I don't really like problems like these. I love Advent of Code and have got 50/50 stars on Christmas day this year -- but this type of problem grinds my gears.
The intended solution only works because the input is more constrained than what the problem statement says (the problem in full generality is PSPACE-hard). If you give me a problem to solve, I'd rather have all the hypotheses, all at once.
The same is true with part 2 of problem 8 and problem 21 which both are generally much harder problems which are made easier by carefully constructed test cases. I also find it somewhat annoying, but par for the course with advent given the same test data for part 1 & 2; and the fact that you actually see the only test case -- it's not hidden test cases like in other programming competitions.
I'm behind this year, so I haven't looked at any postmortems, but I found an interesting solution to problem 8 part 2[1] after watching the brute force methods stall out. Could you explain what you mean by carefully constructed test cases?
It's easy to construct inputs for which the correct solution would not be the LCM of the cycle lengths. (Just take the input you got, and insert a few extra steps between the "start" point for one of the "ghosts" and the point where it enters its otherwise unaltered cycle.)
A more general solution is possible -- but it does require a trick that people may not be aware of unless they've studied a little number theory.
It gets even more interesting if you start worrying about ghosts that could finish at multiple points within each cycle. There are potentially too many combinations to try out each one individually.
I actually wrote the code to solve it in full generality (https://github.com/edoannunziata/jardin/blob/master/misc/Aoc...) -- CRT is the idea but it's actually the "generalized" case of CRT, because you could have moduli that are not coprime. So you have to use the CRT "in reverse" to break each linear congruence into its components (using the fact that \mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn} \iff \gcd(m, n) = 1) and then use the CRT again to find the answer.
Annoying to code, but I've done it a thousand times by hand for math competitions, so it's kind of carved into my skull.
There might be multiple cases, but I don't think there's anything better (please let me know if there is, I'm very interested) -- as a matter of fact, an instance of the original problem can be used to encode an instance of a system of linear congruences, so finding a better algorithm for the latter would imply a better algorithm for the former.
It's possible to answer "find the minimum n such that n%a is in s1 and n%b is in s2, assuming a and b are coprime" in runtime around O(n1+log(n1)*n2) where n1 is the size of s1 and n2 is the size of s2.
n1*n2 can get at least as big as a*b/8 without a naive linear search being guaranteed to be fast, so this roughly square roots the runtime of 'CRT on all pairs' if n1 is close to n2.
Here's the code, sorry it's not as well commented as yours, and I haven't integrated it into a solution to day 8.
For more than 2 moduli, the best I can think of is to separate them into 2 roughly equal parts, generate the full lists of allowed remainders for each part, then use this algorithm to combine the parts. There may be better things you can do here (and perhaps you can show that for large sets of allowed remainders across enough different moduli, the solutions are spread out evenly enough there's one you can find by naive linear search).
Dealing with non-coprime moduli is left as an exercise to the reader :).
So in Question 8 part 2 you had multiple cycles through a graph. You started on all "A" nodes and you had to find the first time at which you were entirely on "Z" nodes.
The intended solution was seemingly to calculate the length of the cycle (as in, until you were on a "Z" node) for each starting node separately. Once you'd done this, you could find the LCM of these cycle lengths to find the overall period of the cycle (~10^13).
This solution might not work if your cycles weren't constant length -- for example imagine when walking your graph you found your i_th Z-node after {6, 7, 6, 7, 6, 7, 6, 7 ...} steps; or perhaps {6, 7, 7, 7, ...}. In the test data, this didn't occur - it was always {n, n, n, n, ...}.
I can give another example perhaps - consider you're asked to find the last 3 digits of 2^n for n >= 1. You start off calculating: 002, 004, 008, ... eventually you get to 2^103 which ends in 008. This means that the cycle length would be {103, 101, 101, 101, 101, ...} since it'll never get back to 002 or 004. Solving this is a bit more difficult than the constant cycle lengths, since it's not a simple LCM.
Is there any theorem that puts limits on how irregular the cycles can be in a general case of tape length L with N options corresponding to N outputs from each graph node?
What has to be constant is the cycle length for (node, tape instruction #) pairs, because all the state to determine every future step is based on the current node and tape position; if both are the same, the path will be the same as before. I think, at least without advanced math, the only thing to rely on for "true" loops is identical pairs of (instr #, node), not just instr # or tape steps or node individually.
I haven't spent that much time thinking about it, but my guess is that there may or may not be:
* a "tail" -- some initial part of the path before you get into a true cycle (like the 002, 004 in the example above)
* an inner repeating cycle before you reach a node you've seen before.
