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Math.sqrt Method: java.lang.Math.sqrt
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Math.sqrt. In mathematics, a square root maps the area of a square to the length of its side. In Java we use the Math class to compute square roots.
Math
With this method, we receive a double. For programs that call Math.sqrt many times, a lookup table cache can be used. This approach (memoization) helps many programs.
Square root. Consider this program. It does not import java.lang.Math at its top. Instead we directly access Math.sqrt using its composite name. It shows the output of sqrt.
Result The square root of 4 is 2. We see 2.0 because Math.sqrt returns a double.
Tip For cases when we do not need fractional values, we can cast the result of Math.sqrt to an int.
Cast
public class Program { public static void main(String[] args) { // ... Test the sqrt method. double value = java.lang.Math.sqrt(4); double value2 = java.lang.Math.sqrt(9); System.out.println(value); System.out.println(value2); } }
2.0 3.0
Benchmark, lookup table. Memoization is a powerful programming technique. In it, we avoid computing a data we previously computed. We store values in a structure like an array.
Version 1 In this version of the code we call Math.sqrt repeatedly—no caching is used here.
Version 2 We use a double array cache of 100 Math.sqrt values. We access those values many times.
Array
Result Accessing array elements is more than ten times faster than calling Math.sqrt each time.
public class Program { public static void main(String[] args) { // A lookup table for square roots. double[] cache = new double[100]; long t1 = System.currentTimeMillis(); // Version 1: use Math.sqrt each time. for (int i = 0; i < 1000000; i++) { for (int x = 1; x < 100; x++) { double result = Math.sqrt(x); if (result == 0) { return; } } } long t2 = System.currentTimeMillis(); // Version 2: use lookup table after first time. for (int i = 0; i < 1000000; i++) { for (int x = 1; x < 100; x++) { double result; if (cache[x] != 0) { result = cache[x]; } else { result = Math.sqrt(x); cache[x] = result; } if (result == 0) { return; } } } long t3 = System.currentTimeMillis(); // ... Benchmark times. System.out.println(t2 - t1); System.out.println(t3 - t2); } }