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