2. Bc 2: M v ng file AVI thao tc: ................................................................... 127 3. Bc 3: M dng d liu thao tc................................................................................. 128 4. Bc 4: Trng hp thao tc vi d liu hnh ca phim.............................. 128 5. Bc 5: Thao tc vi frame........................................................................................................... 128 Ph lc 3: MT S MODUL CHNG TRNH............................................................. 129 1. Nhm c, ghi v hin th nh................................................................................................... 129 1.1. Nhm c nh.................................................................................................................................. 129 1.2. Nhm ghi nh.................................................................................................................................... 137 1.3. Nhm hin th nh....................................................................................................................... 139 2. Nhm pht hin gc nghing vn bn................................................................................ 144 TI LIU THAM KHO............................................................................................................................. 1579 Chng 1:TNG QUAN V X L NH 1.1. X L NH, CC VN C BN TRONG X L NH1.1.1. X l nh l g?Conngithunhnthngtinquaccgicquan,trongthgic ng vai tr quan trng nht. Nhng nm tr li y vi s pht trin ca phn cng my tnh, x l nh v ho pht trin mt cch mnh m v c nhiu ng dng trong cuc sng. X l nh v ho ng mt vai tr quan trng trong tng tc ngi my. Qu trnh x l nh c xem nh l qu trnh thao tc nh u vo nhm cho ra kt qu mong mun. Kt qu u ra ca mt qu trnh x l nh c th l mt nh tt hn hoc mt kt lun. Hnh 1.1. Qu trnh x l nh nh c th xem l tp hp cc im nh v mi im nh c xem nh 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 bin P(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: Hnh 1.2. Cc bc c bn trong mt h thng x l nh Lu tr Thu nhn nh (Scanner, Camera,Sensor) Tinx l Trch chn c im H quyt nh i snh rt ra kt lun Hux l X L NH nh nh Tt hn Kt lun 10 1.1.2. Cc vn c bn trong x l nh1.1.2.1. Mt s khi nim c bn* nh v im nh:im nh c xem nh l du hiu hay cng sng ti 1 to trongkhnggiancaitngvnhcxemnhl1tphpccim 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 in t. nh thu nhn nh mong mun Hnh 1.3. nh thu nhn v nh mong munkhcphcngitasdngccphpchiu,ccphpchiu thng c xy dng trn tp cc im iu khin.Gi s (Pi, Pi) i =n , 1c n cc tp iu khinTm hm f: Pi f (Pi) sao chomin ) (2'1 =i iniP P f Gi s nh b bin i ch bao gm: Tnh tin, quay, t l, bin dng bc nht tuyn tnh. Khi hm f c dng:f (x, y) = (a1x + b1y + c1, a2x + b2y + c2) Ta c:( ) ( ) [ ] = = + + + + + = =nii i i i i iniy c y b x a x c y b x a Pi Pi f12'2 2 22'1 1 12 '1) ) ( ( PiPif(Pi) 11 cho min= + += + += + +=== = = == = = == = = =nininii i inininii inii i i inininii inii i i ix nc y b x ax y y c y b y x ax x x c y x b x acba1 1 1'1 1 11 1 1'1121 11 1 1'11 121111000 Gii h phng trnh tuyn tnh tm c a1, b1, c1

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Xc nh c hm f 1.1.2.3. Kh nhiuC 2 loi nhiu c bn trong qu trnh thu nhn nhNhiuhthng:lnhiucquylutcthkhbngccphpbin i Nhiungunhin:vtbnkhngrnguynnhnkhcphc bng cc php lc1.1.2.4. Chnh mc xm Nhmkhcphctnhkhngngucahthnggyra.Thng thng c 2 hng tip cn: Gimsmcxm:Thchinbngcchnhmccmcxmgn nhauthnhmtb.Trnghpchc2mcxmthchnhl chuynvnhentrng.ngdng:Innhmuramyinen trng. Tngsmcxm:Thchinnisuyraccmcxmtrunggian bngkthutnisuy.Kthutnynhmtngcngmncho nh 1.1.2.5. Phn tch nhLkhuquantrngtrongqutrnhxlnhtintihiunh. Trong phn tch nh vic trch chn c im l mt bc quan trng. Cc c im ca i tng c trch chn tu theo mc ch nhn dng trong qu trnh x l nh. C th nu ra mt s c im ca nh sau y: c im khng gian: Phn b mc xm, phn b xc sut, bin , im un v.v.. 12 c im bin i: Cc c im loi ny c trch chn bng vic thc hin lc vng (zonal filtering). Cc b vng c gi l mt n c im (feature mask) thng l cc khe hp vi hnh dng khc nhau (ch nht, tam gic, cung trn v.v..) cimbinvngbin:ctrngchongbincai tngvdovyrthuchtrongvictrchtrnccthuctnhbtbin cdngkhinhndngitng.Cccimnycthctrch chnnhtontgradient,tontlabn,tontLaplace,tontcho khng (zero crossing) v.v.. Vic trch chn hiu qu cc c im gip cho vic nhn dng cc itngnhchnhxc,vitctnhtoncaovdunglngnh lu tr gim xung. 1.1.2.6. Nhn dngNhndngtng(automaticrecognition),mtitng,phn loi 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 cu hi t ra l: mu (pattern) l g? Watanabe, mt trong nhng ngi i u trong lnh vc ny nh ngha: Ngc li vi hn lon (chaos), mu l mt thc th (entity), c xc nh mt cch ang ng (vaguely defined) v cthgnchonmttngino.Vdmucthlnhcavn tay,nhcamtvtnocchp,mtchvit,khunmtngi hoc mt k tn hiu ting ni. Khi bit mt mu no , nhn dng hoc phn loi mu c th:Hocphnloicmu(supervisedclassification),chnghnphn tch phn bit (discriminant analyis), trong mu u vo c nh danh nh mt thnh phn ca mt lp xc nh. Hocphnloikhngcmu(unsupervisedclassificationhay clustering) trong cc mu c gn vo cc lp khc nhau da trn mt tiu chun ng dng no . Cc lp ny cho n thi im phn loi vn cha bit hay cha c nh danh.Hthngnhndngtngbaogmbakhutngngvibagiai on 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.13 2o. Phn loi thng k.3o. i snh cu trc.4o. Phn loi da trn mng n-ron nhn to.Trongccngdngrrnglkhngthchdngcmtcchtip cnnlphnloitiudovycnsdngcngmtlcnhiu phngphpvcchtipcnkhcnhau.Dovy,ccphngthcphn loi t hp hay c s dng khi nhn dng v nay c nhng kt qu c trin vng da trn thit k cc h thng lai (hybrid system) bao gm nhiu mhnhkt hp. Vicgiiquytbitonnhndngtrongnhngngdngmi,ny sinh trong cuc sng khng ch to ra nhng thch thc v thut gii, m cntranhngyucuvtctnhton.cimchungcattc nhng 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 th tc phn tch d liu. 1.1.2.7. Nn nhNhm gim thiu khng gian lu tr. Thng c tin hnh theo c haicchkhuynhhnglnncbotonvkhngbotonthngtin. Nn khng bo ton th thng c kh nng nn cao hn nhng kh nng phc hi th km hn. Trn c s hai khuynh hng, c 4 cch tip cn c bn trong nn nh: Nnnhthngk:Kthutnnnydavovicthngktn xut xut hin ca gi tr cc im nh, trn c s m c chin lcmhathchhp.Mtvdinhnh cho k thut m ha nyl *.TIF Nnnhkhnggian:Kthutnydavovtrkhnggianca ccimnhtinhnhmha.Kthutlidngsging nhaucaccimnhtrongccvnggnnhau.Vdchok thut ny l m nn *.PCX Nnnhsdngphpbini:ylkthuttipcntheo hngnnkhngbotonvdovy,kthutthngnnhiu 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 cnlutrphngcnhvquylutsinhranhtheonguynl Fractal 14 1.2. THU NHN V BIU DIN NH1.2.1. Mu sc Mt ngi c th phn bit c vi chc mu nhng ch c th cm nhnchngngnmu.Bathuctnhcamtmul:Sc(Hue), thun khit (Saturation), v sng hay chi (Itensity).Trongxlnhvha,mhnhmulmtchsk thut ca mthtamu3chiuvitpccmunhthnhphncthtrng thy c trong h thng ta mu thuc mt gam mu c trng. V d nhmhnhmuRGB(Red,Green,Blue):lmtnvtpccmu thnh phn sp xp theo hnh lp phng ca h trc ta cc. Mcchcamhnhmulcho php cc ch s k thut quy c camtsloimuscthchhpviccmusccamtsgammu khc. Chng ta c th nhn thy trong m hnh mu ny, khng gian mu l mt tp hp nh hn ca khng gian cc mu c th nhn thy c, v vy mtmhnhmukhngthcsdngnhrttccthnhn thy. Sau y, ta xem xt mt s m hnh hay c s dng nht. 1.2.1.1. M hnh mu RGB (Red, Green, Bule) Mu,lcxanhlcy,lamxanhdatri(RGB)csdng ph bin nht. Nhng mu gc RGB c thm vo nhng mu gc khc iu to nn s ng gp ring ca tng mu gc c thm cng nhau mangliktqa.Tphpmunhthnhphnspxptheokhilp phng n v. ng cho chnh ca khi lp phng vi s cn bng v s lng tng mu gc tng ng vi cc mc xm vi en l (0,0,0) v trng (1,1,1). Hnh 1.4. M hnh mu RGB Blue(0,255) (0, 0, 1) (0,0,0) (1,0,0) Red (0,1,0)green 15 1.2.1.2. M hnh mu CMY (Cyan, Magenta, Yellow) L phn b tng ng cho cc mu , lc, lam v cng c s dng nh nhng b lc loi tr cc mu ny t nh sng trng. V vy CMY cn cgilccphnbloitrcamugc.Tphpmuthnhphn biu din trong h ta -cc cho m hnh mu CMY cng ging nh cho m hnh mu RGB ngoi tr mu trng (nh sng trng), c thay th mu en (khng c nh sng) ti ngun sng. Cc mu thng c to thnh bng cch loi b hoc c b t nh sng trng hn l c thm vo nhng mu ti. Hnh 1.5. Cc mu gc b v s pha trn gia chng Khi b mt c bao ph bi lp mc mu xanh tm, s khng c tia muphnchiutbmt.Muxanhtmloibphnmu phn x khi c tia sng trng, m bn cht l tng ca 3 mu , lc, lam. VthtacthcoimuCyanlmutrngtrimuvcngl mulamcngmulc.Tngtnhvytacmuthm(magenta) hp th mu lc, v th n tng ng vi mu cng mu lam. V cui cng mu vng (yellow) hp th mu lam, n s bng mu cng vi lc. Khi b mt ca thc th c bao ph bi xanh tm v vng, chng s hpthhtccphnmuvxanhlamcabmt.Khichtnti duynhtmulcbphnxtschiusngcanhsngtrng.Trong trnghpkhibmtcbaophbic3muxanhtm,vng, thm, hin tng hp th xy ra trn c 3 mu , lc v lam. Do , mu en s mu ca b mt. Nhng mi lin h ny c th c miu t bi: ((((

