Day 8: Playground
Megathread guidelines
- Keep top level comments as only solutions, if you want to say something other than a solution put it in a new post. (replies to comments can be whatever)
- You can send code in code blocks by using three backticks, the code, and then three backticks or use something such as https://topaz.github.io/paste/ if you prefer sending it through a URL
FAQ
- What is this?: Here is a post with a large amount of details: https://programming.dev/post/6637268
- Where do I participate?: https://adventofcode.com/
- Is there a leaderboard for the community?: We have a programming.dev leaderboard with the info on how to join in this post: https://programming.dev/post/6631465
Haskell
import Control.Arrow import Control.Monad import Data.Char import Data.List import Data.Maybe import Data.Ord import Text.ParserCombinators.ReadP import Data.Array.Unboxed qualified as A import Data.Map.Strict qualified as M parse = fst . last . readP_to_S (endBy (sepBy (read <$> munch1 isDigit) (char ',')) (char '\n')) sortedPairs l = sortOn dist [(x, y) | (x : ys) <- tails l, y <- ys] where dist = uncurry $ (sum .) . zipWith (\a b -> (b - a) ^ 2) merge l = scanl' f (initialAssocs, initialSizes) where f s@(assocs, sizes) (a, b) = case compare ia ib of GT -> f s (b, a) LT -> ( M.map (\x -> if x == ib then ia else x) assocs , sizes A.// [(ib, 0), (ia, (sizes A.! ia) + (sizes A.! ib))] ) EQ -> s where (ia, ib) = (assocs M.! a, assocs M.! b) initialAssocs = M.fromList $ zip l [1 ..] initialSizes = A.listArray (1, length l) $ repeat 1 :: A.UArray Int Int main = do contents <- parse <$> getContents let pairs = sortedPairs contents merged = merge contents pairs n = findIndex ((== length contents) . (A.! 1) . snd) merged print $ product . take 3 . sortBy (comparing Down) . A.elems . snd <$> merged !? 1000 print $ uncurry (*) . (head *** head) . (pairs !!) . pred <$> nAnother factor of 10 in problem size would have required more efficient data structures and the “correct” algorithm. But as it is, we can get by with sets-as-lists and just updating a hash table, if we’re willing to wait a few seconds for the answer.
(ql:quickload :str) (defun parse-line (line) (coerce (mapcar #'parse-integer (str:split "," line)) 'vector)) (defun read-inputs (filename) (let ((input-lines (uiop:read-file-lines filename))) (make-array (list (length input-lines) 3) :initial-contents (mapcar #'parse-line input-lines)))) (defun distance (points p q) (flet ((square (x) (* x x))) (sqrt (+ (square (- (aref points p 0) (aref points q 0))) (square (- (aref points p 1) (aref points q 1))) (square (- (aref points p 2) (aref points q 2))))))) (defun all-labeled-edges (points) (loop for j from 0 to (1- (car (array-dimensions points))) nconcing (loop for i from 0 to (1- j) collect (list (distance points i j) i j)))) (defun short-labeled-edges (points nedges) (subseq (sort (all-labeled-edges points) #'< :key #'first) 0 nedges)) (defun adjacency-map (labeled-edges) (let ((result (make-hash-table))) (loop for (nil v w) in labeled-edges do (setf (gethash v result) (cons w (gethash v result))) (setf (gethash w result) (cons v (gethash w result)))) result)) (defun components (n adjacency-map) (let ((remaining (loop for i from 0 to (1- n) collect i)) (result nil)) (loop for v = (car remaining) until (null v) do (let ((this-component nil) (next-component (list v))) (loop until (subsetp next-component this-component) do (progn (setf this-component next-component) (setf next-component (reduce #'union (cons this-component (mapcar #'(lambda (w) (gethash w adjacency-map)) this-component)))))) (setf result (cons this-component result)) (loop for w in this-component do (setf remaining (delete w remaining))))) result)) (defun main-1 (filename) (let* ((points (read-inputs filename)) (adjacency (adjacency-map (short-labeled-edges points 1000))) (components (components (car (array-dimensions points)) adjacency)) (lengths (sort (mapcar #'length components) #'>))) (* (car lengths) (cadr lengths) (caddr lengths)))) (defun fusing-edge (n labeled-edges) (let ((sorted-edges (sort labeled-edges #'< :key #'first)) (components (make-hash-table))) (loop for i from 0 to (1- n) do (setf (gethash i components) (list i))) (loop for (nil v w) in sorted-edges do (let ((new-component (union (gethash v components) (gethash w components)))) (cond ((= (length new-component) n) (return (list v w))) ((not (subsetp new-component (gethash v components))) (loop for x in new-component do (setf (gethash x components) new-component)))))))) (defun main-2 (filename) (let* ((points (read-inputs filename)) (n (car (array-dimensions points)))) (destructuring-bind (v w) (fusing-edge n (all-labeled-edges points)) (* (aref points v 0) (aref points w 0)))))C++
Both parts in 40 ms. As always, the real AoC with C++ is parsing the input, although having a number that changes in the puzzle messes up my test/ solution runners.
Uses Boost’s implementations of sets, since they’re just plain faster than the STL ones.
