Notes_456_Lines_Drawing2_4 (1).pdf

1/3/2023
1
EI-OE-401 Open Elective IV
(Computer Graphics and CAD/CAM)
Module - 2
Point and Lines
Prof. Pardeep Kumar
Department of Instrumentation,
Kurukshetra University, Kurukshetra-136119
pardeepk@kuk.ac.in
• Points and lines, Line drawing algorithms, midpoint circle and ellipse
algorithms. Filled area primitives: scan line polygon fill algorithm, boundary-
fill and flood fill algorithms.
• Translation, scaling, rotation, reflection and shear transformations, matrix
representations and homogeneous coordinates, composite transforms,
transformation between coordinate systems. 2-D Viewing: The viewing
pipeline, viewing coordinate reference frame, window to viewport coordinate
transformation, viewing functions, Cohen-Sutherland and Cyrus beck line
clipping algorithms
• Reference Books:
Computer Graphics by Hearn & Donald, PHI
Computer Graphic by Plastock, McGraw Hill
Principle of Interactive graphics by Newman.W Spraul R.F., McGraw Hill
Procedural Elements of computer graphics by Rogers D.F., McGraw Hill
Module - I Syllabus
2
January 3, 2023 DOI, Kurukshetra University
There will be
continuous
evaluation in the
form of
Assignments,
quizzes etc.
After
completion
there will be a
class test

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• How line-drawing routine works?
– need to calculate which pixels need to be colored
– line to draw: from A = (ax, ay) to B = (bx, by) (integers)
• ideal vs. actual line?
Line Drawing
3
• We are going to analyze how this process is
achieved.
• Some useful definitions:
– Rasterization: Process of determining which pixels
provide the best approximation to a desired line on the
screen.
– Scan Conversion: Combining the rasterization and
generating the picture in scan line order.
• General requirements:
– Straight lines must appear as straight lines.
– They must start and end accurately
– Lines should have constant brightness along their length
– Lines should be drawn quickly/rapidly
Line Drawing Algorithms
4
?
?
?
?

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• For horizontal, vertical and 45º
lines, the choice of raster
elements is obvious.
• These lines exhibit constant
brightness along the length:
• However, for any other
orientation the choice is more
difficult:
5
?
?
?
?
?
Rasterization of straight lines.
6
• Rasterization yields uneven brightness: Horizontal
and vertical lines appear brighter than the 45º lines.
For doing so, we would need:
1. Calculation of square roots
(increasing CPU time)
2. Multiple brightness levels
Compromises:
1. Calculate only an approximate line
2. Use integer arithmetic
3. Use incremental methods
OR

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• A line in Computer graphics typically refers to line segment,
• which is a portion of straight line that extends indefinitely in opposite
direction.
• It is defined by its two end points & the slope intercept equation for a line:
y = mx + b (1)
where, m = Slope of the line
b = the y intercept of a line
7
x
y
P1(x1,y1)
P2(x2,y2)
b
0
• The two endpoints of a line
segment are specified at
positions (x1,y1) and (x2,y2).
• We can determine the value for slope m & b intercept as
m = y2-y1/x2-x1 = Δy/ Δx (2)
And, b = y1 – mx1 (3)
• For a given x interval Δx along a line, we can compute the corresponding y
interval Δy from
Δy = m Δx (4)
• Similarly, we can obtain x interval Δx by Δy:
Δx = Δy/m (5)
Contd. ……
8

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Conditions….
1. If |m| < 1,
– then for every integer value of x between and excluding x1 and x2, calculate the
corresponding value of y using equation
– Δy = m Δx & scan convert (x, y).
2. If |m| > 1,
– then for every integer value of y between and excluding y1 and y2, calculate the
corresponding value of x using equation
– Δx = Δy / m & scan convert (x, y).
3. If |m| = 1 i.e Δx = Δy.
– In each case, a smooth line with slope m is generated between the specific endpoints.
January 3, 2023 DOI, Kurukshetra University 9
Line Algorithm example
• Example 1: The endpoints of line are(0,0) & (6,18). Compute each value of y
as x steps from 0 to 6 and plot the result.
• Solution : Equation of line is y= mx +b
m = y2-y1/x2-x1 = 18-0/6-0 = 3
Next the y intercept b is found by plugging y1& x1 into the equation
y = 3x + b, 0 = 3(0) + b.
Therefore, b=0, so the equation for the line is
y= 3x.

