gcc/libjava/testsuite/libjava.lang/Primes.java
Anthony Green b3967ec43e Primes.java: New file.
* libjava.lang/Primes.java: New file.
	* libjava.lang/Primes.out: New file.

From-SVN: r28613
1999-08-09 06:35:56 +00:00

214 lines
6.5 KiB
Java

// Primes.java
/** Copyright 1998
* Roedy Green
* Canadian Mind Products
* 5317 Barker Avenue
* Burnaby, BC Canada V5H 2N6
* tel: (604) 435-3016
* mailto:roedy@mindprod.com
* http://mindprod.com
*/
// May be freely distributed for any purpose but military
import java.util.BitSet;
/**
* @author Roedy Green
* @version 1.10 1998 November 10
* Calculate primes using Eratostheses Sieve.
* Tell if a given number is prime.
* Find a prime just below a given number.
* Find a prime just above a given number.
*/
/*
* version 1.1 1998 November 10 - new address and phone.
*/
class Primes
{
/**
* constructors
*/
Primes()
{
ensureCapacity(1000);
}
/**
* @param capacity - largest number you will be asking if prime.
* If give too small a number, it will automatically grow by
* recomputing the sieve array.
*/
Primes (int capacity)
{
ensureCapacity(capacity);
}
/**
* @param candidate - is this a prime?
*/
public boolean isPrime(int candidate)
{
ensureCapacity(candidate);
if (candidate < 3) return candidate != 0;
if (candidate % 2 == 0 ) return false;
return !b.get(candidate/2);
}
/**
* @return first prime higher than candidate
*/
public int above(int candidate)
{
do
{
// see what we can find in the existing sieve
for (int i=candidate+1; i<= sieveCapacity; i++)
{
if (isPrime(i)) return i;
}
// Keep building ever bigger sieves till we succeed.
// The next prime P' is between P+2 and P^2 - 2.
// However that is a rather pessimistic upper bound.
// Ideally some theorem would tell us how big we need to build
// to find one.
ensureCapacity(Math.max(candidate*2, sieveCapacity*2));
} // end do
while (true);
} // end above
/**
* @param return first prime less than candidate
*/
public int below (int candidate)
{
for (candidate--; candidate > 0; candidate--)
{
if (isPrime(candidate)) return candidate;
}
// candidate was 1 or 0 or -ve
return 0;
}
/**
* calc all primes in the range 1..n,
* not the first n primes.
* @param n, highest candidate, not necessarily prime.
* @return list of primes 1..n in an array
*/
public final int[] getPrimes(int n)
{
// calculate the primes
ensureCapacity(n);
// pass 1: count primes
int countPrimes = 0;
for (int i = 0; i <= n; i++)
{
if (isPrime(i)) countPrimes++;
}
// pass 2: construct array of primes
int [] primes = new int[countPrimes];
countPrimes = 0;
for (int i = 0; i <= n; i++)
{
if (isPrime(i)) primes[countPrimes++] = i;
}
return primes;
} // end getPrimes
/**
* calculate the sieve, bit map of all primes 0..n
* @param n highest number evalutated by the sieve, not necessarily prime.
*/
private final void sieve ( int n )
{
// Presume BitSet b set is big enough for our purposes.
// Presume all even numbers are already marked composite, effectively.
// Presume all odd numbers are already marked prime (0 in bit map).
int last = (int)(Math.sqrt(n))+1;
for (int candidate = 3; candidate <= last; candidate += 2)
{
// only look at odd numbers
if (!b.get(candidate/2) /* if candidate is prime */)
{
// Our candidate is prime.
// Only bother to mark multiples of primes. Others already done.
// no need to mark even multiples, already done
int incr = candidate*2;
for ( int multiple = candidate + incr; multiple < n; multiple += incr)
{
b.set(multiple/2); // mark multiple as composite
} // end for multiple
} // end if
} // end for candidate
// at this point our sieve b is correct, except for 0..2
} // end sieve
/**
* Ensure have a sieve to tackle primes as big as n.
* If we don't allocate a sieve big enough and calculate it.
* @param n - ensure sieve big enough to evaluate n for primality.
*/
private void ensureCapacity (int n)
{
if ( n > sieveCapacity )
{
b = new BitSet((n+1)/2);
// starts out all 0, presume all numbers prime
sieveCapacity = n;
sieve(n);
}
// otherwise existing sieve is fine
} // end ensureCapacity
private int sieveCapacity;
// biggest number we have computed in our sieve.
// our BitSet array is indexed 0..N (odd only)
private BitSet b; /* true for each odd number if is composite */
/**
* Demonstrate and test the methods
*/
public static void main (String[] args)
{
// print primes 1..101
Primes calc = new Primes(106);
int[] primes = calc.getPrimes(101);
for (int i=0; i<primes.length; i++)
{
System.out.println(primes[i]);
}
// demonstrate isPrime, above, below
System.out.println(calc.isPrime(149));
System.out.println(calc.below(149));
System.out.println(calc.above(149));
// print all the primes just greater than powers of 2
calc = new Primes(10000000);
for (int pow=8; pow < 10000000; pow*=2)
System.out.println(calc.above(pow));
// Validate that isPrime works by comparing it with brute force
for (int i=3; i<=151; i++)
{
boolean prime = true;
for (int j=2; j<i; j++)
{
if (i % j == 0 )
{
prime = false;
break;
}
} // end for j
if ( calc.isPrime(i) != prime ) System.out.println(i + " oops");
} // end for i
} // end main
} // end Primes