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   "source": [
    "# Square root convergents (Euler Problem 57)"
   ]
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   "source": [
    "[https://projecteuler.net/problem=57](https://projecteuler.net/problem=57)\n",
    "\n",
    "It is possible to show that the square root of two can be expressed as an infinite continued fraction.\n",
    "\n",
    "√ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...\n",
    "\n",
    "By expanding this for the first four iterations, we get:\n",
    "\n",
    "1 + 1/2 = 3/2 = 1.5\n",
    "\n",
    "1 + 1/(2 + 1/2) = 7/5 = 1.4\n",
    "\n",
    "1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...\n",
    "\n",
    "1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...\n",
    "\n",
    "The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion, 1393/985, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.\n",
    "\n",
    "In the first one-thousand expansions, how many fractions contain a numerator with more digits than denominator?"
   ]
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    "def get_digit_count(n):\n",
    "    c = 1\n",
    "    while n:\n",
    "        n += 1\n",
    "        n //= 10\n",
    "    return d"
   ]
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  {
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   "execution_count": 44,
   "metadata": {
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   "source": [
    "def gcd(a, b):\n",
    "    if b > a:\n",
    "        a, b = b, a\n",
    "    while a % b != 0:\n",
    "        a, b = b, a % b\n",
    "    return b\n",
    "        \n",
    "assert(gcd(100, 35) == 5)\n",
    "\n",
    "def add_fractions(n1, d1, n2, d2):\n",
    "    d = d1 * d2\n",
    "    n1 = n1 * (d // d1)\n",
    "    n2 = n2 * (d // d2)\n",
    "    n = n1 + n2\n",
    "    p = gcd(n, d)\n",
    "    return (n // p, d // p)"
   ]
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   "cell_type": "code",
   "execution_count": 54,
   "metadata": {
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   "source": [
    "def next_expension(n, d):\n",
    "    n, d = add_fractions(1, 1, n, d)\n",
    "    return add_fractions(1, 1, d, n)\n",
    "\n",
    "c = 0\n",
    "\n",
    "n, d = (3, 2)\n",
    "for i in range(1000):\n",
    "    if get_digit_count(n)> get_digit_count(d):\n",
    "        c += 1\n",
    "    n, d = next_expension(n, d)"
   ]
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   "cell_type": "code",
   "execution_count": 55,
   "metadata": {},
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     "output_type": "stream",
     "text": [
      "153\n"
     ]
    }
   ],
   "source": [
    "s = c\n",
    "print(s)"
   ]
  },
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   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
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