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Topic >>>> Scan Conversion. CSE5280 - Computer Graphics. Graphics Display Devices. Frame Buffer – a region of memory sufficiently large to hold all of the pixel values for the display. Graphics Display Devices - cont. - PowerPoint PPT Presentation
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1/1/2000 1
Topic >>>> Scan Conversion
CSE5280 - Computer Graphics
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Graphics Display Devices
Frame Buffer – a region of memory sufficiently large to hold all of the pixel values for the display
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Graphics Display Devices - cont
How each pixel value in the frame buffer is sent to the right place on the display surface
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Graphics Devices – cont
Each pixel has a 2D address (x,y)For each address (x,y) there is a specific memory location
that holds the value of the pixel (I.e. mem[136][252])The scan controller sends the logical address (136, 252) to
the frame buffer, which emits the value mem[136][252]The value mem[136][252] is converted to a corresponding
intensity or color in the conversion circuit, and that intensity or color is sent to the proper physical position, (136, 252), on the display surface
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Scan Converting Lines
Line DrawingDraw a line on a raster screen between 2 pointsWhat’s wrong with the statement of the problem?
• It does not say anything about which pts are allowed as end pts
• It does not give a clear meaning to “draw”• It does not say what constitutes a “line” in the raster world• It does not say how to measure the success of the
proposed algorithm
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Scan Converting Lines - cont
Problem StatementGiven 2 points P and Q in the plane, both with
integer coordinates, determine which pixels on a raster screen should be “on” in order to make a picture of a unit-width line segment starting at point P and ending at point Q
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Finding the next pixel
Special Case:Horizontal Line:
• Draw pixel P and increment the x coordinate value by one to get the next pixel.
Vertical Line:• Draw the pixel P and increment the y coordinate value by one to get
the next pixelDiagonal Line:
• Draw the pixel P and increment both the x and y coordinate values by one to get the next pixel
What should we use in the general case?
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Vertical Distance
Why can we use the vertical distance as a measure of which point is closer?Because vertical distance is proportional to the
actual distanceHow do we show this?Congruent Triangles
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Vertical Distance – cont
By similar triangles we can see that the true distances to the line (in blue) are directly proportional to the vertical distances to the line (in black) for each point.Therefore the point with the smaller vertical distance to the line is the closest to the line
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Strategy 1 – Incremental Algorithm
The Basic AlgorithmFind the equation of the line that connects the 2
points P and QStarting with the leftmost point P, increment by 1
to calculate where A = slope, and B = y intercept
Intensify the pixel at
This computation selects the closest pixel, the pixel whose distance to the “true” line is smallest
ixBAxy ii
)5.0()(,)(, iiii yFlooryRoundwhereyRoundx
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Strategy 1 – Incremental Algorithm
The Incremental AlgorithmEach iteration requires a floating-point multiplication
therefore, modify
If , then Thus, a unit change in x changes y by slope A, which
is the slope of the lineAt each step, we make incremental calculations
based on the preceding step to find the next y value
xAyBxxABAxY iiii 11
1x Ayy ii 1
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Strategy 1 – Incremental Aglo
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Example Code
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Problem with the Incremental Algorithm
Rounding integers takes timeReal variables have limited precision, summing an inexact slope (A) repetitively introduces a cumulative error buildupVariables y and A must be a real or fractional binary because the slope is a fractionSpecial case needed for vertical lines
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Strategy 2 – Midpoint Line Algorithm Assume that the line’s slope is shallow and positive ( 0 < slope < 1); other slopes can be handled by suitable reflections about the principle axesCall the lower left endpoint and the upper right endpoint Assume that we have just selected the pixel P at
Next, we must choose between the pixel to the right (pixel E), or one right and one up (pixel NE)Let Q be the intersection point of the line being scan-converted with the grid line
0,0 yx 11, yx
pp yx ,
1 pxx
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Strategy 2 – Midpoint Line Algorithm
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Strategy 2 – Midpoint Line AlgorithmThe line passes between E and NEThe point that is closer to the intersection point Q must be chosenObserve on which side of the line the midpoint M lies:E is closer to the line if the midpoint lies above the line (I.e. the line
crosses the bottom half)NE is closer to the line if the midpoint lies below the line, I.e., the
line crosses the top halfThe error, the vertical distance between the chosen pixel and the actual line is always <= ½The algorithm chooses NE as the next pixel for the line shownNow, find a way to calculate on which side of the line the midpoint lies
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The LineThe line equation as a function f(x): f(x) = A*x + B = dy/dx * x + B
Line equation as an implicit function:F(x,y) = a * x + b * y + c = 0 for coefficients a, b, c where a, b != 0;
from above, y *dx = dy*x + B*dx, so a = dy, b = -dx, c=B *dx, a>0 for y(0) < y(1)
Properties (proof by the case analysis): when any point M is on the line when any point M is above the line when any point M is below the lineOur decision will be based on the value of the function at the
midpoint M at
0)( , mm yxF0),( mm yxF
0),( mm yxF
21,1 pp yx
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Decision Variable
Decision Variable d:We only need the sign of to see
where the line lies, and then pick the nearest pixel
• If d > 0 choose pixel NE• If d < 0 choose pixel E• If d = 0 choose either one consistently
How to update d:On the basis of picking E or NE, figure out the location of the
M for that pixel, and the corresponding value of d for the next grid line
)21,1( pp yxF
),1( 21 pp yxFD
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Example Code
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Scan Conversion Summary