Code
#include <boost/log/trivial.hpp> #include <boost/unordered/unordered_flat_set.hpp> #include <filesystem> #include <fstream> #include <stdexcept> namespace { struct Point { int x, y, z; }; auto distance(const Point &a, const Point &b) { auto &[ax, ay, az] = a; auto &[bx, by, bz] = b; ssize_t dx = ax - bx; ssize_t dy = ay - by; ssize_t dz = az - bz; return size_t(dx * dx + dy * dy + dz * dz); } using PList = std::vector<Point>; using DMap = std::vector<std::pair<std::pair<size_t, size_t>, size_t>>; using DSet = boost::unordered::unordered_flat_set<size_t>; auto read() { auto rval = PList{}; auto ih = std::ifstream{"08.txt"}; auto line = std::string{}; while (std::getline(ih, line)) { auto c1 = line.find(','); auto c2 = line.find(',', c1 + 1); rval.emplace_back( std::stoi(line.substr(0, c1)), std::stoi(line.substr(c1 + 1, c2 - c1 - 1)), std::stoi(line.substr(c2 + 1)) ); } return rval; } auto dmap(const PList &t) { auto rval = DMap{}; for (auto a = size_t{}; a < t.size() - 1; ++a) for (auto b = a + 1; b < t.size(); ++b) rval.emplace_back(std::make_pair(a, b), distance(t.at(a), t.at(b))); return rval; } auto find(size_t c, std::vector<DSet> &sets) { for (auto i = sets.begin(); i != sets.end(); ++i) if (i->contains(c)) return i; throw std::runtime_error{"can't find"}; } auto part1(const PList &t, size_t count, const DMap &d) { auto circuits = std::vector<DSet>{}; for (auto i = size_t{}; i < t.size(); ++i) { auto c = DSet{t.size()}; c.insert(i); circuits.push_back(std::move(c)); } for (auto i = size_t{}; i < count; ++i) { auto a = d.at(i).first.first; auto b = d.at(i).first.second; auto ac = find(a, circuits); auto bc = find(b, circuits); if (ac == bc) continue; ac->insert(bc->begin(), bc->end()); circuits.erase(bc); } std::sort(circuits.begin(), circuits.end(), [](auto &a, auto &b) { return a.size() > b.size(); }); auto total = size_t{1}; for (auto i = size_t{}; i < 3; ++i) total *= circuits.at(i).size(); return total; } auto part2(const PList &t, const DMap &d) { auto circuits = std::vector<DSet>{}; for (auto i = size_t{}; i < t.size(); ++i) { auto c = DSet{t.size()}; c.insert(i); circuits.push_back(std::move(c)); } for (auto i = size_t{}; i < d.size(); ++i) { auto a = d.at(i).first.first; auto b = d.at(i).first.second; auto ac = find(a, circuits); auto bc = find(b, circuits); if (ac == bc) continue; ac->insert(bc->begin(), bc->end()); circuits.erase(bc); if (circuits.size() == 1) { return ssize_t(t.at(a).x) * ssize_t(t.at(b).x); } } throw std::runtime_error{"never joins?"}; } } // namespace auto main() -> int { auto t = read(); BOOST_LOG_TRIVIAL(info) << "Day 8: read " << t.size(); auto test = std::filesystem::current_path().filename().string() == std::string{"test"}; auto d = dmap(t); std::sort(d.begin(), d.end(), [](auto &a, auto &b) { return a.second < b.second; }); BOOST_LOG_TRIVIAL(info) << "1: " << part1(t, test ? 10 : 1000, d); BOOST_LOG_TRIVIAL(info) << "2: " << part2(t, d); }Futhark
As always, futhark does not support arbitrary inputs, so I have a sed script to transform the input to something readable.
input transformer
it produces a textual representation of
[][3]u32, try it on your example or input :]1i [ 1,$ { s/^/[/ s/$/]/ } 2,$i, $i ] $dCalculate all the distances (even the redundant ones, I had no idea on how to filter them out). Sort them, keep only the first 1000 for part 1. Keep all for part two. Initialize all boxes to be in no component. Add them to components as time goes on. When connecting two boxes already in a component. Mark all boxes in the second component as part of the first one. Stop when everything is connected.
After improving my implementation of
concatMap(preallocate the entire array), the overall performance improved greatly. My end stats are- Time: 7s -> 0.35s
- Memory: 2GB -> 66MB
Program Source
Basic
import "lib/github.com/diku-dk/sorts/radix_sort" type position = (u32, u32, u32) def positionFromArray (p: [3]u32): position = (p[0], p[1], p[2]) def pair_iota (n: i64): [n](i64, i64) = map (\ j -> (n, j)) (iota n) def gaussian_sum (n: i64) = n * (n + 1) / 2 def euclidean_distance (a: position) (b: position): f64 = f64.sqrt ( (f64.u32 a.0 - f64.u32 b.0) ** 2 + (f64.u32 a.1 - f64.u32 b.1) ** 2 + (f64.u32 a.2 - f64.u32 b.2) ** 2 ) def distance_table [n] (positions: [n]position): [n][n]f64 = let distance_function = \ i j -> euclidean_distance positions[i] positions[j] in tabulate_2d n n distance_function def existsLength 'a 'b (f: a -> ?[l].[l]b) (x: a): i64 = length (f x) def concatMap [n] 'a 'b (f: a -> ?[l].[l]b) (placeholder: b) (xs: [n]a): *[]b = let totalLength = reduce (+) 0 <| map (\ x -> length (f x)) xs in ( loop (results, offset) = (replicate totalLength placeholder, 0) for x in xs do let bs = f x in let scatterIndices = indices bs |> map (+offset) in (scatter results scatterIndices bs, offset + length bs) ).0 def distance_array [n] (positions: [n]position): []((i64, i64), f64) = let table = distance_table positions in let triangle_indices = concatMap pair_iota (i64.highest, i64.highest) (iota n |> drop 1) in map (\ (i, j) -> ((i, j), table[i, j])) triangle_indices def sort_distances (distances: []((i64, i64), f64)): []((i64, i64), f64) = radix_sort_float_by_key (.1) f64.num_bits f64.get_bit distances type option 'a = #Empty | #Present a def empty 'a : option a = #Empty def overrideWith (old: u16) (new: u16) (x: option u16): option u16 = match x case #Empty -> #Empty case #Present inner -> if inner == old then #Present new else #Present inner def orElse 'a (o: option a) (d: a): a = match o case #Empty -> d case #Present x -> x def is_present 'a (o: option a): bool = match o case #Empty -> false case #Present _ -> true def connect (circuits: *[](option u16)) (newCircuitId: u16) (connection: (i64, i64)): (u16, *[](option u16)) = let circuitA = circuits[connection.0] in let circuitB = circuits[connection.1] in match (circuitA, circuitB) case (#Empty, #Empty) -> ( newCircuitId + 1 , scatter circuits [connection.0, connection.1] (rep (#Present newCircuitId)) ) case (#Present a, #Empty) -> ( newCircuitId , scatter circuits [connection.1] [#Present a] ) case (#Empty, #Present b) -> ( newCircuitId , scatter circuits [connection.0] [#Present b] ) case (#Present a, #Present b) -> ( newCircuitId , map (b `overrideWith` a) circuits ) def countCircuit (counts: *[]u64) (o: option u16): *[]u64 = match o case #Empty -> counts case #Present i -> scatter counts [i64.u16 i] [counts[i64.u16 i] + 1] def countCircuits (c: u16) (circuits: [](option u16)): *[i64.u16 c]u64 = let circuitCounts = replicate (i64.u16 c) 0 in loop counts = circuitCounts for circuit in circuits do countCircuit counts circuit def exampleConnectionCount = 10i64 def inputConnectionCount = 1000i64 def part1 (positions: i64) (connectionCount: i64) (distances: []((i64, i64), f64)) = let connections = take connectionCount distances |> map (.0) in let circuitMap: *[positions](option u16) = replicate positions empty in ( loop (circuitCount, circuits) = (0, circuitMap) for connection in connections do connect circuits circuitCount connection ) |> uncurry countCircuits |> radix_sort u64.num_bits u64.get_bit |> reverse |> take 3 |> foldl (*) 1 def part2 (positionCount: i64) (distances: []((i64, i64), f64)) (positions: []position) = let circuitMap: *[positionCount](option u16) = replicate positionCount empty in ( loop (circuitCount, connectionIndex, circuits) = (0, 0, circuitMap) while not ( and (map is_present circuits) && and (map (== circuits[0]) circuits) ) do let connection = distances[connectionIndex].0 in let (newCircuitId, circuits') = connect circuits circuitCount connection in (newCircuitId, connectionIndex+1, circuits') ).1 |> \ i -> distances[i-1].0 |> \ (a, b) -> positions[a].0 * positions[b].0 def main [n] (position_array: [n][3]u32) = let positions = map positionFromArray position_array in let unsorted_distances = distance_array positions in let sorted_distances = sort_distances unsorted_distances in ( part1 n inputConnectionCount sorted_distances , part2 n sorted_distances positions )C
Got stuck for a bit on part 1 on a silly mistake in the group (circuit) merging code, where the nodes from one group are reassigned to the other:
for (i=0; i < nnodes; i++) if (nodes[i].group == pair->b->group) nodes[i].group = pair->a->group;At some point in the loop
pair->b.groupitself is updated, from then on the check is against the new group value. Oops.In the end, my solution’s runtime on my (10 year old!) PC is about 160 ms for both parts, which is more than I would like, so maybe I’ll look into better set representations.
