94 lines
2.0 KiB
Plaintext
94 lines
2.0 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 7\n",
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"\n",
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"By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.\n",
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"\n",
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"What is the 10 001st prime number?"
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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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"We reuse our prime functions and use a trick from [stackoverflow](https://stackoverflow.com/questions/2300756/get-the-nth-item-of-a-generator-in-python) to get the nth element."
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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": 7,
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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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"104743\n"
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]
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}
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],
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"source": [
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"import itertools\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 get_nth_prime(n):\n",
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" ps = prime_generator_function()\n",
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" return next(itertools.islice(ps, n - 1, n))\n",
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"\n",
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"assert(get_nth_prime(6) == 13)\n",
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"print(get_nth_prime(10001))"
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]
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}
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],
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"metadata": {
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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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"prime",
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"nth prime",
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"nth",
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"generator"
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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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