1. Zhen Jiang West Chester University
Computer Graphics
Zhen Jiang
West Chester University

2. Topics Introduction Display and OpenGL Point (pixel), line, polygon
Curves
2D and 3D
Clipping
Projection and shadow
Animation

3. Introduction Computer Graphics Sample programs 1 2 3 4 5 6 7 8 9 10
and their execution file
Objective
Evaluation
Resource

4. Display and OpenGL
CRT (cathode-ray tube)

5. Display and OpenGL

6. Display and OpenGL

7. Display and OpenGL 3 2 1 1 2 3

8. Display and OpenGL 3 2
RGB 1 1 2 3

9. Display and OpenGL
1280x1024
75 refreshes / second (from top to bottom in left right order)
Memory elements called pixels

10. Display and OpenGL OO programming
Introduced but not required in this class
Application Programmer’s Interfaces (API)
Is shielded from the details of both the hardware and software implementation of the graphics library.
!! Match the conceptual model that we will introduce in this class

11. Point , line, polygon Point Line Polygon Introduction
(0,0) -> (dx,dy)
(0,cy)->(dx,cy+dy)
(cx,cy)->(dx+cx,dy+cy)
Brensenham’s algorithm
Complete Brensenham’s algorithm
Polygon
Vertices and edges
Simple polygon
What’s inside?
fill

12. Point , line, polygon (point)
1023 … 2 1 1 2 3 …
1279

13. Point , line, polygon (point)1
1023/1023 1
1279/1279

14. Point , line, polygon (point)
640
639*1023
/1023
768
767*1279/1279

15. Point , line, polygon (line)

16. Point , line, polygon (line)
Introduction
Integer vertices
From start point to end point
y = m x + c, where 0<m<1

17. Point , line, polygon (line)
Upper_y d2 d1
Lower_y

18. Point , line, polygon (line)
(0,0) -> (dx,dy)
//int dx, dy
int x, Upper_y, Lower_y;
float y, d1, d2;
for ( x = 0 ; x <= dx ; x++) { y=
Upper_y=
Lower_y=
d1=
d2=
glBegin(GL_POINTS);
if(d1<d2)
glVertex2i(x, Lower_y);
else
glVertex2i(x, Upper_y);
glEnd(); } dy y x
float (dy*x) / float(dx);
int(y)+1;
int(y);
y-Lower_y;
Upper_y-y;
(0,0) dx
For example, when x=0;
y=0;
Upper_y=1;
Lower_y=0;
d1=0;
d2=1;
d1<d2
glVertex2i(0,0);

19. Point , line, polygon (line)
//int dx, dy
int x, Upper_y, Lower_y;
float y, d1, d2;
for ( x = 0 ; x <= dx ; x++) {
y=float (dy*x) / float(dx);
Upper_y=int(y)+1;
Lower_y=int(y);
d1=y-Lower_y;
d2=Upper_y-y;
glBegin(GL_POINTS);
if(d1<d2)
glVertex2i(x, Lower_y);
else
glVertex2i(x, Upper_y);
glEnd(); }
Check (0,0) – (7,2)

20. Point , line, polygon (line)
(0,cy)->(dx,cy+dy)
//int dx, dy
int x, Upper_y, Lower_y;
float y, d1, d2;
for ( x = 0 ; x <= dx ; x++) {
y=float (dy*x) / float(dx);
Upper_y=int(y)+1+cy;
Lower_y=int(y)+cy;
d1=y+cy-Lower_y;
d2=Upper_y-(y+cy);
glBegin(GL_POINTS);
if(d1<d2)
glVertex2i(x, Lower_y);
else
glVertex2i(x, Upper_y);
glEnd(); } dy y x
(0,cy) dx

21. Point , line, polygon (line)
//int dx, dy, cy
int x, Upper_y, Lower_y;
float y, d1, d2;
for ( x = 0 ; x <= dx ; x++) {
y=float (dy*x) / float(dx);
Upper_y=int(y)+1+cy;
Lower_y=int(y)+cy;
d1=y+cy-Lower_y;
d2=Upper_y-(y+cy);
glBegin(GL_POINTS);
if(d1<d2)
glVertex2i(x, Lower_y);
else
glVertex2i(x, Upper_y);
glEnd(); }
Check (0,-2) – (7,1)

