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Bresenhams Bresenhams Presentation Transcript

  • Computer Graphics 4: Bresenham Line Drawing Algorithm, Circle Drawing & Polygon Filling Course Website: http://www.comp.dit.ie/bmacnamee
  • 2of39 Contents In today’s lecture we’ll have a look at: – Bresenham’s line drawing algorithm – Line drawing algorithm comparisons – Circle drawing algorithms • A simple technique • The mid-point circle algorithm – Polygon fill algorithms – Summary of raster drawing algorithms
  • 3of39 The Big Idea Move across the x axis in unit intervals and at each step choose between two different y coordinates For example, from 5 (x +1, y +1) position (2, 3) we have to choose k k 4 (x , y ) k k between (3, 3) and 3 (3, 4) (x +1, y ) We would like the k k 2 point that is closer to 2 3 4 5 the original line
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  • 6of39 Deriving The Bresenham Line Algorithm At sample position xk+1 yk+1 the vertical separations dupper y from the mathematical dlower yk line are labelled dupper and dlower xk+1 The y coordinate on the mathematical line at xk+1 is: y = m( xk + 1) + b
  • 7of Deriving The Bresenham Line Algorithm39 (cont…) So, dupper and dlower are given as follows: d lower = y − yk = m( xk + 1) + b − yk and: d upper = ( yk + 1) − y = yk + 1 − m( xk + 1) − b We can use these to make a simple decision about which pixel is closer to the mathematical line
  • 8of Deriving The Bresenham Line Algorithm39 (cont…) This simple decision is based on the difference between the two pixel positions: d lower − d upper = 2m( xk + 1) − 2 yk + 2b − 1 Let’s substitute m with ∆y/∆x where ∆x and ∆y are the differences between the end-points: ∆y ∆x(d lower − d upper ) = ∆x(2 ( xk + 1) − 2 yk + 2b − 1) ∆x = 2∆y ⋅ xk − 2∆x ⋅ yk + 2∆y + ∆x(2b − 1) = 2∆y ⋅ xk − 2∆x ⋅ yk + c
  • 9of Deriving The Bresenham Line Algorithm39 (cont…) So, a decision parameter pk for the kth step along a line is given by: pk = ∆x(d lower − d upper ) = 2∆y ⋅ xk − 2∆x ⋅ yk + c The sign of the decision parameter pk is the same as that of dlower – dupper If pk is negative, then we choose the lower pixel, otherwise we choose the upper pixel
  • 10of Deriving The Bresenham Line Algorithm39 (cont…) Remember coordinate changes occur along the x axis in unit steps so we can do everything with integer calculations At step k+1 the decision parameter is given as: pk +1 = 2∆y ⋅ xk +1 − 2∆x ⋅ yk +1 + c Subtracting pk from this we get: pk +1 − pk = 2∆y ( xk +1 − xk ) − 2∆x( yk +1 − yk )
  • 11of Deriving The Bresenham Line Algorithm39 (cont…) But, xk+1 is the same as xk+1 so: pk +1 = pk + 2∆y − 2∆x( yk +1 − yk ) where yk+1 - yk is either 0 or 1 depending on the sign of pk The first decision parameter p0 is evaluated at (x0, y0) is given as: p0 = 2∆y − ∆x
  • 12of39 The Bresenham Line Algorithm BRESENHAM’S LINE DRAWING ALGORITHM (for |m| < 1.0) 1. Input the two line end-points, storing the left end-point in (x0, y0) 2. Plot the point (x0, y0) 3. Calculate the constants Δx, Δy, 2Δy, and (2Δy - 2Δx) and get the first value for the decision parameter as: p0 = 2∆y − ∆x 4. At each xk along the line, starting at k = 0, perform the following test. If pk < 0, the next point to plot is (xk+1, yk) and: pk +1 = pk + 2∆y
