Advent of Code 2018, Day 7

This might be my favourite of the exercises so far, probably because it reminds me of the constraint programming and first-order logic stuff; it's a close tie maybe with day 5.

I again did the first pass in Python and I'm working on the follow-up C++ version as I have time.

The first thing I do is parse the file into a bunch of lines, and use a regex to pick out the task name and its dependency. One thing here that I thought is that maybe this was going to be a slowdown, so I did some testing around this. It turns out that it only adds about 200µs to the run time for part 1 - using ASAN was a much larger performance impact. Removing that brought my run times for part 1 from about 2500µs to about 750µs on average.

The heart of both of my part 1 solvers is a map of task name to a set of its dependencies. In C++, I was able to rely on the fact that maps maintain a weak ordering of keys, so the initial parsing was all the sorting that I needed to do. In Python 3, dictionaries are insertion ordered, so I had to sort the .keys() list in order to do this. I chose a set because it's a type focused on membership; Python supports the in keyword, but semantically, we're thinking about sets.

As I look at each item in the list, I have to figure if it has any unresolved dependencies - if it has none, I remove it from the map, and iterate over all the other constraints, removing as needed from the set.

def mark_done(steps, name):
    for step in steps.keys():
        if name in steps[step]:

def next_task(steps):
    for step in sorted(steps.keys()):
        if len(steps[step]) == 0:
            return step

def sequence(steps):
    result = ""
    while len(steps) > 0:
        step = next_task(steps)
        result += step
        mark_done(steps, step)
    return result

As with all of these, there's a fair amount of cleaning up that I can do, but it's not bad as it stands.

complete(char dependency)
    for (auto step : constraints) {
        if (step.second.count(dependency) == 0) {

    string  result;
    while (constraints.size() > 0) {
        for (auto it : constraints) {
            if (it.second.size() != 0) {


    return result;

The self checks are pretty handy: for example, I write C++ in Vim and I have a shortcut to just run make - a fast way to build and test in one sweep. I don't have that quite set up in Emacs (where I write Python), but it works pretty well still. I like that it catches regressions right away, and makes sure not to even bother with the larger input if it can't get the small test input right.

Part 2 involved some subtlety (a fairly natural consequence of concurrency problems); it's important to note that while the test input will give you some confidence, there's a corner case it doesn't cover that was crucial to solving my particular set of inputs. I imagine it's a relatively common scenario.

In particular, what can happen is the following:

  • At least one worker is idle in a given time step.
  • A later worker in the iteration finishes a task that unblocks another.
  • The earlier worker should pick up the unblocked task - but if you iterate over them in order (like I was) you'll miss that, and you'll calculate at least a second later than you should.

In Python, I used two dictionaries for this: one to keep track of what task each worker was assigned to, and one to keep track of when the worker would complete their task.

The solution looked like this:

  • I look to see if any work is available, which boils down to figuring out if I have workers available and tasks for them to do:

    • If there aren't any workers available, then we're going say there's no work available.

    • I copy the two dictionaries and remove all the assignments that the workers were working on. Then I look for the next task that would be available given the copied dictionaries.

    • Return whether any tasks are available.

  • I get the next available worker (this is a duplicate of the first step before...).

  • If that worker was assigned to another task, that means it's finished; I mark this task as complete the same way I did before.

  • Then I get the next task to be completed. If there aren't any, it's time to move on. This is also a duplicate of work done above.

  • Assign the task to the worker and mark the time it'll be completed.

  • Repeat the above for as long as there is work available.

  • Increment the time step.

  • Repeat all of the above for as long as there are still tasks remaining to be done.

There's some duplicated effort and lots of what feels like redundant copying of dictionaries. If I get around to working on the C++ version of part 2, I'd like to revisit this.