Code
#include <stdio.h> #include <stdlib.h> #include <inttypes.h> #include <assert.h> #define LEN(a) (sizeof(a)/sizeof(*(a))) #define NPAIRS(x) ((x)*((x)-1)/2) #define MAXN 1024 struct node { int x,y,z, group; }; struct pair { struct node *a, *b; int64_t dist_sq; }; struct group { int count; }; static struct node nodes[MAXN]; static struct pair pairs[NPAIRS(MAXN)]; static struct group groups[MAXN]; static int nnodes; static int ngroups; static int64_t node_dist_sq(const struct node *a, const struct node *b) { return (int64_t)(a->x - b->x) * (a->x - b->x) + (int64_t)(a->y - b->y) * (a->y - b->y) + (int64_t)(a->z - b->z) * (a->z - b->z); } static int cmp_pairs(const void *va, const void *vb) { const struct pair *a = va; const struct pair *b = vb; return a->dist_sq < b->dist_sq ? -1 : a->dist_sq > b->dist_sq ? 1 : 0; } static int cmp_groups_asc(const void *va, const void *vb) { const struct group *a = va; const struct group *b = vb; return b->count - a->count; } static void merge_groups(int group_a, int group_b) { int i; if (group_a == group_b) return; groups[group_a].count += groups[group_b].count; groups[group_b].count = 0; for (i=0; i<nnodes; i++) if (nodes[i].group == group_b) nodes[i].group = group_a; ngroups--; } int main() { int p1=0,p2=0, p1_limit, i,j, n,p; for (; ; nnodes++) { assert(nnodes < MAXN); n = scanf(" %d,%d,%d", &nodes[nnodes].x, &nodes[nnodes].y, &nodes[nnodes].z); if (n < 3) break; nodes[nnodes].group = nnodes; groups[nnodes].count = 1; } ngroups = nnodes; for (p=0, i=0; i<nnodes-1; i++) for (j=i+1; j<nnodes; j++, p++) { pairs[p].a = &nodes[i]; pairs[p].b = &nodes[j]; pairs[p].dist_sq = node_dist_sq(&nodes[i], &nodes[j]); } qsort(pairs, NPAIRS(nnodes), sizeof(*pairs), cmp_pairs); p1_limit = nnodes <= 100 ? 10 : 1000; for (i=0; ngroups > 1; i++) { merge_groups(pairs[i].a->group, pairs[i].b->group); if (ngroups == 1) p2 = pairs[i].a->x * pairs[i].b->x; if (i == p1_limit) { qsort(groups, LEN(groups), sizeof(*groups), cmp_groups_asc); p1 = groups[0].count * groups[1].count * groups[2].count; } } printf("08: %d %d\n", p1, p2); }I think we basically did the same bug, but I did it in rust.
160ms is respectable, I am at 300ms. Probably should remove my sqrt.
Rust
Part 1 took a decent while, partly life getting in the way, partly as I was struggling with some Rust things like floats not being sortable without mucking around, plus some weird bugs trying to get
collectto do everything that I eventually just rewrote to avoid.I found the noisy_float crate which let me wrap f64 as a “real” r64 which meant no NaN which meant Ord which meant
sort_by_cached_key()and the rest worked.I’d planned how to partition the closest neighbour search, but the whole thing runs in 24ms so I didn’t bother.
other types
type Id = usize; type Connection = (Id, Id); type Circuit = HashSet<Id>; #[derive(PartialEq)] struct Point { x: usize, y: usize, z: usize, } impl FromStr for Point { type Err = Report; fn from_str(s: &str) -> Result<Self> { let mut parts = s.split(','); Ok(Point { x: parts.next().ok_or_eyre("missing x")?.parse()?, y: parts.next().ok_or_eyre("missing y")?.parse()?, z: parts.next().ok_or_eyre("missing z")?.parse()?, }) } } impl Point { fn distance(&self, other: &Point) -> R64 { let dist = (( self.x.abs_diff(other.x).pow(2) + self.y.abs_diff(other.y).pow(2) + self.z.abs_diff(other.z).pow(2)) as f64) .sqrt(); r64(dist) } }struct Boxes(Vec<Point>); impl Boxes { fn closest(&self) -> Vec<Connection> { let mut closest = (0..self.0.len()) .flat_map(|a| ((a + 1)..self.0.len()).map(move |b| (a, b))) .collect::<Vec<_>>(); closest.sort_by_cached_key(|&(a, b)| self.0[a].distance(&self.0[b])); closest } fn connect_all(&self, p1_threshold: usize) -> Result<(usize, usize)> { let mut circuits: Vec<Circuit> = (0..self.0.len()) .map(|id| HashSet::from_iter(iter::once(id))) .collect(); let mut closest = self.closest().into_iter(); let mut p1 = 0; let mut n = 0; loop { n += 1; let (a, b) = closest.next().ok_or_eyre("All connected already")?; let a_circ = circuits.iter().position(|c| c.contains(&a)); let b_circ = circuits.iter().position(|c| c.contains(&b)); match (a_circ, b_circ) { (None, None) => { circuits.push(vec![a, b].into_iter().collect()); } (None, Some(idx)) => { circuits[idx].insert(a); } (Some(idx), None) => { circuits[idx].insert(b); } (Some(a_idx), Some(b_idx)) if a_idx != b_idx => { let keep_idx = a_idx.min(b_idx); let rem_idx = a_idx.max(b_idx); let drained = circuits.swap_remove(rem_idx); // this position is still valid since we removed the later set circuits[keep_idx].extend(drained); } _ => { /* already connected to same circuit */ } }; if n == p1_threshold { circuits.sort_by_key(|set| set.len()); circuits.reverse(); p1 = circuits.iter().take(3).map(|set| set.len()).product(); } if circuits.len() == 1 { let p2 = self.0[a].x * self.0[b].x; return Ok((p1, p2)); } } } }Kotlin
First tricky puzzle today, instantly leading me to more or less brute force for part2. I took a lot of time to understand which data structures I needed today, and how to compute my matrix.