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BGRYMC111 Hnh 1.6. S bin i t RGB thnh CMY Black Green Blue Yellow Magenta CyanRed 16 1.2.1.3. M hnh mu HSV (Hue, Saturation, Value) CcmhnhmuRGB,CMYcnhhngchophncngtri ngc vi m hnh mu HSV ca Smith hay cn c gi l mu HSB vi B l Brightness ( sng), c nh hng ngi s dng da trn c s nn tng v trc gic v tng mu, sc v sc thi m thut. Hthngtacdnghnhtrvtpmuthnhphncakhng gian bn trong m hnh mu c xc nh l hnh nn hoc hnh chp su cnh nh trong hnh 1.7. nh hnh chp l su cnh khi V= 1 cha ng miquanhgiaccmusngvnhngmutrnmtphngviV=1 u c mu sng. Hnh 1.7. M hnh mu HSV Sc mu (hue) hoc H c o bi gc quanh trc ng vi mu l 0o,mulcl120o,mulaml240o(xemhnh1.7).Ccmubsung tronghnhchpHSV180oidinvimukhc.Gitr ca S l mt tp cc gi tr i t 0 trn ng trc tm (trc V) n 1 trn cc mt bn tinhcahnhchpsucnh.Sbohacotngichogam mu tng ng vi m hnh mu ny. M hnh mu dng hnh chp su cnh ny ng cao V vi nh l im gc ta (0,0). im nh l mu en c gi tr ta mu V= 0, ti cc im ny gi tr ca H v S l khng lin quan vi nhau. Khi im c S= 0 v V= 1 l im mu trng, nhng gi tr trung gian ca V i vi S=0(trnngthngquatm)lcc mu xm. Khi S= 0 gi tr ca H ph thuc c gi bi cc quy c khng xc nh, ngc li khi S khc 0 gi tr ca H s l ph thuc. NhvymtmunoV=1,S=1lgignhmuthunkhit trongmthutcsdngnhimkhiutrongccmuphatrn. Thmmutrngphhpgim S (khng c s thay i V) to nn s thay i sc thi ca gam mu. S chuyn mu c to ra bi vic gi S= 1vgim V to nn s thay i sc v tng mu to thnh bi vic thay i c hai S v V. 17 Chuyn i t RGB sang HSV Hm RGB_HSV_Conversion H: Sc mu [0-360] vi mu ti im 0 S: bo ha [0-1] V: Gi tr cng sng [0-1] Max: Hm ly gi tr cc i Min: Hm ly gi tr nh nht { //Xc nh gi tr cng sng V= Max(R,G,B) //Xc nh bo ha Temp= Min(R,G,B) If V=0 than S= 0 Else S= (V-Temp)/V End //Xc nh sc mu IF s=0 THEN H= Undefined Else Cr= (V-R)/(V-Temp); Cg= (V-G)/(V-Temp); Cb= (V-B)/(V-Temp); //Munmtrongkhonggiavng(Yellow)vta (Magenta) If R=V thenH= Cb-Cg // Mu nm trong khong gia xanh tm (cyan) v vng (yellow) If G= V thenH= 2+Cr-Cb 18 // Mu nm trong khong gia ti (magenta) v xanh (cyan) If B=V then H= 4+ Cg Cr H= 60*H // Chuyn sang //Loi cc gi tr m If H < 0 thenH= H+360 } Chuyn i t HSV sang RGB Hm HSV_RGB_Conversion() H: Sc mu [0-360] vi mu ti im 0 S: bo ha [0-1] V: Gi tr cng sng [0-1] { //Kim tra trng hp nh sng khng mu If S=0 then If H=Undifined then R= V G= V B= V Endif Else If H=360 then H= 0 Else H= H/60 endif I= Floor(H) F= H-I M= V*(1-S) N= V*(1-S*F) 19 K= V*(1-S*(1-F)) //(R,G,B)=(V,K,M) R= V; C= K; B= M If I=0 then (R,G,B)=(V,K,M); If I=1 then (R,G,B)=(N,V,M); If I=2 then (R,G,B)=(M,V,K); If I=3 then (R,G,B)=(M,N,V); If I=4 then (R,G,B)=(K,M,V); If I=5 then (R,G,B)=(V,M,N); } 1.2.1.4. M hnh mu HLS M hnh mu HLS c xc nh bi tp hp hnh chp su cnh i cakhnggianhnhtr.Scmulgcquanhtrcngcuhnhchp su cnh i vi mu ti gc 0o. Cc mu s xc nh theo th t ging nh trong biu CIE khi ranh gii ca n b xoay ngc chiu kim ng h: Mu , mu vng, mu lc, mu xanh tm, mu lam v thm. iu nycnggingnhthtscxptrongmuhnhchpsucnhn HSV. 20 Hnh 1.8. M hnh mu HLS Chngtacthxem mu HLS nh mt s bin dng cu mu HSV mtrongmunymutrngckohnglnhnhchpsucnh pha trn t mt V= 1. Nh vi mu hnh chp su cnh n, phn b sung camtmuscc t v tr 180o hn l xunh quanh hnh chp su cnhi,sbohacoxungquanhtrcng,t0trntrcti1 trn b mt. sng bng khng cho mu en v bng mt cho mu trng. Chuyn i t RGB sang HLS Hm RGB_HLS_Conversion() H: Sc mu [0-360] vi mu ti im 0 S: bo ha [0-1] V: Gi tr cng sng [0-1] Max: Hm ly gi tr cc i Min: Hm ly gi tr nh nht { //Xc nh sng M1= Max(R,G,B) M2= Min(R,G,B) L= (M1+M2) //Xc nh bo ha If M1=M2 //Trng hp khng mu S= 0 H= Undefined Else If L 240 and H = 2fxmax fy >= 2fymax Trong fxmax, fymax l tn s cao nht ca tn hiu theo trc x, y. Dng tn hiu nh nh cha tn hiu quang hc 24 1.2.2.2. Lng t hanh sau khi ly mu s c dng f(m,n) vi m, nl nguyn nhng gi trf(m,n)vnlgitrvtllintc.Qutrnhbinigitrf(m,n) thnh mt s nguyn thch hp lu tr gi l lng t ho. y l qu trnh nh x mt bin lin tc u vo bin ri rc u* thuc tp hu hn [u1, u2,..uL] xc nh trc, L l mc lng t ho c to ra. V d: + To nh a cp xm th L=256, f(m,n) = g[ ] 255 , 0 + To nh 224 th L=224, f(m, n) = g[ ] 1 2 , 024 1.2.3. Biu din nhnhtrnmytnhlktquthu nhn theo cc phng php s ho cnhngtrongccthitbkthutkhcnhau.Qutrnhlutrnh nhm 2 mc ch: Tit kim b nhGim thi gian x lViclutrthngtintrongbnhcnhhngrtlnnvic hin th, in n v x l nh c xem nh l 1 tp hp cc im vi cng kchthcnusdngcngnhiuimnhthbcnhcngp,cng mn v cng 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 dng v c trng ca mi nh c th, trn c s cc nh thng c biu din theo 2 m hnh c bn1.2.3.1. M hnh Raster ylcchbiudinnhthngdngnhthinnay,nhcbiu dindidngmatrnccim(imnh).Thngthunhnquacc thit b nh camera, scanner. Tu theo yu cu thc th m mi im nh c biu din qua 1 hay nhiu bt M hnh Raster thun li cho hin th v in n. Ngy nay cng ngh phn 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 vic hinthtrongmitrngWindowslMicrosoftarakhundngnh DIB(DeviceIndependentBitmap)lmtrunggian.Hnh1.4thhnhquy trnh chung hin th nh Raster thng qua DIB. 25 Mttrongnhnghngnghincucbntrnmhnhbiudin nylkthutnnnhcckthutnnnhlichiaratheo2khuynh hnglnnbotonvkhngbotonthngtinnnbotonckh nngphchihontondliubanucnnukhngbotonchc khnngphchisaischophpno.Theocchtipcnny 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 gm c trong cc k thut nn c kh nng phc hi d liu 100% v nn c kh nng phc hi vi sai s nhn c. Hnh 1.9. Qu trnh hin th v chnh sa, lu tr nh thng qua DIB 1.2.3.2. M hnh Vector Biudinnhngoimcchtitkimkhnggianlutrddng chohinthvinncnmboddngtronglachnsaochpdi chuyn tm kim Theo nhng yu cu ny k thut biu din vector t ra u vit hn.Trongmhnhvectorngitasdnghnggiaccvectorca imnhlncnmhovtitohnhnhban u nh vector c thu nhn trc tip t cc thit b s ho nh Digital hoc c chuyn i t nh Raster thng qua cc chng trnh s hoCng ngh phn cng cung cp nhng thit b x l vi tc nhanh v 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 chuyn i t nh Raster. Hnh 1.10. S chuyn i gia cc m hnh biu din nh BMP PCC . . . DIB Ca sThay iPaint RASTER VECTOR RASTER Vecter haRaster ha26 Chng 2:CC K THUT NNG CAO CHT LNG NH 2.1. CC K THUT KHNG PH THUC KHNG GIAN 2.1.1. Gii thiu Ccphptonkhngphthuckhnggianlccphptonkhng phc thuc v tr ca im nh.Vd:Phptnggimsng,phpthngktnsut,binitn sut v.v.. Mt trong nhng khi nim quan trng trong x l nh l biu tn sut (Histogram) Biu tn sut ca mc xm g ca nh I l s im nh c gi tr g ca nh I. K hiu l h(g) V d: 1204 1007 I =2210 4121 2011 g01247 h(g)57521 2.1.2. Tng gim sng Gi s ta c I ~ kch thc m n v s nguyn c Khi , k thut tng, gim c sng c th hinfor (i = 0; i < m; i + +) for (j = 0; j < n; j + +) I [i, j] = I [i, j] + c; Nu c > 0: nh sng lnNu c < 0: nh ti i27 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 c ngdngkhiqutvnhndngvnbncthxyrasaistnnthnh nh hoc nh thnh nn dn n nh b t nt hoc dnh.2.1.4. B cm K thut nhm gim bt s mc xm ca nh bng cch nhm li s mc xm gn nhau thnh 1 nhmNu ch c 2 nhm th chnh l k thut tch ngng. Thng thng c nhiu nhm vi kch thc khc nhau.tngqutkhibiningitaslycng1kchthc bunch_size I [i,j] = I [i,j]/ bunch - size * bunch_size (i,j) V d: B cm nh sau vi bunch_size= 3 12467 21345 I =72691 41212 g h(g) 0 28 00366 00333 Ikq =60690 30000 2.1.5. Cn bng histogram nh I c gi l cn bng "l tng" nu vi mi mc xm g, g ta c h(g) = h(g) Gi s, ta c nhI~kch thc m nnew_level~s mc xm ca nh cn bng level newn mTB_= ~s im nh trung bnh ca mi mc xm ca nh cn bng ==gii h g t0) ( ) ( ~s im nh c mc xm g Xc nh hm f: g f(g) Sao cho: )`||

