Course Website: http://www.comp.dit.ie/bmacnamee Computer Graphics 4: Bresenham Line Drawing Algorithm, Circle Drawing & Polygon Filling
2 of 39 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
3 of 39 The Bresenham Line Algorithm The Bresenham algorithm is another incremental scan conversion algorithm The big advantage of this algorithm is that it uses only integer calculations Jack Bresenham worked for 27 years at IBM before entering academia. Bresenham developed his famous algorithms at IBM in the early 1960s
4 of 39 The Big Idea Move across the x axis in unit intervals and at each step choose between two different y coordinates 2 3 4 5 2 4 3 5 For example, from position (2, 3) we have to choose between (3, 3) and (3, 4) We would like the point that is closer to the original line (xk, yk) (xk+1, yk) (xk+1, yk+1)
5 of 39 The y coordinate on the mathematical line at xk+1 is: Deriving The Bresenham Line Algorithm At sample position xk+1 the vertical separations from the mathematical line are labelled dupper and dlower bxmy k ++= )1( y yk yk+1 xk+1 dlower dupper
6 of 39 So, dupper and dlower are given as follows: and: We can use these to make a simple decision about which pixel is closer to the mathematical line Deriving The Bresenham Line Algorithm (cont…) klower yyd −= kk ybxm −++= )1( yyd kupper −+= )1( bxmy kk −+−+= )1(1
7 of 39 This simple decision is based on the difference between the two pixel positions: Let’s substitute m with ∆y/∆x where ∆x and ∆y are the differences between the end-points: Deriving The Bresenham Line Algorithm (cont…) 122)1(2 −+−+=− byxmdd kkupperlower )122)1(2()( −+−+ ∆ ∆ ∆=−∆ byx x y xddx kkupperlower )12(222 −∆+∆+⋅∆−⋅∆= bxyyxxy kk cyxxy kk +⋅∆−⋅∆= 22
8 of 39 So, a decision parameter pk for the kth step along a line is given by: 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 Deriving The Bresenham Line Algorithm (cont…) cyxxy ddxp kk upperlowerk +⋅∆−⋅∆= −∆= 22 )(
9 of 39 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: Subtracting pk from this we get: Deriving The Bresenham Line Algorithm (cont…) cyxxyp kkk +⋅∆−⋅∆= +++ 111 22 )(2)(2 111 kkkkkk yyxxxypp −∆−−∆=− +++
10 of 39 But, xk+1 is the same as xk+1 so: 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: Deriving The Bresenham Line Algorithm (cont…) )(22 11 kkkk yyxypp −∆−∆+= ++ xyp ∆−∆= 20
11 of 39 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: 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: xyp ∆−∆= 20 ypp kk ∆+=+ 21
12 of 39 The Bresenham Line Algorithm (cont…) ACHTUNG! The algorithm and derivation above assumes slopes are less than 1. for other slopes we need to adjust the algorithm slightly Otherwise, the next point to plot is (xk+1, yk+1) and: 5. Repeat step 4 (Δx – 1) times xypp kk ∆−∆+=+ 221
13 of 39 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
14 of 39 Bresenham Example (cont…) 17 16 15 14 13 12 11 10 18 292726252423222120 28 30 k pk (xk+1,yk+1) 0 1 2 3 4 5 6 7 8 9
15 of 39 Bresenham Exercise Go through the steps of the Bresenham line drawing algorithm for a line going from (21,12) to (29,16)
16 of 39 Bresenham Exercise (cont…) 17 16 15 14 13 12 11 10 18 292726252423222120 28 30 k pk (xk+1,yk+1) 0 1 2 3 4 5 6 7 8
17 of 39 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
18 of 39 A Simple Circle Drawing Algorithm The equation for a circle is: 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: 222 ryx =+ 22 xry −±=
19 of 39 A Simple Circle Drawing Algorithm (cont…) 20020 22 0 ≈−=y 20120 22 1 ≈−=y 20220 22 2 ≈−=y 61920 22 19 ≈−=y 02020 22 20 ≈−=y
20 of 39 A Simple Circle Drawing Algorithm (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
21 of 39 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) (y, x) (y, -x) (x, -y)(-x, -y) (-y, -x) (-y, x) (-x, y) 2 R
22 of 39 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 only ever calculate the points for the top right eighth of a circle, and then use symmetry to get the rest of the points The mid-point circle algorithm was developed by Jack Bresenham, who we heard about earlier. Bresenham’s patent for the algorithm can be viewed here.
