1 csce 441 lecture 2: scan conversion of lines jinxiang chai
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CSCE 441 Lecture 2:Scan Conversion of Lines
Jinxiang Chai
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Displays – Cathode Ray Tube
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Displays – Liquid Crystal Displays
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Displays – Plasma
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Displays – Pixels
Pixel: the smallest element of picture
- Integer position (i,j)
- Color information (r,g,b)
x
y
(0,0)
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Digital Scan Conversion
Convert graphics output primitives into a set of pixel values for storage in the frame buffer
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OpenGL Geometric Primitives
All geometric primitives are specified by vertices
GL_QUAD_STRIPGL_QUAD_STRIP
GL_POLYGONGL_POLYGON
GL_TRIANGLE_STRIPGL_TRIANGLE_STRIP
GL_TRIANGLE_FANGL_TRIANGLE_FAN
GL_POINTSGL_POINTSGL_LINESGL_LINES
GL_LINE_LOOPGL_LINE_LOOPGL_LINE_STRIPGL_LINE_STRIP
GL_TRIANGLESGL_TRIANGLES
GL_QUADSGL_QUADS
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OpenGL Geometric Primitives
All geometric primitives are specified by vertices
GL_QUAD_STRIPGL_QUAD_STRIP
GL_POLYGONGL_POLYGON
GL_TRIANGLE_STRIPGL_TRIANGLE_STRIP
GL_TRIANGLE_FANGL_TRIANGLE_FAN
GL_POINTSGL_POINTSGL_LINESGL_LINES
GL_LINE_LOOPGL_LINE_LOOPGL_LINE_STRIPGL_LINE_STRIP
GL_TRIANGLESGL_TRIANGLES
GL_QUADSGL_QUADS
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Outline
How to draw a line?
- Digital Differential Analyzer (DDA)
- Midpoint algorithm
Required readings
- HB 6.1 to 6.8
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Problem
Given two points (P, Q) on the screen (with integer coordinates) determine which pixels should be drawn to display a unit width line
P
Q
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Problem
Given two points (P, Q) on the screen (with integer coordinates) determine which pixels should be drawn to display a unit width line
P
Q
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Problem
Given two points (P, Q) on the screen (with integer coordinates) determine which pixels should be drawn to display a unit width line
P
Q
Line-drawing using “Paint”
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Special Lines - Horizontal
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Special Lines - Horizontal
Increment x by 1, keep y constant
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Special Lines - Vertical
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Special Lines - Vertical
Keep x constant, increment y by 1
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Special Lines - Diagonal
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Special Lines - Diagonal
Increment x by 1, increment y by 1
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How about Arbitrary Lines?
How to draw an arbitrary line?
P
Q
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Slope-intercept equation for a line:
Arbitrary Lines
bmxy m: slope
b: y-intercept
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Slope-intercept equation for a line:
How can we compute the equation from (xL,yL), (xH,yH)?
Arbitrary Lines
bmxy m: slope
)( , HH yx
)( , LL yx
b: y-intercept
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Slope-intercept equation for a line:
How can we compute the equation from (xL,yL), (xH,yH)?
Arbitrary Lines
LL
LH
LH
mxyb
xx
yym
bmxy m: slope
)( , HH yx
)( , LL yx
b: y-intercept
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Arbitrary Lines
Assume
bmxy
10 m
10 m
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Arbitrary Lines
Assume Other lines by swapping x, y or negating
- e.g., reverse the role of x and y if m>1
bmxy
10 m
10 m
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Arbitrary Lines
Assume Other lines by swapping x, y or negating
- e.g., reverse the role of x and y if m>1
- e.g., negate y if
bmxy
10 m
10 m
01 m
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Arbitrary Lines
Assume Other lines by swapping x, y or negating
Take steps in x and determine where to fill y
xmy
bmxy
10 m
10 m
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Digital Differential Analyzer
Start from (xL, yL) and draw to (xH, yH)
where xL< xH
( x0, y0 ) = ( xL, yL )
For ( i = 0; i <= xH-xL; i++ )
DrawPixel ( xi, Round ( yi ) )
xi+1 = xi + 1
yi+1 = m xi+1 + b
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Digital Differential Analyzer
Simple algorithm:
- sample unit intervals in one coordinate
- find the nearest pixel for each sampled point.
