Lines and curves algorithms

WHERE TO DRAW A LINE??

 Line drawing is accomplished by calculating
intermediate positions along the line path between
specified end points.
 Precise definition of line drawing

Given two points P and Q in the plane, both with
integer coordinates, determine which pixels on a
raster screen should be on in order to make a
picture of a unit-width line segment starting from P
and ending at Q.

6 (3,
5 3)
4
3
2
1
0
0 1 2 3 4 5 6

Line drawing (contd.)
 The thinnest line is of one-pixel wide. We will
concentrate on drawing a line of 1 pixel resolution.
The Cartesian slope-intercept equation for a straight

line is
y= m. ∆x + b
m is the slope of the line and b is the y intercept.
Given the endpoints of a line segment.
m = (y2-y1) / (x2-x1)
b= y1-m.x1

Line Drawing (cont)
 Also for any given interval ∆x along a line, we can
compute the corresponding y interval ∆y from
∆y= m. x
Similarly we can obtain the x interval ∆x corresponding

to a specified ∆y as
∆x= ∆y / m
These equations form the basis for determining

deflection voltages in analog devices.

Line Drawing (cont)

 On Raster systems, lines are plotted with pixels, and
step sizes in the horizontal and vertical directions are
constrained by pixel separations. Hence we ought to
“sample” a line at discrete positions and determine
the nearest pixel to the line at each sampled position.

Bresenham’s Line Algorithm
An accurate, efficient raster line drawing algorithm
developed by Bresenham, scan converts lines using
only incremental integer calculations that can be
adapted to display circles and other curves.

Keeping in mind the symmetry property of lines, lets
derive a more efficient way of drawing a line.
Starting from the left end point (x0,y0) of a given line , we step to each
successive column (x position) and plot the pixel whose scan-line y
value closest to the line path
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
xk+1.

Bresenham Line Algorithm
(cont)
Choices are(xk +1, yk) and (xk+1, yK+1)
d1 = y – yk = m(xk + 1) + b – yk
d2 = (yk + 1) – y = yk + 1- m(xk + 1) – b
The difference between these 2 separations is
d1-d2 = 2m(xk + 1) – 2 yK + 2b – 1

 A decision parameter pk for the kth step in the line algorithm
can be obtained by rearranging above equation so that it
involves only integer calculations

Define
Pk = Δx ( d1-d2) = 2Δyxk-2 Δxyk + c

The sign of Pk is the same as the sign of d1-d2, since Δx > 0.
Parameter c is a constant and has the value 2Δy + Δx(2b-1)
(independent of pixel position)

If pixel at yk is closer to line-path than pixel at yk +1
(i.e, if d1 < d2) then pk is negative. We plot lower pixel in such a
case. Otherwise , upper pixel will be plotted.

Bresenham’s algorithm (cont)
At step k + 1, the decision parameter can be evaluated as,
pk+1 = 2Δyxk+1- 2Δxyk+1+ c
Taking the difference of pk+ 1 and pk we get the following.
pk+1 – pk = 2Δy(xk+1- xk)-2Δx(yk+1 – yk)

But, xk+1 = xk +1, so that
pk+1 = pk + 2Δy - 2 Δx(yk+1 – yk)

Where the term yk+1-yk is either 0 or 1, depending on the sign of
parameter pk

The first parameter p0 is directly computed

p0 = 2 Δyx0 - 2 Δxy0 + c = 2 Δyx0 + 2 Δxy0 +2 Δy + Δx (2b-1)

Since (x0,y0) satisfies the line equation , we also have
y0 = Δy/ Δx * x0 + b

Combining the above 2 equations , we will have
p0 = 2Δy – Δx
The constants 2Δy and 2Δy-2Δx are calculated once for each
time to be scan converted

 So, the arithmetic involves only integer addition and subtraction of 2 constants
STEP 1 Input the two end points and store the left end point
in (x0,y0)

STEP 2 Load (x0,y0) into the frame buffer (plot the first point)

STEP 3 Calculate the constants Δx, Δy, 2Δy and 2Δy-2Δx and
obtain the starting value of the decision parameter as
p0 = 2Δy- Δx

STEP 4 At each xk along the line, starting at k=0,
perform the following test:
If pk < 0 , the next point is (xk+1, yk) and
pk+1 = pk + 2Δy

Otherwise
Point to plot is (xk+1, yk+1)
pk+1 = pk + 2Δy - 2Δx

Repeat step 4 (above step) Δx times

Loading the frame buffer

•Calculate frame-buffer addresses by incremental
methods
•For a bilevel system(1 bit per pixel), address for
pixel(x,y)
Addr(x,y) = addr(0,0)+y(xmax +1)+x
Addr(x+1,y) = addr(x,y)+1
Addr(x+1,y+1) = addr(x,y)+ x max +2