23 of 39 Mid-Point Circle Algorithm (cont…) (xk+1, yk) (xk+1, yk-1) (xk, yk) Assume that we have just plotted point (xk, yk) The next point is a choice between (xk+1, yk) 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?
24 of 39 Mid-Point Circle Algorithm (cont…) Let’s re-jig the equation of the circle slightly to give us: The equation evaluates as follows: By evaluating this function at the midpoint between the candidate pixels we can make our decision 222 ),( ryxyxfcirc −+=      > = < ,0 ,0 ,0 ),( yxfcirc boundarycircletheinsideis),(if yx boundarycircleon theis),(if yx boundarycircletheoutsideis),(if yx
25 of 39 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: 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 222 ) 2 1()1( ) 2 1,1( ryx yxfp kk kkcirck −−++= −+=
26 of 39 Mid-Point Circle Algorithm (cont…) To ensure things are as efficient as possible we can do all of our calculations incrementally First consider: or: where yk+1 is either yk or yk-1 depending on the sign of p ( ) ( ) 2 2 1 2 111 2 1]1)1[( 2 1,1 ryx yxfp kk kkcirck −−+++= −+= + +++ 1)()()1(2 1 22 11 +−−−+++= +++ kkkkkkk yyyyxpp
27 of 39 Mid-Point Circle Algorithm (cont…) The first decision variable is given as: Then if pk < 0 then the next decision variable is given as: If pk > 0 then the decision variable is: r rr rfp circ −= −−+= −= 4 5 ) 2 1(1 ) 2 1,1( 22 0 12 11 ++= ++ kkk xpp 1212 11 +−++= ++ kkkk yxpp
28 of 39 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: • Calculate the initial value of the decision parameter as: • 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: ),0(),( 00 ryx = rp −= 4 5 0 12 11 ++= ++ kkk xpp
29 of 39 The Mid-Point Circle Algorithm (cont…) Otherwise the next point along the circle is (xk+1, yk-1) and: 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: 6. Repeat steps 3 to 5 until x >= y 111 212 +++ −++= kkkk yxpp cxxx += cyyy +=
30 of 39 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
31 of 39 Mid-Point Circle Algorithm Example (cont…) 9 7 6 5 4 3 2 1 0 8 976543210 8 10 10 k pk (xk+1,yk+1) 2xk+1 2yk+1 0 1 2 3 4 5 6
32 of 39 Mid-Point Circle Algorithm Exercise Use the mid-point circle algorithm to draw the circle centred at (0,0) with radius 15
33 of 39 Mid-Point Circle Algorithm Example (cont…) k pk (xk+1,yk+1) 2xk+1 2yk+1 0 1 2 3 4 5 6 7 8 9 10 11 12 9 7 6 5 4 3 2 1 0 8 976543210 8 10 10 131211 14 15 13 12 14 11 16 1516
34 of 39 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
35 of 39 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
36 of 39 Scan-Line Polygon Fill Algorithm 2 4 6 8 10 Scan Line 0 2 4 6 8 10 12 14 16
37 of 39 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
38 of 39 Scan-Line Polygon Fill Algorithm (cont…)
39 of 39 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
40 of 39 Anti-Aliasing
41 of 39 Summary Of Drawing Algorithms
42 of 39 Mid-Point Circle Algorithm (cont…) 6 2 3 41 5 4 3
43 of 39 Mid-Point Circle Algorithm (cont…) M 6 2 3 41 5 4 3
44 of 39 Mid-Point Circle Algorithm (cont…) M 6 2 3 41 5 4 3
45 of 39 Blank Grid
46 of 39 Blank Grid
47 of 39 Blank Grid 9 7 6 5 4 3 2 1 0 8 976543210 8 10 10
48 of 39 Blank Grid

bresenham circles and polygons in computer graphics(Computer graphics tutorials)

  • 1.
    Course Website: http://www.comp.dit.ie/bmacnamee ComputerGraphics 4: Bresenham Line Drawing Algorithm, Circle Drawing & Polygon Filling
  • 2.
    2 of 39 Contents In today’s lecturewe’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
  • 3.
