# Phone number

### From HaskellWiki

ColinHorne (Talk | contribs) (Added my own implementation) |
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import System.Random |
import System.Random |
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+ | import Control.Monad |
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import Test.QuickCheck |
import Test.QuickCheck |
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number = do |
number = do |
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len <- elements [0..10] |
len <- elements [0..10] |
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− | sequence $ replicate len $ elements alphabet |
+ | replicateM len $ elements alphabet |

numbers :: Gen [String] |
numbers :: Gen [String] |
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− | numbers = sequence $ replicate 1000 number |
+ | numbers = replicateM 1000 number |

makeTest :: IO () |
makeTest :: IO () |

## Revision as of 05:44, 21 February 2010

## 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 Control.Monad import Test.QuickCheck alphabet = ['0'..'9'] number :: Gen String number = do len <- elements [0..10] replicateM len $ elements alphabet numbers :: Gen [String] numbers = replicateM 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)

## 3 Another solution

By Colin Horne

I wrote my solution without first seeing Paul's; however they turn out to be very similar. It took about 2.75 hours to get a (almost!) working solution, and another 15 minutes, or so, to tidy it up (I have most probably not added sufficient comments, however).

Just before posting this, I realised that I had a subtle bug, in that I ignored the following part of the specification: "If and only if at a particular point no word at all from the dictionary can be inserted, a single digit from the phone number can be copied to the encoding instead."

It took me about an extra hour to find why my program was not producing the correct output, and how to fix it. The resulting total development time is about 4 hours, which is slightly less than the average Lisp/Scheme time.

module Main where import Data.Array import System (getArgs) import Data.Char (digitToInt, intToDigit, toLower, isLetter) import Data.Foldable (foldl') translationsList = [ (0, "e"), (1, "jnq"), (2, "rwx"), (3, "dsy"), (4, "ft"), (5, "am"), (6, "civ"), (7, "bku"), (8, "lop"), (9, "ghz")] {--- Lookup functions for converting from/to digits ---} numToChars :: Int -> String numToChars = (!) arr where arr = array (0,9) translationsList charToNum :: Char -> Int charToNum = (!) arr where arr = array ('a','z') convertAll convert :: (Int, String) -> [(Char, Int)] convert (i,s) = map (\c -> (c, i)) s convertAll = concatMap convert translationsList {--- The Trie data structure ---} data Trie = Node [String] (Array Int Trie) | Empty -- Add a node to the trie insert node str = insert' str node $ map toLower $ filter isLetter str insert' :: String -> Trie -> String -> Trie insert' str (Node strs ts) [] = Node (str:strs) ts insert' str (Node strs ts) (c:cs) = Node strs $! (ts // [(num, updatedNode)]) where updatedNode = insert' str (ts ! num) cs num = charToNum c insert' str Empty cs = insert' str (Node [] $ array (0,9) $ zip [0..9] $ repeat Empty) cs readDict :: FilePath -> IO Trie readDict file = fmap (trie . lines) $ readFile file trie :: [String] -> Trie trie strs = foldl' insert Empty strs {--- Other functions ---} -- Given a trie and a number, find the representations for each of -- the number's prefixes, returning each result as a tuple containing -- the prefix's representation, and its postfix getAllStrings :: Trie -> [Int] -> [([String], [Int])] getAllStrings Empty _ = [] getAllStrings (Node strs _) [] = [(strs, [])] getAllStrings (Node strs ts) xxs@(x:xs) = (getAllStrings (ts ! x) xs) ++ [(strs, xxs)] -- Lists all string-representations of the given number -- No further processing of this function's output is required numToStrings :: Trie -> [Int] -> [String] numToStrings trie num = map (dropWhile (== ' ')) $ loop [] True num where loop :: String -> Bool -> [Int] -> [String] loop prefix _ [] = [prefix] loop prefix allowDigit xxs@(x:xs) = case result of [] -> if allowDigit && noPartialMatch then loop (prefix++" "++[intToDigit x]) False xs else [] xs -> xs where noPartialMatch = and $ flip map allStrings $ (== []) . fst allStrings = getAllStrings trie xxs result = flip concatMap allStrings $ \(strs,rest) -> flip concatMap strs $ \str -> loop (prefix++" "++str) True rest -- Reads a number from the given string, -- returning a list of its digits readNumber :: String -> [Int] readNumber [] = [] readNumber (c:cs) | c >= '0' && c <= '9' = digitToInt c : rest | otherwise = rest where rest = readNumber cs main = do (dictFile:numbersFile:[]) <- getArgs dict <- readDict dictFile numbers <- fmap lines $ readFile numbersFile flip mapM_ numbers $ \num' -> do let num = readNumber num' let strs = numToStrings dict num flip mapM strs $ \str -> do putStrLn $ num' ++ ": " ++ str