Computer Graphics
Line Drawing Techniques
5.1 Line
A line, or straight line, is, roughly speaking, an (infinitely) thin, (infinitely) long, straight geometrical object, i.e. a curve that is long and straight. Given two points, in geometry, one can always find exactly one line that passes through the two points; this line provides the shortest connection between the points and is called a straight line. Three or more points that lie on the same line are called collinear.
A line may have three forms with respect to slope i.e. it may have slope = 1 as shown in following figure (a), or may have slope < 1 as shown in figure (b) or it may have slope > 1 as shown in figure (c). Now if a line has slope = 1 it is very easy to draw the line by simply starting form one point and go on incrementing the x and y coordinates till they reach the second point. So that is a simple case but if slope < 1 or is > 1 then there will be some problem.
figure (a) figure (b) figure (c)
There are three techniques to be discussed to draw a line involving different time complexities that will be discussed later. These techniques are:
Incremental line algorithm DDA line algorithm
Bresenham line algorithm
5.2 Incremental line algorithm
This algorithm exploits simple line equation y = m x + b
Where m = dy / dx
and b = y – m x
Now check if |m| < 1 then starting at the first point, simply increment x by 1 (unit increment) till it reaches ending point; whereas calculate y point by the equation for each x and conversely if |m|>1 then increment y by 1 till it reaches ending point; whereas calculate x point corresponding to each y, by the equation.
Now before moving ahead let us discuss why these two cases are tested. First if |m| is less than 1 then it means that for every subsequent pixel on the line there will be unit increment in x direction and there will be less than 1 increment in y direction and vice versa for slope greater than 1. Let us clarify this with the help of an example:
Suppose a line has two points p1 (10, 10) and p2 (20, 18)
Now difference between y coordinates that is dy = y2 – y1 = 18 – 10 = 8
Whereas difference between x coordinates is dx = x2 – x1 = 20 – 10 = 10
This means that there will be 10 pixels on the line in which for x-axis there will be distance of 1 between each pixel and for y-axis the distance will be 0.8.
Consider the case of another line with points p1 (10, 10) and p2 (16, 20)
Now difference between y coordinates that is dy = y2 – y1 = 20 – 10 = 10 Whereas difference between x coordinates is dx = x2 – x1 = 16 – 10 = 6
This means that there will be 10 pixels on the line in which for x-axis there will be distance of 0.6 between each pixel and for y-axis the distance will be 1.
Now having discussed this concept at length let us learn the algorithm to draw a line using above technique, called incremental line algorithm:
Incremental_Line (Point p1, Point p2) dx = p2.x – p1.x
dy = p2.y – p1.y
m = dy / dx
x = p1.x
y = p1.y
b = y – m * x
if |m| < 1
for counter = p1.x to p2.x drawPixel (x, y)
x = x + 1
y = m * x + b
else
for counter = p1.y to p2.y drawPixel (x, y)
y = y + 1
x = ( y – b ) / m
Discussion on algorithm:
Well above algorithm is quite simple and easy but firstly it involves lot of mathematical calculations that is for calculating coordinate using equation each time secondly it works only in incremental direction.
We have another algorithm that works fine in all directions and involving less calculation mostly only addition; which will be discussed in next topic.
5.3 Digital Differential Analyzer (DDA) Algorithm:
DDA abbreviated for digital differential analyzer has very simple technique. Find difference dx and dy between x coordinates and y coordinates respectively ending points of a line. If |dx| is greater than |dy|, than |dx| will be step and otherwise |dy| will be step.
if |dx|>|dy| then
step = |dx|
else
step = |dy|
Now very simple to say that step is the total number of pixel required for a line. Next step is to divide dx and dy by step to get xIncrement and yIncrement that is the increment required in each step to find next pixel value.
xIncrement = dx/step
yIncrement = dy/step
Next a loop is required that will run step times. In the loop drawPixel and add xIncrement in x1 by and yIncrement in y1.
