Lecture outline Graphics and Interaction Scan Converting Polygons and Lines. Inside or outside a polygon? Scan conversion.

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1 Lecture outine Graphics and Interaction Scan Converting Poygons and Lines Department of Computer Science and Software Engineering The Introduction Scan conversion Scan-ine agorithm Edge coherence Bresenham s agorithm Edge coherence (revisited) Scan conversion Inside or outside a poygon? How are poygons and ines drawn onto a digita image? Aim: understanding poygona fiing and ine drawing agorithms. Reading: Foey Sections 3.2 Scan converting ines and Section 3.5 Fiing poygons. Additiona reading: 3.3 Scan converting circes. Fiing poygons requires a way of determining whether a point is inside or outside a poygon. See if you can think up a technique for determining whether you are inside or outside a two-dimensiona poygon from a competey arbitrary point (imagine you have no way of knowing whether you are inside or out initiay)? Hint: imagine you are standing at a point q and watching another point p that can reside at any vertex of poygon P.

2 Fiing poygons using winding numbers Fiing poygons requires a way of determining whether a point is inside or outside a poygon. The winding number of q with respect to P is the number of revoutions point p makes around poygon P: the tota signed anguar turn divided by 2π. q One technique is known as winding numbers. Imagine you are standing at a point q whie watching another point p traverse a poygon P countercockwise (the point moves from vertex to vertex around the perimeter of the poygon). P If q is inside, you woud turn a fu circe, 2π radians, if q is outside the sum of your tota anguar turn wi be zero 0. For poygon P, exterior points q have winding number 0: a tota anguar turn of 0. Ange θ can be determined from dot or cross project The odd-even test v i v i+1 = v i v i+1 cosθ i and v i v i+1 = v i v i+1 sinθ i P[i+1] e P[i] V i+1 i q v i Ange θ i is the ange subtended by e from q The odd-even test foow an arbitrary ray or aong scan ine and count the number of times a boundary is crossed, either an odd or even number of times.

3 Seection of inside mode Describe each of the two cooured ines and which direction they are going in and what happens to their parity. notice for the top most ine starting outside and skimming the top of the poygon, it stays outside because the singe pixe is a specia case of a horizonta ine (more in next sides). Let s now compare the winding-number mode to the odd-even test mode according to our desidera for issues in numerica methods (numerica precision and computationa compexity). Winding numbers work in case when q ies on the perimeter of P, but the odd-even case needs to treat these as specia case (e.g. when horizonta scan ine intersects a horizonta poygona edge). However, winding numbers require foating point numbers that is the winding ange wi not aways be 0 or 2π because of round-off error. In terms of computationa compexity, both agorithms are of order O(n) in the worst case, where n is the number of poygon vertices for winding numbers of pixes for the odd-even test. However, the winding-number agorithm depends criticay on foating-point computations and trigonometric computations, which is means it is significanty sower on standard hardware (up to 20 times sower according to research by Haines in 1992). For fiing poygons in images, the odd-even test is superior as it ony requires a parity bit and can be easiy impemented by stepping aong the scan ines of image rows and coumns. Scan-ine fiing The scan-ine method is an efficient way of fiing-in poygons, based on the odd-even test by crossing segments. Segments are sorted to aow for an ordery progression that determines whether inside or outside.

4 Figure 1.14 (Rowe), using scan ine to determine the interior points of a poygon scan ine B A Question: does scan-ine fiing work for non-simpe and mutipe-boundary poygons? F E D C The first intersection occurs when the scan ine crosses edge AB, so the parity becomes odd at that point. The second intersection occurs with edge AF, so the parity becomes even. We can therefore say that points aong the scan ine between edges AB and AF are interior, points between edges AF and EF are exterior, and points between EF and DE are interior. Scan-ine fiing does work for both non-simpe and mutipe boundary poygons (provided you start scanning from outside. Basic scan-ine agorithm Pixe extrema 1. Find intersections of scan ines with edges of poygon(s) 2. Sort the intersections by increasing x coordinate 3. fi in-between pixe extrema using odd-even parity rue (a) (b) Span extrema Other pixes in the span

