Computer Science 252
Problem Solving with Java
Spring 2016, The College of Saint Rose
Powers BlueJ Project
Click here to download a BlueJ project for Powers.
Powers Source Code
The Java source code for Powers is below. Click on a file name to download it.
import java.util.Scanner;
/*
* Example Powers: examples of recursive methods to compute
* powers of a number.
*
* Jim Teresco, The College of Saint Rose, CSC 252, Fall 2013
*
* $Id: Powers.java 2230 2013-10-27 02:26:10Z terescoj $
*/
public class Powers {
// A main method to drive the program. It's a Java
// application rather than an applet, so more like what
// you saw in 202.
public static void main(String args[]) {
// read in the base and exponent from a Scanner
Scanner s = new Scanner(System.in);
System.out.println("Let's compute the value of some integer raised to a power.");
System.out.print("First, enter the base: ");
int base = s.nextInt();
int exponent = 0;
do {
System.out.print("Next, enter the exponent (>=0): ");
exponent = s.nextInt();
if (exponent < 0) {
System.out.println("Negative exponents are not allowed");
}
} while (exponent < 0);
// now we compute in three different ways, using the methods below.
System.out.println("" + base + "^" + exponent + ", computed three ways:");
System.out.println("Method using a loop: " + loopPower(base, exponent));
System.out.println("Method using straightforward recursion: " + recPower(base, exponent));
System.out.println("Method using smarter recursion: " + fastRecPower(base, exponent));
}
// compute the power using a good old fashioned loop
public static int loopPower(int base, int exponent) {
int answer = 1;
for (int i=0; i<exponent; i++) {
answer *= base;
}
return answer;
}
// the straightforward recursive approach
public static int recPower(int base, int exponent) {
// our base case is exponent == 0
if (exponent == 0) {
return 1;
}
// otherwise, we have to do some work, b^n = b * b^{n-1}
return base * recPower(base, exponent -1);
}
// a more efficient recursive approach, based on the idea
// that we can compute a power b^{2n} as (b*b)^n
public static int fastRecPower(int base, int exponent) {
// base case is again exponent == 0
if (exponent == 0) {
return 1;
}
// now, see if the exponent is even or odd
if (exponent % 2 == 1) {
// it's odd, so use straightforward recursion to get
// down to an even case
return base * fastRecPower(base, exponent - 1);
}
// if we got here, it's even, so we can do better
return fastRecPower(base * base, exponent / 2);
}
}