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TNG HP HMT

Cu 1

Ti sao cc thut ton da trn ng qut (scanline) li ph hp vi kin trc phn cng ca my tnh

hin i?

Vi cc thut ton vng thng trn kin trc phn cng ca my tnh hin i ch loi ng thng

no v ra mi thc sl ng thng?

Tr li

Hin nay, cc thit b ha l thit b mnh. Mnh l mt ma trn ch nht gm cc im nh (pixel).

Mt dng ca mnh gi l ng qut (scanline). Ton b bc nh sc mt b phn iu khin

qut ln lt tng dng t trn xung di. Mt sng in t s bn vo lp pht pho to nn

cng sng ca im, trong h thng mn hnh mu, ba sng in tc dng tng ng vi ba

mu c bn , xanh l cy v xanh da tri. nh hin thtrn mn hnh c to nn do s kt hp cacc im c thp sng v im khng c thp sng. V th, cc thut ton da trn ng qut l

ph hp.

Vi cc thut ton vng thng trn kin trc phn cng ca my tnh hin i th ch nhng ng

thng song song vi trc ta hoc c gc 45 vi trc ta mi thc sl ng thng.

Cu 2

iu kin ca mt ng thng hon ho (l tng)

Tr li

iu kin ca mt ng thng hon ho:

1. Trng thng v lin tc2. Phi i qua hai im u v cui3. Phi c mt v cng sng u: u trn mt ng thng v u trn tt ccc ng

thng.

4. Thut ton v phi hiu qu v c th thc hin nhanh (ci ny ch bit c phi ni hay khngnh???)

Cu 3

Hy ch ra khuyt im ca cch vng thng da trn phng trnh chnh tc khi h s (slope) ln

hn 1?

Tr li

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Da trn phng trnh chnh tc ca ng thng: y=mx+b

Cch tip cn n gin: tng x, ri tm y => v thphi tnh ton s thc

Thut ton ny chn vi nhng ng thng c h s gc nhhn hoc bng 1, cn i vi nhng

ng thng c h s gc nghing ln hn 1 th trng ng thng s ri rc v sim trn ng

thng l rt t.

Gii php: s dng phng php i xng.C thi tn trc ta hoc tng theo trc x nu dy

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Cu 7

im khc bit gia a gic li v a gic lm

Tr li

a gic li l a gic m ton ba gic nm v mt pha ca ng thng cha bt k cnh no

a gic lm l a gic nm v hai pha ca t nht mt ng thng cha cnh no .

Cu 8

Hy trnh by thut ton kim tra mt im nm trong hay nm ngoi mt a gic lm

Tr li

Dng nh l Jordan

Tim cn kim tra, ta k mt tia song song vi trc ta (Ox) v ct a gic. Tnh sgiao im catia vi a gic, nu sgiao im l l, tc l im nm trong a gic, nu sgiao im l chn, tc

l im nm ngoi a gic. Thut ton ny khng ng vi trng hp a gic t ct.

Vi trng hp a gic t ct, ta s dng chng ca ng thng, t e=0. Nu ct t tri qua phi

th e tng ln 1, nu ct t phi qua tri th e gim i 1, cc trng hp khc e gi nguyn. Nu e !=0 th

im nm trong a gic.

Cng vi tng nh trn, t e=0. Nu on ct c duy nht mt im dng th cng e thm 1. Nu

e chn th im nm ngoi a gic, e lth im nm trong a gic.

Cc trng hp c bit: ct trng ln cnh v ct trng ln nh.

Cu 9

Hy m t thut ton Clipping Cohen-Sutherland

Tr li

Bi ton ct xn on thng trn mt phng c m tnh sau:

Cho mt on (P1,P2), phi ct n vi mt ca s hnh ch nht v tr von ct (Q1,Q2) hoc tr v

rng nu on thng nm ngoi ca s ct.

Thut ton Cohen-Sutherland

N bt u bng vic m ha chn khu vc c phn chia bi ng thng cha cc cnh ca ca s

bng m 4 bt, m m ny c gi l outcode. Vi mt im P bt k, gi c(p)=x3x2x1x0, vi xi=0 hoc

1.

