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5.1 Si23_03 SI23 Introduction to Computer Graphics Lecture 5 – Drawing A Line

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5.2 Si23_03 Course Outline Image Display URL GIMP colour n Vector graphics – Line drawing – Area filling n Graphical interaction SVG Viewer 2D vector graphics URL lines., areas graphics algorithms interaction

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5.3 Si23_03 Line Drawing n Line drawing is a fundamental operation in computer graphics n Demanding applications require very fast drawing speeds - say, 1 million lines per second n Hence the algorithm for converting a line to a set of pixels will be implemented in hardware - and efficiency is vital – Want to use integer arithmetic – Want to use additions rather than multiplications

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5.4 Si23_03 Line Drawing - The Problem n To draw a line from one point to another on a display - which pixels should be illuminated? Suppose pt1 = (0,1) and pt2 = (5,4) Where to draw the line?

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5.5 Si23_03 Line Drawing - By Hand n In a simple example, one can do by eye but we would like an algorithm!

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5.6 Si23_03 Equation of a Line n Line equation is: y = m x + c If line joins (x 1,y 1 ) to (x 2,y 2 ), then m = (y 2 -y 1 )/(x 2 -x 1 ) and c = y 1 - m x 1 Suppose pt1 = (0,1) and pt2 = (5,4) Then m = (4-1)/(5-0) ie 0.6 and c = 1 Thus: y = 0.6 x + 1

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5.7 Si23_03 Drawing a Line From Its Equation n We can use the equation to draw the pixels y = 0.6 x + 1 x = 0 -> y =1y =1 x = 1 -> y = 1.6y = 2 x = 2 -> y = 2.2y = 2 x = 3 -> y = 2.8y = 3 x = 4 -> y = 3.4y = 3 Calculate y for x = 0,1,2,3,4,5 and round to nearest integer

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5.8 Si23_03 The Line Drawn y = 0.6 x + 1 x = 0 -> y =1y =1 x = 1 -> y = 1.6y = 2 x = 2 -> y = 2.2y = 2 x = 3 -> y = 2.8y = 3 x = 4 -> y = 3.4y = 3 Calculate y for x = 0,1,2,3,4,5 and round to nearest integer This gives us (after 1 multiplication, 1 addition, and 1 rounding operation per step): How do we make more efficient?

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5.9 Si23_03 DDA Algorithm n We can do this more efficiently by simply incrementing y at each stage y* = y + q where q = (y 2 -y 1 )/(x 2 -x 1 ) y = 1.0y = 1 y = = 1.6y = 2 y = = 2.2y = 2 y = = 2.8y = 3 y = = 3.4y = 3 Here q = 0.6 One addition, one rounding

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5.10 Si23_03 Slope of Line n We have assumed so far that slope of line (m) lies between -1 and 1 n When slope of line is greater than 1 in absolute value, then we increment y by 1 at each step, and x by: d = (x 2 -x 1 )/(y 2 -y 1 ) Exercise: calculate the pixels to draw line from (1,0) to (4,5)

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5.11 Si23_03 Efficiency n The DDA (Digital Differential Analyser)algorithm just described has the problem: – floating point operations (real numbers) are expensive - add and round at each step Yet really all we need to decide is whether to move horizontally or diagonally (for a line of slope < 1). Can we do this in integer arithmetic?

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5.12 Si23_03 Bresenhams Algorithm n One of the classic algorithms in computer graphics is the Bresenham line drawing algorithm – Published in 1965 – Still widely used – Excellent example of an efficient algorithm Jack Bresenham

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5.13 Si23_03 Bresenhams Algorithm - First Step n Look in detail at first step: 012 We have choice of a move to (1,1) or (1,2) Exact y-value at x=1 is: y* = = 1.6 Since this is closer to 2 than 1, we choose (1,2) d2 d1 Decision based on d1>d2, ie d1-d2>

