# 99 questions/90 to 94

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<description of implementation> |
<description of implementation> |
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− | |||

− | == Problem 95 == |
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− | |||

− | (**) English number words |
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− | |||

− | On financial documents, like cheques, numbers must sometimes be written in full words. Example: 175 must be written as one-seven-five. Write a predicate full-words/1 to print (non-negative) integer numbers in full words. |
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− | |||

− | Example in Haskell: |
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− | <pre> |
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− | > fullWords 175 |
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− | one-seven-five |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | import Data.List |
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− | import Data.Maybe |
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− | |||

− | fullWords :: Integer -> String |
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− | fullWords n = concat . intersperse "-" . map (fromJust . (`lookup` table)) $ show n |
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− | where table = [('0',"zero"), ('1',"one"), ('2',"two"), ('3',"three"), ('4',"four"), |
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− | ('5',"five"), ('6',"six"), ('7',"seven"), ('8',"eight"), ('9',"nine")] |
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− | </haskell> |
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− | |||

− | This solution does a simple table lookup after converting the positive integer into a string. Thus dividing into digits is much simplified. |
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− | |||

− | A minor variant of the above solution: |
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− | <haskell> |
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− | import Data.Char |
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− | import Data.List |
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− | |||

− | fullWords :: Integer -> String |
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− | fullWords n = concat $ intersperse "-" [digits!!digitToInt d | d <- show n] |
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− | where digits = ["zero", "one", "two", "three", "four", |
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− | "five", "six", "seven", "eight", "nine"] |
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− | </haskell> |
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− | |||

− | == Problem 96 == |
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− | |||

− | (**) Syntax checker |
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− | |||

− | In a certain programming language (Ada) identifiers are defined by the syntax diagram below. |
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− | |||

− | http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p96.gif |
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− | |||

− | Transform the syntax diagram into a system of syntax diagrams which do not contain loops; i.e. which are purely recursive. Using these modified diagrams, write a predicate identifier/1 that can check whether or not a given string is a legal identifier. |
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− | |||

− | Example in Prolog: |
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− | <pre> |
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− | % identifier(Str) :- Str is a legal identifier |
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− | </pre> |
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− | |||

− | Example in Haskell: |
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− | <pre> |
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− | > identifier "this-is-a-long-identifier" |
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− | True |
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− | > identifier "this-ends-in-" |
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− | False |
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− | > identifier "two--hyphens" |
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− | False |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | import Data.Char |
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− | syntax_check :: String -> Bool |
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− | syntax_check [] = False |
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− | syntax_check (x:xs) = isLetter x && loop xs |
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− | where loop [] = True |
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− | loop (y:ys) | y == '-' = (not . null) ys && isAlphaNum (head ys) && loop (tail ys) |
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− | | isAlphaNum y = loop ys |
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− | | otherwise = False |
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− | </haskell> |
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− | |||

− | Simple functional transcription of the diagram. |
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− | |||

− | Another direct transcription of the diagram: |
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− | <haskell> |
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− | identifier :: String -> Bool |
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− | identifier (c:cs) = isLetter c && hyphen cs |
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− | where hyphen [] = True |
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− | hyphen ('-':cs) = alphas cs |
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− | hyphen cs = alphas cs |
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− | alphas [] = False |
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− | alphas (c:cs) = isAlphaNum c && hyphen cs |
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− | </haskell> |
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− | |||

− | The functions <tt>hyphen</tt> and <tt>alphas</tt> correspond to states in the automaton at the start of the loop and before a compulsory alphanumeric, respectively. |
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− | |||

− | Here is a solution that parses the identifier using Parsec, a parser library that is commonly used in Haskell code: |
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− | <haskell> |
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− | isRight (Right _) = True |
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− | isRight (Left _) = False |
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− | |||

− | identifier x = isRight $ parse parser "" x where |
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− | parser = letter >> many (optional (char '-') >> alphaNum) |
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− | </haskell> |
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− | == Problem 97 == |
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− | |||

− | (**) Sudoku |
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− | |||

− | Sudoku puzzles go like this: |
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− | |||

− | <pre> |
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− | Problem statement Solution |
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− | |||

