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<!DOCTYPE html>
<html lang="en-us">
<head>
<meta charset="utf-8"/>
<title>Day 9 - Advent of Code 2020</title>
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</head><!--
Oh, hello! Funny seeing you here.
I appreciate your enthusiasm, but you aren't going to find much down here.
There certainly aren't clues to any of the puzzles. The best surprises don't
even appear in the source until you unlock them for real.
Please be careful with automated requests; I'm not a massive company, and I can
only take so much traffic. Please be considerate so that everyone gets to play.
If you're curious about how Advent of Code works, it's running on some custom
Perl code. Other than a few integrations (auth, analytics, social media), I
built the whole thing myself, including the design, animations, prose, and all
of the puzzles.
The puzzles are most of the work; preparing a new calendar and a new set of
puzzles each year takes all of my free time for 4-5 months. A lot of effort
went into building this thing - I hope you're enjoying playing it as much as I
enjoyed making it for you!
If you'd like to hang out, I'm @ericwastl on Twitter.
- Eric Wastl
-->
<body>
<header><div><h1 class="title-global"><a href="/">Advent of Code</a></h1><nav><ul><li><a href="/2020/about">[About]</a></li><li><a href="/2020/events">[Events]</a></li><li><a href="https://teespring.com/stores/advent-of-code" target="_blank">[Shop]</a></li><li><a href="/2020/settings">[Settings]</a></li><li><a href="/2020/auth/logout">[Log Out]</a></li></ul></nav><div class="user">m42e <a href="/2020/support" class="supporter-badge" title="Advent of Code Supporter">(AoC++)</a> <span class="star-count">16*</span></div></div><div><h1 class="title-event">&nbsp;&nbsp;&nbsp;<span class="title-event-wrap">0x0000|</span><a href="/2020">2020</a><span class="title-event-wrap"></span></h1><nav><ul><li><a href="/2020">[Calendar]</a></li><li><a href="/2020/support">[AoC++]</a></li><li><a href="/2020/sponsors">[Sponsors]</a></li><li><a href="/2020/leaderboard">[Leaderboard]</a></li><li><a href="/2020/stats">[Stats]</a></li></ul></nav></div></header>
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<article class="day-desc"><h2>--- Day 9: Encoding Error ---</h2><p>With your neighbor happily enjoying their video game, you turn your attention to an open data port on the little screen in the seat in front of you.</p>
<p>Though the port is non-standard, you manage to connect it to your computer through the clever use of several paperclips. Upon connection, the port outputs a series of numbers (your puzzle input).</p>
<p>The data appears to be encrypted with the eXchange-Masking Addition System (<span title="No relation.">XMAS</span>) which, conveniently for you, is an old cypher with an important weakness.</p>
<p>XMAS starts by transmitting a <em>preamble</em> of 25 numbers. After that, each number you receive should be the sum of any two of the 25 immediately previous numbers. The two numbers will have different values, and there might be more than one such pair.</p>
<p>For example, suppose your preamble consists of the numbers <code>1</code> through <code>25</code> in a random order. To be valid, the next number must be the sum of two of those numbers:</p>
<ul>
<li><code>26</code> would be a <em>valid</em> next number, as it could be <code>1</code> plus <code>25</code> (or many other pairs, like <code>2</code> and <code>24</code>).</li>
<li><code>49</code> would be a <em>valid</em> next number, as it is the sum of <code>24</code> and <code>25</code>.</li>
<li><code>100</code> would <em>not</em> be valid; no two of the previous 25 numbers sum to <code>100</code>.</li>
<li><code>50</code> would also <em>not</em> be valid; although <code>25</code> appears in the previous 25 numbers, the two numbers in the pair must be different.</li>
</ul>
<p>Suppose the 26th number is <code>45</code>, and the first number (no longer an option, as it is more than 25 numbers ago) was <code>20</code>. Now, for the next number to be valid, there needs to be some pair of numbers among <code>1</code>-<code>19</code>, <code>21</code>-<code>25</code>, or <code>45</code> that add up to it:</p>
<ul>
<li><code>26</code> would still be a <em>valid</em> next number, as <code>1</code> and <code>25</code> are still within the previous 25 numbers.</li>
<li><code>65</code> would <em>not</em> be valid, as no two of the available numbers sum to it.</li>
<li><code>64</code> and <code>66</code> would both be <em>valid</em>, as they are the result of <code>19+45</code> and <code>21+45</code> respectively.</li>
</ul>
<p>Here is a larger example which only considers the previous <em>5</em> numbers (and has a preamble of length 5):</p>
<pre><code>35
20
15
25
47
40
62
55
65
95
102
117
150
182
127
219
299
277
309
576
</code></pre>
<p>In this example, after the 5-number preamble, almost every number is the sum of two of the previous 5 numbers; the only number that does not follow this rule is <em><code>127</code></em>.</p>
<p>The first step of attacking the weakness in the XMAS data is to find the first number in the list (after the preamble) which is <em>not</em> the sum of two of the 25 numbers before it. <em>What is the first number that does not have this property?</em></p>
</article>
<p>To begin, <a href="9/input" target="_blank">get your puzzle input</a>.</p>
<form method="post" action="9/answer"><input type="hidden" name="level" value="1"/><p>Answer: <input type="text" name="answer" autocomplete="off"/> <input type="submit" value="[Submit]"/></p></form>
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></span>]</span> this puzzle.</p>
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from aoc.input import get_input
import copy
import itertools
import time
import collections
import re
from aoc.partselector import part_one, part_two
import functools
def pw(line):
return int(line.strip())
def p1():
inp = get_input(pw)
for i in range(25, len(inp)):
possible_values = set()
for sample in itertools.combinations(inp[i-25:i], 2):
possible_values.add(sum(sample))
if inp[i] not in possible_values:
print (inp[i])
return inp
return inp
def p2(inp):
l = 1398413738
i = 636
d = {}
for j in range(2, 25):
possible_values = set()
for sample in range(1,636):
possible_values.add(sum(inp[sample: sample+j]))
d [sum(inp[sample: sample+j])] = inp[sample: sample+j]
if l in possible_values:
print(min(d[l]) + max(d[l]))
return (min(d[l]) + max(d[l]))
return 0
if part_one():
start = time.time()
result1 = p1()
print(round(1000*(time.time() - start), 2), 'ms')
if part_two():
start = time.time()
p2(result1)
print(round(1000*(time.time() - start), 2), 'ms')