TRNG I HC LC HNG

KHOA IN - IN T WX

BO CO NGHIN CU KHOA HC

TI:

NHN DNG BIN S XE

NGUYN PHM ANH TUN

BIN HA, THNG 12 NM 2010

TRNG I HC LC HNG

KHOA IN - IN T WX

BO CO NGHIN CU KHOA HC

TI:

NHN DNG BIN S XE GVHD : Th.S PHAN NH QUN SVTH : NGUYN PHM ANH TUN

BIN HA, THNG 12 NM 2010

Li cm n

Sau bn nm i hc, y l thi im quyt nh thnh qu hc tp ca cc

sinh vin.V ti nghin cu khoa hc chnh l bng bo co cho thnh qu hc tp

.

hon thnh tt ti ny, trc tin, em xin chn thnh cm n cc thy c

v ban gim hiu trng H Lc Hng ni chung v cc thy c trong khoa in

in t ni ring to mi iu kin gip chng em trong qu trnh hc tp v

nghin cu.

K n, em xin chn thnh gi li cm n n thy Phan Nh Qun, l gio

vin hng dn v cng l ngi nhit tnh ch bo cho em trong sut qu trnh

hon thnh ti.

Cui cng xin gi li cm n c bit n gia nh lun l mt ch da vng

chc cng nh lun to mi iu kin tt nht cho em c hc tp.

Mc lc Trang ba ph ........................................................................................................... Li cm n .................................................................................................................. Mc lc .......................................................................................................................... Li m u ............................................................................................................... 1 Chng 1: GII THIU V TI................................................... 2 1.1.L do chn ti ..................................................................................................... 2 1.2.Ni dung ti ........................................................................................................ 2 1.3.Gii hn ti ......................................................................................................... 3 1.4. Mc tiu.................................................................................................................. 4

Chng 2: CHP NH BNG WEBCAM ...................................... 5 Chng 3: TCH BIN S......................................................................... 7 3.1.Tng quan v tch bin s ...................................................................................... 7 3.2. Tm vng mu trng .............................................................................................. 8 3.3.Xc nh vng cha bin s ................................................................................. 10 3.4.Ct vng cha bin s .......................................................................................... 11 3.5. Tm gc nghing v xoay: ................................................................................... 14 3.5.1. Bin i Radon:............................................................................................. 14 3.5.1.1. Tng quan v bin i Radon.................................................................. 14 3.5.1.2. Cc bc thc hin.................................................................................. 15 3.5.2. Tm gc nghing v xoay .............................................................................. 16 3.6.Ct bin s chnh xc............................................................................................ 19

Chng 4: PHN ON K T.......................................................... 21 4.1. Tng quan v phn on k t ............................................................................. 21 4.2. Nh phn bin s xe .............................................................................................. 21 4.3. Chun ha bin s ................................................................................................ 23 4.4.Phn on k t ..................................................................................................... 23 4.4.1. Chng trnh chia i bin s ........................................................................ 26 4.4.2. Chng trnh phn vng tng k t: ............................................................. 26 4.4.2.1.Chng trnh chn 4 phn vng ln nht ................................................ 27 4.4.2.2. Chng trnh phn vng........................................................................... 27 4.4.2.3. Chng trnh con vitricuctieu................................................................... 28

Chng 5: NHN DNG K T ......................................................... 30 5.1. Tng qut nhn dng k t .................................................................................... 30 5.2. La chn phng php:........................................................................................ 31 5.2.1.Phng php nhn dng c in: .................................................................... 31 5.2.2. Phng php ng dng mng neural: ............................................................ 31 5.3. Gii thiu v mng neural ( neural networks):..................................................... 31 5.3.1.Khi nim: ....................................................................................................... 31 5.3.2. M hnh ca mt mng neural nhn to ......................................................... 32 5.3.3. Thit k 1 mng Neural: ................................................................................. 33 5.3.3.1. Thu thp d liu: ...................................................................................... 33 5.3.3.2. Cc bc thc hin: ................................................................................ 33 5.3.4. C s k thuyt v gii thut cho hun luyn mng lan truyn ngc .......... 33 5.3.4.1. Cu trc mng lan tryn ngc ................................................................ 33 5.3.4.2. Hun luyn mng lan truyn ngc ......................................................... 35 5.4. ng dng mng lan truyn ngc vo nhn dng k t : ................................... 37 5.4.1. Thit k mng lan truyn ngc..................................................................... 37 5.4.2. Qu trnh nhn dng ....................................................................................... 37 5.4.4. Phng php tng kh nng tng qut ca mng .......................................... 38 5.4.5. Hun luyn mng lan truyn ngc ............................................................... 39

Chng 6: MCH IU KHIN.......................................................... 41 6.1. S nguyn l.................................................................................................... 41 6.2. Nguyn l hot ng ............................................................................................ 42

KT LUN ............................................................................................................ 43 Ti liu tham kho ............................................................................................ 44 Ph lc 1: Code ca chng trnh MATLAB ............................... 45 Ph lc 2: Code ca vi iu khin 89S52 .......................................... 52

1

Li m u

X l v nhn dng l mt lnh vc t lu c nhiu ngi quan tm. N

c ng dng vo nhiu lnh vc nh:

Trong y hc, ci thin nh X-quang v nhn dng ng bin mch mu t nh chp bng tia X , ng dng vo cc xt nghim lm sang nh pht hin v

nhn dng u no, ni soi ct lp

Trong thin vn hc, h thng chp hnh gn trn tu v tr hn ch v kch thc v trng lng, do cht lng hnh nh nhn c b gim cht lng

nh b m, mo hnh hc v nhiu nn. Cc hnh nh c x l bng my tnh.

Trong cc lnh vc cng nghip, ngi my ngy cng ng vai tr quan trng. Chng thc hin cc cng vic nguy him, i hi c tc v chnh

xc cao vt qu kh nng con ngi. Ngi my s tr nn tinh vi hn v th gic

my tnh ng vai tr quan trng hn. Ngi ta s khng ch i hi ngi my pht

hin v nhn dng cc b phn cng nghip m cn phi hiu c nhng g chng

thy v a ra hnh ng ph hp. X l nh s tc ng n th gic ca my tnh.

Ngoi ra, x l v nhn dng cn c ng dng trong lnh vc khc t c ni n hn. Cng an giao thng thng hay chp nh trong mi trng khng

thun li, nh thng b nhe nn cn c x l v nhn dng c th nhn thy

bin s xe.

Trong lun vn ny ngi thc hin xin trnh by mt ng dng ca x l v

nhn dng s l NHN DNG BIN S XE.

2

Chng 1: GII THIU V TI

1.1.L do chn ti:

Cng vi s pht trin khoa hc k thut, nhu cu i li ca con ngi ngy

cng tng, lu lng giao thng ngy cng ln. Vi s lng phng tin giao thng

ngy cng ln v cn khng ngng tng th vic qun l cc phng tin giao thng

gp rt nhiu kh khn do cn c mt h thng t ng. Mt trong nhng h thng

l h thng nhn dng bin s xe. l mt h thng c kh nng c v

hiucc bin s xe mt cch t ng.

