Program error in data types

[adil_bashir]

hello,
i have written the below piece of code for calculating the power function, but when i run the program, it shows the wrong result and i verified it by calculating manually. so please help me to get the correct result.

public class powerFunction
{
	public static void main(String ar[])
	{
		int c = 11;
		int d = 23;
		int n = 187;
		long dt;
		dt = (long)((Math.pow(c,d)) % n);
		System.out.println("dt =" +dt);
	}
}

The output should be 88, but it is displaying 149.

[JamesCherrill]

Isn't this fun!! I tried a "manual" calculation in both double and long variables:

void calcDouble() {
      double c=11, cPow2 = c*c, cPow4 = cPow2*cPow2, 
             cPow8 = cPow4*cPow4,cPow16 = cPow8*cPow8, 
             cPow23 = cPow16*cPow4*cPow2*c;
      System.out.println(cPow23 + " " + cPow23%187);     
   }

   void calcLong() {
      long   c=11, cPow2 = c*c, cPow4 = cPow2*cPow2, 
             cPow8 = cPow4*cPow4,cPow16 = cPow8*cPow8, 
             cPow23 = cPow16*cPow4*cPow2*c;
      System.out.println(cPow23 + " " + cPow23%187);     
   }

Which gave the following results:
8.954302432552372E23 175.0
6839173302027254275 65
neither of which match yours.
Personally I'd put my money on the long manual version.
Any other offers?

[dantinkakkar]

Funnily...

public class test {
    public static void main(String[] args)
    {
        long hyz = (long) Math.pow(11.0, 23.0);
        hyz = hyz%187;
        System.out.println(hyz);
    }
}

Output = 161?

Also, I edited your code a bit and checked...

int c = 11;
    int d = 23;
    int n = 187;
    long dt;
    dt = (long)((Math.pow(c,d)));
    long rc = dt%n;
    System.out.println(rc);

I put the modulus n part in another line... and whoa! 161

So, you want this, right? (11^23)%187?

[stultuske]

when and where you convert to long also has its influence..

[dantinkakkar]

Yeah, he's right. Another version of my code got 65 as output. I guess that's what's dwindling your answer... :\

If you try writing a piece of code to find out the exponent of a number, and you try solving 11^23, you get 6839173302027254275.Then, if you try finding out it's remainder with 187, you get 65, which is wrong, I guess, as per you.

[adil_bashir]

listen dears..
if we break (11^23) % 187, then it can written as
{(11^1 % 187) * (11^2 % 187) * (11^4 % 187) * (11^8 % 187) * (11^8 % 187)} % 187
= {11 * 121 * 55 * 33 * 33} % 187
= 88

but programatically, itz not 88. so wHat to do.

[JamesCherrill]

No, that's not valid. You must complete the power expression before doing the modulo

[adil_bashir]

mathematically its valid

[stultuske]

mathematically its valid

there are forms of math in which

1 + 1 = 1
and
1 + 0 = 0

are valid and correct too, but that's not really what we're looking for, is it?

[dantinkakkar]

Back to the topic you get 65 if you write the exponent code yourself, instead of using Math.pow(int,int). @adil_bashir, are you sure it's 11^23 and not 23^11? o.O

[adil_bashir]

itz 11^23

[dantinkakkar]

The correct answer is, as you say, 88. The problem - I have recognized it. It's not that we're doing anything wrong with the code. 895430243255237372246531, which is the value of 11^23, is too large a number even for long. :)

It is not computed properly by the compiler. The number is just too big.

[adil_bashir]

that is the title of my problem, so can u please tell me the exact datatype that can be used.

[adil_bashir]

can anyone help me to use BigInteger for it?

[adil_bashir]

Solved by myself..

import java.math.BigInteger;

public class abc3
{
	public static void main(String ar[])
	{
		BigInteger ct = new BigInteger("11");
		BigInteger n = new BigInteger("187");
		BigInteger d = new BigInteger("23");
		BigInteger e  = new BigInteger("7");     
                BigInteger decr = ct.modPow(d,n);
		System.out.println("Decr = "+decr);
	}
}

and now the output is 88.

thanks all

[dantinkakkar]

You're welcome! :)