add problem 78

notebooks/problem0078.ipynb

@@ -0,0 +1,117 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "683859dd",
+   "metadata": {},
+   "source": [
+    "# [Coin Partitions](https://projecteuler.net/problem=78)\n",
+    "\n",
+    "SageMath once again makes this pretty trivial."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "48a703c0",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "55374"
+      ]
+     },
+     "execution_count": 1,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "from itertools import count\n",
+    "\n",
+    "for n in count(1):\n",
+    "    if number_of_partitions(n) % 1000000 == 0:\n",
+    "        break\n",
+    "\n",
+    "n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "ec850c78",
+   "metadata": {},
+   "source": [
+    "Theoretically, we could create a generating function like we did in [problem 31](https://projecteuler.net/problem=76) or [problem 76](https://projecteuler.net/problem=76) to solve this, and could even use $\\mathbb{Z}_{1000000}$ as our base ring to handle the modulus automagically, but since the answer is pretty large, it would be impractical to construct the function with the required precision.\n",
+    "\n",
+    "Instead, if you'd like to implement the [partition function](https://en.wikipedia.org/wiki/Partition_function_(number_theory)) yourself, Euler's [pentagonal number theorem](https://en.wikipedia.org/wiki/Pentagonal_number_theorem) leads to a useful recurrence equation:\n",
+    "$$p(n) = p(n - 1) + p(n - 2) - p(n - 5) - p(n - 7) + p(n - 12) + p(n - 15) - p(n - 22) - p(n - 26) + \\cdots$$\n",
+    "Here, the numbers are the [generalized pentagonal numbers](https://en.wikipedia.org/wiki/Pentagonal_number). As base cases, $p(0) = 1$ and $p(n) = 0$ for negative $n$, so the infinite series eventually converges.\n",
+    "\n",
+    "We can directly translate this equation into an implementation."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "id": "c8dd39e8",
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "from functools import cache\n",
+    "\n",
+    "@cache\n",
+    "def p(n, modulus):\n",
+    "    if n < 0:\n",
+    "        return 0\n",
+    "    \n",
+    "    if n == 0:\n",
+    "        return 1\n",
+    "    \n",
+    "    total = 0\n",
+    "    limit = floor((1 + sqrt(1+24*n)) / 6)\n",
+    "    \n",
+    "    # we reverse the range so smaller p(n) values are calculated first and cached\n",
+    "    # this helps avoid hitting Python's maximum recursion depth\n",
+    "    for k in reversed(range(1, limit + 1)):\n",
+    "        total += (-1)^(k+1) * (p(n - polygonal_number(5, k), modulus) + p(n - polygonal_number(5, -k), modulus))\n",
+    "        total %= modulus\n",
+    "    \n",
+    "    return total"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "df91171b",
+   "metadata": {},
+   "source": [
+    "Note that SageMath uses libraries like [FLINT](https://flintlib.org/doc/partitions.html) with faster - but much more technical - methods, like the Hardy-Ramanujan-Rademacher formula. This implementation is much easier to both understand and write, but is less performant.\n",
+    "\n",
+    "## Relevant sequences\n",
+    "* Partition numbers: [A000041](https://oeis.org/A000041)\n",
+    "* Generalized pentagonal numbers: [A001318](https://oeis.org/A001318)"
+   ]
+  }
+ ],
+ "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
+}