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ipython/EulerProblem009.ipynb

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Euler Problem 9\n",
+    "\n",
+    "A Pythagorean triplet is a set of three natural numbers, $a < b < c$, for which,\n",
+    "\n",
+    "$a^2 + b^2 = c^2$\n",
+    "\n",
+    "For example, $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.\n",
+    "\n",
+    "There exists exactly one Pythagorean triplet for which $a + b + c = 1000$.\n",
+    "Find the product $abc$."
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "We start bruteforcing even though it feels like there may be a smart solution."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "66.96907186399949\n",
+      "31875000\n"
+     ]
+    }
+   ],
+   "source": [
+    "import timeit\n",
+    "\n",
+    "def brute_force():\n",
+    "    for a in range(1, 251):\n",
+    "        for b in range(a + 1, 501):\n",
+    "            for c in range(b + 1, 1001):\n",
+    "                if a + b + c == 1000 and a * a + b * b == c * c:\n",
+    "                    return a * b * c\n",
+    "\n",
+    "print(timeit.timeit(brute_force, number=10))\n",
+    "print(brute_force())"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "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."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "64.4656584559998\n",
+      "31875000\n"
+     ]
+    }
+   ],
+   "source": [
+    "def brute_force():\n",
+    "    for a in range(1, 251):\n",
+    "        for b in range(a + 1, 501):\n",
+    "            if a + b > 600:\n",
+    "                break\n",
+    "            for c in range(b + 1, 1001):\n",
+    "                if a + b + c == 1000 and a * a + b * b == c * c:\n",
+    "                    return a * b * c\n",
+    "                \n",
+    "print(timeit.timeit(brute_force, number=10))\n",
+    "print(brute_force())"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "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."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {
+    "collapsed": false
+   },
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "0.22822808900036762\n",
+      "31875000\n"
+     ]
+    }
+   ],
+   "source": [
+    "def smart_brute_force():\n",
+    "    for a in range(1, 251):\n",
+    "        for b in range(a + 1, 501):\n",
+    "            c = 1000 - a - b\n",
+    "            if a * a + b * b == c * c:\n",
+    "                return a * b * c\n",
+    "       \n",
+    "print(timeit.timeit(smart_brute_force, number=10))\n",
+    "print(smart_brute_force())  "
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {
+    "collapsed": true
+   },
+   "outputs": [],
+   "source": []
+  }
+ ],
+ "metadata": {
+  "completion_date": "Wed, 20 Aug 2014, 17:06",
+  "kernelspec": {
+   "display_name": "Python 3",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.5.4"
+  },
+  "tags": [
+   "brute force",
+   "pythagorean",
+   "triplet"
+  ]
+ },
+ "nbformat": 4,
+ "nbformat_minor": 0
+}