Given Prime Number N, Compute the Next Prime? - algorithm

Question

A coworker just told me that the C# Dictionary collection resizes by prime numbers for arcane reasons relating to hashing. And my immediate question was, "how does it know what the next prime is? do they story a giant table or compute on the fly? that's a scary non-deterministic runtime on an insert causing a resize"
So my question is, given N, which is a prime number, what is the most efficient way to compute the next prime number?

Answer 1

About a year ago I was working this area for libc++ while implementing the
unordered (hash) containers for C++11. I thought I would share
my experiences here. This experience supports marcog's accepted answer for a
reasonable definition of "brute force".
That means that even a simple brute force will be fast enough in most
circumstances, taking O(ln(p)*sqrt(p)) on average.
I developed several implementations of size_t next_prime(size_t n) where the
spec for this function is:
Returns: The smallest prime that is greater than or equal to n.
Each implementation of next_prime is accompanied by a helper function is_prime. is_prime should be considered a private implementation detail; not meant to be called directly by the client. Each of these implementations was of course tested for correctness, but also
tested with the following performance test:
int main()
{
typedef std::chrono::high_resolution_clock Clock;
typedef std::chrono::duration<double, std::milli> ms;
Clock::time_point t0 = Clock::now();
std::size_t n = 100000000;
std::size_t e = 100000;
for (std::size_t i = 0; i < e; ++i)
n = next_prime(n+1);
Clock::time_point t1 = Clock::now();
std::cout << e/ms(t1-t0).count() << " primes/millisecond\n";
return n;
}
I should stress that this is a performance test, and does not reflect typical
usage, which would look more like:
// Overflow checking not shown for clarity purposes
n = next_prime(2*n + 1);
All performance tests were compiled with:
clang++ -stdlib=libc++ -O3 main.cpp
Implementation 1
There are seven implementations. The purpose for displaying the first
implementation is to demonstrate that if you fail to stop testing the candidate
prime x for factors at sqrt(x) then you have failed to even achieve an
implementation that could be classified as brute force. This implementation is
brutally slow.
bool
is_prime(std::size_t x)
{
if (x < 2)
return false;
for (std::size_t i = 2; i < x; ++i)
{
if (x % i == 0)
return false;
}
return true;
}
std::size_t
next_prime(std::size_t x)
{
for (; !is_prime(x); ++x)
;
return x;
}
For this implementation only I had to set e to 100 instead of 100000, just to
get a reasonable running time:
0.0015282 primes/millisecond
Implementation 2
This implementation is the slowest of the brute force implementations and the
only difference from implementation 1 is that it stops testing for primeness
when the factor surpasses sqrt(x).
bool
is_prime(std::size_t x)
{
if (x < 2)
return false;
for (std::size_t i = 2; true; ++i)
{
std::size_t q = x / i;
if (q < i)
return true;
if (x % i == 0)
return false;
}
return true;
}
std::size_t
next_prime(std::size_t x)
{
for (; !is_prime(x); ++x)
;
return x;
}
Note that sqrt(x) isn't directly computed, but inferred by q < i. This
speeds things up by a factor of thousands:
5.98576 primes/millisecond
and validates marcog's prediction:
... this is well within the constraints of
most problems taking on the order of
a millisecond on most modern hardware.
Implementation 3
One can nearly double the speed (at least on the hardware I'm using) by
avoiding use of the % operator:
bool
is_prime(std::size_t x)
{
if (x < 2)
return false;
for (std::size_t i = 2; true; ++i)
{
std::size_t q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
}
return true;
}
std::size_t
next_prime(std::size_t x)
{
for (; !is_prime(x); ++x)
;
return x;
}
11.0512 primes/millisecond
Implementation 4
So far I haven't even used the common knowledge that 2 is the only even prime.
This implementation incorporates that knowledge, nearly doubling the speed
again:
bool
is_prime(std::size_t x)
{
for (std::size_t i = 3; true; i += 2)
