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Phone number

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1 Original Program

By Paul Johnson

This program was written after I read "An Empirical Comparison of Seven Programming Languages" at http://www.cis.udel.edu/~silber/470STUFF/article.pdf.

To run this program, copy and paste each code block into a file with the appropriate name (so module Main goes into Main.hs) and compile. Depending on your OS you may need to modify the dictionary file name. You may also have to run it with "+RTS -k16000000" on the command line in order to increase the stack space. It depends on your dictionary.

First, the PhoneWord module. This contains the meat of the program.

 
module PhoneWord where
 
import Data.Array
import Data.Char
import Data.List
 
 
-- The number encoding, as specified in the problem statement.
numberPairs :: [(Int, String)]
numberPairs = [(0, "e"), (1, "jnq"), (2, "rwx"), (3, "dsy"), (4, "ft"),
   (5, "am"), (6, "civ"), (7, "bku"), (8, "lop"), (9, "ghz")]
 
 
-- Encode a character by looking it up in the inverse of the numberPairs
-- table.  Creating the inverse takes three lines of code (including the
-- 'where'), but we can live with that.  The alternative would be to invert
-- the list manually and then have a longer list of pairs above.
encode :: Char -> Int
encode c = codeTable ! (toLower c)
   where
      codeTable = array ('a', 'z') $ concatMap pairList numberPairs
      pairList (n, cs) = map (\c -> (c, n)) cs
 
-- The dictionary is stored in a denary tree.  A node has sub-trees, a leaf
-- does not.  At each node or leaf is the list of strings that encode to the
-- number of that location.
data DenaryTree =
   Node {strings :: [String], subTrees :: Array Int DenaryTree} |
   Leaf {strings :: [String]}
      deriving (Eq, Show)
 
-- Empty and unit trees.
emptyTree :: DenaryTree
emptyTree = Leaf []
 
unitTree :: String -> DenaryTree
unitTree str = Leaf [str]
 
 
-- Given an existing tree, create a new tree to be its parent.  Put the
-- existing tree at position 'n' in the new parent and put 'strs' in it.
newNode :: [String] -> Int -> DenaryTree -> DenaryTree
newNode strs n subtree = Node strs $ array (0,9) $
   (n, subtree) : [(i, emptyTree) | i <- [0..9], i /= n]
 
 
-- Add a new word to a tree, returning a new tree.
addWord :: String -> DenaryTree -> DenaryTree
 
-- The worker function is addWord1.  addWord just wraps it up
-- in a more convenient form.
addWord newString tree = addWord1 numbers tree
   where
      numbers = map encode newString
      addWord1 [] (Leaf oldStrs) = Leaf (newString : oldStrs)
      addWord1 [] (Node oldStrs arr) = Node (newString : oldStrs) arr
      addWord1 (n:ns) (Leaf oldStrs) = newNode oldStrs n $ newBranch ns
      addWord1 (n:ns) (Node oldStrs arr) =
         Node oldStrs $ arr // [(n, addWord1 ns $ arr ! n)]
      newBranch [] = unitTree newString
      newBranch (n:ns) = newNode [] n $ newBranch ns
 
 
-- Convert a list of words into a NumberTree
wordTree :: [String] -> DenaryTree
wordTree = foldr addWord emptyTree
 
 
-- Find the list of words which match a prefix of the number by descending
-- the tree until you either run out of tree or digits, and accumulating
-- the words as you go.
findNumber :: DenaryTree -> String -> [String]
 
findNumber tree [] = strings tree
 
findNumber (Node strs arr) (c:cs) =
   strs ++ findNumber (arr ! (digitToInt c)) cs
 
findNumber (Leaf strs) _ = strs
 
 
-- Find the list of solutions for a number.  This is rendered slightly
-- messy by the fact that one digit can be inserted if no other progress
-- is possible.  So if "Foo Bar" is a legal solution then "Foo 7 Ar" is not.
mnemonics :: DenaryTree -> String -> [String]
mnemonics tree numbers = map tail $ mnemonics1 True numbers
   where
      mnemonics1 _ "" = return " "
      mnemonics1 digitOK numbers =
         case findNumber tree numbers of
            [] ->
               if digitOK then do
                  nextBit <- mnemonics1 False (tail numbers)
                  return $ ' ' : head numbers : nextBit
               else
                  []
            ls -> do
               item <- ls
               nextBit <- mnemonics1 True $ drop (length item) numbers
               return $ ' ' : item ++ nextBit