Trong lun vn ny, ngi thc hin xy dng h thng qun l bi gi

xe bi v cc bi gi xe hin nay cn c vn bt cp nh:

Tn nhiu nhn cng. An ton cha cao, vn cn xy ra hin tng mt xe. V xe bng giy, d b mt hay nhu nt.

1.2.Ni dung ti:

Cng nh mi h thng khc, h thng ny yu cu c phn cng v phn

mm. Phn cng c phn chnh l WebCam thu nhn hnh nh v phn mm s

phn tch hnh nh ly ra cc k t trn bin s xe.

Qu trnh thu nhn bin s xe [5] c thc hin theo s sau:

Tachbien so

Phanoanky t

Nhandangky tKet qua bien so xe:

66K90434

Ket qua bien so xe: 66K90434

Tachbien so

Phanoanky t

Tachbien so

Phanoanky t

Nhandangky tKet qua bien so xe:

66K90434

Ket qua bien so xe: 66K90434

Hnh 1.1: S qu trnh nhn dng

3

Tch bin s: khi ny c chc nng tch bin s t nh chp bng cc phng php x l nh. Kt qu ca khi l nh mu RBG (Red Green Blue) c ct

ra t nh chp. y l mt cng vic rt kh khn v ta khng bit c v tr

chnh xc ca bin s. Ngoi ra, cn ph thuc vo sng ca nh chp.

Phn on k t: sau khi tch bin s, chng ta bt u phn on k t. Khi ny thc hin tch tng k t c trong bin s, to thnh tp nh ring bit cc

k t phc v vic nhn dng k t. nh ca mi k t l nh trng en.

Nhn dng k t: sau khi phn on, tch c cc k t trong bin s v to thnh mt chui k t. Chui ny a vo khi nhn dng dng tin hnh

nhn dng tng k t trong chui.

1.3.Gii hn ti:

Vic c nhiu bin s xe vi nh dng v sng khc nhau gy kh khn cho

vic nhn dng. Do qu trnh nhn dng da vo phng php x l nh v trch xut

bin s t nh chp nn sng khc nhau lm tng phc tp trong qu trnh nhn

dng.

Do thi gian thc hin ti khng cho php nn ngi thc hin gii hn cc

bin s v iu kin nh sau:

Bin s c ch en, nn trng, c 2 hng, s k t l 8. Bin s phi cn nguyn vn, khng b trc sn hay r st, khng b che

khut.

Gc nghing ca bin s khng qu 450 so vi phng ngang. Hnh chp bin s khng b m, k t bin s cn phn bit, nhn dng

c bng trc quan.

Khng b nhiu bi nh sng lm nh chp b chi.

4

1.4. Mc tiu:

Trong ti ny, em t nhng mc tiu sau:

Hn ch nhng vn bt cp ca bi gi xe hin nay. Xy dng mt m hnh h thng qun l bi gi xe t ng.

t c nhng mc tiu trn, em tin hnh cc cng vic nh sau:

Tm hiu v qu trnh x l v nhn dng nh Tm hiu cc h thng bi gi xe c p dng Vit Nam

5

Chng 2: CHP NH BNG WEBCAM

y l khu quan trng nht ca h thng bi v nu nh chp b m hay nhiu

th khi a vo nhn dng s khng c. chp nh th ta c th s dng camera

hoc WebCam. Khi la chn thit b th ta cn quan tm ti cc thng s quyt nh

ti cht lng nh nh:

phn gii ( resolution) phn gii cng ln th cht lng hnh nh cng nt. Thng th trong cc

ng dng khng cn thit phi quan st tht r nt th phn gii 420 TV Lines l

hon ton c th chp nhn c.

S im nh ( CCD Total Pixels ) Thng s ny ni ln cht lng hnh nh, s im nh cng ln th cht lng

hnh nh cng tt, tuy nhin, cht lng hnh nh cng tt th cng ng ngha vi

dung lng nh cng ln, v s tn b nh lu tr cng nh nh hng n tc

ng truyn.

iu kin hot ng ca thit b: Cng nh sng nh nht ( Minimum Illumination ):Thng c tnh bng

Lux. Thng s ny ni ln rng, WebCam(Camera) ch c th hot ng cng

nh sng ln hn cng nh sng nh nht. Trong iu kin qu ti, nu khng

phi l Camera c chc nng hng ngoi th s khng hot ng c.

o nh nng mt tri: 4000 lux, c my: 1000lux

o nh sng n tup: 500 lux, c my: 300lux

o nh sng n tup 500 lux, trng (300 lux) trng sng 1lux

o m khng trng: 0.0001 Lux

Xin ch n loi Camera quan st c chc nng Auto Iris (T ng

hiu chnh nh sng). c im ca Camera loi ny l ch vi 1 ngun

sng nh, n c th t ng khuych i ngun sng ln c th

quan st c.

6

Ngun cung cp ( Power Supply ): Hin nay a s cc WebCam u s dng

ngun my tnh(Laptop),Camera quan st u dng loi ngun 12VDC, ch mt s t

cc Camera dng ngun khc. Tuy nhin, bn khng phi lo lng n vn ngun

12VDC, v phn ln cc cng ty bn camera quan st u bn b chuyn i ngun, do

bn c th s dng trc tip ngun 220VAC.

o Operatinon Temperature: Di nhit hot ng.

o Phn ln cc Camera quan st u cho php hot ng trong di nhit -100C ~ 500C, nu Camera ca bn c s dng trong nhng

iu kin khc nghit nh trong cng nghip, khu vc c nhit cao th

bn nn s dng cc loi Camera quan st chuyn dng trong cng

nghip.

o Operational Humidity: m cho php.

o Thng thng, m cho php l 90% RH ( m tng i)

V mc tiu ca ti ch l xy dng mt m hnh v h thng gi xe nn khi

thc hin ti, ngi thc hin chn WebCam lm thit b chp nh

Hnh 1.1: nh chp bng Webcam

7

Chng 3: TCH BIN S

3.1.Tng quan v tch bin s:

Tch bin s l mt bc rt quan trng trong qu trnh nhn dng bin s xe.

Khi tch bin s xe c chia lm 2 giai on chnh:

Giai on 1: nh v tr ca bin s trong nh chp t webcam. Giai on 2: dng cc gii thut ct bin s xe ra khi nh chp v

xoay bin s xe v phng ngang.

Hnh 3.1: S tng qut ca khi tch bin s

Vic nh v bin s xe da vo cc tnh cht ca bin s nh: hnh dng, mu

sc ca bin s so vi nn. Bin s xe c hnh ch nht vi kch thc chiu di v

rng khng thay i. T chng ta c th tm cc vng c hnh t gic trong hnh,

tnh t l gia hai chiu ca hnh so vi khong gi tr cho trc xem th vng

c cha bin s hay khng. Ngoi ra, nh v bin s cn da trn mu sc ca bin s.