Runtimes:
part1: 141ms
part2: 45.9 seconds .-.Solution
class Day08 : Puzzle { lateinit var boxes: List<Point3D> lateinit var edges: List<Pair<List<Point3D>, Double>> override fun readFile() { val input = readInputFromFile(2025, 8, false) boxes = input.lines().filter { it.isNotBlank() } .map { it.split(",") } .map { Point3D(it[0].toInt(), it[1].toInt(), it[2].toInt()) } edges = calculateEdges(boxes) } private fun calculateEdges(boxes: List<Point3D>): List<Pair<List<Point3D>, Double>> { val edges = mutableListOf<Pair<MutableList<Point3D>, Double>>() for (i in boxes.indices) { for (j in i + 1 until boxes.size) { edges.add(Pair(mutableListOf(boxes[i], boxes[j]), boxes[i].dist(boxes[j]))) } } return edges.sortedBy { it.second } } override fun solvePartOne(): String { val connections = buildEmptyConnectionMatrix(boxes.size) val mutableEdges = edges.toMutableList() for (i in 0..<1000) { connectNextEdge(mutableEdges, connections) } val connectionSet = buildConnectionSet(boxes, connections) return connectionSet.map { it.size } .sortedByDescending { it } .take(3) .reduce(Int::times) .toString() } override fun solvePartTwo(): String { val connections = buildEmptyConnectionMatrix(boxes.size) val mutableEdges = edges.toMutableList() var result: Long? = null while (result == null) { result = connectNextEdge(mutableEdges, connections, true) } return result.toString() } // size: width & height of (square) matrix private fun buildEmptyConnectionMatrix(size: Int): Array<Array<Boolean>> { val connections = Array(size) { Array(size) { false } } for (i in connections.indices) { connections[i][i] = true } return connections } private fun connectNextEdge(mutableEdges: MutableList<Pair<List<Point3D>, Double>>, connections: Array<Array<Boolean>>, part2: Boolean = false): Long? { if (mutableEdges.isEmpty()) return null val next = mutableEdges[0] val point = next.first[0] val other = next.first[1] connectAll(boxes.indexOf(point), boxes.indexOf(other), connections) mutableEdges.remove(next) // all nodes are connected, assume that this is happening for the first time return if (part2 && connections[0].all { it }) { next.first[0].x.toLong() * next.first[1].x.toLong() } else { null } } private fun connectAll(index: Int, other: Int, connections: Array<Array<Boolean>>) { fun connectHelper(hIndex: Int) { val newConnections = mutableSetOf<Int>() for (i in connections[hIndex].indices) { if (connections[hIndex][i]) newConnections.add(i) } for (boxIndex in newConnections.filter { it != hIndex }) { for (conn in newConnections.filter { it != boxIndex }) { connections[boxIndex][conn] = true } } } connections[index][other] = true connectHelper(index) // update matrix with all values from node at [index] connectHelper(other) // update matrix with all values from node at [other] } // returns 2D-list of all indices of currently active connections private fun buildConnectionSet(boxes: List<Point3D>, connections: Array<Array<Boolean>>): Set<List<Int>> { val connectionSet = mutableSetOf<List<Int>>() val indices = (0 until boxes.size).toMutableList() while (indices.isNotEmpty()) { val list = mutableListOf<Int>() val curr = indices.removeFirst() val array = connections[curr] for (j in array.indices) { if (array[j]) list.add(j) } connectionSet.add(list) } return connectionSet } data class Point3D(val x: Int, val y: Int, val z: Int) { fun dist(other: Point3D): Double { return sqrt( (x - other.x).toDouble().pow(2) + (y - other.y).toDouble().pow(2) + (z - other.z).toDouble().pow(2) ) } } }full code on Codeberg
That is a very unoptimal pt2 time! Have you tried profiling to see what’s causing the slowdown? Mine is 300ms, or worst case, 4s to process all links.
My algorithm is probably pretty horrible, I was purely out to get a solution as fast as possible. I might try improving it after Day12…
That’s interesting. I found part 1 to be the hard part and part 2 was just simplifying part 1. They ran in about the same time for me, too.