\|= 1) (, 0 max ) (TBg tround g f V d:Cn bng nh sau vi new_level= 4 12467 21345 I =72691 41212 gh(g)t(g)f(g) 1550 25101 31111 43142 51152 62172 72193 91203 29 01223 10122 Ikq =31230 20101 Ch : nh sau khi thc hin cn bng cha chc l cn bng "l tng " 2.1.6. K thut tm tch ngng t ng Ngng trong k thut tch ngng thng c cho bi ngi s dng. K thut tm tch ngng t ng nhm tm ra ngng mt cch t ng da vo histogram theo nguyn l trong vt l l vt th tch lm 2 phn nu tng lnh trong tng phn l ti thiu.Gi s, ta c nhI~kch thc m nG ~l s mc xm ca nh k c khuyt thiu t(g) ~s im nh c mc xm g ==gii h ig tg m0) ( .) (1) (~mmen qun tnh TB c mc xm g Hm f:) (g f g [ ]2) 1 ( ) () () () ( = G m g mg t mxng tg fTm sao cho:( ) { } ) (max1 0g f fG g < = V d: Tm ngng t ng ca nh sau 012345 001234 I =000123 000012 000001 Lp bnggh(g)t(g)g.h(g) =gii ih0) ( m(g)f(g) 015150001.35 1520550,251.66 30 24248130,541.54 33279220,811.10 42298301,030.49 51305351,16 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 th tnh c nh kt qu v do tascchistogramcanhbini.Nhngthctnhiukhita ch bit histogram ca nh gc v hm bin i, cu hi t ra l liu ta c thcchistogramcanhbini.Nucnhvytacthhiu chnhhmbinithucnhktqucphnbhistogramnhmong mun. Bitontralbithistogramcanh,bithmbinihyv histogram ca nh mi.V d:g1234 h(g)4212 g + 1nu g 2 f(g)=g nu g = 3 g 1nu g > 3 Bc 1: V Histogram ca nh c g f(g) 0 31 Bc 2: V th hm f(g) Bc 3: V Histogram ca nh mit q = f(g) h(q) = card ({P| I(P) = q}) = card ({P| I(P) = f(g)}) = card ({P| g = f-1 (I(P))}) = ) (1) (q f ii h Histogramcanhmithuacbngcchchnghnhvtnhgi tr theo cc q (= f(g)) theo cng thc tnh trn. Kt qu cui thu c sau php quay gc 90 thun chiu kim ng h. 2.2. CC K THUT PH THUC KHNG GIAN 2.2.1. Php nhn chp v mu Gi s ta c nh I kch thc M N, mu T c kch thc m n khi , nh I nhn chp theo mu T c xc nh bi cng thc.g h(g)f(g) 0 g h(g) 0 32 ( ) ( ) j i T j y i x I y x T Injmi, * , ) , (1010 ==+ + = (2.1) Hoc ( ) ( ) j i T j y i x I y x T Injmi, * , ) , (1010 == = (2.2) VD: 124587 211422 I =455882 121144 722152 T = 10 01 ( ) ( ) ( ) ( ) ( ) ( ) 1 , 1 * 1 , 1 0 , 0 * , , * , ) , (1010T y x I T y x I j i T j y i x I y x T Ij i+ + + = + + = = = ( ) ( ) 1 , 1 , + + + = y x I y x I 238710* 769124*Tnh theo (2.1) I T =6661212* 34266* ****** Tnh theo cng thc 2.2 ****** *238710 I T =*769124 *6661212 *34266 * Nhn xt:-Trongqutrnhthchinphpnhnchpcmtsthaotcra ngoinh,nhkhngcxcnhtinhngvtrdnnnhthu c c kch thc nh hn.-nhthchintheocngthc2.1v2.2chsaikhcnhau1php dch chuyn n gin ta s hiu php nhn chp l theo cng thc 2.12.2.2. Mt s mu thng dng33 - Mu:111 T1 = 111 111 ~ Dng kh nhiu Cc im c tn s caoVD1: 124587 2311422 I =455882 121144 722152 55654546** 52583435** I T1 = 29273535** ****** ****** p dng k thut cng hng s vi c = -27, ta c: 28381819** 253178** Ikq =2088** ****** ****** - Mu: 0-10 T2 =-14-1 0-10 ~ Dng pht hin cc im c tn s caoVD2: 114-400-14** -2251416** I T2 =-1-6-10-2** ****** ****** 34 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 k hiu l Med({xn}), c nh ngha: + Nu n l ((

+ 12nx + Nu n chn: ((

2nxhoc ((

+ 12nx * Mnh 2.1 min1 =niix x ti { } ( )nx Med Chng minh+ Xt trng hp n chnt 2nM= Ta c: =+= = + = Mii MMiiniix x x x x x1 1 1 ( ) =+=+ + =Mii i MMii M ix x x x x x1 1 ( ) ( ) [ ]=+ + =Mii M M Mx x x x11 { } ( ) { } ( ) = =+ + =MiMii i i i Mx Med x x Med x1 1 { } ( )= =nii ix Med x1 + Nu n l:Bsungthmphnt{ } ( )ix Med vody.Theotrnghpnchnta c: { } ( ) { } ( )i iniix Med x Med x x + =1 min ti Med({xn}) 35 =niix x1 min ti Med({xn}) * K thut lc trung v Gi s ta c nh I ngng ca s W(P) v im nh PKhi k thut lc trung v ph thuc khng gian bao gm cc bc c bn sau:+ Bc 1: Tm trung v{I(q)| q W(P)} Med (P) + Bc 2: Gn gi tr=) () () (P MedP IP INguoclaiP Med P I ) ( ) ( V d: 1232 41621 I =4211 2121 W(3 3); = 2 1232 4221 Ikq =4211 2121 Gi tr 16, sau php lc c gi tr 2, cc gi tr cn li khng thay i gi tr. 36 2.2.4. Lc trung bnh * nh ngha 2.2 (Trung bnh) Chodyx1,x2,xnkhitrungbnhcadykhiuAV({xn}) ddc nh ngha:{ } ( ) ||

\|==nii nxnround x AV11 * Mnh 2.2 ( ) min21 =niix x ti { } ( )nx AVChng minh:t: ( )= =niix x x12) ( Ta c:( )= =niix x x12 ) ( 0 ) ('= x ( )== niix x10 { } ( )== = nii ix AV xnx11 Mt khc,0 2 ) (' '> = n x min ti { } ( )ix AV x = K thut lc trung bnh Gi s ta c nh I, im nh P, ca s W(P) v ngng . Khi k thut 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) 37 + Bc 2: Gn gi tr=) () () (P AVP IP INguoclaiP AV P I ) ( ) ( V d: 1232 41621 I =4211 2121 W(3 3); = 2 1232 4321 Ikq =4211 2121 Gi tr 16 sau php lc trung bnh c gi tr 3, cc gi tr cn li gi nguyn 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=) () () (P AVP IP Ik NguoclaiP AV P Ik ) ( ) ( V d: 1232 41621 I =4211 2121 W(3 3); = 2; k = 338 1232 4821 Ikq =4211 2121 * Nhn xt:-Nuklnhnkchthccasthk thut chnh l k thut lc trung 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 HC 2.3.1. Cc php ton hnh thi c bn Hnhthilthutngchsnghincuvcutrchayhnhhc topo ca i tng trong nh. Phn ln cc php ton ca "Hnh thi" c nhnghathaiphptoncbnlphp"ginn"(Dilation)vphp "co" (Erosion).Cc php ton ny c nh ngha nh sau: Gi thit ta c i tng X v phn t cu trc (mu) B trong khng gian Euclide hai chiu. K hiu Bx 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 thuc X. Ta c: X B =X xxB

nh ngha 2.4 (EROSION) Php "co" ca X theo B l tp hp tt c cc im x sao cho Bx nm trong X. Ta c:XB = {x : Bx X} V d: Ta c tp X nh sau: X =|||||||

\|0 00 0 00 0 00 00 0x x xx xx xx x xx x x B = x 39 X B = |||||||

\|x x x xx x x xx x xx x x x xx x x x000 00vXB = |||||||

\|0 0 00 0 0 0 00 0 0 00 0 0 00 0 0 0x xxxx

nh ngha 2.5 (OPEN) Php ton m (OPEN) ca X theo cu trc B l tp hp cc im ca nh X sau khi co v gin n lin lip theo B. Ta c: OPEN(X,B) = (XB) B V d: Vi tp X v B trong v d trn ta c OPEN(X,B) = (XB) B =||||||||

\|0 x x x 00 0 0 0 00 0 x x 00 x x 0 0x x 0 0 0 nh ngha 2.6 (CLOSE) Php ton ng (CLOSE) ca X theo cu trc B l tp hp cc im ca nh X sau khi gin n v co lin tip theo B. Ta c: CLOSE(X,B) = (X B)B Theo v d trn ta c: CLOSE(X,B) = (X B)B = ||||||||

\|0 x x x 00 x x x 00 0 x x 0x x x x xx x x x 0 2.3.2. Mt s tnh cht ca php ton hnh thi * Mnh 2.3 [Tnh gia tng]: (i)X X XB XBB X B X BB (ii) B B' XBX B'X X B X B X40 Chng minh: (i)X B = B X B BX xxX xx = '' XB = { } { } ' / / X B x X B xx x = XB(ii) X B = ' ' B X B BX xxX xx = Theo nh ngha: XB = { } { } X B x X B xx x / ' / = XB . *Mnh 2.4 [Tnh phn phi vi php ]: (i)X (B B') = (X B) (X B') (ii) X (B B') = (XB) (X B') 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 Bxy 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 (BB) = (X B) (X B) (ii) X(B B) = (XB) (XB) Ta c:B B BX (BB) X B(tnh gia tng) Tng t : X (B B) XB X(B B) (XB) ( XB)(2.5) 41 Mt khc, x (XB) (XB) Suy ra,x XB Bx X x XB Bx X ( B B)x X 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:Tacthphntchccmuphctptrthnhccmungin thun 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 Bx X x YB Bx Y 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') 42 Chng minh: (i) (X B) B' = X (B' B) Ta c, (X B) B'= (X xxB) B'= X xxB B ) ('= X xxB B ) (' = X (B' B) (i) (XB)B' = X(B B') Trc ht ta i chng minh: 'xB XB xB B ) (' X Tht vy, do 'xB XB nn y'xB yXB By X X BxB yy ' xB B ) (' X Mt khc, xB B ) (' X ('xB B) X 'xB yyB X y'xBta c By X hay y'xBta c y XB Do , 'xB XB Ta c, (XB)B'={ } X B xx / B'= {x/ 'xB XB} = {x/ xB B ) (' X} (do chng minh trn) = X(B B') . * nh l 2.1 [X b chn bi cc cn OPEN v CLOSE] Gis,Xlmtitngnh,Blmu,khi,Xsbchntrn bi 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 43 Chng minh: Ta c: x X Bx X B(V X B = X xxB) x (X B)B (theo nh ngha php co) (X B)B X(2.9) Mt khc, y (XB) B, suy ra: x XB sao cho y Bx(V (XB) B = B X xxB ) Bx X y X Suy ra: X (XB) B(2.10) T (2.9) v (2.10) Ta c: (X B)B X (XB) B . *H qu 2.1 [Tnh bt bin] : (i)((X B) B) B = X B (ii) ((XB) B)B = XB Chng 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 (XB) B X X Do , 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 ((XB) B)B XB (do tnh cht gia tng)(2.13) Mt khc, cng t nh 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: ((XB) B)B = XB (pcm). 44 Chng 3:BIN V CC PHNG PHP PHT HIN BIN 3.1. GII THIU Binlvnquantrngtrongtrchchncimnhmtinti hiunh.Chonnaychacnhnghachnhxcvbin,trongmi ng dng ngi ta a ra cc o khc nhau v bin, mt trong cc o lovsthayitngtvcpxm.Vd:ivinhen trng,mtimcgilimbin nu n l im en c t nht mt imtrngbncnh.Tphpcc im bin to nn bin hay ng bao ca i tng. Xut pht t c s ny ngi ta thng s dng hai phng php pht hin bin c bn: Pht hin bin trc tip: Phng php ny lm ni bin da vo s binthinmcxmcanh.Kthutchyudngphthinbin y l k thut ly o hm. Nu ly o hm bc nht ca nh ta c cc k thutGradient,nulyohmbchaicanhtackthutLaplace. Ngoi ra cn c mt s cc tip cn khcPhthinbingintip:Nubngcchnotaphncnh thnh cc vng th ranh gii gia cc vng gi l bin. K thut 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 th pht hin c bin. Phng php pht hin bin trc tip t ra kh hiu qu v t chu nh hng ca nhiu, song nu s bin thin sng khng t ngt, phng php t ra km hiu qu, phng php pht hin bin gin tip tuy kh ci t,songlipdngkhtttrongtrnghpny.Skhcbitcbn gia hai phng php ny l: Phng php pht hin bin trc tip cho ta ktqulnhbin,cnphngphpphthinbintrctipchotakt qu l ng bin. 3.2. CC PHNG PHP PHT HIN BIN TRC TIP 3.2.1. K thut pht hin bin Gradient Theo nh ngha, gradient l mt vct c cc thnh phn biu th tc thay i gi tr ca im nh, ta c: 45 Trong , dx, dy l khong cch (tnh bng s im) theo hng x v y. * Nhn xt:Tuy ta ni l ly o hm nhng thc cht ch l m phng v xp x o hm bng cc k thut nhn chp v nh s l tn hiu ri rc nn o hm khng tn ti.V d: Vi dx = dy = 1, ta c: ( ) ( )( ) ( ) + + y x f y x fyfy x f y x fxf, 1 ,, , 1 Do , mt n nhn chp theo hng x l A=( ) 1 1 v hng y lB= |||