    3 of 39 The Bresenham LineAlgorithm The Bresenham algorithm is another incremental scan conversion algorithm The big advantage of this algorithm is that it uses only integer calculations Jack Bresenham worked for 27 years at IBM before entering academia. Bresenham developed his famous algorithms at IBM in the early 1960s
  • 4.
    4 of 39 The Big Idea Moveacross the x axis in unit intervals and at each step choose between two different y coordinates 2 3 4 5 2 4 3 5 For example, from position (2, 3) we have to choose between (3, 3) and (3, 4) We would like the point that is closer to the original line (xk, yk) (xk+1, yk) (xk+1, yk+1)
  • 5.
    5 of 39 The y coordinateon the mathematical line at xk+1 is: Deriving The Bresenham Line Algorithm At sample position xk+1 the vertical separations from the mathematical line are labelled dupper and dlower bxmy k ++= )1( y yk yk+1 xk+1 dlower dupper
  • 6.
    6 of 39 So, dupper anddlower are given as follows: and: We can use these to make a simple decision about which pixel is closer to the mathematical line Deriving The Bresenham Line Algorithm (cont…) klower yyd −= kk ybxm −++= )1( yyd kupper −+= )1( bxmy kk −+−+= )1(1
  • 7.
    7 of 39 This simple decisionis based on the difference between the two pixel positions: Let’s substitute m with ∆y/∆x where ∆x and ∆y are the differences between the end-points: Deriving The Bresenham Line Algorithm (cont…) 122)1(2 −+−+=− byxmdd kkupperlower )122)1(2()( −+−+ ∆ ∆ ∆=−∆ byx x y xddx kkupperlower )12(222 −∆+∆+⋅∆−⋅∆= bxyyxxy kk cyxxy kk +⋅∆−⋅∆= 22
  • 8.
    8 of 39 So, a decisionparameter pk for the kth step along a line is given by: 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 Deriving The Bresenham Line Algorithm (cont…) cyxxy ddxp kk upperlowerk +⋅∆−⋅∆= −∆= 22 )(
  • 9.
    9 of 39 Remember coordinate changesoccur 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: Subtracting pk from this we get: Deriving The Bresenham Line Algorithm (cont…) cyxxyp kkk +⋅∆−⋅∆= +++ 111 22 )(2)(2 111 kkkkkk yyxxxypp −∆−−∆=− +++
  • 10.
    10 of 39 But, xk+1 isthe same as xk+1 so: 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: Deriving The Bresenham Line Algorithm (cont…) )(22 11 kkkk yyxypp −∆−∆+= ++ xyp ∆−∆= 20
  • 11.
    11 of 39 The Bresenham LineAlgorithm 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: 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: xyp ∆−∆= 20 ypp kk ∆+=+ 21
  • 12.
    12 of 39 The Bresenham LineAlgorithm (cont…) ACHTUNG! The algorithm and derivation above assumes slopes are less than 1. for other slopes we need to adjust the algorithm slightly Otherwise, the next point to plot is (xk+1, yk+1) and: 5. Repeat step 4 (Δx – 1) times xypp kk ∆−∆+=+ 221
  • 13.
    13 of 39 Bresenham Example Let’s havea 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
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    15 of 39 Bresenham Exercise Go throughthe steps of the Bresenham line drawing algorithm for a line going from (21,12) to (29,16)
  • 16.
  • 17.
    17 of 39 Bresenham Line AlgorithmSummary 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
  • 18.
    18 of 39 A Simple CircleDrawing Algorithm The equation for a circle is: 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: 222 ryx =+ 22 xry −±=
  • 19.
    19 of 39 A Simple CircleDrawing Algorithm (cont…) 20020 22 0 ≈−=y 20120 22 1 ≈−=y 20220 22 2 ≈−=y 61920 22 19 ≈−=y 02020 22 20 ≈−=y
  • 20.
    20 of 39 A Simple CircleDrawing Algorithm (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
  • 21.
    21 of 39 Eight-Way Symmetry The firstthing we can notice to make our circle drawing algorithm more efficient is that circles centred at (0, 0) have eight-way symmetry (x, y) (y, x) (y, -x) (x, -y)(-x, -y) (-y, -x) (-y, x) (-x, y) 2 R
  • 22.
