我的代码:
import java.math.BigDecimal;
public class ScienceFair {
private static long NewtonMethod()
{
BigDecimal TWO = new BigDecimal(2);
BigDecimal SQRT_TWO = new BigDecimal("1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727");
BigDecimal TOLERANCE = BigDecimal.ONE.scaleByPowerOfTen(-100);
long start = System.nanoTime();
BigDecimal a = new BigDecimal(1);
while(a.subtract(SQRT_TWO).abs().compareTo(TOLERANCE) >= 0) {
a = a.add(TWO.divide(a, 100, BigDecimal.ROUND_HALF_UP)).divide(TWO);
}
return System.nanoTime() - start;
}
private static long MidpointMethod()
{
BigDecimal TWO = new BigDecimal(2);
BigDecimal SQRT_TWO = new BigDecimal("1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727");
BigDecimal TOLERANCE = BigDecimal.ONE.scaleByPowerOfTen(-100);
long start = System.nanoTime();
BigDecimal a = new BigDecimal(1);
BigDecimal b = new BigDecimal(2);
while(a.add(b).divide(TWO).subtract(SQRT_TWO).abs().compareTo(TOLERANCE) >= 0) {
if(a.multiply(a).subtract(TWO).abs().compareTo(b.multiply(b).subtract(TWO).abs()) == 1)
{
a = a.add(b).divide(TWO);
}
else
{
b = a.add(b).divide(TWO);
}
}
return System.nanoTime() - start;
}
private static long SecantMethod()
{
BigDecimal TWO = new BigDecimal(2);
BigDecimal SQRT_TWO = new BigDecimal("1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727");
BigDecimal TOLERANCE = BigDecimal.ONE.scaleByPowerOfTen(-100);
long start = System.nanoTime();
BigDecimal a = new BigDecimal(1);
BigDecimal b = new BigDecimal(2);
BigDecimal b_old = new BigDecimal(2);
while(a.add(b).divide(TWO).subtract(SQRT_TWO).abs().compareTo(TOLERANCE) >= 0) {
b_old = b;
b = a.multiply(b).add(TWO).divide(a.add(b), 100, BigDecimal.ROUND_HALF_UP);
a = b_old;
}
return System.nanoTime() - start;
}
public static void main(String[] args) {
double a = 0;
int trials = 100;
for(int i=1; i<= trials; i++)
{
a += (NewtonMethod() / 10e6);
}
System.out.printf("Newton's Method: %f\n", a/trials);
a = 0;
for(int i=1; i<= trials; i++)
{
a += (MidpointMethod() / 10e6);
}
System.out.printf("Midpoint Method: %f\n", a/trials);
a = 0;
for(int i=1; i<= trials; i++)
{
a += (SecantMethod() / 10e6);
}
System.out.printf("Secant Method: %f\n", a/trials);
}
}
旨在运行牛顿方法,中点方法和正割方法,以查找近似2到100个小数位的平方根所需的时间。
它每个都运行100次试验,并将它们平均输出所用的毫秒数。
中点法总是大约1.5秒。然而,Newton方法和Secant方法变化很大,范围从0.08到0.14秒(大约两倍)。为什么会这样?
编辑:这是我现在尝试的(仅适用于NewtonMethod)
public class ScienceFairTwo {
public static void main(String[] args) throws Exception {
Runnable NewtonMethod = new Runnable()
{
public void run()
{
BigDecimal TWO = new BigDecimal(2);
BigDecimal SQRT_TWO = new BigDecimal("1.4142135623730950488016887242096980785696718753769480731766797379907324784621070388503875343276415727");
BigDecimal TOLERANCE = BigDecimal.ONE.scaleByPowerOfTen(-100);
long start = System.nanoTime();
BigDecimal a = new BigDecimal(1);
while(a.subtract(SQRT_TWO).abs().compareTo(TOLERANCE) >= 0) {
a = a.add(TWO.divide(a, 100, BigDecimal.ROUND_HALF_UP)).divide(TWO);
}
}
};
System.out.println("Newton's Method: " + new Benchmark(NewtonMethod));
}
}
答案 0 :(得分:4)
通常,对Java代码进行基准测试很困难。您的简单基准测试不够可靠。
我可以看到你的方法有三个问题(可能会有更多):
java.math.BigDecimal
的第一种方法是不吉利的,因为它浪费了将该类加载到内存中的时间。其他方法没有这种惩罚 - BigDecimal
已经加载。我建议你做两件事:
(当你正确地做基准测试时 - 在评论中发布一个链接。我想看看你的方法是如何表现的)