Part 1 run time:
Input IO: 0m 0s 10ms Input Parse: 0m 0s 21ms Algorithm: 0m 0s 250ms Total: 0m 0s 281msPart 2 run time:
Input IO: 0m 0s 11ms Input Parse: 0m 0s 22ms Algorithm: 0m 0s 255ms Total: 0m 0s 288msCode:
const val connectionsLimit = 1000 const val circuitsLimit = 3 fun part1() { val input = getInput(8) val circuits: MutableList<MutableSet<Triple<Int, Int, Int>>> = parseInput1(input) val distances: MutableList<Pair<Pair<Triple<Int, Int, Int>, Triple<Int, Int, Int>>, Double>> = mutableListOf() val tempList = circuits.toList() for (i in tempList.indices) { val left = tempList[i].first() for (j in i + 1..<tempList.size) { val right = tempList[j].first() val distance = calculateDistance(left, right) val coordinates = left to right distances.add(coordinates to distance) } } distances.sortBy { it.second } // Parts 1 and 2 are the same until here for (i in 0..<connectionsLimit) { val distance = distances[i] val leftCircuit = circuits.first { it.contains(distance.first.first) } val rightCircuit = circuits.first { it.contains(distance.first.second) } if (leftCircuit != rightCircuit) { leftCircuit.addAll(rightCircuit) circuits.remove(rightCircuit) } } val sizes = circuits.map { it.size.toLong() }.sortedDescending().slice(0..<circuitsLimit) val total = sizes.reduce { acc, i -> acc * i } println(total) } fun part2() { val input = getInput(8) val circuits: MutableList<MutableSet<Triple<Int, Int, Int>>> = parseInput1(input) val distances: MutableList<Pair<Pair<Triple<Int, Int, Int>, Triple<Int, Int, Int>>, Double>> = mutableListOf() val tempList = circuits.toList() for (i in tempList.indices) { val left = tempList[i].first() for (j in i + 1..<tempList.size) { val right = tempList[j].first() val distance = calculateDistance(left, right) val coordinates = left to right distances.add(coordinates to distance) } } distances.sortBy { it.second } // Part 2 differs starting here var answer = 0 for (distance in distances) { val leftCircuit = circuits.first { it.contains(distance.first.first) } val rightCircuit = circuits.first { it.contains(distance.first.second) } if (leftCircuit != rightCircuit) { leftCircuit.addAll(rightCircuit) circuits.remove(rightCircuit) if (circuits.size == 1) { answer = distance.first.first.first * distance.first.second.first break } } } println(answer) } fun parseInput1(input: String): MutableList<MutableSet<Triple<Int, Int, Int>>> { return input.lines() .filter { it.isNotBlank() } .map { val split = it.split(",") mutableSetOf(Triple(split[0].toInt(), split[1].toInt(), split[2].toInt())) } .toMutableList() } fun calculateDistance(left: Triple<Int, Int, Int>, right: Triple<Int, Int, Int>): Double { val dx = (left.first - right.first).toDouble() val dy = (left.second - right.second).toDouble() val dz = (left.third - right.third).toDouble() val distanceSquared = dx.pow(2) + dy.pow(2) + dz.pow(2) return sqrt(distanceSquared) }Oh wow, I see you went the Kotlin way of hacking together an object with Pairs and Triples. I used to do that too but then it got too confusing lol
It can be confusing, for sure. But I really like Pairs and Triples.
answer = distance.first.first.first * distance.first.second.firstis not very elegant, lol.
Go
God damn it, I thought I would never see the end of part 1. I had a hard time finding a good representation and then I failed at not eliminating valid distances etc. The code is quite messy, but I did it!!!
day08.go
package main import ( "aoc/utils" "cmp" "math" "slices" "strconv" "strings" ) type pos3D struct { x, y, z int } func (p pos3D) Compare(other pos3D) int { dx := cmp.Compare(p.x, other.x) if dx != 0 { return dx } dy := cmp.Compare(p.y, other.y) if dy != 0 { return dy } return cmp.Compare(p.z, other.z) } func (p pos3D) distance(other pos3D) float64 { dx := float64(other.x - p.x) dy := float64(other.y - p.y) dz := float64(other.z - p.z) d2 := math.Pow(dx, 2.0) + math.Pow(dy, 2.0) + math.Pow(dz, 2.0) return math.Sqrt(d2) } func getPosChannel(input chan string) chan pos3D { ch := make(chan pos3D, cap(input)) go func() { for line := range input { parts := strings.Split(line, ",") x, _ := strconv.Atoi(parts[0]) y, _ := strconv.Atoi(parts[1]) z, _ := strconv.Atoi(parts[2]) ch <- pos3D{x, y, z} } close(ch) }() return ch } type circuits struct { circuits map[pos3D]int nextID int } func (cc *circuits) newCircuit() (id int) { id = cc.nextID cc.nextID++ return id } func (cc *circuits) mergeCircuits(id1, id2 int) { for p, id := range cc.circuits { if id == id2 { cc.circuits[p] = id1 } } } func createSingletonCircuits(points []pos3D) (cc circuits) { cc.circuits = make(map[pos3D]int) for _, p := range points { id := cc.newCircuit() cc.circuits[p] = id } return cc } func (cc circuits) reverseMap() (m map[int][]pos3D) { m = make(map[int][]pos3D) for p, id := range cc.circuits { if _, ok := m[id]; !ok { m[id] = []pos3D{} } m[id] = append(m[id], p) } return m } func (cc circuits) sizeMap() map[int]int { circuitSizeMap := make(map[int]int) for _, id := range cc.circuits { circuitSizeMap[id]++ } return circuitSizeMap } type targetedDistance struct { distance float64 p1, p2 pos3D } func (p pos3D) distanceWithAll( points []pos3D, alreadyVisited *map[pos3D]any, ) (tds []targetedDistance) { tds = []targetedDistance{} for _, op := range points { if _, ok := (*alreadyVisited)[op]; ok { continue } var td targetedDistance td.distance = math.MaxFloat64 td.p1 = p d := p.distance(op) if d < td.distance { td.distance = d td.p2 = op } tds = append(tds, td) } return tds } type pointPair [2]pos3D func (pp pointPair) equals(other pointPair) bool { pp1 := pp[0] pp2 := pp[1] o1 := other[0] o2 := other[1] return (pp1 == o1 && pp2 == o2) || (pp1 == o2 && pp2 == o1) } var IterationCount = 1000 func stepOne(input chan string) (int, error) { ch := getPosChannel(input) points := []pos3D{} for p := range ch { points = append(points, p) } slices.SortFunc(points, pos3D.Compare) cc := createSingletonCircuits(points) alreadyVisited := make(map[pos3D]any) tds := []targetedDistance{} for _, p := range points { alreadyVisited[p] = nil dsts := p.distanceWithAll(points, &alreadyVisited) tds = append(tds, dsts...) } slices.SortFunc(tds, func(a, b targetedDistance) int { return cmp.Compare(a.distance, b.distance) }) for idx := range IterationCount { td := tds[idx] cc.mergeCircuits(cc.circuits[td.p1], cc.circuits[td.p2]) } circuitSizeMap := cc.sizeMap() circuitSizes := []int{} for _, v := range