Cc bit c ngha nh sau

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X0=1 khi v ch khi P nm hon ton bn phi ca ng bin bn phi

X1=1 khi v ch khi P nm hon ton bn trn ca ng bin bn trn

X2=1 khi v ch khi P nm hon ton bn tri ca ng bin bn tri

X3=1 khi v ch khi P nm hon ton bn di ca ng bin bn di

Thut ton c 3 bc:

Bc 1:M ha P1 v P2. t c1=c(P1) v c2=c(P2)

Bc 2: kim tra nu mt on b loi b mt cch d dng s dng ton t OR v AND bit

1. C1 or c2 =0 hoc2. C1 and c2 khc 0

trng hp (a), on thng nm hon ton trong ca s ct v chai u ca on nm trongca s v ca s ct l li. Tr v Q1=P1, Q2=P2

trng hp (b), on thng nm hon ton bn ngoi ca s ct, v hai u ca on thng

nm trn mt na mt phng khng cha ca s ct. Tr von rng

Bc 3: Nu mt on thng m khng b loi b mt cch ddng th ta chia on ra. Sau ,

chng ta quay li bc 1 vi quy trnh nh sau:

1. Trc ht, ta xc nh im u mt P xc nh on s s dng ct.Nu c1=0000, th P1 khng cn phi ct, ta t P l P2 v Q l P1

Nu c1!=0000 th P=P1 v Q=P2

2. ng thng dng ct c xc nh bng bit bn tri nht ca m bng 1 trong c(P). Ta tA l giao im ca PQ vi ng thng dng ct

3. Lp li cc bc t1 n 3 vi on AQThut ton s kt thc bc 2.

Lu , vi thut ton ny vic m ha l rt d dng, ta ch cn so snh ta ca P1 v P2 vi cc

hng s, vi gi thit l cc ng bin ca ca s ct song song vi cc trc ta . Chbc 3mi thc s cn tnh ton.

Cu 10

Thut ton clipping ng thng Cyrus and Beck

Tr li

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Tm cc im vo im ra

Cu 11

M t thut ton Clipping theo tng cnh ca hcn to nn khung nhn (thut ton Sutherland

Hodgman)

Tr li

Ct a gic bng cch dng ln lt cc cnh ca ca s ct a gic.

Ta xt ng thng e cha cnh ca ca s ct v nh u vo P, nh ny ssinh ra cc im mi, c

th l 0,1 hay 2, ty thuc vo v ca P vi e.

C bn trng hp (v hnh)

i vng quanh cc nh ca a gic, kim tra vi cnh ang dng ct ca ca s. Chy thut ton ny

vi a gic mi thu c vi cnh tip theo ca ca s.

u im: khng cn lu tr nhiu, d dng ci t.

Cu 12

Ngi ta lm thno th hin php tnh tin qua ma trn sao cho ma trn ny c th ghp vi ma

trn th hin cc php bin i khc qua php nhn. Hy vit ra ma trn ca php tnh tin trong khng

gian 3 chiu.

Tr li

Dng ta ng nht:

4. Thm mt thnh phn ta na, W, cho mt im P(x,y,W)5. Hai ta ng nht cng biu din mt im nu chng l bi s ca nhau. V d (1,2,3) v

(3,6,9)

6. Phi c t nht mt thnh phn khc 0 => (0,0,0) l khng xc nh7. Nu W khc 0, chia cc ta cn li cho n c ta cc (x/W,y/W,1)8. Nu W=0, im coi nh v cng.

Ma trn bin i xem sch ;))

Cu 13

Thno l im bin mt trong php chiu phi cnh (Vanishing point)

Tr li

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Qua php chiu phi cnh, cc ng thng song song s khng song song vi nhau m hi t ti mt

im, im ny gi l im bin mt.

Cu 14

Thno c gi l php chiu phi cnh 1 im, 2 im, v 3 im?

Tr li

Php chiu phi cnh c phn loi theo slng im bin mt theo trc, cng c ngha l theo s

lng giao im gia cc trc ta v mt phng chiu.

