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)
Code block support is not fully rolled out yet but likely will be in the middle of the event. Try to share solutions as both code blocks and using something such as https://topaz.github.io/paste/ , pastebin, or github (code blocks to future proof it for when 0.19 comes out and since code blocks currently function in some apps and some instances as well if they are running a 0.19 beta)
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
Actually, all my comments in Part 2 make it too big for a single post. The full solution consists of two functions (that could be combined with a bit more work...).
The first function is a brute-force approach, that has a similar termination condition to the first part. It halts, once all paths are walking in circles. This function does not find a solution yet. Then there is a second function that looks for solutions that appear later, while all paths are already walking in circles. That function writes all goals as equations of the form x = start + n * period. The target condition is that all paths have the same value for x. This was then solved by looking at all permutations between two paths, and solving the Diophantine equations, creating a new, combined, path.
In the end, the closest resulting valid goal was also the final result.
Part2, the Brute Force function
--------------------------------------------------------------------------------------------------------
-- okay, part 2 is nasty.
-- what do we know?
-- o) All paths we follow simultaneously have the same path length, as they have common instructions.
-- x) This means that once we walk in circles on all of them, we are lost.
-- -> That's the way to convince the compiler this program terminates.
-- o) We could use the cycle detection to rule out short, cycled paths.
-- x) Once a path is in a cycle, the targets repeat at cycle-lenght interval.
-- x) I doubt that this would be much faster than brute-force though.
-- let's try brute force first.
private def parallelPass (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (alreadyDoneSteps : Nat) (currentPositions : List WaypointId) (instructions : List Instruction) : Except (Option (Nat × (List WaypointId))) Nat := do
if currentPositions.all λw ↦ w.endsWith "Z" then
return alreadyDoneSteps
match instructions with
| [] => throw $ some (alreadyDoneSteps, currentPositions)
| a :: as =>
let currentWaypoint := currentPositions.mapM waypoints.find?
match currentWaypoint with
| none => throw none -- should not happen
| some currentWaypoints =>
let nextWaypoints := currentWaypoints.map $ flip ConnectedWaypoints.get a
parallelPass waypoints (alreadyDoneSteps + 1) nextWaypoints as
private def totalRemainingStarts (s : List (List WaypointId)) : Nat :=
s.foldl (·+·.length) 0
private def part2_impl (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (instructions : List Instruction) (alreadyDoneSteps : Nat) (possibleStarts : List (List WaypointId)) (currentPositions : List WaypointId) : Option Nat :=
let remainingStarts := (possibleStarts.zip currentPositions).map λs ↦ s.fst.filter λt ↦ t != s.snd
-- I _really_ don't want to prove stuff by hand... Luckily this works.
if totalRemainingStarts remainingStarts < totalRemainingStarts possibleStarts then
let passResult := parallelPass waypoints alreadyDoneSteps currentPositions instructions
match passResult with
| Except.ok n => some n
| Except.error none => none -- dead end on map. Should not be possible.
| Except.error $ some n => part2_impl waypoints instructions n.fst remainingStarts n.snd
else
none -- walking in circles
termination_by part2_impl a b c d e => totalRemainingStarts d
And last, but not least, here's the part that actually finds the result:
Part 2, the part for after the brute force approach fails (don't be alarmed, it's mostly comments, very few actual code lines)
-- Okay, tried Brute Force, it did NOT work. Or rather, it might work, but I won't be able to prove
-- termination for it. Not that it wouldn't be possible to prove, just I won't manage to.
-- The problem is that the termination condition in part2_impl is too soon.
-- You can actually see this in the example (for which part2_impl works, but by chance).
-- While the goals in each part repeat, they repeat at different rates.
-- Soo, we would need to continue even after each part has started walking in circles.
-- However, just doing that runs for a very long time without finding a result.
-- Sooo, let's be smarter.
--
-- Every path consist of 2 segments. The part that leads up to a cycle, and the cycle.
-- Both parts can contain goals, but once the cycle starts, the goals repeat with cycle-length.
-- A cycle is at least one pass, but can (and does...) consist of multiple passes too.