− | . . 4 | 8 . . | . 1 7 9 3 4 | 8 2 5 | 6 1 7 |
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− | | | | | |
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− | 6 7 . | 9 . . | . . . 6 7 2 | 9 1 4 | 8 5 3 |
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− | | | | | |
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− | 5 . 8 | . 3 . | . . 4 5 1 8 | 6 3 7 | 9 2 4 |
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− | --------+---------+-------- --------+---------+-------- |
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− | 3 . . | 7 4 . | 1 . . 3 2 5 | 7 4 8 | 1 6 9 |
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− | | | | | |
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− | . 6 9 | . . . | 7 8 . 4 6 9 | 1 5 3 | 7 8 2 |
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− | | | | | |
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− | . . 1 | . 6 9 | . . 5 7 8 1 | 2 6 9 | 4 3 5 |
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− | --------+---------+-------- --------+---------+-------- |
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− | 1 . . | . 8 . | 3 . 6 1 9 7 | 5 8 2 | 3 4 6 |
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− | | | | | |
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− | . . . | . . 6 | . 9 1 8 5 3 | 4 7 6 | 2 9 1 |
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− | | | | | |
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− | 2 4 . | . . 1 | 5 . . 2 4 6 | 3 9 1 | 5 7 8 |
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− | </pre> |
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− | |||

− | Every spot in the puzzle belongs to a (horizontal) row and a (vertical) column, as well as to one single 3x3 square (which we call "square" for short). At the beginning, some of the spots carry a single-digit number between 1 and 9. The problem is to fill the missing spots with digits in such a way that every number between 1 and 9 appears exactly once in each row, in each column, and in each square. |
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− | |||

− | Solutions: see [[Sudoku]] |
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− | |||

− | == Problem 98 == |
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− | |||

− | (***) Nonograms |
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− | |||

− | Around 1994, a certain kind of puzzle was very popular in England. The "Sunday Telegraph" newspaper wrote: "Nonograms are puzzles from Japan and are currently published each week only in The Sunday Telegraph. Simply use your logic and skill to complete the grid and reveal a picture or diagram." As a Prolog programmer, you are in a better situation: you can have your computer do the work! Just write a little program ;-). |
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− | |||

− | The puzzle goes like this: Essentially, each row and column of a rectangular bitmap is annotated with the respective lengths of its distinct strings of occupied cells. The person who solves the puzzle must complete the bitmap given only these lengths. |
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− | |||

− | Problem statement: Solution: |
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− | |||

− | |_|_|_|_|_|_|_|_| 3 |_|X|X|X|_|_|_|_| 3 |
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− | |_|_|_|_|_|_|_|_| 2 1 |X|X|_|X|_|_|_|_| 2 1 |
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− | |_|_|_|_|_|_|_|_| 3 2 |_|X|X|X|_|_|X|X| 3 2 |
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− | |_|_|_|_|_|_|_|_| 2 2 |_|_|X|X|_|_|X|X| 2 2 |
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− | |_|_|_|_|_|_|_|_| 6 |_|_|X|X|X|X|X|X| 6 |
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− | |_|_|_|_|_|_|_|_| 1 5 |X|_|X|X|X|X|X|_| 1 5 |
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− | |_|_|_|_|_|_|_|_| 6 |X|X|X|X|X|X|_|_| 6 |
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− | |_|_|_|_|_|_|_|_| 1 |_|_|_|_|X|_|_|_| 1 |
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− | |_|_|_|_|_|_|_|_| 2 |_|_|_|X|X|_|_|_| 2 |
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− | 1 3 1 7 5 3 4 3 1 3 1 7 5 3 4 3 |
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− | 2 1 5 1 2 1 5 1 |
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− | |||

− | |||

− | For the example above, the problem can be stated as the two lists [[3],[2,1],[3,2],[2,2],[6],[1,5],[6],[1],[2]] and [[1,2],[3,1],[1,5],[7,1],[5],[3],[4],[3]] which give the "solid" lengths of the rows and columns, top-to-bottom and left-to-right, respectively. Published puzzles are larger than this example, e.g. 25 x 20, and apparently always have unique solutions. |
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− | |||

