Day 8: use cycle & add a section about different ways to choose direc…

…tion

src/day_8.ipynb

@@ -86,14 +86,14 @@
  "source": [
  "To solve this puzzle, I start with node `AAA` and loop until I have stepped into `ZZZ`.\n",
  "\n",
- "To keep `instructions` running even if it runs out, I use _math_. Using modulo operator (`%`) with a length of a list is a good way to keep the index bound between 0 and length of list - 1.\n",
+ "To keep `instructions` running even if it runs out, I use [itertools.cycle](https://docs.python.org/3/library/itertools.html#itertools.cycle).\n",
  "\n",
  "I them pattern match against the instructions and either take the first or the second of the next values. In part 2 I realized that pattern matching was bit too much here since we just have two values but I wanted to leave it here so you can compare it with the if/else choice in part 2."
  ]
  },
  {
  "cell_type": "code",
- "execution_count": 2,
+ "execution_count": 9,
  "id": "fd4c01ea",
  "metadata": {},
  "outputs": [
@@ -106,13 +106,14 @@
  }
  ],
  "source": [
+ "from itertools import cycle\n",
+ "\n",
  "def solve_part_1(instructions, the_map):\n",
  " current = 'AAA'\n",
  " end = 'ZZZ'\n",
  "\n",
  " steps = 0\n",
- " while True:\n",
- " instruction = instructions[steps % len(instructions)]\n",
+ " for instruction in cycle(instructions):\n",
  " steps += 1\n",
  " match instruction:\n",
  " case 'L':\n",
@@ -191,7 +192,7 @@
  },
  {
  "cell_type": "code",
- "execution_count": 3,
+ "execution_count": 12,
  "id": "0d499914",
  "metadata": {},
  "outputs": [
@@ -205,18 +206,18 @@
  ],
  "source": [
  "import math\n",
+ "from itertools import cycle\n",
  "\n",
  "\n",
- "def find_loop(node, the_map):\n",
+ "def find_loop(node, the_map, instructions):\n",
  " sequence = []\n",
  " step = 0\n",
- " while True:\n",
- " mod_step = step % len(instructions)\n",
- " \n",
- " instruction = instructions[mod_step]\n",
+ " instructions_length = len(instructions)\n",
+ " for instruction in cycle(instructions):\n",
  " idx = 0 if instruction == 'L' else 1\n",
  " node = the_map[node][idx]\n",
  " \n",
+ " mod_step = step % instructions_length\n",
  " if (mod_step, node) in sequence:\n",
  " return step, sequence\n",
  " \n",
@@ -232,7 +233,7 @@
  "start_nodes = [key for key in the_map if key.endswith('A')]\n",
  "last_z_in_each_loop = []\n",
  "for node in start_nodes:\n",
- " loop_step, seq = find_loop(node, the_map)\n",
+ " loop_step, seq = find_loop(node, the_map, instructions)\n",
  " last_z_offset = find_last_z_node(seq)\n",
  " last_z_in_each_loop.append(loop_step - last_z_offset)\n",
  "\n",
@@ -241,6 +242,53 @@
  "assert part_2 == 12357789728873"
  ]
  },
+ {
+ "cell_type": "markdown",
+ "id": "9397c8c5",
+ "metadata": {},
+ "source": [
+ "## Few ways to get the next step\n",
+ "\n",
+ "There are multiple different ways to map our next step based on an instruction and a map.\n",
+ "\n",
+ "In each of these, the `node` refers to current position and `the_map` is a dictionary that looks like `{ \"key\": (\"left_node\", \"right_node\") }`.\n",
+ "\n",
+ "Which one to use depends on the case and how many different options there are. The if/else in either form is great if there's only 2 options. With more options, I'd use either a dictionary or pattern matching approach. Pattern matching gives the best \"bang for buck\" as it improves readability and enables doing more than just simple value mapping like guards or more complex patterns.\n",
+ "\n",
+ "### Using if/else:\n",
+ "\n",
+ "```python\n",
+ "if instruction == 'L':\n",
+ " node = the_map[node][0]\n",
+ "else:\n",
+ " node = the_map[node][1]\n",
+ "```\n",
+ "\n",
+ "or with ternary operation:\n",
+ "\n",
+ "```python\n",
+ "index = 0 if instruction == 'L' else 1\n",
+ "node = the_map[node][idx]\n",
+ "```\n",
+ "\n",
+ "### Using pattern matching:\n",
+ "\n",
+ "```python\n",
+ "match instruction:\n",
+ " case 'L':\n",
+ " node = the_map[node][0]\n",
+ " case 'R':\n",
+ " node = the_map[node][1]\n",
+ "```\n",
+ "\n",
+ "### Using a dictionary\n",
+ "\n",
+ "```python\n",
+ "left_right = { 'left': 0, 'right': 1 }\n",
+ "node = the_map[node][left_right[instruction]]\n",
+ "```"
+ ]
+ },
  {
  "cell_type": "markdown",
  "id": "b5af38ca",