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5.14 Si23_03 Bresenhams Algorithm n In general, suppose we are at (x k,y k ) - slope of line, m, is between 0 and 1 n Next move will be to (x k +1, y k ) or (x k +1, y k +1) d2 d1 xkxk x k +1 ykyk y k +1 d1 = [m(x k + 1)+c] - y k d2 = y k [m(x k + 1)+c] d1 - d2 = 2m(x k + 1) - 2y k + 2c -1 >0 indicates (x k +1, y k +1) <0 indicates (x k +1, y k )

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5.15 Si23_03 Bresenhams Algorithm d1 - d2 = 2m(x k + 1) - 2y k + 2c -1 : d1-d2 >0 diagonal Let m = Dy/Dx and multiply both sides by Dx, setting Dx(d1 – d2) = p k p k = 2Dy x k - 2Dx y k + b (b a real constant) Decision is still p k >0 - but can we work only in integers? How does p k change at next step to p k+1 ? We know that x k increases by 1, and y k either increases by 1 or stays the same. So ….

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5.16 Si23_03 Bresenhams Algorithm p k = 2Dy x k - 2Dx y k + b p k+1 = p k + 2Dy - 2Dx - if y increases p k+1 = p k + 2Dy - if y stays the same Thus we have a decision strategy that only involves Integers, no multiplication and no rounding.. we start things off with p 0 = 2Dy - Dx

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5.17 Si23_03 Bresenhams Algorithm - A Summary n To draw from A to B with slope 0

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5.18 Si23_03 Bresenhams Algorithm - Example –(x 0,y 0 ) = (0,1) –Dx = 5, Dy = 3 - p 1 = p 0 + 2Dy -2Dx = -3 <0 hence plot (2,2) –p 0 = 2Dy - Dx = 1 > 0 –hence plot (1,2) And so on …

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5.19 Si23_03 Bresenhams Algorithm - Other Slopes and Other Shapes n There are straightforward variations for lines with slopes: m > 1 -1 < m < 0 m < -1 n There are similar algorithms for drawing circles and ellipses efficiently - see Hearn & Baker textbook

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5.20 Si23_03 Other Ways of Drawing Lines n Over past 38 years, researchers have strived to improve on Bresenhams incremental algorithm – for example, think how to parallelise the algorithm n Alternative structural approach comes from recognition that horizontal (0) and diagonal (1) moves come in runs n We had: – n One of the moves occurs singly, and is evenly spread within the sequence - the other occurs in up to two lengths (consecutive numbers) – – ie 1 occurs singly, 0 occurs in twos and threes – decision is now which length of run

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5.21 Si23_03 Aliasing n Lines on raster displays suffer from aliasing effects - especially noticeable with lines near to horizontal This is caused because we are `sampling the true straight line at a discrete set of positions. Called `aliasing because many straight lines will have the same raster representation. More noticeable on screen than laser printer - why?

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5.22 Si23_03 Supersampling – Bresenham on Finer Grid Each pixel is divided into 9 sub pixels. Either 0,1,2 or 3 sub-pixels can be set in any pixel. This can be used to set the intensity level of each pixel.

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5.23 Si23_03 Supersampling This shows the intensity levels for the pixels in the previous slide. We are really using intensity levels to compensate for lack of resolution.

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5.24 Si23_03 Antialiased Line The end result is a thicker line, with the intensity spread out over a larger area. To the eye, the line appears much sharper and straighter.

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5.25 Si23_03 d1 - d2 = 2m(x k + 1) - 2y k + 2c -1 : d1-d2 >0 diagonal Let m = Dy/Dx and multiply both sides by Dx, setting Dx(d1 – d2) = p k p k = 2Dy x k - 2Dx y k + b (b a constant) Decision is still p k >0 - but can we work only in integers? p k+1 = 2Dy x k+1 - 2Dx y k+1 + b... So p k+1 = p k + 2Dy - 2Dx(y k+1 - y k ) - where (y k+1 - y k ) = 0 or 1 Thus we have a decision strategy that only involves integers.. we start things off with p 0 = 2Dy - Dx Bresenhams Algorithm

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