{
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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Reflections on symmetry\n",
    "\n",
    "- https://adventofcode.com/2023/day/13\n",
    "\n",
    "This is a very easy problem to solve with numpy arrays; we can use `numpy.split()` to divide the image in two parts, and `numpy.fliy()` to mirror one of them. I do so for the first split half, as that's easier to then slice (from zero to the smaller width).\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
   "outputs": [],
   "source": [
    "import typing as t\n",
    "from dataclasses import dataclass\n",
    "\n",
    "import numpy as np\n",
    "from numpy.typing import NDArray\n",
    "\n",
    "\n",
    "@dataclass\n",
    "class ValleyPattern:\n",
    "    pattern: NDArray[np.bool_]\n",
    "\n",
    "    @classmethod\n",
    "    def from_text(cls, text: str) -> t.Self:\n",
    "        return cls(np.array([[c == \"#\" for c in line] for line in text.splitlines()]))\n",
    "\n",
    "    def _reflection(self, axis: t.Literal[0, 1], difference: int = 0) -> int:\n",
    "        patt, size = self.pattern, self.pattern.shape[axis]\n",
    "        for i in range(1, size):\n",
    "            rev, fwd = np.split(patt, [i], axis=axis)\n",
    "            idx = range(min(i, size - i))\n",
    "            if (\n",
    "                np.sum(\n",
    "                    fwd.take(idx, axis=axis)\n",
    "                    ^ np.flip(rev, axis=axis).take(idx, axis=axis)\n",
    "                )\n",
    "                == difference\n",
    "            ):\n",
    "                return i\n",
    "        return 0\n",
    "\n",
    "    def summary(self, difference: int = 0) -> int:\n",
    "        return (\n",
    "            self._reflection(1, difference)\n",
    "            or 100 * self._reflection(0, difference)\n",
    "            or 0\n",
    "        )\n",
    "\n",
    "\n",
    "test_patterns = \"\"\"\\\n",
    "#.##..##.\n",
    "..#.##.#.\n",
    "##......#\n",
    "##......#\n",
    "..#.##.#.\n",
    "..##..##.\n",
    "#.#.##.#.\n",
    "\n",
    "#...##..#\n",
    "#....#..#\n",
    "..##..###\n",
    "#####.##.\n",
    "#####.##.\n",
    "..##..###\n",
    "#....#..#\n",
    "\"\"\".split(\n",
    "    \"\\n\\n\"\n",
    ")\n",
    "test_valley_patterns = [ValleyPattern.from_text(blk) for blk in test_patterns]\n",
    "assert sum(p.summary() for p in test_valley_patterns) == 405"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Part 1: 33520\n"
     ]
    }
   ],
   "source": [
    "import aocd\n",
    "\n",
    "pattern_blocks = aocd.get_data(day=13, year=2023).split(\"\\n\\n\")\n",
    "valley_patterns = [ValleyPattern.from_text(blk) for blk in pattern_blocks]\n",
    "print(\"Part 1:\", sum(p.summary() for p in valley_patterns))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 1 spot makes all the difference\n",
    "\n",
    "Insteod of looking for a simple equality, you can find the reflection with a single smudge by looking at the _difference_ between the two parts being equal to 1. Simple!\n",
    "\n",
    "I refactored part 1 to sum the XOR (boolean difference) of the two parts, and for part 2 the difference needs to be 1 instead of 0.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "assert sum(p.summary(1) for p in test_valley_patterns) == 400"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Part 2: 34824\n"
     ]
    }
   ],
   "source": [
    "print(\"Part 2:\", sum(p.summary(1) for p in valley_patterns))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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