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Digital Differential Analyzer (DDA)
Start from (xL, yL) and draw to (xH, yH)
where xL< xH
( x0, y0 ) = ( xL, yL )For ( i = 0; i <= xH-xL; i++ ) //sample unit intervals in x coordinate
DrawPixel ( xi, Round ( yi ) ) //find the nearest pixel for each sampled point.
xi+1 = xi + 1
yi+1 = m ( xi + 1 ) + b
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DDA
Start from (xL, yL) and draw to (xH, yH)
where xL< xH
( x0, y0 ) = ( xL, yL )
For ( i = 0; i <= xH-xL; i++ )
DrawPixel ( xi, Round ( yi ) )
xi+1 = xi + 1
yi+1 = m xi + m + b
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DDA
Start from (xL, yL) and draw to (xH, yH)
where xL< xH
( x0, y0 ) = ( xL, yL )
For ( i = 0; i <= xH-xL; i++ )
DrawPixel ( xi, Round ( yi ) )
xi+1 = xi + 1
yi+1 = m xi + b + m
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DDA
Start from (xL, yL) and draw to (xH, yH)
where xL< xH
( x0, y0 ) = ( xL, yL )
For ( i = 0; i <= xH-xL; i++ )
DrawPixel ( xi, Round ( yi ) )
xi+1 = xi + 1
yi+1 = m xi + b + m
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DDA
Start from (xL, yL) and draw to (xH, yH)
where xL< xH
( x0, y0 ) = ( xL, yL )
For ( i = 0; i <= xH-xL; i++ )
DrawPixel ( xi, Round ( yi ) )
xi+1 = xi + 1
yi+1 = yi + m
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DDA - Example
)2,0(
)4,6(
bmxy
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DDA - Example
)2,0(
)4,6(
bmxy
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
2
0
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
37
1
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
37
1
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
38
2
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
38
2
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
3
3
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
3
3
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
310
4
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
310
4
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
311
5
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
311
5
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
4
6
y
x
31
0624
LH
LH
xxyym
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DDA - Example
)2,0(
)4,6(
bmxy
4
6
y
x
31
0624
LH
LH
xxyym
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DDA - Problems
Floating point operations Rounding Can we do better?
( x0, y0 ) = ( xL, yL )
For ( i = 0; i <= xH-xL; i++ )
DrawPixel ( xi, Round ( yi ) )
xi+1 = xi + 1
yi+1 = yi + m
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Outline
How to draw a line?
- Digital Differential Analyzer (DDA)
- Midpoint algorithm
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How to Improve It?
)4,6(
bmxy
31
0624
LH
LH
xxyym
Question: if we draw the current pixel, where is the next pixel?
)2,0(
(xk,yk) (xk+1,yk+1)
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How to Improve It?
)4,6(
bmxy
31
0624
LH
LH
xxyym
Question: if we draw the current pixel, where is the next pixel?
)2,0(
(xk,yk) (xk+1,yk+1)
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How to Improve It?
)4,6(
bmxy
31
0624
LH
LH
xxyym
Question: if we draw the current pixel, where is the next pixel?
)2,0(
(xk,yk) (xk+1,yk+1)
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How to Improve It?
)2,0(
)4,6(
bmxy
31
0624
LH
LH
xxyym
Question: if we draw the current pixel, where is the next pixel?
Next column: E or NE direction; why?
xk+1 = xk+1
yk+1 = yk or yk+1
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Midpoint Algorithm
Given a point just drawn, determine whether we move E or NE on next step
xk+1 = xk+1
yk+1 = yk or yk+1
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Midpoint Algorithm
Given a point just drawn, determine whether we move E or NE on next step
Is the line above or below midpoint ?
Below: move E
Above: move NE
),1( 21 yx
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Above/Below the Line?
)2,0(
)4,6(
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Two Issues
How to evaluate if the midpoint is above or below the line?
How to incrementally evaluate this?
How to avoid floating point operations?
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line? bmxy
x
y
)2,0(
)5,6(
)7,3(
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line? bmxy
x
y
)2,0(
)5,6(
)7,3(Properties
y=mx+b (x,y) on the line
y<mx+b (x,y) below the line
y>mx+b (x,y) above the line
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line? bmxy
x
y
)2,0(
)5,6(
)7,3(Properties
y=mx+b (x,y) on the line
y<mx+b (x,y) below the line
y>mx+b (x,y) above the line
But this still requires floating point operations!
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line? bmxy
bmxyyxf ),(
Properties
f(x,y)=0 (x,y) on the line
f(x,y)<0 (x,y) below the line
f(x,y)>0 (x,y) above the line
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line?
bxyLH
LH
xxyy
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line?
