130 lines
5.6 KiB
Haskell
Executable File
130 lines
5.6 KiB
Haskell
Executable File
#! /usr/bin/env -S"GHCRTS=-N4" nix-shell
|
||
#! nix-shell -p ghcid
|
||
#! nix-shell -p "haskellPackages.ghcWithPackages (p: with p; [pretty-simple attoparsec arithmoi])"
|
||
#! nix-shell -i "ghcid -c 'ghci' -T main"
|
||
|
||
{-# OPTIONS_GHC -Wall -Wincomplete-uni-patterns #-}
|
||
{-# LANGUAGE OverloadedStrings #-}
|
||
|
||
import Control.Applicative
|
||
import Control.Monad (foldM)
|
||
import Data.Attoparsec.Text (Parser)
|
||
import Data.Euclidean (gcdExt)
|
||
import Data.List (find,sortOn)
|
||
import Data.Maybe (fromMaybe,catMaybes)
|
||
import Text.Pretty.Simple
|
||
import qualified Data.Attoparsec.Text as A
|
||
import qualified Data.Text as T
|
||
|
||
exampleData :: String
|
||
exampleData = "939\n7,13,x,x,59,x,31,19"
|
||
|
||
numOrXParser :: Parser (Maybe Int)
|
||
numOrXParser = (Just <$> A.decimal) <|> ("x" *> pure Nothing)
|
||
|
||
inputParser :: Parser (Int,[Int])
|
||
inputParser = do
|
||
n <- A.decimal
|
||
A.skipSpace
|
||
xs <- numOrXParser `A.sepBy` ","
|
||
pure (n,catMaybes $ xs)
|
||
|
||
parseInput :: String -> Either String (Int,[Int])
|
||
parseInput = (A.parseOnly inputParser) . T.pack
|
||
|
||
solvePart1 :: String -> Either String Int
|
||
solvePart1 str = do
|
||
(n,xs) <- parseInput str
|
||
let (bus, time) = head
|
||
$ sortOn (snd)
|
||
$ map (\x -> (x,fromMaybe (-1) $ find (> n) [0,x..])) xs
|
||
pure $ bus * (time - n)
|
||
|
||
inputParser2 :: Parser (Int,[Maybe Int])
|
||
inputParser2 = do
|
||
n <- A.decimal
|
||
A.skipSpace
|
||
xs <- numOrXParser `A.sepBy` ","
|
||
pure (n,xs)
|
||
|
||
parseInput2 :: String -> Either String (Int,[Maybe Int])
|
||
parseInput2 = (A.parseOnly inputParser2) . T.pack
|
||
|
||
-- -------------------------------------------------------------------------- --
|
||
-- Here I sneak around a bit, and realize the problem is well defined --
|
||
-- (and solved!) already: it's called the Chinese Remainder Theorem --
|
||
-- https://en.wikipedia.org/wiki/Chinese_remainder_theorem --
|
||
-- -------------------------------------------------------------------------- --
|
||
-- Apllying the theorem allows to reduce a system of equation on x: --
|
||
-- --
|
||
-- x ≡ a1 (mod n1) --
|
||
-- · --
|
||
-- · --
|
||
-- · --
|
||
-- x ≡ ak (mod nk) --
|
||
-- --
|
||
-- to a single equation: --
|
||
-- --
|
||
-- x ≡ as (mod ns) --
|
||
-- --
|
||
-- It relates to the buses schedules in the following way: t is x, the bus --
|
||
-- number is the modulo factor (since a bus comes *every* ni) and subsequent --
|
||
-- additions to t (for other buses) is (-ai), so, for a but coming at --
|
||
-- t+ai, one would write x ≡ -ai (mod ni) --
|
||
-- --
|
||
-- I chose to encode ai and ni as a tuple (ai,ni), named startAndIds --
|
||
-- --
|
||
-- Basically, we're creating a “chinese” function: --
|
||
-- --
|
||
-- chinese :: (Int,Int) -> (Int,Int) -> (Int,Int) --
|
||
-- --
|
||
-- Then, given a list [(Int, Int)] we can fold over it to obtain the solution --
|
||
-- --
|
||
|
||
chinese :: (Integer,Integer) -> (Integer,Integer) -> Maybe (Integer,Integer)
|
||
chinese (0,n1) (0,n2) = chinese (n1,n1) (n2,n2)
|
||
chinese v (0,n2) = chinese v (n2,n2)
|
||
chinese (0,n1) v = chinese (n1,n1) v
|
||
chinese (a1,n1) (a2,n2) = do
|
||
-- Computes a solution such that: n1×c1 + n2×c2 = g, for some c2
|
||
-- n1×c1 - g = - n2×c2, for some c2
|
||
-- 1/n2 (n1×c1 - g) = - c2, for some c2 (n2 is > 0)
|
||
-- - 1/n2 (n1×c1 - g) = c2, for some c2 (n2 is > 0)
|
||
-- n1 and n2 must be coprimes for this to work (g must be 1), fail otherwise
|
||
(m1,m2) <- case gcdExt n1 n2 of
|
||
(1,c1) -> Just ( c1, negate ((n1 * c1) - 1) `div` n2 )
|
||
_ -> Nothing
|
||
let x = a1 * m2 * n2 + a2 * m1 * n1
|
||
let a12 = x `mod` (n1 * n2)
|
||
pure $ (a12, n1 * n2)
|
||
|
||
e2m :: Either e a -> Maybe a
|
||
e2m (Right v) = Just v
|
||
e2m _ = Nothing
|
||
|
||
solvePart2 :: String -> Maybe (Integer,Integer)
|
||
solvePart2 str = do
|
||
(_,xs) <- e2m $ parseInput2 str
|
||
let startAndIds = catMaybes $ sequence <$> zip [0..] (map (fmap fromIntegral) xs)
|
||
let chineseEqs = fmap (\(a,n) -> ((-a) `mod` n, n)) startAndIds
|
||
foldM chinese (1,1) chineseEqs
|
||
|
||
main :: IO ()
|
||
main = do
|
||
putStrLn ":: Test"
|
||
pPrint $ A.parseOnly inputParser $ T.pack exampleData
|
||
pPrint $ take 3 ((\n -> [1,(n::Integer)..]) 59)
|
||
putStrLn ":: Day 13 - Part 1"
|
||
input <- readFile "day13/input"
|
||
pPrint $ solvePart1 exampleData
|
||
pPrint $ solvePart1 input
|
||
putStrLn ":: Test 2"
|
||
print $ solvePart2 exampleData
|
||
print $ solvePart2 "1\n17,x,13,19"
|
||
print $ solvePart2 "1\n67,7,59,61"
|
||
print $ solvePart2 "1\n67,x,7,59,61"
|
||
print $ solvePart2 "1\n67,7,x,59,61"
|
||
print $ solvePart2 "1\n1789,37,47,1889"
|
||
putStrLn ":: Day 13 - Part 2"
|
||
print $ solvePart2 input
|