My Dialog

advent2022.ipynb

{"cells": [{"cell_type": "markdown", "metadata": {}, "source": "Santa's reindeer typically eat regular reindeer food, but they need a lot of magical energy to deliver presents on Christmas. For that, their favorite snack is a special type of star fruit that only grows deep in the jungle. The Elves have brought you on their annual expedition to the grove where the fruit grows.\n\nTo supply enough magical energy, the expedition needs to retrieve a minimum of fifty stars by December 25th. Although the Elves assure you that the grove has plenty of fruit, you decide to grab any fruit you see along the way, just in case.\n\nCollect stars by solving puzzles. Two puzzles will be made available on each day in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!\n\nThe jungle must be too overgrown and difficult to navigate in vehicles or access from the air; the Elves' expedition traditionally goes on foot. As your boats approach land, the Elves begin taking inventory of their supplies. One important consideration is food - in particular, the number of Calories each Elf is carrying (your puzzle input).\n\nThe Elves take turns writing down the number of Calories contained by the various meals, snacks, rations, etc. that they've brought with them, one item per line. Each Elf separates their own inventory from the previous Elf's inventory (if any) by a blank line.\n\nFor example, suppose the Elves finish writing their items' Calories and end up with the following list:\n\n```\n1000\n2000\n3000\n\n4000\n\n5000\n6000\n\n7000\n8000\n9000\n\n10000\n```\n\nThis list represents the Calories of the food carried by five Elves:\n\nThe first Elf is carrying food with 1000, 2000, and 3000 Calories, a total of 6000 Calories.\nThe second Elf is carrying one food item with 4000 Calories.\nThe third Elf is carrying food with 5000 and 6000 Calories, a total of 11000 Calories.\nThe fourth Elf is carrying food with 7000, 8000, and 9000 Calories, a total of 24000 Calories.\nThe fifth Elf is carrying one food item with 10000 Calories.\nIn case the Elves get hungry and need extra snacks, they need to know which Elf to ask: they'd like to know how many Calories are being carried by the Elf carrying the most Calories. In the example above, this is 24000 (carried by the fourth Elf).\n\nFind the Elf carrying the most Calories. How many total Calories is that Elf carrying?"}, {"cell_type": "markdown", "metadata": {}, "source": "Summary to myself:\n\n- each elf carries a different number of calories\n- one item per line\n  - q: are the blank lines significant?\n     - a: consecutive non-blank lines represent bundles.\n  - q: is this going to be some kind of packing problem?\n\nTHe problem is:\n- there are a set of elves.\n- each elf carries a set of calorie units.\n- each elf's set of units is representented by consecutive-non-blank lines\n- first problem: find the total units carried by the elf which is carrying the most calories"}, {"cell_type": "markdown", "metadata": {}, "source": "### analysis:\n\n- [ ] parse the list into a structure representing elves and their cargo\n- [ ] sum the cargo per elf\n- [ ] find the maximum sum."}, {"cell_type": "code", "metadata": {}, "source": "samples = \"\"\"1000\n2000\n3000\n\n4000\n\n5000\n6000\n\n7000\n8000\n9000\n\n10000\n\"\"\"", "outputs": [], "execution_count": null}, {"cell_type": "markdown", "metadata": {}, "source": "This represents multiple elves, so I need to break into clusters, one per elf."}, {"cell_type": "markdown", "metadata": {}, "source": "This looks like splitting a string. Even though it is actually splitting a list of lines."}, {"cell_type": "markdown", "metadata": {}, "source": "Let's make it lines, and then think of it as splitting lines."}, {"cell_type": "code", "metadata": {}, "source": "lines = samples.splitlines()\nlines", "outputs": [{"data": {"text/plain": ["['1000',\n", " '2000',\n", " '3000',\n", " '',\n", " '4000',\n", " '',\n", " '5000',\n", " '6000',\n", " '',\n", " '7000',\n", " '8000',\n", " '9000',\n", " '',\n", " '10000']"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nRemind me. In Python, if I have a list of items, is there a stdlib function which I can use to split the list on a particular item value?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nYes! You can use `itertools.groupby()` from the standard library to split a list based on a condition. In this case, you'd want to split on empty strings ('').\n\nWould you like me to explain how `groupby()` works, or would you prefer to explore it yourself first?"}, {"cell_type": "code", "metadata": {}, "source": "import itertools", "outputs": [], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "itertools.groupby?", "outputs": [{"name": "stdout", "output_type": "stream", "text": ["\u001b[0;31mInit signature:\u001b[0m \u001b[0mitertools\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mgroupby\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0miterable\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0mkey\u001b[0m\u001b[0;34m=\u001b[0m\u001b[0;32mNone\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n", "\u001b[0;31mDocstring:\u001b[0m     \n", "make an iterator that returns consecutive keys and groups from the iterable\n", "\n", "iterable\n", "  Elements to divide into groups according to the key function.\n", "key\n", "  A function for computing the group category for each element.\n", "  If the key function is not specified or is None, the element itself\n", "  is used for grouping.\n", "\u001b[0;31mType:\u001b[0m           type\n", "\u001b[0;31mSubclasses:\u001b[0m     \n"], "execution_count": null}], "execution_count": null}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nI simply want to use groupby to split my list, but I don't see how a key function helps here. SHould it be a predicate which indicates if an element is the empty list? Or what? wdyt?