eulerbooks/notebooks/problem0073.ipynb

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    "# [Counting Fractions in a Range](https://projecteuler.net/problem=73)\n",
    "\n",
    "As mentioned in [problem 71](https://projecteuler.net/problem=71), it's easy to compute the next value to appear between two neighbors in a Farey sequence by computing their mediant. Using this fact, we can can repeatedly calculate mediants to get *every* value between two neighbors in the sequence, stopping when the denominators get too big - in fact, for this problem, we *only* need to calculate the denominators.\n",
    "\n",
    "This essentially amounts to a depth-first search of the [Stern-Brocot tree](https://en.wikipedia.org/wiki/Stern%E2%80%93Brocot_tree) - check it out if you'd like a visualization.\n",
    "\n",
    "Theoretically, we could count using a recursive function, but Python will hit its recursion limit if we try to use it to solve this problem."
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    "def stern_brocot_count(d, low, high):\n",
    "    middle = low + high\n",
    "    if middle > d:\n",
    "        return 0\n",
    "    \n",
    "    return 1 + stern_brocot_count(d, low, middle) + stern_brocot_count(d, middle, high)"
   ]
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    "Instead, we'll solve the problem with the same principle, just implemented with a stack."
   ]
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       "7295372"
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    "d = 12000\n",
    "total = 0\n",
    "stack = [(3, 2)]\n",
    "while stack != []:\n",
    "    low, high = stack.pop()\n",
    "    middle = low + high\n",
    "    \n",
    "    if middle > d:\n",
    "        continue\n",
    "    total += 1\n",
    "    \n",
    "    stack.extend([(low, middle), (middle, high)])\n",
    "    \n",
    "total"
   ]
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   "source": [
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
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