− | Example in Haskell: |
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− | <pre> |
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− | Nonogram> putStr $ nonogram [[3],[2,1],[3,2],[2,2],[6],[1,5],[6],[1],[2]] [[1,2],[3,1],[1,5],[7,1],[5],[3],[4],[3]] |
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− | |_|X|X|X|_|_|_|_| 3 |
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− | |X|X|_|X|_|_|_|_| 2 1 |
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− | |_|X|X|X|_|_|X|X| 3 2 |
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− | |_|_|X|X|_|_|X|X| 2 2 |
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− | |_|_|X|X|X|X|X|X| 6 |
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− | |X|_|X|X|X|X|X|_| 1 5 |
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− | |X|X|X|X|X|X|_|_| 6 |
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− | |_|_|_|_|X|_|_|_| 1 |
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− | |_|_|_|X|X|_|_|_| 2 |
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− | 1 3 1 7 5 3 4 3 |
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− | 2 1 5 1 |
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− | </pre> |
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− | |||

− | Solutions: |
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− | The first solution is a simple backtracking algorithm, but is quite slow for larger problems. |
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− | <haskell> |
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− | data Square = Blank | Cross deriving (Eq) |
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− | instance Show Square where |
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− | show Blank = " " |
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− | show Cross = "X" |
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− | |||

− | -- create all possibilities of arranging the given blocks in a line of "n" elements |
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− | rows n [] = [replicate n Blank] |
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− | rows n (k:ks) | n < k = [] |
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− | rows n (k:ks) = |
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− | [Blank : row | row <- rows (n-1) (k:ks)] ++ |
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− | if null ks then [replicate k Cross ++ replicate (n-k) Blank] |
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− | else [replicate k Cross ++ Blank : row | row <- rows (n-k-1) ks] |
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− | |||

− | -- contract a given line into the block format |
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− | -- i.e. contract [Cross,Blank,Cross] == [1,1] |
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− | contract = map length . filter (\(x:_) -> x==Cross) . group |
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− | |||

− | -- create all solutions by combining all possible rows in all possible ways |
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− | -- then pick a solution and check whether its block signature fits |
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− | solver horz vert = filter fitsVert possSolution |
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− | where possSolution = sequence $ map (rows (length vert)) horz |
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− | fitsVert rs = map contract (transpose rs) == vert |
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− | |||

− | -- output the (first) solution |
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− | nonogram horz vert = printSolution $ head $ solver horz vert |
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− | where printSolution = putStr . unlines . map (concatMap show) . transpose |
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− | </haskell> |
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− | |||

− | This is a solution done for simplicity rather than performance. It's SLOOOOW. |
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− | |||

− | It builds all combinations of blocks in a row (stolen from solution 2 :) and then builds all combinations of rows. The resulting columns are then contracted into the short block block form and the signature compared to the target. |
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− | |||

− | We can make the search much faster (but more obscure) by deducing the values of as many squares as possible before guessing, as in this solution: |
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− | <haskell> |
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− | module Nonogram where |
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− | |||

− | import Control.Monad |
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− | import Data.List |
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− | import Data.Maybe |
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− | |||

− | data Square = Filled | Blank | Unknown |
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− | deriving (Eq, Show) |
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− | type Row = [Square] |
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− | type Grid = [Row] |
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− | |||

− | -- Print the first solution (if any) to the nonogram |
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− | nonogram :: [[Int]] -> [[Int]] -> String |
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− | nonogram rs cs = case solve rs cs of |
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− | [] -> "Inconsistent\n" |
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− | (grid:_) -> showGrid rs cs grid |
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− | |||

− | -- All solutions to the nonogram |
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− | solve :: [[Int]] -> [[Int]] -> [Grid] |
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− | solve rs cs = [grid' | |
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− | -- deduce as many squares as we can |
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− | grid <- maybeToList (deduction rs cs), |
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− | -- guess the rest, governed by rs |
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− | grid' <- zipWithM (rowsMatching nc) rs grid, |
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− | -- check each guess against cs |
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− | map contract (transpose grid') == cs] |
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− | where nc = length cs |
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− | contract = map length . filter (\(x:_) -> x==Filled) . group |
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− | |||

− | -- A nonogram with all the values we can deduce |
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− | deduction :: [[Int]] -> [[Int]] -> Maybe Grid |
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− | deduction rs cs = converge step init |
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− | where nr = length rs |
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− | nc = length cs |
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− | init = replicate nr (replicate nc Unknown) |
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− | step = (improve nc rs . transpose) <.> (improve nr cs . transpose) |
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− | improve n = zipWithM (common n) |
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− | (g <.> f) x = f x >>= g |
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− | |||