\| 11 Chng hn: 0000 0333 I =0333 0333 Ta c, 000*033* I A = 300*; I B=000* 300*000* ******** dyy x f dy y x ffyyy x fdxy x f y dx x ffxxy x f) , ( ) , ( ) , () , ( ) , ( ) , ( + = + =46 000* I A + I B=300* 300* **** 3.2.1.1. K thut Prewitt Kthutsdng2mtnnhpchpxpxohmtheo2hngxv y l: -101 Hx =-1 01 -1 01 -1-1-1 Hy =000 111 Cc bc tnh ton ca k thut Prewitt + Bc 1: Tnh I Hx v I Hy + Bc 2: Tnh I Hx + I Hy V d:000000 555500 555500 I =555500 000000 000000 00-10-10** 00-15-15** I Hx =00-10-10** 00-5-5** ****** ****** 47 1515105** 0000** -15-15-10-5** I Hy =-15-15-10-5** ****** ****** 15150-5** 00-15-15** I Hx + I Hy =-15-15-20-15** -15-15-15-10** ****** ****** 3.2.1.2. K thut SobelTng t nh k thut Prewitt k thut Sobel s dng 2 mt n nhn chp theo 2 hng x, y l: -101 Hx =-2 02 -1 01 -1-2-1 Hy =000 121 Cc bc tnh ton tng t Prewitt + Bc 1: Tnh I Hx v I Hy + Bc 2: Tnh I Hx + I Hy 3.2.1.3. K thut la bnK thut s dng 8 mt n nhn chp theo 8 hng 00, 450, 900, 1350, 1800, 2250, 2700, 3150 55-3555 H1 =50-3H2 =-30-3 -3-3-3-3-3-3 -355-3-35 H3 =-305H4 =-305 -3-3-3-3-35 48 -3-3-3-3-3-3 H5 =-305H6 =-30-3 -355555 -3-3-35-3-3 H7 =50-3H8 =50-3 55-35-3-3 Cc bc tnh ton thut ton La bn + Bc 1: Tnh I Hi ; i = 1,8 + Bc 2: =81 iiH I3.2.2. K thut pht hin bin Laplace Cc phng php nh gi gradient trn lm vic kh tt khi m sngthayirnt.Khimcxmthayichm,minchuyntiptri rng, phng php cho hiu qu hn l phng php s dng o hm bc hai Laplace.Ton t Laplace c nh ngha nh sau: Ta c: ( ) ) , ( ) , 1 (22y x f y x fx xfx xf +||

\|= [ ] [ ]) , 1 ( ) , ( 2 ) , 1 () , 1 ( ) , ( ) , ( ) , 1 (y x f y x f y x fy x f y x f y x f y x f + + + Tng t, ( ) ) , ( ) 1 , (22y x f y x fy yfy yf +|||

\|= [ ] [ ]) 1 , ( ) , ( 2 ) 1 , () 1 , ( ) , ( ) , ( ) 1 , ( + + + y x f y x f y x fy x f y x f y x f y x f Vy: 2 f= f(x+1,y) + f(x,y+1) - 4f(x,y) + f(x-1,y) + f(x,y-1) 22222yfxff+= 49 Dn ti:

0 1 01 4 10 1 0H||||

\| = Trong thc t, ngi ta thng dng nhiu kiu mt n khc nhau xpxrircohmbchaiLaplace.Diylbakiumtnthng dng: VD:000000 555500 I =555500 555500 000000 000000 3.2.3. K thut Canny ylmtthuttontngitt,ckhnngarangbin mnh, v pht hin chnh xc im bin vi im nhiu. Thut ton + Bc 1: Lm trn nh Tnh I H, vi: (((((((

=2 4 5 4 24 9 12 9 45 12 15 12 54 9 12 9 42 4 5 4 21151HGi G l kt qu lc nhiu: G= I H ||||