  • 13of39 The Bresenham Line Algorithm (cont…) Otherwise, the next point to plot is (xk+1, yk+1) and: pk +1 = pk + 2∆y − 2∆x 5. Repeat step 4 (Δx – 1) times ACHTUNG! The algorithm and derivation above assumes slopes are less than 1. for other slopes we need to adjust the algorithm slightly
  • 14of39 Bresenham Example Let’s have a go at this Let’s plot the line from (20, 10) to (30, 18) First off calculate all of the constants: – Δx: 10 – Δy: 8 – 2Δy: 16 – 2Δy - 2Δx: -4 Calculate the initial decision parameter p0: – p0 = 2Δy – Δx = 6
  • 15of39 Bresenham Example (cont…)18 k pk (xk+1,yk+1)17 016 1 215 314 413 512 611 710 8 20 21 22 23 24 25 26 27 28 29 30 9
  • 16of39 Bresenham Exercise Go through the steps of the Bresenham line drawing algorithm for a line going from (21,12) to (29,16)
  • 17of39 Bresenham Exercise (cont…)18 k pk (xk+1,yk+1)17 016 115 2 314 413 512 611 710 8 20 21 22 23 24 25 26 27 28 29 30
  • 18of39 Bresenham Line Algorithm Summary The Bresenham line algorithm has the following advantages: – An fast incremental algorithm – Uses only integer calculations Comparing this to the DDA algorithm, DDA has the following problems: – Accumulation of round-off errors can make the pixelated line drift away from what was intended – The rounding operations and floating point arithmetic involved are time consuming
  • 19of39 A Simple Circle Drawing Algorithm The equation for a circle is: x +y =r 2 2 2 where r is the radius of the circle So, we can write a simple circle drawing algorithm by solving the equation for y at unit x intervals using: y = ± r 2 − x2
  • 20of A Simple Circle Drawing Algorithm39 (cont…) y0 = 20 2 − 0 2 ≈ 20 y1 = 20 2 − 12 ≈ 20 y2 = 20 2 − 2 2 ≈ 20 y19 = 20 2 − 19 2 ≈ 6 y20 = 20 2 − 20 2 ≈ 0
  • 21of A Simple Circle Drawing Algorithm39 (cont…) However, unsurprisingly this is not a brilliant solution! Firstly, the resulting circle has large gaps where the slope approaches the vertical Secondly, the calculations are not very efficient – The square (multiply) operations – The square root operation – try really hard to avoid these! We need a more efficient, more accurate solution
  • 22of39 Eight-Way Symmetry The first thing we can notice to make our circle drawing algorithm more efficient is that circles centred at (0, 0) have eight-way symmetry (-x, y) (x, y) (-y, x) (y, x) R 2 (-y, -x) (y, -x) (-x, -y) (x, -y)
  • 23of39 Mid-Point Circle Algorithm Similarly to the case with lines, there is an incremental algorithm for drawing circles – the mid-point circle algorithm In the mid-point circle algorithm we use eight-way symmetry so The mid-point circle only ever calculate the points algorithm was for the top right eighth of a developed by Jack Bresenham, who we circle, and then use symmetry heard about earlier. to get the rest of the points Bresenham’s patent for the algorithm can be viewed here.
  • 24of39 Mid-Point Circle Algorithm (cont…) Assume that we have (xk+1, yk) just plotted point (xk, yk) (xk, yk) The next point is a choice between (xk+1, yk) (xk+1, yk-1) and (xk+1, yk-1) We would like to choose the point that is nearest to the actual circle So how do we make this choice?