To sum-up all above in the algorithm, we will get,
DDA_Line (Point p1, Point p2) dx = p2.x – p1. X
dy = p2.y – p1. Y
x1=p1.x
y1=p1.y
if |dx|>|dy| then
step = |dx|
else
step = |dy|
xIncrement = dx/step yIncrement = dy/step
for counter = 1 to step drawPixel (x1, y1)
x1 = x1 + xIncrement y1 = y1 + yIncrement
Criticism on Algorithm:
There is serious criticism on the algorithm that is use of floating point calculation. They say that when we have to draw points that should have integers as coordinates then why to use floating point calculation, which requires more space as well as they have more computational cost.
Therefore there is need to develop an algorithm which would be based on integer type calculations. Therefore, work is done and finally we will come up with an algorithm “Bresenham Line Drawing algorithm” which will be discussed next.
5.4 Bresenham's Line Algorithm
Bresenham's algorithm finds the closest integer coordinates to the actual line, using only integer math. Assuming that the slope is positive and less than 1, moving 1 step in the x direction, y either stays the same, or increases by 1. A decision function is required to resolve this choice.
If the current point is (xi, yi), the next point can be either (xi+1,yi) or (xi+1,yi+1) . The actual position on the line is (xi+1, m(xi+1)+c) . Calculating the distance between the true point, and the two alternative pixel positions available gives:
d1 = y - yi
= m * (x+1)+b-yi
d2 = yi+ 1 - y
= yi+ 1 – m ( xi + 1 ) - b
Let us magically define a decision function p, to determine which distance is closer to the true point. By taking the difference between the distances, the decision function will be positive if d1 is larger, and negative otherwise. A positive scaling factor is added to ensure that no division is necessary, and only integer math need be used.
pi = dx (d1-d2)
pi = dx (2m * (xi+1) + 2b – 2yi-1 )
pi = 2 dy (xi+1) –2 dx yi + dx (2b-1 ) ------------------ (i) pi = 2 dy xi – 2 dx yi + k ------------------ (ii)
where k=2 dy + dx (2b-1)
Then we can calculate pi+1 in terms of pi without any xi ,yi or k .
pi+1 = 2 dy xi+1 – 2 dx yi+1 + k
pi+1 = 2 dy (xi + 1) - 2 dx yi+1 + k since xi+1= xi + 1 pi+1 = 2 dy xi + 2 dy- 2 dx yi+1 + k ------------------ (iii)
Now subtracting (ii) from (iii), we get pi+1 - pi = 2 dy- 2 dx (yi+1 - yi ) pi+1 = pi + 2 dy- 2 dx (yi+1 - yi )
If the next point is: (xi+1,yi) then
d1<d2 => d1-d2<0
=> pi<0
=> pi+1= pi + 2 dy
If the next point is: (xi+1,yi+1) then
d1>d2 => d1-d2>0
=> pi>0
=> pi+1= pi + 2 dy - 2 dx
The pi is our decision variable, and calculated using integer arithmetic from precomputed constants and its previous value. Now a question is remaining how to calculate initial value of pi. For that use equation (i) and put values (x1, y1)
pi = 2 dy (x1+1) – 2 dx yi + dx (2b-1 ) where b = y – m x implies that
pi = 2 dy x1 +2 dy – 2 dx yi + dx ( 2 (y1 – mx1) -1 ) pi = 2 dy x1 +2 dy – 2 dx yi + 2 dx y1 – 2 dy x1 - dx pi = 2 dy x1+2 dy– 2 dx yi+ 2 dx y1– 2 dy x1 - dx
there are certain figures will cancel each other shown in same different colour
pi = 2 dy - dx
Thus Bresenham's line drawing algorithm is as follows:
dx = x2-x1
dy = y2-y1
p = 2dy-dx
c1 = 2dy
c2 = 2(dy-dx)
x = x1
y = y1
plot (x,y,colour)
while (x < x2 )
x;
if (p < 0) p = p + c1 else
p = p + c2
y
plot (x,y,colour)
Again, this algorithm can be easily generalized to other arrangements of the end points of the line segment, and for different ranges of the slope of the line.