Topic >>>> Scan Conversion

1. Topic >>>> Scan Conversion CSE5280 - Computer Graphics

2. Graphics Display Devices • Frame Buffer – a region of memory sufficiently large to hold all of the pixel values for the display

3. Graphics Display Devices - cont • How each pixel value in the frame buffer is sent to the right place on the display surface

4. 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

5. Scan Converting Lines • Line Drawing • Draw a line on a raster screen between 2 points • What’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

6. Scan Converting Lines - cont • Problem Statement • Given 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

7. 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 pixel • Diagonal 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?

8. 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 distance • How do we show this? • Congruent Triangles

9. 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

10. Strategy 1 – Incremental Algorithm • The Basic Algorithm • Find the equation of the line that connects the 2 points P and Q • Starting 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

11. Strategy 1 – Incremental Algorithm • The Incremental Algorithm • Each 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 line • At each step, we make incremental calculations based on the preceding step to find the next y value

12. Strategy 1 – Incremental Aglo

13. Example Code

14. Problem with the Incremental Algorithm • Rounding integers takes time • Real variables have limited precision, summing an inexact slope (A) repetitively introduces a cumulative error buildup • Variables y and A must be a real or fractional binary because the slope is a fraction • Special case needed for vertical lines

15. 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 axes • Call 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

16. Strategy 2 – Midpoint Line Algorithm

17. Strategy 2 – Midpoint Line Algorithm • The line passes between E and NE • The point that is closer to the intersection point Q must be chosen • Observe 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 half • The error, the vertical distance between the chosen pixel and the actual line is always <= ½ • The algorithm chooses NE as the next pixel for the line shown • Now, find a way to calculate on which side of the line the midpoint lies

18. The Line • The 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 line • Our decision will be based on the value of the function at the midpoint M at

19. 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

20. Example Code

21. Scan Conversion Summary