1-1. Application of CG.
(1) Computer-Aided Design
(2) Presentation Graphics
(3) Computer Art
(4) Entertainment
(5) Education and Training
(6) Visualization
(7) Image Processing
(8) Graphical User Interfaces
1-2. CG and IP
Although methods used in computer graphics and image processing overlap, the two areas are
concerned with fundamentally different operations. In computer graphics, a computer is used to
create a picture. Image processing, on the other hand, applies techniques to modify or interpret
existing pictures, such as photographs and TV scans.
2-1. CRT 组成及工作原理。
2-2. Input Devices and Output Devices.
Input Devices: keyboard、mouse、trackball and spaceball、joysticks、data glove、touch panel、
image scanner、voice system、light pen、digitizer
Output Devices: screen、printer、plotter.
3-1. An example: Draw a line from point (0, 0) to point (4, 4.8).
1 (0, 0) → (0, 0)
Solution: (1)○
(8, 4.8) → (8, 5)
2 m = (5-0) / (8-0) = 0.625, b = 0
○
3 ∵m<1 ∴ xi+1 = xi+1
○
yi+1 = mxi+1+b = 0.625 xi+1
(2) Data Table
i
1
2
3
4
5
6
7
8
xi
1
2
3
4
5
6
7
8
yi
0.625
1.25
1.875
2.5
3.125
3.75
4.375
5
[yi]
1
1
2
3
3
4
4
5
(xi, yi)
(1, 1)
(2, 1)
(3, 2)
(4, 3)
(5, 3)
(6, 4)
(7, 4)
(8, 5)
(3) Plot the Figure
3-2. An example: Draw a line from point (0, 0) to point (4, 7) with the DDA algorithm.
Solution: (1) Write the DDA algorithm.
begin
if abs (x2′-x1′)≥abs(y2′-y1′)
then length = abs (x2′-x1′)
else length = abs(y2′-y1′)
endif
△x = (x2′-x1′) / length
△y = (y2′-y1′) / length
k = 1;
x = x1′;
y = y1′;
while (k ≤ length + 1)
putpixel(x, y)
k = k + 1;
x = x + △x;
y = y + △y;
endwhile
end
(2) Operation and data table
(0, 0) → (0, 0)
(4, 7) → (4, 7)
abs(4-0) = 4
abs(7-0) = 7
putpixel
(x, y)
Δx=4/7
length = 7
Δy =7/7=1.0
(0, 0)
(1, 1)
(1, 2)
(2, 3)
(2, 4)
(3, 5)
(3, 6)
(4, 7)
k
1
2
3
4
5
6
7
8
9
x
0
4/7
8/7
12/7
16/7
20/7
24/7
4
32/7
y
0
1
2
3
4
5
6
7
8
Outside
while loop
Inside while loop
Outside
while loop
(3) Plot the figure
3-3. An example: Draw a line from the Point P1(0, 0) to Point P2(10, 6) with the Bresenham
Algorithm.
Solution: (1) Write the Bresenham algorithm.
int x = x’1, y = y’1；
int dx = x’2 - x’1, dy = y’2 - y’1, dT = 2(dy - dx), dS = 2dy;
int d = 2dy - dx;
setPixel(x, y);
while (x < x’2) {
x++;
if (d < 0)
d = d + dS;
else {
y++;
d = d + dT;
}
setPixel(x, y);
}
(2) Data Table.
x=0, y=0, dx=10-0=10, dy=6-0=6, dT=2*(6-10)=-8, dS=2*6=12, d=2*6-10=2, setPixel(0,0)
while
(x≥x’2)
while (x< x’2)
x
1
2
3
y
1
d
-6
6
-2
(x, y)
(1, 2)
(2, 1)
(3, 2)
4
5
6
3
4
10
2
-6
6
-2
10
2
(4, 2)
(5, 3)
(6, 4)
(7, 4)
(8, 5)
(9, 5)
(10, 6)
2
7
8
9
5
10
6
(3) Plot the figure.
3-4. An example: Indicate which raster locations would be chosen by Bresenham’s algorithm
when scan-converting a circle centered at the origin with the radius of 8 pixels (from 90° to 45°).
Solution: (1) Write the algorithm.
int x = 0, y = r, d = 3 - 2r;
while (x <= y) {
setPixel(x, y);
if (d < 0)
d = d + 4x + 6;
else {
d = d + 4(x - y) +10;
y--;
}
x++;
}
(2) Data Table.
x = 0, y = 8, d = 3-2*8 = -13
while (x>y)
while (x≤y)
(x, y)
(0, 8)
(1, 8)
(2, 8)
(3, 7)
(4, 7)
(5, 6)
d
-7
3
-11
7
5
11
6
5
5
6
y
x