    22 of 39 Mid-Point Circle Algorithm Similarlyto 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 only ever calculate the points for the top right eighth of a circle, and then use symmetry to get the rest of the points The mid-point circle algorithm was developed by Jack Bresenham, who we heard about earlier. Bresenham’s patent for the algorithm can be viewed here.
  • 23.
    23 of 39 Mid-Point Circle Algorithm(cont…) (xk+1, yk) (xk+1, yk-1) (xk, yk) Assume that we have just plotted point (xk, yk) The next point is a choice between (xk+1, yk) 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?
  • 24.
    24 of 39 Mid-Point Circle Algorithm(cont…) Let’s re-jig the equation of the circle slightly to give us: The equation evaluates as follows: By evaluating this function at the midpoint between the candidate pixels we can make our decision 222 ),( ryxyxfcirc −+=      > = < ,0 ,0 ,0 ),( yxfcirc boundarycircletheinsideis),(if yx boundarycircleon theis),(if yx boundarycircletheoutsideis),(if yx
  • 25.
    25 of 39 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: 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 222 ) 2 1()1( ) 2 1,1( ryx yxfp kk kkcirck −−++= −+=
  • 26.
    26 of 39 Mid-Point Circle Algorithm(cont…) To ensure things are as efficient as possible we can do all of our calculations incrementally First consider: or: where yk+1 is either yk or yk-1 depending on the sign of p ( ) ( ) 2 2 1 2 111 2 1]1)1[( 2 1,1 ryx yxfp kk kkcirck −−+++= −+= + +++ 1)()()1(2 1 22 11 +−−−+++= +++ kkkkkkk yyyyxpp
  • 27.
    27 of 39 Mid-Point Circle Algorithm(cont…) The first decision variable is given as: Then if pk < 0 then the next decision variable is given as: If pk > 0 then the decision variable is: r rr rfp circ −= −−+= −= 4 5 ) 2 1(1 ) 2 1,1( 22 0 12 11 ++= ++ kkk xpp 1212 11 +−++= ++ kkkk yxpp
  • 28.
    28 of 39 The Mid-Point CircleAlgorithm 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: • Calculate the initial value of the decision parameter as: • 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: ),0(),( 00 ryx = rp −= 4 5 0 12 11 ++= ++ kkk xpp
  • 29.
    29 of 39 The Mid-Point CircleAlgorithm (cont…) Otherwise the next point along the circle is (xk+1, yk-1) and: 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: 6. Repeat steps 3 to 5 until x >= y 111 212 +++ −++= kkkk yxpp cxxx += cyyy +=
  • 30.
    30 of 39 Mid-Point Circle AlgorithmExample To see the mid-point circle algorithm in action lets use it to draw a circle centred at (0,0) with radius 10
  • 31.
    31 of 39 Mid-Point Circle AlgorithmExample (cont…) 9 7 6 5 4 3 2 1 0 8 976543210 8 10 10 k pk (xk+1,yk+1) 2xk+1 2yk+1 0 1 2 3 4 5 6
  • 32.
    32 of 39 Mid-Point Circle AlgorithmExercise Use the mid-point circle algorithm to draw the circle centred at (0,0) with radius 15
  • 33.
    33 of 39 Mid-Point Circle AlgorithmExample (cont…) k pk (xk+1,yk+1) 2xk+1 2yk+1 0 1 2 3 4 5 6 7 8 9 10 11 12 9 7 6 5 4 3 2 1 0 8 976543210 8 10 10 131211 14 15 13 12 14 11 16 1516
  • 34.
    34 of 39 Mid-Point Circle AlgorithmSummary 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
  • 35.
    35 of 39 Filling Polygons So wecan 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
  • 36.
    36 of 39 Scan-Line Polygon FillAlgorithm 2 4 6 8 10 Scan Line 0 2 4 6 8 10 12 14 16
  • 37.
    37 of 39 Scan-Line Polygon FillAlgorithm 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
  • 38.
  • 39.
    39 of 39 Line Drawing Summary Overthe 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
  • 40.
  • 41.
  • 42.
    42 of 39 Mid-Point Circle Algorithm(cont…) 6 2 3 41 5 4 3
  • 43.
    43 of 39 Mid-Point Circle Algorithm(cont…) M 6 2 3 41 5 4 3
  • 44.
    44 of 39 Mid-Point Circle Algorithm(cont…) M 6 2 3 41 5 4 3
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  • 48.