{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "d84ea0fd",
   "metadata": {},
   "source": [
    "# [Passcode Derivation](https://projecteuler.net/problem=79)\n",
    "\n",
    "Gotta read the data first."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "56df44fb",
   "metadata": {},
   "outputs": [],
   "source": [
    "with open(\"txt/0079_keylog.txt\") as f:\n",
    "    keylog = {tuple(int(c) for c in line.strip()) for line in f}"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "043f31a9",
   "metadata": {},
   "source": [
    "If we assume that no characters are repeated in the passcode, then each entry in the keylog gives a [partial ordering](https://en.wikipedia.org/wiki/Partially_ordered_set) of the characters in the passcode. In math jargon, we are then looking for a [linear extension](https://en.wikipedia.org/wiki/Linear_extension) of the partial ordering. This can be found with a [topological sorting](https://en.wikipedia.org/wiki/Topological_sorting) algorithm.\n",
    "\n",
    "(If characters *are* repeated, we would instead have a [preordering](https://en.wikipedia.org/wiki/Preorder), and a topological sort wouldn't work. Fortunately for us, our assumption turns out to be correct.)\n",
    "\n",
    "Conveniently, and somewhat interestingly, Python has [graphlib](https://docs.python.org/3/library/graphlib.html) in its standard library, which (as of the time of writing) *only* implements topological sorting - but that's all we need for this problem, so I guess don't look a gift horse in the mouth."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "4e742b06",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(7, 3, 1, 6, 2, 8, 9, 0)"
      ]
     },
     "execution_count": 2,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "from graphlib import TopologicalSorter\n",
    "\n",
    "ts = TopologicalSorter()\n",
    "for attempt in keylog:\n",
    "    for (idx, d) in enumerate(attempt):\n",
    "        predecessors = attempt[:idx]\n",
    "        ts.add(d, *predecessors)\n",
    "\n",
    "tuple(ts.static_order())"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "242e35e6",
   "metadata": {},
   "source": [
    "#### Copyright (C) 2025 filifa\n",
    "\n",
    "This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
   ]
  }
 ],
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   "codemirror_mode": {
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