circuitSizeMap { circuitSizes = append(circuitSizes, v) } slices.Sort(circuitSizes) largestThree := circuitSizes[len(circuitSizes)-3:] product := 1 for _, v := range largestThree { product *= v } return product, nil } func stepTwo(input chan string) (int, error) { ch := getPosChannel(input) points := []pos3D{} for p := range ch { points = append(points, p) } slices.SortFunc(points, pos3D.Compare) cc := createSingletonCircuits(points) alreadyVisited := make(map[pos3D]any) tds := []targetedDistance{} for _, p := range points { alreadyVisited[p] = nil dsts := p.distanceWithAll(points, &alreadyVisited) tds = append(tds, dsts...) } slices.SortFunc(tds, func(a, b targetedDistance) int { return cmp.Compare(a.distance, b.distance) }) idx := 0 var lastConnection pointPair for { td := tds[idx] idx++ cc.mergeCircuits(cc.circuits[td.p1], cc.circuits[td.p2]) circuitSizeMap := cc.sizeMap() if len(circuitSizeMap) == 1 { lastConnection = pointPair{td.p1, td.p2} break } } return lastConnection[0].x * lastConnection[1].x, nil } func main() { inputFile := utils.FilePath("day08.txt") utils.RunStep(utils.ONE, inputFile, stepOne) utils.RunStep(utils.TWO, inputFile, stepTwo) }Nim
view code
type AOCSolution[T,U] = tuple[part1: T, part2: U] Vec3 = tuple[x,y,z: int] Node = ref object pos: Vec3 cid: int proc dist(a,b: Vec3): float = sqrt(float((a.x-b.x)^2 + (a.y-b.y)^2 + (a.z-b.z)^2)) proc solve(input: string, p1_limit: int): AOCSolution[int, int] = let boxes = input.splitLines().mapIt: let parts = it.split(',') let pos = Vec3 (parseInt parts[0], parseInt parts[1], parseInt parts[2]) Node(pos: pos, cid: -1) var dists: seq[(float, (Node, Node))] for i in 0 .. boxes.high - 1: for j in i+1 .. boxes.high: dists.add (dist(boxes[i].pos, boxes[j].pos), (boxes[i], boxes[j])) var curcuits: Table[int, HashSet[Node]] var curcuitID = 0 dists.sort(cmp = proc(a,b: (float, (Node, Node))): int = cmp(a[0], b[0])) for ind, (d, nodes) in dists: var (a, b) = nodes let (acid, bcid) = (a.cid, b.cid) if acid == -1 and bcid == -1: # new curcuit a.cid = curcuitID b.cid = curcuitID curcuits[curcuitId] = [a, b].toHashSet inc curcuitID elif bcid == -1: # add to a b.cid = acid curcuits[acid].incl b elif acid == -1: # add to b a.cid = bcid curcuits[bcid].incl a elif acid != bcid: # merge two curcuits for node in curcuits[bcid]: node.cid = acid curcuits[acid].incl curcuits[bcid] curcuits.del bcid if ind+1 == p1_limit: result.part1 = curcuits.values.toseq.map(len).sorted()[^3..^1].prod if not(acid == bcid and acid != -1): result.part2 = a.pos.x * b.pos.xRuntime: 364 ms
Part 1:
I compute all pairs of Euclidean distances between 3D points, sort them, then connect points into circuits, using, what I think is called a Union‑Find algorithm (circuits grow or merge). After exactly 1000 connections (including redundant ones), I take the three largest circuits and multiply their sizes.Part 2:
While iterating through the sorted connections, I also calculate the product of each pair x‑coordinates. The last product is a result for part 2.Problems I encountered while doing this puzzle:
- I’ve changed a dozen of data structures, before settled on curcuitIDs and
ref objects stored in a HashTable (~ 40 min) - I did a silly mistake of mutating the fields of an object and then using new fields as keys for the HashTable (~ 20 min)
- I am stil confused and don’t understand why do elves count already connected junction boxes (~ 40 min, had to look it up, otherwise it would be a lot more)
Time to solve Part 1: 1 hour 56 minutes
Time to solve Part 2: 4 minutesFull solution at Codeberg: solution.nim
Upvote for mentioning union-find. It’s the little algorithm that could, and almost nobody has heard about it.
- I’ve changed a dozen of data structures, before settled on curcuitIDs and
Rust
It’s getting spicier, luckily part 2 wasn’t really much additional complexity this time. There exists a pretty fancy union-find data structure which would have made representing the subcircuits much faster, but I went with a lazy approach.
Code
use euclid::default::Point3D; use euclid::point3; fn parse_input(input: &str) -> Vec<Point3D<i64>> { input .lines() .map(|l| { let mut parts = l.split(',').map(|p| p.parse::<i64>().unwrap()); let (x, y, z) = ( parts.next().unwrap(), parts.next().unwrap(), parts.next().unwrap(), ); point3(x, y, z) }) .collect() } // Distances between all points. Reflexive and symmetric pairs are skipped, // so the Vec's have increasing size, starting at 0. fn dists(points: &[Point3D<i64>]) -> Vec<Vec<i64>> { points .iter() .enumerate() .map(|(idx, &p1)| { points .iter() .take(idx) .map(|&p2| (p2 - p1).square_length()) .collect::<Vec<i64>>() }) .collect() } fn sorted_distances(dists: &[Vec<i64>]) -> Vec<(usize, usize, i64)> { let mut sorted: Vec<(usize, usize, i64)> = dists .iter() .enumerate() .flat_map(|(i, row)| row.iter().enumerate().map(move |(j, d)| (i, j, *d))) .collect(); sorted.sort_by_key(|(_, _, d)| *d); sorted } fn part1(input: String) { let points = parse_input(&input); let d = dists(&points); let sorted = sorted_distances(&d); let mut circuits: Vec<u32> = (0..points.len() as u32).collect(); for (i, j, _) in sorted.into_iter().take(1000) { let new_circuit = circuits[i]; let old_circuit = circuits[j]; if new_circuit != old_circuit { for c in circuits.iter_mut() { if *c == old_circuit { *c = new_circuit; } } } } let mut sizes: Vec<u32> = vec![0; points.len()]; for c in circuits { sizes[c as usize] += 1 } sizes.sort_unstable(); let result = sizes.iter().rev().take(3).product::<u32>(); println!("{result}"); } fn part2(input: String) { let points = parse_input(&input); let d = dists(&points); let sorted = sorted_distances(&d); let mut circuits: Vec<u32> = (0..points.len() as u32).collect(); for (i, j, _) in sorted.into_iter() { let new_circuit = circuits[i]; let old_circuit = circuits[j]; if new_circuit != old_circuit { let mut all_connected = true; for c in circuits.iter_mut() { if *c == old_circuit { *c = new_circuit; } if *c != new_circuit { all_connected = false; } } if all_connected { let result = points[i].x * points[j].x; println!("{result}"); return; } } } } util::aoc_main!();Just a messy part1 so far. Plus a moment of pure rage when I realised the trap. (I have no idea what algorithm Lemmy uses to highlight Uiua code, but it’s always a surprise.)
(edit: part 2 now added. I had a stupid error earlier. Takes 12s native or 20+s in the pad.)