))(()( LHxxyy
LH xxbxyxxLH
LH
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line?
bxxxyyyxx LHLHLH )()()(
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line?
0)()()( bxxxyyyxx HLHLLH
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line?
edycxyxf ),(HL yyc LH xxd )( HL xxbe
0)()()( bxxxyyyxx HLHLLH
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line?
edycxyxf ),(HL yyc LH xxd )( HL xxbe
Properties
f(x,y)=0 (x,y) on the line
f(x,y)<0 (x,y) below the line
f(x,y)>0 (x,y) above the line
0)()()( bxxxyyyxx HLHLLH
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Midpoint Algorithm – Implicit Forms
How do we tell if a point is above or below the line?
edycxyxf ),(HL yyc LH xxd )( HL xxbe
Properties
f(x,y)=0 (x,y) on the line
f(x,y)<0 (x,y) below the line
f(x,y)>0 (x,y) above the line
0)()()( bxxxyyyxx HLHLLH
LL
LH
LH
mxyb
xx
yym
Note that e is an integer because
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Midpoint Algorithm – Implicit Forms
x
y
)2,0(
)5,6(
263 xy
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Midpoint Algorithm – Implicit Forms
x
y
)2,0(
263 xy
1236),( xyyxf)5,6(
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Midpoint Algorithm – Implicit Forms
x
y
)2,0(
263 xy
1236),( xyyxf 18)5,0( f)5,6(
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Midpoint Algorithm – Implicit Forms
x
y
)2,0(
263 xy
1236),( xyyxf
18)1,4( f
)5,6(
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Midpoint Algorithm – Implicit Forms
x
y
)2,0(
263 xy
1236),( xyyxf
0)4,4( f)5,6(
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Midpoint Algorithm – Implicit Forms
x
y
)2,0(
263 xy
1236),( xyyxf)5,6(
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Two Issues
How to evaluate if the midpoint is above or below the line?
How to incrementally evaluate this?
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Above/Below the Line?
)2,0(
)4,6(
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Midpoint Algorithm
What about starting value?
What is the middle point?
(xL,yL)
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Midpoint Algorithm
What about starting value?
What is the middle point? (xL+1,yL+1/2)
(xL,yL)
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Midpoint Algorithm
What about starting value?
What is the middle point?
eydxcyxf LLLL )()1(),1( 21
21
(xL+1,yL+1/2)
(xL,yL)
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Midpoint Algorithm
What about starting value?
eydxcyxf LLLL )()1(),1( 21
21
dcyxfyxf LLLL 21
21 ),(),1(
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Midpoint Algorithm
What about starting value?
eydxcyxf LLLL )()1(),1( 21
21
dcyxfyxf LLLL 21
21 ),(),1(
is on the line!),( LL yx
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Midpoint Algorithm
What about starting value?
eydxcyxf LLLL )()1(),1( 21
21
dcyxf LL 21
21 ),1(
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Midpoint Algorithm
What about starting value?
eydxcyxf LLLL )()1(),1( 21
21
dcyxf LL 2),1( 21
Multiplying coefficients by 2 doesn’t change sign of f
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Midpoint Algorithm
Need value of to determine E or NE
Build incremental algorithm Assume we have value of
Find value of if E chosenFind value of if NE chosen
),1( 21 yxf
),1( 21 yxf
),2( 21 yxf
),2( 23 yxf
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Midpoint Algorithm
Need value of to determine E or NE
Build incremental algorithm Assume we have value of
Find value of if E chosen
),1( 21 yxf
),1( 21 yxf
),2( 21 yxf
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Midpoint Algorithm
Need value of to determine E or NE
Build incremental algorithm Assume we have value of
Find value of if E chosenFind value of if NE chosen
),1( 21 yxf
),1( 21 yxf
),2( 21 yxf
),2( 23 yxf
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Midpoint Algorithm
If E was chosen, find ),2( 21 yxf
eydxcyxf )()2(),2( 21
21
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Midpoint Algorithm
If E was chosen, find ),2( 21 yxf
eydxcyxf )()2(),2( 21
21
),1(),2( 21
21 yxfcyxf
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Midpoint Algorithm
If NE was chosen, find ),2( 23 yxf
eydxcyxf )()2(),2( 23
23
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Midpoint Algorithm
If NE was chosen, find ),2( 23 yxf
eydxcyxf )()2(),2( 23
23
),1(),2( 21
23 yxfdcyxf
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Midpoint Algorithm – Summary
x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+d draw(x,y)
while ( x < xH)if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
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Midpoint Algorithm – Summary
x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+d //initial valuedraw(x,y)
while ( x < xH) if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
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Midpoint Algorithm – Summary