\| =||||

\| =||||

\| =1 2 12 4 21 2 1H 1 1 11 8 11 1 1H 0 1 01 4 10 1 0H3 2 150 +Bc 2: Tnh gradient ca nh bng mt n PreWitt, kt qu t vo Gx,Gy. Gx = G Hx, Gy = G Hy + Bc3: Tnh gradient hng ti mi im (i,j) ca nh. Hng ny s c nguyn ha nm trong 8 hng [0..7],tng ng vi 8 ln cn ca mt im nh. +Bc4:Dngrngbucloibnhngimkhngphilcc i xa b nhng im khng l bin. Xt (i,j), l gradient hng ti (i,j). I1, I2 l hai im ln cn ca (i,j) theo hng . Theo nh ngha im binccbth(i,j)lbinnuI(i,j)cciaphngtheohng gradientNu I(i,j) > I1 v I(i,j) > I2 th mi gi li I(i,j), ngc li xa I(i,j) v im nh nn. +Bc5:Phnngng.Viccimcgili,thchinly ngng gradient bin ln cui xc nh cc im bin thc s. 3.3. PHT HIN BIN GIN TIP 3.3.1 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. nhcgilnhnhphnnuccgitrIijchnhngitr0hoc 1. ytachxttinhnhphnvnhbtkcthavdng nh 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 gingGis(i,j)lmtimnh,ccim4-lngginglccimk trn, 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 3.1 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. 51 Hnh 3.1.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-linthng)trongEnutntitpccimcgilngi (io,jo)...(in,jn)saocho(io,jo)=Ps,(in,jn)=Pe,(ir,jr)Ev(ir,jr)l8-lng ging (hoc 4-lng ging tng ng) ca (ir-1,jr-1) vi r = 1,2,...,n Nhn xt: Quan h k-lin thng trong E (k=4,8) l mt quan h phn x, ixngvbccu.Bivylmtquanhtngng.Milp tng ng c gi l mt thnh phn k-lin thng ca nh. V sau ta s gi 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 nh P1,,Pn sao cho Pi v Pi+1 l cc 8-lng ging ca nhau (i=1,...,n-1) v P1 l 8-lnggingcaPn,iQkhngthucitngnhvQl4-lng ging ca 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 ca chu tuyn v k hiu Len(C) v hng PiPi+1 l hng chn nu Pi v Pi+1 l cc 4 lng ging (trng hp cn li th PiPi+1 l hng l).Hnh3.2diybiudinchutuyncanh,trong,Plim khi u chu tuyn. Hnh 3.2. V d v chu tuyn ca i tng nh P5 P7 P6 P4 P0 P P3P1 P2 P 52 nh ngha 3.2 [Chu tuyn i ngu] HaichutuynC=vC=cgili ngu ca 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 3.3a) nu v ch nu (i) Chu tuyn i ngu C l chu tuyn ca cc im nn (ii) di ca C nh hn di C nh ngha 3.4 [Chu tuyn trong] Chu tuyn C c gi l chu tuyn trong (Hnh 3.3b) 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 C Chu tuyn CChu tuyn C a) Chu tuyn ngoib) Chu tuyn trong Hnh 3.3. Chu tuyn trong, chu tuyn ngoi nh ngha 3.5 [im trong v im ngoi chu tuyn] Gi s C= l chu tuyn ca mt i tng nh v P l mt im 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 ca E, C= l chu tuyn i ngu tng ng. Khi : (i) Nu C l chu tuyn trong th in(Qi,C) i (i=1,....,m) 53 (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] GisElmtitngnhvCElchu 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 thut d bin. C hai k thut d bin c bn. K thut th nht xt nh bin thu ctnhvngsaumtlnduytnhmtth,saupdngcc thuttonduytcnhth.Kthutthhaidatrnnhvng,kthp ng thi qu trnh d bin v tch bin. y ta quan tm cch tip cn th hai. Trcht,gisnhcxtchbaogmmtvngnh8-lin thng,cbaobc bi mt vnh ai cc im nn. D thy l mt vng 4-lin thng ch l mt trng ring ca trng hp trn. Vcbn,ccthuttondbintrnmtvngubaogmccbc 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 Doxutphttnhngtiuchunvnhnghakhcnhauvim bin, v quan h lin thng, cc thut ton d bin cho ta cc ng bin mang cc sc thi rt khc nhau. Kt qu tc ng ca ton t d bin ln mt im bin ri l im bin ri+1 (8-lng ging ca ri). Thng thng cc ton t ny c xy dng nh mthmisBooleantrncc8-lnggingcari.Micchxydng cc ton t u ph thuc vo nh ngha quan h lin thng v im bin. Dosgykhkhnchovickhostcctnhchtcangbin. Ngoi ra, v mi bc d bin u phi kim tra tt c cc 8-lng ging ca mi im nn thut ton thng km hiu qu. khc phc cc hn ch trn, thay v s dng mt im bin ta s dng cp im bin (mt thuc 54 , mt thuc), cc cp im ny to nn tp nn vng, k hiu l NV v phn tch ton t d bin thnh 2 bc: Xc nh cp im nn vng tip theo.La chn im bin Trong bc th nht thc hin chc nng ca mt nh x trn tp NV 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: Nugplicpxutphtthdng,nukhngquaylibc 2. Vicxcnhcpnn-vngxut pht c thc hin bng cch duyt nh ln lt t trn xung di v t tri qua phi ri kim tra iu kin la chn cp nn-vng. Do vic chn im bin ch mang tnh cht quy c, nn ta 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:NVNV (b,r)(b,r) Gi T l mt ton t d bin c s nu n tho mn iu kin: b,r l cc 8-lng ging ca r. Gis(b,r)NV;giK(b,r)lhmchnimbin.Bincamt dng 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. Cchnhnghathbatngngmicpnn-vngvi mt im bin.Cnivicchnhnghathnhtvthhaimtscpnn-vng c th c chung mt im bin. Bi vy, qu trnh chn im bin c thc hin nh sau: i:= 1; (bi,ri):= (bo,ro); While K(bi,ri)K(bn,rn) and i8 do 55 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) * Xc nh cp nn vng xut phtCp nn vng xut pht c xc nh bng cch duyt nh ln lt t trn xung di v t tri sang phi im em u tin gp c cng vi im 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 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 ; 56 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:pdir im trng trc Trng so vi en mi012 124 234 346 456 560 670 702 Do cng thc tnh hng tip theo s l : dir= ((pdir+3)/ 2 * 2)%8 ; 3.4.PHT HIN BIN DA VO TRUNG BNH CC B 3.4.1. Bin v bin i v mc xmNhtrnhbytrn,trongthctngitathngdnghai phngphpphphinbincbnl:Phthinbintrctipvgin tip. Phn ny cp n k thut mi da vo trung bnh cc b trn c snhgichnhlchvgitrmcxmcaimnhsovicc imlncndokthpcuimcachaikhuynhhngtrc tip v gin tip. i vi cc nh mu theo m hnh no u c th chuyn sang m hnh gm 3 thnh phn mu R, G, B. Sau d dng chuyn cc nh mu sang dng nh a cp xm. Chng hn: Gray = ( R + G + B ) / 3 Vic x l, thao tc trn cc nh xm c mt u im l d x l hn cc nh mu m vn gi c cc c tnh ca nh. Cc nh trng en tuy d x l nht nhng s b mt nhiu chi tit sau khi chuyn i. Mtcchltngthbinthinmcxmcaimnhkhiqua bin phi c dng: x 0 Mc xm 57 Trongthctdngthnychgptrongccnhtrngen(nh xm c hai mu), cn vi cc nh thc th th ca n c dng: Kh khn cho vic phn tch cc nh thc l ch do s bin thin v mc xm ca im nh khng phi ch c th hin theo mt hng duy nhtmphixttheoctmhngcaccimnh lng ging, ti cc vng bin v ln cn bin s bin thin mc xm ca cc im nh thng khng t ngt m tri qua mt khong bin thin khng u nhng c tc binthinnhanh.Chngtacthxcnhcccngbinnh thnybngkthutLaplacenhngnhtrnnikthutnyrt nhy cm vi nhiu m nhiu hu nh li l vn m trong bc nh no cng c. Ngoi ra, trong thc t khi d bin cho cc nh xm ty theo mc chxlsaunymngitacthmunlybincattccci tng trong nh hoc ch mt s i tng chnh trong nh. Cc k thut ohmdosdngccmtnlccmatrnnhnchpnnkhiu chnh chi tit ca nh bin thu c. Mun lm c iu ny li phi tnhtonliccgitrcaccphnttrongmatrntheocccng thc nht nh, rt phc tp v tn km. Khng nhng th nh thu c sau khi lc khng lm mt i c tt c cc im khng thuc ng bin m ch lm ni ln cc im nm trn bin v mun nhn dng c cc i tng th ta cn phi x l thm mt vi bc na th mi thu c nh bin thc s.Cthnhnthylccthuttondbintruynthngmchngta hay dng vn cha t c s hon thin nh mong mun [3,8]. 3.4.2. Pht hin bin da vo trung bnh cc btngchnhcathuttoncxutl:Xcnhttccc imnmtrnbinkhngtheohngtmkimvsdngccmatrn lc, thng qua vic so snh chnh lch v mc xm ca n so vi mc xm chung ca cc im nh ln cn (mc xm nn). Trc ht gi tr xm trung bnh ca cc im nh nm trong phm vi ca ma trn 33 hoc 55 c tm l im nh ang xt s c tnh ton. Nu nh chnh lch mc xm gia im ang xt vi gi tr xm trung bnh tha mn ln hn mt mc ti thiu 1 no (PTB+ 1< P) th chng ta s coi n l im bin x 0 Mc xm 58 v ghi nhn li, cn cc im khng tha mn iu kin trn s c coi l im nn. a) Ma trn im nh trc khi lcb) Ma trn im nh sau khi lcHnh 3.4. Ma trn im nh trc v sau lc Thut ton c th c m t nh sau: for (i=0; i< biHeight; i++) for (j=0; j< biWidth; j++) { tt_GrayScale=0;for (ii=i-1; ii T} (4.8) : l hm hiu chnh. D thy nu ngng T cng ln th cng th s lng im tham gia trong 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 nh phn. (b) S Voronoi. (c) Hiu chnh bi hm Potential, T=9.0.