Stupid error
I added elements just until the set first hit size 1, rather than continuing until total number of elements equalled number of points.
Run it here(for what it’s worth)
# AOC 2025 Day 08 "162,817,812\n57,618,57\n906,360,560\n592,479,940\n352,342,300\n466,668,158\n542,29,236\n431,825,988\n739,650,466\n52,470,668\n216,146,977\n819,987,18\n117,168,530\n805,96,715\n346,949,466\n970,615,88\n941,993,340\n862,61,35\n984,92,344\n425,690,689" N ← 10 L ← 20 # &fras"AOC2025day08.txt"◌ # Uncomment these three lines, # N ← 1000 # drop your file onto this pane and correct that # L ← 1000 # filename, to run this against your data. # You will need to change settings>execution time to 30s or more. Row ← ( ⊙(⊃⊢↘₁) ⊚◡≡⌟◇(/+˜∊) ⨬(⊂⊙□◌ # Not found: new circuit | ⍜⊡(⍜°□(◴⊂))⊢ # One found: add | ▽>₀⊸≡◇⧻⍜(⊏|⊂{[]}{∘}◴/◇⊂) # Connect two )⊸⧻◴ ) Parse ← ⍆⋕°csv Sort ← ⊏⍏/+ⁿ2/-⊸⍉⊸⧅>2 IndexPrep ← {[°⊟]}⊸°⊂⊸≡⌟₁˜⨂ Part₁ ← ⊙◌/×↙3⇌⍆≡◇⧻⍥Row(-1N) Part₂ ← /×⊢⍉⊏⊣↘¯⧻◌⍢(Row|<L/+≡◇⧻) ⊃(&p⍜nowPart₁)(&p⍜nowPart₂) &p⍜now(IndexPrep Sort Parse)the trap
Re. what connections to process or not? That seems to be like one of those that’s either completely obvious when you implement your solution one way, and a nasty pitfall when you do it another. In this case, pre-computing the list of pairs to process vs. finding the next one when you need it.
spoiler
Stupider than that: I’d just got halfway through writing a complicated divide and conquer algorithm to calculate nearest pairs when I realised that I could just brute-force it. :-)
But yes, precomputing was definitely the way to go.
Haskell
They’re getting interesting now!
import Control.Monad import Data.List import Data.List.Split import Data.Ord import Data.Set qualified as Set readPos = (\([x, y, z] :: [Int]) -> (x, y, z)) . map read . splitOn "," pairs = init . tails >=> (\(x : xs) -> map (x,) xs) dist (x1, y1, z1) (x2, y2, z2) = (x2 - x1) ^ 2 + (y2 - y1) ^ 2 + (z2 - z1) ^ 2 connect circuits (a, b) = let (joined, rest) = partition (\c -> a `Set.member` c || b `Set.member` c) circuits in Set.unions joined : rest main = do boxes <- map readPos . lines <$> readFile "input08" let steps = (zip <*> tail . scanl' connect (map Set.singleton boxes)) $ sortOn (uncurry dist) (pairs boxes) print . product . take 3 . sortOn Down . map Set.size $ (snd . last . take 1000 $ steps) let Just (((x1, _, _), (x2, _, _)), _) = find ((== 1) . length . snd) steps in print $ x1 * x2This is crazy concise and fast! Impressive.
Thank you! ☺️
Python
Both of my solutions make use of a Heap and a Disjoint-Set-Union / Union-Find data structure
Part 1: Use a heap to keep 1000 shortest connections, then union them all in DSU and get the 3 largest connected components.
Part 2: Use a heap to keep all connections, then union the shortest ones until all components are connected.click to view code
import heapq as hq from itertools import combinations from collections import Counter # Disjoint Set Union (or Union-Find) data structure # with path compression and union by rank class DSU: def __init__(self, size: int) -> None: self.parent = list(range(size)) # ith value is the parent node to node i self.rank = [1] * size # max depth of subtree rooted here (used for union by rank) def find(self, x: int) -> int: # if the node is not its own parent, we need to recurse on parent if x != self.parent[x]: # path compression self.parent[x] = self.find(self.parent[x]) return self.parent[x] def isConnected(self, x: int, y: int) -> bool: return self.find(x) == self.find(y) # returns a boolean whether or not union was needed def union(self, x: int, y: int) -> bool: rootX = self.find(x) rootY = self.find(y) if rootX == rootY: # no union needed return False if self.rank[rootX] > self.rank[rootY]: # rootX has deeper subtree, therefore set it as parent to rootY (and its subtree) self.parent[rootY] = rootX elif self.rank[rootX] < self.rank[rootY]: # rootY has deeper subtree, therefore set it as parent to rootX (and its subtree) self.parent[rootX] = rootY else: # both subtrees are of equal depth, therefore choose either one of them as the parent of the other # here we chose rootX as the parent of rootY, therefore rootX depth increases by 1 self.parent[rootY] = rootX self.rank[rootX] += 1 # union complete return True # parse puzzle input into junction coordinates (x, y, z) def parse_junctions(data: str): return [tuple(map(int, line.split(','))) for line in data.splitlines()] # get distance between two points in 3D space # skipping the sqrt because we only care about relative distances def get_dist(p1, p2): x1, y1, z1 = p1 x2, y2, z2 = p2 return (x2 - x1) ** 2 + (y2 - y1) ** 2 + (z2 - z1) ** 2 def part1(data: str, max_connections: int = 1000): junctions = parse_junctions(data) n = len(junctions) # <max_connections> shortest connections closest_connections = [] # iterate over all pairs of junctions # combinations is just some syntactic sugar to avoid a nested for loop (still O(n^2)) for i, j in combinations(range(n), 2): dist = get_dist(junctions[i], junctions[j]) # keep <max_connections> closest connections # have to use -dist to simulate max heap if len(closest_connections) < max_connections: hq.heappush(closest_connections, (-dist, i, j)) else: hq.heappushpop(closest_connections, (-dist, i, j)) # union all the closest connections dsu = DSU(n) for _, i, j in closest_connections: dsu.union(i, j) # count all circuit sizes circuit_sizes = Counter() for i in range(n): circuit_sizes[dsu.find(i)] += 1 # multiply the sizes of the 3 largest circuits ans = 1 for _, f in circuit_sizes.most_common(3): ans *= f return ans def part2(data: str): junctions = parse_junctions(data) n = len(junctions) # all n^2 junction connections all_connections = [] # combinations is just some syntactic sugar to avoid a nested for loop (still O(n^2)) for i, j in combinations(range(n), 2): dist = get_dist(junctions[i], junctions[j]) all_connections.append((dist, i, j)) # create min heap of all connections # Heapify later to benefit from the heapify algorithm (O(n) instead of O(n log n)) hq.heapify(all_connections) # connect junctions until all are connected dsu = DSU(n) unions = 0 while all_connections: _, i, j = hq.heappop(all_connections) unions += dsu.union(i, j) # if we have n-1 unions, that means all n junctions are connected if unions == n-1: return junctions[i][0] * junctions[j][0] assert False, "It's not possible that all junctions never connect" sample = """162,817,812 57,618,57 906,360,560 592,479,940 352,342,300 466,668,158 542,29,236 431,825,988 739,650,466 52,470,668 216,146,977 819,987,18 117,168,530 805,96,715 346,949,466 970,615,88 941,993,340 862,61,35 984,92,344 425,690,689""" assert part1(sample, max_connections=10) == 40 assert part2(sample) == 25272It seems like you forgot the backticks around the code. It’s very hard to read this way. Also python comments look like markdown headlines :]
Fixed, thanks!
