algorithm - Given Prime Number N, Compute the Next Prime? - Stack Overflow Given Prime Number N, Compute the Next Prime? up vote 54 down vote favorite 24 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? This really belongs on mathoverflow. – Kirk Broadhurst Dec 18 '10 at 1:05 2 Maybe your colleague is incorrect, or maybe it uses a few pre-computed primes rather than finding the next prime. – James K Polk Dec 18 '10 at 1:11 16 @Kirk: I disagree -- this is an algorithm question, not a math question.

– Jim Lewis Dec 18 '10 at 1:15 4 @Kirk It all falls under theoretical computer science, which is very much in the middle of programming and math. So I honestly don't see a problem posting this question on either site. – marcog Dec 18 '10 at 1:27 10 @Kirk: This definitely does not belong on MathOverflow, which is for research-level questions only. I also disagree that it needs to be on math.stackexchange.com , but it would at least be suitable there as well. – Antal Spector-Zabusky Dec 18 '10 at 2:19 up vote 30 down vote accepted 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. I wouldn't call that "fast" in the context of growing a dictionary. – James K Polk Dec 18 '10 at 1:13 Agree that the complexity isn't excessive but each of those operations is a relatively heavy remainder check; & the complexity of the check itself increases as p increases. – Kirk Broadhurst Dec 18 '10 at 1:19 @GregS See my edit. @Kirk For sure, and coming to realise these expenses is one of the things that make an experienced programmer. – marcog Dec 18 '10 at 1:24 @marcog Unless I am still asleep, for p = 10000, ln(p) = 9.2 and sqrt(p) = 100, => O(920).

– Kirk Broadhurst Dec 18 '10 at 1:30 up vote 51 down vote 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. 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. up vote 3 down vote 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 of course 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 7

1 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.

The First 1,000 Primes (the 1,000th is 7919) For more information on primes see 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 

311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691 701 709 

719 727 733 739 743 751 757 761 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 983 991 997 1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069 1087 1091 1093 1097 1103 1109 11

17 1123 1129 1151 1153 1163 1171 1181 1187 1193 1201 1213 1217 1223 1229 1231 1237 1249 1259 1277 1279 1283 1289 1291 1297 1301 1303 1307 1319 1321 1327 1361 1367 1373 1381 1399 1409 1423 1427 1429 1433 1439 1447 1451 1453 1459 1471 1481 1483 1487 1489 149

3 1499 1511 1523 1531 1543 1549 1553 1559 1567 1571 1579 1583 1597 1601 1607 1609 1613 1619 1621 1627 1637 1657 1663 1667 1669 1693 1697 1699 1709 1721 1723 1733 1741 1747 1753 1759 1777 1783 1787 1789 1801 1811 1823 1831 1847 1861 1867 1871 1873 1877 1879

 1889 1901 1907 1913 1931 1933 1949 1951 1973 1979 1987 1993 1997 1999 2003 2011 2017 2027 2029 2039 2053 2063 2069 2081 2083 2087 2089 2099 2111 2113 2129 2131 2137 2141 2143 2153 2161 2179 2203 2207 2213 2221 2237 2239 2243 2251 2267 2269 2273 2281 2287 

2293 2297 2309 2311 2333 2339 2341 2347 2351 2357 2371 2377 2381 2383 2389 2393 2399 2411 2417 2423 2437 2441 2447 2459 2467 2473 2477 2503 2521 2531 2539 2543 2549 2551 2557 2579 2591 2593 2609 2617 2621 2633 2647 2657 2659 2663 2671 2677 2683 2687 2689 2

693 2699 2707 2711 2713 2719 2729 2731 2741 2749 2753 2767 2777 2789 2791 2797 2801 2803 2819 2833 2837 2843 2851 2857 2861 2879 2887 2897 2903 2909 2917 2927 2939 2953 2957 2963 2969 2971 2999 3001 3011 3019 3023 3037 3041 3049 3061 3067 3079 3083 3089 31