Php chiu phi cnh 1 im, tc l mt phng chiu ch ct 1 trc, hay ch c mt im bin mt theo

trc.

Php chiu phi cnh 2 im, tc l mt phng chiu ch ct 2 trc, hay c hai im bin mt theo trc.

Php chiu phi cnh 3 im, tc l mt phng chiu ch ct 3 trc, hay c ba im bin mt theo trc.

Cu 15

Hy vhnh v a ra c ma trn bin i ca php chiu phi cnh?

Tr li

Hnh v c trong sch

Ma trn bin i

1 0 0 0

0 1 0 0

0 0 1 0

10 0 0

M

d

Cu 16

Thng thng, ngi ta lu trcc a gic trong my tnh nh thno? Ti sao dng triangle strip v

triangle fan li tng c tc tnh ton trong ha my tnh?

Tr li

Khi cc i tng c m hnh s dng a gic, cca gic k nhau c th c chung cnh. m bo

rng khi kt xut khng c cc khong trng c to ra gia cc a gic lin k do qu trnh tnh ton

s thc, cc a gic c cnh chung nn dng cng gi tr ta cho hai u mt ca cc cnh chung.

Chnh v th, ngi ta thng dng cu trc cha cc cnh d liu tham chiu n cc im va tit

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kim b nh, va gii quyt c vn im chung v cnh chung. Ta to ra ba danh sch: danh sch

cc nh, danh sch cc cnh v danh sch a gic b mt.

9. Danh sch cc nh cha cc nh ca cc a gic v ta tng ng10. Danh sch cc cnh tham chiu ti cc nh l u mt ca cnh11. Danh sch a gic b mt tham chiu ti cc cnhThng thng, ngi ta c th b danh sch cc cnh v chlu tr danh sch cc nh v danh sch b

mt.

tng tc trong chng trnh ha my tnh, ng thi tit kim khng gian lu tr, ngi ta

bin cc tam gic c lp thnh mt qut tam gic hay mt chui tam gic. Nh vy, ngoi tam gic u

tin, mi tam gic mi ch cn lu thm 1 nh. ng thi, cc php bin i trn tam gic cng ch cn

thc hin trn mt nh, trtam gic u tin.

Cu 17

To ra cy BSP xy dng th t v ca cc mt cho v dsau, sau a ra th t v cc phn ca

cc mt

Tr li

Thut ton BSP nguyn bn gm 2 bc

12. Bc tin x l s chun danh sch a gic u vo sang dng cu trc cy nh phn gi l cyBSP

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13. Mt thut ton duyt s duyt qua cy BSP v vcc a gic ra bm khung theo th t tsau ra trc.

Ch : mt phng phn tch phi tha mn: khng c a gic no nm na khng gian cha im

nhn li b mt a gic na khng gian kia che khut.

Th t v: 2a,3,5,1b,2b,4,1a.

Mt phng phn cch: 3, 2b, 4, 5.

Cu 18

To ra cy BSP xy dng th t v ca cc mt cho v dsau, sau a ra th t v cc phn ca

cc mt

Tr li

Th t v: 2b, 3, 2a, 1

Cu 19

To ra BSP tree xy dng th t v ca cc mt cho v dsau, sau a ra th t v cc phn ca

cc mt

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Tr li

Hnh ny xu qu, ch lm ;))

Cu 20

i vi cc on cong, th no l lin tc bc 0, 1 v 2 (C0, C1, C2)? Hy trnh by cch hiu ca bn v

cc hm c s (basic function) ca mt ng cong tham s?

Tr li

Lin tc bc 0 tc l hai u ng cong ti nh phi ni vi nhau, hay ni cch khc f(x)=g(x).

Lin tc bc 1 l lin tc bc 0 v tip tuyn ti im phi trng nhau, hay ni cch khc o hm bc

1 ca hai hm phi bng nhau

Lin tc bc 2 l lin tc bc 1 v un ti im l nh nhau, hay ni cch khc o hm bc hai

ca hai hm bng nhau.