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Line Drawing Algorithm
• Algorithm 1: Direct Scan Conversion
1. Start at the pixel from the left-hand endpoint xl
2. Step along the pixels horizontally until we
reach the right-hand end of the line, xr
3. For each pixel compute the corresponding y
value
4. round this value to the nearest integer to select
the nearest pixel
The equation of a straight line is given by: y = m.x + b
x = xl;
while (x <= xr)
{
ytrue = m*x + b;
y = Round (ytrue);
PlotPixel (x, y);
/* Set the pixel at (x,y) on */
x = x + 1;
}
The algorithm performs a floating-point multiplication for every step in x.
This method therefore requires an enormous number of floating-point
multiplications, addition and rounding and is therefore expensive.
© 2001 Andrés Iglesias. See: http://personales.unican.es/iglesias
Simple Line Drawing- explained
• Equation for the line: y = m (x – ax) + ay
• What's the line slope m ?
– m = (by – ay)/(bx – ax) calculated once
– increment x, from ax to bx in steps of one pixel
• for each pixel
– calcúlate y = round (m (x – ax) + ay)
– color the resulting pixel (x, y)
• Requires a multiplication, a subtraction, an addition, and a
round for each pixel –
• Expensive! Slow!

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Drawbacks….
• While this approach is mathematically sound, it involves floating-point
computation (multiplication & addition) in every step that uses the line
equation since m & b are generally real numbers.
• The challenge is to find a way to achieve the same goal as quickly as
possible.
Algorithm 2: Digital Differential Analyzer (DDA)
– The digital differential analyzer (DDA) algorithm is an
incremental scan-conversion method.
– Such an approach is characterized by performing
calculations at each step using results from the
preceding step.
– Suppose, at step i we have calculated (xi, yi) to be a
point on the line.
– Since the next point (xi+1,yi+1) should satisfy
Δy/Δx= m
where Δy = yi+1 – yi &
Δx = xi+1 – xi.
We have, yi+1 = yi + mΔx
yi+1 = yi + Δy (1)
or xi+1 = xi + Δy/m (2)
Line Drawing Algorithm
14
DDA uses repeated addition
𝑑𝑦
𝑑𝑥
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
or
∆𝑦
∆𝑥
=
𝑦2−𝑦1
𝑥2−𝑥1
We need only compute m
once, as the start of the scan-
conversion.
The DDA algorithm runs rather
slowly because it requires real
arithmetic (floating-point
operations).

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1. DDA algorithm for lines with -1 < m < 1
x = xl;
ytrue = yl;
while (x <= xr)
{
ytrue = ytrue + m;
y = Round (ytrue);
PlotPixel (x, y);
x = x + 1;
}
Example: Third quadrant
• Switching the roles of x and y when m >1
– Gaps occur when m > 1
– Reverse the roles of x and y using a unit step in y,
and 1/m for x.
Contd…
16
Limitations of DDA:
(1) The rounding operation & floating point arithmetic are time
consuming procedures.
(2) The accumulation of round-off error in successive addition of
floating point increment can cause the calculated pixel
position to drift away from the true line path for long line
segment.

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Algorithm: (x1,y1) (x2,y2) are the end points
and dx, dy are the float variables.
(i) If abs(x2-x1) > abs(y2-y1) then
length = abs(x2-x1)
else
length = abs(y2-y1)
endif
(ii) dx = (x2-x1)/length
dy = (y2-y1)/length
(iii) x = x1 + 0.5
y = y1 + 0.5
(iv) i = 0
(v) Plot (trunc(x), trunc(y))
(vi) x = x + dx
y = y + dy
i = i + 1
(viii) If i < length then go to step (v)
(ix) Stop
Digital Differential Analyzer (DDA)
17
Example 2 Scan convert a line having end points
(3,2) & (4,7) using DDA.
Solution: x2 - x1 = 4-3 = 1
y2 - y1 = 7-2 = 5
As, abs(x2-x1) < abs(y2-y1) then
length = y2-y1 = 5
dx = (x2-x1)/ length = 1/5 = .2
dy = (y2-y1)/ length = 5/5 = 1
x1 y1 x2 y2 L dx dy i x y Result Plot
3 2 4 7 5 .2 1 0 3.5 2.5 3.5, 2.5 3,2
1 3.7 3.5 3.7,3.5 3,3
2 3.9 4.5 3.9,4.5 3,4
3 4.1 5.5 4.1,5.5 4,5
4 4.3 6.5 4.3,6.5 4,6
5 4.5 7.5 4.5,7.5 4,7
• Algorithm 3: Bresenham’s algorithm (1965)
– Bresenham, J.E. Algorithm for computer control of a digital plotter, IBM Systems Journal, January
1965, pp. 25-30.
– This algorithm uses only integer arithmetic, and runs significantly faster.
18
,0)
(0 0)
,
(1
,
(1 1)
(0,1)
X
Y
/
1 2
?
?
1/2 ≤ m ≤ 1
0 ≤ m ≤ 1/2
Plot (1,1)
Plot (1,0)
Key idea: distance between
the actual line and the nearest
grid locations (error).
Initialize error:
e = -1/2
Error is given by:
e = e + m
Reinitialize error:
when e > 0