7
1
2
3
4
(3) Plot the figure.
3-5. An example: (1) Write 4_connected boundary algorithm and the order in which pixel are
filled in the following Figs. The number “1” represents a seed.
(2) Write 8_connected boundary algorithm
Solution: (1) Write the algorithm.
void boundaryFill4 (int x, int y, int fill, int boundary)
{
int current;
current = getpixel (x, y);
if ((current != boundary) && (current != fill))
{
setcolor (fill);
setpixel (x, y);
boundaryFill4 (x+1, y, fill, boundary);
boundaryFill4 (x-1, y, fill, boundary);
boundaryFill4 (x, y+l, fill, boundary);
boundaryFill4 (x, y-1, fill, boundary);
}
}
(2) Plot the result.
4-1. Give the coordinates of three points P0(4,4), P1(24,4), P2(36,3) respectively and their slopes
D0(8.832, 5.547), D1(8.832, -5.547), D2(8.832, 5.547). Generate a Hermit spline, compute the
coordinates of points at the spline when u=0, 0.25, 0.5, 0.75 and 1, and plot the curve.
Solution: A Hermit spline can be described as
P(u)= pkH0(u) + pk+1H1(u) + DpkH2(u) + Dpk+1H3(u)
where H0(u) =2u3 – 3u2 + 1
H1(u)=-2u3 + 3u
H2(u) =u3 – 2u2 + u
H3(u)= u3 – u2
(1) For the line P0, P1:
P0=[4 4] D0(8.832, 5.547)
P1=[24 4] D1(8.832, -5.547)
At u=0 and u=1 we have
P(0)= P0 =[4 4]
P(1)= P1 =[24 4]
At u=0.5 H0(0.5)=2*0.53-3*0.52+1=0.5
H1(0.5)=0.5
H2(0.5)=0.125
H3(0.5)= -0.125
Therefore we obtain:
P(0.5)= H0(0.5) P0+ H1(0.5) P1+ H2(0.5) D0+ H3(0.5) D1
=0.5[4 4]+0.5[24 4]+0.125[8.832 5.547]-0.125[8.832 -5.547]
=[14 5.386]
Similarly
At u=0.25 P(0.25)=[7.953 5.04]
At u=0.75 P(0.75)=[20.04 5.04]
(2) For the line P1P2:
P1=[24 4] D1 (8.832, -5.547)
P2=[36 3] D2 (8.832, 5.547)
At u=0 and u=1 we have
P(0)= P1 =[24 4]
P(1)= P2 =[36 3]
At u=0.5 H0(0.5) = 0.5, H1(0.5)=0.5, H2(0.5)=0.125, H3(0.5)= -0.125
P(0.5)=[30 2.11]
At u=0.25 P(0.25)=[26.7 2.8]
At u=0.75 P(0.75)=[33.3 2.12]
Y
5
4
P1
P0
P2
3
2
1
0
4
8
12 16 20 24 28 32 36
X
4-2. Construct a cubic Bezier curve with four points P1(0, 0, 0), P2(1, 1, 1), P3 (2, -1, -1), P4(3, 0,
0), and compute the values of coordinates at u=0, 1/3, 2/3, 1.
Solution: A Cubic Bezier Curve can be described as
P(u)=B0,3(u)P1+B1,3(u)P2+B2,3(u)P3+B3,3(u)P4
where: B0,3(u)=(1-u)3
B1,3(u)=3u(1-u)2
B2,3(u)=3u2(1-u)
B3,3(u)=u3
At u=0 P(0)=P1=[0 0 0]
At u=1 P(1)=P4=[3 0 0]
At u=1/3 B0,3(1/3)=(1-1/3)3=8/27
B1,3(1/3)=3*(1/3)*(1-1/3)2=4/9
B2,3(1/3)=3*(1/3)2*(1-1/3)=2/9
B3,3(1/3)=(1/3)3=1/27
Thus P(1/3)=8/27[0 0 0]+4/9[1 1 1]+2/9[2 -1 -1]+1/27[3 0 0]=[1 2/9 2/9]
At u=2/3 B0,3(2/3)=(1-2/3)3=1/27
B1,3(2/3)=3*(2/3)*(1-2/3)2=2/9
B2,3(2/3)=3*(2/3)2*(1-2/3)=4/9
B3,3(2/3)=(2/3)3=8/27
Thus P(2/3)=1/27[0 0 0]+2/9[1 1 1]+4/9[2 -1 -1]+8/27[3 0 0]=[2 -2/9 -2/9]
4-3. Construct a cubic B-spline curves with 5 points P0(1, 0), P1(0, 2), P2(2, 3), P3 (5, 2.5), P4(6, 1),
and compute the values of coordinates at u=0, 1/2, 1, draw the curves.
Solution: A Cubic B-spline Curve can be described as
P(u)=F0,3(u)P0+F1,3(u)P1+F2,3(u)P2+F3,3(u)P3
where: F0,3(u)=1/6(1-u)3
F1,3(u)= 1/6 (3u3 – 6u2 + 4)
F2,3(u)= 1/6 (– 3u3 + 3u2 + 3u +1)
F3,3(u)= 1/6 u3
(1) For P0~P3:
At u=0 F0,3(0)=1/6 F1,3(0)= 4/6 F2,3(0)= 1/6 F3,3(0)=0
P(0)=[1/6 4/6 1/6
1

