euler-001-multiples-of-3-and-5-20160629.ipynb

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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This notebook has a minor addition to\n",
    "euler-001-multiples-of-3-and-5-20160319.ipynb."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9.\n",
    "The sum of these multiples is 23.\n",
    "\n",
    "Find the sum of all the multiples of 3 or 5 below 1000."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "from __future__ import print_function"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "..."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "I will have another go at the union thing, this time as a function."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def union(sets):\n",
    "    u = set([])\n",
    "    for s in sets:\n",
    "        u |= s\n",
    "    return u"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 30,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def foo(n, divisors):\n",
    "    return sum(union(set(range(0, n, d)) for d in divisors))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "10000 loops, best of 3: 70.6 µs per loop\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "233168"
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     "execution_count": 31,
     "metadata": {},
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   "source": [
    "n=1000\n",
    "divisors = (3, 5)\n",
    "%timeit foo(n, divisors)\n",
    "foo(n, divisors)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "It works, but is not as readable, so forget it."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "2016-06-29: I have another way of writing the union function from above, this time using reduce() instead of iterating through a loop.\n",
    "Why did I not think of reduce() earlier?\n",
    "\n",
    "The code and output above is preserved as it ran on 2016-03-19 in Python 2.\n",
    "The code and output below ran on 2016-06-29 in Python 3."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "from functools import reduce\n",
    "from operator import or_"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def union(sets):\n",
    "    return reduce(or_, sets)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def foo(n, divisors):\n",
    "    return sum(union(set(range(0, n, d)) for d in divisors))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "10000 loops, best of 3: 59.1 µs per loop\n"
     ]
    },
    {
     "data": {
      "text/plain": [
       "233168"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "n=1000\n",
    "divisors = (3, 5)\n",
    "%timeit foo(n, divisors)\n",
    "foo(n, divisors)"
   ]
  }
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