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7/31/2019 CE-201 Lect 4-6 _ZA
http://slidepdf.com/reader/full/ce-201-lect-4-6-za 1/69
Lecture Outline
• Scan Conversion
– Point
– Line
– Circle
7/31/2019 CE-201 Lect 4-6 _ZA
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Point Scan
• Point plotting is furnished by converting a single
coordinate position into appropriate operations for
the output device in use.
• Random scan uses stored point-plotting instructions(coordinates) and convert them in deflection voltage
that positions the electron beam at the screen
locations to be plotted during each refresh cycle.
7/31/2019 CE-201 Lect 4-6 _ZA
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Point Scan…1
• Raster scan depends on setting the bit value
corresponding the specified screen position within
the frame buffer to 1. The electron beam when
sweeps across such position in a line emits burst of electrons with intensities corresponding to colour
codes to be displayed at that location.
7/31/2019 CE-201 Lect 4-6 _ZA
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Point Scan…2
• Let as scan-convert a point (x, y) to a pixel location
(x’, y’)
• x’ = Floor(x), and y’ = Floor(y)
• Where Floor is a function that returns largest integer thatis less than or equal to the argument.
• In such a case, the origin is placed at the lowest left
corner of the pixel grid in the image space
• All points that satisfy x’ x < x’+1 and y’ y < y’ + 1
will be mapped to pixel location (x’, y’) (refer Fig a)
7/31/2019 CE-201 Lect 4-6 _ZA
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Point Scan…3
7/31/2019 CE-201 Lect 4-6 _ZA
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Point Scan…4
• If the origin of coordinate system for (x, y) is to
be shifted to new location say (x+0.5, y+0.5),
then
– x’ = Floor(x+0.5), and y’ = Floor(y+0.5)
– This will place the origin of the coordinate system
(x, y) at the centre of pixel (0, 0)
– Now, all points that satisfy x’-0.5 x < x’+0.5 andy’-0.5 y < y’+0.5 will be mapped to pixel location
(x’, y’) (refer Fig b)
7/31/2019 CE-201 Lect 4-6 _ZA
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Line Scan-Conversion
• Line plotting is done by calculating intermediate
positions between the two specified endpoint positions.
• Linearly varying horizontal and vertical deflection
voltages are generated in proportion to the requiredchanges in x, y directions to produce the smooth line.
• A straight line segment is displayed by plotting the
discrete points between endpoint positions.
– These intermediate positions are calculated from the equationof the line
– Colour intensity is loaded into the frame buffer at the
corresponding pixel coordinate
7/31/2019 CE-201 Lect 4-6 _ZA
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Line Scan-Conversion…1
• Screen locations are referenced with integer values. E.g.
(10.48, 20.51) will be converted to pixel position (10, 20)
• This rounding effect causes lines to be displayed with a
stairstep appearance . This is noticeable on lowresolution systems.
• Better smoothing can be achieved by adjusting light
intensities along the line paths.
7/31/2019 CE-201 Lect 4-6 _ZA
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Line Scan-Conversion…2
• Pixel column number is basically pixel position across a
scan line. These are increasing from left to right.
• Scan line number are increasing from bottom to top.
• Loading a specified colour into framebuffer at a pixel position defined by
column ‘x’ along scan line ‘y’
setPixel (x, y)
• Current frame-buffer intensity for a
specified location can be accomplished
getPixel (x, y)Pixel
column No
S c a n
L i n e
N o
0
0 1 2
12
7/31/2019 CE-201 Lect 4-6 _ZA
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Line Scan-Conversion…3
• Line is defined by two end points P1, P2 and the equation of
line as shown in the Figure
• These can be plotted or scan-converted using different line
plotting algorithms.
• The slope-intercept equation is
not suitable for vertical lines.
Horizontal, vertical and diagonal
(|m| = 1) lines can be handled
as special cases and can bemapped to image space
straightforward.
y
b
y = m.x + b
P1 (x1, y1)
P2(x2, y2)
x
7/31/2019 CE-201 Lect 4-6 _ZA
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Line Drawing Algorithms
• Direct use of line equation
• DDA algorithm
• Bresenham’s Line Algorithm
7/31/2019 CE-201 Lect 4-6 _ZA
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Direct use of Line equation
• Direct approach
– The Cartesian slope
intercept equation for a
straight line:
y = m . x + b (1)
– ‘m’ represents slope of
the line and ‘b’ is
intercept on y-axis, suchthat
(2)
(3)11
12
.