Where do we draw a circle???
Properties of a circle:
A circle is defined as a set of points that are all the given distance (x ,y ).
c c
This distance relationship is expressed by the pythagorean theorem in
Cartesian coordinates as
(x – xc)2 + (y – yc) 2 = r2
We could use this equation to calculate the points on the circle

circumference by stepping along x-axis in unit steps from xc-r to xc+r and
calculate the corresponding y values at each position as
y = yc +(- ) (r2 – (xc –x )2)1/2
This is not the best method:

Considerable amount of computation
Spacing between plotted pixels is not uniform

Polar co-ordinates for a circle
We could use polar coordinates r and θ,
x = xc + r cosθ y = yc + r sinθ
 A fixed angular step size can be used to plot equally spaced points
along the circumference
A step size of 1/r can be used to set pixel positions to approximately

1 unit apart for a continuous boundary
But, note that circle sections in adjacent octants within one quadrant

are symmetric with respect to the 45 deg line dividing the two octants
Thus we can generate all pixel positions around a circle by calculating

just the points within the sector from x=0 to x=y
This method is still computationally expensive

Bresenham to Midpoint
 Bresenham requires explicit equation
 Not always convenient (many equations are
implicit)
 Based on implicit equations: Midpoint
Algorithm (circle, ellipse, etc.)
 Implicit equations have the form F(x,y)=0.

Midpoint Circle Algorithm
We will first calculate pixel positions for a circle centered around the
origin (0,0). Then, each calculated position (x,y) is moved to its proper
screen position by adding xc to x and yc to y

Note that along the circle section from x=0 to x=y in the first octant,
the slope of the curve varies from 0 to -1

Circle function around the origin is given by
fcircle(x,y) = x2 + y2 – r2

Any point (x,y) on the boundary of the circle satisfies the equation
and circle function is zero

 For a point in the interior of the circle, the circle function is negative and
for a point outside the circle, the function is positive
 Thus,

 f (x,y) < 0 if (x,y) is inside the circle boundary
circle
 fcircle(x,y) = 0 if (x,y) is on the circle boundary
 fcircle(x,y) > 0 if (x,y) is outside the circle boundary

Assuming we have just plotted the pixel at (xk,yk) , we next need to
determine whether the pixel at position (xk + 1, yk-1) is closer to the circle
Our decision parameter is the circle function evaluated at the midpoint
between these two pixels

pk = fcircle (xk +1, yk-1/2) = (xk +1)2 + (yk -1/2)2 – r2

If pk < 0 , this midpoint is inside the circle and the pixel on the scan line yk
is closer to the circle boundary. Otherwise, the
mid position is outside or on the circle boundary, and we select the pixel
on the scan line yk-1

Successive decision parameters are obtained using incremental
calculations
Pk+1 = fcircle(xk+1+1, yk+1-1/2)
= [(xk+1)+1]2 + (yk+1 -1/2)2 –r2

OR
Pk+1 = Pk+2(xK+1) + (yK+12 – yk2) – (yK+1- yk)+1
Where yk+1 is either yk or yk-1, depending on the sign of pk
Increments for obtaining Pk+1:
2xk+1+1 if pk is negative
2xk+1+1-2yk+1 otherwise

Midpoint circle algorithm
Note that following can also be done incrementally:
2xk+1 = 2xk +2
2 yk+1 = 2yk – 2
At the start position (0,r) , these two terms have the values 2 and 2r-2
respectively
Initial decision parameter is obtained by evaluating the circle function at

the start position (x0,y0) = (0,r)
p0 = fcircle(1, r-1/2) = 1+ (r-1/2)2-r2
OR
P0 = 5/4 -r
If radius r is specified as an integer, we can round p0 to
p0 = 1-r

The actual algorithm
1: Input radius r and circle center (xc,yc) and obtain the first point
on the circumference of the circle centered on the origin as
(x0,y0) = (0,r)

2: Calculate the initial value of the decision parameter as
P0 = 5/4 - r
3: At each xk position starting at k = 0 , perform the following
test:
If pk < 0 , the next point along the circle centered on (0,0) is
(xk+1, yk) and
pk+1 = pk + 2xk+1 + 1

The algorithm
Otherwise the next point along the circle is (xk+1, yk-1) and
pk+1 = pk + 2xk+1 +1 -2yk+1
Where 2xk+1 = 2xk+2 and 2yk+1 = 2yk-2

4: Determine symmetry points in the other seven octants

5: Move each calculated pixel position (x,y) onto the
circular path centered on (x,yc) and plot the coordinate
values
x = x+ xc , y= y+ yc

6: Repeat steps 3 through 5 until x >= y

Lines and curves algorithms

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