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mjpietersmjpieters
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Clean up 2020
- Reformatted using ruff - Warnings addressed No outputs have changed (but matplotlib plots are a little larger, timeit results are faster)
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‎2020/Day 01.ipynb

+46-36
Original file line numberDiff line numberDiff line change
@@ -1,38 +1,17 @@
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{
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"metadata": {
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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": "3.9.0-final"
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},
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"orig_nbformat": 2,
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"kernelspec": {
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"name": "python3",
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"display_name": "Python 3"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2,
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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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"# Welcome to AoC 2020, time for a holiday!\n",
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"\n",
28-
"* https://adventofcode.com/2020/day/1\n",
9+
"- https://adventofcode.com/2020/day/1\n",
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"\n",
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"Another year, another set of puzzles! This time, we are 'not doing christmas'; Eric Wastl has torn up the [AoC Bingo card](https://www.reddit.com/r/adventofcode/comments/k3q7tr/my_advent_of_code_2020_bingo_card_fun_little_side/) and decided to send us to a tropical island instead. :-P\n",
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"\n",
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"As always, we start with a warm-up. Rather than iterate over all combinations ($O(n^2)$), put the numbers in a set, loop over the set for the first number, and see if `2020 - first` is a member of the set. That gives us a $O(n)$ runtime!"
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],
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"cell_type": "markdown",
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"metadata": {}
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"As always, we start with a warm-up. Rather than iterate over all combinations ($O(n^2)$), put the numbers in a set, loop over the set for the first number, and see if `2020 - first` is a member of the set. That gives us a $O(n)$ runtime!\n"
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]
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},
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{
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"cell_type": "code",
@@ -41,6 +20,7 @@
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"outputs": [],
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"source": [
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"import aocd\n",
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"\n",
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"coins = set(map(int, aocd.get_data(day=1, year=2020).splitlines()))"
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]
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},
@@ -52,25 +32,33 @@
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"source": [
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"from operator import mul\n",
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"\n",
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"\n",
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"def find_sum(coins, target=2020):\n",
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" for first in coins:\n",
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" second = target - first\n",
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" if second in coins:\n",
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" return first, second\n",
60-
" \n",
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"\n",
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" raise ValueError(\"Not solvable\")\n",
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"\n",
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"\n",
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"def sum_coins(coins):\n",
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" return mul(*find_sum(coins))\n",
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"\n",
66-
"test = set(map(int, '''\\\n",
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"\n",
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"test = set(\n",
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" map(\n",
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" int,\n",
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" \"\"\"\\\n",
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"1721\n",
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"979\n",
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"366\n",
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"299\n",
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"675\n",
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"1456\n",
73-
"'''.splitlines()))\n",
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"\"\"\".splitlines(),\n",
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" )\n",
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")\n",
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"assert sum_coins(test) == 514579"
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]
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},
@@ -80,8 +68,8 @@
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Part 1: 73371\n"
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]
@@ -92,13 +80,13 @@
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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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"# Part 2\n",
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"\n",
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"To find the third coin, just pop one of the values of the set of coins, run the first puzzle solution to see if there is a combination that sums to `2020 - selected`, and continue until we found a combination. That's an $O(N^2)$ solution:\n"
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],
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"cell_type": "markdown",
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"metadata": {}
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]
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},
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{
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"cell_type": "code",
@@ -115,6 +103,7 @@
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" except ValueError:\n",
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" continue\n",
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"\n",
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"\n",
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"assert sum_three(test) == 241861950"
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]
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},
@@ -124,8 +113,8 @@
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Part 2: 127642310\n"
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]
@@ -135,5 +124,26 @@
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"print(\"Part 2:\", sum_three(coins))"
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]
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}
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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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": "3.12.0"
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},
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"orig_nbformat": 2
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