Hu ht cc bin s xe Vit Nam u nn trng, ch en. Kt hp hai tnh cht trn,

chng ta xc nh c vng cha bin s.

Sau khi nh v bin s xe, chng ta tin hnh ct bin s xe. Bin s xe c

ct theo 2 bc. Bc u tin l ct vng rng hn vng cha bin s. Sau tm gc

nghing ca bin s v thc hin xoay bin s v phng thng ng. Bc hai l ct

bin s ra khi vng trn. Thc hin vic ct bin s qua hai bc nh trn lm tng

chnh xc, bin s c ct nguyn vn, khng ct phm ch, tr trng hp nh b

chi, tng phn khng u hoc b che khut th kt qu ca vic tch bin s mi

khng chnh xc.

Phn tch bin s s dng rt nhiu gii thut v phng php x l i vi nh

s nh s sau y:

8

Hnh 3.2: S chi tit ca khi tch bin s

3.2. Tm vng mu trng:

V bin s mu trng nn ngi thc hin s nh phn ha nh bng cch gn

gi tr cho cc pixel trng l 1, cn ngc li l 0. u tin, chng ta s bin i nh

gc thnh nh xm (c mc sng t 0 n 255), sau nh phn ha vi mt ngng

thch hp. Nu nh c chp vo ban m hay ban ngy nhng t nh sng th mc

ngng s l 120. Cn ban ngy, nhiu nh sng l 190.

Vn y l lm sao chng ta nhn bit c l nh sng hay nh ti?

Ngi thc hin da vo lc mc xm ( histogram tn s xut hin ca mc

xm ) ca nh [8]. Nu tn s xut hin cc pixel c gi tr

9

V d cho hai nh gc c bin s nh sau:

Hnh 3.3: nh ti v nh sng

nh sau khi nh phn ha s nh sau:

Hnh 3.4: nh sau khi nh phn

y l s thut gii ca bc tm v tch vng mu trng:

10

Hnh 3.5: S thut gii tm v tch vng mu trng

3.3.Xc nh vng cha bin s:

nh nhn c sau khi nh phn ha s c nhiu vng mu trng, v bin s s

nm trong vng mu trng tha iu kin:

0.75 < chiu cao / chiu ngang < 0.91 16000 S_pixel_trng 61000 S_pixel_trng / S_bin s 0.7

Vi:

Chiu cao l H { H = start(i).Boundingbox(4) }

Chiu ngang l W { W = start(i).Boundingbox(3) }

Din tch vng trng l S_pixel_trng { S_pixel_trng = start(i).Area }

Din tch bin s l S_bin s { S_bin s = W H }

11

T s 2 kch thc ca bin s l 0.75 nhng khi b nghing th l 0.91

Hnh 3.6: Hnh dng bin s khi b nghing

Sau iu kin th nht, s c nhng vng trng khng cha bin s nhng c t

s gia chiu cao v chiu ngang ph hp th vn c chn. Ta s loi b nhng

vng ny bng iu kin th hai din tch vng trng. V khong cch t webcam ti

xe c nh (ta ly trc khong cch chp ) nn s lng pixel trng cha trong bin

s c nh.

i vi iu kin th ba, nu bin s nm ngay ngn th t s l 0.7 cn nu

bin s b nghing, s xut hin pixel en nn t s ny gim cn 0.38.

Hnh 3.7: Bin s sau khi nh phn

3.4.Ct vng cha bin s:

Sau khi xc nh c vng no cha bin s, ta tin hnh ct bin s trn nh

mu RBG. Do s tng ng ca nh nh phn v nh mu RBG nn to mt pixel

trn nh nh phn tng ng ta trn nh mu RBG. Do khi xc nh c ta

trn nh nh phn, ta dng ta ny ct trn nh mu RBG.

12

Khi ta ct bin s ra khi nh m bin s b nghing th ta tin hnh ct theo

hnh ch nht ln hn hnh ch nht ln thc s mt lng bin an ton trnh vic

mt thng tin bin s.

Hnh 3.8: Hnh th hin vng nh cn tch ra vi bin an ton

Trong chng trnh, ngi thc hin chn vng bin an ton l 30 pixel.

Trong bc ny, ta khng ct bin s ra khi nh ngay m ch ct vng cha

bin s. Vic ct ny c th xem nh l ct th.

Hnh 3.9: nh chp ban u

Hnh 3.10: nh sau khi ct th

13

y, chng ta c th thy cc iu kin trn cha cht ch, v vy ta lu tt c

cc thng s x, y, W, H ca vng cha bin s vo bin r .

V sau y l thut gii ca bc xc nh vng cha bin s v ct th:

Hnh 3.11: S thut gii ca bc xc nh vng cha bin s v ct th

14

3.5. Tm gc nghing v xoay:

3.5.1. Bin i Radon:

3.5.1.1. Tng quan v bin i Radon:

Dng bin i cc nh trong khng gian 2 chiu vi cc ng thng thnh

min Radon, trong mi ng thng trong nh s cho 1 im trong min Radon.

Cng thc ton hc ca bin i Radon:

( ) ( )+

+= dsssAR cossin,sincos,

Phng trnh trn biu din vic ly tch phn dc theo ng thng s trn nh,

trong l khong cch ca ng thng so vi gc ta O, v l gc lch so vi

phng ngang.

Hnh 3.12: Phng php bin i Radon

Trong x l nh s, bin i Radon tnh ton hnh chiu ca ma trn nh dc

theo 1 hng xc nh. Hnh chiu ca 1 hm s 2 chiu l f(x,y) la tp hp cc tch

phn ng. Hm Radon tnh ton tch phn ng dc theo cc tia song song theo

cc phng khc nhau ( bng cch xoay h trc ta xung quanh O theo cc gi tr

khc nhau ), chiu rng ca cc tia l 1 pixel. Hnh di y biu din 1 hnh chiu

n gin theo 1 gi tr ca gc .

Hnh 3.13: Hnh chiu n gin theo gc

15

Cng thc tng qut trn c th vit li nh sau:

+= ')cos'sin',sin'cos'( dyyxyxfR

Vi

=

yx

yx

cossinsincos

''

Hnh sau s biu din phng php bin i Radon di dng hnh hc:

Hnh 3.13: Phng php bin i Radon di dng hnh hc

3.5.1.2. Cc bc thc hin:

a. Bin i nh v nh nh phn.

16

b. Thc hin bin i Radon trn nh bin vi = 0:179

Gi R ca bin i Radon c biu din nh sau:

c. Tm gi tr ln nht ca R trong bin i Radon.

V tr cc gi tr ln nht ny tng ng vi cc gi tr ca ng thng trong

nh ban u.

3.5.2. Tm gc nghing v xoay:

Bin s c chp vi nhiu gc nghing khc nhau, do ta phi tm gc

nghing v xoay v phng thng. y l vic rt quan trng, v nu khng quay v

phng thng th khi ct bin s s b phm vo bin s. Chng ta xc nh gc

nghing bng phng php bin i Radon.