-- We can use part2_impl still - to verify that we do not reach the goals before all our paths start
-- cycling. That allows us to only care about cycling paths in the second part of the solution, which
-- we only reach if part2_impl does not yield a result (we know it doesn't, but that would be cheating).
-- Soo, how can the second part look like?
-- For simplicity let's not do this in parallel. Rather, let's find the goals for each start individually
-- So, let's just consider a single path (like the one from part1)
-- We need to write down the number of steps at which we reach a goal.
-- Whenever we remove a remaining start from the possible starts list, we need to note it down, and
-- how many steps it took us to get there.
-- Once we detect a circle, we can look up
-- how many steps we took in total till we startec cycling
-- and how many steps it took us to reach the cycle start for the first time
-- that's the period of each goal in the cycle.
-- For each goal that was found between cycle-start and cycle-end, we can write down an equation:
-- x = steps_from_start + cycle_length * n
-- n is a Natural number here, not an integer. x is the number of steps at which we pass this goal
-- Once we have that information for all goals of all starts, we can combine it:
-- That's a set of Diophantine equations.
--
-- Or, rather, several sets of Diophantine equations...
-- For each combination of goals that are reached in the cycles of the participating paths, we need to
-- solve the following system:
--
-- We can write each goal for each run in the form x = g0 + n * cg
-- Where x is the solution we are looking for, g0 is the number of steps from the start until
-- we hit the goal for the first time **in the cycle**, and cg is the cycle length
--
-- Once we have those equations, we can combine them pairwise: https://de.wikipedia.org/wiki/Lineare_diophantische_Gleichung
-- This allows us to reduce all paths to a single one, which has multiple equations that
-- describe when a goal is reached.
-- For each of those equations we need to find the first solution that is larger than
-- the point where all paths started to cycle. The smallest of those is the result.
-- a recurring goal, that starts at "start" and afterwards appears at every "interval".
private structure CyclingGoal where
start : Nat
interval : Nat
deriving BEq
instance : ToString CyclingGoal where
toString g := s!"`g = {g.start} + n * {g.interval}`"
private def CyclingGoal.nTh (goal : CyclingGoal) (n : Nat) : Nat :=
goal.start + n * goal.interval
-- Combine two cycling goals into a new cycling goal. This might fail, if they never meet.
-- This can for instance happen if they have the same frequency, but different starts.
private def CyclingGoal.combine (a b : CyclingGoal) : Option CyclingGoal :=
-- a.start + n * a.interval = b.start + m * b.interval
-- n * a.interval - m * b.interval = b.start - a.start
-- we want to do as much as possible in Nat, such that we can easily reason about which numbers are
-- positive. Soo
let (a, b) := if a.start > b.start then (b,a) else (a,b)
let (g, u, _) := Euclid.xgcd a.interval b.interval
-- there is no solution if b.start - a.start is not divisible by g
let c := (b.start - a.start)
let s := c / g
if s * g != c then
none
else
let deltaN := b.interval / g
let n0 := s * u -- we can use u directly - v would need its sign swapped, but we don't use v.
-- a.start + (n0 + t * deltaN)*a.interval
-- a.start + n0*a.interval + t * deltaN * a.interval
-- we need the first value of t that yields a result >= max(a.start, b.start)
-- because that's where our cycles start.
let x := ((max a.start b.start : Int) - a.interval * n0 - a.start)
let interval := a.interval * deltaN
let t0 := x / interval
let t0 := if t0 * interval == x || x < 0 then t0 else t0 + 1 -- int division rounds towards zero, so for x < 0 it's already ceil.