5 Specia cases 1. Given an intersection with an arbitrary, fractiona x vaue, how do we determine which pixe on either side of the intersection is interior? 2. How do we dea with intersections at integer pixe coordinates? 3. How do we dea with shared vertices? 4. How do we dea with horizonta edges? Specia case soutions 1. If inside and approaching fraction intersection to the right round x coordinate down; ese if outside round up. 2. If eftmost pixe (has integer coordinate) define as interior; ese if right most coordinate (has integer coordinate) define to be exterior. 3. Count the y min vertex of an edge in parity but not the y max vertex (therefore vertex y max is ony drawn if it is the y min vertex for the adjacent edge) For exampe vertex A in 3.14, is counted one in the parity cacuation because it is the y min vertex for edge FA but the y max vertex for edge AB (Thus, both edges and spans are treated as intervas that are cosed at their minimum vaue and open at their maximum vaue). Horizonta edges The bucket-sorted edge tabe right for the poygon on the eft (Foey Figures 3.14 and 3.18). I J G H C F D E D F Scan ine a b c d E C A B y coordinate EF DE CD AB FA BC A B y max 3 7 x min m

6 Active edge tabe for scan ine 9 (above) and 10 (beow) (Foey Figures 3.14 and 3.19) D F Scan ine a b c d E C A B AET pointer FA y max x 1 m EF DE CD AET pointer DE (a) (b) CD Need to fi discreet pixe ocations, but don t want to have to check a edges from a poygons for arge scenes - woud take too much computationay! Therefore sort edge ocations based on x and y coordinates and use a hash-tabe ike approach of retrieving them, reative to what ocation you are presenty scanning (or fiing). Foey s Figure 3.18 is shown aong side figure 3.14 to indicate the associated bucket-sorted edge tabe generated (ahead of actua fiing operation) for this poygon. Scan-ine poygon fiing agorithm 1 Create a data structure for edges, containing the minimum and maximum y vaues (y min and y max, respectivey) for each edge, and the x vaue corresponding to y min, caed x min. 2 Create an edge tabe in which the edges are isted, sorted in ascending order by their y min vaue. Omit any horizonta edges from the tabe. 3 Set scan ine vaue y scan to the minimum y min vaue for a edges (that is, to the owest y vaue reached by the poygon). 4 Create an active edge tabe, containing a the edges currenty intersected by the scan ine. 5 Add to the active edge tabe (and deete from the master edge tabe) any edges whose y min vaue is equa to the current y scan. Scan-ine poygon fiing agorithm 6 Sort the active edge tabe in ascending order on x. 7 Traverse active edge tabe, taking pairs of edges and drawing a pixes between x vaue for the first edge in a pair and the x vaue in the second edge in the same pair. 8 Move to the next scan ine by incrementing y scan by 1. 9 Deete any edges from the active edge tabe if their y max vaue is equa to y scan. 10 Determine where scan ine intersects edge (see edge coherence side in next ecture). 11 Whie either master edge tabe or the active edge tabe has edges, goto step 5.

7 Define a data structure for representing an edge. It must be remembered that x min is not necessariy the minimum x vaue on the edge it is the x vaue that corresponds to the point with minimum y vaue. The edge tabe ists edges, as represented by the data structures defined in step 1. In practice, an array or inked ist is used for the edge tabe. (Omitting horizonta edges, ike in previous diagram, treat as a specia outside case). This is easiy done by taking the y min vaue from the first entry in the edge tabe, since the tabe was sorted on y min. We begin scanning the poygon at this point (ony start scanning at top of poygon). Like the origina edge tabe, this second active edge tabe is usuay impemented as an array or inked ist (ony need to do this once). Edges are therefore distributed between active edge tabe and master edge tabe (oops from step 11, so do this at beginning of each scan ine). Sorting the edges in the active edge tabe means they are isted in the order that they are intersected by the scan ine (as it traves eft to right). Traversing active edge tabe For exampe, with the scan ine positioned as shown in Fig 1.14, pixes aong this ine between edges AB and AF woud be drawn from the first pair of edges in the active edge tabe, then pixes between EF and ED. (Note that the active edge tabe wi aways contain an even number of edges). Move to next scan ine. Edge coherence Deete edges from active edge tabe Note that we do not draw the pixe corresponding to the y max vaue before deeting the edge this is consistent with the rue ed in determining the parity of a pixe. Determine edge coherence Repeat agorithm by going back to step 5 and repeating, whie either master edge tabe or the active edge tabe has edges. Many edges that intersect scan ine i aso intersect scan ine i + 1 Can incrementay cacuate intersection x i+1 = x i + a m, where m is the sope of the edge Can avoid fractiona arithmetic by computing an integer decision variabe and checking ony its sign to decide when to increment x (as we wi see ater in ecture)

