From edcd9b60d1b4cb40db2f4f3020752e243a1ad52c Mon Sep 17 00:00:00 2001
From: filifa <filifa.breath652@8alias.com>
Date: Sun, 15 Jun 2025 23:45:01 -0400
Subject: [PATCH] add problem 80

---
 notebooks/problem0080.ipynb | 109 ++++++++++++++++++++++++++++++++++++
 1 file changed, 109 insertions(+)
 create mode 100644 notebooks/problem0080.ipynb

diff --git a/notebooks/problem0080.ipynb b/notebooks/problem0080.ipynb
new file mode 100644
index 0000000..c0d149a
--- /dev/null
+++ b/notebooks/problem0080.ipynb
@@ -0,0 +1,109 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "3aeab8bf",
+   "metadata": {},
+   "source": [
+    "# [Square Root Digital Expansion](https://projecteuler.net/problem=80)\n",
+    "\n",
+    "Unfortunately, just calling Python's `sqrt` won't cut it - floats don't have nearly enough digits of precision.\n",
+    "\n",
+    "The easiest approach here is to use SageMath's arbitrary precision routines, or Python's [decimal](https://docs.python.org/3/library/decimal.html) package. We need to watch for rounding issues, so we'll calculate slightly more than 100 digits and truncate."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "b1d58be8",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "40886"
+      ]
+     },
+     "execution_count": 1,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "def digital_sum(n):\n",
+    "    s = sqrt(n).n(digits=110)\n",
+    "    digits = (int(d) for d in str(s)[:101] if d != '.')\n",
+    "    return sum(digits)\n",
+    "    \n",
+    "\n",
+    "sum(digital_sum(n) for n in range(1, 101) if not is_square(n))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ad2b9aee",
+   "metadata": {},
+   "source": [
+    "If you'd rather not use those tools for some reason, another simple approach is to take advantage of Python's arbitrary precision integers and calculate the [integer square root](https://en.wikipedia.org/wiki/Integer_square_root) of $n \\times 10^k$ for a sufficiently large $k$ ($k \\geq 200$ for this problem)."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "6bd0cb0d",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "40886"
+      ]
+     },
+     "execution_count": 2,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "def digital_sum(n):\n",
+    "    s = isqrt(n * 10^200)\n",
+    "    digits = (int(d) for d in str(s)[:100])\n",
+    "    return sum(digits)\n",
+    "    \n",
+    "\n",
+    "sum(digital_sum(n) for n in range(1, 101) if not is_square(n))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "c193ba51",
+   "metadata": {},
+   "source": [
+    "If you want to go the extra mile, there are several [algorithms for computing square roots](https://en.wikipedia.org/wiki/Square_root_algorithms) that you can implement yourself, such as Heron's method, which is a special case of [Newton's method](https://en.wikipedia.org/wiki/Newton%27s_method) for solving $x^2 - n = 0$. The method works by starting with an initial estimate $x_0$ (such as $\\frac{n}{2}$), then repeatedly calculating\n",
+    "$$x_{k+1} = \\frac{1}{2}\\left(x_k + \\frac{n}{x_k}\\right)$$\n",
+    "until $|x_{k+1} - x_k|$ is sufficiently small. For computing the integer square root, this can be when $|x_{k+1} - x_k| < 1$."
+   ]
+  }
+ ],
+ "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
+}