x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+d //initial valuedraw(x,y)
while ( x < xH) // iterate until reaching the end pointif ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
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Midpoint Algorithm – Summary
x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+d //initial valuedraw(x,y)
while ( x < xH) // iterate until reaching the end pointif ( sum < 0 ) // below the line and choose NE
sum += 2dy++
x++sum += 2cdraw(x,y)
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Midpoint Algorithm – Summary
x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+d //initial valuedraw(x,y)
while ( x < xH) // iterate until reaching the end pointif ( sum < 0 ) // below the line and choose NE
sum += 2dy++
x++sum += 2cdraw(x,y)
- Update the current pixel
- Update the function value of midpoint
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Midpoint Algorithm – Summary
x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+d //initial valuedraw(x,y)
while ( x < xH) // iterate until reaching the end pointif ( sum < 0 ) // below the line and choose NE
sum += 2dy++
x++sum += 2cdraw(x,y)
- Draw the pixel based on the function value
of midpoint
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x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+ddraw(x,y)
while ( x < xH)if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
Midpoint Algorithm – Example
)2,0(
)4,6(
0x 2y 2sum
d =6 c = -2
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Midpoint Algorithm – Example
)2,0(
)4,6(
0x 2y 2sumx=xL y=yL
d=xH - xL c=yL - yH
sum=2c+ddraw(x,y)
while ( x < xH)if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
d =6 c = -2
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Midpoint Algorithm – Example
)2,0(
)4,6(
1x 2y 2sumx=xL y=yL
d=xH - xL c=yL - yH
sum=2c+ddraw(x,y)
while ( x < xH)if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
d =6 c = -2
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x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+ddraw(x,y)
while ( x < xH)if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
Midpoint Algorithm – Example
)2,0(
)4,6(
2x 3y 6sum
d =6 c = -2
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Midpoint Algorithm – Example
)2,0(
)4,6(
3x 3y 2sumx=xL y=yL
d=xH - xL c=yL - yH
sum=2c+ddraw(x,y)
while ( x < xH)if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
d =6 c = -2
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Midpoint Algorithm – Example
)2,0(
)4,6(
4x 3y 2sumx=xL y=yL
d=xH - xL c=yL - yH
sum=2c+ddraw(x,y)
while ( x < xH)if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
d =6 c = -2
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x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+ddraw(x,y)
while ( x < xH)if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
Midpoint Algorithm – Example
)2,0(
)4,6(
5x 4y 6sum
d =6 c = -2
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x=xL y=yL
d=xH - xL c=yL - yH
sum=2c+ddraw(x,y)
while ( x < xH)if ( sum < 0 )
sum += 2dy++
x++sum += 2cdraw(x,y)
Midpoint Algorithm – Example
)2,0(
)4,6(
6x 4y 2sum
d =6 c = -2
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Midpoint Algorithm
Only integer operations Exactly the same as the more commonly
found Bresenham’s line drawing algorithm Extends to other types of shapes (circles)
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Midpoint Algorithm: Circle
0)()( 220
20 ryyxx
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Midpoint Algorithm: Circle
0)()( 220
20 ryyxx
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Midpoint Algorithm: Circle
0)()( 220
20 ryyxx
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Midpoint Algorithm: Circle
0)()( 220
20 ryyxx
Two issues:- What are two possible solutions?
- What is the evaluation function?
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Midpoint Algorithm: Circle
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Midpoint Algorithm: Circle
(xk+1,yk)(xk+1,yk-1)
Two issues:- What are two possible solutions?
- What is the evaluation function?
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Midpoint Algorithm: Circle
(xk+1,yk) Midpoint: (xk+1,yk-1/2)(xk+1,yk-1)
Two issues:- What are two possible solutions?
- What is the evaluation function?
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Midpoint Algorithm: Circle
(xk+1,yk) Midpoint: (xk+1,yk-1/2)(xk+1,yk-1)
Two issues:- What are two possible solutions?
- What is the evaluation function?
220
20 )()(),( ryyxxyxf
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Circle, Ellipse, etc
Need implicit forms
- circle:
- ellipse:
Equations for incremental calculations Initial positions More details in section 6.4-6.6 of the
textbook
022 FEyDxCxyByAx
0)()( 220
20 ryyxx
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Next Lecture
How to draw a polygon?
- section 6.10- 6.13