12
xm yb
x xm
y y
y1
y2
x1 x2
y1
y2
x1 x2
7/31/2019 CE-201 Lect 4-6 _ZA
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Direct use of Line equation..1
• For a small change in ‘x’
along the line, the
corresponding change
in ‘y’ can be given by y = m . x (4)
or x = y/m (5)
– These equations form
the basis for determiningthe deflection voltages
in analog devices.
• For lines with slope
magnitudes |m|<1, x
is set proportional to
small horizontaldeflection voltage and
the corresponding
vertical deflection is
then set proportional toy as calculated from
the equation (4).
7/31/2019 CE-201 Lect 4-6 _ZA
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Direct use of Line equation…2
• For lines with slope
magnitudes |m|>1, y
is set proportional to a
small vertical deflectionvoltage and the
corresponding
horizontal deflection is
then set proportional tox as calculated from
the equation (5).
• For lines with slope
magnitudes m=1, x =
y and the horizontal
and vertical deflectionvoltage are equal
7/31/2019 CE-201 Lect 4-6 _ZA
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Direct use of Line equation…3
• This means:
– First scan convert end points to pixel locations (x1’, y1’) and
(x2’, y2’)
– Set ‘m’ and ‘b’ using equations – For |m| 1, for integer value of x between and excluding
x1’, x2’, calculate the corresponding value of y and scan-
convert (x, y)
– For |m|> 1, for integer value of y between and excludingy1’, y2’, calculate the corresponding value of x and scan-
convert (x, y)
7/31/2019 CE-201 Lect 4-6 _ZA
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Direct use of Line equation…4
• Example:
– End points are (0, 0) and (6, 18). Compute each value of x
as y steps from 0 to 18 and plot.
• For equation y = m.x + b, compute slope as
– m = y/x
• Using value of ‘m’ compute ‘b’
• Place y = 1 to 17 and compute corresponding value of ‘x’
• Plot
7/31/2019 CE-201 Lect 4-6 _ZA
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Direct use of Line equation…5
• Example:
– M = (18 – 0) /(6 – 0) = 3 m > 1 b = 0
– Change y: compute for y = 1 the value of x = ∆y/m = 1/3
• Cartesian coordinate point (1, 0.3)• Pixel coordinate point (Case-I): (1, 0)
• Pixel coordinate point (Case-II): (1, 0)
– Compute for y = 2 the value of x = 2/3
• Cartesian coordinate point (1, 0.67)
• Pixel coordinate point (Case-I): (1, 0)
• Pixel coordinate point (Case-II): (1, 1)
– Plot the three graphs
– Continue …..
7/31/2019 CE-201 Lect 4-6 _ZA
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Digital Differential Analyzer (DDA)
• DDA is an incremental scan-conversion line algorithm
based on calculating either x or y as mentioned in
line primitives.
• Suppose at step ‘i’ the point is (xi, yi) on the line.• For next point (xi+1, yi+1)
– yi+1 = yi + y, where y = m x
– xi+1
= xi
+ x, where x = y/m
– The process continues until x reaches x2’ (for |m| 1 case)
or y reaches y2’ (for |m|> 1 case)
7/31/2019 CE-201 Lect 4-6 _ZA
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Digital Differential Analyzer…1
• Consider a line with positive slope. Let the slope be
less than or equal to 1. For unit interval x=1,
yk+1 = yk + m, where k = 1, 2, …., end point (6)
– The value of yk+1 should be rounded to the nearest integer• For slope greater than 1, increasing y by 1
xk+1 = xk + 1/m, where k = 1, 2, …., end point (7)
– The value of xk+1 should be rounded to the nearest integer
7/31/2019 CE-201 Lect 4-6 _ZA
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Digital Differential Analyzer…2
• The assumption here is that lines are processed from
left to right. If it is processed from right to left then
yk+1 = yk – m (8)
xk+1 = xk – 1/m (9)
x
y
7/31/2019 CE-201 Lect 4-6 _ZA
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Digital Differential Analyzer…3
• For negative slope of line as shown in figure:
• If moving from left to right:
– Set x = +1 and use equation (8) for m<1 to calculate y values
– Set y = -1 and use equation (7) for m>1 to calculate x values
• If moving from right to left:
– If m < 1 then set x = -1 and obtain y
values from equation (6)
– If m > 1 then set y = +1 and obtain x
values from equation (9)
x
y
7/31/2019 CE-201 Lect 4-6 _ZA
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Digital Differential Analyzer…4
• This algorithm is faster in calculating pixel positions
• It eliminates multiplication by making use of raster
characteristics
• But rounding off in successive additions and itsaccumulation may cause drifting of the pixel position
away from the true line
• This rounding off procedure is also time consuming• By converting m or 1/m into integer or fractional
part, the time consumed can be reduced.
