Solved 2015/13 P1+P2

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FrederikBaerentsen 2024-12-11 23:40:08 +01:00
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commit 9d38fa31d7
5 changed files with 206 additions and 2 deletions

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2015/13/13.md Normal file
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## \-\-- Day 13: Knights of the Dinner Table \-\--
In years past, the holiday feast with your family hasn\'t gone so well.
Not everyone gets along! This year, you resolve, will be different.
You\'re going to find the *optimal seating arrangement* and avoid all
those awkward conversations.
You start by writing up a list of everyone invited and the amount their
happiness would increase or decrease if they were to find themselves
sitting next to each other person. You have a circular table that will
be just big enough to fit everyone comfortably, and so each person will
have exactly two neighbors.
For example, suppose you have only four attendees planned, and you
calculate
their potential happiness as follows:
Alice would gain 54 happiness units by sitting next to Bob.
Alice would lose 79 happiness units by sitting next to Carol.
Alice would lose 2 happiness units by sitting next to David.
Bob would gain 83 happiness units by sitting next to Alice.
Bob would lose 7 happiness units by sitting next to Carol.
Bob would lose 63 happiness units by sitting next to David.
Carol would lose 62 happiness units by sitting next to Alice.
Carol would gain 60 happiness units by sitting next to Bob.
Carol would gain 55 happiness units by sitting next to David.
David would gain 46 happiness units by sitting next to Alice.
David would lose 7 happiness units by sitting next to Bob.
David would gain 41 happiness units by sitting next to Carol.
Then, if you seat Alice next to David, Alice would lose `2` happiness
units (because David talks so much), but David would gain `46` happiness
units (because Alice is such a good listener), for a total change of
`44`.
If you continue around the table, you could then seat Bob next to Alice
(Bob gains `83`, Alice gains `54`). Finally, seat Carol, who sits next
to Bob (Carol gains `60`, Bob loses `7`) and David (Carol gains `55`,
David gains `41`). The arrangement looks like this:
+41 +46
+55 David -2
Carol Alice
+60 Bob +54
-7 +83
After trying every other seating arrangement in this hypothetical
scenario, you find that this one is the most optimal, with a total
change in happiness of `330`.
What is the *total change in happiness* for the optimal seating
arrangement of the actual guest list?
Your puzzle answer was `733`.
## \-\-- Part Two \-\-- {#part2}
In all the commotion, you realize that you forgot to seat yourself. At
this point, you\'re pretty apathetic toward the whole thing, and your
happiness wouldn\'t really go up or down regardless of who you sit next
to. You assume everyone else would be just as ambivalent about sitting
next to you, too.
So, add yourself to the list, and give all happiness relationships that
involve you a score of `0`.
What is the *total change in happiness* for the optimal seating
arrangement that actually includes yourself?
Your puzzle answer was `725`.
Both parts of this puzzle are complete! They provide two gold stars:
\*\*
At this point, you should [return to your Advent calendar](/2015) and
try another puzzle.
If you still want to see it, you can [get your puzzle
input](13/input).

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2015/13/solution.py Normal file
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#!/bin/python3
import sys,time,re
from pprint import pprint
sys.path.insert(0, '../../')
from fred import list2int,get_re,nprint,lprint,loadFile,TSP
start_time = time.time()
input_f = 'input'
#########################################
# #
# Part 1 #
# #
#########################################
def parseInput():
inst = []
graph = {}
with open(input_f) as file:
for line in file:
inst = line.rstrip().split(' happiness units by sitting next to ')
t = inst[0].split(' would ')
inst = [t[0]] + [inst[1].replace('.','')] + [int(t[1].replace('lose ','-').replace('gain ',''))]
if inst[0] not in graph:
graph[inst[0]] = []
#if inst[1] not in graph:
# graph[inst[1]] = []
graph[inst[0]].append((inst[1],inst[2]))
#graph[inst[1]].append((inst[0],inst[2]))
return graph
def addBothWays(graph):
n_graph = {}
for _,a in enumerate(graph):
for b in graph[a]:
for c in graph[b[0]]:
if c[0] == a:
if a not in n_graph:
n_graph[a] = []
n_graph[a].append((b[0],b[1]+c[1]))
return n_graph
def part1():
graph = addBothWays(parseInput())
pprint(graph)
longest = 0
people = set(graph.keys())
for neighbors in graph.values():
for neighbor, _ in neighbors:
people.add(neighbor)
for c in people:
route,distance = TSP(graph, c,'longest',True)
if longest < distance:
longest = distance
return longest
start_time = time.time()
print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
#########################################
# #
# Part 2 #
# #
#########################################
def part2():
# Just add 'Myself' to the input file with gain of 0 towards everyone
return
start_time = time.time()
print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')

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from functools import cache
def rule2str(number):
num_str = str(number)
length = len(num_str)
middle = length // 2
left_part = num_str[:middle]
right_part = num_str[middle:]
return [str(int(left_part)), str(int(right_part))]
values = {}
def calc(start,end):
if start == '0':
result = calculateNumber('1',end)
else:
if len(start) % 2 == 0:
x = rule2str(start)
result = 0
result += calculateNumber(x[0],end)
result += calculateNumber(x[1],end)
else:
result = calculateNumber(str(int(start)*2024),end)
return result
def calculateNumber(start,end):
if end == 0:
return 1
end -= 1
if (start,end) not in values:
values[(start,end)] = calc(start,end)
result = values[(start,end)]
return result
numbers = ['125','17']
# 2097446912 14168 4048 2 0 2 4 40 48 2024 40 48 80 96 2 8 6 7 6 0 3 2
result = 0
for i in numbers:
result += calculateNumber(i,75)
print(result)

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@ -479,7 +479,7 @@ def dijkstra(graph, start, end):
return path, distances[path[-1]] return path, distances[path[-1]]
def TSP(graph, start, path_type='shortest'): def TSP(graph, start, path_type='shortest',circle=None):
""" """
Solves TSP (Traveling Salesperson Problem) using brute force by generating all possible paths. Solves TSP (Traveling Salesperson Problem) using brute force by generating all possible paths.
Handles missing edges by skipping invalid paths. Handles missing edges by skipping invalid paths.
@ -517,6 +517,9 @@ def TSP(graph, start, path_type='shortest'):
# Generate all permutations of the remaining cities # Generate all permutations of the remaining cities
for perm in permutations(cities): for perm in permutations(cities):
# Create a complete path starting and ending at the start node # Create a complete path starting and ending at the start node
if circle is not None:
path = [start] + list(perm) + [start]
else:
path = [start] + list(perm) path = [start] + list(perm)
# Calculate the total distance of this path # Calculate the total distance of this path