09 3119 3121 3137 3163 3167 3169 3181 3187 3191 3203 3209 3217 3221 3229 3251 3253 3257 3259 3271 3299 3301 3307 3313 3319 3323 3329 3331 3343 3347 3359 3361 3371 3373 3389 3391 3407 3413 3433 3449 3457 3461 3463 3467 3469 3491 3499 3511 3517 3527 3529 353

3 3539 3541 3547 3557 3559 3571 3581 3583 3593 3607 3613 3617 3623 3631 3637 3643 3659 3671 3673 3677 3691 3697 3701 3709 3719 3727 3733 3739 3761 3767 3769 3779 3793 3797 3803 3821 3823 3833 3847 3851 3853 3863 3877 3881 3889 3907 3911 3917 3919 3923 3929

 3931 3943 3947 3967 3989 4001 4003 4007 4013 4019 4021 4027 4049 4051 4057 4073 4079 4091 4093 4099 4111 4127 4129 4133 4139 4153 4157 4159 4177 4201 4211 4217 4219 4229 4231 4241 4243 4253 4259 4261 4271 4273 4283 4289 4297 4327 4337 4339 4349 4357 4363 

4373 4391 4397 4409 4421 4423 4441 4447 4451 4457 4463 4481 4483 4493 4507 4513 4517 4519 4523 4547 4549 4561 4567 4583 4591 4597 4603 4621 4637 4639 4643 4649 4651 4657 4663 4673 4679 4691 4703 4721 4723 4729 4733 4751 4759 4783 4787 4789 4793 4799 4801 4

813 4817 4831 4861 4871 4877 4889 4903 4909 4919 4931 4933 4937 4943 4951 4957 4967 4969 4973 4987 4993 4999 5003 5009 5011 5021 5023 5039 5051 5059 5077 5081 5087 5099 5101 5107 5113 5119 5147 5153 5167 5171 5179 5189 5197 5209 5227 5231 5233 5237 5261 52

73 5279 5281 5297 5303 5309 5323 5333 5347 5351 5381 5387 5393 5399 5407 5413 5417 5419 5431 5437 5441 5443 5449 5471 5477 5479 5483 5501 5503 5507 5519 5521 5527 5531 5557 5563 5569 5573 5581 5591 5623 5639 5641 5647 5651 5653 5657 5659 5669 5683 5689 569

3 5701 5711 5717 5737 5741 5743 5749 5779 5783 5791 5801 5807 5813 5821 5827 5839 5843 5849 5851 5857 5861 5867 5869 5879 5881 5897 5903 5923 5927 5939 5953 5981 5987 6007 6011 6029 6037 6043 6047 6053 6067 6073 6079 6089 6091 6101 6113 6121 6131 6133 6143

 6151 6163 6173 6197 6199 6203 6211 6217 6221 6229 6247 6257 6263 6269 6271 6277 6287 6299 6301 6311 6317 6323 6329 6337 6343 6353 6359 6361 6367 6373 6379 6389 6397 6421 6427 6449 6451 6469 6473 6481 6491 6521 6529 6547 6551 6553 6563 6569 6571 6577 6581 

6599 6607 6619 6637 6653 6659 6661 6673 6679 6689 6691 6701 6703 6709 6719 6733 6737 6761 6763 6779 6781 6791 6793 6803 6823 6827 6829 6833 6841 6857 6863 6869 6871 6883 6899 6907 6911 6917 6947 6949 6959 6961 6967 6971 6977 6983 6991 6997 7001 7013 7019 7

027 7039 7043 7057 7069 7079 7103 7109 7121 7127 7129 7151 7159 7177 7187 7193 7207 7211 7213 7219 7229 7237 7243 7247 7253 7283 7297 7307 7309 7321 7331 7333 7349 7351 7369 7393 7411 7417 7433 7451 7457 7459 7477 7481 7487 7489 7499 7507 7517 7523 7529 75

37 7541 7547 7549 7559 7561 7573 7577 7583 7589 7591 7603 7607 7621 7639 7643 7649 7669 7673 7681 7687 7691 7699 7703 7717 7723 7727 7741 7753 7757 7759 7789 7793 7817 7823 7829 7841 7853 7867 7873 7877 7879 7883 7901 7907 7919 end.

Question: Which is the next prime number that is greater than 90 ?
Answer: 97