add problem 53

notebooks/problem0053.ipynb

@@ -0,0 +1,78 @@
+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "id": "59fa6654",
+   "metadata": {},
+   "source": [
+    "# [Combinatoric Selections](https://projecteuler.net/problem=53)\n",
+    "\n",
+    "Python/SageMath's [unlimited precision integers](https://docs.python.org/3/library/stdtypes.html#numeric-types-int-float-complex) make this trivial."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "id": "abf93cda",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "4075"
+      ]
+     },
+     "execution_count": 1,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "binoms = (binomial(n, r) for n in range(1, 101) for r in range(0, n + 1))\n",
+    "len([b for b in binoms if b > 1000000])"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "6d7f4409",
+   "metadata": {},
+   "source": [
+    "If it weren't for unlimited precision, we would have a problem, since a lot of these [binomial coefficients](https://en.wikipedia.org/wiki/Binomial_coefficient) overflow a 64 bit integer - the largest one we encounter here is $\\binom{100}{50} = 100891344545564193334812497256$.\n",
+    "\n",
+    "However, even in that case there's a pretty simple workaround: finding $n$ and $r$ such that\n",
+    "$$\\binom{n}{r} > 1000000$$\n",
+    "is the same as finding $n$ and $r$ such that\n",
+    "$$\\log{\\binom{n}{r}} > \\log{1000000}$$\n",
+    "By applying the definition of the binomial coefficients and [logarithmic identities](https://en.wikipedia.org/wiki/List_of_logarithmic_identities), this turns to\n",
+    "$$\\log{n!} - \\log{r!} - \\log{(n-r)!} > \\log{1000000}$$\n",
+    "$\\log{100!} \\approx 363.739$, nowhere close to overflowing a float.\n",
+    "\n",
+    "$\\log{n!}$ can be computed in a number of ways. You can use the [log-gamma function](https://en.wikipedia.org/wiki/Gamma_function), or you can implement a function yourself, either by applying logarithmic identities again:\n",
+    "$$\\log{n!} = \\log{1} + \\log{2} + \\log{3} \\cdots + \\log{n}$$\n",
+    "or by using [Stirling's approximation](https://en.wikipedia.org/wiki/Stirling%27s_approximation):\n",
+    "$$\\log{n!} \\approx \\left(n + \\frac{1}{2}\\right)\\log{n} - n + \\frac{1}{2}\\log{2\\pi}$$"
+   ]
+  }
+ ],
+ "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
+}