Scan Converting Ellipse

7/30/2019 Scan Converting Ellipse

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Scan Converting Ellipse

General equation of an ellipse:

d1+ d2= constant

Or,

However, we will only consider standard ellipse:

(x, y)

F1

F2

d1

d2

constant222

2

2

1

2

1 yyxxyyxx

1

22

b

yy

a

xx cc

a

b


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Formally An ellipse is defined as the locus of points

satisfying following equation: r from a fixed point (h,k).

This distance is described In the Pythagoras theorem as :

where (h,k) is the centre, a and b are the length of major

and minor axis respectively

The standard equation for the ellipse at centre (0,0) is:

1)(

2

2

2

2

b

ky

a

h)(x

12

2

2

2

b

y

a

x


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So, we can write a simple circle drawing algorithm by

solving the equation fory at unitx intervals using:

Which is not a good solution. (you know that)

))/(1.( 22 axby


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Four-Way Symmetry Like circle, ellipse centred at (0, 0) follows symmetry. But

now it isfour symmetry.

Thus ellipse points are computed

in the first quadrant,

rest three of the ellipse points

are found by symmetry.

Centre can be shifted while

plotting

(x,y)(-x,y)

(-x,-y)(x,-y)

X-axis

Y-axis

o


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So we define a routine that plots ellipse pixels with centre

(h,k) in all the four quadrants

put_ellipse_pixel(x,y,h,k)

{

Put_pixel(h+x, k+y)

Put_pixel(h-x, k+y)

Put_pixel(h+x, k-y)

Put_pixel(h-x, k-y)}


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Ellipse Drawing Algorithms

1. Polar Domain Algorithms

2. Midpoint Ellipse Algorithm3. Bresenham Ellipse Algorithm


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Polar Domain Algorithm

We could use polar coordinates r and ,

x =h + a cos y = k + b sin

where (h,k) is the centre, a and b are the length of major and minor axis

respectively

A fixed angular step size can be used to plot equally spaced

points along the circumference

A step size of 1/a or 1/b, whichever is smaller can be used to

set pixel positions to approximately 1 unit apart for a

continuous boundary.


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We could also use incremental form. Let P(xk,yk) be currentlyplotted point,

xk = a cos()

yk= b sin()

At next iteration k = k+1 ; = +Dxk+1 = a cos(+ D)

= acos() cos(D) - a sin() sin(D)

= xkcos(D) - yk(a/b)sin(D)

yk+1 = b sin(+D)

= b sin() cos(D) - b cos() sin(D)

= ykcos(D) + xk(b/a)sin(D)


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STEPS

1. Input centre (h,k),anda,b

2. If (a>b) D = 1/a else D = 1/b ;

3. C = cos(D);S = sin(D); t = a/b;

4. Initialize x = 0;y = b;= 0;

5. WHILE ( < /2)

{

xtemp = x

x = x*C-t*y*S;

y = y*C+(1/t)*xtemp*S;put_ellipse_pixel(x,y,h,k);

= + D;

}

This method is still computationally expensive


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2. Midpoint Ellipse Algorithm

3. Bresenham Ellipse Algorithm


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Midpoint Ellipse Algorithm

Reconsider an ellipse centered at the origin:

The discriminator function?

and its properties:

fe(x,y) < 0 for a point inside the ellipse

fe(x,y) > 0 for a point outside the ellipse

fe(x,y) = 0 for a point on the ellipse

222222, bayaxbyxfe

1

22

b

y

a

x


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Ellipse is different from circle.

Similar approach with circle, different

is sampling direction.

Region 1: Sampling is atx direction

Choose between (xk+1, yk), or (xk+1, yk-1)

Midpoint: (xk+1, yk-0.5)

Region 2:

Sampling is aty direction

Choose between (xk, yk-1), or (xk+1, yk-1)

Midpoint: (xk+0.5, yk-1)

Slope =

-1

Region 1

Region 2


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Region 1:

p1kve:

midpoint is inside

choose pixel (xk+1

, yk)

p1k

+ve:

midpoint is outside

choose pixel (xk+1

, yk-1)

21,11 kkek yxfp

Region 2:

p2kve:

midpoint is inside

choose pixel (xk+1, y

k-1)

p2k+ve:

midpoint is outside

choose pixel (xk, y

k-1)

1,2 21 kkek yxfp

Decision Parameters


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Derivation for Region 1:

Let us assume that P(xk, yk )

is the currently plotted pixel.

