JAVA JAVA JAVA PLEASE INDLUDE COMMENTS TO HALP MEH Objective: Write a program in

Question

JAVA JAVA JAVA

PLEASE INDLUDE COMMENTS TO HALP MEH

Objective:

Write a program in which draws (yes it actually makes a picture) a triangular fractal using recursion. This is best if done using a java applet.

Suggested Methodology

The idea for it is this

First draw a filled equilateral triangle

Next draw another filled equilateral triangle of a different color that’s upside down in the middle of that triangle

Using the other triangles formed repeat step 2 until a pixel limit of 4 is reached

HINTS:

It may be a good idea to look at the examples I gave 02/12/2015

The method fillPolygon(int[] xPoints, int[] yPoint, numberOfPoints) as called by the graphics device is important

The method setColor(Color aColor) is important for picking different colors to draw things.

Example Image of Results:

Explanation / Answer

Can help you with this:

import java.awt.*;
import java.util.*;
public class Traingl
{
public static final int SIZE = 512;
public static void main(String[] args)
{
Scanner console = new Scanner(System.in);
System.out.print("What level do you want? ");
int level = console.nextInt();
DrawingPanel p = new DrawingPanel(SIZE, SIZE);
p.setBackground(Color.CYAN);
Graphics g = p.getGraphics();
int triangleHeight = (int) Math.round(SIZE * Math.sqrt(3.0) / 2.0);
Point p1 = new Point(0, triangleHeight);
Point p2 = new Point(SIZE / 2, 0);
Point p3 = new Point(SIZE, triangleHeight);
drawFigure(level, g, p1, p2, p3);
}
public static void drawFigure(int level, Graphics g,Point p1, Point p2, Point p3)
{
if (level == 1)
{
Polygon p = new Polygon();
p.addPoint(p1.x, p1.y);
p.addPoint(p2.x, p2.y);
p.addPoint(p3.x, p3.y);
g.fillPolygon(p);
}
else
{
Point p4 = midpoint(p1, p2);
Point p5 = midpoint(p2, p3);
Point p6 = midpoint(p1, p3);
drawFigure(level - 1, g, p1, p4, p6);
drawFigure(level - 1, g, p4, p2, p5);
drawFigure(level - 1, g, p6, p5, p3);
}
}
public static Point midpoint(Point p1, Point p2)
{
return new Point((p1.x + p2.x) / 2, (p1.y + p2.y) / 2);
}
}

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