- If it wasn't clear (and it wasn't,) the algorithm I just posted was

one to speed up exponentiation. I use it as a wrapper around

BigInteger.pow(BigInteger big, int exponent) in Java.

--

Alan Eliasen | "It is not enough to do your best;

eliasen@... | you must know what to do and THEN

http://futureboy.homeip.net/ | do your best." -- W. Edwards Deming - Alan,

Yes is clear .. Thanks.

But, my problem still the same when the exponent is a prime > 2...

I'm trying to find out the best (fastest) way to calculate it ...

I welcome any new ideas...

Regards

Faysal

-----Original Message-----

From: primenumbers@yahoogroups.com [mailto:primenumbers@yahoogroups.com] On

Behalf Of Alan Eliasen

Sent: Sunday, August 07, 2005 3:34 PM

To: fyatim

Cc: 'Jan van Oort'; 'Prime Number'

Subject: Re: [PrimeNumbers] big numbers library

fyatim wrote:> I wrote a small method using a well known algorithm "Exponentiation by

That's because Sun's Java implementation still uses horrible O(n^2)

> Squaring", but it still take horrible execution time.

algorithms for multiplication. Their exponentiation routine already

does exponentiation by squaring, so you probably can't improve on it

without fixing multiplication.

> Where can I find the algorithm you are mentioning i.e. "bit shifting" ??

"Bit shifting" works when your base contains powers of 2. This is

because you can do powers of 2 by simply left-shifting the binary

representation by the appropriate number of bits. If you're doing

something like calculating large Mersenne numbers, this makes it about a

thousand times faster or more than Sun's implementation. It's an easy

and obvious optimization that Sun missed.

In short, here's a code snippet that does it. The base is expected

to be in a BigInteger called "big", and the exponent in an int called

"exponent". It factors out powers of two quickly by the call to

getLowestSetBit(), and does the exponentiation for powers of two rapidly

with shiftLeft() and then multiplies it by the remaining part that isn't

a power of 2.

This only helps if your base contains powers of 2.

if (big.signum() > 0)

{

// Get factor of two

int bit = big.getLowestSetBit();

if (bit > 0)

{

big = big.shiftRight(bit);

BigInteger twoPower = FrinkBigInteger.ONE.shiftLeft(bit*exponent);

if (big.equals(FrinkBigInteger.ONE))

return FrinkInteger.construct(twoPower);

else

{

big = big.pow(exponent);

return FrinkInteger.construct(big.multiply(twoPower));

}

}

}

--

Alan Eliasen | "It is not enough to do your best;

eliasen@... | you must know what to do and THEN

http://futureboy.homeip.net/ | do your best." -- W. Edwards Deming

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