‎2020/Day 02.ipynb

+50-42
Original file line numberDiff line numberDiff line change
@@ -1,36 +1,15 @@
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{
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"metadata": {
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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": "3.9.0-final"
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},
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"orig_nbformat": 2,
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"kernelspec": {
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"name": "python3",
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"display_name": "Python 3"
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}
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},
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"nbformat": 4,
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"nbformat_minor": 2,
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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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"# Counting letters\n",
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"\n",
28-
"* https://adventofcode.com/2020/day/2\n",
9+
"- https://adventofcode.com/2020/day/2\n",
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"\n",
30-
"I like to use a dataclass for parsing tasks like these. A single regex to read out each line, and methods on the class to implement the password rule checks."
31-
],
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"cell_type": "markdown",
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"metadata": {}
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"I like to use a dataclass for parsing tasks like these. A single regex to read out each line, and methods on the class to implement the password rule checks.\n"
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]
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},
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{
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"cell_type": "code",
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"import re\n",
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"from dataclasses import dataclass\n",
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"\n",
44-
"_line = re.compile(r'^(?P<min_>\\d+)-(?P<max_>\\d+) (?P<letter>[a-z]):\\s*(?P<password>[a-z]+)$')\n",
23+
"_line = re.compile(\n",
24+
" r\"^(?P<min_>\\d+)-(?P<max_>\\d+) (?P<letter>[a-z]):\\s*(?P<password>[a-z]+)$\"\n",
25+
")\n",
26+
"\n",
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"\n",
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"@dataclass\n",
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"class PWRule:\n",
@@ -51,27 +33,31 @@
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" password: str\n",
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"\n",
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" @classmethod\n",
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" def from_line(cls, line: str) -> 'PWRule':\n",
36+
" def from_line(cls, line: str) -> \"PWRule\":\n",
5537
" match = _line.search(line)\n",
56-
" min_, max_ = int(match['min_']), int(match['max_'])\n",
57-
" return cls(min_, max_, match['letter'], match['password'])\n",
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" min_, max_ = int(match[\"min_\"]), int(match[\"max_\"])\n",
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" return cls(min_, max_, match[\"letter\"], match[\"password\"])\n",
5840
"\n",
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" def is_valid(self) -> bool:\n",
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" return self.min_ <= self.password.count(self.letter) <= self.max_\n",
6143
"\n",
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" def is_valid_toboggan_policy(self) -> bool:\n",
63-
" return (self.password[self.min_ - 1], self.password[self.max_ - 1]).count(self.letter) == 1\n",
45+
" return (self.password[self.min_ - 1], self.password[self.max_ - 1]).count(\n",
46+
" self.letter\n",
47+
" ) == 1\n",
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"\n",
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"\n",
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"def read_passwords(lines):\n",
67-
" return [PWRule.from_line(l) for l in lines]\n",
51+
" return [PWRule.from_line(line) for line in lines]\n",
6852
"\n",
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"\n",
70-
"test = read_passwords('''\\\n",
54+
"test = read_passwords(\n",
55+
" \"\"\"\\\n",
7156
"1-3 a: abcde\n",
7257
"1-3 b: cdefg\n",
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"2-9 c: ccccccccc\n",
74-
"'''.splitlines())\n",
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"\"\"\".splitlines()\n",
60+
")\n",
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"\n",
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"assert sum(pwr.is_valid() for pwr in test) == 2\n",
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"assert sum(pwr.is_valid_toboggan_policy() for pwr in test) == 1"
@@ -83,8 +69,9 @@
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"metadata": {},
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"outputs": [],
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"source": [
86-
" import aocd\n",
87-
" rules = read_passwords(aocd.get_data(day=2, year=2020).splitlines())"
72+
"import aocd\n",
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"\n",
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"rules = read_passwords(aocd.get_data(day=2, year=2020).splitlines())"
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]
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},
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{
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Part 1: 591\n"
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]
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}
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],
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"source": [
104-
"print('Part 1:', sum(pwr.is_valid() for pwr in rules))"
91+
"print(\"Part 1:\", sum(pwr.is_valid() for pwr in rules))"
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]
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},
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{
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"metadata": {},
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"outputs": [
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{
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"output_type": "stream",
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"Part 2: 335\n"
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]
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}
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],
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"source": [
121-
"print('Part 2:', sum(pwr.is_valid_toboggan_policy() for pwr in rules))"
108+
"print(\"Part 2:\", sum(pwr.is_valid_toboggan_policy() for pwr in rules))"
122109
]
123110
}
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3",
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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": "3.12.0"
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},
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"orig_nbformat": 2
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},
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"nbformat": 4,
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"nbformat_minor": 2
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}

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