10CSL67 CG LAB PROGRAM 5
- 1. Program 5 Program to implement Cohen-Suderland Line Clipping Algorithm. Make provision to specify the input line, window for clipping and view port for displaying the clipped image.
- 2. Cohen-Sutherland Line-Clipping: • It is a very good algorithm for clipping pictures that are larger than the screen. • This algorithm divides a 2D space into 9 parts, of which only the middle part (view port) is visible. • The 9 regions can be uniquely identified using a 4 bit code. This 4 bit code is called outcode.
- 3. Cohen-Sutherland Line-Clipping Outcode 9 8 10 1 0 2 5 4 6 We use the order Left, Right, Bottom, Top for these 4 bits.
- 4. Cohen-Sutherland Line-Clipping Outcodes 1001 1000 1010 0001 0000 0010 0101 0100 0110 We use the order Left, Right, Bottom, Top for these 4 bits.
- 5. Cohen-Sutherland Line-Clipping Region Outcodes 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax Xmin Xmax Ymin Ymax
- 6. Cohen-Sutherland Line-Clipping Region Outcodes Bit Number 1 0 Fourth (LSB) Left of left edge X<Xmin Right of left edge X>Xmin Third Right of right edge X>Xmax Left of right edge X<Xmax Second Below bottom edge Y<Ymin Above bottom edge Y>Ymin First (MSB) Above top edge Y>Ymax Below top edge Y<Ymax 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax
- 7. Line-Clipping Algorithm Assume two endpoints are p0 and p1 1. If code(p0) OR code(p1) is 0000, • the line can be trivially accepted. • the line is drawn. 2. If code(p0) AND code(p1) is NOT 0000, • the line can be trivially rejected. • the line is not drawn at all. 3. Otherwise, compute the intersection points of the line segment and window boundary lines (make sure to check all the boundary lines)
- 8. Line-Clipping 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax L1 L2 L3 Line L1 code1 = 0000, code2 = 0000 Is code1|code2==0? Trivial accept Is code1 & code2!=0? Trivial reject Else clip the line until one code becomes 0 code1 | code2 = 0000 | 0000 = 0000 Trivial accept completely – no need to clip.
- 9. Line-Clipping 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax L1 L2 L5 Line L2 code1 = 0101, code2 = 0100 Is code1|code2==0? Trivial accept Is code1 & code2!=0? Trivial reject Else clip the line until one code becomes 0 code1 | code2 = 0101 | 0100 = 0101 (NOT 0000) : No Trivial accept Code1 & code2 = 0101 & 0100 = 0100 (NOT 0000) : Trivial reject
- 10. Line-Clipping 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax L1 L2 L5 Line L5 code1 = 0000, code2 = 1010 Is code1|code2==0? Trivial accept Is code1 & code2!=0? Trivial reject Else clip the line until one code becomes 0 code1 | code2 = 0000 | 1010 = 1010 (NOT 0000) : No Trivial accept code1 & code2 = 0000 & 1010 = 0000 : No Trivial reject Clip the line
- 11. Line-Clipping Example 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax a c f e d b x0,y0 x1,y1 Intersection computation Line equation y = y0 + m(x - x0) where m = y1 - y0 x1 - x0 x=x min x=x max y=y max y=y min
- 12. Line-Clipping Example 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax a c f e d b x0,y0 x1,y1 Line intersection with the left vertical boundary Assume the intersection is c x = xmin y = y0 +m(xmin – x0) Line ab is clipped w.r.t. x= xmin now ab becomes cb Intersection computation Line equation y = y0 +m(x – x0) where m = y1 – y0 x1 – x0 x=x max y=y max y=y min x=x min x=x max y=y max y=y min
- 13. Line-Clipping Example 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax c f e d b x0,y0 x1,y1 Line intersection with the bottom boundary Assume the intersection is f y = ymin x = (1/m) (ymin – y0) + x0 Line cb is clipped w.r.t. y= ymax Line cb becomes fb Intersection computation Line equation y = y0 +m(x – x0) where m = y1 – y0 x1 – x0 x=x min x=x max y=y max y=y min
- 14. Line-Clipping Example 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax f e x0,y0 Line intersection with the top boundary Assume the intersection is d y = ymax x = 1/m (ymax – y0) + x0 Line fb is clipped w.r.t. y=ymax line fb becomes fd Intersection computation Line equation y = y0 +m(x – x0) where m = y1 – y0 x1 – x0 x=x min x=x max y=y max y=y min f e d x1,y1 b
- 15. Line-Clipping Example 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax f e x0,y0 Line intersection with the right boundary Assume the intersection is e x = xmax y = y0 + m(xmax – x0) Line fd is clipped w.r.t. x=xmax line fd becomes fe Intersection computation Line equation y = y0 +m(x – x0) where m = y1 – y0 x1 – x0 x=x min x=x max y=y max y=y min d
- 16. Line-Clipping Example 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax f e x0,y0 x1,y1 FINALLY……………. ab -----> cb cb -----> fb fb -----> fd fd -----> fe x=x min x=x max y=y max y=y min
- 17. #include<GL/glut.h> #define outcode int double xmin=50,ymin=50, xmax=100,ymax=100; // Window boundaries double xvmin=200,yvmin=200,xvmax=300,yvmax=300; // Viewport boundaries double m; //bit codes for the right, left, top, & bottom const int LEFT = 1; const int RIGHT = 2; const int BOTTOM = 4; const int TOP = 8; //used to compute bit codes of a point outcode ComputeOutCode (double x, double y); //Cohen-Sutherland clipping algorithm clips a line from //P0 = (x0, y0) to P1 = (x1, y1) against a rectangle with //diagonal from (xmin, ymin) to (xmax, ymax).