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm
• This raster-line generating algorithm is more accurate
and efficient
• It uses only incremental integer calculations
• It can be adapted to display circles and other curves.• The method is based on identifying the nearest pixel
position closer to the line path at each sample step.
10
11
12
13
10 11 12 13
50
51
52
53
50 51 52 53
Fig-1 Fig-2
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…1
• In Fig 1 the line segment
starts from scan line 11
and pixel position
(column) 10. The next
point along the line may
be at pixel position (11,
11) or (11, 12)
• Similarly, in Fig 2 the linesegment starts from (50,
52) and the next pixel
position may be (51, 52)
or (51, 51)
Fig-1
Fig-2
10
11
12
13
10 11 12 13
50
51
52
53
50 51 52 53
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…2
• Bresenham’s line algorithm tests the sign of an
integer parameter and the one whose value is
proportional to the difference between the
separations of the two pixel positions from the actualline path is selected.
• Let the line starts from left-end
point (xo, yo) of a given line. Now,
the pixel whose scan-line y valueis closest to the line path is plotted.
Fig-3
yk
yk+1
yk+2
yk+3
xk xk+1 xk+2 xk+3
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…3
• Let the pixel displayed is (xk, yk). The next pixel to
plot will be in column xk+1 and may be represented as
(xk+1, yk) or (xk+1, yk+1)
• The difference between estimatedvalue of y and new y value along the
line is to be computed.
For first point yest =m . (xk+1) + b (10)
Difference d1 = yest – yk (11)
For second point d2 = (yk+1) – yest (12)
Difference between the two separations = d1 – d2
Fig-3
yk
yk+1
yk+2
yk+3
xk xk+1 xk+2 xk+3
d1
d2
y = m.x + b
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…4
• Difference (d1 – d2) = 2m.(xk+1) – 2 yk + 2b – 1 (13)
• Let m = y/x denoting vertical and horizontal
separations. Let px is defined as
px = x (d1 – d2)px = 2 y . xk - 2 x . yk + c (14)
Where, c = 2 y + x (2b – 1) (15)
• Decision parameter px
is –ve i.e. d1
<d2
– Pixel at yk is closer to line than that at yk+1
– Lower point will be plotted
– Vice-Versa
Fig-3
yk
yk+1
yk+2
yk+3
xk xk+1 xk+2 xk+3
d1
d2
y = m.x + b
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…5
• The change in coordinate along the line takes place
in unit steps. The decision parameter at next step will
be:
pk+1 = 2y . xk+1 - 2 x . yk+1 + cAnd pk+1 – pk = 2 y (xk+1 – xk) - 2 x (yk+1 – yk) (16)
But xk+1 = xk+1, so that
pk+1 = pk + 2 y - 2 x (yk+1 – yk) (17)
(yk+1 – yk) is either 0 or 1 depending upon the sign of pk.
The initial value of parameter will be
po = 2 y - x (18)
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…6
• Algorithm for |m|<1: (steps)
– Input and store (xo, yo). Plot it (load on buffer).
– Calculate x, y, 2 y - 2 x
– Compute decision parameter, po (= 2 y - x) – Assuming pk < 0, the next point will be (xk+1, yk) and
pk+1 = pk + 2y
Otherwise, plot point (xk +1, yk +1) and
pk+1 = pk + 2 y - 2 x
– Repeat step x times
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…7
• E.g. Plot a line with end points (20, 10) and (30, 18).
The slope of line is 0.8.
– Calculate x (=10), y (=8)
– Compute decision parameter, po (= 2 y - x = 6) – Compute 2 y (=16), 2 y - 2 x (= -4)
– Assuming pk < 0, the next point will be (xk+1, yk) and
pk+1 = pk + 2y
Otherwise, plot point (xk +1, yk +1) and
pk+1 = pk + 2 y - 2 x
– Repeat step x times
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…8
• It is generalized to lines with arbitrary slopes.
