Class 8 - Recursive Pictures

1. Class 8 - Recursive Pictures • Recursively-defined curves • The Hilbert curve

2. Recursive drawings • Many shapes are “self-similar” - their overall structure is repeated in sub-pictures. • Peano curve:

3. Recursive drawings (cont.) “Koch snowflake”:

4. Recursive drawings (cont.)

5. The Hilbert Curve • Created by David Hilbert (1862-1943), this is a “space-filling” curve. • Hilbert curve of order n is constructed from four copies of the Hilbert curve of order n-1, properly oriented and connected.

6. Hilbert curves of order 1 & 2 HC(1): HC(2):

7. Hilbert curve of order 3 HC(3):

8. Hilbert curve of order 4

9. Hilbert curve of order 5

10. n-1 n-1 n-1 n-1 n-1 The Hilbert Curve Pattern Hilbert curves have “open” side on left: Form H.C. of order n by combining four copies of H.C. of order n-1, plus three connecting lines (of length sl):

11. Recursive definition of H.C. This leads to the basic recursive pattern for defining HC(order, sl): int diam = size of HC of order n-1; hcsub1 = HC(order-1, sl); hcul = rotate hcsub1 by -90 degrees; hcur = translate hcsub1 to (diam+sl, 0); hclr = translate hcsub1 to (diam+sl, diam+sl); hcll = rotate hcsub1 by 90 degrees, then translate to (0, diam+sl); hc = append(hcul, append(hcur, append(hclr, hcll))); return hc, with three lines added;

12. Translation We have already seen how to translate a LineList: static LineList translate (LineList L, int x, int y) { if (L.empty()) return L; else { Line ln = L.hd(); Line transLine = new Line(ln.x0()+x, ln.y0()+y, ln.x1()+x, ln.y1()+y); return LL.cons(transLine, translate(L.tl(), x, y)); } }

13. Rotation Rotation is more complicated. Consider rotating one point around the origin by angle : (x’,y’) 1. Calculate m and  m = x2+y2  = tan-1(y/x) 2.  =  +  3. (x’,y’) = point of length m, at angle     (x,y)

14. Rotation (cont.) We’ll rotate shapes (i.e. LineList’s) about the origin by rotating each line: static LineList rotateShapeAboutOrigin (LineList L, double theta) { if (L.empty()) return LL.nil; else return LL.cons(rotateLine(L.hd(), theta), rotateShapeAboutOrigin(L.tl(), theta)); }

15. Rotation (cont.) Rotating individual lines around the origin is a matter of rotating both endpoints, as we have indicated. Some added complexity comes from two factors: • Math.atan returns angles in the range -/2 to  /2 (i.e. only angles in the right half-plane) • The graphics coordinate system is “upside-down” relative to ordinary Cartesian coordinates.

16. Rotation (cont.) static Line rotateLine (Line ln, double theta) { int x0 = ln.x0(), y0 = -ln.y0(), // turn coordinates rightside-up x1 = ln.x1(), y1 = -ln.y1(); // turn coordinates rightside-up double currangle0 = Math.atan((double)y0/x0); double newangle0 = currangle0+theta; if (x0<0) newangle0 = newangle0 + Math.PI; double mag0 = Math.sqrt(x0*x0+y0*y0); int newx0 = (int)Math.round(mag0*Math.cos(newangle0)); int newy0 = -(int)Math.round(mag0*Math.sin(newangle0));

17. Rotation (cont.) double currangle1 = Math.atan((double)y1/x1); double newangle1 = currangle1+theta; if (x1<0) newangle1 = newangle1 + Math.PI; double mag1 = Math.sqrt(x1*x1+y1*y1); int newx1 = (int)Math.round(mag1*Math.cos(newangle1)); int newy1 = -(int)Math.round(mag1*Math.sin(newangle1)); return new Line(newx0,newy0,newx1,newy1); }

18. Rotation (cont.) One remaining technicality: when we say “rotate a shape”, we usually mean, “rotate it around its center”. However, so far we know only how to rotate a shape around the origin. The rotateShape method takes a shape and an angle and a point (x,y) which is taken to be the center of the shape.

19. Rotation (cont.) It does so by translating the shape, then rotating, then translating back: static LineList rotateShape (LineList L, double theta, int x, int y) { LineList transL = translate(L, -x, -y); LineList rotateL = rotateShapeAboutOrigin(transL, theta); return translate(rotateL, x, y); }

20. The Hilbert Curve Recall again the abstract version of HC(order, sl): int diam = size of HC of order n-1; hcsub1 = HC(order-1, sl); hcul = rotate hcsub1 by -90 degrees; hcur = translate hcsub1 to (diam+sl, 0); hclr = translate hcsub1 to (diam+sl, diam+sl); hcll = rotate hcsub1 by 90 degrees, then translate to (0, diam+sl); hc = append(hcul, append(hcur, append(hclr, hcll))); return hc, with three lines added;

21. Hilbert Curve (cont.) We can now say that the rotation steps should rotate the shape around the center of the Hilbert curve of order n-1. That center is at (diam/2, diam/2). How do we calculate diam? A review of the Hilbert curve of various orders show that HC(n-1) has diameter ( 2n-1-1)*sl. This leads to our solution:

22. The HC Method static LineList HC (int order, int sl) { if (order == 1) return LL.cons(new Line(0,0,sl,0), LL.cons(new Line(sl,0,sl,sl), LL.cons(new Line(sl,sl,0,sl), LL.nil))); else { int diam = sl*(int)(Math.pow(2,order-1)-1); // diameter of HC(order-1) int radius = diam/2; LineList hcsub1 = HC(order-1, sl);

23. The HC Method (cont.) LineList hcul = rotateShape(hcsub1, -Math.PI/2, radius, radius); LineList hcur = translate(hcsub1, diam+sl, 0); LineList hclr = translate(hcsub1, diam+sl, diam+sl); LineList hcll = translate( rotateShape(hcsub1,Math.PI/2, radius, radius), 0, diam+sl); LineList hc = append(hcul, append(hcur, append(hclr, hcll)));

24. The HC Method (cont.) hc = LL.cons(new Line(diam,0,diam+sl,0), LL.cons(new Line(diam+sl,diam,diam+sl,diam+sl), LL.cons(new Line(diam+sl,2*diam+sl, diam,2*diam+sl), hc))); return hc; } }