"The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.\n",
"\n",
"Find the sum of all the primes below two million."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Okay, reuse prime generator from 7 and go."
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"61.8695481220002\n",
"142913828922\n"
]
}
],
"source": [
"import timeit\n",
"\n",
"def is_prime(n, smaller_primes):\n",
" for s in smaller_primes:\n",
" if n % s == 0:\n",
" return False\n",
" if s * s > n:\n",
" return True\n",
" return True\n",
"\n",
"def prime_generator_function():\n",
" primes = [2, 3, 5, 7]\n",
" for p in primes:\n",
" yield p\n",
" while True:\n",
" p += 2\n",
" if is_prime(p, primes):\n",
" primes.append(p)\n",
" yield p\n",
" \n",
"def brute_force(): \n",
" ps = prime_generator_function()\n",
" s = 0\n",
" p = next(ps)\n",
" while p < 2000000:\n",
" s += p\n",
" p = next(ps)\n",
" return s\n",
" \n",
"print(timeit.timeit(brute_force, number=10))\n",
"assert(brute_force() == 142913828922)\n",
"print(brute_force())"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Okay, here it may actually be way smarter to use a sieve. We implent it because the old one was shitty. Okay, I am actually interested in the time difference. So we use the old one first and then implement the optimization."
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"def get_primes_smaller(number):\n",
" primes = []\n",
" prospects = [n for n in range(2, number)]\n",
" while prospects:\n",
" p = prospects[0]\n",
" prospects = [x for x in prospects if x % p != 0]\n",
" primes.append(p)\n",
" return primes\n",
"\n",
"def brute_force():\n",
" return sum(get_primes_smaller(2000000))\n",
"\n",
"#print(timeit.timeit(brute_force, number=1))\n",
"#print(brute_force())"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"This did not even terminate. We optimize the sieve by stopping when $p^2 > number$."
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"74.5315607270004\n",
"142913828922\n"
]
}
],
"source": [
"def sieve_of_eratosthenes(number):\n",
" primes = []\n",
" prospects = [n for n in range(2, number)]\n",
" while prospects:\n",
" p = prospects[0]\n",
" prospects = [x for x in prospects if x % p != 0]\n",
