AoC 2017 Day 6: Tightrope

Source: Memory Reallocation

Part 1: Start with n stacks of different sizes. Take the largest block and distribute its items starting with n+1 and looping around. How many iterations of this does it take before you see a state you’ve seen before?

The core of this problem is getting the balancing function right:

def balance(blocks):
    to_distribute = max(*blocks)
    index = blocks.index(to_distribute)

    for_each = to_distribute // len(blocks)
    left_over = to_distribute - (for_each * len(blocks))
    give_left_over_to = {i % len(blocks) for i in range(index + 1, index + 1 + left_over)}

    return tuple(
        (
            (amount if i != index else 0)
            + for_each
            + (1 if i in give_left_over_to else 0)
        )
        for i, amount in enumerate(blocks)
    )

The blocks are defined as a tuple since that allows it to be used as an element in a set (lists don’t work since they are mutable).

To actually distribute the blocks, we will start by dividing the amount and keeping the remainder. The quotient for_each goes to all of the blocks and the remainder left_over is the odd bits that will only go to some of the blocks. give_left_over_to is me figuring out which indexes are going to get the remainder. Finally, we create a new tuple by adding the current amount (except for the block we’re redistributing) plus for_each plus 1 for those left_over.

Then we’ll loop over those until we see a duplicate:

blocks = tuple(
    int(node)
    for line in lib.input()
    for node in line.split()
)

seen = set()
for cycle in itertools.count():
    if blocks in seen:
        break
    else:
        seen.add(blocks)

    blocks = balance(blocks)

print(cycle)

Part 2: How many iterations are there in the loop?

Interesting. To do this, what we want to do is store the index we see each arrangement of blocks at, rather than just that we saw them. Then the difference between the current iteration and that first one is the length of the cycle:

seen = {}
for cycle in itertools.count():
    if blocks in seen:
        break
    else:
        seen[blocks] = cycle

    blocks = balance(blocks)

print(cycle, cycle - seen[blocks])

This will run both parts in a single command:

$ python3 run-all.py day-06

day-06  python3 tightrope.py input.txt  0.10088086128234863     3156 1610
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