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:sparkles: soving problems with Haskell
Problems solved were Reverse Polish Notation and Heathrow to London
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Haskell/solvingProblems.org
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Haskell/solvingProblems.org
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* Reverse Polish Notation Calculator
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[[https://en.wikipedia.org/wiki/Reverse_Polish_notation][Link for the wikipedia for the explanation]]
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#+begin_src haskell
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import Data.List
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#+end_src
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#+RESULTS:
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#+begin_src haskell
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:{
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solveRPN :: String -> Float
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solveRPN = head . foldl foldingFunction [] . words
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where foldingFunction (x:y:ys) "*" = (x * y):ys
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foldingFunction (x:y:ys) "+" = (x + y):ys
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foldingFunction (x:y:ys) "-" = (y - x):ys
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foldingFunction (x:y:ys) "/" = (y / x):ys
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foldingFunction (x:y:ys) "^" = (y ** x):ys
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foldingFunction (x:xs) "ln" = log x:xs
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foldingFunction xs "sum" = [sum xs]
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foldingFunction xs numberString = read numberString:xs
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:}
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solveRPN "10 10 10 10 sum 4 /"
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#+end_src
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#+RESULTS:
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: Prelude Data.List> 10.0
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* Heathrow to London
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Your plane just landed in England and you rent a car.
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You have a meeting really soon and you have to get fro Heathrow Airport to London as fast as you can (but safely!).
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There are two main roads going from Heathrow to London and there's a number of regional roads crossing them.
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It takes you a fixed amount of time to travel from one crossroads to another.
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It's up to you to find the optimal path to take so that you get to London as fast as you can!
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You start on the lest side and can either cross to the other main road or go forward.
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Our job is to make a program that takes input that represents a road system and print out what the shortert path across it is.
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Here's what the input would look like for this care:
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#+begin_example
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50
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10
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30
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5
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90
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40
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2
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25
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10
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8
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0
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#+end_example
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To mentally parse the input file, read it in threes and mentally split the road system into sections.
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Each section is comprised of a road A, road B and a crossing road.
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To have it neatly fit into threes, we say that there's a last crossing section that takes 0 minutes to drive over.
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That's because we don't care where we arrive in London, as long as we're in London.
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To get the best path we do this: first we see what the best path to the next crossroads on main road A is.
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The two options are going directly forward or starting at the opposite road, going forward and then crossing over.
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We remember the cost and the path. We use the same method to see what the best path to the next crossroads on main road B is and remember that.
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Then, we see if the path to the next crossroads on A is cheaper if we go prom the previous A crossroads or if we to from the previous B crossroads and then cross over.
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We remember the cheaper path and then we to the same for the crossroads opposite of it.
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We do this for every section until we reach the end.
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Once we've reachead the end, the cheapest of the two paths that we have is our optimal path!
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So in essence, we keep one shortest path on the road A and one path on the B road and when we reach the end, the shorter of those two ir our path.
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#+begin_src haskell
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:{
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data Section = Section { getA :: Int, getB :: Int, getC :: Int } deriving (Show)
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type RoadSystem = [Section]
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heathrowToLondon :: RoadSystem
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heathrowToLondon = [Section 50 10 30, Section 5 90 20, Section 40 2 25, Section 10 8 0]
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data Label = A | B | C deriving (Show)
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type Path = [(Label, Int)]
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roadStep :: (Path, Path) -> Section -> (Path, Path)
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roadStep (pathA, pathB) (Section a b c) =
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let priceA = sum $ map snd pathA
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priceB = sum $ map snd pathB
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forwardPriceToA = priceA + a
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crossPriceToA = priceB + b + c
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forwardPriceToB = priceB + b
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crossPriceToB = priceA + a + c
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newPathToA = if forwardPriceToA <= crossPriceToA
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then (A,a):pathA
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else (C,c):(B,b):pathB
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newPathToB = if forwardPriceToB <= crossPriceToB
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then (B,b):pathB
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else (C,c):(A,a):pathA
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in (newPathToA, newPathToB)
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optimalPath :: RoadSystem -> Path
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optimalPath roadSystem =
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let (bestAPath, bestBPath) = foldl roadStep ([], []) roadSystem
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in if sum (map snd bestAPath) <= sum (map snd bestBPath)
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then reverse bestAPath
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else reverse bestBPath
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resolveProblem :: RoadSystem
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resolveProblem roadSystem =
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let path = optimalPath roadSystem
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pathString = concat $ map (show . fst) path
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pathPrice = sum $ map snd path
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putStrLn $ "The best path to take is: " ++ pathString
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putStrLn $ "The price is: " ++ pathPrice
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:}
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resolveProblem heathrowToLondon
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#+end_src
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#+RESULTS:
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: Prelude Data.List>
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: <interactive>:208:1-14: error:
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: Variable not in scope: resolveProblem :: RoadSystem -> t