{ "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "# Euler Problem 24 \n", "\n", "A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:\n", "\n", "012 021 102 120 201 210\n", "\n", "What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "I tried to solve this by thinking and failed to be hones. So we implement a generator and get the millionth element." ] }, { "cell_type": "code", "execution_count": 19, "metadata": { "collapsed": false }, "outputs": [], "source": [ "from collections import deque\n", "\n", "def permutation_generator(xs):\n", " if not xs:\n", " yield deque()\n", " r = []\n", " for i in range(len(xs)):\n", " x = xs[i]\n", " ys = permutation_generator(xs[:i] + xs[i+1:])\n", " for y in ys:\n", " y.appendleft(x)\n", " yield y\n", " raise StopIteration\n", "\n", "assert(list(map(lambda s: \"\".join(s), permutation_generator(\"012\"))) == ['012', '021', '102', '120', '201', '210'])" ] }, { "cell_type": "markdown", "metadata": { "collapsed": true }, "source": [ "All right, now we reuse the function to get the nth element from problem 7 and give ourselves a solution." ] }, { "cell_type": "code", "execution_count": 26, "metadata": { "collapsed": false }, "outputs": [ { "name": "stdout", "output_type": "stream", "text": [ "2783915460\n" ] } ], "source": [ "def get_nth(generator, n):\n", " import itertools\n", " return next(itertools.islice(generator, n - 1, n))\n", "\n", "ps = permutation_generator(\"0123456789\")\n", "s = int(\"\".join(get_nth(ps, 1000000)))\n", "assert(s == 2783915460)\n", "print(s)" ] } ], "metadata": { "completion_date": "Thu, 5 Nov 2015, 16:04", "kernelspec": { "display_name": "Python 3", "language": "python", "name": "python3" }, "language_info": { "codemirror_mode": { "name": "ipython", "version": 3 }, "file_extension": ".py", "mimetype": "text/x-python", "name": "python", "nbconvert_exporter": "python", "pygments_lexer": "ipython3", "version": "3.5.4" }, "tags": [ "permutation" ] }, "nbformat": 4, "nbformat_minor": 0 }