22. Point , line, polygon (line)
(cx,cy)->(dx+cx,dy+cy)
//int dx, dy, cx, cy
int x, Upper_y, Lower_y;
float y, d1, d2;
for ( x = cx ; x <= dx+cx ; x++) {
y=float (dy*(x-cx)) / float(dx);
Upper_y=int(y)+1+cy;
Lower_y=int(y)+cy;
d1=y+cy-Lower_y;
d2=Upper_y-(y+cy);
glBegin(GL_POINTS);
if(d1<d2)
glVertex2i(x, Lower_y);
else
glVertex2i(x, Upper_y);
glEnd(); } dy y x
(cx,cy) dx

23. Point , line, polygon (line)
//int dx, dy, cx, cy
int x, Upper_y, Lower_y;
float y, d1, d2;
for ( x = cx ; x <= cx+dx ; x++) {
y=float (dy*(x-cx)) / float(dx);
Upper_y=int(y)+1+cy;
Lower_y=int(y)+cy;
d1=y+cy-Lower_y;
d2=Upper_y-(y+cy);
glBegin(GL_POINTS);
if(d1<d2)
glVertex2i(x, Lower_y);
else
glVertex2i(x, Upper_y);
glEnd(); }
Check (3,-1) – (10,1)

24. Point , line, polygon (line)
Brensenham’s Algorithm
Given (x,y), must choose from either (x+1, y+1) or (x+1, y)
(x+1,y+1)
(x,y) d2
y+1 y d1 x
x+1
(x+1,y)

25. Point , line, polygon (line)
Brensenham’s Algorithm (m=dy/dx)
d1-d2 = 2m(x+1)-2y+2c-1
For integer calculations, use
p = 2dy x-2dx y+2dy+2c dx-dx
d1<d2 =>p negative =>y
d1>d2 =>p positive=>y+1

26. Point , line, polygon (line)
//int dx, dy, cx, cy
int x, p,y;
y=cy;
for ( x = cx ; x <= cx+dx ; x++) {
p=2*dy*x -2*dx*y+2*dy+2*c*dx-dx;
if(p>=0) y++;
glBegin(GL_POINTS);
glVertex2i(x, y);
glEnd(); }
Check (3,-1) – (10,1)

27. Point , line, polygon (line)
Complete Algorithm for (x1, y1) to (x2, y2)?
M<1
M>1
M=0
M=

28. Point , line, polygon glBegin(GL_POINTS); If (x1==x2) {
if(y2>y1) {cy=y1; dy=y2-y1;}
else {cy=y2; dy=y1-y2;}
for (i=0;i<=dy;i++)
glVertex2i(x1, cy+i); }
Else if (y1==y2){
// ? Leave for your exercises
Else {
dx=x2-x1, dy=y2-y1;
m=dy/dx;
if(m>-1 && m < 0){
if(y2>y1) {cy=y1; cx=x1;}
else {cy=y2;cx=x2;dx=-dx; dy=-dy;}
c=y2-x2(y1-y2)/(x1-x2);
y=cy;
for ( x = cx ; x >= cx+dx ; x--) {
p=2*dy*x -2*dx*y-2*dy+2*c*dx-dx;
if(p>=0) y++;
glVertex2i(x, y);
if(m>0 && m < 1){
for ( x = cx ; x <= cx+dx ; x++) {
p=2*dy*x -2*dx*y+2*dy+2*c*dx-dx;
if(m>1){
if(x2>x1) {cx=x1; cy=y1;}
else {cx=x2;cy=y2;dx=-dx; dy=-dy;}
c=y2-x2(y1-y2)/(x1-x2);
x=cx;
for ( y = cy ; y >= cy+dy ; y++) {
p=2*dx*y -2*dy*x+2*dx-2*c*dx-dy;
if(p>=0) x++;
glVertex2i(x, y); }
if(m<-1){
for ( y = cy ; y <= cy+dy ; y--) {
p=2*dx*y -2*dy*x-2*dx-2*c*dx-dy;
glEnd();
Point , line, polygon

29. Point , line, polygon (polygon)
Vertices and edges
Simple polygon
No zero-winding?
No exterior?
Parity?
What’s inside?
Given a pixel, which polygon does it lie in?
Given a pixel, on edge?
Given a pixel, on vertex?

30. 2 intersections ->inside or not?
1 intersection ->inside or not?

31. Even number of intersections ->inside or not?
Odd number of intersections ->inside or not?

32. Draw a line start from that pixel (point), even number of intersections -> outside; odd number of intersections -> inside.
0 intersection ? (need another line?)