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19
Example: m = 3/8
If e < 0 below
else above below below above above
Reinitialize
error:
e = 1/4 -1
= -3/4
error
0
Error: e = e + m
Initial value:
e = - 1/2
e = -1/2+3/8
=-1/8
e = -1/8+3/8
=1/4
e = -3/4 +3/8
=-3/8
What is the question?
– Which pixel (L or U) do we set next?
– uses no floating point arithmetic and no rounding.
– Incremental approach
• Assume a line with 0 < slope < 1
• Define extents of the line segment in x and y:
– ∆x = bx – ax, and ∆y = by – ay.
– 0 < ∆y < ∆x. Why?
• Line equation: ∆y/∆x = (y-ay)/(x-ax)
– –∆x*(y-ay) + ∆y*(x – ax) = 0
• Define F(x, y) = –2 ∆x*(y - ay) + 2 ∆y*(x – ax)
– equation of line is F(x, y) = 0
Bresenham Technique
20
1
U
M’’
Ideal line
M’
M
L
Q
P=(px, py )

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• Grid: Points on the grid are centers of pixels
• Let P a pixel which is currently set (chosen)
• Next step pixel: Candidates L (low) and U (up)
• Midpoint between them M = (px+1, py+1/2)
• if ideal line below M
– Set next pixel L
– (px+1, py)
• Else set next pixel U
– (px+1, py+1)
• how do we decide?
Bresenham Technique
21
1
U
M’’
Ideal line
M’
M
L
Q
P=(px, py )
• Case 1: line below M F(Mx, My) < 0
– Next M is M’ = (px+2, py+1/2)
– next F: F(px+2, py+1/2) = -2∆x(py+1/2 – ay) + 2∆y(px+2-ax)
• Calculate the change in F, from this step to next step:
– ∆F = F(px+2, py+1/2) - F(px+1, py+1/2)
– ∆F = 2∆y(px+2-ax) - 2∆y(px+1-ax) = 2∆y
• F(M’x, M’y) = F(Mx,My) + 2∆y (constant integer)
• Case 2: line above M F(Mx, My) > 0
– Next M is M” = (px+2, py+3/2)
– F = F(px+2, py+3/2) - F(px+1, py+1/2) = -2(∆x-∆y)
– F(M”x, M”y) = F(Mx, My) – 2(∆x-∆y)
• In either case, F changes by a constant integer:
– 2∆y if y was not incremented (negative F)
– -2(∆x-∆y) if y was incremented (positive F)
Bresenham Technique
22
1
U
M’’
Ideal line
M’
M
L
Q
P=(px, py )

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Calculate the initial point value
• M = (ax+1, ay+1/2)
– F(ax, ay) = -2∆x(ay+1/2-ay) + 2∆y(ax+1-ax) = 2∆y-∆x
– Initial value: F = 2∆y-∆x
• For (i = ax; x <= bx; x++)
{ Set pixel (x, y)
If F< 0 {
F += 2∆y; // no change in y
else
{ y++;
F += 2(∆y-∆x)
}}
}
Bresenham algorithm
23
• Until now: ax < bx and slope < 1
– Other cases: (slope >1; negative slopes)
• Exercises
– tricks: replace roles of X and Y, and ∆x with
–∆x
• Drawing Patterned lines:
– Patterned lines (dotted or dashed lines)
• store the desired line pattern in a bit mask
• When a pixel is selected, consult the bit
mask to check whether it should be drawn.
• Another approach
Bresenham Line Algorithm
24

1/3/2023
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25
for (0<m<1)
d2
d1
x
Xk Xk+1 Xk+2
yk+2
yk+1
yk
If d2 > d1
Plot yk
else
Plot yk +1
➢ Is it possible to compute and compare
d1 and d2 using only integer operations?
y = m (xk+ 1) + b
d2 = (yk + 1) − y = yk +1 − m (xk + 1) − b
d1 = y − yk= m (xk + 1) + b − yk
d1−d2 = 2m (xk +1)− 2yk + 2b − 1
d1−d2 = 2 dy/dx (xk +1)− 2yk + 2b − 1
dx(d1−d2) = 2 dy (xk +1)− 2 dx yk + dx (2b − 1)
dx(d1−d2) = 2 dy xk +2 dy− 2 dx yk + dx (2b − 1)
Let 2dy + dx(2b−1) = c
We have:
dx(d1−d2) = 2 dy xk - 2 dx yk + c
26
d2
d1
x
Xk Xk+1 Xk+2
yk+2
yk+1
y
k

