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Lect3cg2011

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• Important algorithm.
• Very important.
• Only principle of this method important.
• Lect3cg2011

1. 1. Computer Graphics Computer Graphics Lecture 3 Line & Circle Drawing
2. 2. Computer Graphics Towards the Ideal Line • We can only do a discrete approximation • Illuminate pixels as close to the true path as possible, consider bi-level display only – Pixels are either lit or not lit 2
3. 3. Computer Graphics What is an ideal line • Must appear straight and continuous – Only possible axis-aligned and 45o lines • Must interpolate both defining end points • Must have uniform density and intensity – Consistent within a line and over all lines – What about antialiasing? • Must be efficient, drawn quickly – Lots of them are required!!! 3
4. 4. Computer Graphics Simple Line Based on slope-intercept algorithm from algebra: y = mx + b Simple approach: increment x, solve for y Floating point arithmetic required 4
5. 5. Computer Graphics Does it Work?It seems to work okay for lines witha slope of 1 or less,but doesn’t work well for lines withslope greater than 1 – lines becomemore discontinuous in appearanceand we must add more than 1 pixelper column to make it work.Solution? - use symmetry. 5
6. 6. Computer Graphics Modify algorithm per octantOR, increment along x-axis if dy<dx else increment along y-axis 6
7. 7. Computer Graphics DDA algorithm • DDA = Digital Differential Analyser – finite differences • Treat line as parametric equation in t : Start point - ( x1 , y1 ) x(t ) = x1 + t ( x2 − x1 ) End point - ( x2 , y2 ) y (t ) = y1 + t ( y2 − y1 ) 7
8. 8. Computer Graphics DDA Algorithm x(t ) = x1 + t ( x2 − x1 ) y (t ) = y1 + t ( y2 − y1 ) dx • Start at t = 0 xnew = xold + dt • At each step, increment t by dt ynew = yold + dy dt • Choose appropriate value for dt dx = x2 − x1 • Ensure no pixels are missed: dy = y2 − y1 – Implies: dx and dy <1 <1 dt dt • Set dt to maximum of dx and dy 8
9. 9. Computer Graphics DDA algorithm line(int x1, int y1, int x2, int y2) { n - range of t. float x,y; int dx = x2-x1, dy = y2-y1; int n = max(abs(dx),abs(dy)); float dt = n, dxdt = dx/dt, dydt = dy/dt; x = x1; y = y1; while( n-- ) { point(round(x),round(y)); x += dxdt; y += dydt; } } 9
10. 10. Computer Graphics DDA algorithm • Still need a lot of floating point arithmetic. – 2 ‘round’s and 2 adds per pixel. • Is there a simpler way ? • Can we use only integer arithmetic ? – Easier to implement in hardware. 10
11. 11. Computer Graphics Observation on lines. while( n-- ) { draw(x,y); move right; if( below line ) move up; } 11
12. 12. Computer Graphics Testing for the side of a line. • Need a test to determine which side of a line a pixel lies. • Write the line in implicit form: F ( x, y ) = ax + by + c = 0 •If (b<0) F<0 for points above the line, F>0 for points below. 12
13. 13. Computer Graphics Testing for the side of a line. F ( x, y ) = ax + by + c = 0 • Need to find coefficients a,b,c. • Recall explicit, slope-intercept form : dy y = mx + b and so y = x+b dx • So: F ( x, y ) = dy.x − dx. y + c = 0 13
14. 14. Computer Graphics Decision variable. Let’s assume dy/dx < 0.5 (we can use symmetry) Evaluate F at point M 1 d = F ( x p + 1, y p + ) 2 Referred to as decision variable NE M E Previous Choices for Choices for Pixel Current pixel Next pixel (xp,yp)
15. 15. Computer Graphics Decision variable.Evaluate d for next pixel, Depends on whether E or NE Is chosen :If E chosen : 1 1d new = F ( x p + 2, y p + ) = a ( x p + 2) + b( y p + ) + c 2 2 But recall : 1 d old = F ( x p + 1, y p + ) NE 2 1 M = a ( x p + 1) + b( y p + ) + c 2 E Previous Choices for So : d new = d old + a Pixel Choices for Next pixel (xp,yp) Current pixel = d old + dy 15
16. 16. Computer Graphics Decision variable.If NE was chosen : 3 3d new = F ( x p + 2, y p + ) = a ( x p + 2) + b( y p + ) + c 2 2 M So : NE d new = d old + a + b E = d old + dy − dx Previous Choices for Pixel Choices for Next pixel (xp,yp) Current pixel 16
17. 17. Computer Graphics Summary of mid-point algorithm • Choose between 2 pixels at each step based upon the sign of a decision variable. • Update the decision variable based upon which pixel is chosen. • Start point is simply first endpoint (x1,y1). • Need to calculate the initial value for d 17
