From f5604613cfc26a08975cf713b817e52ef6a4768f Mon Sep 17 00:00:00 2001
From: filifa <filifa.breath652@8alias.com>
Date: Sat, 28 Jun 2025 22:34:48 -0400
Subject: [PATCH] add problem 85

---
 notebooks/problem0085.ipynb | 153 ++++++++++++++++++++++++++++++++++++
 1 file changed, 153 insertions(+)
 create mode 100644 notebooks/problem0085.ipynb

diff --git a/notebooks/problem0085.ipynb b/notebooks/problem0085.ipynb
new file mode 100644
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+++ b/notebooks/problem0085.ipynb
@@ -0,0 +1,153 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "f6ad1b47",
+   "metadata": {},
+   "source": [
+    "# [Counting Rectangles](https://projecteuler.net/problem=85)\n",
+    "\n",
+    "Let $R(m, n)$ give the number of rectangles contained in an $m \\times n$ grid. We will derive a simple formula for $R$. First, observe that an $m \\times 1$ grid contains $\\frac{m(m+1)}{2}$ rectangles, i.e.\n",
+    "$$R(m, 1) = \\frac{m(m+1)}{2}$$\n",
+    "(You could prove this [inductively](https://en.wikipedia.org/wiki/Mathematical_induction) if you really wanted to).\n",
+    "\n",
+    "Then, we can derive the following recursive formula for $R$.\n",
+    "$$R(m, n) = R(m, n - 1) + n R(m, 1)$$\n",
+    "To make sense of this formula, consider that the number of rectangles in an $m \\times n$ grid must be the sum of the number of rectangles in an $m \\times (n-1)$ grid (i.e. $R(m, n-1)$) and the number of rectangles in an $m \\times n$ grid *that are at least partially in the bottom row* - these are [disjoint sets](https://en.wikipedia.org/wiki/Disjoint_sets), so there's no double counting with this approach. We already know the number of rectangles that lie *entirely* in the bottom row - it's just $R(m, 1)$. We can then make any of those rectangles any height we want between 1 and $n$, giving the formula $n R(m, 1)$.\n",
+    "\n",
+    "At this point, we could easily translate this into a recursive function to compute $R$, but we can also simplify this into a closed formula. Substituting the formula for $R(m,1)$ from above, we get\n",
+    "$$R(m, n) = R(m, n - 1) + \\frac{mn(m+1)}{2}$$\n",
+    "Then, by repeatedly substituting $R(m, n-1)$ with our recursive formula, we get\n",
+    "$$R(m, n) = \\frac{mn(m+1)}{2} + \\frac{m(n-1)(m+1)}{2} + \\frac{m(n-2)(m+1)}{2} + \\cdots + \\frac{m(m+1)}{2} = \\sum_{k=1}^n \\frac{mk(m+1)}{2}$$\n",
+    "Applying some [summation identities](https://en.wikipedia.org/wiki/Summation) (also see [problem 1](https://projecteuler.net/problem=1)), this simplifies to\n",
+    "$$R(m, n) = \\frac{mn(m+1)(n+1)}{4}$$\n",
+    "Interestingly, this formula is really just the product of $m$th and $n$th [triangular numbers](https://en.wikipedia.org/wiki/Triangular_number)."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "2c2038a4",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def R(m, n):\n",
+    "    return m * n * (m + 1) * (n + 1) // 4"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "f29eaf82",
+   "metadata": {},
+   "source": [
+    "Figuring out the formula's the hard part - now we can just iterate over grid sizes to calculate how many rectangles they have."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "3409e88c",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "totals = dict()\n",
+    "\n",
+    "for x in range(1, 2001):\n",
+    "    for y in range(1, 2001):\n",
+    "        r = R(x, y)\n",
+    "        totals[(x,y)] = r\n",
+    "        if r > 2000000:\n",
+    "            break"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "d9ce5480",
+   "metadata": {},
+   "source": [
+    "We'll figure out which grid is the closest to containing 2000000 rectangles."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "id": "dbc72aaa",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "(36, 77)"
+      ]
+     },
+     "execution_count": 3,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "m, n = min(totals, key=lambda x: abs(totals[x] - 2000000))\n",
+    "m, n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "e69c5c48",
+   "metadata": {},
+   "source": [
+    "The area of that grid is our answer."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 4,
+   "id": "f230a0d0",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "2772"
+      ]
+     },
+     "execution_count": 4,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "m * n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "d30d6d4f",
+   "metadata": {},
+   "source": [
+    "## Relevant sequences\n",
+    "* Number of rectangles in an $m \\times n$ grid: [A098358](https://oeis.org/A098358)"
+   ]
+  }
+ ],
+ "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
+}