(d) Hiu chnh bi hm Potential, T=18.0 a)b) c)d) 68 4.3.4. Thut ton tm xng Trong mc ny s trnh by tng c bn ca thut ton tm xng v m t bng ngn ng ta Pascal. Tng trng: Vic tnh ton s Voronoi c bt u t mt im sinh trong mt phng. Sau im sinh th hai c thm vo v qu trnh tnh ton tip tc vi a gic Voronoi tm c vi im va c thm vo . C nh th, qu trnh tnh ton s Voronoi c thc hin cho nkhikhngcnimsinhnocthmvo.Nhcimcachin lc ny l mi khi mt im mi c thm vo, n c th gy ra s phn vng ton b cc a gic Voronoi c tnh. Chia tr: Tp cc im bin u tin c chia thnh hai tp im c kch c bng nhau. Sau thut ton tnh ton s Voronoi cho c hai tp 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 cc im bin thnh hai phn khng phi c thc hin mt ln, m c lp li nhiu ln cho n khi vic tnh ton s Voronoi tr nn n gin. V th, vic tnh s Voronoi tr thnh vn lm th no trn hai s Voronoi li vi nhau. Thuttonstrnhbyylskthpcahaitngtrn. Tuy nhin, n s mang nhiu dng dp ca thut ton chia tr. Hnh4.3minhhotngcathuttonny.Mi mt im bin cchiathnhhaiphn(bntri:1-6,bnphi:7-11)binggp khc , v hai s Voronoi tng ng Vor(SL) v Vor(SR). thu c s Vornonoi Vor(SL SR), ta thc hin vic trn hai s trn v xc nh li mt s a gic s b sa i do nh hng ca cc im bn cnh thuc s kia. Mi phn t ca s l mt b phn ca ng trung trc ni hai im m mt im thuc Vor(SL) v mt thuc Vor(SR). Trc khi xydng,tatmraphntuvcuican.Nhnvohnhtrn,ta nhn thy rng cnh 1 v 5 l cc tia. D nhn thy rng vic tm ra cc cnh u v cui ca tr thnh vic tm cnh vo t v cnh ra t. Hnh 4.3. Minh ho thut ton trn hai s Voronoi 1 2 4 3 6 5 7 9 8 11 10 CH(SR) CH(SL) 1 5 t t 69 Saukhitmctvt,ccimcuicatcsdng xy dng phn t u tin ca (1 trong hnh trn). Sau thut ton tm im giao ca vi Vor(SL) v Vor(SR). Trong v d trn, u tin giao vi V(3). K t y, cc im nm trn phn ko di s gn im 6 hn im3.Do,phnttiptheo2casthucvongtrungtrc caim6vim7.Sauimgiaotiptheocasthucv Vor(SL); by gi s i vo V(9) v 2 s c thay th bi 3. Qu trnh ny s kt thc khi gp phn t cui 5. Trn y ch l minh ho cho thut trn hai s Voronoi trong chin lcchiatr.Tuynhin,trongthuttonstrnhbyythsthc hin c khc mt cht. Tp cc im nh khng phi c a vo ngay t umscqutvotngdngmt.Gistibcthi,tathu c mt s Voronoi gm i-1 hng cc im sinh Vor(Si-1). Tip theo, ta qut ly mt hng Li cc im nh t tp cc im bin cn li. Thc hin victnhsVoronoiVor(Li)chohngny,sautrnVor(Si-1)vi Vor(Li). Kt qu ta s c mt s mi, v li thc hin vic qut hng Li+1 cc im 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 LickimthVor(Li)sgmk-1ngthngng),nnvictrn Vor(Si-1) vi Vor(Li) c phn n gin hn. Hnh 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;73 index = i; } } 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 DouglasPeucker int DouglasPeucker(POINT *pLINE, int n, float ) { inti, j; BOOL chiso [MAX_PT];for(i = 0; i < m; i++)//Tt c cc im c gi li 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. tngTrong thut ton Band Width, ta hnh dung c mt di bng di chuyn tumtngcongdctheongcongsaochongcongnm trong di bng cho n khi c im thuc ng cong chm vo bin ca di bng, im ny s c gi li. Qu trnh ny c thc hin vi phn 74 cnlicangcongbtutimvatmcchonkhiht ng cong. C th nh sau: 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 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. Ban u khong cch t im trung gian n on thng ni im cht v im ng c tnh tonvkimtra.Nukhongcchtnhc ny nh hn mt ngng cho trc th im trung gian c th b i, tin trnh tip tc vi im cht l im cht c, im trung gian l im ng c v im ng l im k tip sau im ng c. Trong trng hp ngc li, khong cch tnh c ln hn ngng cho trc th im trung gian s c gi li, tin trnh tip tc vi im cht l in trung gian, im trung gian l im ng c vimnglimktipsauimngc.Tintrnhclpcho n 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 mtimcht(P1).imthba(P3)ccoilim ng.imgiaimchtvimng(P2)lim 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 chotrcthimtrunggiancthbi.Trong trnghpngcliimchtchuynnimtrung gian. Bc 4:Chutrnhclplithimtrunggiancchuyn nimngvimktipsauimngcch nh lm im ng mi.. Nhnxt:Thuttonnytngtctrongtrnghpngngcha nhiu im, iu c ngha l lch gia cc im trong ng thng l nh, hay dy nt ca ng c vct ho l mnh. di P3 P2 P4 dk P5P1 75 5.1.3.2. Chng trnh//Hm tnh ng cao t nh n on thng ni hai im dau, cuoi float Tinhduongcao(POINT dau, POINT cuoi, POINT dinh) { floot h;tnh ng caoreturm h ; } //Hm quy nhm nh du loi b cc 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; elsechot = tg;tg = tg + 1 if(tg < n - 1)BWSimple (pLINE, chot, tg, chiso, , n) ; } //Hm rt gn s lng im BandWidth int 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 li BWSimple(pLINE, 0, 1, chiso, , n); for(i= j= 0; i < n; i++) if(chiso [i]== TRUE) 76 pLINE [j ++1] = pLINE [i]; return j; } 5.1.4. Thut ton Angles 5.1.4.1. tngTngtnhthuttonBandWidthnhngthayvictnhton khongcchbitnhgc.Cththuttonbtuviimung cong (P1) l im cht. Hnh 5.3. n gin ha ng cong vi thut ton Angles imth3cangcong(P3)limng, im gia im cht v im ng (P2) l im trung gianGctobiimcht,trunggian, ng vi im trung gian l nh vic tnh ton v kim traNu th im trung gian c th b i trong trng hp ngc li im cht s l im trung gian c v qu trnh lp vi im trung gian l im ng c, im ng mi l im k tip sau im ng c. Tin trnh thc hin cho n ht ng cong.5.1.4.2. Chng trnh//Hm tnh ng cao t nh n on thng ni hai im dau, cuoi float Tinhgoc(POINT dau, POINT cuoi, POINT dinh) { float ;tinhgoc (t vit) return ;} //Hm quy nhm nh du loi b cc im trong ng cong void ALSimple(POINT *pLINE,int chot,int tg,BOOL *chiso,float ,int n) P1 i P3 P2 P4 k P5 77 { if(Tinhgoc(pLINE[chot], pLINE[tg], pLINE[tg+1]) > ) chiso[tg] = FALSE;elsechot = tg;tg = tg + 1;if(tg < n - 1)ALSimple(pLINE, chot, tg, chiso, , n); } //Hm rt gn s lng im Angles int 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 im gia thng hng. Thut ton Angles phi c = 180o b i cc im gia thng 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 thut d 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 ho nh Douglas Peucker, Band Width, Angle v.v.. ta s thu c mt polyline hay ni khc i l thu c mt a gic xc nh i tng du. Vn l tacnphixcnhxemitngcphilitngcntchhay 78 khng? Nh ta bit mt a gic c th c hnh dng ta nh mt hnh c s, c th c nhiu cch tip cn xp x khc nhau. Cch xp x da trn cc c trng c bn sau: c trng ton cc: Cc m men thng k, s o hnh hc nh chu vi, din tch, tp ti u cc hnh ch nht ph hay ni tip a gic v.v.. ctrngaphng:Ccsoctrngcangcongnh gc, im li, lm, un, cc tr v.v.. Hnh 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 bit nhtamgic,ngtrn,hnhchnht,hnhvung,hnhellipse,hnh trn 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 mu MtagicviccnhV0,..,Vm-1cxpxviagicmu U0,..,Un-1 vi o xp x nh sau: E V Und md( , ) min = 0 1,Nhn dng i tng Bt bin ng dng Bt binAphin ng trn Ellipse Hnh ch nht Tam gic u Ellipse Tam gic T gic a gic 79 Trong dRj j d mjndkR U a V = + +=min,( ) mod0 20122 ,karea V Varea U Umn=( )( )0 10 1,viR l php quay quanh gc to mt gc .Trong , d c tnh hiu qu bng cng thc sau: d j d m j d mjnj j j d mjnjnjndVnV k U k U V = + + +=+=== | | | | | | | |( ) mod ( ) mod ( ) mod2012 2 201010112yUj,Vjchiulccsphcticcnhtngng.Khim>>nthphctptnhtonrtln.Vicchnhcbitnhhnh trn,ellipse,hnhchnht,hnhxcnhduynhtbitmvmtnh (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 x y dxdypqp q= Pg. Trong 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 . O(0,0) f x y x y dxdysign x y x yf x y x y dxdyp qi i i iinp qOU Ui i( , )( )( , )Pg= + +=+1 1011 Hnh 5.6. Phn tch min a gic thnh tng i s cc min tam gic U0 U1 U2 U3 Un- - - 80 5.2.1.1. Xp x a gic bng ng trn Dng phng php bnh phng ti thiu, ta c o xp x: E(Pg,Cr)= min ( ), , a b c Ri i i iinnx y ax by c=+ + + +12 2 21