let start := a.start + n0 * a.interval + t0 * deltaN * a.interval
assert! (start ≥ max a.start b.start)
let start := start.toNat
some {start, interval }
private def findCyclingGoalsInPathPass (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (alreadyDoneSteps : Nat) (currentPosition : WaypointId) (instructions : List Instruction) (visitedGoals : List Nat) : Option (Nat × WaypointId × (List Nat)) := do
let visitedGoals := if currentPosition.endsWith "Z" then
alreadyDoneSteps :: visitedGoals
else
visitedGoals
match instructions with
| [] => some (alreadyDoneSteps, currentPosition, visitedGoals)
| a :: as =>
let currentWaypoint := waypoints.find? currentPosition
match currentWaypoint with
| none => none -- should not happen
| some currentWaypoint => findCyclingGoalsInPathPass waypoints (alreadyDoneSteps + 1) (currentWaypoint.get a) as visitedGoals
private def findCyclingGoalsInPath_impl (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (instructions : List Instruction) (possibleStarts : List WaypointId) (visitedGoals : List Nat) (visitedStarts : List (WaypointId × Nat)) (currentPosition : WaypointId) (currentSteps : Nat) : List CyclingGoal :=
let remainingStarts := possibleStarts.filter λs ↦ s != currentPosition
if remainingStarts.length < possibleStarts.length then -- written this way to make termination_by easy
let visitedStarts := (currentPosition, currentSteps) :: visitedStarts
let passResult := findCyclingGoalsInPathPass waypoints currentSteps currentPosition instructions visitedGoals
match passResult with
| none => [] -- should not happen. Only possible if there's a dead end
| some (currentSteps, currentPosition, visitedGoals) => findCyclingGoalsInPath_impl waypoints instructions remainingStarts visitedGoals visitedStarts currentPosition currentSteps
else
let beenHereWhen := visitedStarts.find? λs ↦ s.fst == currentPosition
let beenHereWhen := beenHereWhen.get!.snd --cannot possibly fail
let cycleLength := currentSteps - beenHereWhen
visitedGoals.filterMap λ n ↦ if n ≥ beenHereWhen then
some {start := n, interval := cycleLength : CyclingGoal}
else
none -- goal was reached before we started to walk in cycles, ignore.
termination_by findCyclingGoalsInPath_impl a b c d e f g => c.length
private def findCyclingGoalsInPath (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (instructions : List Instruction) (possibleStarts : List WaypointId) (startPosition : WaypointId) : List CyclingGoal :=
findCyclingGoalsInPath_impl waypoints instructions possibleStarts [] [] startPosition 0
-- returns the number of steps needed until the first _commmon_ goal that cycles is found.
private def findFirstCommonCyclingGoal (waypoints : Lean.HashMap WaypointId ConnectedWaypoints) (instructions : List Instruction) (possibleStarts : List WaypointId) (startPositions : List WaypointId) : Option Nat :=
let cyclingGoals := startPositions.map $ findCyclingGoalsInPath waypoints instructions possibleStarts
let combinedGoals : List CyclingGoal := match cyclingGoals with
| [] => []
| g :: gs => flip gs.foldl g λc n ↦ c.bind λ cc ↦ n.filterMap λ nn ↦ nn.combine cc
let cyclingGoalStarts := combinedGoals.map CyclingGoal.start
cyclingGoalStarts.minimum?
open Lean in
def part2 (input : List Instruction × List Waypoint) : Option Nat :=
let possibleStarts := input.snd.map Waypoint.id
let waypoints : HashMap WaypointId ConnectedWaypoints := HashMap.ofList $ input.snd.map λw ↦ (w.id, {left := w.left, right := w.right : ConnectedWaypoints})
let instructions := input.fst
let positions : List WaypointId := (input.snd.filter λ(w : Waypoint) ↦ w.id.endsWith "A").map Waypoint.id
part2_impl waypoints instructions 0 (positions.map λ_ ↦ possibleStarts) positions
<|> -- if part2_impl fails (it does), we need to dig deeper.
findFirstCommonCyclingGoal waypoints instructions possibleStarts positions
Actually, all my comments in Part 2 make it too big for a single post. The full solution consists of two functions (that could be combined with a bit more work...).
The first function is a brute-force approach, that has a similar termination condition to the first part. It halts, once all paths are walking in circles. This function does not find a solution yet. Then there is a second function that looks for solutions that appear later, while all paths are already walking in circles. That function writes all goals as equations of the form
x = start + n * period
. The target condition is that all paths have the same value forx
. This was then solved by looking at all permutations between two paths, and solving the Diophantine equations, creating a new, combined, path.In the end, the closest resulting valid goal was also the final result.
Part2, the Brute Force function
And last, but not least, here's the part that actually finds the result:
Part 2, the part for after the brute force approach fails (don't be alarmed, it's mostly comments, very few actual code lines)