0] 0
2

5


 =[0.5


2.5
0
2
3
1.83]
At u=1 F0,3(1)=0 F1,3(0)= 1/6 F2,3(1)= 4/6 F3,3(1)=1/6
1

1/6] 0
2

5
P(1) [0 1/6 4/6


 =[2.17


2.5
0
2
3
2.75]
At u=0.5 F0,3(0.5)=0.021 F1,3(0.5)= 0.479 F2,3(0.5)= 0.479 F3,3(0.5)=0.021
1

P(0.5)=[0.021 0.479 0.479 0.021] 0
2

5


 =[1.08


2.5
0
2
3
2.45]
(2) For P1~P4:
P(0)=[2.17 2.75]
P(1)=[4.67 2.33]
P(0.5)=[3.479 2.698]
P2
3
2
P3
P1
1
0
P4
P0
1
2
3
4
5
6
5-1. The three vertices a triangle are expressed by A(0, 20), B(40, 40)), and C(0, 60). We translate
the triangle x direction and y direction respectively 30, 20. Compute and draw the new position of
the triangle.
Solution:
 xA  t x   x A  30   0  30 
 y    t    y   20  20  40
 A  y  A      
 xB  t x   x B  30  40 70
 y    t    y   20  40  60
 B  y  B      
 xC  t x   xC  30   0  30
 y    t    y   20  60  80 
 C  y  C      
5-2. A triangle shown in the following figure is rotated by 30° about the origin. Compute and draw
the new triangle.
Solution:


 x1  cos 30  sin 30  0 0
P1     
  

cos 30   0 0
 y1  sin 30


 x2  cos 30  sin 30  60 52

P2     
  

cos 30    0  30
 y 2  sin 30
 x3  cos 30   sin 30   30  4
P3     
 6    

cos 30   0  67 
 y3  sin 30
Y
P3(30,60)
60
30
P2(60,0)
0 P1(0,0)
30
60
X
5-3. Show that the compositiom of two rotations is additive by concatenating the matrix
representations for R(θ1) and R(θ2) to obtain R(θ1)*R(θ2) = R(θ1+θ2).
Solution:
cos1  sin 1 0
cos 2  sin  2 0