8 Scan converting ines Invoves turning on the correct pixes, either in dispay or in [ an ] x0 image. Assume ine segment is described by its start point y [ ] 0 x1 and end point. y 1 First attempt Choose function y = mx + b, and use the foowing basic iterative agorithm that steps through the range of x vaues [x 0..x 1 ]: For each x compute corresponding (rea) y vaue from the ine equation. Round that y vaue to the nearest [ ] pixe (grid) position and x switch on, or set pixe at that position. y Exampe: Drawing y = 7x + 2 (Rowe fig 1.4). (0,40) Increasing x The probem: a ine with sope m > 1 wi have gaps in it. Increasing y First, the probem that foating point arithmetic is computationay compex therefore sow to compute, particuary since this needs to be done very frequenty, and Second, the probem of discretisation in determining which pixe to turn on, one or both? Soution: m > 1 draw x as function of y m 1 draw y as function of x (as above)

9 Refinement Reduce ine generation to one very specia case, ike 0 m 1 Hande a other cases as symmetries of the specia case, by parameterised subroutines expicit transformations dupicating code Leads to kind of ine generation agorithm where step is aways 0or1. Further refinement: Bresenham s agorithm and variants ony integer arithmetic, ony addition/ subtraction. Bresenham s agorithm Bresenham s insight was to step horizontay, accumuating error unti error gets big enough to force a step upwards. For integer coordinates a quantities are ratios with 2 or Δx = x 1 x 0 in the denominator, mutipying everything by 2Δx, makes everything integer. Essentia insight: For integer coordinates, a quantities are ratios with 2 or Δx = x 1 x 0 in the denominator. Mutipying everything by 2Δx, makes everything integer. Step horizontay, accumuating error, unti error gets big enough to force a step upwards. The recursive probem At each step, choose between two pixes based on the sign of the decision variabe cacuated in the previous iteration: then it updates the decision variabe by adding the error to the od vaue, depending on the choice of pixe. Assuming starting point (x 0, y 0 ) is provided, the recursive probem is given (x k, y k ) to find y vaue for subscript position x k + 1, y = m(x k + 1)+b y is either y k or y k + 1, assuming that agorithm is drawing ine from eft to right as in first case (positive sope < 1). Initiay, y = mx k + b, nextrowweget y = m(x k + 1)+b and either y = y + k or y = y k + 1.

10 (x k, y k ) d 2 d 1 y k + 1 d 1 = y y k = m(x k + 1)+b y k d 2 = y k + 1 y = y k + 1 m(x k + 1) b d 1 d 2 = 2m(x k + 1)+2b 2y k 1 The sign of the difference d 1 d 2 determines which pixe to choose: if difference is negative then y k is coser to actua ine (y), if it is positive y k + 1 is coser. How are non-integer cacuations eiminated? y k Bresenham requires that the endpoints of ine segments (x 0, y 0 ) and (x n, y n ) are provided, therefore the sope m can be expressed as a ratio of integers, m = Δy Δx, substitution gives d 1 d 2 = 2m(x k + 1)+2b 2y k 1 Δx(d 1 d 2 ) = 2Δyx k 2Δxy k +[2Δy +Δx(2b 1)] As Δx is aways 1 and [2Δy +Δx(2b 1)] does not depend on pixe index k, we can write C = 2Δy +Δx(2b 1). The pixe choice parameter p k can now be defined as p k =Δx(d 1 d 2 )=2Δyx k 2Δxy k + C If p k is positive, we choose y k + 1; if negative, y k. Now how do we get rid of C? The answer is to use recursion for cacuating a series of p k vaues, the pixe choice at the next ocation is determined by p k+1, and C can be eiminated by subtraction, as foows p k = Δx(d 1 d 2 )=2Δyx k 2Δxy k + C p k+1 = 2Δyx k+1 2Δxy k+1 + C p k+1 p k = 2Δy(x k+1 x k ) 2Δx(y k+1 y k ) For initiaisation, a starting vaue for the first pixe, p 0,is required, using equation (see earier) Δx(d 1 d 2 )=2Δyx k 2Δxy k +[2Δy +Δx(2b 1)] and using starting point of the ine (x 0, y o ) and substituting the foowing for b and since agorithm steps aong x axis, x k+1 = x + k + 1, can recursivey define a reationship for choosing pixes as p k+1 = p k + 2Δy 2Δx(y k+1 y k ) we obtain b = y 0 mx 0 = y 0 (Δy/Δx)x 0 p 0 = 2Δy Δx Note that a quantities in this equation are integers.

