RogerBW's Blog

Everybody Codes 2025 week 4 16 December 2025

I've been doing Everybody Codes, a set of programming challenges in the style of Advent of Code, now in its second year.

All of my code is on Codeberg.

Day 16

Premature optimisation was still premature, but probably worth it this time.

Part A: yeah, I could count up individual cells, but the number of blocks from element N is length (integer divide) N, so I'll just do that.

Part B: OK, I do actually need a wall array for this. For each key number N in ascending order, if index (N-1) is greater than zero, it must be because N is in the recipe. So subtract 1 from all affected locations in the wall array, increment N, and try again.

Part C starts with part B (to derive the recipe), then uses my reasonably quick part A implementation to get a mapping from wall length to blocks. Then I do a search, in three stages:

  • if the number of blocks is too loe, multiple the candidate length by 10.

  • binary search between the last two candidate lengths (one is too low, one is too high) until we're nearly at the target.

  • start with the known too-high value from that, and drop it by 1 until the number of blocks is too low. That's the answer.

Day 17

This one went interestingly sideways.

Part A: fairly straightforward; the grids are small enough to scan cell by cell. (A custom subtraction function so that I can get abs(usize - usize) without messing about. And a custom integer exponentiation function for the squaring.)

Part B: again, once I'd worked out the trick of setting destroyed grid values to zero so that they don't contribute to the next total, easy enough.

Part C: all right I should have convered this to use my standard grid and coordinate types. But I didn't. And I made an assumption which didn't work on one test case, but did on my actual data.

How do I define "a loop" with specific points so that I can pathfind? Well, all the examples hug the destroyed area as closely as possible to west, south and east. And I can find those coordinates during the destruction scan. So in the end I do four separate pathfinding calculations: start to west, west to south, south to east, and east back to start. (It is trivial to envision a grid layout in which this doesn't give you the quickest overall loop.)

Then there's a custom "neighbours" function which avoids the cost zero ("destroyed") cells, and one thing that tripped me up is that the start point is a cost zero cell, so there's an exception for that. (Yeah, I should have used a separate HashSet of destroyed cells. Or indeed put the whole grid in a HashSet as I usually do, instead of a Vec<Vec>. But never mind, it works.)

So I do my four pathfindings, and if they all found paths and the total cost is low enough, return the answer. As I say, this doesn't work on testC3-3180 (giving 9 × 300 = 2700), but before I went to track down the bug I thought I'd try it on the contest data, and that worked. If you find out why it doesn't work on testC3-3180 feel free to let me know.

Later, on reflection: I should have split the map into all but bottom left, and all but bottom right, with the start column below the volcano in common, then tone a dijkstra_all to the start point on each side, then taken the lowest total to/from the same cell. But I didn't. Too bad.

Day 18

Fun in the end, but a bit of a slog. I found the description remarkably unclear, or perhaps I was just tired. But realising that this is a tree rather than a potentially looping network helped a lot.

As usual structures are important, and really getting the right one for part A was most of the solution. The key is to reverse the links: the link from A to B is listed under B, but I'll be using it when I've got a value for A. (I seem to recall some dependency tree function in the pathfinding crate which can automate this.)

Part B introduces a input mask, and I moved the energy field out of the Node structure into a separate array since I'll be rebuilding it for each mask.

For part C I cheated to find the maximum value. I'm not proud of it, or of the problem that makes it amenable to this solution, but with 81 bits of input this obviously wouldn't be amenable to an exhaustive search.

I did think about doing a limited depth-first search starting from all-on and all-off configurations, but tried a short cut first: any input node which produces at least one negative weight in its outgoing links is set to off, and any other is set to on. This turns out to produce the correct answer on the full data, though not on the test data. It doesn't feel like a "fair" programming problem.

Day 19

That was an interesting progression. The core of it was getting a reasonably efficient calculation method for part A.

The order of individual flaps or not doesn't matter. There is a diagonal up or down phase, and if the horizontal distance is long enough, a bobbing-up-and-down phase.

For part A I built a struct of X offset and Y range (using std::ops::Range as usual, though I didn't actually need the range intersection tests this time).

Then I store a rolling list of heights and number of flaps taken to each the latest gate at that height.

For each new gate, I double loop: the outer loop is the list of valid entry heights (at the previous date), and the inner loop is the list of possible exit heights (at the next gate). Half of those exit heights can be thrown away because the absolute height gain over distance X, mod 2, must be the same as distance X mod 2. Also only check reachable heights: maximum height gain or loss is one per horizontal step.

Then the number of flaps taken to achieve that particular exit height is the sum of:

  • the maximum of 0 and the height gain (the diagonal upward or downward phase of the flight)
  • half the difference between the distance X and the absolute height gain (the up-and-down phase of the flight).

(Plus the number of flaps needed to get to the starting height.)

A hash entry for each exit height holds the list of flaps needed, and the rolling list is reinitialised with the minimum cumulative flap value for each height.

The output is the minimum value in the final list of flaps, because any valid height at the final gate is equally acceptable.

For part B I dropped the custom gate structure, because I realised I was ending up with a HashMap. Key is the x value of the wall, and value is the HashSet of Ranges indicating the holes in it.

Then we sort the keys and work as before, except that rather than iterate over one range I'm pulling each one out of the inner HashSet and working on it separately. Some of the gates are 1,500+ units high, but the calculation I'm doing on each possible height is pretty quick.

As a result, I didn't need to modify part B at all to do the final part in reasonable time. (About 0.8 seconds on my test machine.) I'm sure this could be done more efficiently, but sub-second is certainly Good Enough.

Day 20

A frustrating final part, but I got it in the end.

A triangular grid is just the dual of a hex grid, and my usual technique for pathfinding on those is to expand the coordinates. So for example the cells in the first row would be assigned to coordinates

(0, 0) (1, 1) (2, 0) (3, 1) ...

and the second row (noting that x=0 is unused) to

(1, 2) (2, 3) (3, 2) (4, 3) ...

etc. I drop these into a HashSet, and then finding neighbours is a matter of scanning x = {-1, 0, 1}, y = {-1, 1} (half of which can innately not exist anyway, and I could find out which half by looking at coordinate values, but I don't).

Part A also ends up defining an ordering for coordinates so that I can store unique pairs. HashSets are not innately hashable.

Part B throws that nonsense away and gets on with the pathfinding I was expecting. Fairly straightforward.

Part C was more of a pain. This coordinate system is not readily susceptible to rotation. (I suspect nothing with rational coordinates would be, but my X and Y scales here are not only different but non-uniform.)

So I ended up writing a rotate method for the Grid which iterates over the whole space, working out old and new coordinate pairs; it's inefficient, but it works.

Coordinates expand to three dimensions (with three layers, and each layer can reach only to cells in the next), which accounts for one grid rotation per step. The exit location rotates too (not in the example but in my real data) so that becomes another HashSet with the valid exit on each layer.

Conclusions

What I learned:

I still haven't got the hang of dynamic programming, but I didn't run into any problems with inability to use Rust. I may of course be missing tricks I could have used to make my code more efficient.

Crates used:

  • pathfinding, three times
  • regex, three times
  • counter, twice
  • permutator, once

If anything the parsing was easier than last year. Nothing that was a challenge in regex. Not as much preponderance of pathfinding this year as last, hurrah. (Not that I mind, but it started to feel a bit samey.)

I'll be back next year.

Thanks to Emil Kaczyński, and the EC team.

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