Trc khi bin i Radon, nh cha bin s c bin i thnh nh c tch bin

nh phn [1],[2].

17

Hnh 3.14: nh c tch bin bin

Sau , ta tin hnh bin i Radon tm gc xoay. Thc hin bin i Radon

vi gc chy trong khong ( 0: 180), ta s c mt ma trn vi cc im R() vi

tng gc v ta pixel tng ng.

Sau khi bin i Radon, chng ta xc nh c gc Rmax, ng vi Rmax th ta c

c max v gc lch l ( 90o - max ). Sau ta s dng hm Rotate trong

MATLAB xoay nh vi gc lch tm c.

Hnh 3.15: nh bin s sau khi xoay v phng thng ng.

18

V y l thut gii ca bc ny:

Hnh 3.16: Thut gii tm gc nghing bin s

19

3.6.Ct bin s chnh xc:

Sau khi xoay bin s v phng thng ng ta thc hin vic ct bin s. y

l mt vic rt quan trng, kt qu ca n quyt nh ti kt qu ca h thng nhn

dng bin s.

Ta thc hin li vic chn vng bin s nhng vi bin l 0 pixel v 3 iu kin

chnh xc hn:

1600 < din tch vng trng < 6100 0.73 < width/height < 0.77 0.6 < s pixel trng / s pixel en < 0.7

Hnh 3.17: Bin s sau khi ct hon chnh

Di y l thut gii :

20

Hnh 3.18: Thut gii ct bin s chnh xc

Sau khi ct c bin s ta tin hnh phn on k t .

21

Chng 4: PHN ON K T

4.1. Tng quan v phn on k t:

Kt qu ca khi tch bin s l mt nh mu RBG c cha bin s xe. nhn

dng cc k t trong bin s, ta tin hnh phn on k t trong bin s. Phn on k

t l vic ct cc k t trong bin s xe .

Sau khi nhn kt qu ca khi tch bin s, khi phn on k t bt u tin

hnh tch tng k t trong bin s. Trc khi phn on k t, nh ca bin s c

chuyn thnh nh nh phn. nh nh phn c chun ha v kch chun, sau tin

hnh ct cc k t. Kt qu ca qu trnh phn on l mt ma trn cha cc nh en

trng ca k t.

Hnh 4.1: S khi phn on k t

4.2. Nh phn bin s xe:

y l bc quan trng nhn dng bin s xe. Bc ny s tm mc ngng

ti u, sau tin hnh nh phn ha nh vi ngng va tm c ( nhm lm tng

tng phn ca k t vi nn bin s ).

Hnh 4.2: nh sau khi c nh phn.

22

Hnh 4.3: Gii thut nh phn bin s

23

4.3. Chun ha bin s

Bin s c chun ha v kch thc [50 150], sau c ly b.

Hnh 4.4: nh bin s sau khi c chun ha

Hnh 4.5: Thut gii chun ha bin s

4.4.Phn on k t:

Ma trn binary ca bin s chnh l ng vo ca chng trnh phn vng k t.

Trc khi phn vng k t, ta chia ma trn nh bin s thnh tng hng v ln lt

a tng hng vo chng trnh phn vng. Tuy nhin, y, ngi thc hin ch tin

hnh nhn dng k t ca hng 2 nn ta ch ng vo ca chng trnh phn vng k t

l ma trn ca hng 2.

phn chia thnh nhiu ma trn k t t ma trn bin s, ta da vo tng s

pixel mc 1 ( mc 1 l mu trng- mu ca k t, mc 0 l mu en mu ca

24

nn).Vi ma trn ca hng 2 sau khi chia i, gia 2 k t c rt t pixel c mc 1

(trong trng hp l tng, th s l 0 ). Nh vy khi cng gi tr cc pixel theo tng

ct, nh hnh sau, ta thy ga tr ti cc vng gia 2 k t rt thp ( y cng l tng

s pixel mc 1). T , gii thut phn vng s nhng vng ny da vo gi tr ca n

nh hn nhng vng ln cn v s phn chia thnh tng vng. y, ta s tm 4 phn

vng tng ng vi 4 k t.

Hnh 4.6: Tng s cc bt theo 1 hng ca bin s

Chng trnh c la chn 2 thng s: Min_area v digit_width.

Min_area l din tch cho php nh nht ca 1 k t, l tch ca gi tr ct ln

nht vi rng ca phn vng .

Hnh 4.7: Hnh th hin thng s Min_area

25

Digit_width l rng ti a cho php ca 1 phn vng k t.

Hnh 4.8: Hnh th hin thng s Digit_width

Sau khi phn vng c cc k t ta tin hnh ct cc k t ra khi bin s.

Hnh 4.9: cc k t c ct khi bin s

Di y l s thut gii:

26

Hnh 4.10: Gii thut phn vng k t

4.4.1. Chng trnh chia i bin s:

y l chng trnh n gin, bin s sau khi c chun ha v kch thc

[50 150] th kch thc tng hng sau khi chia i l [25 150].

4.4.2. Chng trnh phn vng tng k t:

Chng trnh con ny thc hin vic tm vng ranh gii gia 2 k t, tng ng

gia 2 vng ranh gii lin tip s l vng k t, chng trnh s tr v v tr ca vng

k t. Tuy nhin, chng trnh cng c th tr v s vng k t ln hn 4 do nh

hng ca mi trng. Trng hp ny ta vn c th ti u chung trnh tng

chnh xc.

Vi kt qu tr v nh hn 4, ta s xa cc pixel ti cc vng ranh gii ( gn cc

gi tr bng 0 ). iu ny lm cho vic thc hin chng trnh chnh xc hn.

27

Vi kt qu tr v ln hn 4 ( ngha l c 1 hay nhiu vng khng cha k t ),

ta s ly 4 vng c din tch ln nht ( v thng thng, cc vng khng c k t nh

hn cc vng c k t ).

Hnh 4.11: thut gii phn vng tng k t

4.4.2.1.Chng trnh chn 4 phn vng ln nht:

Trc tin chng trnh s sp xp cc phn vng theo th t din tch t nh

n ln, sau s gi li 4 phn vng c din tch ln nht nhng vn gi nguyn v

tr ca chng trc khi sp xp.

4.4.2.2. Chng trnh phn vng:

Chng trnh ny s tm ra v tr ranh gii gia cc k t v tr v s phn vng

tng ng vi s k t.

28

Hnh 4.12: gii thut chng trnh phn vng

Chng trnh con vitricuctieu: tm v tr cc ct trong ma trn u vo m c gi tr ngng nh hn gi tr ta t.

Chng trnh con clean: xa nhng v tr ( tm c trong Chng trnh con vitricuctieu ) m khong cch ti v tr k tip qu nh.

4.4.2.3. Chng trnh con vitricuctieu:

y l chng trnh con quan trng nht trong chng trnh phn vng.

Chng trnh qut ton b ma trn nh u vo, tm v tr kh nghi, xa cc v tr m

khong cch vi v tr k tip kh ln ( ln hn khong cch cho php ) th s thc

hin li chng trnh ny trong khong 2 v tr tip tc phn thnh nhiu vng

nh hn.

