Program to create a cylinder and Parallelepiped by extruding Circle and Quadrilateral respectively. Allow the user to specify the circle and quadrilateral.

1. PROGRAM 6
Program to create a cylinder and Parallelepiped
by extruding Circle and Quadrilateral respectively.
Allow the user to specify the circle and
quadrilateral.

2. Output :

3. Midpoint Circle Algorithm
In the mid-point circle algorithm, we use eight-way
symmetry.
Calculate the points for the top right eighth of a
circle, and then use symmetry to get the rest of the
points.

4. Eight-Way Symmetry
(1, 3)
(3, 1)
(3, -1)
(1, -3)(-1, -3)
(-3, -1)
(-3, 1)
(-1, 3)
2
R
Assume x=1, y=3
(x, y)(-x, y)
(-y, x)
(-y,-x)
(y, x)
(y, -x)
(x, -y)(-x, -y)

5. The equation of the circle:
The equation evaluates as follows:
If we choose a point inside a circle, x2+y2<r2
If we choose a point outside a circle, x2+y2>r2
If we choose a point on the circle, x2+y2=r2
222
),( ryxyxfcirc
,0
,0
,0
),( yxfcirc
boundarycircletheinsideis),(if yx
boundarycircleon theis),(if yx
boundarycircletheoutsideis),(if yx
Midpoint Circle Algorithm

6. (xk+1, yk)
(xk+1, yk-1)
(xk, yk)
E
SE
(xk+1, yk- ½ )
Assuming we have just plotted the
pixel at (xk, yk) so we need to choose
between (xk+1,yk) and (xk+1,yk-1)
Our decision variable can be defined
as:
If pk < 0, the pixel at yk is closer to the
circle.
Otherwise yk-1 is closer.
222
)
2
1()1(
)
2
1,1(
ryx
yxfp
kk
kkcirck
Midpoint Circle Algorithm

7. (xk+1, yk)
(xk+1, yk-1)
(xk, yk)
E
SE
(xk+1, yk- ½ )
Let pk=d
dold=F(M)
dold=F(Xk+1,Yk- ½ )
If d<0, choose E
x=x+1 which gives dnew
dnew=(Xk+1)+1, (Yk- ½ )
( d)E = dnew- dold
F(Xk+2, Yk- ½ ) – F(Xk+1, Yk- ½ )
( d)E = 2Xk+3
Midpoint Circle Algorithm

8. (xk+1, yk)
(xk+1, yk-1)
(xk, yk)
E
SE
(xk+1, yk- ½ )
Let pk=d
dold=F(M)
dold=F(Xk+1,Yk- ½ )
If d>=0, choose SE
x=x+1
y=y-1 which gives dnew
dnew=(Xk+1)+1, (Yk- ½)-1
( d)SE = dnew-dold
F(Xk+2, Yk- 3/2) – F(Xk+1, Yk- ½)
( d)SE = 2Xk-2Yk+5
Midpoint Circle Algorithm

9. 222
)
2
1()1(
)
2
1,1(
ryx
yxfp
kk
kkcirck
For the boundary condition, x=0, y=r
Substitute in the equation
p0=d
d=(0+1)2 + (r- ½ )2 – r2
d = 5/4 – r 1-r
For integer values of pixel coordinates, we
can approximate p0= d =1-r

10. #include<GL/glut.h>
void draw_pixel(GLint cx, GLint cy)
{
glColor3f(1.0,0.0,0.0);
glBegin(GL_POINTS);
glVertex2i(cx,cy);
glEnd();
}

11. void plotpixels(GLint h, GLint k, GLint x, GLint y)
{
draw_pixel(x+h,y+k);
draw_pixel(-x+h,y+k);
draw_pixel(x+h,-y+k);
draw_pixel(-x+h,-y+k);
draw_pixel(y+h,x+k);
draw_pixel(-y+h,x+k);
draw_pixel(y+h,-x+k);
draw_pixel(-y+h,-x+k);
}
x and y of the
window
x and y of the
circle

12. // Midpoint Circle Drawing Algorithm
void Circle_draw(GLint h, GLint k, GLint r)
{
GLint d =1-r, x=0, y=r;
while(y > x)
{
plotpixels(h,k,x,y);
if(d < 0) // choose E, ( d)E = 2Xk+3
d+=2*x+3;
else // choose SE, ( d)SE = 2Xk-2Yk+5
{
d+=2*(x-y)+5;
--y;
}
++x;
}
plotpixels(h,k,x,y);
}

13. void Cylinder_draw()
{
GLint xc=100, yc=100, r=50, i,n=50;
for(i=0;i<n;i+=3)
Circle_draw(xc,yc+i,r);
}
void parallelepiped(int x1,int x2,int y1,int y2)
{
glColor3f(0.0, 0.0, 1.0);
glBegin(GL_LINE_LOOP);
glVertex2i(x1,y1);
glVertex2i(x2,y1);
glVertex2i(x2,y2);
glVertex2i(x1,y2);
glEnd();
}

14. void parallelepiped_draw()
{
int x1=200,x2=300,y1=100,y2=175, i, n=40;
for(i=0;i<n;i+=2)
parallelepiped(x1+i,x2+i,y1+i,y2+i);
}
void init(void)
{
glClearColor(1.0,1.0,1.0,0.0);
glMatrixMode(GL_PROJECTION);
gluOrtho2D(0.0,400.0,0.0,300.0);
}

15. void display(void)
{
glClear(GL_COLOR_BUFFER_BIT);
glColor3f(1.0,0.0,0.0);
Cylinder_draw();
parallelepiped_draw();
glFlush();
}

16. void main(int argc, char **argv)
{
glutInit(&argc,argv);
glutInitDisplayMode(GLUT_SINGLE | GLUT_RGB);
glutInitWindowPosition(50,50);
glutInitWindowSize(400,300);
glutCreateWindow("Cylinder,parallelePiped Disp by Extruding
Circle &Quadrilaterl ");
init();
glutDisplayFunc(display);
glutMainLoop();
}