Explain day 24 part 2 solution

day24/src/Day24.hs

@@ -165,11 +165,47 @@ main = do
  [path] -> do
  raw <- readFile' path
 
- let input = mapMaybe parseHailstone (lines raw)
- xings = mapMaybe (uncurry intersect2) (pairs (fmap projectXY <$> input))
- bounds = Rect2 (both 200000000000000) (both 400000000000000)
- part1 = length (filter (inRect bounds) xings)
- p i = (input !! (i - 1)).pos
+ -- Part 1 explanation:
+ --
+ -- - Parse lines
+ -- - Compute 2D line intersections
+
+ let input = mapMaybe parseHailstone (lines raw)
+ xings = mapMaybe (uncurry intersect2) (pairs (fmap projectXY <$> input))
+ bounds = Rect2 (both 200000000000000) (both 400000000000000)
+ part1 = length (filter (inRect bounds) xings)
+
+ putStrLn $ "Part 1: " ++ show part1
+
+ -- Part 2 explanation:
+ --
+ -- - Let each hailstone i have position p[i] and velocity v[i], effectively a 3D line
+ -- - We're looking for a new line with position p and v, such that for i = 1, ..., n and some t[i]:
+ --
+ -- p + t[i] * v = p[i] + t[i] * v[i] (note that t[i] is the same on both sides due to the time constraint)
+ -- <=> p - p[i] = t[i] * (v[i] - v) (i.e. (p - p[i]) and (v[i] - v) are collinear...)
+ -- <=> 0 = (p - p[i]) x (v[i] - v) (...therefore we can rewrite this with the cross product)
+ -- = (p - p[i]) x (v - v[i])
+ -- = (p x v) - (p x v[i]) - (p[i] x v) + (p[i] x v[i])
+ --
+ -- - Note that (p x v) is the only nonlinear term (because it contains two unknowns)
+ -- - The clever trick is now that since (p x v) is the same for every i,
+ -- we can just take two of these vector equations and subtract the
+ -- nonlinear term (p x v) to get rid of it (i >= 2 henceforth):
+ --
+ -- (p x v) - (p x v[1]) - (p[1] x v) + (p[1] x v[1])
+ -- - ((p x v) - (p x v[i]) - (p[i] x v) + (p[i] x v[i]))
+ -- = - (p x v[1]) - (p[1] x v) + (p[1] x v[1])
+ -- + (p x v[i]) + (p[i] x v) - (p[i] x v[i])
+ --
+ -- - Remembering that each vector equation expands to three scalar equations and that
+ -- we have 6 unknowns (px, py, pz, vx, vy, vz), we only have to pick two of these
+ -- remaining equations (we'll go with i = 2 and i = 3) to obtain 6 equations.
+ -- - These 6 equations form a linear system of equations, which we can solve via Gaussian elimination.
+ -- 
+ -- Credits for helping me crack this problem go to u/evouga: https://www.reddit.com/r/adventofcode/comments/18pnycy/comment/kepu26z
+
+ let p i = (input !! (i - 1)).pos
  v i = (input !! (i - 1)).vel
  [px,py,pz,vx,vy,vz] = map (round @_ @Integer) . fromRight' $ solveLinearSystem @Double LinearSystem
  -- Unknowns: px py pz vx vy vz
@@ -188,9 +224,10 @@ main = do
  , (p 1).z * (v 1).y - (p 1).y * (v 1).z + (p 3).y * (v 3).z - (p 3).z * (v 3).y
  ]
  }
+ part2 = px + py + pz
+
+ putStrLn $ "Part 2: " ++ show part2 ++ " (p = " ++ show [px, py, pz] ++ ", v = " ++ show [vx, vy, vz] ++ ")"
 
- putStrLn $ "Part 1: " ++ show part1
- putStrLn $ "Part 2: " ++ show (px + py + pz) ++ " (p = " ++ show [px, py, pz] ++ ", v = " ++ show [vx, vy, vz] ++ ")"
  _ -> do
  putStrLn "Usage: day24 <path to input>"
  exitFailure