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Class 8 - Recursive PicturesPowerPoint Presentation

Class 8 - Recursive Pictures

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Class 8 - Recursive Pictures

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- Recursively-defined curves
- The Hilbert curve

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

“Koch snowflake”:

- 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.

HC(1):

HC(2):

HC(3):

n-1

n-1

n-1

n-1

n-1

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):

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;

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));

}

}

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)

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));

}

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.

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));

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);

}

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.

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);

}

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;

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:

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);

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)));

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;

}

}