Advent of Code 2023 Day 8

Author

Nathan Moore

— Day 8: Haunted Wasteland —

There are ghosts, and you have to follow some Left/Right instructions on a map.

Starting at AAA, follow the left/right instructions. How many steps are required to reach ZZZ?

with open('data-2023-08.txt', 'r') as f:
    inp = f.read().splitlines()

Create a dictionary with L: 0, R: 1 for directions i.e. list members; Split the instructions into a list; create a dictionary with the coordinates.

from itertools import cycle

# first line is left/right directions
inst = list(inp[0])

# for converting left/right into nodes
id = {'L': 0, 'R': 1}

# function for dictionary
def pairwise(y): 
    return y.split(' = ')[1].replace('(', '').replace(')', '').split(', ')

# nodes 
nodes = {x.split(' = ')[0]: pairwise(x) for x in inp[2:]}

# initialise
steps = 0
lmnt = 'AAA'
trgt = 'ZZZ'

# let us start by assuming less than one trip through the instructions
# I'm not sure how many trips it was but 1000 to 1001 didn't change the answer
# and then I found itertools.cycle
for k in cycle(inst): 
    steps += 1
    # print(k)
    # print(lmnt)
    # print(nodes[lmnt])
    lmnt = nodes[lmnt][id[k]]
    if lmnt == trgt: break
    
steps

Part one: 16343

— Part Two —

It’s not one start AAA and one end ZZZ there are actually more starts and ends.

Simultaneously start on every node that ends with A. How many steps does it take before you’re only on nodes that end with Z?

# for instructions, for elements
lmnts = []

# get our starting elements, rather than just one element
for k in nodes.keys():
    if k[2] == 'A':
        lmnts.append(k)

# check on this        
# lmnts

steps = 0

# checking function, if they all end in 'Z'
def checker(lst):
    # if any([j[2] == 'Z' for j in lst]):
    #     print(lst)
    return all([j[2] == 'Z' for j in lst])
  

for k in cycle(inst): 
    steps += 1
    for e, l in enumerate(lmnts): 
        lmnts[e] = nodes[l][id[k]]
    if checker(lmnts): break
    if steps == 10**6: break
    # print(lmnts)
    
steps

It’s more than one million, and I don’t know how much more.

Don’t do this, do them individually and multiply the results

# for instructions, for elements
lmnts = []

# get our starting elements, rather than just one element
for k in nodes.keys():
    if k[2] == 'A':
        lmnts.append(k)

# check on this        
lmnts

for l in lmnts: 
    steps = 0
    for k in cycle(inst): 
        steps += 1
        l = nodes[l][id[k]]
        if l[2] == 'Z': break
    print(steps)
    print(l)

21883
TBZ
13019
LQZ
19667
QDZ
16343
ZZZ
18559
FNZ
14681
FQZ

Lowest Common Multiple

from math import lcm

lcm(21883, 13019, 19667, 16343, 18559, 14681)

15299095336639

Part two: 15299095336639