Line Drawing by Algorithm
Line Drawing Algorithms
• Line drawn as pixels
• Graphics system
– Projects the endpoints to their pixel locations in the
frame buffer (screen coordinates as integers)
– Finds a path of pixels between the two
– Loads the color
– Plots the line on the monitor from frame buffer (video
controller)
– Rounding causes all lines except horizontal or vertical
to be displayed as jigsaw appearance (low resolution)
– Improvement: high resolution, or adjust pixel intensities
on the line path.
Line Drawing Algorithms
• Line equation
– Slope-intercept form
y=m.x+b
slope m
Y-intercept b
y
end
– Slope
m=
– Y-intercept
- y0 δ y
=
==
x end - x 0 δ x
b = y0 - mx0
Line Drawing Algorithms
• DDA (Digital Differential Analyzer)
– Scan conversion line algorithm
– Line sampled at regular intervals of x, then
corresponding y is calculated
– From left to right
if m ≤ 1 ,
if m > 1 .0 ,
=
y k +1 = y k + m
x k +1 = x k
1
+
m
(δ x = 1)
(δ y = 1)
Line Drawing Algorithms
void lineDDA (int x0, int y0, int xEnd, int yEnd)
{
int dx = xEnd - x0, dy = yEnd - y0, steps, k;
float xIncrement, yIncrement, x = x0, y = y0;
if (fabs (dx) > fabs (dy))
steps = fabs (dx);
else
steps = fabs (dy);
xIncrement = float (dx) / float (steps);
yIncrement = float (dy) / float (steps);
setPixel (round (x), round (y));
for (k = 0; k < steps; k++) {
x += xIncrement;
y += yIncrement;
setPixel (round (x), round (y));
}
}
Line Drawing Algorithms
• Advantage
– Does not calculate coordinates based on the
complete equation (uses offset method)
• Disadvantage
– Round-off errors are accumulated, thus line
diverges more and more from straight line
– Round-off operations take time
• Perform integer arithmetic by storing float as
integers in numerator and denominator and
performing integer artithmetic.
Observation on lines.
while( n-- )
{
draw(x,y);
move right;
if( below line )
move up;
}
Line Drawing Algorithms
• Bresenham’s line drawing
– Efficient line drawing algorithm using only incremental
integer calculations
– Can be adapted to draw circles and other curves
• Principle
– Vertical axes show scan line positions
– Horizontal axes show pixel columns
– At each step, determine the best next pixel based on the
sign of an integer parameter whose value is
proportional to the difference between the vertical
separations of the two pixel positions from the actual
line.
Bresenham’s Algorithm
• Improve upon DDA algorithm to use integer
arithmetic only.
• Applicable to circle drawing too. We discuss only
the line drawing here.
• An Accurate and efficient raster line generating
Algo. And scan converts lines using only
incrementa1 integer calculations that can be
adapted to display circles and other curves
Testing for the side of a line.
• Need a test to determine which side of a line
a pixel lies.
• Write the line in implicit form:
F ( x, y)  ax  by  c  0
• Easy to proveY=m*X+Cnts below.
Decision Variables
• Variables a and b record the distances to the
pixel centers of (i+1, j) and (i+1, j+1)
• If d1 < d2, then y=j
• If d1 > d2, then y= j+1
• The vertical axes show-scan-line
positions, and the horizontal axes
identify pixel columns.
• Sampling at unit x intervals in these
examples, we need to decide which of
two possible pixel positions is closer
to the line path at each sample step.
Starting from the left endpoint we
need to determine at the next sample
• position whether to plot the pixel at
position (11, 11) or the one at (11, 12)
Line with negative slope-line path
starting from the left endpoint at pixel
position (50, 50). In this one, do we
select the next pixel position as
(51,50)or as (51,49)? These questions
are answered with Bresenham's line
algorithm by testing the sign of an
integer parameter, whose value is
proportional to the difference between
the separations of the two pixel
positions from the actual line path.
• we- first consider the scanconversion process for lines with
positive slope less than 1. Pixel
positions along a line path are then
determined by sampling at unit x
intervals. Starting from the left end
• point (x0, yo) of a given line, we
step to each successive column ( x
position) and plot the pixel whose
scan-line y value is closest to the
line path.
• demonstrates the Kh step in this
process. Assuming we have
determined that the pixel at (xk, yk)
is to be displayed, we next need to
decide which pixel to plot in
• column x k + 1, . Our choices are
the pixels at positions ( Xk+l, yk)
and (xk+l, yk+l).
• At sampling position xk+l, we label
vertical pixel separations from the
• mathematical line path as d1 and d2
• Y co-ordinate at column Xk+1 can be calculated as
Y=m(Xk+1)+b then
• D1=y-yk=m(xk+1)+b-yk
• D2=(Yk+1) – y=Yk+1 –m(Xk+1)-b
• Difference in these two sepration is
• D1-D2=2m(Xk+1)-2Yk+2b-1
• Decision Parameter Pk( Kth iteration)
Pk=DX(d1-d2)=2Dy * Xk-2Dx * Yk+C
Note Pk has Same Sign as d1-d2Sine DX>1,C is
constant & C=2Dy+Dx(2b-1) is Independent of
Pixel Position ,Ignored In recursive cal of Pk
Pk+1 can be calculated then Recursive Pk+1 be as
Pk+1=Pk+2Dy-2Dx(Yk+1- Yk) then Po=2Dy-Dx
void lineares (int xa ,Int ya , int x b , int yb)
(int dx = a b s ( x a - x b) , dy = abs (ya - yb);
int p = 2 * dy - d x ;
int twoDy = 2 * dy, twoDyDx = 2 *(dy - dx ) ;
int x , y, xEnd:
/ * Determine which point to use a s start, which as end * /
if ( x a > x b ) {x = x b ;Y = yb;xEnd = x a ;}
else {x = xa;Y = ya;xEnd = xb;}
setpixel ( x , y);
while (x < xEnd) {
x++;
if (p < 0 )
p+= twoDy;
else {y + + ;P += twoDyDx;}
setpixel ( x , y);}}