Now the Main module with the IO in it:

module Main where
 
import Data.Char
import System.Environment
import System.IO
 
import PhoneWord
 
 
-- File containing the words.  This is the standard Unix dictionary.
dictFile = "/usr/share/dict/words"
 
-- Read the words file.  Return only those "words" that consist entirely
-- of letters and are at least three letters long.
dictWords :: IO [String]
dictWords =
   do
      text <- readFile dictFile
      return $ filter allowed $ lines text
   where
      allowed wrd =
         (and . (map isAlpha)) wrd &&
         length wrd >= 4
 
-- Read the numbers file.
numbers :: String -> IO [String]
numbers fileName = do
   text <- readFile fileName
   return $ map (filter isDigit) $ lines text
 
 
-- The Main Function executed when the program runs.
main :: IO ()
main =
   do
      args <- getArgs
      nums <- numbers $ head args
      dict <- dictWords
      let
         tree = wordTree dict
         results = map (\n -> (n, mnemonics tree n)) nums
      sequence_ $ concatMap printResult results
   where
      printResult (num, strings) =
         map (\str -> putStrLn $ num ++ ": " ++ str) strings

Finally, a little module to generate a random test file:

module MakeTest where
 
import System.Random
import Test.QuickCheck
 
alphabet = ['0'..'9']
 
number :: Gen String
number = do
   len <- elements [0..10]
   sequence $ replicate len $ elements alphabet
 
numbers :: Gen [String]
numbers = sequence $ replicate 1000 number
 
makeTest :: IO ()
makeTest = do
   rnd <- getStdGen
   writeFile "testData" $ unlines $ generate 1 rnd numbers

2 Shorter solution

By John Hamilton.

I heard about this problem from Peter Norvig's page Lisp as an Alternative to Java. (You should also check out http://www.flownet.com/ron/papers/lisp-java/ for more info.) I've been learning Haskell, and to see how it would compare with Lisp, I recently wrote the following program.

import Data.Char
import Data.List
import Data.Map (fromListWith, findWithDefault)
import System.Environment
 
encodeWord = map f . filter (/= '"') where
  f x = head $ [d | (s, d) <- ps, (toLower x) `elem` s]
  ps = zip ["e", "jnq", "rwx", "dsy", "ft", "am", "civ", "bku", "lop", "ghz"]
           ['0'..'9']
 
translate wordMap _ "" = [""]
translate wordMap digit xs@(x:xs') =
  if all null ys && digit
    then combine [[x]] (translate wordMap False xs')
    else concat $ zipWith combine ys zs
  where
    ys = [findWithDefault [] s wordMap | s <- (tail . inits) xs]
    zs = [translate wordMap True s | s <- tails xs']
    combine [] _ = []
    combine ys [""] = ys
    combine ys zs = [y ++ " " ++ z | y <- ys, z <- zs]
 
process wordMap n = [n ++ ": " ++ x | x <- xs] where
  xs = translate wordMap True $ filter (`notElem` "-/") n
 
main = do
  [dictionary, input] <- getArgs
  words <- readFile dictionary
  let wordMap = fromListWith (++) [(encodeWord w, [w]) | w <- lines words]
  numbers <- readFile input
  mapM_ putStrLn $ lines numbers >>= process wordMap

27 lines of code. Nice work. You have handily beaten both Common Lisp and Scheme.

I took 88 lines, of which 22 implemented a denary tree rather than using Data.Map. It's some time since I wrote the program, and I can't remember now why I did this. Some of the other 66 were module declarations and repeated imports (for two modules), and type declarations for all top-level functions. Finally there was the fact that I was using /usr/dict/words because I didn't have the original data set, and this required a certain amount of extra filtering. But even allowing for all that I think your solution is indeed shorter. PaulJohnson 20:05, 24 April 2006 (UTC)