• Considers symmetry between octants and quadrants of
the xy plane.
• Algorithm allows moving along pixel position from eitherend of the line.
• For positive slope and start from left end point, both the
increments are positive. It is vice-versa for start from
right end point.• For negative slope, one increment will be positive and
other negative depending upon the start point.
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…9
• When the two vertical separations from the line path are
equal (d1 = d2), we choose upper (or lower) of the two
candidate pixels.
• Special cases like
y = 0 (horizontal line);
x = 0 (verticalline) and |x| = |y| (diagonal line) can be loaded
directly into the frame buffer without processing them
through the line –plotting algorithm.
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…10
• For negative slope of a line (|m|<1)
– When moving from left to right
– Po = - 2∆y - ∆x
– If po
is – ve point will be (xk+1
, yk
)
and pk+1 = po - 2 ∆y
– If po is + ve point will be (xk+1, yk-1)
and pk+1 = po - 2 ∆y - 2 ∆x
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…11
• For negative slope of a line (|m|<1)
– When moving from right to left
– Po = 2∆y + ∆x
– If po
is – ve point will be (xk-1
, yk
)
and pk+1 = po - 2 ∆y
– If po is + ve point will be (xk-1, yk+1)
and pk+1 = po + 2 ∆y + 2 ∆x
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…12
• For negative slope of a line (|m|>1)
– Increment is ∆y
– When moving from left to right
– ∆x = ∆y/m or (1/m) = ∆x/ ∆y
– d1 = xest - xk
– d1 = (1/m)(yk+1 – b) – xk
– d2 = xk+1 – xest
– d2
= xk+1
– (1/m)(yk+1
– b)
– d1 – d2 = (2/m)(yk + 1) – 2xk – (2b/m) – 1
– pk = ∆y . (d1 – d2)
– pk = 2 ∆x . yk - 2 ∆y . xk + 2 ∆x (1-b) - ∆y
d1 d2
y = m. x + b
yk
yk+1
yk+2
yk+3
xk xk+1 xk+2 xk+3
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…13
• For negative slope of a line (|m|>1)
– When moving from left to right
– pk = 2 ∆x . yk - 2 ∆y . xk + 2 ∆x (1-b) - ∆y
– pk = 2 ∆x . yk - 2 ∆y . xk + c – Where c = 2 ∆x (1-b) - ∆y
– When pk ≤ 0, then d1 < d2 point (xk, yk+1)
– When pk > 0, then d1 > d2 point (xk+1, yk+1)
– Now, pk+1 = 2 ∆x . yk+1 - 2 ∆y . xk+1 + c – pk+1 – pk = 2 ∆x . (yk+1 – yk) - 2 ∆y . (xk+1 –xk)
– pk+1 = pk+2 ∆x . (yk + 1 – yk) - 2 ∆y . (xk+1 –xk)
d1 d2
y = m. x + b
yk
yk+1
yk+2
yk+3
xk xk+1 xk+2 xk+3
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…14
• For negative slope of a line (|m|>1)
– When moving from left to right
– pk+1 = pk + 2 ∆x . (yk + 1 – yk) - 2 ∆y . (xk+1 –xk)
– pk+1 = pk + 2 ∆x - 2 ∆y . (xk+1 –xk) – Now, if pk+1 ≤ 0, then d1 < d2 point (xk, yk+1)
– Decision parameter pk+1 = pk + 2 ∆x
– When pk+1 > 0, then d1 > d2 point (xk+1, yk+1)
– Decision parameter pk+1 = pk + 2 ∆x - 2∆y – Initial decision parameter po = 2 ∆x - ∆y
d1 d2
y = m. x + b
yk
yk+1
yk+2
yk+3
xk xk+1 xk+2 xk+3
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…15
• For negative slope of a line (|m|>1)
– When moving from left to right
– Steps:
– Initial decision parameter po
= 2 ∆x - ∆y
– If ≤ 0, then point (xk, yk+1) and
Decision parameter pk+1 = pk + 2 ∆x
– If > 0, then point (xk+1, yk+1) and
Decision parameter pk+1 = pk + 2 ∆x - 2∆y – For other orientations or start from right to left, change the sign
of ∆x and ∆y based on change in value of respective x or y
d1 d2
y = m. x + b
yk
yk+1
yk+2
yk+3
xk xk+1 xk+2 xk+3
7/31/2019 CE-201 Lect 4-6 _ZA
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Bresenham’s Line Algorithm…16
• Example: Line 6x + 5y = 30
– Points are (0, 6) and (5, 0) (|m|>1) increment ∆y
– ∆x = 5; ∆y = -6; 2 ∆x = 10; 2 ∆y = -12
– po = 2 ∆x + ∆y = 10 - 6 = 4 point (1, 5) – p > 0; pk+1 = pk + 2 ∆x + 2∆y = 4 + 10 – 12 = 2
point (2, 4)
– p2 = 2 + 10 -12 = 0 point (2, 3)
– p3 = pk + 2 ∆x = 0 + 10 = 10 point (3, 2) – p4 = 10 + 10 - 12 = 8 point (4, 1)
– p5 = 8 + 10 - 12 = 6 point (5, 0)
7/31/2019 CE-201 Lect 4-6 _ZA
