Question
By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10001st prime number?
Clojure
#!/usr/bin/env clojure
(defn primes [n]
(defn improve [p nums]
(filter #(or
(not (= (rem % p) 0))
(= % p))
nums))
(defn prime-iter [p nums i]
(if (> (* p p) n)
nums
(prime-iter (nth nums (+ i 1)) (improve (nth nums (+ i 1)) nums) (+ i 1))))
(prime-iter 2 (range 2 (+ n 1)) -1))
(println (nth (primes 1000000) 10000))
Go
package main
import "fmt"
func iSqrt(n int64) int64 {
var r1, r int64 = n, n + 1
for r1 < r {
r, r1 = r1, (r1+n/r1)>>1
}
return r
}
func PrimeSieve(n int64) []int64 {
result := make([]int64, 0, n)
sieve := make([]bool, n+1)
sn := iSqrt(n)
var i, j int64
for i = 2; i <= sn; i++ {
if !sieve[i] {
for j = i * i; j <= n; j += i {
sieve[j] = true
}
}
}
for i = 2; i <= n; i++ {
if !sieve[i] {
result = append(result, i)
}
}
return result
}
func main() {
primes := PrimeSieve(1000000)
fmt.Println(primes[10000])
}
$ go build -o 10001st-prime 10001st-prime.go
$ time ./10001st-prime
real 0m0.005s
user 0m0.000s
sys 0m0.006s
Haskell
primes :: [Integer]
primes = 2 : sieve primes [3,5..] where
sieve (p:ps) xs = h ++ sieve ps [x | x <- t, rem x p /= 0]
where (h, t) = span (< p*p) xs
main :: IO ()
main = print $ primes !! 10000
$ ghc -O2 -o 10001st-prime 10001st-prime.hs
$ time ./10001st-prime
real 0m0.025s
user 0m0.016s
sys 0m0.008s
JavaScript
const sieve = {}
let n = 0
for (var q = 2; n < 10001; q++) {
if (sieve[q]) {
sieve[q].forEach((p) => {
const list = sieve[p + q] || []
list.push(p)
sieve[p + q] = list
})
} else {
sieve[q * q] = [q]
n++
}
}
console.log(q - 1)
Python
#!/usr/bin/env python
def eratosthenes():
'''Yields the sequence of prime numbers via the Sieve of Eratosthenes.'''
D = {} # map composite integers to primes witnessing their compositeness
q = 2 # first integer to test for primality
while 1:
if q not in D:
yield q # not marked composite, must be prime
D[q*q] = [q] # first multiple of q not already marked
else:
for p in D[q]: # move each witness to its next multiple
D.setdefault(p+q,[]).append(p)
del D[q] # no longer need D[q], free memory
q += 1
def nth_prime(n):
for i, prime in enumerate(eratosthenes()):
if i == n - 1:
return prime
print(nth_prime(10001))
Ruby
Rust
fn eratosthenes(limit: usize) -> Vec<usize> {
let mut sieve = vec![true; limit];
let mut p = 2;
loop {
// Eliminate multiples of p.
let mut i = 2 * p - 1;
while i < limit {
sieve[i] = false;
i += p;
}
// Find the next prime.
if let Some(n) = (p..limit).find(|&n| sieve[n]) {
p = n + 1;
} else {
break;
}
}
sieve
.iter()
.enumerate()
.filter(|&(_, &is_prime)| is_prime)
.skip(1)
.map(|(i, _)| i + 1)
.collect()
}
fn main() {
let primes = eratosthenes(1000000);
println!("{}", primes[10000]);
}