  • 25of39 Mid-Point Circle Algorithm (cont…) Let’s re-jig the equation of the circle slightly to give us: f circ ( x, y ) = x 2 + y 2 − r 2 The equation evaluates as follows: < 0, if ( x, y ) is inside the circle boundary  f circ ( x, y ) = 0, if ( x, y ) is on the circle boundary > 0, if ( x, y ) is outside the circle boundary  By evaluating this function at the midpoint between the candidate pixels we can make our decision
  • 26of39 Mid-Point Circle Algorithm (cont…) Assuming we have just plotted the pixel at (xk,yk) so we need to choose between (xk+1,yk) and (xk+1,yk-1) Our decision variable can be defined as: pk = f circ ( xk + 1, yk − 1 ) 2 = ( xk + 1) 2 + ( yk − 1 ) 2 − r 2 2 If pk < 0 the midpoint is inside the circle and and the pixel at yk is closer to the circle Otherwise the midpoint is outside and yk-1 is
  • 27of39 Mid-Point Circle Algorithm (cont…) To ensure things are as efficient as possible we can do all of our calculations incrementally First consider: ( pk +1 = f circ xk +1 + 1, yk +1 − 1 2 ) = [( xk + 1) + 1] + yk +1 − 1 2 ( 2 ) 2 −r 2 or: pk +1 = pk + 2( xk + 1) + ( y 2 k +1 − y ) − ( yk +1 − yk ) + 1 2 k where yk+1 is either yk or yk-1 depending on the sign of p
  • 28of39 Mid-Point Circle Algorithm (cont…) The first decision variable is given as: p0 = f circ (1, r − 1 ) 2 = 1 + (r − 1 ) 2 − r 2 2 = 5 −r 4 Then if pk < 0 then the next decision variable is given as: pk +1 = pk + 2 xk +1 + 1 If pk > 0 then the decision variable is: pk +1 = pk + 2 xk +1 + 1 − 2 yk + 1
  • 29of39 The Mid-Point Circle Algorithm MID-POINT CIRCLE ALGORITHM • Input radius r and circle centre (xc, yc), then set the coordinates for the first point on the circumference of a circle centred on the origin as: ( x0 , y0 ) = (0, r ) • Calculate the initial value of the decision parameter as: p0 = 5 − r 4 • Starting with k = 0 at each position xk, perform the following test. If pk < 0, the next point along the circle centred on (0, 0) is (xk+1, yk) and: pk +1 = pk + 2 xk +1 + 1
  • 30of39 The Mid-Point Circle Algorithm (cont…) Otherwise the next point along the circle is (xk+1, yk-1) and: pk +1 = pk + 2 xk +1 + 1 − 2 yk +1 4. Determine symmetry points in the other seven octants 5. Move each calculated pixel position (x, y) onto the circular path centred at (xc, yc) to plot the coordinate values: x = x + xc y = y + yc 6. Repeat steps 3 to 5 until x >= y
  • 31of39 Mid-Point Circle Algorithm Example To see the mid-point circle algorithm in action lets use it to draw a circle centred at (0,0) with radius 10
  • 32of Mid-Point Circle Algorithm Example39 (cont…)10 k pk (xk+1,yk+1) 2xk+1 2yk+198 07 16 254 33 421 50 6 0 1 2 3 4 5 6 7 8 9 10
  • 33of39 Mid-Point Circle Algorithm Exercise Use the mid-point circle algorithm to draw the circle centred at (0,0) with radius 15
  • 34of Mid-Point Circle Algorithm Example39 (cont…)16 pk (xk+1,yk+1) k 2xk+1 2yk+11514 013 11211 210 3 9 4 8 5 7 6 6 7 5 4 8 3 9 2 10 1 11 0 12 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
  • 35of39 Mid-Point Circle Algorithm Summary The key insights in the mid-point circle algorithm are: – Eight-way symmetry can hugely reduce the work in drawing a circle – Moving in unit steps along the x axis at each point along the circle’s edge we need to choose between two possible y coordinates
  • 36of39 Filling Polygons So we can figure out how to draw lines and circles How do we go about drawing polygons? We use an incremental algorithm known as the scan-line algorithm
  • 37of39 Scan-Line Polygon Fill Algorithm 10 Scan Line 8 6 4 2 0 2 4 6 8 10 12 14 16
  • 38of39 Scan-Line Polygon Fill Algorithm The basic scan-line algorithm is as follows: – Find the intersections of the scan line with all edges of the polygon – Sort the intersections by increasing x coordinate – Fill in all pixels between pairs of intersections that lie interior to the polygon
  • 39of Scan-Line Polygon Fill Algorithm39 (cont…)
  • 40of39 Line Drawing Summary Over the last couple of lectures we have looked at the idea of scan converting lines The key thing to remember is this has to be FAST For lines we have either DDA or Bresenham For circles the mid-point algorithm
  • 41of39 Anti-Aliasing
  • 42of39 Summary Of Drawing Algorithms
  • 43of39 Mid-Point Circle Algorithm (cont…) 6 5 4 3 1 2 3 4
  • 44of39 Mid-Point Circle Algorithm (cont…) 6 M 5 4 3 1 2 3 4
  • 45of39 Mid-Point Circle Algorithm (cont…) 6 M 5 4 3 1 2 3 4
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  • 48of39 Blank Grid 10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10
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