Scan Conversion. Lines and Circles
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- Miles Townsend
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1 Scan Conversion Lines and Circles (Chapter 3 in Foley & Van Dam) 2D Line Implicit representation: αx + βy + γ = 0 Explicit representation: y y = mx+ B m= x Parametric representation: x P= y P = t y P + ( P P ) 0 t y x 0 0 [0..1] y1 P1 x0 B x1 x P0 y0
2 Scan Conversion - Lines y = mx + B (x 1,y 1 ) (x 0,y 0 ) slope = m = y 1 - y 0 x 1 - x 0 offset= B = y 1 -mx 1 Assume m 1 Assume x 0 x 1 Basic Algorithm For x = x 0 to x 1 y = mx + B PlotPixel(x,round(y)) For each iteration: 1 float multiplication, 1 addition, 1 Round Incremental Algorithm: y i+1 = mx i+1 + B = m(x i + x) + B = y i + m x if x = 1 then y i+1 = y i + m Algorithm y=y 0 For x = x 0 to x 1 PlotPixel(x,round(y)) y = y + m
3 ( x i +1,Round(y i +m) ) ( x i, y i ) ( x i +1, y i +m ) ( x i,round(y i ) ) Pseudo Code for Basic Line Drawing: Assume x 1 >x 0 and line slope absolute value is 1 Line(x 0,y 0,x 1,y 1 ) begin float dx, dy, x, y, slope; dx := x 1 -x 0 ; dy := y 1 -y 0 ; slope := dy/dx; y := y 0 ; for x:=x 0 to x 1 do begin PlotPixel( x,round(y) ); y := y+slope;
4 Symmetric Cases: m 1 x = x 0 For y = y 0 to y 1 PlotPixel(round(x),y) x = x + 1/m Special Cases: m = ± 1 (diagonals) m = 0, (horizontal, vertical) Symmetric Cases: if x 0 > x 1 for m 1 or y 0 > y 1 for m 1 swap((x 0,y 0 ),(x 1,y 1 )) Basic Line Drawing: For each iteration: 1 addition, 1 Round. Drawback: Accumulated error float arithmetic Round operations Midpoint (~Bresenham) Line Drawing Assumptions: x 0 < x 1, y 0 < y 1 0 < slope < 1 Q NE M (x p,y p ) E Given (x p,y p ): next pixel is E = (x p +1,y p ) or NE = (x p +1,y p +1) Bresenham: sign(m-q) determines NE or E M = (x p +1,y p +1/2)
5 Bresenham s Line Algorithm yi+1 yi y d2 d1 xi xi+1 y = m(xi + 1) + b d1 = y - yi = m(xi +1) + b - yi d2 = (yi + 1) - y = yi m (xi + 1) - b d1 - d2 > 0? Bresenham s Line Algorithm d1 - d2 = 2m(xi + 1) - 2yi + 2b -1 d1 - d2 = 2(dy/dx)(xi + 1) - 2yi + 2b -1 dx(d1-d2) = 2dy*xi + 2dy - 2dx*yi + 2dx*b - dx fi = dx(d1-d2) fi+1 - fi = 2dy(xi+1 - xi) - 2dx(yi+1 - yi) If y is incremented then fi+1 = fi + 2dy-2dx else fi+1 = fi + 2dy
6 Bresenham s Line Algorithm Const1 = 2dy; Const2 = 2dy - 2dx; f = 2dy - dx; set_pixel(x1,y1); x = x1; y = y1; while (x++ < x2){ if (f < 0) f += Const1; else { f += Const2; y++; } set_pixel(x,y); } Bresenham s Line Algorithm Const1 = 2dy; Const2 = 2dy - 2dx; p = A + n*y + x; offset_h = sign(dx); offset_d = sign(dx) + n*sign(dy) f = 2dy - dx; *p = color; d8 = dx; while (d8--){ if (f < 0){ f += Const1; p += offset_h; } else { f += Const2; p += offset_d; } *p = color; }
7 Midpoint Line Drawing (cont.) dy y = x + B dx Implicit form of a line: f(x,y) = ax + by + c = 0 f(x,y) = dy x - dx y + B dx = 0 f(x,y) < 0 f(x,y) = 0 f(x,y) > 0 Decision Variable : d = f(m) = f(x p +1,y p +1/2) = a(x p +1) + b(y p +1/2) + c choose NE if d > 0 choose E if d 0 Incremental Algorithm: What happens at x p + 2? Q NE M M M (x p,y p ) E If E was chosen at x p + 1 M = (x p +2,y p +1/2) d new = f(x p +2,y p +1/2) = a(x p +2) + b(y p +1/2) + c d old = f(x p +1,y p +1/2) = a(x p +1) + b(y p +1/2) + c d new = d old + a = d old + dy d new = d old + E If NE was chosen at x p + 1 M = (x p +2,y p +3/2) d new = f(x p +2,y p +3/2) = a(x p +2) + b(y p +3/2) + c d old = f(x p +1,y p +1/2) = a(x p +1) + b(y p +1/2) + c d new = d old + a + b = d old + dy - dx d new = d old + NE
