Euler猜想

这是从http://duodaa.com/blog/index.php/archives/538/截得图，以下是代码

package math;

import java.math.BigDecimal;
import java.util.function.BiConsumer;

public class TestEuler {
    public static void main(String[] args) {
        boolean flg=true;
        
        for(long x=1;flg;x++){
            for(long y=1;flg&&(y<x);y++){
                for(long z=1;flg&&(z<y);z++){
                    for(long w=1;true;w++){
                        int r=power4Long(w).compareTo(sum(power4Long(x),power4Long(y),power4Long(z)));
                        System.out.print(x+":"+power4Long(x).toString()+",");
                        System.out.print(y+":"+power4Long(y).toString()+",");
                        System.out.print(z+":"+power4Long(z).toString()+",");
                        System.out.println(w+":"+power4Long(w).toString()+";");
                        if(r==1){
                            break;
                        }
                        if(r==0){
                            flg=false;
                            break;
                        }
                    }
                }
            }
        }
    }
    public static boolean checkEuler(long x,long y,long z,long w){
        return power4Long(w).compareTo(sum(power4Long(x),power4Long(y),power4Long(z)))==0;
    }
    public static BigDecimal power4Long(Long b){
        return  power4(new BigDecimal(b));
    }
    
    public static BigDecimal power4(BigDecimal b){
        return b.multiply(b).multiply(b).multiply(b);
    }
    public static BigDecimal sum(BigDecimal... bs){
        BigDecimal reB=new BigDecimal(0);
        for(BigDecimal b:bs){
            reB=reB.add(b);
        }
        return reB;
    }
}

事实上这样的四层循环极大的消耗着计算机的性能计算很慢，要考我的这些代码来验证欧拉猜想估计得跑到我死都没结果

所以一下代码直接验证下结果

package math;

public class TestEuler2 {
	public static void main(String[] args) {
		long x=2682440L;
		long y=15365639L;
		long z=18796760L;
		long w=20615673L;
		System.err.println(x+"的四次方是"+TestEuler.power4Long(x).toString());
		System.err.println(y+"的四次方是"+TestEuler.power4Long(y).toString());
		System.err.println(z+"的四次方是"+TestEuler.power4Long(z).toString());
		System.err.println(w+"的四次方是"+TestEuler.power4Long(w).toString());
		System.out.println(TestEuler.checkEuler(x, y, z, w));
	}
}

　　此代码结果如下

2682440的四次方是51774995082902409832960000
15365639的四次方是55744561387133523724209779041
18796760的四次方是124833740909952854954805760000
20615673的四次方是180630077292169281088848499041
true

　　有人证明这个方程式有无穷的解，真是让人惊叹数学的深邃伟大。

以下测试运行用时

package math;

import java.math.BigDecimal;
import java.util.function.BiConsumer;

import org.jgroups.tests.perf.Data;

/**
 * @author zxl
 * @jdk 1.8
 * @Date 2016年10月13日上午10:04:24
 */
public class TestEuler {
	public static void main(String[] args) {
		long currTime=System.currentTimeMillis();
		
		boolean flg=true;
		
		for(long x=1;flg&&(x<10L);x++){
			for(long y=1;flg&&(y<x);y++){
				for(long z=1;flg&&(z<y);z++){
					for(long w=1;true;w++){
						int r=power4Long(w).compareTo(sum(power4Long(x),power4Long(y),power4Long(z)));
						System.out.print(x+":"+power4Long(x).toString()+",");
						System.out.print(y+":"+power4Long(y).toString()+",");
						System.out.print(z+":"+power4Long(z).toString()+",");
						System.out.println(w+":"+power4Long(w).toString()+";");
						if(r==1){
							break;
						}
						if(r==0){
							flg=false;
							break;
						}
					}
				}
			}
		}
		System.out.println("用时共计："+(System.currentTimeMillis()-currTime));
	}
	public static boolean checkEuler(long x,long y,long z,long w){
		return power4Long(w).compareTo(sum(power4Long(x),power4Long(y),power4Long(z)))==0;
	}
	public static BigDecimal power4Long(Long b){
		return  power4(new BigDecimal(b));
	}
	
	public static BigDecimal power4(BigDecimal b){
		return b.multiply(b).multiply(b).multiply(b);
	}
	public static BigDecimal sum(BigDecimal... bs){
		BigDecimal reB=new BigDecimal(0);
		for(BigDecimal b:bs){
			reB=reB.add(b);
		}
		return reB;
	}
}

　该代码计算到10用时163毫秒,因为w在小于x的时候等式恒不成立

for(long w=x;true;w++)

所以w从x开始循环有效的降低了运行时间大概达到原先的四分之一耗时。