29

Hnh 4.13: gii thut chng trnh con con vitricuctieu

Tm v tr nh hn ngng: tm cc v tr trong ma trn u vo c gi tr nh hn ngng. Nu khng tm c v tr no hoc ch tm c 1 v tr th ta s

tng gi tr ngng ln 1.

on chng trnh thm vo 2 v tr u tin v kt thc nu c gi tr ln hn ngng: nhng v tr tm c c th c v tr hai bin ca ma trn u vo

hoc l khng. Tuy nhin, nu bin vn c gi tr ln hn ngng th ta phi thm v

tr ny.

30

Chng 5: NHN DNG K T 5.1. Tng qut nhn dng k t:

Sau khi thc hin phn vng ta s c 4 ma trn tng ng vi 4 k t trn 1

hng bin s. Ln lt tng ma trn k t s c a vo chng trnh nhn dng.

Kt qu cui cng s l 4 k t s v chng trnh s hin th k t ny di dng

text.

Thc cht, qu trnh nhn dng l qu trnh i ma trn im nh ca cc k t

thnh m ASCII tng ng vi k t . lm c iu ny ngi ta em so snh

ma trn ca k t vi tt c cc ma trn trong tp mu, ma trn mu no c kh nng

ging nhiu nht th c chnh l k t cn tm.

Hnh 5.1: thut gii nhn dng k t

Trong lnh vc nhn dng, c 2 phng php nhn dng l phng php c

in v phng php s dng mng neural.

31

5.2. La chn phng php:

5.2.1.Phng php nhn dng c in:

Gii thiu phng php: Phng php ny s c 1 tp ma trn k t mu. Phng php ny kh n

gin: m trn k t cn nhn dng kh ging vi ma trn k t trong tp mu. V d,

ta nhn dng cc s t 0 n 9 thi trong tp mu, ta s to ra cc ma trn k t t 0 n

9. Gi s ma trn cn nhn dng l s 1 th ma trn ny nhn bng mt thy cng kh

ging sao vi ma trn s 1 trong tp mu.

u nhc im ca phng php: Phng php ny tuy n gin nhng hiu qu trong trng hp tp nh nhn

dng r nt, t b nhiu. Nu nh b nhiu th ma trn u vo thay i, lc ny tp mu

s khng cn chnh xc. Trong 1 vi trng hp bin s ban u b nghing hoc lch,

khi ta xoay va chun ha kch thc th cc k t b nhiu v khi so snh vi tp mu

cng khng cn chnh xc.

5.2.2. Phng php ng dng mng neural:

Cng nh phng php c in, ta nhn dng cc k t nh vo v tr v hng

ca k t trn bin s, nhng y ta s dng nhiu tp mu so snh qua lm

tng chnh xc khi nhn dng. Hn na trong MATLAB, c TOOLBOX h tr v

phn neural lm cho vic nhn dng tr nn d dng hn.

V l do trn m khi tin hnh nhn dng k t, ngi thc hin chn phng php nhn dng dng mng neural [3],[4].

5.3. Gii thiu v mng neural ( neural networks):

5.3.1.Khi nim:

Neural networks pht trin t cc nghin cu v tr tu nhn to, da trn vic

m phng cp thp h thng Neural sinh hc c gng bt chc kh nng hc v

chp nhn sai ca b no c cu trc thp.

32

Hnh 5.2: M t ton hc tng qut ca mng Neural

Tn hiu ng vo sau khi qua mng neural s c tnh ton v ng ra ca mng

s c so snh vi tn hiu ch mong mun. Mng s tip tc cp nht v iu chnh

trng s v ngng cho n khi tha mn ng ra yu cu.

5.3.2. M hnh ca mt mng neural nhn to:

Neural nhn to c 1 s ng vo ( t cc d liu gc hau t ng ra ca cc

neural khc). Mi kt ni n ng vo u c 1 cng ( hay trng s ). Ng vo ca

neural c th v hng hay hu hng. Mi neural c 1 gi tr ngng. Tn hiu c

truyn qua hm kch hot ( hay l hm truyn) to gi tr ng ra ca neural.

f

Hnh 5.3: neural 1 ng vo

f

Hnh 5.4: neural nhiu ng vo

33

5.3.3. Thit k 1 mng Neural:

5.3.3.1. Thu thp d liu:

Trc ht ta cn xc nh c tp d liu l tp bao gm 1 s trng hp,

mi tp hp cha gi tr ca ng vo v ng ra khc nhau.Sau xc nh nhng bin

no s c s dng, bao nhiu trng hp cn thu thp.

Vic la chn cc bin c s dung thng do trc gic quyt nh v thng

ph thuc vo cng vic chuyn mn v lnh vc m n c ng dng.

5.3.3.2. Cc bc thc hin:

Chn cu hnh ban u (thng l 1 lp n c s neural n bng na tng s

neural ng vo v ng ra).

Thc hin lp i lp li s th nghim ca mi cu hnh, gi li mng tt nht

(thng da vo sai s).

Trong mi ln th nghim nu xy ra vic hc cha ( kt qu th nghim

khng t c xc sut nh yu cu ) th th tng s neural trong lp n. Nu xy ra

vic hc qu mc ( sai s ban u tng ln ) th hy b bt 1 vi neural n ( c th b

lp n)

5.3.4. C s k thuyt v gii thut cho hun luyn mng lan truyn ngc:

5.3.4.1. Cu trc mng lan tryn ngc:

Cc loi neural: (tansig, logsig, purelin) Mt loi neural c R ng vo th hin nh sau:

Hnh 5.5: Cu to 1 neural

Mi ng vo tng ng vi 1 trng s w. Tng ca gi tr ngng b vi tch ca

tch ca cc ng v v cc trng s s l ng vo ca hm truyn f ( f(Wp+b) ). Neural

dng nhiu hm truyn khc nhau tnh ton ng ra. Mng nhiu lp thng dng

hm truyn log-signoid (logsig).

34

Hnh 5.6: Hm truyn logsig

Hm logsig tnh ton gi tr ng ra nm trong khong 0 v 1 trong khi gi tr

ng vo t - n +.

Tip n l hm truyn tan-signiod ( tansig)

Hnh 5.7: Hm truyn tansig

Him khi hm truyn tuyn tnh ( purelin ) c s dng trong mng lan truyn

ngc ( thng ch s dng ph bin trong mng tuyn tnh ).

Hnh 5.8: Hm truyn purelin ( tuyn tnh )

Cu trc cac lp trog mng lan truyn ngc: Mng 1 lp vi S neural logsig c R ng vo c th hin nh hnh bn tri,

cn bn phi l s khi cc lp

Hnh 5.9: Cu trc mng 1 lp

35

Mng lan truyn ngc vi cc neural c hm truyn t phi tuyn cho php

neural hc c nhng mi lien quan tuyn tnh v phi tuyn gia ng vo v ng ra.

Nu gii hn ng ra l 0 v 1 th nn dng hm truyn logsig.

