183 lines
4.3 KiB
Plaintext
183 lines
4.3 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Euler Problem 10\n",
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"\n",
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"The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.\n",
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"\n",
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"Find the sum of all the primes below two million."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Okay, reuse prime generator from 7 and go."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {
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"collapsed": false
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"61.8695481220002\n",
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"142913828922\n"
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]
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}
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],
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"source": [
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"import timeit\n",
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"\n",
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"def is_prime(n, smaller_primes):\n",
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" for s in smaller_primes:\n",
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" if n % s == 0:\n",
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" return False\n",
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" if s * s > n:\n",
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" return True\n",
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" return True\n",
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"\n",
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"def prime_generator_function():\n",
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" primes = [2, 3, 5, 7]\n",
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" for p in primes:\n",
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" yield p\n",
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" while True:\n",
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" p += 2\n",
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" if is_prime(p, primes):\n",
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" primes.append(p)\n",
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" yield p\n",
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" \n",
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"def brute_force(): \n",
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" ps = prime_generator_function()\n",
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" s = 0\n",
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" p = next(ps)\n",
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" while p < 2000000:\n",
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" s += p\n",
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" p = next(ps)\n",
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" return s\n",
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" \n",
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"print(timeit.timeit(brute_force, number=10))\n",
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"assert(brute_force() == 142913828922)\n",
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"print(brute_force())"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"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."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {
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"collapsed": false
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},
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"outputs": [],
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"source": [
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"def get_primes_smaller(number):\n",
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" primes = []\n",
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" prospects = [n for n in range(2, number)]\n",
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" while prospects:\n",
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" p = prospects[0]\n",
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" prospects = [x for x in prospects if x % p != 0]\n",
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" primes.append(p)\n",
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" return primes\n",
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"\n",
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"def brute_force():\n",
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" return sum(get_primes_smaller(2000000))\n",
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"\n",
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"#print(timeit.timeit(brute_force, number=1))\n",
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"#print(brute_force())"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"This did not even terminate. We optimize the sieve by stopping when $p^2 > number$."
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {
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"collapsed": false
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"74.5315607270004\n",
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"142913828922\n"
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]
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}
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],
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"source": [
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"def sieve_of_eratosthenes(number):\n",
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" primes = []\n",
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" prospects = [n for n in range(2, number)]\n",
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" while prospects:\n",
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" p = prospects[0]\n",
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" prospects = [x for x in prospects if x % p != 0]\n",
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" primes.append(p)\n",
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" if p * p > number:\n",
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" break\n",
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" return primes + prospects\n",
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"\n",
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"def brute_force():\n",
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" return sum(sieve_of_eratosthenes(2000000))\n",
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"\n",
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"assert(sieve_of_eratosthenes(10) == [2, 3, 5, 7,])\n",
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"print(timeit.timeit(brute_force, number=10))\n",
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"assert(brute_force() == 142913828922)\n",
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"print(brute_force())"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Okay, I honestly did not expect this to be slower than our generator. Gotta keep that in mind for future prime related problems."
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]
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}
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],
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"metadata": {
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"completion_date": "Wed, 20 Aug 2014, 20:28",
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"kernelspec": {
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"display_name": "Python 3",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.5.4"
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},
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"tags": [
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"primes",
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"brute force",
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"timeit"
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]
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},
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"nbformat": 4,
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"nbformat_minor": 0
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}
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