Java 具有g2d的毕达哥拉斯树

我正在尝试构建我的第一个分形（毕达哥拉斯树）：

在Java中使用Graphics2D。以下是我现在拥有的：

import java.awt.*;
import java.awt.geom.*; 
import javax.swing.*;
import java.util.Scanner;

public class Main {

public static void main(String[] args) {

    int i=0;
    Scanner scanner = new Scanner(System.in);

    System.out.println("Give amount of steps: ");
    i = scanner.nextInt();

    new Pitagoras(i);
    }
}

class Pitagoras extends JFrame {

private int powt, counter;

public Pitagoras(int i) {
    super("Pythagoras Tree.");
    setSize(1000, 1000);
    setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    setVisible(true);
    powt = i;
}

private void paintIt(Graphics2D g) {

    double p1=450, p2=800, size=200;

    for (int i = 0; i < powt; i++) {
        if (i == 0) {
            g.drawRect((int)p1, (int)p2, (int)size, (int)size);
            counter++;
        }
        else{
            if( i%2 == 0){
                //here I must draw two squares
            }
            else{
                //here I must draw right triangle
            }
        }
    }
}

@Override
public void paint(Graphics graph) {

    Graphics2D g = (Graphics2D)graph;
    paintIt(g);

}

import java.awt.*；
导入java.awt.geom.*；
导入javax.swing.*；
导入java.util.Scanner；
公共班机{
公共静态void main（字符串[]args）{
int i=0；
扫描仪=新的扫描仪（System.in）；
System.out.println（“给出步数：”）；
i=scanner.nextInt（）；
新皮塔戈拉（一）；
}
}
类Pitagoras扩展了JFrame{
私人内部电源、计数器；
公共皮塔戈拉（int i）{
超级（“毕达哥拉斯树”）；
设置大小（10001000）；
setDefaultCloseOperation（JFrame.EXIT_ON_CLOSE）；
setVisible（真）；
功率=i；
}
专用void paintIt（图形2d g）{
双p1=450，p2=800，尺寸=200；
对于（int i=0；i

---

基本上我设定了步数，然后画第一个正方形（p1，p2和大小）。如果步长是奇数，我需要在正方形的顶部建立直角三角形。如果步长是偶数，我需要在三角形的自由边上建立两个正方形。现在我应该选择什么方法来绘制三角形和正方形？我正在考虑用简单的线绘制三角形，用仿射变换它们，但我不确定这是否可行它不解决绘制正方形的问题。

您不必绘制三角形，只需在该树中绘制正方形（正方形的边就是三角形）

您可以更轻松地查看递归（这些类型的分形是递归的标准示例）：

伪码

drawSquare(coordinates) {
    // Check break condition (e.g. if square is very small)
    // Calculate coordinates{1|2} of squares on top of this square -> Pythagoras
    drawSquare(coordinates1)
    drawSquare(coordinates2)
}

由于我经常编程分形，提示：在BuffereImage中绘制分形本身，并且只在paint方法中绘制图像。paint方法可能每秒调用几次，因此它必须是FaaaaAt

---

也不要直接在JFrame中绘制，而是使用画布（如果您想使用awt）或JPanel（如果您使用swing）。

您不必绘制三角形，只需在此树中绘制正方形（正方形的边是三角形）

您可以更轻松地查看递归（这些类型的分形是递归的标准示例）：

伪码

drawSquare(coordinates) {
    // Check break condition (e.g. if square is very small)
    // Calculate coordinates{1|2} of squares on top of this square -> Pythagoras
    drawSquare(coordinates1)
    drawSquare(coordinates2)
}

由于我经常编程分形，提示：在BuffereImage中绘制分形本身，并且只在paint方法中绘制图像。paint方法可能每秒调用几次，因此它必须是FaaaaAt

也不要直接在JFrame中绘制，而是使用画布（如果要使用awt）或JPanel（如果使用swing）。

我的最终解决方案：

import java.awt.*;
import java.util.Scanner;
import javax.swing.*;

public class Main extends JFrame {;

    public Main(int n) {
        setSize(900, 900);
        setTitle("Pythagoras tree");
        add(new Draw(n));
        setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        setVisible(true);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Give amount of steps: ");
        new Main(sc.nextInt());
    }
}

class Draw extends JComponent {
    private int height = 800;
    private int width = 800;
    private int steps;

    public Draw(int n) {
        steps = n;

        Dimension d = new Dimension(width, height);
        setMinimumSize(d);
        setPreferredSize(d);
        setMaximumSize(d);
    }

    @Override
    public void paintComponent(Graphics g) {
        super.paintComponent(g);
        g.setColor(Color.white);
        g.fillRect(0, 0, width, height);
        g.setColor(Color.black);

        int x1, x2, x3, y1, y2, y3;
        int base = width/7;

        x1 = (width/2)-(base/2);
        x2 = (width/2)+(base/2);
        x3 = width/2;
        y1 = (height-(height/15))-base;
        y2 = height-(height/15);
        y3 = (height-(height/15))-(base+(base/2));

        g.drawPolygon(new int[]{x1, x1, x2, x2, x1}, new int[]{y1, y2, y2, y1, y1}, 5);

        int n1 = steps;
        if(--n1 > 0){
            g.drawPolygon(new int[] {x1, x3, x2}, new int[] {y1, y3, y1}, 3);
            paintMore(n1, g, x1, x3, x2, y1, y3, y1);
            paintMore(n1, g, x2, x3, x1, y1, y3, y1);
        }
    }

