{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Odd period square roots (Euler Problem 64)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {
    "collapsed": true
   },
   "source": [
    "[https://projecteuler.net/problem=64](https://projecteuler.net/problem=64)\n",
    "\n",
    "The first ten continued fraction representations of (irrational) square roots are:\n",
    "\n",
    "√2=[1;(2)], period=1\n",
    "\n",
    "√3=[1;(1,2)], period=2\n",
    "\n",
    "√5=[2;(4)], period=1\n",
    "\n",
    "√6=[2;(2,4)], period=2\n",
    "\n",
    "√7=[2;(1,1,1,4)], period=4\n",
    "\n",
    "√8=[2;(1,4)], period=2\n",
    "\n",
    "√10=[3;(6)], period=1\n",
    "\n",
    "√11=[3;(3,6)], period=2\n",
    "\n",
    "√12= [3;(2,6)], period=2\n",
    "\n",
    "√13=[3;(1,1,1,1,6)], period=5\n",
    "\n",
    "Exactly four continued fractions, for N ≤ 13, have an odd period.\n",
    "\n",
    "How many continued fractions for N ≤ 10000 have an odd period?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "import math\n",
    "\n",
    "def get_floor_sqrt(n):\n",
    "    return math.floor(math.sqrt(n))\n",
    "\n",
    "assert(get_floor_sqrt(5) == 2)\n",
    "assert(get_floor_sqrt(23) == 4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(3, 3, 2)"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "def next_expansion(current_a, current_nominator, current_denominator, original_number):\n",
    "    # Less typing\n",
    "    cn = current_nominator\n",
    "    cd = current_denominator\n",
    "\n",
    "    # Step 1: Multiply the fraction so that we can use the third binomial formula.\n",
    "    # Make sure we can reduce the nominator.\n",
    "    assert((original_number - cd * cd) % cn == 0)\n",
    "    # The new nominator is the denominator since we multiply with (x + cd) and then\n",
    "    # reduce the previous nominator.\n",
    "    # The new denominator is calculated by applying the third binomial formula and\n",
    "    # then by divided by the previous nominator.\n",
    "    cn, cd = cd, (original_number - cd * cd) // cn\n",
    "    \n",
    "    # Step 2: Calculate the next a by finding the next floor square root.\n",
    "    next_a = math.floor((math.sqrt(original_number) + cn) // cd)\n",
    "    \n",
    "    # Step 3: Remove next a from the fraction by substracting it.\n",
    "    cn = cn - next_a * cd\n",
    "    cn *= -1\n",
    "    \n",
    "    return next_a, cn, cd\n",
    "\n",
    "next_expansion(1, 7, 3, 23)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "def get_continued_fraction_sequence(n):\n",
    "    \n",
    "    # If number is a square number we return it.\n",
    "    floor_sqrt = get_floor_sqrt(n)\n",
    "    if n == floor_sqrt * floor_sqrt:\n",
    "        return ((floor_sqrt), [])\n",
    "    \n",
    "    # Otherwise, we calculate the next expansion till we\n",
    "    # encounter a step a second time.    \n",
    "    a = floor_sqrt\n",
    "    cn = a\n",
    "    cd = 1\n",
    "    sequence = []\n",
    "    previous_steps = []\n",
    "    \n",
    "    while not (a, cn, cd) in previous_steps :\n",
    "        #print(\"a: {} cn: {} cd: {}\".format(a, cn, cd))\n",
    "        previous_steps.append((a, cn, cd))\n",
    "        a, cn, cd = next_expansion(a, cd, cn, n)\n",
    "        sequence.append(a)\n",
    "    sequence.pop()\n",
    "    return ((floor_sqrt), sequence)\n",
    "\n",
    "assert(get_continued_fraction_sequence(1) == ((1), []))\n",
    "assert(get_continued_fraction_sequence(4) == ((2), []))\n",
    "assert(get_continued_fraction_sequence(25) == ((5), []))\n",
    "assert(get_continued_fraction_sequence(2) == ((1), [2]))\n",
    "assert(get_continued_fraction_sequence(3) == ((1), [1,2]))\n",
    "assert(get_continued_fraction_sequence(5) == ((2), [4]))\n",
    "assert(get_continued_fraction_sequence(13) == ((3), [1,1,1,1,6]))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "def get_period(n):\n",
    "    _, sequence = get_continued_fraction_sequence(n)\n",
    "    return len(sequence)\n",
    "\n",
    "assert(get_period(23) == 4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1322\n"
     ]
    }
   ],
   "source": [
    "s = len([n for n in range(1, 10001) if get_period(n) % 2 != 0])\n",
    "print(s)\n",
    "assert(s == 1322)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
  }
 ],
 "metadata": {
  "completion_date": "Tue, 22 Jan 2019, 05:10",
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python3.6",
   "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": "3.6.5"
  },
  "tags": [
   "fractions",
   "square roots",
   "sequence",
   "periodic"
  ]
 },
 "nbformat": 4,
 "nbformat_minor": 2
}