{
std::size_t q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
}
return true;
}
std::size_t
next_prime(std::size_t x)
{
if (x <= 2)
return 2;
if (!(x & 1))
++x;
for (; !is_prime(x); x += 2)
;
return x;
}
21.9846 primes/millisecond
Implementation 4 is probably what most people have in mind when they think
"brute force".
Implementation 5
Using the following formula you can easily choose all numbers which are
divisible by neither 2 nor 3:
6 * k + {1, 5}
where k >= 1. The following implementation uses this formula, but implemented
with a cute xor trick:
bool
is_prime(std::size_t x)
{
std::size_t o = 4;
for (std::size_t i = 5; true; i += o)
{
std::size_t q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
o ^= 6;
}
return true;
}
std::size_t
next_prime(std::size_t x)
{
switch (x)
{
case 0:
case 1:
case 2:
return 2;
case 3:
return 3;
case 4:
case 5:
return 5;
}
std::size_t k = x / 6;
std::size_t i = x - 6 * k;
std::size_t o = i < 2 ? 1 : 5;
x = 6 * k + o;
for (i = (3 + o) / 2; !is_prime(x); x += i)
i ^= 6;
return x;
}
This effectively means that the algorithm has to check only 1/3 of the
integers for primeness instead of 1/2 of the numbers and the performance test
shows the expected speed up of nearly 50%:
32.6167 primes/millisecond
Implementation 6
This implementation is a logical extension of implementation 5: It uses the
following formula to compute all numbers which are not divisible by 2, 3 and 5:
30 * k + {1, 7, 11, 13, 17, 19, 23, 29}
It also unrolls the inner loop within is_prime, and creates a list of "small
primes" that is useful for dealing with numbers less than 30.
static const std::size_t small_primes[] =
{
2,
3,
5,
7,
11,
13,
17,
19,
23,
29
};
static const std::size_t indices[] =
{
1,
7,
11,
13,
17,
19,
23,
29
};
bool
is_prime(std::size_t x)
{
const size_t N = sizeof(small_primes) / sizeof(small_primes[0]);
for (std::size_t i = 3; i < N; ++i)
{
const std::size_t p = small_primes[i];
const std::size_t q = x / p;
if (q < p)
return true;
if (x == q * p)
return false;
}
for (std::size_t i = 31; true;)
{
std::size_t q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
i += 6;
q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
i += 4;
q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
i += 2;
q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
i += 4;
q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
i += 2;
q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
i += 4;
q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
i += 6;
q = x / i;
if (q < i)
return true;
if (x == q * i)
return false;
i += 2;
}
return true;
}
std::size_t
next_prime(std::size_t n)
{
const size_t L = 30;
const size_t N = sizeof(small_primes) / sizeof(small_primes[0]);
// If n is small enough, search in small_primes
if (n <= small_primes[N-1])
return *std::lower_bound(small_primes, small_primes + N, n);
// Else n > largest small_primes
// Start searching list of potential primes: L * k0 + indices[in]
const size_t M = sizeof(indices) / sizeof(indices[0]);
// Select first potential prime >= n
// Known a-priori n >= L
size_t k0 = n / L;
size_t in = std::lower_bound(indices, indices + M, n - k0 * L) - indices;
n = L * k0 + indices[in];
while (!is_prime(n))
{
if (++in == M)
{
++k0;
in = 0;
}
n = L * k0 + indices[in];
}
return n;
}
This is arguably getting beyond "brute force" and is good for boosting the
speed another 27.5% to:
41.6026 primes/millisecond
Implementation 7
It is practical to play the above game for one more iteration, developing a
formula for numbers not divisible by 2, 3, 5 and 7:
210 * k + {1, 11, ...},
The source code isn't shown here, but is very similar to implementation 6.
This is the implementation I chose to actually use for the unordered containers
of libc++, and that source code is open source (found at the link).
This final iteration is good for another 14.6% speed boost to:
47.685 primes/millisecond
Use of this algorithm assures that clients of libc++'s hash tables can choose
any prime they decide is most beneficial to their situation, and the performance
for this application is quite acceptable.

