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day08: replace static foreach with workaround
author Jordi Gutiérrez Hermoso <>
date Tue, 16 Jan 2018 11:28:55 -0500
parents 049fb8e56025
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--- Day 17: Spinlock ---

Suddenly, whirling in the distance, you notice what looks like a
massive, pixelated hurricane: a deadly spinlock. This spinlock isn't
just consuming computing power, but memory, too; vast, digital
mountains are being ripped from the ground and consumed by the vortex.

If you don't move quickly, fixing that printer will be the least of
your problems.

This spinlock's algorithm is simple but efficient, quickly consuming
everything in its path. It starts with a circular buffer containing
only the value 0, which it marks as the current position. It then
steps forward through the circular buffer some number of steps (your
puzzle input) before inserting the first new value, 1, after the value
it stopped on. The inserted value becomes the current position. Then,
it steps forward from there the same number of steps, and wherever it
stops, inserts after it the second new value, 2, and uses that as the
new current position again.

It repeats this process of stepping forward, inserting a new value,
and using the location of the inserted value as the new current
position a total of 2017 times, inserting 2017 as its final operation,
and ending with a total of 2018 values (including 0) in the circular

For example, if the spinlock were to step 3 times per insert, the
circular buffer would begin to evolve like this (using parentheses to
mark the current position after each iteration of the algorithm):

    (0), the initial state before any insertions.

    0 (1): the spinlock steps forward three times (0, 0, 0), and then
    inserts the first value, 1, after it. 1 becomes the current

    0 (2) 1: the spinlock steps forward three times (0, 1, 0), and
    then inserts the second value, 2, after it. 2 becomes the current

    0 2 (3) 1: the spinlock steps forward three times (1, 0, 2), and
    then inserts the third value, 3, after it. 3 becomes the current

And so on:

    0  2 (4) 3  1
    0 (5) 2  4  3  1
    0  5  2  4  3 (6) 1
    0  5 (7) 2  4  3  6  1
    0  5  7  2  4  3 (8) 6  1
    0 (9) 5  7  2  4  3  8  6  1

Eventually, after 2017 insertions, the section of the circular buffer
near the last insertion looks like this:

1512  1134  151 (2017) 638  1513  851

Perhaps, if you can identify the value that will ultimately be after
the last value written (2017), you can short-circuit the spinlock. In
this example, that would be 638.

What is the value after 2017 in your completed circular buffer?

Your puzzle answer was 1306.

--- Part Two ---

The spinlock does not short-circuit. Instead, it gets more angry. At
least, you assume that's what happened; it's spinning significantly
faster than it was a moment ago.

You have good news and bad news.

The good news is that you have improved calculations for how to stop
the spinlock. They indicate that you actually need to identify the
value after 0 in the current state of the circular buffer.

The bad news is that while you were determining this, the spinlock has
just finished inserting its fifty millionth value (50000000).

What is the value after 0 the moment 50000000 is inserted?

Your puzzle answer was 20430489.

Both parts of this puzzle are complete! They provide two gold stars: **