First question, I have tried to calculate the expression, di+1=di+2*Δy−2*Δx(yi+1−yi)
for the four quadrants. Irrespective of the quadrant, the expression was found to be the same, including signs.
Am I right, or, there has been some mistakes in my calculations (hence, I am wrong)?
Second question, if this expression is only applicable for the first octet, how can I apply this to other octets? To me, there is no way to determine which octet I am working on. Coz, the value of m always represent two opposite octets. For example, if 0<m<1
, it represents 1st and 5th octet. Right?
Thirdly, how can we determine the initial/starting value of di?
#include <iostream>
#include "utils.h"
void BresenhamLine(double x1, double y1, double x2, double y2, int color)
{
if(x1>x2 || y1>y2)
{
Swap(x1, x2);
Swap(y1, y2);
}
double x = x1;
double y = y1;
double dx = x2 - x1;
double dy = y2 - y1;
double dt = 2 * (dy - dx);
double ds = 2 * dy;
double d = 2*dy - dx;
PlotPixel(x, y, color);
if(dx>=dy)
{
while(x<=x2)
{
x++;
if(d<0)
{
d = d + ds;
}
else
{
y++;
d = d + dt;
}
PlotPixel(x, y, color);
}
}
else
{
while(y<=y2)
{
y++;
if(d<0)
{
x++;
d = d + dt;
}
else
{
d = d + ds;
}
PlotPixel(x, y, color);
}
}
}
int main()
{
int gm = DETECT;
int gd = DETECT;
initgraph(&gm, &gd, "");
double x1 = 0;
double y1 = 0;
double r = 50;
double x2 = 0;
double y2 = 0;
double signx = 0;
double signy = 0;
for(int theta=0 ; theta<=360 ; theta++)
{
x2 = r * cos(DegreeToRad((double) theta));
y2 = r * sin(DegreeToRad((double) theta));
x1 = 5 * cos(DegreeToRad((double) theta));
y1 = 5 * sin(DegreeToRad((double) theta));
BresenhamLine(x1, y1, x2, y2, YELLOW);
}
getch();
closegraph();
return 0;
}
The lines that go through 2nd and 4th quadrant are not showing up.
How to fix that with some minor changes in my code?
With this input: x1: 100 y1: -100 x2: -100 y2: 100
this logic:
if(x1>x2 || y1>y2)
{
Swap(x1, x2);
Swap(y1, y2);
}
fails.