8 Initialization: Incremental Algorithm: First point = (x 0,y 0 ), first MidPoint = (x 0 +1,y 0 +1/2) d start = f(x 0 +1,y 0 +1/2) = a(x 0 +1) + b(y 0 +1/2) +c = ax 0 + by 0 + c + a + b/2 = f(x 0,y 0 ) + a + b/2 = a + b/2 d start =dy - dx/2 Enhancement: To eliminate fractions, define: f(x,y) = 2(ax + by + c) = 0 d start =2dy - dx Midpoint Line Drawing - Summary The sign of f(x 0 +1,y 0 +1/2) indicates whether to move East or North-East. At the beginning d=f(x 0 +1,y 0 +1/2)=2dy-dx. The increment in d (after this step) is: If we moved East: E =2dy If we moved North-East: ΝΕ =2dy-2dx Comments: Integer arithmetic (dx and dy are integers). One addition for each iteration. No accumulated errors. By symmetry, we deal with 0>slope>-1
9 Pseudo Code for Midpoint Line Drawing: Assume x 1 >x 0 and 0 < slope 1 Line(x 0,y 0,x 1,y 1 ) begin int dx, dy, x, y, d, E, ΝE ; x:= x 0 ; y=y 0 ; dx := x 1 -x 0 ; dy := y 1 -y 0 ; d := 2*dy-dx; E := 2*dy; ΝΕ := 2*(dy-dx); PlotPixel(x,y); while(x < x 1 ) do if (d < 0) then begin d:=d+ E ; x:=x+1; else begin d:=d+ ΝE ; x:=x+1; y:=y+1; PlotPixel(x,y); Drawing Circles Implicit representation (centered at the origin with radius R): x + y R = 0 Explicit representation: y = ± R x Parametric representation: x cos = R y Rsin () t () t 2 2 t [0..2π ] R x
10 Scan Conversion - Circles Basic Algorithm For x = -R to R y = sqrt(r 2 -x 2 ) PlotPixel(x,round(y)) PlotPixel(x,-round(y)) Comments: square-root operations are expensive. Float arithmetic. Large gap for x values close to R. Exploiting Eight-Way Symmetry For a circle centered at the origin: If (x,y) is on the circle then - (y,x) (y,-x) (x,-y) (-x,-y) (-y,-x) (-y,x) (-x,y) are on the circle as well. Therefore we need to compute only one octant (45 o ) segment. (-x,y) (x,y) (-y,x) (y,x) (-y,-x) (y,-x) (-x,-y) (x,-y)
11 Circle Midpoint (for one octant) (The circle is located at (0,0) with radius R) We start from (x 0,y 0 )=(0,R). One can move either East or South-East. Again, d(x,y) will be a threshold criteria at the midpoint. Threshold Criteria d(x,y)=f(x,y) = x 2 + y 2 -R 2 = 0 f(x,y) = 0 f(x,y) < 0 f(x,y) > 0
12 Circle Midpoint (cont.) y0 x 0 At the beginning d start =d(x 0 +1,y 0-1/2) = d(1,r-1/2) = 5/4 - R If d<0 we move East: E = d(x 0 +2,y 0-1/2) - d(x 0 +1,y 0-1/2)=2x 0 +3 if d>0 we move South-East: SE = d(x 0 +2,y 0-3/2) - d(x 0 +1,y 0-1/2) = = 2(x 0 -y 0 )+5 E and SE are not constant anymore. Since d is incremented by integer values, we can use d start = 1-R, yielding an integer algorithm. This has no affect on the threshold criteria. Midpoint Circle Algorithm Circle Octant2 (R) begin int x, y, d; x := 0; y :=R; d := 1-R; PlotPixel(x,y); while ( y>x ) do if ( d<0 ) then /* East */ begin d := d+2x+3; x := x+1; else begin d := d+2(x-y)+5; x := x+1; y := y-1; PlotPixel( x,y ); (*) One may mirror (*), creating the other seven octans.
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