Q(xk+1, yk+1 ) (xk+1, y ) is

the next point along the actualcircle path. We need to decide

next pixel to be plotted from

among candidate positions

Q1(xk+1, yk) orQ2(xk+1, yk-1)


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Our decision parameter is the circle function evaluated at the

midpoint between these two pixels

p1k= fe (xk+1, yk-1/2) = b2(xk+1)

2 + a2(yk-1/2)2a2 b2

Ifp1k< 0 ,

this midpoint is inside the ellipse and

the pixel on the scan lineykis closer to the ellipse boundary.

Otherwise, the mid position is outside or on the ellipse boundary,

and we select the pixel on the scan lineyk-1


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Successive decision parameters are obtained using incrementalcalculations

p1k = b2 (xk+1)

2 + a2 (yk )2a2b2

Put k = k+1p1k+1 = b

2 [(xk+1)+1]2 + a2(yk+1 )

2a2b2

= b2 (xk+2)2 + a2(yk+1 )

2a2b2

subtracting pk from pk+1

p1k+1p1k= b2(xk+2)2 + a2(yk+1 )2[b2(xk+1)2 + a2(yk )2]or

p1k+1 = p1k+2 b2(xk+1) + a

2 [(yk+1)2(yk)

2]+b2

Whereyk+1is eitherykoryk-1, depending on the sign of p1k


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Ifp1k< 0

Q1(xk+1, yk) was the next choice

yk+1 = yk

(yk+1

)

2

(yk

)

2

= 0p1k+1 = p1k+b

2(2.xk+3)

else

Q2(xk+1, yk-1) was the next choice

yk+1 = yk1(yk+1)

2(yk)2=2.yk+2

p1k+1 = p1k+b2(2.xk+3) + a

2(2.yk+ 2)


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Midpoint Ellipse algorithm

Initial decision parameter is obtained by evaluating the circle

function at the start position (x0,y0) = (0,b)

p10 = fe(1, b1/2)

= b2a2b + a2


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Derivation for Region 2:

Let us assume that P(xk, yk )

is the currently plotted pixel.

Q(xk+1, yk+1 ) (xk+1, y ) is

the next point along the actualcircle path. We need to decide


among candidate positions

Q1(xk, yk-1) orQ2(xk+1, yk-1)


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Our decision parameter is the circle function evaluated at the

midpoint between these two pixels

p2k= fe (xk+1/2, yk-1) = b2(xk+1/2)

2 + a2(yk1)2a2 b2

Ifp2k


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Successive decision parameters are obtained using incrementalcalculations

p2k = b2 (xk+)

2 + a2 (yk1 )2a2b2

Put k = k+1

p2k+1 = b2 [(xk+1)+]

2 + a2(yk+11 )2a2b2

= b2 [(xk+1)+]2 + a2(yk2 )

2a2b2

subtracting pk from pk+1

p2k+1

p2k= b2(x

k+1+)2 + a2(y

k2 )2[b2(x

k+)2 + a2(y

k1 )2]

or

p2k+1 = p2k+b2[(xk+1+)

2(xk+)2 ]a2 (2.yk

3)

Wherexk+1is eitherxkorxk+1, depending on the sign of p2k


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Ifp2k< 0

Q2(xk+1, yk-1) was the next choice

xk+1 = xk+1

(xk+1+)2

(xk+)2

= 2.xk+2p2k+1 = p2k+b

2(2.xk+2)a2(2.yk3)

else

Q1(xk, yk1) was the next choice

xk+1 = xk(xk+1+)

2(xk+)2= 0

p2k+1 = p2ka2(2.yk3)


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Initial decision parameter is obtained by evaluating the circlefunction at the start position (x0,y0) the last point plotted afterdrawing region 1

p20 = fe(x0+, y01)

= b2(x0+)2 + a2(y01)

2a2b2


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Condition to move from one region to another

Starting at point (0,b) in the region 1 where slope is 1. At each

step we need to check slope.