In the case given |t| = 0 and |c| = 1, but it's easy to construct a more complex example with nodes (A, B, C, D), edges (A->B, B->C, C->D, D->B), and 'ending nodes' being B and D. In this case left and right paths go to the same node. This case would have a tail of length 1, and then the inner cycles would be of length {2, 1, 2, 1 ...}.
As a result, valid 'ending states' (Z-nodes) for this graph would be after {1, 3, 4, 6, 7, 9, 10, ...} steps.
I can see the case of an aperiodic first cycle length, but given that each vertex in the graph created by the input -> output nodes of the traversal path has only one outgoing edge, is it possible for the cyclic graph from a1 to some z1 to have an inconsistent length? And is there ever going to be some z2 for which the aperiodic cycle length from a1 to either z1 or z2 results in a better answer to the problem given that the periodic cycle length of a1 to z1 is shorter than the cycle length of a1 to z2? Please forgive me if I am not using the correct graph theory terms.
I don't necessarily mind these "reverse engineering" type problems, I think you can consider your input part of the problem statement. In earlier years it was common, particularly in early days, for your input to be on the page itself IIRC, like "Your input is 103053439" rather than a link.
One thing that bothers me more though is when the example input is nonsense for Part II or can be solved but in a radically different way because it has very different properties than the inputs people are given for real. Contrast day 19, where the example input has a rational answer, which you are told about for the Part II, against day 20 where the example input is completely irrelevant for Part II and good luck.
Yeah, it's always rough when you write a solution for part 2, and it passes all the samples, but then has issues with the real input.
One thing I've tried to do when I run into those kinds of problems is to write a little benchmark to illustrate the difference between approaches. It's always kinda fun to watch your initial brute force solution start chugging while your shiny new solution seems to handle whatever you toss at it - great lesson in choosing effective algorithms.
I use Elixir to do the puzzles, so I've used Benchee for this, and it works very well. So easy to set up too.
Here's an example benchmark, which also has the output. As the input size increases you start getting some pretty crazy ratios between the two algorithms (the "smart" version was 200,000 times faster than the "naive" one on the largest test input!)
> I think you can consider your input part of the problem statement
That would be fine if there was only one input. But we have seen countless cases of a solution working for some and not others (including/especially the test input). So the cases and patterns I see in my input may be part of the hidden problem statement, or they may be just an artifact of my particular input. I at least feel more satisfied when I know my solution will solve all inputs.
For that reason, I side with the desire for a bit more purity and closure, vs having to guess, and not being sure about your guess until you go to reddit. Making educated guesses about patterns is a valuable ability, but I don't
like the aesthetics of it here.
This is definitely something where tastes vary, for example you can see below several people assumed you can multiply any numbers, and that actually worked because the numbers are co-prime, but strictly you need LCM (and since I had LCM in my toolkit I used it)
I try to write solutions which would work for inputs like mine but I'm not fussed if, for example, my code panics on hypothetical "valid" inputs that I didn't consider.
So e.g I have panics for nonsense input like, "Pipe networks with more than one Start can't be solved" and "Uneveen seed list cannot work as described in problem" [Yes that's a typo] but I also have "Should not send signals unexpectedly" and "Surely not all the stones have X velocity 0"
I like them a lot because they inject a little bit of empirical or heuristic thinking into problems. For this particular problem I immediately reached for graphviz to get an idea of what the circuit looks like and it made it obvious what the solution was.
It's much closer to how real world engineering problems work where you have to do a little bit of investigation and maybe find some creative way to constrain your problem that nobody explicitly tells you about. Much prefer it to the 'here are the exact five keywords so you know what CS class algorithm you need to use' kind of thing.
I _love_ when one of these problems encourages me to reach for a tool like that. (At least, when it's one I'm already aware of and I have time to dedicate to grokking and solving the puzzle).
I used graphviz on one of last years problems and found it pretty helpful. It just feels so cool when you learn a new "spell" to help you feel out the problem.
Another example was from 2020, where I used lexx/yacc to generate a parser for the expressions in the input. This was honestly probably slower than just doing it myself but it felt so cool to solve it with a tool - and it's neat to plant little signposts in your brain that will light up when you run across a similar class of problem in your real work.
Another thing I often do is use vim to munge the input and save myself some parsing code, which is a very generically useful branch of "magic" to have under your fingers.
The intended solution only works because the input is more constrained than what the problem statement says (the problem in full generality is PSPACE-hard). If you give me a problem to solve, I'd rather have all the hypotheses, all at once.