Hm c sc sch

Cu 21

Vit ma trn Hermite

Tr li

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2 2 1 1

3 3 2 1

0 0 1 0

1 0 0 0

hM

Cu 22 [b-spline]

Hy m t cch to ra cc ng cong c bn ca ng cong tham s B-spline. Cho mt ng cong B-

spline c 10 im iu khin tP0 n P9, hy xc nh son cong to nn ng cong . V hnh

minh ha.

Tr li

Cu 23

Gi s gi trcng chiu sng ca im A l 1.0, ca im B l 0.8 v ca C l 0.5. Theo m hnh

chiu Gouraud Shading, hy tnh gi trcng sng ca D v E

Tr li

To bng Gouraud l s m phng b mt nhn nhi v m. y l thut ton to bng ni suy cng

hay to bng ni suy mu sc v n tnh cng ca mi nh a gic v sau ni suy ra ton b

a gic theo cc cnh v theo ng qut.

c c gi tr sng ti cc nh a gic, chng ta cn tnh vecto php tuyn. Gi trny thng

c tnh bng cch ly trung bnh vecto php tuyn ca cc mt k vi nh . Lu rng, cch lytrung bnh ny to ra hiu ng trn ca vt th.

Vn xy ra:

14. Khng th hin c tnh li lm15. Khi quay cc a gic, kt qu to bng c th bthay i.

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Kt qu:

31 5,

40 8D EI I

Cu 24

Trong cng thc

os osn

a a p d p sI I k I k c I k c

Hy m t tng thnh phn ca tng trn tng ng vi cc m hnh nh sng no, gii thch tng h

s?

Tr li

aI : cng ca nh sng mi trng

ak : h s phn quang ca nh sng mi trng, l lng nh sng mi trng c phn quang li

pI : cng ca ngun sng ti im P

dk : h s khuch tn, l mt hng s ph thuc vo vt liu

: l gc gia hng nh sng n L v vecto php tuyn N ti P

sk : h s phn chiu

n: cp s phn chiu

: l gc to bi tia phn xR v hng nhn ca mt V

16. Thnh phn th nht l thnh phn nh sng mi trng17. Thnh phn th hai l thnh phn nh sng khuch tn18. Thnh phn th ba l thnh phn nh sng phn chiu

Cu 25

Trong m hnh nh sng trong ho my tnh, c 3 m hnh nh sng phn li t 1 vt: nh sng xung

quanh (ambient), nh sng phn x(specular) v nh sng tn x(diffusion). Hy gii thch 3 m hnh ny.

Tr li

nh sng mi trng hay nh sng xung quanh m t nh sng l mt thit lp nh sng c cng

khng i trong mt cnh vt, l tng ca tt c cc nh sng gin tip trong cnh vt .

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Khi tia sng chiu ti b mt vt th th s b phn x li, thnh phn ny gi l nh sng phn x.

Khi tia sng chiu ti b mt vt th m b phn x li theo mi hng v cng nh sng l bng nhau

theo mi hng, ta gi l hin tng tn x.

Cu 26

M t tng ca phng php Ray tracing trong ho my tnh. Ch ra u v nhc im caphng php ny. Phng php ny thng c dng trong nhng loi ng dng no?Tr li

Ci ny thy ch dy, nn cng ch hc :D

Cu 27

Lnh glutDisplayFunc dng lm g? Lnh glutIdleFunc dng lm g? Lnh glNormal dng lm g?

Trong lnh glVertext3fv, gii thch ngha ca s 3, ch f v ch v?

Tr li

glutDisplayFunc l mt th tc ng k s kin quan trng nht v cn gi u tin. N gi li th tc th

hin nhng ci m chng ta s nhn thy. Bt k khi no GLUT xc nh ni dung ca ca s hin th

li, hm c khai bo bng glutDisplayFunc c thc thi. Do vy, chng ta cn y tt c nhng g cn

v vo trong hm display callback.

glutIdleFunc gi li hm. Hm ny sc thc hin khi khng c bt c s kin no c xy ra.

glNormal dng thit lp mt vecto php tuyn.

glVertex3fx dng v mt nh, 3 l c 3 ta , f l kiu d liu l float, v tc l kiu vector.