我对以下算法有几个问题来判断一个数字是否是素数,我也知道随着埃拉托斯特尼筛法可以得到更快的响应。
- 为什么计算速度更快
i i * sqrt (n)
次。比sqrt (n)
就一次 ?
- Why
Math.sqrt()
比我的快sqrt()
方法 ?
-
这些算法的复杂度分别是O(n)、O(sqrt(n))、O(n log(n))?
public class Main {
public static void main(String[] args) {
// Case 1 comparing Algorithms
long startTime = System.currentTimeMillis(); // Start Time
for (int i = 2; i <= 100000; ++i) {
if (isPrime1(i))
continue;
}
long stopTime = System.currentTimeMillis(); // End Time
System.out.printf("Duracion: %4d ms. while (i*i <= N) Algorithm\n",
stopTime - startTime);
// Case 2 comparing Algorithms
startTime = System.currentTimeMillis();
for (int i = 2; i <= 100000; ++i) {
if (isPrime2(i))
continue;
}
stopTime = System.currentTimeMillis();
System.out.printf("Duracion: %4d ms. while (i <= sqrt(N)) Algorithm\n",
stopTime - startTime);
// Case 3 comparing Algorithms
startTime = System.currentTimeMillis();
for (int i = 2; i <= 100000; ++i) {
if (isPrime3(i))
continue;
}
stopTime = System.currentTimeMillis();
System.out.printf(
"Duracion: %4d ms. s = sqrt(N) while (i <= s) Algorithm\n",
stopTime - startTime);
// Case 4 comparing Algorithms
startTime = System.currentTimeMillis();
for (int i = 2; i <= 100000; ++i) {
if (isPrime4(i))
continue;
}
stopTime = System.currentTimeMillis();
System.out.printf(
"Duracion: %4d ms. s = Math.sqrt(N) while (i <= s) Algorithm\n",
stopTime - startTime);
}
public static boolean isPrime1(int n) {
for (long i = 2; i * i <= n; i++) {
if (n % i == 0)
return false;
}
return true;
}
public static boolean isPrime2(int n) {
for (long i = 2; i <= sqrt(n); i++) {
if (n % i == 0)
return false;
}
return true;
}
public static boolean isPrime3(int n) {
double s = sqrt(n);
for (long i = 2; i <= s; i++) {
if (n % i == 0)
return false;
}
return true;
}
public static boolean isPrime4(int n) {
// Proving wich if faster between my sqrt method or Java's sqrt
double s = Math.sqrt(n);
for (long i = 2; i <= s; i++) {
if (n % i == 0)
return false;
}
return true;
}
public static double abs(double n) {
return n < 0 ? -n : n;
}
public static double sqrt(double n) {
// Newton's method, from book Algorithms 4th edition by Robert Sedgwick
// and Kevin Wayne
if (n < 0)
return Double.NaN;
double err = 1e-15;
double p = n;
while (abs(p - n / p) > err * n)
p = (p + n / p) / 2.0;
return p;
}
}
这也是我的代码的链接:http://ideone.com/Fapj1P
1. Why is faster to compute i*i, sqrt (n) times. than sqrt (n) just one time ?
看看下面的复杂性。计算平方根的额外成本。
2. Why Math.sqrt() is faster than my sqrt() method ?
Math.sqrt() 委托对 StrictMath.sqrt 的调用,这是在硬件或本机代码中完成的。
3. What is the complexity of these algorithms?
您描述的每个函数的复杂性
i=2 .. i*i<n
O(平方(n))
i=2 .. sqrt(n)
O(sqrt(n)*log(n))
i=2 .. sqrt (by Newton's method)
O(sqrt(n)) + O(log(n))
i=2 .. sqrt (by Math.sqrt)
O(平方(n))
牛顿法的复杂度为
http://en.citizendium.org/wiki/Newton%27s_method#Computational_complexity
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