18. 18. Computer Graphics Initial value of d.Start point is (x1,y1) 1 1 d start = F ( x1 + 1, y1 + ) = a( x1 + 1) + b( y1 + ) + c 2 2 b = ax1 + by1 + c + a + 2 b = F ( x1 , y1 ) + a + 2 But (x1,y1) is a point on the line, so F(x1,y1) =0 d start = dy − dx / 2 Conventional to multiply by 2 to remove fraction ⇒ doesn’t effect sign. 18
19. 19. Computer Graphics Midpoint algorithm void MidpointLine(int x1,y1,x2,y2) while (x < x2) { { if (d<= 0) { int dx=x2-x1; d+=incrE; x++ int dy=y2-y1; } else { int d=2*dy-dx; d+=incrNE; x++; int increE=2*dy; y++; int incrNE=2*(dy-dx); } WritePixel(x,y); x=x1; } y=y1; } WritePixel(x,y); 19
20. 20. Computer Graphics Circle drawing. • Can also use Bresenham to draw circles. • Use 8-fold symmetry E M SE Previous Choices for Choices for Pixel Current pixel Next pixel 20
21. 21. Computer Graphics Circle drawing. • Implicit form for a circle is: f ( x, y ) = ( x − xc ) + ( y − yc ) − r 2 2 2 If SE is chosen d new = d old + (2 x p − 2 y p + 5) If E is chosen d new = d old + (2 x p + 3) • Functions are linear equations in terms of (xp,yp) –Termed point of evaluation 21
22. 22. Computer Graphics Summary of line drawing so far. • Explicit form of line – Inefficient, difficult to control. • Parametric form of line. – Express line in terms of parameter t – DDA algorithm • Implicit form of line – Only need to test for ‘side’ of line. – Bresenham algorithm. – Can also draw circles. 22
23. 23. Computer Graphics Problems with Bresenham algorithm • Pixels are drawn as a single line ⇒ unequal line intensity with change in angle. Pixel density = √2.n pixels/mm Can draw lines in darker colours according to line direction. -Better solution : antialiasing ! -(eg. Gupta-Sproull algorithm) Pixel density = n pixels/mm 23
24. 24. Computer Graphics Gupta-Sproull algorithm.• Calculate the distance of the line and the pixel center• Adjust the colour according to the distance
25. 25. Computer Graphics Gupta-Sproull algorithm.Calculate distance using features of mid-point algorithm D = v cos φ vdx = dx 2 + dy 2 dy NE dx v Angle = ∅ M D E 25
26. 26. Computer Graphics Gupta-Sproull algorithm (cont)Recall from the midpoint algorithm: F ( x, y ) = 2( ax + by + c ) = 0 (doubled to avoid fraction) Line to draw NE yp+1So y = ( ax + c ) m −b D v yp xp E ΘFor pixel E: xp+1 x p +1 = x p + 1 y p +1 = y p v = y − y p +1So: a ( x p + 1) + c v= − yp −b
27. 27. Computer Graphics Gupta-Sproull algorithm (cont) Line to drawFrom previous slide: NE a ( x p + 1) + c v= − yp yp+1 −b mFrom the midpoint computation, y p D v b = − dx xp E Θ xp+1So: vdx = a ( x p + 1) + by p + c = F ( x p + 1, y p ) / 2
28. 28. Computer Graphics Gupta-Sproull algorithm (cont)From the midpoint algorithm, we had the decision variable (remember?) 1 d = F ( M ) = F ( x p +1 , y p + ) 2 Line to drawGoing back to our previous equation: NE yp+1 2vdx = F ( x p + 1, y p ) m = 2a ( x p + 1) + 2by p + 2c py D v Θ = 2a ( x p + 1) + 2b( y p + 1 / 2) − 2b / 2 + 2c xp E xp+1 = F ( x p + 1, y p + 1 / 2) − b = F (M ) − b = d −b = d + dx
29. 29. Computer Graphics Gupta-Sproull algorithm (cont)So, d + dx D= 2 dx 2 + dy 2And the denominator is constantSince we are blurring the line, we also need to compute the distances to points y p – 1 and yp + 1 numerator for y p − 1 ⇒ 2(1 − v)dx = 2dx − 2vdx numerator for y p + 1 ⇒ 2(1 + v)dx = 2dx + 2vdx
30. 30. Gupta-Sproull algorithm (cont)Computer GraphicsIf the NE pixel had been chosen: 2vdx = F ( x p + 1, y p + 1) = 2a( x p + 1) + 2(by p + 1) + 2c = 2a( x p + 1) + 2b( y p + 1 / 2) + 2b / 2 + 2c = F ( x p + 1, y p + 1 / 2) + b = F (M ) + b = d +b = d − dx numerator for y p + 2 ⇒ 2(1 − v)dx = 2dx − 2vdx numerator for y p ⇒ 2(1 + v)dx = 2dx + 2vdx
31. 31. Computer Graphics Gupta-Sproull algorithm (cont) • Compute midpoint line algorithm, with the following alterations: • At each iteration of the algorithm: – If the E pixel is chosen, set numerator = d + dx – If the NE pixel is chosen, set numerator = d – dx – Update d as in the regular algorithm – Compute D = numerator/denominator – Color the current pixel according to D – Compute Dupper = (2dx-2vdx)/denominator – Compute Dlower = (2dx+2vdx)/denominator – Color upper and lower accordingly