The ellipse slope can be calculated as At the boundary of region 1 and 2

dy/dx = -1

2b2x = 2a2y

We move out of region 1 if 2b2x 2a2y

ya

xb

dx

dy2

2

2

2


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1. Input a, b and ellipse center (xc, y

c). First point on

the similar ellipse centered at the origin is (0, b).

2. Initial value for decision parameter at region 1:

3. At each xk in region 1, starting from k = 0, testp1k :Ifp1

k< 0, next point (x

k+1, y

k) and

else, next point (xk+1, y

k-1) and

4. Determine symmetry points in the other 3 octants.

5. Get the actual point for ellipse centered at (xc, y

c)

that is (x+ xc, y+ y

c).

6. Repeat step 3 - 6 until 2b2x 2a2y.

2

4

122

01 ababp

,3211 221 bxbpp kkk

.2322112222

1 abyaxbpp kkkk


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7. Initial value for decision parameter in region 2:

8. At each ykin region 2, starting from k = 0, testp2

k:

Ifp2k> 0, next point is (x

k, y

k-1) and

else, next point is (xk+1, y

k-1) and


10.Get the actual point for ellipse centered at (xc, y

c)


c).

11.Repeat step 8 - 10 until y< 0.

22

1 3222 ayapp kkk

222022

21

02

0 12 bayaxbp

.2322222222

1bayaxbpp kkkk


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2. Midpoint Ellipse Algorithm

3. Bresenham Ellipse Algorithm


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Bresenhams Ellipse Algorithm

Reconsider an ellipse centered at the origin:

or

We will define the distance from the actual value to two

candidate positions and proceed as in circle algorithm.

1

22

b

y

a

x

)2)/(1.(22 axby


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Derivation for Region 1

Let us assume that P(xk, yk ) is

the currently plotted pixel.

Q(xk+1, yk+1 ) (xk+1, y ) is the

next point along the actual

circle path. We need to decide


two candidate positions

Q1(xk+1, yk) or Q2(xk+1, yk-1)

Q1

Q2

P


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Bresenhams Ellipse AlgorithmThus actual value ofy atx = x

k+1is given by

y2 = b2 (1((xk+1) /a)

2 )

We define the distance measure in squares to simplify calculations

Let d1 =yk2y

2

=yk2

[b

2

(1((xk+1) /a)2

)]d2 =y2

(yk-1)2

= [b2 (1((xk+1) /a)2 )](yk-1)

2

The difference in d1 and d2 determine the next pixel to be plottedif(d1-d2)


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Bresenhams Ellipse AlgorithmWe can define a decision parameterp

kfor the kth step as follows:

p1k= ( d1-d2)

=2. b2(1(xk+1)2/a2 )+yk

2 + (yk1)2

To putpk in recurrence relation replace k = k+1

p1k+1 =

2. b

2

(1

(xk+1 +1)2

/a

2

)+yk+12

(yk+1

1)2

=2. b2(1(xk+2)2/a2 )+yk+1

2(yk+11)2

From above two equations

p1k+1p1k= ?

p1k+1 = p1k+ (b2/a2)(4xk+6) +[y

2k+1y

2k][(yk+1-1)

2(yk-1)2]


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Ifp1k< 0

Q1(xk+1, yk) was the next choice

yk+1 = yk[y

k+1

2yk

2][(yk+1

-1)

2(yk

-1)

2 ] = 0

p1k+1 = p1k+ (b2/a2)(4.xk+6)

else

Q2(xk+1, yk1) was the next choice

yk+1 = yk1[yk+1

2yk2][(yk+1 -1)

2(yk-1)2 ] =4.yk+4

p1k+1 = p1k+ (b2/a2)(4.xk+6)(4.yk4)


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The first parameterp0 is directly computed from:

Pk=2. b2(1(xk+1)

2/a2 )+yk2 + (yk1)

2

By putting k = 0,xk= 0 andyk= bp0 =2. b

2(1(0+1)2/a2 )+b2 + (b1)

2

= 2.b2/a22.b + 1

Let us derive it for region 2


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Derivation for Region 2

Let us assume that P(xk, yk ) is

the currently plotted pixel.