- 18. void CohenSutherlandLineClipAndDraw(double x0, double y0,double x1, double y1) { //Outcodes for P0, P1, and whatever point lies outside the clip rectangle outcode outcode0, outcode1, outcodeOut; bool accept = false, done = false; outcode0 = ComputeOutCode (x0, y0); outcode1 = ComputeOutCode (x1, y1); m= (y1-y0)/(x1-x0); do { if (!(outcode0 | outcode1)) //logical or is 0 Trivially accept & exit { accept = true; done = true; } else if (outcode0 & outcode1) //logical and is not 0.Trivially reject &exit done = true;
- 19. else { /* failed both tests, so calculate the line segment to clip from an outside point to an intersection with clip edge */ double x, y; outcodeOut = outcode0? outcode0: outcode1; if (outcodeOut& TOP) //point is above the clip rectangle { x = x0 + (1/m) * (ymax – y0); y = ymax; } else if (outcodeOut& BOTTOM) //point is below the clip rectangle { x = x0 + (1/m) * (ymin – y0); y = ymin; }
- 20. else if(outcodeOut& RIGHT) //point is to right of clip rectangle { y = y0 +m*(xmax – x0); x = xmax; } else // Left //point is to the left of clip rectangle { y = y0 +m*(xmin – x0); x = xmin; }
- 21. //Now we move outside point to intersection point to clip //and get ready for next pass. if (outcodeOut == outcode0) { x0 = x; y0 = y; outcode0 = ComputeOutCode (x0, y0); } else { x1 = x; y1 = y; outcode1= ComputeOutCode (x1, y1); } } }while (!done);
- 22. if (accept) { double sx=(xvmax-xvmin)/(xmax-xmin); double sy=(yvmax-yvmin)/(ymax-ymin); double vx0=xvmin+(x0-xmin)*sx; double vy0=yvmin+(y0-ymin)*sy; double vx1=xvmin+(x1-xmin)*sx; double vy1=yvmin+(y1-ymin)*sy; glColor3f(1.0, 0.0, 0.0); // new view port in red color glBegin(GL_LINE_LOOP); glVertex2f(xvmin, yvmin); glVertex2f(xvmax, yvmin); glVertex2f(xvmax, yvmax); glVertex2f(xvmin, yvmax); glEnd(); glColor3f(0.0,0.0,1.0); // clipped line in blue color glBegin(GL_LINES); glVertex2d (vx0, vy0); glVertex2d (vx1, vy1); glEnd(); } }
- 23. outcode ComputeOutCode (double x, double y) { outcode code = 0; if (y >ymax) //above the clip window code |= TOP; else if (y <ymin) //below the clip window code |= BOTTOM; if (x >xmax) //to the right of clip window code |= RIGHT; else if (x <xmin) //to the left of clip window code |= LEFT; return code; } Bit Number 1 0 Fourth (LSB) Left of left edge X<Xmin Right of left edge X>Xmin Third Right of right edge X>Xmax Left of right edge X<Xmax Second Below bottom edge Y<Ymin Above bottom edge Y>Ymin First (MSB) Above top edge Y>Ymax Below top edge Y<Ymax
- 24. void display() { double x0=60,y0=20,x1=80,y1=120; glClear(GL_COLOR_BUFFER_BIT); glColor3f(1.0,0.0,0.0); //draw the line with red color glBegin(GL_LINES); glVertex2d (x0, y0); glVertex2d (x1, y1); glEnd(); glColor3f(0.0, 0.0, 1.0); //draw a blue colored window glBegin(GL_LINE_LOOP); glVertex2f(xmin, ymin); glVertex2f(xmax, ymin); glVertex2f(xmax, ymax); glVertex2f(xmin, ymax); glEnd(); CohenSutherlandLineClipAndDraw(x0,y0,x1,y1); glFlush(); }
- 25. void myinit() { glClearColor(1.0,1.0,1.0,1.0); glColor3f(1.0,0.0,0.0); glMatrixMode(GL_PROJECTION); glLoadIdentity(); gluOrtho2D(0.0,499.0,0.0,499.0); } void main(int argc, char** argv) { glutInit(&argc,argv); glutInitDisplayMode(GLUT_SINGLE|GLUT_RGB); glutInitWindowSize(500,500); glutCreateWindow("Cohen Suderland Line Clipping Algorithm"); glutDisplayFunc(display); myinit(); glutMainLoop(); }
- 27. Cohen-Sutherland Line-Clipping 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax Algorithm Is code1|code2==0? Trivial accept Is code1 & code2!=0? Trivial reject Else clip the line until one code becomes 0
- 28. Intersection computation Line equation y = y0 +m(x – x0) where m = y1 – y0 x1 – x0
- 29. Line intersection with the left boundary x = xmin y = y0 +m(xmin – x0) Line intersection with the right boundary x = xmax y = y0 + m(xmax – x0) Line intersection with the bottom boundary y = ymin x = (1/m) (ymin – y0) + x0 Line intersection with the top boundary y = ymax x = 1/m (ymax – y0) + x0 1001 1000 1010 0001 0000 0010 0101 0100 0110 Xmin Xmax Ymin Ymax
- 30. Bit Number 1 0 Fourth (LSB) Left of left edge X<Xmin Right of left edge X>Xmin Third Right of right edge X>Xmax Left of right edge X<Xmax Second Below bottom edge Y<Ymin Above bottom edge Y>Ymin First (MSB) Above top edge Y>Ymax Below top edge Y<Ymax Checking the Edges