5.2.1.2. Xp x a gic bng ellipse Cng nh i vi ng trn phng trnh xp x i vi ellipse c cho bi cng thc: E(Pg,El)=min ( ), , , , a b c d ei i i i i iinnx ay bx y cx dy e=+ + + + +R12 2 21 Mt bin th khc ca phng php bnh phng ti thiu khi xp x cc ng cong bc hai c a ra trong [7]. 5.2.1.3. Xp x a gic bi hnh ch nht S dng tnh cht din tch bt bin qua php quay, xp x theo din tch nh sau: Gi 11 20 02, ,l cc m men bc hai ca a gic (tnh theo din tch). Khi gc quay c tnh bi cng thc sau: tg2 =2- .1120 02 Gidintch ca hnh ch nht nh nht c cc cnh song song vi cc trc qun tnh v bao quanh a gic Pg l S.K hiu E(Pg, Rect)= S area Pg ( ) Hnh 5.8. Xp x a gic bng hnh ch nht x y Hnh 5.7. Xp x a gic bng ng trn 81 5.2.1.4. Xp x a gic bi a gic u n cnh GiM(x0,y0)ltrngtmcaagic,lymtnhQtucaa gic, 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 chng tacth chuyn bi ton xp x i tng bi bt bin aphin v bi ton xp x mu trn cc dng chun tc. Nh vy c th a vic i snh cc i tng vi mu bi cc bt bin ng dng, chng hn vic xp x bi tamgic,hnhbnhhnh,ellipsetngngvixpxtamgicu, hnh vung, hnh trn v.v... Th tc xp x theo bt bin aphin mt a gic vi hnh c s c thc hin tun t nh sau: + Bc 0:Phn loi bt bin aphin cc dng hnh c s Dng hnh c sDng chun tcTam gic Tam gic u Hnh bnh hnhHnh vung Ellipseng trn + Bc 1:Tm dng chun tc c s Pg' tho mn iu kin: m mm mm m01 1002 2013 31010= == == = (php tnh tin) (php co dn theo hai trc x, y) (**) + Bc 2: XcnhbiniaphinTchuynagicthnhagicPgdng chun 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).82 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 (**) lP1=(0,-2),P2=) , ( 3, P3= ) , ( 3,33 28 4= .V d 2:Tn ti hai hnh vung P1P2 P3 P4 tho mn tnh cht (**) Hnhvungthnhtc4nhtngngl(-p,-p),(-p,p),(p,-p),(p,p), vi p=344 Hnhvungthhaic4nhtngngl(-p,0),(p,0),(0,-p),(0,p), vi p=34.5.3. BIN I HOUGH 5.3.1. Bin i Hongh cho ng thngBng cch no ta thu c mt s im vn t ra l cn phi kim tra xem cc im c l ng thng hay khngBi ton: Chonim(xi;yi)i=1,nvngnghykimtranimc to thnh ng thng hay khng?* tng Gisnimnmtrncngmtngthngvngthngc phng trnh y = ax + bV (xi, yi) i = 1, n thuc ng thng nn y1 = ax1 + b, i = 1, n b = - xia + y1; i = 1, nNh vy, mi im (xi; yi) trong mt phng s tng ng vi mt s ng thng b = - xia + yi trong mt phng tham s a, b. n im (xi; yi) i = 1,nthucngthngtrongmtphngtngngvinngthng trong mt phng tham s a, b giao nhau ti 1 im v im giao chnh l a, b.Chnhlhsxcnhphngtrnhcangthngmccimnm vo.83 * Phng php:- Xy dng mng ch s [a, b] v gn gi tr 0 ban u cho tt c cc phn t ca mng-Vimi(xi;yi)va,blchscaphntmngthomnb = - 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 so vi s phn t ln hn hoc bng ngng cho trc th ta c th kt lun ccimnmtrncng1ngthngvngthngcphngtrnhy = ax + b trong a, b tng ng l ch s ca phn t mng c gi tr ln nht tm c:V d:Cho 5 im (0, 1); (1, 3); (2, 5); (3, 5); (4, 9) v = 80%. Hy kim tra xem 5 im cho c nm trn cng mt ng thng hay khng? Hy cho 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 nhtc gi tr 4 4/5 = 80%- Kt lun: 5 im ny nm trn cng 1 ng thngPhng trnh: y = 2x + 1 84 OH.HA=0 5.3.2. Bin i Hough cho ng thng trong ta 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 bi mt cp (r,) trong ta cc vi r l khong cch t gc ta ti ng thngvlgctobitrc0Xvingthng vung gc vi n, hnh 5.9 biu din ng thng hough trong ta Decard. Ngcli,mimtcp(r,)trongtocccngtngngbiu dim mt ng thng trong mt phng. GisM(x,y)lmimthucngthngcbiudinbi (r,), gi H(X,Y) l hnh chiu ca gc to O trn ng thng ta c: X= r. cos v Y= r.sinMt khc, ta c: T ta c mi lin h gia (x,y) v (r,) nh sau: x*cos+y*sin= r. Xtnimthnghngtrongtacccphngtrnhx*cos0+y*sin0=r0.BiniHoughnhxnimnythnhnng sin trong ta cc m cc ng ny u i qua (r0,0). Giao im (r0,0) ca n 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 c bao nhiu ngqua (x,y) s c by nhiu cp gi tr (r,). y 0 H x x.cos +y.sin =r Hnh 5.9. ng thng Hough trong to cc r 85 Chng 6: NG DNG X L NH 6.1. PHT HIN GC NGHING VN BN DA VO CHU TUYN Gcnghingvnbnlmtbitonkinhintrongxlnhvn bn. Mt h thng x l nh vn bn thng phi gii quyt bi ton pht hin gc nghing nh mt bc u tin v tt yu. Chnh v vy, cng vi s pht trin ca x l nh ni chung v x l nh vn bn ni ring, bi tongcnghingvnbncngcquantmngycngnhiuvdi nhiugckhcnhau.Crtnhiuhngtipcnchobitongc nghingvnbnttrctinay.Ccthuttonphthingcnghing thng c xy dng cho cc h thng phn tch nh vn bn khc nhau nn ch gii quyt cho nhng loi nh vn bn c th. C th chia ra mt s hng tip cn c bn cho bi ton gc nghing vn bn nh sau: Cc thut ton da vo phn tch hnh chiu (Projection Profile) Cc thut ton da vo bin i Hough (Hough Transform) Ccthuttonphntchlngging(NearestNeighbour Clustering) Phng php dng cc php ton hnh thi Da vo tnh cht mi i tng nh c duy nht mt chu tuyn ngoi vquannimcon ngi nhn ra nghing ca vn bn da vo c ch chimchotrongvnbn.Mcnycpnvictnhtonkch thc ch o ca cc i tng nh trong vn bn thng qua k thut tnh biutnxutkchthchnhchnhtnhnhtbaoquanhitng nh.Vicxcnhgcnghingvnbnscxcnh nh php bin i Hough cho nhng im gia y ca hnh ch nht nh nht bao quanh i tng nh cho cc i tng nh c kch thc ch o. 6.1.1. Tnh ton kch thc ch o ca cc i tng nh Nh ni t cc phn trn, gc nghing c xc nh da vo bin i Hough. y, chng ta ch p dng bin i Hough cho nhng im gia y ca cc hnh ch nht ngoi tip cc i tng c kch thc ch o trong nh. Nh vy, cng vic u tin cn thc hin l xc nh c cc hnh ch nht ngoi tip cc i tng hay ni cch khc l xc nh bin cc i tng. 86 Hnh 6.1. Cc hnh ch nht ngoi tip i tng nh y, ta dng thut ton d bin c ci tin trong chng 2 xcnhbinchoccitngtrongnhvnbn.Hnhchnhtngoi tip i tng s c xc nh ngay sau khi d c bin cho i tng .Mt cch trc tip ging nh mt s thut ton khc, ta c th dng bin i Hough p dng ln y ca cc hnh ch nht ngoi tip cc i tngnyvclnggc nghing cho vn bn. Tuy nhin, y bin i Hough c p dng sau khi loi bt i mt s i tng bng cc ngng kch thc. Hnh 6.2. V d v mt nh vn bn nghing vi nhiu loi i tng Mcchcavicdngngngldavothcokchthc phn loi i tng. Ni cch khc, dng ngng phn loi ta c th phn bit c mt cch tng i nhng i tng l k t v i tng phi k t.Nhbitphnbititng,taschlmvicviccitngc 87 kch thc ch o trong nh do chnh xc ca thut ton c ci thin ng k. Gi s ta c mt nh vn bn nghing nh hnh v trn y. R rng y l mt nh vn bn phc tp vi nhiu i tng phi k t v s k t ch ci trong nh trn l rt t. Mc d vy, chng ta vn cho rng nh b nghing. Vy ta cn c vo u khi kt lun nh b nghing? Trong mt nh vn bn, thng thng cc i tng k t chim nhiu hn nhng i tng khc. Xut pht t quan im nhn nhn s vt ca mt ngi v c th trn y ca nh vn bn, gi cho chng ta mt hng gii quyt bi ton gc nghing l xc nh cc i tng ch o trong nh chnh l cc k t, v cn c vo chng c lng gc nghing. tng xc nh cc i tng c kch thc ch o trong nh l dng k thut lp biu tn xut hay Histogram kch thc c lng mtktctnsxuthinnhiunhttrongvnbnmtagili tngchun.Vimimtnhuvo,tasxcnhmtitng chun ring v t ng trong chng trnh. Sau , ly i tng ny lm chunvsosnhccitngcnlivin.Nhngitngckch thc xp x bng kch thc ca i tng ny s c chn p dng biniHough.Mtitngcxemlxpxbngkchthcca i tng khc nu chnh lch kch thc gia chng b hn mt ngng c nh ngha trc. 6.1.2. Bin i Hough v pht hin gc nghing vn bn6.1.2.1. p dng bin i Hough trong pht hin gc nghing vn bn tngcavicpdngbiniHoughtrongphthingc nghing vn bn l dng mt mng tch lu m s im nh nm trn mtngthngtrongkhnggiannh.Mngtchlulmtmnghai chiu vi ch s hng ca mng cho bit gc lch ca mt ng thng v ch s 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 theo cc gc lch thay i. Gc nghing vn bn tng ng vi gc c tng gi tr mng tch lu cc i.TheobiniHough,mimtngthngtrongmtphngtng ngcbiudinbimtcp(r,).Gistacmtimnh(x,y) trongmtphng.Vquaim nh ny c v s ng thng, mi ng thng li cho mt cp (r,) nn vi mi im nh ta s xc nh c mt s cp (r,) tho mn phng trnh Hough. HnhvdiyminhhocchdngbiniHoughphthin gcnghingvnbn.Gistacmtsimnh.ylnhngim giaycchnhchnhtngoitipccitngclachnt 88 y x.cos +y.sin = r1 Hough[ ][r1] = 3 x 0 x.cos +y.sin = rHough[ ][r1] = 4 ccbctrc.y,tathytrnmtphngchaingthngsong song nhau. ng thng th nht c ba im nh nn gi tr mng tch lu bng 3. ng thng th hai c gi tr mng tch lu bng 4. Do , tng gi tr mng tch ly cho cng gc trng hp ny bng 7. Hnh 6.3. Bin i Hough pht hin gc nghing6.1.2.2. Thut ton pht hin v hiu chnh gc nghing vn bn a) X l ngoi l Sau giai on tin x l nh ta thu c nh trung gian TempImage. Thuttonphthingcnghingslmvicvinhtrunggianny tmragcnghingchovnbnvsaudngthuttonquaynh quay nh ban u vi gc nghing va tm c. Tuy nhin, mt im cn c xt n trong thut ton pht hin gc nghinglxlnhngnhvnbnphctphoccctrnghp ngoi l. Ta s ln lt a ra cc phng n x l cho cc trng hp ny. nh c qu t k t Hnh 6.4. V d v mt nh nghing c t k t ch ci 89 Trng hp th nht l trong nh c qu t k t ch ci cha xc nh c gc nghing. Cc i tng trong nh ch yu l nh hoc nhiu, c bit cc k t nghing cc gc khc nhau do c th ring ca nh. Hnh v di y minh ho mt nh vn bn nghing vi s k t rt t.Vvy,tacharaktlunvgcnghingchovnbntrong trng hp s lng cc i tng ny phi ln hn mt ngng no . Trong chng trnh s lng ny c chn bng 70 i tng. Cc i tng bao nhau Trng hp ngoi l khc l cc i tng bao nhau. y l mt cn tr i vi nhng thut ton xc nh gc nghing khc c bit l nhng thut ton theo phng php phn tch lng ging thn cn nh c cp trn. Mc d s k t trong vn bn c th rt nhiu nhng cc k t hu ht b cha trong cc i tng khc ln hn nhiu chng hn nh nh hay bngbiu.Hnh5.5diyminhhachochotrnghpccktb bao bi i tng nh. Khi , nhim v l phi nhn ra c s bao hm gia cc i tng v tch, ly c cc i tng k t b bao bi cc i tngln hn. Hnh 6.5. V d v vn bn nghing c cc i tng bao nhau y, ta dng mt k thut bc dn nhng i tng ln ngoi c xcnhnhngkttrong.Mtitngcgilckchthc ngoiccquyclitngcchiurngvchiucaolnhn 200 pixel. Nu trong qu trnh d bin ta gp mt i tng nh vy, ta s 90 cch ly n ra khi tp i tng ang xt. Cc i tng ny s c dng n nu cui cng s i tng c chn p dng bin i Hough b hn70.Taxemnhitngnylmtnhvtiptcduytcci tng bn trong n ly ra nhng i tng k t.b) Thut ton pht hin gc nghing vn bn da vo bin Gi s nh u vo l nh mu (Image). Thut ton pht hin v chnh sa gc nghing vn bn c thc hin theo cc bc chnh sau: Bc 1: Tin x l nh mu Image c nh trung gian TempImageBc 2: Xc nh chu tuyn ngoi cho cc i tng: Duytnhttrnxungdi,ttrisangphi,imnhhintil (x,y): Nu (x,y) c mu khc mu nn v cha xt Label [x][y]=0 : - Tng gi tr nhn ln mt n v: label=label+1. -GihmxcnhchutuynDetectAnObjectviimxut pht (x,y), rec dng lu hnh ch nht cha i tng, hm tr v -1 nu i tng c lp, 1 nu i tng c kch thc bnh thng v 0 trong trng hp ngc li c kch thc k l. - Nu hm chu tuyn tr v 1 : + Tng s i tng: Id=Id+1. + Lu li Rec[Id] =rec. +Duyttphisangtri,tmimcnghngcnhnbng label v nhy ti . - Ngc li nu hm DetectAnObject tr v 0: +Nurec.Wid>200vrec.Hei>200(kchthcquln) thit lp mu nn cho cc im bin ly im (x,y+1) lm im xt tip theo. +Ngcli,duyttphisangtritmimcnghngu tin c nhn bng label v nhy ti . Nu (x,y) c mu khc mu nn v xt, Label [x][y] > 0, duyt t phisangtritmimutincnghngcnhnbngLabel[x][y]v nhy ti .Bc3:DngmngRec[N]xcnhccgitrngngtrungbnh WidAvr, HeiAvr v PrmAvr. Bc 4: p dng bin i HoughVi mi phn t Rec[i] ca mng Rec:91 NuRec[i].Pmr = ) ,M Segments(c c CGrid k k MGrid( ) ( )= ==h vninjGrid k Grid kj i M c M c Segments1 1) , ( , Intersec ,( )( )|||

