Scan Conversion or Rasterization
• Drawing lines, circles, and etc. on a grid implicitly
involves approximation.
• The general process: Scan Conversion or Rasterization
• Ideally, the following properties should be considered
–
–
–
–
–
smooth
continuous
pass through specified points
uniform brightness
efficient
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Line Drawing and Scan Conversion
• There are three possible choices which are potentially
useful.
• Explicit: y = f(x)
– y = m (x - x0) + y0 where m = dy/dx
• Parametric: x = f(t), y = f(t)
– x = x0 + t(x1 - x0),
– y = y0 + t(y1 - y0)
t in [0,1]
• Implicit: f(x, y) = 0
– F(x,y) = (x-x0)dy - (y-y0)dx
– if F(x,y) = 0 then (x,y) is on line
–
F(x,y) > 0 then (x,y) is below line
–
F(x,y) < 0 then (x,y) is above line
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Line Drawing - Algorithm 1
A Straightforward Implementation
DrawLine(int x1,int y1, int x2,int y2, int color)
{
float y;
int x;
}
for (x=x1; x<=x2; x++)
{
y = y1 + (x-x1)*(y2-y1)/(x2-x1)
WritePixel(x, Round(y), color );
}
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Line Drawing - Algorithm 2
A Better Implementation
DrawLine(int x1,int y1,int x2,int y2, int color)
{
float m,y;
int dx,dy,x;
dx = x2 - x1;
dy = y2 - y1;
m = dy/dx;
y = y1 + 0.5;
for (x=x1; x<=x2; x++)
{
WritePixel(x, Floor(y), color );
y = y + m;
}
}
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Line Drawing Algorithm Comparison
• Advantages over Algorithm 1
– eliminates multiplication
– improves speed
• Disadvantages
–
–
–
–
–
–
round-off error builds up
get pixel drift
rounding and floating point arithmetic still time consuming
works well only for |m| < 1
need to loop in y for |m| > 1
need to handle special cases
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Line Drawing - Midpoint Algorithm
• The Midpoint or Bresenham’s Algorithm
– The midpoint algorithm is even better than the above
algorithm in that it uses only integer calculations. It treats
line drawing as a sequence of decisions. For each pixel that
is drawn the next pixel will be either N or NE, as shown
below.
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Midpoint Algorithm
• The midpoint algorithm makes use of
the implicit definition of the line,
F(x,y) =0. The N/NE decisions are
made as follows.
• d = F(xp + 1, yp + 0.5)
– if d < 0 line below midpoint choose E
– if d > 0 line above midpoint choose NE
• if E is chosen
– dnew = F(xp + 2, yp + 0.5)
– dnew- dold = F(xp + 2, yp + 0.5) F(xp + 1, yp + 0.5)
– Delta = d new -d old = dy
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Midpoint Algorithm
• If NE is chosen
– dnew = F(xp+2, yp+1.5)
– Delta = dy-dx
• Initialization
– dstart = F(x0+1, y0+0.5)
= (x0+1-x0)dy - (y0+0.5-y0)dx
= dy-dx/2
• Integer only algorithm
– F’(x,y) = 2 F(x,y) ; d’ = 2d
– d’start = 2dy - dx
– Delta’ = 2Delta
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Midpoint Algorithm for x1 < x2 and slope <= 1
DrawLine(int x1, int y1,
int x2, int y2, int color)
{
int dx, dy, d, incE, incNE, x, y;
dx = x2 - x1;
dy = y2 - y1;
d = 2*dy - dx;
incE = 2*dy;
incNE = 2*(dy - dx);
y = y1;
}
for (x=x1; x<=x2; x++)
{
WritePixel(x, y, color);
if (d>0) {
d = d + incNE;
y = y + 1;
} else {
d = d + incE;
}
}
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General Bressenham’s Algorithm
• To generalize lines with arbitrary slopes
– consider symmetry between various octants and quadrants
– for m > 1, interchange roles of x and y, that is step in y
direction, and decide whether the x value is above or below
the line.
– If m > 1, and right endpoint is the first point, both x and y
decrease. To ensure uniqueness, independent of direction,
always choose upper (or lower) point if the line go through
the mid-point.
– Handle special cases without invoking the algorithm:
horizontal, vertical and diagonal lines
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Scan Converting Circles
• Explicit: y = f(x)
y   R2  x2
Usually, we draw a quarter circle by
incrementing x from 0 to R in unit steps
and solving for +y for each step.
• Parametric:
x  R cos 
y  R sin 
- by stepping the angle from 0 to 90
- avoids large gaps but still insufficient.
• Implicit: f(x) = x2+y2-R2
If f(x,y) = 0 then it is on the circle.
f(x,y) > 0 then it is outside the circle.
f(x,y) < 0 then it is inside the circle.
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Eight-way Symmetry
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Midpoint Circle Algorithm
• dold = F(xp+1, yp+0.5)
• If dold < 0, E is chosen
– dnew = F(xp+2, yp-0.5)
= dold+(2xp+3)
– DeltaE = 2xp+3
• If dold >= 0, SE is chosen
– dnew = F(xp+2, yp-1.5)
= dold+(2xp-2yp+5)
– DeltaSE = 2xp-2yp+5
• Initialization
– dinit
= 5/4 – R
=1-R
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Midpoint Circle Algorithm (cont.)
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Scan Converting Ellipses
F ( x, y)  b2 x 2  a 2 y 2  a 2b2  0
•
•
•
•
2a is the length of the major axis along the x axis.
2b is the length of the minor axis along the y axis.
The midpoint can also be applied to ellipses.
For simplicity, we draw only the arc of the ellipse that lies in the
first quadrant, the other three quadrants can be drawn by
symmetry
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Scan Converting Ellipses: Algorithm
j > i in region 1
j < i in region 2
• Firstly we divide the quadrant into two regions
• Boundary between the two regions is
– the point at which the curve has a slope of -1
– the point at which the gradient vector has the i and j components of
equal magnitude
grad F ( x, y)  F / x i  F / y j  2b2 x i  2a 2 y j
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Ellipses: Algorithm (cont.)
• At the next midpoint, if a2(yp-0.5)<=b2(xp+1), we switch region 1=>2
• In region 1, choices are E and SE
– Initial condition: dinit = b2+a2(-b+0.25)
– For a move to E, dnew = dold+DeltaE with DeltaE = b2(2xp+3)
– For a move to SE, dnew = dold+DeltaSE with
DeltaSE = b2(2xp+3)+a2(-2yp+2)
• In region 2, choices are S and SE
– Initial condition: dinit = b2(xp+0.5)2+a2((y-1)2-b2)
– For a move to S, dnew = dold+Deltas with Deltas = a2(-2yp+3)
– For a move to SE, dnew = dold+DeltaSE with
DeltaSE = b2(2xp+2)+a2(-2yp+3)
• Stop in region 2 when the y value is zero.
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Midpoint Ellipse Algorithm
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Midpoint Ellipse Algorithm (cont.)
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