11 Bresenham s agorithm Impementation of Bresenham s agorithm 1. Provide endpoints (x 0, y 0 ) and (x n, y n ) of ine. 2. Cacuate constants 2Δy and 2Δy 2Δx, where Δy = y n y 0 and Δx = x n x Pot the origin (x 0, y 0 ). 4. Cacuate the pixe choice parameter p k, and if pk < 0, pot (x k + 1, y k ) and p k+1 = p k + 2Δy Impementation of Bresenham s agorithm for drawing ines based on mathematica equation y = mx + b for either (i) positive sope < 1 (ony this case covered here) (ii) negative sope > 1, or (iii) positive sope 1. For the discrete case, subscript k is added to pixe x k, the equation for a ine then becomes otherwise pot (x k + 1, y k + 1) and p k+1 = p k + 2Δy 2Δx 5. Repeat step 4, unti x n is reached. y = mx k + b however, in genera y wi not be an integer. / / Bresenham s agorithm for ine drawing void drawbresenhamline ( Graphics g, Point start, Point end){ / / Impements case for sope < 1 if ( m_sopelt1 ) { x = m_x0 + 1 ; y = m_y0 ; whie ( x < m_x1 ) { if ( m_param < 0) { m_param += m_twodetay ; } ese { m_param += m_twodydx ; y += m_sopesign ; } drawpixe (g, x, y ); x++; } Edge coherence (revisited) In principe, finding the intersection of the point between a scan ine and a poygon edge just invoves substituting y = y scan into equation for a ine and soving for s. However, it suffers from same probem as drawing ines, in that it invoves foating point arithmetic. It turns out that there is a more efficient agorithm which aows us to cacuate the x vaue on a scan ine directy from the x vaue on the previous scan ine (ike Bresenham s agorithm), based on the edge coherence property of a poygon.

12 The idea behind coherence is that most geometric figures have regions where properties are reativey consistent. In edge coherence, this means that non-horizonta edge in a poygon wi be intersected by more than one scan ine and therefore shoud be abe to generate intersection points iterativey, rather than using the ine equation at each step. Suggests a recursive agorithm cosey reated to Bresenham s ine drawing agorithm. Edge coherence agorithm 1. When a scan ine is added to the active edge tabe, initiaise x to x min and the increment to 0. Cacuate and store Δx and Δy. 2. When a scan ine is increased by 1, add Δx to the increment. 3. If the magnitude of the sope is greater than or equa to 1, compare the magnitude of the increment with Δy. Ifitis ess than Δy, eave x as it is. Otherwise, increase x by 1, and subtract (add) Δy to the increment if the increment is positive (negative). 4. If magnitude of sope is ess than 1, determine number of pixes to increase x by dividing magnitude of increment by Δy (ignoring remainder). The remainder (positive if increment is positive, and negative if it is negative) becomes new vaue of the increment. Summary Poygons can be fied using a scan-ine approach based on the odd-even to determine whether inside or outside. Efficient scan-ine agorithms rey on bucket (hash) sorting edges and maintaining active-edge tabes. This aows ony edges of reevant poygons to be checked. Efficient ine drawing agorithms typicay utiise the edge coherence property to cacuate the vaue of the pixe on the current scan ine from the vaue on previous scan ine. Bresenham s insight was how to rey ony on integer operations, thus avoiding foating point arithmetic. Edge-coherence properties can be utiised in poygon fiing agorithms, to decide whether x vaues in the active-edge tabe need incrementing. Can use same approach for drawing circes and eipses

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