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# Computer Graphics Implementation I - PowerPoint PPT Presentation

Computer Graphics Implementation I. Algorithms for Drawing 2D Primitives. Line Drawing Algorithms DDA algorithm Midpoint algorithm Bresenham’s line algorithm Circle Generating Algorithms Bresenham’s circle algorithm Extension to Ellipse drawing. Line Drawing. Line Drawing

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ImplementationI

• Line Drawing Algorithms

• DDA algorithm

• Midpoint algorithm

• Bresenham’s line algorithm

• Circle Generating Algorithms

• Bresenham’s circle algorithm

• Extension to Ellipse drawing

Draw a line on a raster screen between two points

What’s wrong with statement of problem?

doesn’t say anything about which points are allowed as endpoints

doesn’t give a clear meaning of “draw”

doesn’t say what constitutes a “line” in raster world

doesn’t say how to measure success of proposed algorithms

Problem Statement

Given two points P and Q in XY plane, both with integer coordinates, determine which pixels on raster screen should be on in order to make picture of a unit-width line segment starting at P and ending at Q

Scan Converting Lines

Horizontal Line:

Draw pixel P and increment x coordinate value by 1 to get next pixel.

Vertical Line:

Draw pixel P and increment y coordinate value by 1 to get next pixel.

Diagonal Line:

Draw pixel P and increment both x and y coordinate by 1 to get next pixel.

What should we do in general case? => Fast

- Digital differential analyzer (DDA) algorithm

- Mid-point algorithm

- Bresenham algorithm

Finding next pixel:

• Start with line segment in window coordinates with integer values for endpoints

• Assume implementation has a write_pixel function

y = mx + h

Find equation of line that connects two points P(x1,y1) and Q(x2,y2)

Starting with leftmost point P, increment xi by 1 to calculate yi= m*xi+ h

where m = slope, h = y intercept

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

Round (yi) = Floor (0.5 + yi)

• Digital Differential Analyzer

• DDA was a mechanical device for numerical solution of differential equations

• Line y=mx+ h satisfies differential equation

dy/dx = m = Dy/Dx = (y2-y1)/(x2-x1)

• Along scan line Dx = 1

For(x=x1; x<=x2,x++) {

y+=m;

write_pixel(x, round(y), line_color)

}

// 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;

}

}

• DDA = for each x plot pixel at closest y

• Problems for steep lines

• Use for 1  m  0

• For m > 1, swap role of x and y

• For each y, plot closest x

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

Assume that line’s slope is shallow and positive (0 < slope < 1); other slopes can be handled by suitable reflections about principle axes

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

Assume that we have just selected pixel P at (xp, yp)

Next, we must choose between pixel to right (E pixel) (xp+1, yp), or one right and one up (NE pixel) (xp+1, yp+1)

Let Q be intersection point of line being scan-converted and vertical line x=xp+1

Strategy 2 – Midpoint Line

Algorithm (1/3)

Strategy 2 – Midpoint Line slope < 1); other slopes can be handled by suitable reflections about principle axes

Algorithm (2/3)

NE pixel

Q

Midpoint M

E pixel

Previous pixel

Choices for current pixel

Choices for next pixel

Line passes between E and NE slope < 1); other slopes can be handled by suitable reflections about principle axes

Point that is closer to intersection point Q must be chosen

Observe on which side of line midpoint M lies:

E is closer to line if midpoint M lies above line, i.e., line crosses bottom half

NE is closer to line if midpoint M lies below line, i.e., line crosses top half

Error (vertical distance between chosen pixel and actual line) is always <= ½

NE pixel

Q

M

E pixel

Strategy 2 – Midpoint Line

Algorithm (3/3)

• Algorithm chooses NE as next pixel for line shown

• Now, need to find a way to calculate on which side of line midpoint lies

Line equation as function slope < 1); other slopes can be handled by suitable reflections about principle axesf(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 M is on line

f(xm, ym) < 0 when any point M is above line

f(xm, ym) > 0 when any point M is below line

Our decision will be based on value of function at midpoint M at (xp + 1, yp + ½)

Line

void MidpointLine(int x0, int y0, int x1, int y1) { slope < 1); other slopes can be handled by suitable reflections about principle axes

int dx = x1 - x0;

int dy = y1 - y0;

int x = x0;

int y = y0;

float c = y0 * dx – dy * x0;

writePixel(x, y);

x++;

while (x < x1) {

d = dy * x – (y + 0.5) * dx + c;

if (d > 0) { // Northeast Case

y++;

}

writePixel(x, y);

x++;

} /* while */

}

Decision Variable slope < 1); other slopes can be handled by suitable reflections about principle axesd:

We only need sign of f(xp+ 1, yp + ½) to see where line lies, and then pick nearest pixel

d = f(xp + 1, yp + ½)

- if d > 0 choose pixel NE

- if d < 0 choose pixel E

- if d = 0 choose either one consistently

How do we incrementally update d?

