Computer Graphics_Chapter 2 Line Draing Algorithms

7/24/2019 Computer Graphics_Chapter 2 Line Draing Algorithms

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COMPUTER GRAPHICSLINE DRAWING ALGORITHM

Dr. Zeeshan Bhatti

BSIT-PIV

Chapter 2

Institute of Information and Communication TechnologyUniversity of Sindh, Jamshoro BY: ZEESHAN BHATTI 1


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POLYNOMIAL ALGORITHM

BY: ZEESHAN BHATTI 2


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DDA (DIGITAL DIFFERENCE ANALYSER)FOR LINE

Let endpoints coordinates are

(x0, y0) and (x1, y1) respectively.

Se we have

=

0

0

=

=

=

As already mentioned, in digital systems the coordinates of points arerepresented by pixels positions. The pixels are considered to beplaced at a distance of unity from each other.



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DDA FOR LINE

A very simple procedure for scan converting a straight line is given asfallows:

Step 1: Plot the first pixel at (x,y) = (x0, y0)

Step 2: Increment the x and y values by dx and dy respectively to getthe next pixel. Since in digital systems pixel are a unit distance away,we can consider either dx or dy to be unity and compute the other.

Plot pixel at ( x+dx, y+dy).

Step 3: Continue the above step until the final pixel (x1, y1) isreached.



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Depending on weather we select dx or dy to be unity , two differentsituations can appear.

If we consider dx = 1, the following relations are used to compute thenext y-value.

yi = mxi + b

yi+1 = mxi+1 + b

=m (xi + x) + b

= yi +mx

If x = 1 , then

yi+1 = yi + m



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Similarly, if we consider dy = 1, the following relations are used tocompute the next x-value.

If y = 1, then

Xi+1 = xi +

The above algorithm is called Digital Difference Analyzer (DDA). TheDDA is actually a mechanical device that solves differential equationsby simultaneously incrementing x and y by small steps proportional tothe first derivative of x and y.

In our algorithm when x is incremented by dx, y is incremented by dy.



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USING SYMMETRY



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For lines with |m| < 1, we are to increment the xvalue, at eachsteps, by unity. For other lines, we increment the y-0values.

If x = 1,

then y = m, where

= m.

If y = 1,

then x =

, where

=

.



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DDA ALGORITHM

Digital Differential Analyzer (DDA) algorithm is the simple line generation

algorithm which is explained step by step here.

Step 1 Get the input of two end points (X0,Y0) and (X1,Y1).

Step 2 Calculate the difference between two end points.

dx = X1 - X0

dy = Y1 - Y0

Step 3 Based on the calculated difference in step-2, you need to identify thenumber of steps to put pixel. If dx > dy, then you need more steps in x

coordinate; otherwise in y coordinate.if (dx > dy)

Steps = absolute(dx);else

Steps = absolute(dy);



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Step 4 Calculate the increment in x coordinate and y coordinate.Xincrement = dx / (float) steps;

Yincrement = dy / (float) steps;

Step 5 Put the pixel by successfully incrementing x and ycoordinates accordingly and complete the drawing of the line.

for(int v=0; v < Steps; v++)

{

x = x + Xincrement;

y = y + Yincrement;

putpixel(x,y);

}



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PROBLEM: DRAW A LINE USING DDA ALGORITHM FORTHE POINTS BETWEEN 2,5 AND 6, 9. DETERMINE THE

VALUES OF PIXEL AND PLOT THE GRAPH.

Step 1:

x0 = 2 x1 = 6

y0 = 5 y1 = 9

Step 2:

dx = x1 - x0

dx = 6-2 = 4 dx= 4

dy = y1 - y0

dy = 9 - 5 = 4 dy=4



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Step 3:

Because dy is greater than & equal to dx i.e. 4=4, so we use the value of dy

Steps = dySteps = 4

Step 4:

xInc = dx / steps

yInc = dy / steps

xInc = 4 / 4 = 1

yInc = 4 / 4 = 1

Step 5:

x = x + xInc y = y + yInc

2 +1 = 3 5+1 = 6

3 + 1 = 4 6+1 = 7

4+1 = 5 7+ 1 = 8

5 +1 = 6 8+1 = 9



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BRESENHAMS LINE GENERATION

The Bresenham algorithm is anotherincremental scan conversion algorithm.The big advantage of this algorithm isthat, it uses only integer calculations.

Moving across the x axis in unitintervals and at each step choosebetween two different y coordinates.

For example, as shown in the followingillustration, from position (2, 3) youneed to choose between (3, 3) and (3,4). You would like the point that iscloser to the original line.



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DDA requires one floating point addition per step

We can eliminate all fp through Bresenhams algorithm

Consider only 1 m 0

- Other cases by symmetry

Assume pixel centers are at half integers

If we start at a pixel that has been written, there are only twocandidates for the next pixel to be written into the frame buffer



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CANDIDATE PIXELS



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DECISION VARIABLE



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SOLVING BRESENHAM

Let at any moment the Pixel I(xi, yi) as initial pixel and the next pixel is(xi+1, yi+1).

Since the pixel are at unit distance, therefore xi+1 = xi +1, but thevalue of yi+1 can be P(xi+1, yi) or Q (xi+1, yi+1) depending on thenearest point of interaction denoted by R(xi+1, yi+1 ), which is theactual location of the point on the line.



