166 lines
3.9 KiB
Plaintext
166 lines
3.9 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 9\n",
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"\n",
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"A Pythagorean triplet is a set of three natural numbers, $a < b < c$, for which,\n",
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"\n",
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"$a^2 + b^2 = c^2$\n",
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"\n",
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"For example, $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.\n",
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"\n",
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"There exists exactly one Pythagorean triplet for which $a + b + c = 1000$.\n",
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"Find the product $abc$."
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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 start bruteforcing even though it feels like there may be a smart solution."
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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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"66.96907186399949\n",
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"31875000\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 brute_force():\n",
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" for a in range(1, 251):\n",
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" for b in range(a + 1, 501):\n",
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" for c in range(b + 1, 1001):\n",
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" if a + b + c == 1000 and a * a + b * b == c * c:\n",
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" return a * b * c\n",
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"\n",
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"print(timeit.timeit(brute_force, number=10))\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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"Let's do some optimization by cancelling the loops when we exceed the boundaries. Actually, I have also realized that choosing 251 and 501 is a little bit arbitrary. For example if a, b, c where something like $332, 333, 335$ that could be a solution and the first range was too low. Then again, we would have realized that as soon as we get back None, so it is okay."
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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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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"64.4656584559998\n",
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"31875000\n"
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]
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}
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],
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"source": [
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"def brute_force():\n",
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" for a in range(1, 251):\n",
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" for b in range(a + 1, 501):\n",
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" if a + b > 600:\n",
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" break\n",
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" for c in range(b + 1, 1001):\n",
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" if a + b + c == 1000 and a * a + b * b == c * c:\n",
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" return a * b * c\n",
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" \n",
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"print(timeit.timeit(brute_force, number=10))\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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"Big time save. Kappa. Okay, I am stupid. If I have a and b I can calculate c and check if it is a solution."
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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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"0.22822808900036762\n",
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"31875000\n"
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]
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}
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],
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"source": [
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"def smart_brute_force():\n",
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" for a in range(1, 251):\n",
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" for b in range(a + 1, 501):\n",
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" c = 1000 - a - b\n",
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" if a * a + b * b == c * c:\n",
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" return a * b * c\n",
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" \n",
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"print(timeit.timeit(smart_brute_force, number=10))\n",
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"print(smart_brute_force()) "
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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": null,
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"metadata": {
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"collapsed": true
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},
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"outputs": [],
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"source": []
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}
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],
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"metadata": {
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"completion_date": "Wed, 20 Aug 2014, 17:06",
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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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"brute force",
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"pythagorean",
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"triplet"
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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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