On basis of picking E or NE, figure out location of M for that pixel, and corresponding value d for next grid line

We can derive d for the next pixel based on our current decision

Decision Variable

Increment slope < 1); other slopes can be handled by suitable reflections about principle axesM by one in x direction

dnew = f(xp + 2, yp + ½)

= a(xp + 2) + b(yp + ½) + c

dold = a(xp + 1)+ b(yp + ½)+ c

dnew - dold is the incremental difference E

dnew= dold+ a

E= a = dy

We can compute value of decision variable at next step incrementally without computing F(M) directly

dnew= dold + E= dold+ dy

E can be thought of as correction or update factor to take dold to dnew

It is referred to as forward difference

Strategy 3 – Bresenham algorithm

If E was chosen:

Increment M slope < 1); other slopes can be handled by suitable reflections about principle axes by one in both x and y directions

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

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

NE = dnew – dold

dnew = dold + a + b

NE = a + b = dy – dx

Thus, incrementally,

dnew = dold + NE = dold + dy – dx

If NE was chosen:

At each step, algorithm chooses between 2 pixels based on sign of decision variable calculated in previous iteration.

It then updates decision variable by adding either E or NE to old value depending on choice of pixel. Simple additions only!

First pixel is first endpoint (x0, y0), so we can directly calculate initial value of d for choosing between E and NE.

Summary (1/2)

First midpoint for first sign of decision variable calculated in previous iteration.d = dstartis at

(x0 + 1, y0 + ½)

f(x0 + 1, y0 + ½)

= a(x0 + 1) + b(y0 + ½) + 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 = dy – dx/2

use dstart to choose second pixel, etc.

To eliminate fraction in dstart :

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

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

Summary (2/2)

Example Code sign of decision variable calculated in previous iteration.

void Bresenham(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 <= 0) { // East Case

d = d + incrE;

} else { // Northeast Case

d = d + incrNE;

y++;

}

x++;

writePixel(x, y);

} /* while */

} /* MidpointLine */

Drawing Circles sign of decision variable calculated in previous iteration.

Bresenham’s Circle Algorithm sign of decision variable calculated in previous iteration.

• Circle Equation

• (x-xc)2 + (y- yc)2 = R2

• Problem simplifying

• Making the circle origin (xc, yc) in the coordinate origin

• Symmetry points(y,x), (y,-x),(x,-y),(-x,-y),(-y,-x),(-y,x),(-x,y)

Scan Converting Circles sign of decision variable calculated in previous iteration.

(0, 17)

(0, 17)

(17, 0)

(17, 0)

• Version 1: really bad

• For x = – R to R

• y = sqrt(R *R – x *x);

• Pixel (round(x), round(y));

• Pixel (round(x), round(-y));

• Version 2: slightly less bad

• For x = 0 to 360

• Pixel (round (R • cos(x)), round(R • sin(x)));

Suppose circle origin is ( sign of decision variable calculated in previous iteration.xc,yc), radius is R.

Let D(x, y) = (x-xc)2+(y-yc)2–R2

If point(x,y) is outside the circle, D(x,y)>0。

If point(x,y) is inside the circle, D(x,y)<0。

(xc, yc)

(x sign of decision variable calculated in previous iteration.i, yi)

(xi+1, yi)

(xi+1, yi-1)

(xi, yi-1)

(xi, yi)

(xi+1, yi)

(xi+1, yi-1)

(xi, yi-1)

Bresenham’s Circle Algorithm

Two choice: (xi+1, yi) or (xi+1, yi-1)

Use Symmetry sign of decision variable calculated in previous iteration.

(x0 + a, y0 + b)

R

(x0, y0)

(x-x0)2 + (y-y0)2 = R2

• Symmetry: If (x0 + a, y0 + b) is on circle

• also (x0± a, y0± b) and (x0± b, y0± a); hence 8-way symmetry.

• Reduce the problem to finding the pixels for 1/8 of the circle

Scan top right 1/8 of circle of radius sign of decision variable calculated in previous iteration.R

Circle starts at (x0, y0 + R)

Let’s use another incremental algorithm with decision variable evaluated at midpoint

Using the Symmetry

(x0, y0)

x = x sign of decision variable calculated in previous iteration.0; y = y0 + R; Pixel(x, y);

for (x = x0+1; (x – x0) < (y – y0); x++) {

if (decision_var < 0) {

/* move east */

update decision_var;

}

else {

/* move south east */

update decision_var;

y--;

}

Pixel(x, y);

}

Sketch of Incremental Algorithm

E

SE

Decision variable sign of decision variable calculated in previous iteration.

negative if we move E, positive if we move SE (or vice versa).

Follow line strategy: Use implicit equation of circle

f(x,y) = x2 + y2 – R2 = 0

f(x,y) is zero on circle, negative inside, positive outside

If we are at pixel (x, y)

examine (x + 1, y) and (x + 1, y – 1)

Compute f at the midpoint

What we need for Incremental Algorithm

Decision Variable sign of decision variable calculated in previous iteration.

Evaluate f(x,y) = x2 + y2 – R2

at the point

We are asking: “Is

positive or negative?” (it is zero on circle)

If negative, midpoint inside circle, choose E

vertical distance to the circle is less at

(x + 1, y) than at (x + 1, y–1).

