Pixels Pixels Row and column indicates a PIXEL not a POINT. A - PowerPoint PPT Presentation

Pixels

Pixels ● Row and column indicates a PIXEL not a POINT. A pixel can theoretically contain infinitely many points ● Drawing lines with pixels: fill pixels along the trajectory of a line from point p1 to p2 Aliasing

Antialiasing ● Antialiasing is a term for techniques that are designed to mitigate the effects of aliasing ● The idea is that when a pixel is only partially covered by a shape, the color of the pixel should be a mixture of the color of the shape and the color of the background.

Line algorithm ● A naive line-drawing algorithm dx = x2 - x1 dy = y2 - y1 for x from x1 to x2 { y = y1 + dy * (x - x1) / dx plot(x, y) } Problems with this? Assume order is OK, x2>x1

Naive line-drawing ● Inefficient, slow due to floating point computations ● If dx<dy, the line becomes sparse with lots of gaps ● What happens at dx=0?

Bresenham’s Line Algorithm ● Solve inefficiency due to floating point arithmetic by using ONLY integer arithmetic ● Key idea: – when focusing on one octant, say 0 – start point of line is origin, slope is < 1 – loop x over integers from x1 to x2 – y will go from y1 to y2 and at every point we have to decide to increase y by 1 or not

Breshenman contd. A point (x, y) is on the line when f(x, y) = 0

Bresenham contd. ● At each point, base decision to increase y on where the integer x is w.r.t. true line

Line w.r.t. halfway point ● Points on the line ● How can we evaluate the line at next integer x 0 +1 and midway between two vertical integers? ● How about ● But the whole point of this was to avoid floating point arithmetic… so why all the complications? ● Let’s do a little more algebra

Integer Arithmetic ● Decision at x 0 ● Simplifying ● If D is positive then increase y, otherwise leave y alone ● What about x 0 +1?

Second point ● If the difference is positive, then increase y by 1 ● The error D accumulates as x increases ● But there are still fractions…? We only care about sign so we can multiply the equation by 2 with no consequence

Bresenham Algorithm plotLine(x0,y0, x1,y1) dx = x1 - x0 dy = y1 - y0 D = 2*dy - dx y = y0 for x from x0 to x1 plot(x,y) if D > 0 y = y + 1 D = D - 2*dx end if D = D + 2*dy Homework: implement Bresenham’s line algorithm in Java. The above algorithm sketch only works for one of the 8 octants.

Coordinate Systems ● We will talk about orthogonal coordinate systems: axes are perpendicular to each other ● We have to be flexible when we think about coordinate systems and transforming from one to another ● The display is a rectangle of pixels, the last destination of the output of our algorithms, however, we may want to specify other intermediary rectangles with their own origins, e.g., window or part of a window

Transforming between real-number Coordinate Systems newX = newLeft + ((oldX - oldLeft) / (oldRight - oldLeft)) * (newRight - newLeft)) newY = newTop + ((oldY - oldTop) / (oldBottom - oldTop)) * (newBotom - newTop)

Aspect Ratio