From ecf37375b1ed59886d69ebabfbe06b7f777bf1a7 Mon Sep 17 00:00:00 2001 From: filifa Date: Sun, 25 May 2025 20:47:21 -0400 Subject: [PATCH] add problem 78 --- notebooks/problem0078.ipynb | 117 ++++++++++++++++++++++++++++++++++++ 1 file changed, 117 insertions(+) create mode 100644 notebooks/problem0078.ipynb diff --git a/notebooks/problem0078.ipynb b/notebooks/problem0078.ipynb new file mode 100644 index 0000000..253d957 --- /dev/null +++ b/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 +}