From 2ca02f9aba02af2760a75aee3a8748a466d7bbe1 Mon Sep 17 00:00:00 2001
From: filifa <filifa.breath652@8alias.com>
Date: Thu, 17 Jul 2025 23:58:53 -0400
Subject: [PATCH] add problem 73

---
 notebooks/problem0073.ipynb | 96 +++++++++++++++++++++++++++++++++++++
 1 file changed, 96 insertions(+)
 create mode 100644 notebooks/problem0073.ipynb

diff --git a/notebooks/problem0073.ipynb b/notebooks/problem0073.ipynb
new file mode 100644
index 0000000..dfe0167
--- /dev/null
+++ b/notebooks/problem0073.ipynb
@@ -0,0 +1,96 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "e8fb07d9",
+   "metadata": {},
+   "source": [
+    "# [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."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "81408bb4",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "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)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ef9c62b5",
+   "metadata": {},
+   "source": [
+    "Instead, we'll solve the problem with the same principle, just implemented with a stack."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "489b8afb",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "7295372"
+      ]
+     },
+     "execution_count": 2,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "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"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "SageMath 9.5",
+   "language": "sage",
+   "name": "sagemath"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.11.2"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 5
+}