Source: Wikipedia Bertrand’s postulate (actually a theorem) states that if n > 3 is an integer, then there always exists at least one prime number p with n < p < 2n − 2. A weaker but more elegant formulation is: for every n > 1 there is always at least one prime p such … Read more
There are a couple of small optimizations for your version. By reversing the roles of True and False, you can change “if flags[i] is False:” to “if flags[i]:“. And the starting value for the second range statement can be i*i instead of i*3. Your original version takes 0.166 seconds on my system. With those changes, … Read more
A simple fast solution in Java. Uses approach described by Anon.. Here TreeSet is just a container capable of returning smallest element in it. (No duplicates stored.) int n = 20; SortedSet<Long> next = new TreeSet<Long>(); next.add((long) 1); long cur = 0; for (int i = 0; i < n; ++i) { cur = next.first(); … Read more
To calculate the n-th prime, I know two main variants. The straightforward way That is to count all the primes starting from 2 as you find them until you have reached the desired nth. This can be done with different levels of sophistication and efficiency, and there are two conceptually different ways to do it. … Read more
Here’s another approach that celebrates Clojure’s Java interop. This takes 374ms on a 2.4 Ghz Core 2 Duo (running single-threaded). I let the efficient Miller-Rabin implementation in Java’s BigInteger#isProbablePrime deal with the primality check. (def certainty 5) (defn prime? [n] (.isProbablePrime (BigInteger/valueOf n) certainty)) (concat [2] (take 10001 (filter prime? (take-nth 2 (range 1 Integer/MAX_VALUE))))) … Read more
RSA doesn’t pick from a list of known primes: it generates a new very large number, then applies an algorithm to find a nearby number that is almost certainly prime. See this useful description of large prime generation): The standard way to generate big prime numbers is to take a preselected random number of the … Read more
Here’s another way: boolean isPrime(long n) { if(n < 2) return false; if(n == 2 || n == 3) return true; if(n%2 == 0 || n%3 == 0) return false; long sqrtN = (long)Math.sqrt(n)+1; for(long i = 6L; i <= sqrtN; i += 6) { if(n%(i-1) == 0 || n%(i+1) == 0) return false; } … Read more
All numbers (when hashed) are still going to be the p lowest-order bits of k for 127 too. That is wrong (or I misunderstood..). k % 127 depends on all bits of k. k % 128 only depends on the 7 lowest bits. EDIT: If you have a perfect distribution between 1 and 10,000. 10,000 … Read more
Here’s an example of a sieve implementation in JavaScript: function getPrimes(max) { var sieve = [], i, j, primes = []; for (i = 2; i <= max; ++i) { if (!sieve[i]) { // i has not been marked — it is prime primes.push(i); for (j = i << 1; j <= max; j += … Read more
I stumbled across a comment that sheds some light on what DJB is up to: /* * DJBX33A (Daniel J. Bernstein, Times 33 with Addition) * * This is Daniel J. Bernstein’s popular `times 33′ hash function as * posted by him years ago on comp.lang.c. It basically uses a function * like “hash(i) = … Read more