\| =GridM kGridC cGrid kGrid k MM c Segmentsj i M c j i N,1) , ( , Intersec ) , (= ==h vGrid GridninjM Mj iN N1 1) , (99 lch ca vn bn so vi mu c nh gi bng t s gia tng lch ca cc vng trong vn bn v mu i vi tng ca li kt hp gia hai li c xy dng t cc vng ca vn bn v mu trn tng s vng ca vn bn v mu: Trong : MGGrid- l li kt hp t hai li c xy dng t cc hnh ch nht vng ca vn bn v mu ,,c cn n- l s vng ca vn bn v s vng ca mu N NGrid GridMG MGj i'), , (- l lnh ca vn bn v mu so vi li (i,j) *V d minh ho nh gi lch vn bn so vi muCutrcvnbn,cutrcmuvlitahnhchnhtxydng tng ng Li ta hnh ch nht tng ng '1 1') , ( ) , (c cninjMG MGn nj i j iSh vGrid GridN N+== =100 Li xy dng kt hp t cc li ta vng ch nht vn bn v mu Khi , gi tr lch ca vn bn v mu so vi cc li c tnh theo cng thc ) , ( j iNGridM l: 1/4 1/41/4 1/41/21/21/8 1/81/8 1/81/81/81/81/80 0 00000 0 00 01/4 1/41/4 1/41/3 1/3 1/31/81/81/81/81/81/81/81/8000 0000 000 v do , lch ca vn bn so vi mu c tnh theo cng thc l: 6.2.3.2. Thut ton phn tch trang vn bn da vo mu Diy,chngtitrnhbythuttonphntchtrangvnbn pageANALYSIS*davomunhkthutphntchtrangvnbn pageANALYSIS [1] theo tip cn di ln nh s dng quan h Q v vic nh gi lch cu trc vn bn theo mu mc trn. Vo:nh vn bn I cn phn tch,Tp cu trc vn bn mu tempStructs Ngng Tolerance Ra:Cu trc trang vn bn cu phn tch pageStruct Phng php: Thut ton gm cc bc c bn sau Tnh biu tn xut theo khong cch Hausdorff 3125 , 01654 43 / 2 6 / 1 2 / 1 4 / 1 * 4= =++ + += S101 + Tch cc i tng da vo chu tuyn ngoi + Tnh khong cch Hausdorff gia cc i tng + Xy dng biu tn xut theo khong cch tnh Vi biu tn xut xy dng la chn ngng PhntchtrangvnbntheothuttonpageANALYSIStheo quanhQvingnglachndavobiutnxut bc 2 nhgilchcacutrctrangvnbnvacphntch bc 3 vi cc cu trc trang vn bn mu v tm ra cu trc trang tng ng c lch nh nht. Lplibc2nbc4chngnocnlachnctheo ccnhbiutnxuttheokhongcchHausdorffgiacc itng nh. Chnramuclchnhnhttrongscclchnhnht tm c trong bc 4 ng vi cc la chn. Kimtranulchnhnhttmctrongbc6nhhn ngngTolerancethcthktlunvnbncnphntchc dng l mu c lch nh nht tng ng v cu trc trang phn tchthuccutrctngngthucbc2saubc phntchtheothuttonpageANALYSIStheoquanhQ. Trongtrnghpngclicthktlunvnbnkhngnm trongccmuvnbnchotrc,nngcaochtlngcho bc sau c th b sung thm vn bn vi cc cu trc tm c tng ng vo tp mu cu trc vn bn. Mnh6.2:ThuttonphntchtrangvnbnpageANALYSIS*da vo mu l dng v cho kt qu ng.Chng minh:V s im ca chu tuyn v i tng xc nh bi chu tuyn l hu hn nn bc xt duyt chu tuyn l dng do bc c lp cc i tng sdng.Sccitngthuclhuhnnnvictnhbiutn xuttheokhongcchHausdorffldng.Do,ccbclachn ngng da vo cc nh ca biu tn xut l hu hn. V thut ton phn tch trang vn bn pageANALYSIS* da vo mu nhkthutphntchtrangvnbnpageANALYSIStheotipcn di ln nh s dng quan h Q v vic nh gi lch cu trc vn bn theo mu.Tnh ng n ca thut ton pageANALYSIS c ch ra trong 102 [1] v t mc 6.2.3.1 ta thy tnh ng n ca vic nh gi lch vn bn theo mu dn n tnh ng n ca thut ton pageANALYSIS*. Tng hp cc bc trn ta c thut ton pageANALYSIS* l dng v cho kt qu ng. Cc bc tin hnh phn phn vng v i snh mu Hnh 6.10: Cc bc tin hnh phn vng v i snh mu 6.3. CT CH IN DNH DA VO CHU TUYN Mimtthaymtcmchdnhcthcxemnhmti tng nh. Mc ny xut mt thut ton ct ch Vit in dnh nh vic sdngtnhchtcachutuynvvtrtnghgiachng.Vics dngtnhchtcachutuyn,vtrtnghgiachngvmtstnh cht khc lin quan trong vic ct ch dnh, qua thc nghim th hin hiu quhnivilpccchVitindnhcpdngccphng php truyn thng. 6.3.1. t vn Mttrongnhngimmuchtcavnnhndngchinni chung v ch Vit in ni ring l phi c lp c chng. Vic c lp cc ch c tin hnh sau khi tin hnh phn tch trang tch cc khi, t cc khi tch ra cc dng, t cc dng tch ra cc t v cc t tch ra cc k t. 103 Phng php chiu ngang VPP (Vertical Project Profile) hay biu tnsutngangVH(VerticalHistogram)lphngphpkhphbin cpdngtrongcchOCR(OpticalCharacterRecognition).Nhng phngphpnythngbhnchkhiccktbdnh(touch)hayb chn (overlap) [35,36,54], c bit i vi ch Vit vi cc tng m v du iu ny thng hay xy ra. Hnh 6.11. V d v ch Vit in b dnh Hnh 6.11 l mt s v d v k t ch Vit in b dnh v b chn m phn mm VnDOCR(*) phin bn 2.0 s dng php ct theo VPP nn php ctkhngchnhxc.ivinhngchchtlngxunhth,nhng ch dnh c din tch tip xc ln hn din tch tip xc ca cc phn camt ch. VickhcphcnhcimnycaccphngphpVPPvVH cnhiutcgigiiquytchonnhngthiimhinnaythhin tnh thi s ca vn trong c vic s dng chu tuyn kt hp vi cc phng php cho ch dnh v c bit l ch vit tay. Mtkhc,trongccnghincutrccctcgichrarngmi mt i tng nh c gii hn bi duy nht mt chu tuyn ngoi v cc chu tuyn trong. Trong phm vi nghin cu ny mi mt t hay mt cm (*) Phn mm ot gii nht Sng to khoa hc k thut Vit Nam (gii VIFOTECH trc y) nm 1999. 104 chdnhlmtthnhphnlinthngtrongnhvdonlmtitng nh. Xutphtthoncnh,mcnytrnhbymtkthutctch Vit in dnh, nh vic s dng cc tnh cht ca chu tuyn v v tr tng hgiachng,viitngtrctiplccchindnhmphnmm VnDOCR2.0 sau khi s dng phng php ct ch VPP khng ct c. 6.3.2. Mt s khi nim c bn Cp im tht: L cp im nm trn chu tuyn ngoi ca mt t hay mtcmchdnhmkhongcchgiachngnhhnnhiusovi khong cch dc theo chu tuyn gia chng. Khong cch dc theo chu tuyn ngoi gia hai im pi v pj c nh ngha l: distB(pi, pj) l khong cch nh nht t pi n pj dc theo chu tuyn. GitrwijcngcthctnhtonbihmnnglngbinW:B(E)RdctheobinB(E)caE.Tac:distB(pi,pj)=min(|W(pi)-W(pj)}, W-|W(pi)-W(pj)|) y, W tng nng lng bin dc theo B(E). Ngng tht (L): L t s khong cch ti a gia 2 im dc theo chu tuyn v khong cch gia chng trong cp im thuc chu tuyn ang xt c tm coi l cp im tht. Ngngqut(S):Lsimcchimangxtv2phatrn ng i ang xt. Th d: S= 24, v im qut ti im A th ngng qut c m t nh hnh di y: Hnh 6.12. S im cch v hai pha ti im qut A Mt ct (M): L khong cch tng i gia 2 cp im tht lin nhau v 2 pha. Th d M = 5 th tt c cc im ct lin k nhau s cch nhau 1 khong ti thiu l 5 im nh. Mt ct rt quan trng. Tht vy, trong mt t c th c nhiu im tht v cc im tht c th rt gn nhau (cch nhau khong 1,2,3,5,6 pixel) iu ny lm cho nt ct s khng mnh v vy cn phi c khong cch gia cc im tht. V d: Khi M = 0 thchngtrnhsnhnra1lotccimthtvara cc quyt nh ct nh trong Hnh 5.13.b. Khi M= 5 th chng trnh a ra quyt nh ct nh trong Hnh 5.13.c A 2S=48 Start End 105 Hnh 5.13. V d v ng ct v mt ct 6.3.3. Thut ton ct ch in dnh da vo chu tuyn 6.3.3.1. Phn tch bi ton tng chnh ca phn ny v vic ct ch in dnh l gn nh, xa ng, da vo cc c im ca cc cp im tht ra quyt nh ct.Vi mi mt t hay mt cm ch dnh, trc tin ta tin hnh tch ra cc chu tuyn, sau da vo chu tuyn ngoi tm ra cc cp im tht. Mi chu tuyn u tn ti mt chu tuyn i ngu. Trong cc nghin cu trc cc tc gi ch ra rng mi i tng nh, y l mt t hay mt cm ch dnh, tn ti duy nht mt chu tuyn ngoi. Ch trc khi d bin Ch sau khi d bin Chu tuyn trong Chu tuyn ngoi Qu trnh tm cp im tht im ang xt im phn na trn v na di chu tuyn ngoi Hnh 6.14. Qu trnh tm chu tuyn v cp im xt duyt Xut pht t t tng trn, vi mi mt t hay cm ch dnh ban u s c tin hnh d bin tm ra cc chu tuyn v tnh cht tng ng, ccthngtinvchutuynngoivchutuyntrongtmccngcc thng tin v ngng qut, ngng tht, mt qut c lu li. Qutrnhtmcccpimthtcaitngbtutvicxt duyt cc im thuc chu tuyn ngoi. Cp im xt duyt c xc nh bi mt im nm na phn trn chu tuyn v im cn li nm phn na di chu tuyn ngoi (Hnh 6.14). Vic phn loi na trn v na di chu tuyn xc nh bi im bn tri nht v im bn phi nht ca i tng. Na chu tuyn di Na chu tuyn trn c) ng Tp M=0 M=5 b) a) 106 Vi mi im xt duyt nm na trn chu tuyn ngoi ta tm mt im tngngnmnadichutuynckhongcchnhnhtnim angxtvthamnngngqut(S).iunynhmtrnhvicxt nhng im bin lin st vi im ang xt dc theo chu tuyn. Nu khong cch t im ang xt n im di chu tuyn tm c c t s so vi khong cch hai im ny tnh dc theo chu tuyn nh hn ngng (L) th cp im ang xt c xem l cp im tht v c lutr li. Tiptheo,cccpimtht tm c s c kim tra xem c tha mniukinlcpimcthaykhng?haynicchkhcngni gia 2 im c l ng phn ranh gii ca cc ch dnh hay khng? Trc ht, v tr ca tng cp im tht s c xt duyt, nu lch honhcachngquxahayqugn(nhmtrnhtrnghpct ngang)thcpimsbloi.Trongtrnghpngcli,cpim bin(khiulDb)sctruyntheoligihmFindCutPoint(...). Hm FindCutPoint(...) s thc hin tm m s im thuc na di ca chutuynngoichonhgnvihonhcaimthucnatrn chu tuyn ngoi trong cp im ang xt (k hiu s im bin tm c l B).NusimBtmc>1th hm FindCutPoint() tr v gi tr FALSE, do cp im tht Db s b loi khi qu trnh xt duyt sau . Ngcli,hmFindCutPoint()tnhcaol1ca2imbin trong Db, v tnh cao l2 ca im chun trong Db vi im B va tm c. Nu l2-ll>ll th hm FindCutPoint(...) cng s tr v gi tr FALSE. Ngcli,cpimDbsckimtraxemcgnvicnhtriv cnhphicahnhchnhtbaoquanhchutuynngoiangxthay khng.NuchngqugnthhmFindCutPoint(..)trvgitr FALSE,ngclihmFindCutPoint(..)trvgitrTRUE.Tipna, hmFindCutPoint()sdngnmtMnhmxemxtkhong cch tng i gia cc cp im tht theo honh . Nu chng ln hn iu kin mt ct (M) hm FindCutPoint() tr v gi tr TRUE. Do c im ca ch Vit l cc ch c ru u c ru nm v bn phi nn th t u tin i vi mt ct l t phi sang tri trnh trng hp khi ct cc ch v.v.. li nt cho ch vit sau. Hnh 6.15. Mt s hnh nh minh ho v cc iu kin trn Li Tng s P 1 Khng kim tra l2-ll>ll Khng kim tra s tng h vi hnh ch nht bao quanh chu tuyn ngoi vi Db Li 107 Cpimnycgilcpimct.tngchnhxccho qu trnh ct, honh ca 2 im trong cp im tht c ly trung bnh, sau,honhminyshpvitungcaimchuntrongcp im tht hnh thnh ln mt im mi (gi l im ct tm thi) v n c lu gi nhm phc v cho cc bc tip theo trong qu trnh ct ch. Cui cng, sau khi xc nh c tp im ct tm thi trong tp cccpimbingnnhau,hmVerifyCutPoint()scginhm thm tra li da theo qu trnh phn tch rng ca k t trong t v v tr tng h gia chu tuyn trong v chu tuyn ngoi trn c s phn tch cc mu c bn nh s dng cc quy tc, cc du hiu nh: l: S chu trnh (loops) j: S im khp (junction points) e: S im ngot (turning points) f: S im kt thc (end points) t: Hng (trn, di, tri, phi) biu din cc k t, t tm ra im ct ph hp, chng hn cc imctthngcgiihnbitrcvsauchutuyntrongcai tng (Hnh 6.16). a) Cm ch dnhb) Cc cp im thttm c c) Cc cp im ct sau khi tnh ti v tr tng h ca chu tuyn trong Hnh 6.16. Thc hin VerifyCutPoint c tnh n v tr tng h cachu tuyn trong 6.3.3.2. Thut ton CutCHARACTER ct ch in