euler/ipython/EulerProblem008.ipynb

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    "# Euler Problem 8\n",
    "\n",
    "The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.\n",
    "\n",
    "~~~\n",
    "73167176531330624919225119674426574742355349194934\n",
    "96983520312774506326239578318016984801869478851843\n",
    "85861560789112949495459501737958331952853208805511\n",
    "12540698747158523863050715693290963295227443043557\n",
    "66896648950445244523161731856403098711121722383113\n",
    "62229893423380308135336276614282806444486645238749\n",
    "30358907296290491560440772390713810515859307960866\n",
    "70172427121883998797908792274921901699720888093776\n",
    "65727333001053367881220235421809751254540594752243\n",
    "52584907711670556013604839586446706324415722155397\n",
    "53697817977846174064955149290862569321978468622482\n",
    "83972241375657056057490261407972968652414535100474\n",
    "82166370484403199890008895243450658541227588666881\n",
    "16427171479924442928230863465674813919123162824586\n",
    "17866458359124566529476545682848912883142607690042\n",
    "24219022671055626321111109370544217506941658960408\n",
    "07198403850962455444362981230987879927244284909188\n",
    "84580156166097919133875499200524063689912560717606\n",
    "05886116467109405077541002256983155200055935729725\n",
    "71636269561882670428252483600823257530420752963450\n",
    "~~~\n",
    "\n",
    "Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?"
   ]
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   "cell_type": "markdown",
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   "source": [
    "We need the number as a list of integers first."
   ]
  },
  {
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   "execution_count": 1,
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   "source": [
    "def remove_newlines(s):\n",
    "    import re\n",
    "    return re.sub(r'\\n', '', s)\n",
    "\n",
    "digits_string = \"\"\"73167176531330624919225119674426574742355349194934\n",
    "96983520312774506326239578318016984801869478851843\n",
    "85861560789112949495459501737958331952853208805511\n",
    "12540698747158523863050715693290963295227443043557\n",
    "66896648950445244523161731856403098711121722383113\n",
    "62229893423380308135336276614282806444486645238749\n",
    "30358907296290491560440772390713810515859307960866\n",
    "70172427121883998797908792274921901699720888093776\n",
    "65727333001053367881220235421809751254540594752243\n",
    "52584907711670556013604839586446706324415722155397\n",
    "53697817977846174064955149290862569321978468622482\n",
    "83972241375657056057490261407972968652414535100474\n",
    "82166370484403199890008895243450658541227588666881\n",
    "16427171479924442928230863465674813919123162824586\n",
    "17866458359124566529476545682848912883142607690042\n",
    "24219022671055626321111109370544217506941658960408\n",
    "07198403850962455444362981230987879927244284909188\n",
    "84580156166097919133875499200524063689912560717606\n",
    "05886116467109405077541002256983155200055935729725\n",
    "71636269561882670428252483600823257530420752963450\"\"\"\n",
    "\n",
    "digits = [int(d) for d in remove_newlines(digits_string)]"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    " Then we can use slicing to do a brute force."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "23514624000\n"
     ]
    }
   ],
   "source": [
    "def product(xs):\n",
    "    from operator import mul\n",
    "    from functools import reduce\n",
    "    return reduce(mul, xs, 1)\n",
    "\n",
    "def get_largest_product_of_n_digits(digits, n):\n",
    "    return max((product(digits[i:i + n]) for i in range(len(digits))))\n",
    "\n",
    "assert(get_largest_product_of_n_digits(digits, 4) == 5832)\n",
    "assert(get_largest_product_of_n_digits(digits, 13) == 23514624000)\n",
    "print(get_largest_product_of_n_digits(digits, 13))"
   ]
  }
 ],
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  "completion_date": "Wed, 20 Aug 2014, 16:03",
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