 R (1 )  sin 1 cos1 0  R( 2 )  sin  2 cos 2 0 
 0
 0
0
1 
0
1 
cos1 cos 2  sin 1 sin  2  cos1 sin  2  sin 1 cos 2 0
R (1 )  R ( 2 )  sin 1 cos 2  cos1 sin  2  sin 1 sin  2  cos1 cos 2 0 

0
0
1 
cos(1   2 )  sin( 1   2 ) 0 
 sin( 1   2 ) cos(1   2 ) 0

0
0
1 
cos(1   2 )  sin( 1   2 ) 0 
R (1   2 )  sin( 1   2 ) cos(1   2 ) 0

0
0
1 
 R (1 )  R ( 2 )  R (1   2 )
6-1. Let R be the rectangular window whose lower left-hand corner is at L(-1, -1) and upper
right-hand corner is at R(1, 1). The points P1, P2, P5 are represented with (-3/2, 1/6), (1/2, 3/2) and
(3/2, 3/4). Clip the line segment P1P2, P2P5.
Solution: (1) The line segment P1P2: P1: 0001, P2: 1000, 0001and1000=0000
So the line segment P1P2 is a candidate for clipping.
To find the intersection points with the boundaries of window.
The line slop: m=(y2-y1)/(x2-x1)=(3/2-1/6)/(1/2+3/2)=2/3
The line equation: y=1/6+2/3*(x+3/2)
The left: x=-1, y=1/6+2/3(-1+3/2)=1/2: saved
The right x=1, y=1/6+2/3*(1+3/2)=11/6>ymax: discard
The top: y=1, 1=1/6+2/3*(x+3/2)----------x=-1/4, y=1: saved
The bottom: y=-1, -1=1/6+2/3*(x+3/2)-----------x=-13/4<xmin, y=-1: discard
So the visible portion of line P1P2 is from (-1, 1/2) to (-1/4, 1).
(2) The line segment P2P5: P5: 0010, P2: 1000, 0010and1000=0000
So the line segment P2P5 is a candidate for clipping.
To find the intersection points with the boundaries of window.
The line slop: m=(y2-y1)/(x2-x1)=(3/4-3/2)/(3/2-1/2)=-3/4
The line equation: y=3/2-3/4*(x-1/2)
The left: x=-1, y=2.625>ymax: discard
The right x=1, y=1.125>ymax: discard
The top: y=1, x=1.167>xmax: discard
The bottom: y=-1, x=3.833>xmax: discard
So the line P2P5 should be deleted(invisible).
Y
P2
R(1,1)
P5
P1
X
L(-1,-1)
6-2. Use the Liang-barsky algorithm to clip the Line AB in the following figure.
Solution：△x=x2-x1=4, △y=y2-y1=1
Calculate pk, qk, rk:
p1=-△x, q1=-1, r1=1/4
p2=△x=4, q2=3, r2=3/4
p3=-△y=-1, q3=0, r3=0
p4=△y=1, q4=2, r4=2
Since p1<0, p3<0: u1=max{0, r1, r3}=max{0, 1/4, 0}=1/4
p2>0, p4>0: u2=min{1, r2, r4}=min{1, 3/4, 2}=3/4
To calculate the visible portion of line
u1=1/4 x=x1+△x*u1=1+4*(1/4)=2
y=y1+△y*u1=2+1*(1/4)=9/4
u2=3/4 x=x1+△x*u2=1+4*(3/4)=4
y=y1+△y*u2=2+1*(3/4)=11/4
So the line segment from (2, 9/4) to (4, 11/4) is visible.
X
(4,4)
4
3
2
B(5,3)
A(0,2)
(2,2)
1
0
1
2
3
4
5
X