M hnh mng 2 lp nh sau:

Hnh 5.10: Cu trc mng 2 lp

5.3.4.2. Hun luyn mng lan truyn ngc:

Mng lan truyn ngc c hun luyn xp x 1 hm phi tuyn , 1 m hnh

lin kt hay phn lp. Trong sut qu trnh hun luyn, cc trng s v gi tr ngng

s c iu chnh 1 cch hp l lm ti thiu gi tr hm li ca mng. Hm li

mc nh trong mng lan truyn ngc l trung bnh phng li (mean square error

hay mse ) li trung bn phng gia ng ra thc t v ng ra c tnh ton. Sau y

l 1 vi thut ton hun luyn mng lan truyn ngc. Tt c cc thut ton u dng

dc (gradient) ca hm li iu chnh cc trng s sao cho ti thiu c gi tr

hm li. dc c tnh ton da vo k thut lan truyn ngc.

STEEPEST DECENT GRADIENT: Trong thut ton lan truyn ngc, vector gradient ca mt phng sai s s

c tnh ton. Vector ny ch ra ng dc nht v v tr hin ti, v th nu ta di

chuyn theo n 1 khong ngn ta c th t gi tr nh nht.

Tuy nhin kh khn y l quyt nh lp ca tng bc di chuyn. Bc

ln c th hi t nhanh hn nhng c th vt qu im cn n hay hay ra khi vng

c cc tiu ( nu mt phng sai s b lch tm). Ngc li bc nh c th i n ng

hng nhng phi thc hin lp li nhiu ln.

QUY TC HC THCH NGHI: Phng php thc ra rt n gin. Mi phng php c 1 h s hc e khc

nhau.Khi cp nht trng s, nu hng li hin hnh cng bc vi hng trc, cho e

ln, cn ngc hng, cho e nh.

36

Hng li c xc nh l du ca dm, l o hm ring ca hm li theo trng

s bc m. Nu dm dng, li gim khi trng s gim, nu dm m, li gim khi trong

s tng.

fm+1= fm + (1- ) dm Nu ta cho f l trung bnh trng s ca cc o hm hin ti v qu kh, l

trng s ca cc hm qu kh , (1- ) l trng s ca cc hm hin ti. Nu f dng th y l li gim khi trng s gim v ngc li cng nh i vi o hm.

Da vo f v ta c th o chnh xc c hng ca li ang gim ln hng ca

li va mi gim. Nu chng cng du, vic gim li xy ra theo hng c, chng

khc du, ngc hng vi hng c.

CONJUGATE GRADIENT DESCENT: y l thut ton nh hng ng, bng cch ly ra 1 hng nhy di

chuyn theo chiu ngang a chiu, ri chiu ng thng theo hng xc nh

im nh nht v lp li. Hng nhy l hng c dc ln nht . tng y l

khi thut ton c cc tiu ha dc theo 1 hng c th ha no , th o hm

bc 2 dc theo hng phi gi zero. Cc hng lin hp c gi o hm bc 2

ny vi gi thit mt phng ny l parabol. Nu iu kin ny c gi, N epoch s

t c gi tr cc tiu.

Levenberg Marquardt: L k thut m phng vng tin cy: thay v theo 1 nh hng c nh, ta gi s

mt phng c hnh dng n gin sao cho cc tiu c th nh v trc ti, nu gi thit

ng. Th m hnh v xem xt mc tt ca im c chn. M hnh c gi s

rng mt phng c hnh dng tt s ng nu gn t c cc tiu. Ngoi im gi

thit c th b vi phm, v m hnh c th chn nhng im sai di chuyn. M hnh

ch c th c tin cy trong 1 vng ca im hin ti v kch thc ca vng ny th

cha bit. Do , chn cc im mi kim tra vi im c chn. Nu im mi

tt, di chuyn n im v tng cng vai tr ca m hnh trong vic la chn im

mi, cn nu xu, khng di chuyn v tng cng vai tr ca bc dc gtradient trong

vic la chn im mi.

37

5.4. ng dng mng lan truyn ngc vo nhn dng k t :

Mng lan truyn ngc c ng dng rng ri trong cc vn thc t lien

quan n mng neural. Vic thit k n gin cng nhng gii thut hun luyn hi t

nhanh th hin mng lan truyn ngc l 1 cng c rt mnh v ph bin, c bit

trong vn nhn dng k t c cng font, do vic ng dng n cho 1 kt qu kh

quan.

5.4.1. Thit k mng lan truyn ngc:

Cc k t s sau khi c phn vng s c nh chun vi kch thc 2010.

Sau ma trn nh ca k t s chuyn i thnh ma trn mt ct v tr thnh ng

vo ca mng neural. Nh vy, ta dng v tr ca gi tr ln nht ca neural ng ra, t

c th bit c gi tr ca k t s . Tp mu hun luyn cng nhiu th

chnh xc cng ln.

Mng nui tuyn lan truyn ngc gm 2 lp neural trong :

Lp vo : 200 ng vo ( tng ng vi ma trn k t 2010 sau khi chuyn

thnh ma trn ct ).

Lp n: 20 neural

Lp ng ra : 10 neural ( tng ng nhn dng 10 k t s t 0 n 9)

Dng hm truyn lgsig co neural lp n v lp ng ra.

Hnh 5.11: cu trc mng dng nhn dng k t s.

5.4.2. Qu trnh nhn dng:

Mng sau khi c hun luyn a vo s dng s thc hin nhn dng i vi

cc ma trn k t s. V d, ma trn u vo lc ny l ma trn s 6. Nh vy, nu

mng hun luyn c kt qu tt th kt qu lan truyn ma trn ny trong mng l

neural tng ng vi v tr s 6 ( trong hnh ny l neural th 6) s c gi tr ln nht.

38

Hnh 5.12: Hnh minh ha hot ng ca mng trong nhn dng

5.4.4. Phng php tng kh nng tng qut ca mng:

Mt trong nhng vn xut hin trong vic hun luyn mng neural c gi

l qu khp. Li trong tp hun luyn c gi tr rt nh nhng khi a d liu mi vo

mng th li li qu ln. Ngha l mng c kh nng nh nhng tp hun luyn rt tt

nhng cha c kh nng tng qut i vi d liu mi.

Hnh sau th hin p ng ca mng neural 1-20-1 c hun luyn xp x 1

hm sin. Mng nh vy s qu khp vi d liu v khng c kh nng tng qut.

Hnh 5.13: Trng hp qu khp

C rt nhiu phng php gii quyt vn ny, trong phng php

ngng hc sm c xem l phng php hiu qu nht.

39

Phng php ngng hc sm Phng php ny dng 3 tp mu, mt tp dng hun luyn, mt tp dng

kim tra v mt tp dng th li. Trong qu trnh hc tp mu hun luyn th li ca

tp mu kim tra cng c gim st. Trong qu trnh hc tp th li ca tp hun

luyn v kim tra u gim nhng n mt lc no th li ca tp hun luyn bt

u tng ln, lc ny xy ra vic mng neural bt u qu khp vi tp hun luyn.

