Question

use recursive drawing to solve this problem

A figure even more remarkable than the Koch curve described in the previous project is the Koch snowflake or Koch star. You can create this snowflake by applying the three-step recursive process used to create a Koch curve to each side of an equilateral triangle rather than to only one line segment: △★本 Use java to write a recursive program that creates a Koch snowflake.

Answer 1

import java.awt.*;

import javax.swing.*;

public class Snowflake extends JApplet

{

int level = 0;

public void init()

{

setSize(640, 480);

String levelStr = JOptionPane.showInputDialog

("Enter the recursion depth: ");

level = Integer.parseInt(levelStr);

}

public void paint(Graphics g)

{

Point pointOne = new Point(120, 320);

Point pointTwo = new Point(440, 320); // reusing Points from example program * 2;

Point pointThree = new Point(280, 40);

drawSnowflake(g, level, pointOne, pointTwo, pointThree);

}

private void drawSnowflake(Graphics g, int lev,

Point p1, Point p2, Point p3)

{

drawSegment(g, lev, p1, p2); // draw 3 lines to make a triangle

drawSegment(g, lev, p2, p3);

drawSegment(g, lev, p3, p1);

}

private void drawSegment(Graphics g, int lev, Point pOne, Point pTwo)

{

if (lev == 0)

{

g.drawLine(pOne.x, pOne.y, pTwo.x, pTwo.y);

}

if (lev >= 1){

Point distance = new Point( (pTwo.x-pOne.x)/3, (pTwo.y-pOne.y)/3 );

Point pA = new Point( pOne.x+distance.x, pOne.y+distance.y);

Point pB = new Point( pTwo.x-distance.x, pTwo.y-distance.y);

double sin60 = -0.866025403784438646763723170752936183471402626905190;

Point pTip = new Point(

pA.x + (int)(distance.x* 0.5 + distance.y*sin60),

pA.y + (int)(distance.y* 0.5 - distance.x*sin60)

);

drawSegment(g, lev - 1, pOne, pA); // if level 1 or higher,

drawSegment(g, lev - 1, pA, pTip); // recursively call self 4 times

drawSegment(g, lev - 1, pTip, pB);

drawSegment(g, lev - 1, pB, pTwo);

}

}

}