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Thread: Program to compute powers

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    Default Program to compute powers

    Hi there folks.

    I'm trying to create a program that takes 2 numbers from command line arguments and use the second value as a power so for e.g. java Power 10 2 should give 100. But i want to do this without using the Math.pow method, and using a for loop. I have the following so far but i'm not sure what the problem is and it;s only seem to be doinng square numbers (i.e. to the power of 2)
    public class Power
    {
      public static void main(String [] args)
      {
        int mantissa = Integer.parseInt(args[0]);
        int exponent = Integer.parseInt(args[1]);
        int answer;
     
     
        for(int i=0; (mantissa*mantissa) < exponent; i++)
          exponent=exponent + i;
     
        answer = mantissa*mantissa;
        System.out.println("Result is " + answer);
     
     
      }
    }
    Last edited by copeg; October 25th, 2010 at 04:30 PM.


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    Member Darryl.Burke's Avatar
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    Default Re: Program to compute powers

    Well, all you ever set the value of answer to is mantissa*mantissa so what did you expect?

    Put your code aside. Take a paper and write down the steps to get a power in simple English (or your native language). Check your steps with 3 or 4 sample data sets. When it's all correct (and not before!), convert those steps into Java code.

    db

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    Default Re: Program to compute powers

    I know how to get powers, i wrote pseudocode and designed it, but im not sure what the case is for like cubics and powers more than 2.

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    Default Re: Program to compute powers

    You know there already is a class called Math that has a method called power that takes two numbers and raises the first number to the power of the second number.

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    Default Re: Program to compute powers

    public power(int num, int power)
    {
    int num2;
    if (power == 0 && num!=0)
    {
    num2 = 1;
    }
    else if (power == 1)
    {
    num2 = num;
    }
     
    else if (1 < power)
    {
    num2 = num * power(num, power -1);
    }
     
    else if(0 > power && num!=0)
    {
    num2 = 1/power(num, - power);
    }
    }

    Well...that may not quite work, haven't tried it out yet, but it's on the right track.
    Last edited by javapenguin; October 25th, 2010 at 04:55 PM.

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    Default Re: Program to compute powers

    You're algorithm doesn't compute x^y.

    Get out a piece of paper and "execute" the code as if you were the computer (pick some sample inputs):

    ex. mantissa = 4, exponent = 3

    4^3 = 64

        for(int i=0; (mantissa*mantissa) < exponent; i++)
          exponent=exponent + i;
    1. initialize i = 0
    2. 4 * 4 = 16
    3. exponent = 3
    4. 16 < 3?
    5. false, exit the for loop

    answer = mantissa*mantissa;
    System.out.println("Result is " + answer);
    6. answer = 4 * 4 = 16
    7. Print out "result is 16"

    Re-visit your pseudo-code and make sure that that is correct.

    As far as I can tell, your code at best will give you mantissa^2, and at worst will keep you stuck indefinately inside the for-loop (well, not really indefinately, integers wrap around at 2^31 so eventually your code will stop and give you mantissa^2)

    edit:

    public power(int num, int power)
    {
     
    if (power == 0 && num!=0)
    {
    num2 = 1;
    }
    else if (power == 1)
    {
    num2 = num;
    }
     
    else if (1 < power)
    {
    num2 = num * power(num, power -1);
    }
     
    else if(0 > power)
    {
    num2 = 1/power(num, - power);
    }
    }
    This isn't quite correct (for various reasons).

    1. This isn't proper Java syntax (no return type, num2 is never declared).
    2. 0^0 = 1. In your code, as far as I can work out it's un-determined
    3. Try running power(2,-1). You'll get a rather interesting result (correct result = 0.5). I actually can't think of a case off the top of my head where this code would give you correct results.
    4. No non-integer inputs for power (or number, but this is easier to fix than non-integer powers), i.e. power(2.5, -1.346)
    5. (minor issue) Assuming this code did actually work, this gives an O(n) solution at best (I didn't perform a thourough analysis, but it can only be worse). The best solution for computing x^y is O(1) (granted, there is a bit of "cheating" by using logarithm tables)
    6. We're here to help people find the answers to their problems, not give the answer to them
    Last edited by helloworld922; October 25th, 2010 at 05:05 PM.

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