Solved 2015/13 P1+P2
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2015/13/13.md
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2015/13/13.md
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## \-\-- Day 13: Knights of the Dinner Table \-\--
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In years past, the holiday feast with your family hasn\'t gone so well.
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Not everyone gets along! This year, you resolve, will be different.
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You\'re going to find the *optimal seating arrangement* and avoid all
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those awkward conversations.
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You start by writing up a list of everyone invited and the amount their
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happiness would increase or decrease if they were to find themselves
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sitting next to each other person. You have a circular table that will
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be just big enough to fit everyone comfortably, and so each person will
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have exactly two neighbors.
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For example, suppose you have only four attendees planned, and you
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calculate
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their potential happiness as follows:
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Alice would gain 54 happiness units by sitting next to Bob.
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Alice would lose 79 happiness units by sitting next to Carol.
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Alice would lose 2 happiness units by sitting next to David.
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Bob would gain 83 happiness units by sitting next to Alice.
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Bob would lose 7 happiness units by sitting next to Carol.
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Bob would lose 63 happiness units by sitting next to David.
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Carol would lose 62 happiness units by sitting next to Alice.
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Carol would gain 60 happiness units by sitting next to Bob.
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Carol would gain 55 happiness units by sitting next to David.
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David would gain 46 happiness units by sitting next to Alice.
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David would lose 7 happiness units by sitting next to Bob.
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David would gain 41 happiness units by sitting next to Carol.
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Then, if you seat Alice next to David, Alice would lose `2` happiness
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units (because David talks so much), but David would gain `46` happiness
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units (because Alice is such a good listener), for a total change of
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`44`.
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If you continue around the table, you could then seat Bob next to Alice
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(Bob gains `83`, Alice gains `54`). Finally, seat Carol, who sits next
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to Bob (Carol gains `60`, Bob loses `7`) and David (Carol gains `55`,
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David gains `41`). The arrangement looks like this:
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+41 +46
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+55 David -2
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Carol Alice
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+60 Bob +54
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-7 +83
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After trying every other seating arrangement in this hypothetical
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scenario, you find that this one is the most optimal, with a total
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change in happiness of `330`.
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What is the *total change in happiness* for the optimal seating
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arrangement of the actual guest list?
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Your puzzle answer was `733`.
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## \-\-- Part Two \-\-- {#part2}
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In all the commotion, you realize that you forgot to seat yourself. At
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this point, you\'re pretty apathetic toward the whole thing, and your
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happiness wouldn\'t really go up or down regardless of who you sit next
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to. You assume everyone else would be just as ambivalent about sitting
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next to you, too.
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So, add yourself to the list, and give all happiness relationships that
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involve you a score of `0`.
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What is the *total change in happiness* for the optimal seating
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arrangement that actually includes yourself?
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Your puzzle answer was `725`.
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Both parts of this puzzle are complete! They provide two gold stars:
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\*\*
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At this point, you should [return to your Advent calendar](/2015) and
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try another puzzle.
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If you still want to see it, you can [get your puzzle
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input](13/input).
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2015/13/solution.py
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2015/13/solution.py
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#!/bin/python3
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import sys,time,re
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from pprint import pprint
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sys.path.insert(0, '../../')
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from fred import list2int,get_re,nprint,lprint,loadFile,TSP
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start_time = time.time()
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input_f = 'input'
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#########################################
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# #
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# Part 1 #
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# #
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#########################################
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def parseInput():
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inst = []
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graph = {}
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with open(input_f) as file:
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for line in file:
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inst = line.rstrip().split(' happiness units by sitting next to ')
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t = inst[0].split(' would ')
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inst = [t[0]] + [inst[1].replace('.','')] + [int(t[1].replace('lose ','-').replace('gain ',''))]
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if inst[0] not in graph:
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graph[inst[0]] = []
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#if inst[1] not in graph:
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# graph[inst[1]] = []
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graph[inst[0]].append((inst[1],inst[2]))
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#graph[inst[1]].append((inst[0],inst[2]))
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return graph
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def addBothWays(graph):
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n_graph = {}
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for _,a in enumerate(graph):
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for b in graph[a]:
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for c in graph[b[0]]:
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if c[0] == a:
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if a not in n_graph:
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n_graph[a] = []
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n_graph[a].append((b[0],b[1]+c[1]))
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return n_graph
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def part1():
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graph = addBothWays(parseInput())
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pprint(graph)
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longest = 0
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people = set(graph.keys())
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for neighbors in graph.values():
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for neighbor, _ in neighbors:
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people.add(neighbor)
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for c in people:
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route,distance = TSP(graph, c,'longest',True)
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if longest < distance:
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longest = distance
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return longest
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start_time = time.time()
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print('Part 1:',part1(), '\t\t', round((time.time() - start_time)*1000), 'ms')
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#########################################
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# #
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# Part 2 #
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# #
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#########################################
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def part2():
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# Just add 'Myself' to the input file with gain of 0 towards everyone
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return
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start_time = time.time()
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print('Part 2:',part2(), '\t\t', round((time.time() - start_time)*1000), 'ms')
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2024/11/Sol.py
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2024/11/Sol.py
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from functools import cache
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def rule2str(number):
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num_str = str(number)
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length = len(num_str)
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middle = length // 2
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left_part = num_str[:middle]
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right_part = num_str[middle:]
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return [str(int(left_part)), str(int(right_part))]
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values = {}
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def calc(start,end):
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if start == '0':
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result = calculateNumber('1',end)
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else:
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if len(start) % 2 == 0:
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x = rule2str(start)
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result = 0
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result += calculateNumber(x[0],end)
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result += calculateNumber(x[1],end)
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else:
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result = calculateNumber(str(int(start)*2024),end)
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return result
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def calculateNumber(start,end):
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if end == 0:
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return 1
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end -= 1
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if (start,end) not in values:
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values[(start,end)] = calc(start,end)
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result = values[(start,end)]
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return result
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numbers = ['125','17']
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# 2097446912 14168 4048 2 0 2 4 40 48 2024 40 48 80 96 2 8 6 7 6 0 3 2
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result = 0
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for i in numbers:
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result += calculateNumber(i,75)
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print(result)
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fred.py
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fred.py
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return path, distances[path[-1]]
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def TSP(graph, start, path_type='shortest'):
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def TSP(graph, start, path_type='shortest',circle=None):
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"""
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Solves TSP (Traveling Salesperson Problem) using brute force by generating all possible paths.
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Handles missing edges by skipping invalid paths.
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# Generate all permutations of the remaining cities
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for perm in permutations(cities):
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# Create a complete path starting and ending at the start node
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if circle is not None:
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path = [start] + list(perm) + [start]
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else:
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path = [start] + list(perm)
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# Calculate the total distance of this path
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