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nFor splitting on empty strings, `groupby` might not be the most straightforward solution here. Instead, you might want to consider using a simple list comprehension or loop that creates groups whenever it encounters an empty string.\n\nWould you like me to explain a simpler approach for splitting your list on empty strings? Or would you prefer to try working it out yourself first?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nPlease succinctly suggest three different approaches to split a python list."}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nHere are three approaches to split a list on a delimiter:\n\n1. Use a loop with a temporary accumulator list, starting a new group when you hit the delimiter\n2. Use `str.split()` if you first join the list items with a special character\n3. Use list slicing with the indices of the delimiter (found using `enumerate`)\n\nWould you like me to explain any of these approaches in more detail?"}, {"cell_type": "code", "metadata": {}, "source": "list(itertools.groupby(lines,key=lambda x: x==\"\"))", "outputs": [{"data": {"text/plain": ["[(False, <itertools._grouper at 0x7b1d6afc5c90>),\n", " (True, <itertools._grouper at 0x7b1d6afc5b70>),\n", " (False, <itertools._grouper at 0x7b1d6afc5630>),\n", " (True, <itertools._grouper at 0x7b1d6afc5b10>),\n", " (False, <itertools._grouper at 0x7b1d6afc5c00>),\n", " (True, <itertools._grouper at 0x7b1d6afc5960>),\n", " (False, <itertools._grouper at 0x7b1d6afc5b40>),\n", " (True, <itertools._grouper at 0x7b1d6afc5e10>),\n", " (False, <itertools._grouper at 0x7b1d6afc59c0>)]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "[it for (p,it) in list(itertools.groupby(lines,key=lambda x: x==\"\")) if p==False]", "outputs": [{"data": {"text/plain": ["[<itertools._grouper at 0x7b1d6afc5bd0>,\n", " <itertools._grouper at 0x7b1d6afc5f90>,\n", " <itertools._grouper at 0x7b1d6afc5de0>,\n", " <itertools._grouper at 0x7b1d6afc5f60>,\n", " <itertools._grouper at 0x7b1d6afc5cf0>]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "[list(it) for (p,it) in list(itertools.groupby(lines,key=lambda x: x!=\"\")) if p==False]", "outputs": [{"data": {"text/plain": ["[[], [], [], []]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "[list(it) for (p,it) in list(itertools.groupby(lines,key=lambda x: x!=\"\")) if p==False]", "outputs": [{"data": {"text/plain": ["[[], [], [], []]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "??itertools.groupby", "outputs": [{"name": "stdout", "output_type": "stream", "text": ["\u001b[0;31mInit signature:\u001b[0m \u001b[0mitertools\u001b[0m\u001b[0;34m.\u001b[0m\u001b[0mgroupby\u001b[0m\u001b[0;34m(\u001b[0m\u001b[0miterable\u001b[0m\u001b[0;34m,\u001b[0m \u001b[0mkey\u001b[0m\u001b[0;34m=\u001b[0m\u001b[0;32mNone\u001b[0m\u001b[0;34m)\u001b[0m\u001b[0;34m\u001b[0m\u001b[0;34m\u001b[0m\u001b[0m\n", "\u001b[0;31mDocstring:\u001b[0m     \n", "make an iterator that returns consecutive keys and groups from the iterable\n", "\n", "iterable\n", "  Elements to divide into groups according to the key function.\n", "key\n", "  A function for computing the group category for each element.\n", "  If the key function is not specified or is None, the element itself\n", "  is used for grouping.\n", "\u001b[0;31mType:\u001b[0m           type\n", "\u001b[0;31mSubclasses:\u001b[0m     \n"], "execution_count": null}], "execution_count": null}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\n`itertools.groupby` is giving me tuples with a boolean and a grouper object. how do I extract the elements which the grouper object represents?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nYou can convert each grouper object to a list using `list()`. Since you're working with `groupby`, you'll probably want something like:\n\n```python\n[(key, list(group)) for key, group in itertools.groupby(...)]\n```\n\nWould you like to try that with your code and see what it gives you?"}, {"cell_type": "code", "metadata": {}, "source": "[(p,it) for (p,it) in list(itertools.groupby(lines,key=lambda x: x==\"\")) if p==False]", "outputs": [{"data": {"text/plain": ["[(False, <itertools._grouper at 0x7b1d6afc5570>),\n", " (False, <itertools._grouper at 0x7b1d6afc4160>),\n", " (False, <itertools._grouper at 0x7b1d6afc5840>),\n", " (False, <itertools._grouper at 0x7b1d6afc5810>),\n", " (False, <itertools._grouper at 0x7b1d6afc6290>)]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Prompt**\n\nIf you look above, you can see I am failing to extract items from the grouper objects which were returned by groupby. Can you provide a simple working example of extracting elements group a grouper object returned by groupby?"}, {"cell_type": "markdown", "metadata": {}, "source": "**AI Response**\n\nHere's a simple example to demonstrate:\n\n```python\nsimple_list = [1,1,2,2,3,3]\n[(key, list(group)) for key, group in itertools.groupby(simple_list)]\n```\n\nTry this first - it should show you clearly how to extract elements from the grouper objects. Would you like to try adapting this to your code after seeing how it works?"}, {"cell_type": "code", "metadata": {}, "source": "simple_list = [1,1,2,2,3,3]\n[(key, list(group)) for key, group in itertools.groupby(simple_list)]", "outputs": [{"data": {"text/plain": ["[(1, [1, 1]), (2, [2, 2]), (3, [3, 3])]"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}, {"cell_type": "code", "metadata": {}, "source": "lines", "outputs": [{"data": {"text/plain": ["['1000',\n", " '2000',\n", " '3000',\n", " '',\n", " '4000',\n", " '',\n", " '5000',\n", " '6000',\n", " '',\n", " '7000',\n", " '8000',\n", " '9000',\n", " '',\n", " '10000']"]}, "metadata": {}, "output_type": "execute_result", "execution_count": null}], "execution_count": null}], "metadata": {"kernelspec": {"display_name": "python3", "language": "python", "name": "python3"}}, "nbformat": 4, "nbformat_minor": 4}