33. Overall Algorithm: Line y=py, start from the point (px,py);
Has segment (x1,y1) (x2,y2) an intersection with line y=py?
For each edge in a polygon, see if it has an intersection with y=py.
For each polygon, see if it has odd number of intersections with y=py.
For all polygons in the area, see how many polygons contain this point (px,py).

34. // detect how many polygons in p[number_of_polygons] contain the pixel point.
For (i=0;i<number_of_polygons;i++){
p[ i ]=get_polygon (i);
count=0;
for(j=0;j<p[i].size_of_edge;j++){
start=get_xy(p[ i ][ j ]); // (x1,y1)
end=get_xy(p[ i ][ (j+1) % p[ i ].size_of_edge) // (x2,y2)
count+=how_many_intersections (start, end, point); }
if(count%2) //
it’s inside;
else
it’s outside;

35. Has segment (x1,y1) (x2,y2) an intersection with line y=py?
if(x1==x2) {
if ((y1-py)*(y2-py)<0)
it has one intersection at (x1,py). }
if (y1==y2) {
if (py==y1)
it has two intersections (x1,py) and (x2,py); one for joining the edge and the other for leaving from this edge.
X=(py-y1)(x2-x1)/(y2-y1)+x1
if ((x2-x)*(x1-x)<0)
it has an intersection at (x,py). or or or

36. Question?

37. Has segment (x1,y1) (x2,y2) an intersection with line y=py?
if(x1==x2) {
if ((y1-py)*(y2-py)<=0 && y2!=py) => 1
it has one intersection at (x1,py). }
if (y1==y2) {
if (py==y1) => 1 (even) or 0 (odd)
it has two intersections (x1,py) and (x2,py). (x2,py) will be counted on the edge leaving from y=py. (x1,py) (where the edges join y=py) is counted to make even when joining edge and leaving edge are on the same side of y=py; otherwise, 0 for odd intersection.
X=(py-y1)(x2-x1)/(y2-y1)+x1
if ((x2-x)*(x1-x)<=0 && x2!=x) =>1
it has an intersection at (x,py).
(x2,y2)
(x1,y1)
(x1,y1)
(x2,y2)
(x2,y2)
(x1,y1)

38. Point , line, polygon (polygon)
Fill (interior)
Sweep fill
Flood fill
Boundary fill
Pattern fill (openGL)

39. Point , line, polygon (polygon)
Sweep fill (interior)
Xmin, xmax
Fill the bottom horizontal span of pixels; move up and keep filling
Keep edge information for any change on xmin and xmax
AEL (active edge list) (x, 1/m, ymax)

40. Overall Algorithm: This algorithm maintains one line based on AEL structure
Get all the structures at the bottom line (filling procedure on this bottom line will be skipped).
Get the structure for the next line (add 1/m easily to x for each old AEL structure, add new one if any, delete the expired one). If no AEL unit, stop!
Find filling segments with xmin and xmax (sorting)
Filling all segments on the current line, the endpoints of each segment are excluded (no painting)
Go to step 2.

41. 3 2 1 1 3 6 3

42. 1 3 6 3 * 1 1 3 6 3 * 2 1 3 6 3 3 1 3 6 3 3
End!

43. 4 3 2 4 4 1 1 2 6 4

44. 1 2 6 4 * 1 1 2 6 4 2 1 2 6 4 4 4 * 2 4 4 6 4 * 3 4 4 6 4 4 4 4 6 4 4
End!

45. 6 5 4 3 3 -1 6 5 1 6 2 1 1 3 8 -1 3

46. 1 3 8 -1 3 * 1 1 1 3 7 -1 3 * 2 2 1 3 6 -1 3 3 3 1 3 5 -1 3 3 -1 6 5 1 6 * 3 3 -1 6 5 1 6 * 4 2 -1 6 6 1 6 * 5 1 -1 6 7 1 6 6 -1 6 8 1 6 6
End!

47. Why? 5 4 3 3 4 6 7 2 1 7 1 4

48. 7 1 4 1 7 1 1 4 * 2 7 2 1 4 * 3 7 3 1 4 3 4 6 7 * 4 7 6 7 * 5 7 6 7 * 6 7 6 7 7
End!

49. Quiz preparation
How about a sweep fill for the interior and the edges?

50. Point , line, polygon (polygon)
Flood fill
4 connected fill
8 connected fill
Spans fill ? 1 2

51. Point , line, polygon (polygon)
Boundary fill
All neighbors
Pattern fill (openGL)