Chnh v vy ta cho dng qu trnh hun luyn, th ti thi im ny, ta c mng neural

c kh nng tng qut nht, lc ny li ca tp th li ( khng c trong tp hun luyn

v tp kim tra) l li ca mt d liu mi.

Hnh 5.14: Hm xp x khi mng ngng hc sm.

5.4.5. Hun luyn mng lan truyn ngc:

Vic hun luyn c ngha quyt nh n chnh xc cng nh s thnh

cng ca chng trnh. Kch thc ca tp mu, s neural ca lp n, kh nng tng

qut ha trnh trng hp qu khp ca mng cng cn c cn nhc v tnh ton

k lng tng chnh xc ca qu trnh nhn dng k t.

tng kh nng tng qut ca neural, ta s c 3 tp mu: mt tp hun

luyn, mt tp kim tra, mt tp th li. Trong qu trnh hc, khi hm li ca tp

kim tra bt u tng th cho dng hc, thi im ny, neural c kh nng tng qut,

v li ca tp mu th li cng chnh l li ca 1 i tng bt k no a vo mng.

40

S lng mu ca cc tp mu nh sau:

Tp mu hun luyn gm 30 mu cho mi k t s t 0 n 9. Tp mu dng kim tra gm 5 mu cho mi k t. Tp mu dng th li gm 5 mu cho mi k t.

Hnh 5.15: Hm li khi ngng hc sm hun luyn mng.

41

Chng 6: MCH IU KHIN 6.1. S nguyn l:

42

6.2. Nguyn l hot ng:

Sau khi nhn dng c k t trn bin s xe ra, chng trnh s tin hnh so

snh chui k t vi d liu bin s xe vo c lu t trc. My tnh s truyn

tn hiu xung vi iu khin 89S52 lm sng led, nu ng th led green sng, nu sai

th led blue sng.

Tn hiu t chn s 2 ca cng COM vo chn 8 ca Max232, sau tn hiu

t chn 9 ca Max232 vo chn 10 ca vi iu khin 89S52.

Vi iu khin s kim tra tn hiu truyn xung. Nu l ng th s sng led

green, cn sai th sng led blue.

Khi led green sng, tn hiu truyn xung, lm quay ng c. 4s sau, n Green

tt, ng c tr v v tr ban u.

43

KT LUN chnh xc ca chng trnh nhn dng bin s xe ph thuc vo nh sng

mi trng. Tuy nhin yu t ny ta c th iu chnh c.

Vic nhn dng k t cn nhiu sai st do khng c c mt tp mu hon

chnh.

Chng trnh c vit bng MATLAB gii quyt c bi ton nhn dng

vi dung lng ln trong thc t, qua gii quyt c yu cu ca ti.

gii quyt cc kh khn ca ti, ngi thc hin xin xut 1 s kin

sau y:

S dng camera chuyn dng. Thit lp mi trng n nh xung quanh lm tng chnh xc ca

nh chp ( xc nh 1 v tr c nh chp nh).

Thu thp nhiu mu k t tng tp mu qua tng chnh xc ca chng trnh.

S dng th m vch lm v gi xe.

44

TI LIU THAM KHO

[1] Amin Sarafraz (2004), Detects lines in a binary image using common

computer vision operation known as the Hough Transform, University of Tehran,

Iran.

[2] [Beal72] Beale, E. M. L., "A derivation of conjugate gradients," in F. A.

Lootsma, ed., Numerical methods for nonlinear optimization, London: Academic

Press, 1972.

[3] [Caud89] Caudill, M., Neural Networks Primer, San Francisco, CA: Miller

Freeman Publications, 1989.

[4] [Cabu92] Caudill, M., and C.Butler, Understanding Neural Networks:

Computer Explorations, vols. 1 and 2, cambridge, ma: the mit press, 1992.

[5] Ondrej martinsky, Algorithmic and mathematical principles of automatic

number plate recognition systems , brno 2007.

[6] Otsu, N. (1979), A Threshold Selection Method from Gray-Level

Histograms, IEEE Transactions on Systems, Man, and Cybernetics, Vol. 9, No. 1, pp.

62-66.

[7] www.Mathworks.com

45

Ph lc 1: Code ca chng trnh MATLAB 1.CHP V LU NH T WEBCAM:

obj = videoinput('winvideo', 1,'YUY2_640x480'); set(obj,'ReturnedColorSpace','rgb'); src_obj = getselectedsource(obj); get(src_obj); vidRes = get(obj, 'VideoResolution'); nBands = get(obj, 'NumberOfBands'); hImage = image( zeros(vidRes(2), vidRes(1), nBands) ); preview(obj, hImage); pause(2); x=getsnapshot(obj); x=imresize(x,[480 640]); imwrite(x,'D:\NCKH\bai lam\x.jpg','Quality', 100);

2.CHNG TRNH TCH BIN S:

function [bienso,biensomau]=laybiensohoanchinh(pic1); XSIZE = 50; YSIZE = 150; white_pic=laybienso(pic1); [r]=lp_area_extraction_lap(white_pic,30); n=length(r)/4 for i=1:n x=r(4*i-3); x2=r(4*i-2); y=r(4*i-1); y2=r(4*i); lp_area = pic1(y:y2, x:x2, :); angle = find_angle(lp_area); pic = imrotate(white_pic(y:y2, x:x2), angle, 'bilinear'); [small_pic, xx, xx2, yy, yy2] = improved_lp_area(pic, angle); [image, RECTx, RECTy] = crop_lp(small_pic, lp_area, xx, xx2, yy, yy2, angle); end image = imrotate(lp_area, angle, 'bilinear'); image=image(yy:yy2,xx:xx2,:); cao = length(image(:,1,1)) ; rong=length(image(1,:,1)); if cao > rong image = imrotate(image, -90, 'bilinear'); end biensomau = image; [grayImage, quantImage, bw] = quantizeImage(image); quantImage = imadjust(grayImage, stretchlim(grayImage), [0 1]); bienso=bw; bienso = normalized_lp_contour(bienso, [XSIZE, YSIZE]);

46

return %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [vungbienso]=laybienso(x) %CT TH mucxam=rgb2gray(x); mucxam=medfilt2(mucxam,[3 3]); threshold=graythresh(mucxam); bw = im2bw(mucxam,threshold); bw = bwareaopen(bw,5000); vungbienso = bw; return %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [angle] = find_angle(rgb_image); %TM GC XOAY gray_image = rgb2gray(rgb_image); theta = (0:179)'; [R, xp] = radon(edge(gray_image), theta); i = find(R > (max(R(:)) - 25)); [foo, ind] = sort(-R(i)); [y, x] = ind2sub(size(R), i); t = -theta(x)*pi/180; r = xp(y); [r,c] = find(R == max(R(:))); thetap = theta(c(1)); angle = 90 - thetap; return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [pic, x1, x2, y1, y2] = improved_lp_area(image, angle); [x1,x2,y1,y2] = detect_lp_area(image, 0); pic = image(y1:y2, x1:x2); return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [x, x2, y, y2] = detect_lp_area (white_pic, spacing); LP_MIN_AREA = 15000; LP_MAX_AREA=74200; LP_MAX_RATIO = 0.8; LP_MIN_RATIO = 0.7; dilated_pic = imdilate(white_pic, strel('diamond', 5)); stat = imfeature(bwlabel(dilated_pic)); depth = -1; for i = 1 : length([stat.Area]) if stat(i).BoundingBox(2) >= depth && stat(i).Area > LP_MIN_AREA && ... stat(i).BoundingBox(4) = (LP_MIN_RATIO)*stat(i).BoundingBox(3) && stat(i).Area >= max([stat.Area])/3.5 depth = stat(i).BoundingBox(2); end; end;