If positive, opposite is true, choose SE

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Decision based on vertical distance sign of decision variable calculated in previous iteration.

Ok for lines, since d and dvertare proportional

For circles, not true:

Which d is closer to zero? (i.e. which of the two values below is closer to R):

The right decision variable?

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We could ask instead: sign of decision variable calculated in previous iteration.

“Is (x + 1)2 + y2 or (x + 1)2 + (y – 1)2 closer to R2?”

The two values in equation above differ by

Alternate Phrasing (1/3)

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(0, 17)

fE = 12 + 172 = 290

(1, 17)

E

fSE = 12 + 162 = 257

(1, 16)

SE

fE – fSE = 290 – 257 = 33

2y – 1 = 2(17) – 1 = 33

The second value, which is always less, sign of decision variable calculated in previous iteration.

is closer if its difference from R2 is less than

i.e., if

then

so

so

so

Alternate Phrasing (2/3)

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The sign of decision variable calculated in previous iteration.radial distance decision is whether

is positive or negative

And the vertical distance decision is whether

is positive or negative; d1 and d2 are apart.

The integer d1 is positive only if d2 + is

positive (except special case where d2 = 0).

Alternate Phrasing (3/3)

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How to compute the value of sign of decision variable calculated in previous iteration.

at successive points?

is just

and that

is just

Incremental Computation, Again

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(1/2)

If we move E, update by adding 2 sign of decision variable calculated in previous iteration.x + 3

If we move SE, update by adding 2x + 3 – 2y + 2.

Forward differences of a 1st degree polynomial are constants and those of a 2nd degree polynomial are 1st degree polynomials

this “first order forward difference,” like a partial derivative, is one degree lower

Incremental Computation(2/2)

F

F’

F’’

The function is linear, hence amenable to incremental computation, viz:

Similarly

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SE

SE

Second Differences(1/2)

For any step, can compute new hence amenable to incremental computation, viz:ΔE(x, y) from old ΔE(x, y) by adding appropriate second constant increment – update delta terms as we move.

This is also true of ΔSE(x, y)

Having drawn pixel (a,b), decide location of new pixel at (a + 1, b) or (a + 1, b – 1), using previously computed d(a, b).

Having drawn new pixel, must update d(a, b) for next iteration; need to find either d(a + 1, b) or d(a + 1, b – 1) depending on pixel choice

Must add ΔE(a, b)or ΔSE(a, b) to d(a, b)

So we…

Look at d(i) to decide which to draw next, update x and y

Update d using ΔE(a,b) or ΔSE(a,b)

Update each of ΔE(a,b) and ΔSE(a,b) for future use

Draw pixel

Second Differences(2/2)

Bresenham’s hence amenable to incremental computation, viz: Eighth Circle Algorithm

MEC (R) /* 1/8th of a circle w/ radius R */

{ int x = 0, y = R;

int delta_E, delta_SE;

float decision;

delta_E = 2*x + 3;

delta_SE = 2(x-y) + 5;

decision = (x+1)*(x+1) + (y - 0.5)*(y - 0.5) –R*R;

Pixel(x, y);

while( y > x ) {

if (decision < 0) {/* Move east */

decision += delta_E;

delta_E += 2; delta_SE += 2; /*Update delta*/

}

else {/* Move SE */

y--;

decision += delta_SE;

delta_E += 2; delta_SE += 4; /*Update delta*/}

x++;

Pixel(x, y);

}

}

Uses floats! hence amenable to incremental computation, viz:

1 test, 3 or 4 additions per pixel

Initialization can be improved

Multiply everything by 4 No Floats!

Makes the components even, but sign of decision variable remains same

Questions

Are we getting all pixels whose distance from the circle is less than ½?

Why is y > x the right stopping criterion?

What if it were an ellipse?

Analysis

Equation is hence amenable to incremental computation, viz:

i.e,

Computation of and is similar

Only 4-fold symmetry

When do we stop stepping horizontally and switch to vertical?

2

2

x

y

+

=

1

2

2

a

b

+

=

2

2

2

2

2

2

b

x

a

y

a

b

D

D

SE

E

Aligned Ellipses

When absolute value of slope of ellipse is more than 1, viz: hence amenable to incremental computation, viz:

How do you check this? At a point (x,y) for which f(x,y) = 0, a vector perpendicular to the level set is f(x,y) which is

This vector points more right than up when

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In our case, hence amenable to incremental computation, viz:

and

so we check for

i.e.

This, too, can be computed incrementally

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b

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Direction Changing Criterion (2/2)

Patterned primitives hence amenable to incremental computation, viz:

Non-integer primitives

General conics

Other Scan Conversion Problems

Patterned line from hence amenable to incremental computation, viz:P to Q is not same as patterned line from Q to P.

Patterns can be geometric or cosmetic

Cosmetic: Texture applied after transformations

Geometric: Pattern subject to transformations

Cosmetic patterned line

Geometric patterned line

Patterned Lines

Q

P

Q

P

Geometric Pattern vs. Cosmetic Pattern hence amenable to incremental computation, viz:

+

Geometric (Perspectivized/Filtered)

Cosmetic