Computer Science 523
Advanced Programming
Summer 2014, 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.math.BigInteger;
import java.util.Scanner;
/*
* Example Powers: examples of recursive methods to compute
* powers of a number.
*
* Updated to use long for exponents and BigIntegers for base and answer
* values to be able to compute much larger problem sizes.
*
* Jim Teresco, The College of Saint Rose, CSC 252, Fall 2013
* CSC 523, Summer 2014
*
* $Id: Powers.java 2388 2014-07-03 18:54:49Z terescoj $
*/
public class Powers {
// a main method to try out our power-computing methods below
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: ");
long base = s.nextLong();
long exponent = 0L;
do {
System.out.print("Next, enter the exponent (>=0): ");
exponent = s.nextLong();
if (exponent < 0L) {
System.out.println("Negative exponents are not allowed");
}
} while (exponent < 0L);
// 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(BigInteger.valueOf(base), exponent));
System.out.println("Method using straightforward recursion: " + recPower(BigInteger.valueOf(base), exponent));
System.out.println("Method using smarter recursion: " + fastRecPower(BigInteger.valueOf(base), exponent));
}
// compute the power using a good old fashioned loop
public static BigInteger loopPower(BigInteger base, long exponent) {
BigInteger answer = BigInteger.ONE;
for (long i=0; i<exponent; i++) {
answer = answer.multiply(base);
}
return answer;
}
// the straightforward recursive approach
public static BigInteger recPower(BigInteger base, long exponent) {
// our base case is exponent == 0
if (exponent == 0L) {
return BigInteger.ONE;
}
// otherwise, we have to do some work, b^n = b * b^{n-1}
return base.multiply(recPower(base, exponent-1L));
}
// a more efficient recursive approach, based on the idea
// that we can compute a power b^{2n} as (b*b)^n
public static BigInteger fastRecPower(BigInteger base, long exponent) {
// base case is again exponent == 0
if (exponent == 0L) {
return BigInteger.ONE;
}
// now, see if the exponent is even or odd
if (exponent % 2L == 1L) {
// it's odd, so use straightforward recursion to get
// down to an even case
return base.multiply(fastRecPower(base, exponent-1L));
}
// if we got here, it's even, so we can do better
return fastRecPower(base.multiply(base), exponent/2L);
}
}