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## \-\-- Day 13: Packet Scanners \-\--
You need to cross a vast *firewall*. The firewall consists of several
layers, each with a *security scanner* that moves back and forth across
the layer. To succeed, you must not be detected by a scanner.
By studying the firewall briefly, you are able to record (in your puzzle
input) the *depth* of each layer and the *range* of the scanning area
for the scanner within it, written as `depth: range`. Each layer has a
thickness of exactly `1`. A layer at depth `0` begins immediately inside
the firewall; a layer at depth `1` would start immediately after that.
For example, suppose you\'ve recorded the following:
0: 3
1: 2
4: 4
6: 4
This means that there is a layer immediately inside the firewall (with
range `3`), a second layer immediately after that (with range `2`), a
third layer which begins at depth `4` (with range `4`), and a fourth
layer which begins at depth 6 (also with range `4`). Visually, it might
look like this:
0 1 2 3 4 5 6
[ ] [ ] ... ... [ ] ... [ ]
[ ] [ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ]
Within each layer, a security scanner moves back and forth within its
range. Each security scanner starts at the top and moves down until it
reaches the bottom, then moves up until it reaches the top, and repeats.
A security scanner takes *one picosecond* to move one step. Drawing
scanners as `S`, the first few picoseconds look like this:
Picosecond 0:
0 1 2 3 4 5 6
[S] [S] ... ... [S] ... [S]
[ ] [ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ]
Picosecond 1:
0 1 2 3 4 5 6
[ ] [ ] ... ... [ ] ... [ ]
[S] [S] [S] [S]
[ ] [ ] [ ]
[ ] [ ]
Picosecond 2:
0 1 2 3 4 5 6
[ ] [S] ... ... [ ] ... [ ]
[ ] [ ] [ ] [ ]
[S] [S] [S]
[ ] [ ]
Picosecond 3:
0 1 2 3 4 5 6
[ ] [ ] ... ... [ ] ... [ ]
[S] [S] [ ] [ ]
[ ] [ ] [ ]
[S] [S]
Your plan is to hitch a ride on a packet about to move through the
firewall. The packet will travel along the top of each layer, and it
moves at *one layer per picosecond*. Each picosecond, the packet moves
one layer forward (its first move takes it into layer 0), and then the
scanners move one step. If there is a scanner at the top of the layer
*as your packet enters it*, you are *caught*. (If a scanner moves into
the top of its layer while you are there, you are *not* caught: it
doesn\'t have time to notice you before you leave.) If you were to do
this in the configuration above, marking your current position with
parentheses, your passage through the firewall would look like this:
Initial state:
0 1 2 3 4 5 6
[S] [S] ... ... [S] ... [S]
[ ] [ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ]
Picosecond 0:
0 1 2 3 4 5 6
(S) [S] ... ... [S] ... [S]
[ ] [ ] [ ] [ ]
[ ] [ ] [ ]
[ ] [ ]
0 1 2 3 4 5 6
( ) [ ] ... ... [ ] ... [ ]
[S] [S] [S] [S]
[ ] [ ] [ ]
[ ] [ ]
Picosecond 1:
0 1 2 3 4 5 6
[ ] ( ) ... ... [ ] ... [ ]
[S] [S] [S] [S]
[ ] [ ] [ ]
[ ] [ ]
0 1 2 3 4 5 6
[ ] (S) ... ... [ ] ... [ ]
[ ] [ ] [ ] [ ]
[S] [S] [S]
[ ] [ ]
Picosecond 2:
0 1 2 3 4 5 6
[ ] [S] (.) ... [ ] ... [ ]
[ ] [ ] [ ] [ ]
[S] [S] [S]
[ ] [ ]
0 1 2 3 4 5 6
[ ] [ ] (.) ... [ ] ... [ ]
[S] [S] [ ] [ ]
[ ] [ ] [ ]
[S] [S]
Picosecond 3:
0 1 2 3 4 5 6
[ ] [ ] ... (.) [ ] ... [ ]
[S] [S] [ ] [ ]
[ ] [ ] [ ]
[S] [S]
0 1 2 3 4 5 6
[S] [S] ... (.) [ ] ... [ ]
[ ] [ ] [ ] [ ]
[ ] [S] [S]
[ ] [ ]
Picosecond 4:
0 1 2 3 4 5 6
[S] [S] ... ... ( ) ... [ ]
[ ] [ ] [ ] [ ]
[ ] [S] [S]
[ ] [ ]
0 1 2 3 4 5 6
[ ] [ ] ... ... ( ) ... [ ]
[S] [S] [S] [S]
[ ] [ ] [ ]
[ ] [ ]
Picosecond 5:
0 1 2 3 4 5 6
[ ] [ ] ... ... [ ] (.) [ ]
[S] [S] [S] [S]
[ ] [ ] [ ]
[ ] [ ]
0 1 2 3 4 5 6
[ ] [S] ... ... [S] (.) [S]
[ ] [ ] [ ] [ ]
[S] [ ] [ ]
[ ] [ ]
Picosecond 6:
0 1 2 3 4 5 6
[ ] [S] ... ... [S] ... (S)
[ ] [ ] [ ] [ ]
[S] [ ] [ ]
[ ] [ ]
0 1 2 3 4 5 6
[ ] [ ] ... ... [ ] ... ( )
[S] [S] [S] [S]
[ ] [ ] [ ]
[ ] [ ]
In this situation, you are *caught* in layers `0` and `6`, because your
packet entered the layer when its scanner was at the top when you
entered it. You are *not* caught in layer `1`, since the scanner moved
into the top of the layer once you were already there.
The *severity* of getting caught on a layer is equal to its *depth*
multiplied by its *range*. (Ignore layers in which you do not get
caught.) The severity of the whole trip is the sum of these values. In
the example above, the trip severity is `0*3 + 6*4 = 24`.
Given the details of the firewall you\'ve recorded, if you leave
immediately, *what is the severity of your whole trip*?
To begin, [get your puzzle input](13/input).
Answer:

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#!/bin/python3
import sys,re
from pprint import pprint
sys.path.insert(0, '../../')
from fred import list2int
input_f = 'test'
part = 1
#########################################
# #
# Part 1 #
# #
#########################################
if part == 1:
with open(input_f) as file:
for line in file:
#########################################
# #
# Part 2 #
# #
#########################################
if part == 2:
exit()

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solution.py
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#!/bin/python3
import sys,re
from pprint import pprint
sys.path.insert(0, '../../')
from fred import list2int
input_f = 'test'