    public void paintMore(int n1, Graphics g, double x1_1, double x2_1, double x3_1, double y1_1, double y2_1, double y3_1){
        int x1, x2, x3, y1, y2, y3;

        x1 = (int)(x1_1 + (x2_1-x3_1));
        x2 = (int)(x2_1 + (x2_1-x3_1));
        x3 = (int)(((x2_1 + (x2_1-x3_1)) + ((x2_1-x3_1)/2)) + ((x1_1-x2_1)/2));
        y1 = (int)(y1_1 + (y2_1-y3_1));
        y2 = (int)(y2_1 + (y2_1-y3_1));
        y3 = (int)(((y1_1 + (y2_1-y3_1)) + ((y2_1-y1_1)/2)) + ((y2_1-y3_1)/2));

        g.setColor(Color.green);
        g.drawPolygon(new int[] {x1, x2, (int)x2_1, x1}, new int[] {y1, y2, (int)y2_1, y1}, 4);
        g.drawLine((int)x1, (int)y1, (int)x1_1, (int)y1_1);
        g.drawLine((int)x2_1, (int)y2_1, (int)x2, (int)y2);
        g.drawLine((int)x1, (int)y1, (int)x2, (int)y2);

        if(--n1 > 0){
            g.drawLine((int)x1, (int)y1, (int)x3, (int)y3);
            g.drawLine((int)x2, (int)y2, (int)x3, (int)y3);
            paintMore(n1, g, x1, x3, x2, y1, y3, y2);
            paintMore(n1, g, x2, x3, x1, y2, y3, y1);
        }
    }
}

我的最终解决方案：

import java.awt.*;
import java.util.Scanner;
import javax.swing.*;

public class Main extends JFrame {;

    public Main(int n) {
        setSize(900, 900);
        setTitle("Pythagoras tree");
        add(new Draw(n));
        setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        setVisible(true);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        System.out.print("Give amount of steps: ");
        new Main(sc.nextInt());
    }
}

class Draw extends JComponent {
    private int height = 800;
    private int width = 800;
    private int steps;

    public Draw(int n) {
        steps = n;

        Dimension d = new Dimension(width, height);
        setMinimumSize(d);
        setPreferredSize(d);
        setMaximumSize(d);
    }

    @Override
    public void paintComponent(Graphics g) {
        super.paintComponent(g);
        g.setColor(Color.white);
        g.fillRect(0, 0, width, height);
        g.setColor(Color.black);

        int x1, x2, x3, y1, y2, y3;
        int base = width/7;

        x1 = (width/2)-(base/2);
        x2 = (width/2)+(base/2);
        x3 = width/2;
        y1 = (height-(height/15))-base;
        y2 = height-(height/15);
        y3 = (height-(height/15))-(base+(base/2));

        g.drawPolygon(new int[]{x1, x1, x2, x2, x1}, new int[]{y1, y2, y2, y1, y1}, 5);

        int n1 = steps;
        if(--n1 > 0){
            g.drawPolygon(new int[] {x1, x3, x2}, new int[] {y1, y3, y1}, 3);
            paintMore(n1, g, x1, x3, x2, y1, y3, y1);
            paintMore(n1, g, x2, x3, x1, y1, y3, y1);
        }
    }

    public void paintMore(int n1, Graphics g, double x1_1, double x2_1, double x3_1, double y1_1, double y2_1, double y3_1){
        int x1, x2, x3, y1, y2, y3;

        x1 = (int)(x1_1 + (x2_1-x3_1));
        x2 = (int)(x2_1 + (x2_1-x3_1));
        x3 = (int)(((x2_1 + (x2_1-x3_1)) + ((x2_1-x3_1)/2)) + ((x1_1-x2_1)/2));
        y1 = (int)(y1_1 + (y2_1-y3_1));
        y2 = (int)(y2_1 + (y2_1-y3_1));
        y3 = (int)(((y1_1 + (y2_1-y3_1)) + ((y2_1-y1_1)/2)) + ((y2_1-y3_1)/2));

        g.setColor(Color.green);
        g.drawPolygon(new int[] {x1, x2, (int)x2_1, x1}, new int[] {y1, y2, (int)y2_1, y1}, 4);
        g.drawLine((int)x1, (int)y1, (int)x1_1, (int)y1_1);
        g.drawLine((int)x2_1, (int)y2_1, (int)x2, (int)y2);
        g.drawLine((int)x1, (int)y1, (int)x2, (int)y2);

        if(--n1 > 0){
            g.drawLine((int)x1, (int)y1, (int)x3, (int)y3);
            g.drawLine((int)x2, (int)y2, (int)x3, (int)y3);
            paintMore(n1, g, x1, x3, x2, y1, y3, y2);
            paintMore(n1, g, x2, x3, x1, y2, y3, y1);
        }
    }
}

---

递归的问题是，在分形的某个深度，你会耗尽堆栈空间。因此，虽然它很优雅，但在这里可能是不切实际的。在这种情况下，无论你使用哪种数据结构，你都会耗尽空间。递归的问题是，在分形的某个深度，你会耗尽堆栈空间。所以e优雅，在这里可能不实用。在这种情况下，无论使用哪种数据结构，都会耗尽空间。