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• since 0 < m < 1, dx > 0
– dx(d1−d2) has the same sign as of (d1−d2), so
– if (d1−d2) < 0,
– then yk +1 = yk
– else yk +1 = yk +1
• Let Pk= (d1−d2) = 2dy xk − 2dx yk + c
where c = 2dy + dx(2b−1), hence
• Pk+1= 2dy(xk +1)- 2dx(yk+1)+ c
27
Subtracting
each other as
shown on next
page…….
Selection criterion
for yk +1
• Continued from previous page: To derive Pk+1 from Pk
– Pk+1-Pk= 2dy(xk +1- xk )- 2dx(yk+1-yk)
– Pk+1=Pk + 2dy- 2dx = Pk + 2(dy- dx)
• At initial value when k = 0, let’s consider again the previous equation
– P0= (d1−d2) = 2dy x0 − 2dx y0 + c,
• substitute value of c here
– P0= (d1−d2) = 2dy x0 − 2dx y0 + 2dy + dx(2b−1)
• further at k = 0, form y0 = m x0 + b,
– b = y0 - m x0
• substitute the b in above equation
28

1/3/2023
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– P0= (d1−d2) = 2dy x0 − 2dx y0 + 2dy + dx (2(y0 - m x0 ) −1)
– P0= (d1−d2) = 2dy x0 − 2dx y0 + 2dy + dx (2(y0 - dy/dx x0 ) −1)
– P0= (d1−d2) = 2dy x0 − 2dx y0 + 2dy + 2 dx y0 - 2 dy x0 − dx
– P0= (d1−d2) = 2dy − dx
• (at the starting point of line or initial condition)
29
• Given end points (x0, y0) (x1, y1)
dx = x1−x0, dy=y1−y0
• Starting with an end point (x0, y0):
1. Compute P0 = 2dy −dx
2. For each k, staring with k = 0
if (Pk < 0)
the next point is (Xk+1, Yk),
Pk+1 = Pk + 2 dy,
else
the next point is (Xk+1, Yk+1),
Pk+1 = Pk + 2dy − 2dx
3. Repeat step 2 x1−x0 times
Bresenham’s Line Algorithm
30

1/3/2023
16
• Given end points (x0, y0) (x1, y1)
dx = x1−x0, dy=y1−y0
• Starting with an end point (x0, y0):
1. Compute P0 = 2dy −dx
2. For each k, staring with k = 0
if (Pk < 0)
the next point is (Xk+1, Yk),
Pk+1 = Pk + 2 dy,
else
the next point is (Xk+1, Yk+1),
Pk+1 = Pk + 2dy − 2dx
3. Repeat step 2 x1−x0 times
Bresenham’s Line Algorithm
31
void line(int x0, int y0, int x1, int y1, int value)
{ int dx = x1-x0,dy = y1 – y0;
int d = 2 * dy-dx; /* Initial Value of d
int incrL = 2*dy; /* increment used
incrU = 2*(dy-dx); /* increment used
int x = x0, y = y0;
putpixel (x,y, value) /* start p
while (x < x1)
{ if (d <= ))
{ /*choose L */
d +=incrL;
x++;}
else { d += incrU; /* Choose U */
x++;
y++; }
putpixel (x, y, value); /* selected pixel clo
} /* while */
} /* Midpoint Line */
void line(int x0, int y0, int x1, int y1, int value)
{ int dx = x1-x0,dy = y1 – y0;
int d = 2 * dy-dx; /* Initial Value of d*/
int incrL = 2*dy; /* increment used for move to L*/
incrU = 2*(dy-dx); /* increment used for move to U*/
int x = x0, y = y0;
putpixel (x,y, value) /* start pixel */
while (x < x1) {
if (d <= ))
{ /*choose L */
d +=incrL;
x++;}
else { d += incrU; /* Choose U */
x++;
y++; }
putpixel (x, y, value); /* selected pixel closest to the line */
} /* while */
} /* Midpoint Line */
Bresenham algorithm
32

1/3/2023
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Bresenham algorithm
33
January 3, 2023
DOI, Kurukshetra University
34
Any Question?
Thanks!

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