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Circle Generating Algorithms
• Properties of circle:
– A circle is defined as the set of points that are all at a given
distance r from a center position ( x c ,y c)
– This distance relationship is expressed by the Pythagoreantheorem in Cartesian coordinates as
(x – xc)2 + (y – yc)
2 = r2 (22)
– For position of points on a circle
• Move from (xc – r) to (xc + r) in unit steps• Calculate corresponding y values as:
y = yc ± √(r2 – (xc – x)2 (23) xc
yc
r
7/31/2019 CE-201 Lect 4-6 _ZA
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Circle Generating Algorithms…1
• This is not the best method for generating a circle.
• One problem with this approach is that it involvesconsiderable computation at each step.
• Moreover, the spacing between plotted pixelpositions is not uniform
• The spacing can be adjusted by interchanging x and ywhenever the absolute value of
slope of the curve is greater than 1.• This is computationally more
cumbersome.
7/31/2019 CE-201 Lect 4-6 _ZA
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Circle Generating Algorithms…2
• Another approach is to use polar coordinates r and θ ,
yielding
x = xc + r Cos θ (24)
y = yc + r Sin θ (25)• Larger angular separations along the circumference can
be connected with straight line segments to approximate
the circular path.
• For a more continues boundary on a raster display, thestep size is set at 1/r. This plots pixel positions that are
approximately one unit apart.
7/31/2019 CE-201 Lect 4-6 _ZA
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Circle Generating Algorithms…3
• Computation can be reduced either by
– Considering symmetry w.r.t. y-axis for segment in 1st and
2nd quadrant, and then considering symmetry w.r.t. x-axis
for segment in 3rd and 4th quadrant or
– Considering the symmetry between adjacent octants
within a quadrant w.r.t 45o line
– Therefore calculations are required
for points between x=0 and x=y
and then these are mapped to rest
eight sectors.
450
(y, x)
(-x, y)
(-y, x)
(-x, -y) (x, -y)
(-y, -x) (y, -x)
(x, y)
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Circle Generating Algorithm…4
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Circle Generating Algorithms…5
• All these procedures require a good deal of
computation time
• More efficient circle algorithms are based on
incremental calculations as in the Bresenham linealgorithm
• Bresenham’s line algorithm is adapted to circlegeneration by setting up decision parameters for
finding the closest pixel to the circumference at eachsampling step
• It avoids square-root calculations by comparing thesquares of the pixel separation distances
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Midpoint Circle Algorithm
• The closest pixel position to the specified circle path
is determined by moving unit distance at each step
• Let radius is r and screen center position is ( x c , y c)
• Set up algorithm to calculate pixel positions around acircle path centered at the coordinate origin (0, 0)
• To move each calculated position ( x, y ) to its proper
screen position add x c
to x and y c
to y
• Along the circle section from x = 0 to x = y in the first
quadrant the slope of the curve varies from 0 to -1
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Midpoint Circle Algorithm…1
• To apply the midpoint method, circle function is
defined as:
f circle (x, y) = x2 + y2 – r2 (26)
Such thatf circle (x, y) < 0, if ( x, y ) is inside the circle boundary
= 0, if ( x, y ) is on the circle boundary
> 0, if ( x, y ) outside the circle boundary
• This becomes the decision parameter for midpointalgorithm
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Midpoint Circle Algorithm…2
• Let us consider the midpoint between the two
candidate pixels at sampling position xk +1.