Q(xk+1, yk+1 ) (x, yk+1 ) is the

next point along the actual

circle path. We need to decide


two candidate positions

Q1(xk, yk-1 ) or Q2(xk+1, yk-1)


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Bresenhams Ellipse AlgorithmThus actual value ofx aty = y

k+1= y

k1is given by

x2 = a2 (1((yk1) /b)

2 )

We define the distance measure in squares to simplify calculations

Let d1 =x2

xk2

= [a2 (1((yk1) /b)

2 )]x

k

2

d2 = (xk+ 1)

2x2= (xk+1)

2[a2 (1((yk1) /b)2 )]

The difference in d1

and d2

determine the next pixel to be plotted

if(d1-d2)>0 Q1(xk, yk1 ) is plotted

else Q2(xk+1, yk1) is plotted.


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Bresenhams Ellipse AlgorithmWe can define a decision parameterp2

kfor the kth step as follows:

p2k= ( d1-d2)

= 2. a2(1(yk-1)2/b2 )xk

2(xk+ 1)2

To putp2k in recurrence relation replace k = k+1p2

k+1=2. a2(1(y

k+11)

2/b2 )xk+1

2(xk+1

+ 1)

2

= 2. a2(1(yk2)2/b2 )xk+1

2(xk+1 + 1)2

From above two equations

p2k+1p2k= ?

p2k+1 = p2k(a2/b2)(4yk6)[x

2k+1 +x

2k][(xk+1+1)

2+ (xk+1)2]


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Ifp2k> 0

Q1(xk, yk1) was the next choice

xk+1 = xk[x

k+1

2xk

2][(xk+1

+1)

2(xk

+1)

2 ] = 0

p2k+1 = p2k(a2/b2)(4.yk6)

else

Q2(xk+1, yk1) was the next choice

xk+1 = xk+1[xk+1

2xk2][(xk+1+1)

2(xk+1)2 ] = 4.xk+4

p2k+1 = p2k(a2/b2)(4.yk6) + (4.xk + 4)


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The first parameterp20 is directly computed from:

p2k=2. a2(1(yk-1)

2/b2 )xk2(xk+ 1)

2

Where (xk,yk) is the last point plotted in region .

Write the algorithm, implement in lab and report me how the

output of this algorithm is different from Midpoint Algorithm


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Brsenhams Ellipse Algorithm

1. Input a, b and ellipse center (xc, yc). First point onthe similar ellipse centered at the origin is (0, b).

2. Initial value for decision parameter at region 1:

3. At each xk in region 1, starting from k = 0, testp1k :Ifp1

k< 0, next point (x

k+1, y

k) and

else, next point (xk+1, y

k-1) and


5. Get the actual point for ellipse centered at (xc, y

c)


c).

6. Repeat step 3 - 6 until 2b2x 2a2y.

.21.2122

0babp

).3.2(211 221 kkk xabpp

).1(4)3.2(21122

1 yxabpp

kkk


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7. Initial value for decision parameter in region 2:

8. At each ykin region 2, starting from k = 0, testp2

k:

Ifp2k> 0, next point is (x

k, y

k-1) and

else, next point is (xk+1, y

k-1) and


10.Get the actual point for ellipse centered at (xc, y

c)


c).

11.Repeat step 8 - 10 until y< 0.

).6.4)((22 221 kkk ybapp

222220 )1()1(1.22 xxbyap

).4.4()6.4(22 221 kkkk xyabpp

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Scan Converting Ellipse