Answer 2

Just in case somebody is curious:
Using reflector I determined that .Net uses a static class that contains a hard coded list of ~72 primes ranging up to 7199369 which is scans for the smallest prime that is at least twice the current size, and for sizes larger than that it computes the next prime by trial division of all odd numbers up to the sqrt of the potential number. This class is immutable and thread safe (i.e. larger primes are not stored for future use).

Answer 3

The gaps between consecutive prime numbers is known to be quite small, with the first gap of over 100 occurring for prime number 370261. That means that even a simple brute force will be fast enough in most circumstances, taking O(ln(p)*sqrt(p)) on average.
For p=10,000 that's O(921) operations. Bearing in mind we'll be performing this once every O(ln(p)) insertion (roughly speaking), this is well within the constraints of most problems taking on the order of a millisecond on most modern hardware.

Answer 4

A nice trick is to use a partial sieve. For example, what is the next prime that follows the number N = 2534536543556?
Check the modulus of N with respect to a list of small primes. Thus...
mod(2534536543556,[3 5 7 11 13 17 19 23 29 31 37])
ans =
2 1 3 6 4 1 3 4 22 16 25
We know that the next prime following N must be an odd number, and we can immediately discard all odd multiples of this list of small primes. These moduli allow us to sieve out multiples of those small primes. Were we to use the small primes up to 200, we can use this scheme to immediately discard most potential prime numbers greater than N, except for a small list.
More explicitly, if we are looking for a prime number beyond 2534536543556, it cannot be divisible by 2, so we need consider only the odd numbers beyond that value. The moduli above show that 2534536543556 is congruent to 2 mod 3, therefore 2534536543556+1 is congruent to 0 mod 3, as must be 2534536543556+7, 2534536543556+13, etc. Effectively, we can sieve out many of the numbers without any need to test them for primality and without any trial divisions.
Similarly, the fact that
mod(2534536543556,7) = 3
tells us that 2534536543556+4 is congruent to 0 mod 7. Of course, that number is even, so we can ignore it. But 2534536543556+11 is an odd number that is divisible by 7, as is 2534536543556+25, etc. Again, we can exclude these numbers as clearly composite (because they are divisible by 7) and so not prime.
Using only the small list of primes up to 37, we can exclude most of the numbers that immediately follow our starting point of 2534536543556, only excepting a few:
{2534536543573 , 2534536543579 , 2534536543597}
Of those numbers, are they prime?
2534536543573 = 1430239 * 1772107
2534536543579 = 99833 * 25387763
I've made the effort of providing the prime factorizations of the first two numbers in the list. See that they are composite, but the prime factors are large. Of course, this makes sense, since we've already ensured that no number that remains can have small prime factors. The third one in our short list (2534536543597) is in fact the very first prime number beyond N. The sieving scheme I've described will tend to result in numbers that are either prime, or are composed of generally large prime factors. So we needed to actually apply an explicit test for primality to only a few numbers before finding the next prime.
A similar scheme quickly yields the next prime exceeding N = 1000000000000000000000000000, as 1000000000000000000000000103.

Answer 5

Just a few experiments with inter-primes distance.
This is a complement to visualize other answers.
I got the primes from the 100.000th (=1,299,709) to the 200.000th (=2,750,159)
Some data:
Maximum interprime distance = 148
Mean interprime distance = 15
Interprime distance frequency plot:
Interprime Distance vs Prime Number
Just to see it's "random". However ...

Answer 6

There's no function f(n) to calculate the next prime number. Instead a number must be tested for primality.
It is also very useful, when finding the nth prime number, to already know all prime numbers from the 1st up to (n-1)th, because those are the only numbers that need to be tested as factors.
As a result of these reasons, I would not be surprised if there is a precalculated set of large prime numbers. It doesn't really make sense to me if certain primes needed to be recalculated repeatedly.

Answer 7

As others have already noted, a means of finding the next prime number given the current prime has not been found yet. Therefore most algorithms focus more on using a fast means of checking primality since you have to check n/2 of the numbers between your known prime and the next one.
Depending upon the application, you can also get away with just hard-coding a look-up table, as noted by Paul Wheeler.

Answer 8

For sheer novelty, there’s always this approach:
#!/usr/bin/perl
for $p ( 2 .. 200 ) {
next if (1x$p) =~ /^(11+)\1+$/;
for ($n=1x(1+$p); $n =~ /^(11+)\1+$/; $n.=1) { }
printf "next prime after %d is %d\n", $p, length($n);
}
which produces
next prime after 2 is 3
next prime after 3 is 5
next prime after 5 is 7
next prime after 7 is 11
next prime after 11 is 13
next prime after 13 is 17
next prime after 17 is 19
next prime after 19 is 23
next prime after 23 is 29
next prime after 29 is 31
next prime after 31 is 37
next prime after 37 is 41
next prime after 41 is 43
next prime after 43 is 47
next prime after 47 is 53
next prime after 53 is 59
next prime after 59 is 61
next prime after 61 is 67
next prime after 67 is 71
next prime after 71 is 73
next prime after 73 is 79
next prime after 79 is 83
next prime after 83 is 89
next prime after 89 is 97
next prime after 97 is 101
next prime after 101 is 103
next prime after 103 is 107
next prime after 107 is 109
next prime after 109 is 113
next prime after 113 is 127
next prime after 127 is 131
next prime after 131 is 137
next prime after 137 is 139
next prime after 139 is 149
next prime after 149 is 151
next prime after 151 is 157
next prime after 157 is 163
next prime after 163 is 167
next prime after 167 is 173
next prime after 173 is 179
next prime after 179 is 181
next prime after 181 is 191
next prime after 191 is 193
next prime after 193 is 197
next prime after 197 is 199
next prime after 199 is 211
All fun and games aside, it is well known that the optimal hash table size is rigorously provably a prime number of the form 4N−1. So just finding the next prime is insufficient. You have to do the other check, too.

Answer 9

As far as I remember, it uses prime number next to the double of current size. It doesn't calculate that prime number - there table with preloaded numbers up to some big value (do not exactly, something about around 10,000,000). When that number is reached, it uses some naive algorithm to get next number (like curNum=curNum+1) and validates it using some if these methods: http://en.wikipedia.org/wiki/Prime_number#Verifying_primality