• After plotting of the pixel at (xk , yk ), next to
determine is whether the pixel at position (xk +1, yk )or the one at position (xk +1, yk -1) is closer to the
circle.
• Decision parameter will be:
(27)
Midpoint
x2+y2-r2=0
xk xk + 1 xk + 2
yk-1
yk
along a circular path.
2
1,1 k k circlek y x f p
22
k 2
k r2
1y1x
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Midpoint Circle Algorithm…3
• If pk<0, this midpoint is inside the circle and the pixel
on 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 scanlineyk-1.
• Successive decision parameters are obtained using
incremental calculations. For pixel location (xk+1+ 1)
(28)
• yk+1 is either yk or yk-1 depending on the sign of pk
1)yy()yy()1x(2pp k 1k 2k
21k k k 1k
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Midpoint Circle Algorithm…4
• Increments for obtaining pk+1 are either 2xk+1 +1 (if pk
is negative) or (2xk+1 + 1 - 2yk+1)
• Evaluation of the terms 2xk+1 and 2yk+1 can also be
done incrementally as2xk+1 = 2xk + 2; 2yk+1 = 2yk - 2
• At the start position (0, r) these two terms have the
values 0 and 2r, respectively.
• Each successive value is obtained by adding 2 to the
previous value of 2 x and subtracting 2 from the
previous value of 2y .
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Midpoint Circle Algorithm…5
• The initial decision parameter is obtained by
evaluating the circle function at the start position
( x 0 , y 0) = (0, r) as:
(29)• If the radius r is specified as an integer, we can simply
round to p0 = 1 – r (for r an integer).
r p
r r
r f p cricle
4
5
2
11
21 ,1
0
2
2
0
or
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Steps
• Input radius r, circle center ( x c , y c), and obtain the first pointon the circumference of a circle centered at the origin as (x0,y0) = (0, r)
• Calculate the initial value of the decision parameter as
p0 = 5/4 – r• 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
• Otherwise, the next point along the circle is (xk+1, yk-1) and
pk+1 = pk + 2xk+1 + 1 - 2yk+1
Where 2xk+1 = 2x k + 2 and 2yk+1 = 2y k - 2
• Determine symmetry points in the seven other octants
• Move pixel position (calculated) to screen position by adding
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y = x
x
0 1 2 3 4 5 6 7 8 9 10
0
1
2
3
4
5
6
7
8
9
10
y
Example
• Plot the generated pixel positions for a circle with
radius r = 10 using midpoint circle algorithm.
– po = 1 – r = -9
– Initial point (0, 10) – Decision parameters
2xo = 0; 2yo = 20
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BRESENHAM’S ALGORITHM
• Assume that (xi , yi ) are the coordinates of the last scan-
converted pixel upon entering step i .
• Let the distance from the origin to pixel T squared minus
the distance to the true circle squared = D(t ).
• Then let the distance from the origin to pixel S squared
minus the distance to the true circle squared = D(s).
• As the coordinates of T are (xi + 1, yi ) and those of S are
(xi + 1, yi -1), the following expression can be developed:
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Choosing Pixels in Bresenham’s Circle Algorithm
(xi + 1, yi)
(xi, + 1, yi - 1)
T(xi, yi)
S
r
y
x
2
2
) 1
(
i
i
y
x
22
11 ii y x
BRESENHAM’S ALGORITHM…1
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BRESENHAM’S ALGORITHM…2
D(t ) = (xi + 1)2 + yi2 - r2
D(s) = (xi+1)2 + (yi – 1)2 – r2
• This function D provides a relative measurement of the
distance from the center of a pixel to the true circle.• Since D(t ) will always be positive (T is outside the true
circle) and D(s) will always be negative (S is inside thetrue circle), a decision variable, d i may be defined as
follows:d i = D(t ) + D(s)
Therefore
d i = 2(xi + 1)2 + yi2 + (yi – 1)2 – 2r2
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BRESENHAM’S ALGORITHM…3
• When d i <0, |D(t )|<|D(s)|, then select pixel T.
• When di ≥ 0, |D(t ) ≥ |D(s)|, then select pixel S.