47

r = []; for i = 1 : length([stat.Area]) if stat(i).BoundingBox(2) == depth && stat(i).Area > LP_MIN_AREA && ... stat(i).BoundingBox(4) = (LP_MIN_RATIO)*stat(i).BoundingBox(3) && stat(i).Area >= max([stat.Area])/3.5 r = [r stat(i).Area]; end; end; if(length(r) == 0) index = (find([stat.Area] == max([stat.Area]))); else index = (find([stat.Area] == max(r))); end; x = max(floor(stat(index).BoundingBox(1) - spacing), 1); y = max(floor(stat(index).BoundingBox(2) - spacing), 1); width = ceil(stat(index).BoundingBox(3) + 2*spacing); height = ceil(stat(index).BoundingBox(4) + 2*spacing); y2 = min(y + height, size(white_pic, 1)); x2 = min(x + width, size(white_pic, 2)); return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [image, RECTx, RECTy] = crop_lp(pic, lp_area, x1, x2, y1, y2, angle); rec = find_lp_location(pic); image = imrotate(lp_area, angle, 'bilinear'); image = image(y1:y2, x1:x2, :); RECTy = [rec(2), rec(2), rec(2) + rec(4), rec(2) + rec(4)]; RECTx = [rec(1), rec(1) + rec(3), rec(1) + rec(3), rec(1)]; image = imcrop(image, rec); return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [rec] = find_lp_location(im); p1 = sum(im); [x1, x2] = find_contours(p1); p2 = sum(im'); [y1, y2] = find_contours(p2); rec = [x1, y1, x2-x1, y2-y1]; return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [index1, index2] = find_contours(vec); avg = mean(vec); for j = 1 : length(vec) if(vec(1,j)

48

end; for j = length(vec) : -1 : 1 if(vec(1,j)

49

pic = bw(:, seg(i,1) : seg(i,2), :); area(i) = bwarea(pic); end; area1 = sort(area); seg = seg'; for j = 1:(length(area1)-4) i = find(area == area1(j)); len = length(area); if i == 1 area = [area(2:len)]; seg = [seg(:,2:len)]; elseif i == len area = [area(1:i-1)]; seg = [seg(:,1:i-1)]; else area = [area(1:i-1) area(i+1:len)]; seg = [seg(:,1:i-1) seg(:,i+1:len)]; end; end; seg = seg'; return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [segmentation] = phanvung(im, digit_width, min_area); segmentation = []; t = sum(im); seg = clean(vitricuctieu(t, 2, 1, digit_width), 3); j = 1; for i = 1 : (length(seg) - 1) band_width = seg(i+1) - seg(i); maxi = max(t(1, seg(i):seg(i+1))); if(maxi * band_width > min_area) segmentation(j, 1) = seg(i); segmentation(j, 2) = seg(i+1); j = j + 1; end; end; return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [s] = vitricuctieu(t, val, offset, digit_width); s = find(t < val); if(length(s) < 2) s = vitricuctieu(t, val + 1, offset, digit_width); return; end; if((t(1,1) >= val) && s(1) ~= 1) s = [1 s];

50

end; if((t(1, length(t)) >= val) && s(length(s)) ~= length(t)) s = [s length(t)]; end; s = add(s, offset - 1); s = clean(s, 3); while bad_digit(s, digit_width) == 1 for i = 1: (length(s) - 1) if (s(i + 1) - s(i)) > digit_width sub_vec = t(1, s(i) - offset + 1 : s(i+1) - offset + 1); s = [s(1 : i) vitricuctieu(sub_vec, val + 1, s(i), digit_width) s(i+1 : length(s))]; end; end; end; return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [bool] = bad_digit(s, digit_width); if length(s) == 0 bool = 0; return; end; tmp = s(1); bool = 0; for i = 2 : length(s) if(s(i) - tmp) > digit_width bool = 1; return; end; tmp = s(i); end; return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [t] = clean(s, val); t = []; len = length(s); i = 2; j = 1; while i val t(j) = s(i-1);

51

j = j + 1; end; t(j) = s(i); j = j + 1; i = i + 1; end; return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function [t] = add(s, val); len = length(s); t = []; for i = 1:len t(i) = s(i) + val; end; return;

4.NHN DNG K T:

function number2 = nhandangsonoron2(hang2, seg2, netso) load 'netso.mat'; number2 = []; for i = 1:size(seg2, 1) input = hang2(:, seg2(i,1) : seg2(i,2), :); rec = nhandangsonoron(netso, input); number2 = strcat(number2, rec); end return; %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

function so = nhandangsonoron(net,input ) matranmau=[35 20]; input = imresize(input, matranmau, 'nearest'); input=bwareaopen(input,60); vec = double(im2col(input, size(input), 'distinct')); rslt = sim(net, vec); [Y,I] = max(rslt); num = I-1; so = char('0' + num); return;

52

Ph lc 2: Code ca vi iu khin 89S52 RS BIT P2.3

RW BIT P2.4

EN BIT P2.5

ORG 0000H

LJMP MAIN

ORG 0023H

LJMP INT_SERIAL

ORG 0030H

MAIN:

MOV IE,#10010000B

MOV TMOD,#00100001B

MOV PCON,#00000000B

MOV SCON,#01010000B

MOV TH1,#-3

SETB TR1

MOV R1,#0

MOV R0,#"D"

MOV P2,#0FFh

INT_SERIAL:

JNB RI,EXIT_SERIAL

MOV A,SBUF

CJNE A,#"A",TTO

CPL p2.2

LCALL TRANF

LCALL DELAYHT

CPL p2.2

CLR RI

RETI

TTO:

CJNE A,#"B",EXIT_SERIAL

CPL p2.3

53

LCALL TRANF

LCALL DELAYHT

CPL p2.3

CLR RI

RETI

EXIT_SERIAL:

CLR TI

RETI

TRANF:

CPL p2.4

CLR ES

MOV SBUF,A

JNB TI,$

CLR TI

CPL p2.4

SETB ES

RET

DELAY:

MOV 77H,#255

DJNZ 77H,$

RET

DELAYHT:

MOV 7CH,#16

K:

MOV 7EH,#255

L:

LCALL DELAY

DJNZ 7EH,L

DJNZ 7CH,K

RET

END