I went here for graphs (to find connected components) and binary search (for efficiency)
#!/usr/bin/env python3 from pathlib import Path import numpy as np import networkx as nx cwd = Path(__file__).parent.resolve() def parse_input(file_path): with file_path.open("r") as fp: coords = list(map(lambda x: list(map(int, x.split(','))), fp.readlines())) return np.array(coords) def binary_search(G, dist): Itriu = np.triu_indices(dist.shape[0], k=1) dist = dist[Itriu] components = list(nx.connected_components(G)) Isort = [(Itriu[0][i], Itriu[1][i]) for i in np.argsort(dist)] icurrent = 0 ihist = [] nhist = [] while len(ihist)<2 or ihist[-1]!=ihist[-2]: ihist.append(icurrent) G.add_edges_from(Isort[:icurrent]) components = list(nx.connected_components(G)) nhist.append(len(components)) G.remove_edges_from(Isort[:icurrent]) if len(components)>1: # if >1 component, go between this and closest next index with 1 component I = [ind for ind,n in enumerate(nhist) if n==1] if len(I)==0: inext = len(Isort) else: inext = min(np.array(ihist)[I]) icurrent = icurrent + int(np.ceil((inext-icurrent)/2)) else: # if 1 component, go between nearest previous >1 components and this index I = [ind for ind,n in enumerate(nhist) if n>1] if len(I)==0: iprev = 0 else: iprev = max(np.array(ihist)[I]) icurrent = iprev + int(np.ceil((icurrent-iprev)/2)) return Isort[icurrent-1] def solve_problem(file_name, nconnect): coordinates = parse_input(Path(cwd, file_name)) G = nx.Graph() G.add_nodes_from(range(coordinates.shape[0])) dist = np.sqrt(np.sum((coordinates[None,:] - coordinates[:,None])**2, axis=-1)) if nconnect != -1: # part 1 Itriu = np.triu_indices(dist.shape[0], k=1) dist = dist[Itriu] Isort = [(Itriu[0][i], Itriu[1][i]) for i in np.argsort(dist)[:nconnect]] G.add_edges_from(Isort) components = sorted(list(nx.connected_components(G)), key=len)[::-1] return np.prod(list(map(len, components[:3]))) else: # part 2 indices = binary_search(G, dist) return np.prod(coordinates[indices,:],axis=0)[0] if __name__ == "__main__": assert solve_problem("test_input", 10) == 40 assert solve_problem("input", 1000) == 115885 assert solve_problem("test_input", -1) == 25272 assert solve_problem("input", -1) == 274150525Rust
Not an easy one, mostly because I messed up my merge circuits function (Pop/remove first circuit, load into second, but the pop/remove messed up the indexes, so I was merging the wrong thing).
After getting that it was all good. Pt2 was trivial once pt1 was solved.
spoiler
#[derive(Debug)] struct Circuits { circuits: Vec<Vec<usize>>, } impl Circuits { fn find_circuit(&mut self, i: usize) -> Option<usize> { for (j, c) in &mut self.circuits.iter().enumerate() { if c.contains(&i) { return Some(j); } } None } fn new_circuit(&mut self, i: usize, j: usize) { self.circuits.push(vec![i, j]); } fn join_circuits(&mut self, i: usize, j: usize) { let mut other = self.circuits.get(j).unwrap().clone(); self.circuits.get_mut(i).unwrap().append(&mut other); self.circuits.remove(j); } fn append_circuit(&mut self, c: usize, x: usize) { self.circuits.get_mut(c).unwrap().push(x); } fn top_three(&mut self) -> Vec<usize> { let mut sizes = self .circuits .iter() .map(|c| c.len()) .collect::<Vec<usize>>(); sizes.sort(); sizes.reverse(); sizes[0..3].to_vec() } } type Pole = (isize, isize, isize); fn dist_poles(p1: &Pole, p2: &Pole) -> f32 { sqrt( ((p1.0 - p2.0) * (p1.0 - p2.0)) as f32 + ((p1.1 - p2.1) * (p1.1 - p2.1)) as f32 + ((p1.2 - p2.2) * (p1.2 - p2.2)) as f32, ) } #[test] fn test_y2025_day8_part1() { let input = include_str!("../../input/2025/day_8.txt"); let poles = input .lines() .map(|l| { let [x, y, z] = l .splitn(3, ",") .map(|c| c.parse::<isize>().unwrap()) .collect::<Vec<isize>>()[..] else { panic!(); }; (x, y, z) }) .collect::<Vec<Pole>>(); let len = poles.len(); let mut circuits: Circuits = Circuits { circuits: vec![] }; let mut pairs = vec![]; for i in 0..len { let first = poles.get(i).unwrap(); for j in i + 1..len { if i == j { continue; } let second = poles.get(j).unwrap(); let dist = dist_poles(first, second); pairs.push((dist, i, j)); } } pairs.sort_by(|a, b| { if a.0 < b.0 { Ordering::Less } else { Ordering::Greater } }); for (dist, a, b) in pairs[0..1000].iter() { let first_circuit = circuits.find_circuit(*a); let second_circuit = circuits.find_circuit(*b); match (first_circuit, second_circuit) { (None, None) => { circuits.new_circuit(*a, *b); } (Some(c), None) => { circuits.append_circuit(c, *b); } (None, Some(c)) => { circuits.append_circuit(c, *a); } (Some(c1), Some(c2)) => { if c1 != c2 { circuits.join_circuits(c1, c2); } } } } assert_eq!(circuits.top_three().iter().product::<usize>(), 66640) }








