Example: (7,9) (12,0) Example 2: Point1 V:(7,9 ) C:( 0,255,0) Point2 V:(12,0) C:(0,255,0) (0,0) (18,0) (0,9) What are the problems with this method? Slope>1

Revised Slope Intercept DrawLine (Point point1, Point point2){ m=(point2.y-point1.y) / (point2.x-point2.x); b=point1.y + (-point1.x) * m; if (m>1){ for i = point1.x to < point2.x { SetPixel(i, round(i*m+b)); } else{ for i = point1.y to point2.y { SetPixel(round(i-b/m), i); } Which one should we use if m=1? What is the cost per pixel?

Optimization (DDA Algorithm) Since we increment y by the same amount, we can also inline rounding: New cost: one floating point addition, one integer addition, one cast. DrawLine (Point point1, Point point2){ m=(point2.y-point1.y) / (point2.x-point2.x); j=point1.y + (-point1.x) * m + 0.5; for i=point1.x to point2.x SetPixel(i, (int)j+=m)); }

Optimizations In general processing unit: –Addition and Subtraction are faster than Multiplication which is faster than Division –Integer calculations are faster than Floating and double calculations –Logical operation is lighter the math operations –Bitwise operation is the lightest

Bresenham Line Drawing Algorithm Let us consider a line equation y = mx + b could be written as F (x, y) =ax + by + c F(x,y) is zero on the line, positive above the, and negative below the line M E NE if ( d > 0) then we choose NE If ( d < 0 ) then we choose E If ( d == 0) we choose any of them

Bresenham Line Drawing Algorithm What happens to next grid point after x p +2, that depends on whether we select E or NE. If E was selected then M E NE If NE was selected then Since the first point (x 0,y 0 ) We can directly calculate d to choose NE or E

public void DrawLine(Point point1, Point point2) { int dy = point2.y - point1.y; int dx = point2.x - point1.x; dy <<= 1; // dy is now 2*dy dx <<= 1; // dx is now 2*dx int fraction = dy - (dx >> 1); // same as 2*dy - dx j=point1.y; for (i=point1.x; i<point2.x; i++) { if (fraction >= 0) { j++; fraction -= dx; // same as fraction -= 2*dx } fraction += dy; // same as fraction -= 2*dy SetPixel(i, j); }