• The decision variable di +1 for the next step:
d i +1 = 2(xi +1+1)2 + y2i + 1 + (yi+1-1)2 – 2 r2 Hence
d i+1- d i = 2(xi+1 + 1)2 + y2i+1 + (yi+1 - 1)2
– 2(xi
+ 1)2 – y2
i
– (yi
– 1)2
Since xi+1 = xi + 1 so,
d i+1 = d i + 4xi + 2(y2i+1 – y2
i ) – 2(yi+1 - yi ) + 6
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BRESENHAM’S ALGORITHM…4
• If T is the chosen pixel (di < 0) then yi+1=yi and so
d i+1 = di + 4xi + 6
• If S is the chosen pixel (d i ≥ 0) then yi+1 = yi -1and so
d i+1
= d i
+ 4(xi
–yi
) + 10
• Set (0, r ) to be the starting pixel coordinates andcompute the base case value d i :
d 1 = 2 (0 + 1)2
+ r2
+ (r – 1)2
– 2r2
= 3- 2r
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Example
• Given a radius of 10 units.
• Compute di = 2(xi+1)2 + yi2+(yi-1)2-2r2.
• Set the value as (0,r) as the start point
d0 = 2(0+1)2 + 102 + 10-1)2 - 2x102
= 2+100+81-200
= 183-200
= -17. Or = 3 – 2r = 3 – 20 = -17
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Example…1
• As d0 < 0, yi+1=yi.
• So next point (1,10)
d1 = d0 + 4xi + 6.
= - 17+ 4 + 6= -7
• So next point is (2, 10).
d2
= d1
+ 4x2+6
= -7+8+6.
=+7.
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Example…2
• So next point (3, 9).
d3 = di+4(xi-yi)+10
= 7+4(3-9)+10
= 7-24+10= -7.
• So next point (4, 9)
d4 = -7+16+6 = 15.
• So next point (5, 8).
d5 = 15+4(5-8)+10 = 13.
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C i
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Comparison
Iteration Mid point Bresenham
0 (0,10) (0,10)
1 (1,10) (1,10)
2 (2,10) (2,10)
3 (3,10) (3,9)
4 (4,9) (4,9)
5 (5,9) (5,8)
6 (6,8) (6,7)
7 (7,7) (7,6)
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Coordinate Representations
• General graphics packages are designed to be used with CartesianCoordinate specifications.
• If coordinate values for a picture are specified in some otherreference frame, they must be converted to Cartesian system.
• Several different Cartesian reference frames are used to construct anddisplay a scene:
– Modeling or local coordinates
– World coordinates
– Device or screen coordinates
• Graphics system first converts world coordinates to normalizedcoordinates before final conversion to device coordinates.
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65
Modeling Coordinates
Coordinate Representations
ModelingTransformations
FRAME BUFFER
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FRAME BUFFER
A frame buffer is a video output device that drives a video display from amemory buffer containing a complete frame of data. The information in thememory buffer typically consists of color values for every pixel (point that can bedisplayed) on the screen.
Color values are commonly stored in 1-bit binary (monochrome), 4-bit palettized,8-bit palettized, 16-bit highcolor and 24-bit truecolor formats.
The total amount of the memory required to drive the frame buffer depends onthe resolution of the output signal, and on the color depth and palette size.
Loading Frame Buffer
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g
• When objects are scan converted for display with
raster system, frame buffer positions must becalculated.
• Scan conversion algorithms generate pixel positionsat successive unit intervals, this allows the use of incremental methods to calculate frame buffer
addresses.• This is accomplished with setPixel procedure, which
stores intensity values for the pixels at thecorresponding addresses within the frame-bufferarray.
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Frame-buffer array is addressed in row-major order
Pixel screen positions stored linearly in row-major order within the frame buffer
For 1 bit per pixel, the frame buffer addressfor pixel position (x,y) is calculated as
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)+xmax+2
SOFTWARE STANDARDS
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SOFTWARE STANDARDS
The primary goal is portability among the graphics softwares.
Without standards, programs designed for one hardware system can not betransferred to another system without extensive rewriting of programs.
Several Int. & national standards planning organizations have cooperatedand developed Graphical Kernel System (GKS). The is adopted by Int.Standards organisation (ISO) and American National Standards Institute
(ANSI). Other system is PHIGS (Programmer’s Hierarchial Interactive Graphics
Standard), which is extension of GKS.
PHIGS+ is developed to provide 3D surface-shading.
Standard graphics functions are defined as a set of specifications that is in-dependent to any programming language.
PHIGS polyline(n,x,y)Fortran call gpl(n,x,y)C ppolyline (n pts)