Advent of Code 2024 Day 11

Author

Nathan Moore

— Day 11: Plutonian Pebbles —

There is a weird set of magic pebbles that split and have numbers on them.

Consider the arrangement of stones in front of you. How many stones will you have after blinking 25 times?

with open('data-2024-11.txt', 'r') as f:
    inp = f.read().splitlines()

Create rules for what happens

zero becomes 1
even number of digits, split into two
else, multiply by 2024

peb = inp[0].split()

peb = [int(p) for p in peb]

for i in range(25): 
    new_peb = []
    for j in peb: 
        js = str(j)
        if j == 0: 
            new_peb.append(1)
        elif len(js) % 2 == 0: 
            new_peb.append(int(js[:len(js)//2]))
            new_peb.append(int(js[len(js)//2:]))
        else: 
            new_peb.append(j * 2024)
    # assign it back to work on it next time
    peb = new_peb
    # print(peb)


len(peb)

# 43        (5)
# 346       (10)
# 2901      (15)
# 22872     (20)
# 185894    (25)
# 1503224   (30)
# 12133105  (35)

— Part Two —

75 times, see above.

It’s going to be too long!

How do I take care of this? Keep the counts, not in a list.

from collections import Counter

peb_count = Counter(peb)

len(peb_count)

def apply_rule(j): 
    js = str(j)
    if j == 0: 
        return 1
    elif len(js) % 2 == 0: 
        return int(js[:len(js)//2]),int(js[len(js)//2:])
    else: 
        return j * 2024


for r in range(26, 76):
    new_count = {}
    for k,v in peb_count.items():
        nn = apply_rule(k)
        if isinstance(nn, int):
            if nn in new_count:
                new_count[nn] += v
            else: 
                new_count[nn] = v
        else: 
            for m in nn:
                if m in new_count:
                    new_count[m] += v
                else: 
                    new_count[m] = v
    # assign back and loop again
    peb_count = new_count

len(peb_count)

sum(peb_count.values())

# 221632504974231

221632504974231