More About Recursion - 2

1. More About Recursion - 2 Java

2. Looking at more recursion in Java • A simple math example • Fractals

3. A simple math example • Calculating n! • factorial (n) = n * factorial (n-1) if n>1 • factorial (1) = 1 • How do we code this in Java?

4. Fractals • Sierpinski’s Gasket • Starting with an equilateral triangle, recursively drawing smaller equilateral triangles inside, by building triangles of the midpoints

5. SG(0)

6. SG(1)

7. SG(2)

8. SG(3)

9. SG(4)

10. SG(5)

11. // created Sierpinski's gasket fractal public void sierpinski(int level) {// precondition: there is a blank picture at least // 291 x 281int p1X = 10;int p1Y = 280;int p2X = 290;int p2Y = 280;int p3X = 150;int p3Y = 110;Graphics g = this.getGraphics();g.setColor(Color.RED); drawGasket(g,level,p1X,p1Y,p2X,p2Y,p3X,p3Y); }

12. How do we solve the problem? • Well, draw the line segments for the triangle • Then, if the level is at least 1, calculate the midpoints, and make the 3 recursive calls on the sub-triangles

13. private void drawGasket(Graphics g, int level, int x1, int y1, int x2, int y2, int x3, int y3){ g.drawLine ( x1, y1, x2, y2); g.drawLine ( x2, y2, x3, y3); g.drawLine ( x3, y3, x1, y1); if ( level > 0 ) { int midx1x2 = (x1 + x2) / 2; int midy1y2 = (y1 + y2) / 2; int midx2x3 = (x2 + x3) / 2; int midy2y3 = (y2 + y3) / 2; int midx3x1 = (x3 + x1) / 2; int midy3y1 = (y3 + y1) / 2; drawGasket ( g, level - 1, x1, y1, midx1x2, midy1y2, midx3x1, midy3y1); drawGasket ( g, level - 1, x2, y2, midx2x3, midy2y3, midx1x2, midy1y2); drawGasket ( g, level - 1, x3, y3, midx3x1, midy3y1, midx2x3, midy2y3); }} Note how we decrease the level on the recursive calls Note2: We could also use Points instead of x’s and y’s

14. Assignment • No reading assignment from Media Computation