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BRESENHAM LINE DRAWING ALGORITHM

Step 1:Get the values of x0,y0 and x1,y1dx = x

1

x0dy = y1y0

Step 2:

p = 2dydx

E (Q) = 2dy

NE (P) = 2* (dydx)

Step 3: Plot the Pixel at I(x0, y0)



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Step 4:

while ( x0 < x1)

if ( p < 0) thenp = p + Ex0 = x0 +1

EndIFElse

p = p + NEx

0

= x0

+1y0 = y0+1

EndElse

PlotPixel (x0, y0)

EndWhile



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PROBLEM: SOLVE BRESENHAM LINE DRAWING

ALGORITHM FOR VALUS (2, 3) (8, 7)

Step 1:Get the values of

x0

= 2, y0

=3

X1 = 8 , y1 = 7

dx = x1x0 = 8 - 2 = 6

dy = y1y0 = 7- 3 = 4

Step 2:

p = 2dydx = 2 (4)6 = 2

E (Q) = 2dy = 2(4) = 8

NE (P) = 2* (dydx) = 2(4-6) = -4

Step 3: Plot the Pixel at I(x0, y0) BY: ZEESHAN BHATTI 20

2 3


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Step 4:

while ( x0 < x1)

if ( p < 0) thenp = p + Ex0 = x0 +1

EndIFElse

p = p + NEx

0

= x0

+1y0 = y0+1

EndElse

PlotPixel (x0, y0)

EndWhile

2 3

P P < 0

p = p + E

x0 = x0 +1

y0 = y0

P => 0

p = p + NE

x0 = x0 +1

y0 = y0+1

X Y

2 (step 1) 3 (step 2) 4 (step 3)

2+(-4) = -2(step 4)

-2 4 4

-2 + 8 = 6

6 5 5

6 + (-4) = 2

2 6 6

2 +(-4) = -2

-2 7 6

-2 +8 = 66 8 7

6 + (-4) = 2

2 9 8

2 +(-4) = -2

-2 10 8

p = 2

E = 8NE = -4



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PROBLEM: SOLVE BRESENHAM LINE DRAWING

ALGORITHM FOR VALUS (10, 16) (18, 20)

Step 1:Get the values of

x0

= 10, y0

=16

X1 = 18 , y1 = 20

dx = x1x0 = 18-10 = 8

dy = y1y0 = 20-16 = 4

Step 2:

p = 2dydx = 2 (4)8 = 0

E (Q) = 2dy = 2(4) = 8

NE (P) = 2* (dydx) = 2(4-8) = -8

Step 3: Plot the Pixel at I(x0, y0) = (10, 16) BY: ZEESHAN BHATTI 22

10 16


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Step 4:

while ( x0 < x1)

if ( p < 0) thenp = p + Ex0 = x0 +1

EndIFElse

p = p + NEx0 = x0 +1y0 = y0+1

EndElse

PlotPixel (x0, y0)

EndWhile

p = 0

E = 8

NE = -8

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P P < 0

p = p + E

x0 = x0 +1

y0 = y0

P => 0

p = p + NE

x0 = x0 +1

y0 = y0+1

X Y


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P P < 0

p = p + E

x0 = x0 +1

y0 = y0

P => 0

p = p + NE

x0 = x0 +1

y0 = y0+1

X Y

0 11 17

0 + -8 = -8

-8 12 17

-8 + 8 = 0

0 13 18

0+ -8 = -8

-8 14 18

-8 + 8 = 0

0 15 19

0+ -8 = -8

-8 16 19

-8 + 8 = 0

0 17 20

0+ -8 = -8

-8 18 20

p = 0

E = 8

NE = -8



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POLYNOMIAL METHOD FOR CIRCLES

The polynomial method is the simplest approach.

In polynomial, we solve the equation of a circle for known values on

one of the coordinates, and thus determine the calues of the othercoordinate.

The equation of a circle at centre (xc, yc) and radius r is given as

(x-xc)2 + (y - yc)

2 = r2

Therefore

= 2 2



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SCAN CONVERSION - CIRCLES



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CIRCLE MIDPOINT (FOR ONE OCTANT)



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BRESENHAMS CIRCLE ALGORITHM

For simplicity We Assume that the center of a circle is at the origin, i.e.

xc = yc = 0

So, the equation of circle, i.e. (polynomial equation becomes)

x2 + y2 = r2



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ALGORITHM FOR BRESENHAMS CIRCLE



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PROBLEM: TRACE BRESENHAMS ALGORITHM

FOR DRAWING A CIRCLE OF RADIUS 6.To solve, first calculate the initial values of p0 of the decisionparameter, and then recalculating the p at each successive iteration.

Step 1:

x=0; // assumedy= radius;r = radius;

p0 = 32r = 3 -12 = -9

circlepoints(x,y);

Step 2:

-9 0 6

P P < 0p = p + 4x + 6

x0 = x0 +1

y0 = y0

P => 0p = p + 4(x-y)+10

x0 = x0 +1

y0 = y0 - 1

X Y

-9 + 4(0)+6 = -3

-3 1 6

-3 + 4(1)+6= 7

7 2 5

7 + 4 (2 5) +10 = 5

5 3 4

5 + 4 ( 3-4) +10 = 11

11 4 3BY: ZEESHAN BHATTI 32


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PROBLEM: TRACE BRESENHAMS ALGORITHM

FOR DRAWING A CIRCLE OF RADIUS 11.To solve, first calculate the initial values of p0 of the decisionparameter, and then recalculating the p at each successive iteration.

Step 1:

x=0; // assumedy= radius;r = radius;

p0 = 32r =

circlepoints(x,y);

Step 2:

P P < 0p = p + 4x + 6

x0 = x0 +1

y0 = y0

P => 0p = p + 4(x-y)+10

x0 = x0 +1

y0 = y0 - 1

X Y



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COMPLETING THE CIRCLES BY

MIRRORING THE 8 OCTANTS

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Computer Graphics_Chapter 2 Line Draing Algorithms

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