Computer Graphics

Computer Graphics Zhen Jiang West Chester University

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

Introduction • Computer Graphics • Sample programs 12345678910 and their execution file • Objective • Evaluation • Resource

Display and OpenGL • CRT (cathode-ray tube)

Display and OpenGL

Display and OpenGL 3 2 1 0 0 1 2 3

Display and OpenGL 3 2 RGB 1 0 0 1 2 3

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

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

Point , line, polygon • Point • Line • 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

Point , line, polygon (point) 1023 … 2 1 0 … 0 1 2 3 1279

Point , line, polygon (point) 1 1023/1023 0 1 0 1279/1279

Point , line, polygon (point) 640 639*1023 /1023 0 768 767*1279/1279 0

Point , line, polygon (line)

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

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

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);

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)

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

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)

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

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)

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)

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

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)

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

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){ 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); } } } glEnd(); 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){ 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); } }

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?

2 intersections ->inside or not? Inside? 1 intersection ->inside or not?

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

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

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).

// 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; }

or • 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

Question?

(x2,y2) (x1,y1) • 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). (x1,y1) (x2,y2) (x2,y2) (x1,y1)

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

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) Point , line, polygon (polygon)

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.

3 2 1 0 1 0 3 6 0 3

0 1 0 3 6 0 3 * 1 1 0 3 6 0 3 * 2 1 0 3 6 0 3 3 1 0 3 6 0 3 End! 3

4 3 2 4 0 4 1 0 1 0 2 6 0 4

0 1 0 2 6 0 4 * 1 1 0 2 6 0 4 2 1 0 2 6 0 4 4 0 4 * 2 4 0 4 6 0 4 * 3 4 0 4 6 0 4 4 4 0 4 6 0 4 End! 4

6 5 4 3 3 -1 6 5 1 6 2 1 0 0 1 3 8 -1 3

0 0 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 0 -1 6 8 1 6 End! 6

Why? 5 4 3 3 0 4 6 0 7 2 1 0 0 0 7 0 1 4

0 0 0 7 0 1 4 1 0 0 7 1 1 4 * 2 0 0 7 2 1 4 * 3 0 0 7 3 1 4 3 0 4 6 0 7 * 4 0 0 7 6 0 7 * 5 0 0 7 6 0 7 * 6 0 0 7 6 0 7 End! 7

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

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

Computer Graphics