scan conversion 1

7/30/2019 Scan Conversion 1

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I N T R O D U C T I O N T O C O M P U T E R G R A P H I CS

Andries van Dam September 23, 2008 Scan Conversion 1 #/18

Scan Conversion 1(pages 72 -78 of textbook)


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Line Drawing Draw a line on a raster screen between two

points

Whats wrong with statement of problem?

doesnt say anything about which points areallowed as endpoints

doesnt give a clear meaning of draw

doesnt say what constitutes a line in rasterworld

doesnt say how to measure success ofproposed algorithms

Problem Statement

Given two points Pand Q in XY plane, bothwith integer coordinates, determine whichpixels on raster screen should be on in orderto make picture of a unit-width line segmentstarting at Pand ending at Q

Scan Converting Lines


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Special case: Horizontal Line:

Draw pixel Pand incrementxcoordinatevalue by 1 to get next pixel.

Vertical Line:

Draw pixel Pand increment ycoordinatevalue by 1 to get next pixel.

Diagonal Line:

Draw pixel Pand increment bothxand y

coordinate by 1 to get next pixel.

What should we do in general case?

Increment x coordinate by 1 andchoose point closest to line.

But how do we measure closest?

Finding next pixel:


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Why can we use vertical distance as measureof which point is closer?

because vertical distance is proportional toactual distance

how do we show this?

with similar triangles

By similar triangles we can see that true

distances to line (in blue) are directlyproportional to vertical distances to line (inblack) for each point

Therefore, point with smaller vertical distanceto line is closest to line

Vertical Distance

(x1, y1)

(x2, y2)


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Basic Algorithm

Find equation of line that connects twopoints Pand Q

Starting with leftmost point P, incrementxiby 1 to calculate yi= m*xi+ B

where m = slope, B = yintercept

Draw pixel at (xi, Round(yi)) where

Round (yi) = Floor (0.5 + yi)

Incremental Algorithm:

Each iteration requires a floating-pointmultiplication

Modify algorithm to use deltas (yi+1 yi) = m*(xi+1 - xi) + B B

yi+1 = yi + m*(xi+1 - xi)

Ifx= 1, then yi+1 = yi+ m At each step, we make incremental

calculations based on preceding step tofind next yvalue

Strategy 1 - Incremental

Algorithm (1/2)


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Strategy 1 - Incremental

Algorithm (2/2)

),( ii yx ))(,1( myRoundx ii

))(,( ii yRoundx ),1( myx ii


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Example Code

// Incremental Line Algorithm

// Assume x0 < x1

void Line(int x0, int y0,

int x1, int y1) {int x, y;

float dy = y1 y0;

float dx = x1 x0;

float m = dy / dx;

y = y0;

for (x = x0; x < x1; x++) {

WritePixel(x, Round(y));

y = y + m;

}

}


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Problem with Incremental

Algorithm:

void Line(int x0, int y0,

int x1, int y1) {

int x, y;

float dy = y1 y0;

float dx = x1 x0;

float m = dy / dx;

y = y0;

for (x = x0; x < x1; x++) {

WritePixel(x, Round(y));

y = y + m;

}

}

Rounding takes time

Since slope is fractional, needspecial case for vertical lines


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Assume that lines slope is shallowand positive (0 < slope < 1); otherslopes can be handled by suitable

reflections about principle axes

Call lower left endpoint (x0, y0) andupper right endpoint (x1, y1)

Assume that we have just selected

pixel Pat (xp, yp)

Next, we must choose between pixelto right (E pixel), or one right andone up (NE pixel)

Let Q be intersection point of linebeing scan-converted and verticallinex=xp+1

Strategy 2 Midpoint Line

Algorithm (1/3)


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Strategy 2 Midpoint Line

Algorithm (2/3)

),(pp

yxP 1

pxx

Previouspixel

Choices forcurrentpixel

Choicesfor next

pixel

E pixel

NE pixel

Midpoint M

Q


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Line passes between E and NE

Point that is closer to intersection point Qmust be chosen

Observe on which side of line midpoint M

lies:

E is closer to line if midpoint Mlies aboveline, i.e., line crosses bottom half

NE is closer to line if midpoint Mlies belowline, i.e., line crosses top half

Error (vertical distance between chosenpixel and actual line) is always


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Line equation as function f(x):

Line equation as implicit function:

for coefficients a, b, c, where a, b 0

from above,

Properties (proof by case analysis):

f(xm, ym) = 0 when any point Mis on line

f(xm, ym) < 0 when any point Mis above line f(xm, ym) > 0 when any point Mis below line

Our decision will be based on value offunction at midpoint Mat (xp + 1, yp + )

Line

Bxdx

dyy

Bmxy

0),( cbyaxyxf

dxBcdxbdya

dxBdxyxdy

dxBxdydxy

,,

0


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Decision Variable d:

We only need sign off(xp+ 1, yp + )

to see where line lies, and then pick

nearest pixel

d= f(xp + 1, yp + )- ifd> 0 choose pixel NE

- ifd< 0 choose pixel E

- ifd= 0 choose either one consistently

How do we incrementally update d?

On basis of picking E or NE, figure out

location ofMfor that pixel, and

corresponding value dfor next grid line

We can derive dfor the next pixel

based on our current decision

Decision Variable


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Increment Mby one inxdirection

dnew= f(xp + 2, yp + )

= a(xp + 2) + b(yp + ) + c

dold = a(xp + 1)+ b(yp + )+ c

dnew- dold is the incremental difference E

dnew= dold+ a

E= a = dy(2 slides back)

We can compute value of decision variableat next step incrementally without

computing F(M) directly

dnew= dold+ E= dold+ dy

E can be thought of as correction orupdate factor to take doldto dnew

It is referred to as forward difference

If E was chosen:


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Increment Mby one in bothxand ydirections

dnew= F(xp + 2, yp + 3/2)

= a(xp + 2)+ b(yp + 3/2)+ c

NE = dnewdolddnew= dold+ a + b

NE = a + b = dydx

Thus, incrementally,

dnew= dold+ NE = dold+ dydx

If NE was chosen:


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At each step, algorithm choosesbetween 2 pixels based on sign ofdecision variable calculated inprevious iteration.

It then updates decision variableby adding either E or NE to oldvalue depending on choice of pixel.Simple additions only!

First pixel is first endpoint (x0, y0),so we can directly calculate initialvalue ofdfor choosing between Eand NE.

Summary (1/2)


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First midpoint for first d = dstartis at(x0 + 1, y0 + )

f(x0 + 1, y0 + )= a(x0 + 1) + b(y0 + ) + c

= a *x0 + b * y0 + c+ a + b/2= f(x0, y0) + a + b/2

But (x0, y0) is point on line and f(x0, y0) = 0

Therefore, dstart= a + b/2 = dydx/2

use dstartto choose second pixel, etc.

To eliminate fraction in dstart:

redefine fby multiplying it by 2; f(x,y) =2(ax+ by+ c)

this multiplies each constant and decisionvariable by 2, but does not change sign

Bresenhams line algorithm is same but doesntgeneralize as nicely to circles and ellipses

Summary (2/2)


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Example Code

void MidpointLine(int x0, int y0,

int x1, int y1) {

int dx = x1 - x0;

int dy = y1 - y0;

int d = 2 * dy - dx;

int incrE = 2 * dy;

int incrNE = 2 * (dy - dx);

int x = x0;

int y = y0;

writePixel(x, y);

while (x < x1) {if (d

scan conversion 1

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