− | -- repeatedly apply f until a fixed point is reached |
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− | converge :: (Monad m, Eq a) => (a -> m a) -> a -> m a |
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− | converge f s = do |
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− | s' <- f s |
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− | if s' == s then return s else converge f s' |
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− | |||

− | -- common n ks partial = commonality between all possible ways of |
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− | -- placing blocks of length ks in a row of length n that match partial. |
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− | common :: Int -> [Int] -> Row -> Maybe Row |
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− | common n ks partial = case rowsMatching n ks partial of |
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− | [] -> Nothing |
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− | rs -> Just (foldr1 (zipWith unify) rs) |
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− | |||

− | -- rowsMatching n ks partial = all possible ways of placing blocks of |
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− | -- length ks in a row of length n that match partial. |
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− | rowsMatching :: Int -> [Int] -> [Square] -> [[Square]] |
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− | rowsMatching n [] partial = [replicate n Blank | all (/= Filled) partial] |
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− | rowsMatching n ks [] = [] |
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− | rowsMatching n ks (Unknown:partial) = |
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− | rowsMatching n ks (Filled:partial) ++ |
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− | rowsMatching n ks (Blank:partial) |
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− | rowsMatching n ks (Blank:partial) = |
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− | [Blank : row | row <- rowsMatching (n-1) ks partial] |
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− | rowsMatching n [k] (Filled:partial) = |
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− | [replicate k Filled ++ replicate (n-k) Blank | |
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− | n >= k && all (/= Blank) front && all (/= Filled) back] |
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− | where (front, back) = splitAt (k-1) partial |
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− | rowsMatching n (k:ks) (Filled:partial) = |
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− | [replicate k Filled ++ Blank : row | |
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− | n > k+1 && all (/= Blank) front && blank /= Filled, |
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− | row <- rowsMatching (n-k-1) ks partial'] |
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− | where (front, blank:partial') = splitAt (k-1) partial |
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− | |||

− | unify :: Square -> Square -> Square |
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− | unify Filled Filled = Filled |
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− | unify Blank Blank = Blank |
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− | unify _ _ = Unknown |
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− | |||

− | showGrid :: [[Int]] -> [[Int]] -> Grid -> String |
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− | showGrid rs cs ss = unlines (zipWith showRow rs ss ++ showCols cs) |
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− | where showRow rs ss = concat [['|', name s] | s <- ss] ++ "| " ++ |
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− | unwords (map show rs) |
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− | showCols cs |
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− | | all null cs = [] |
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− | | otherwise = concatMap showCol cs : showCols (map advance cs) |
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− | showCol (k:_) |
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− | | k < 10 = ' ':show k |
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− | | otherwise = show k |
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− | showCol [] = " " |
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− | advance [] = [] |
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− | advance (x:xs) = xs |
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− | |||

− | name :: Square -> Char |
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− | name Filled = 'X' |
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− | name Blank = '_' |
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− | name Unknown = '?' |
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− | </haskell> |
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− | We build up knowledge of which squares must be filled and which must be blank, until we can't make any more deductions. |
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− | Some puzzles cannot be completely solved in this way, so then we guess values by the same method as the first solution for any remaining squares. |
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− | |||

− | == Problem 99 == |
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− | |||

− | (***) Crossword puzzle |
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− | |||

− | Given an empty (or almost empty) framework of a crossword puzzle and a set of words. The problem is to place the words into the framework. |
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− | |||

− | http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p99.gif |
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− | |||

− | The particular crossword puzzle is specified in a text file which first lists the words (one word per line) in an arbitrary order. Then, after an empty line, the crossword framework is defined. In this framework specification, an empty character location is represented by a dot (.). In order to make the solution easier, character locations can also contain predefined character values. The puzzle above is defined in the file [http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p99a.dat p99a.dat], other examples are [http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p99b.dat p99b.dat] and [http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p99d.dat p99d.dat]. There is also an example of a puzzle ([http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p99c.dat p99c.dat]) which does not have a solution. |
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− | |||

− | Words are strings (character lists) of at least two characters. A horizontal or vertical sequence of character places in the crossword puzzle framework is called a site. Our problem is to find a compatible way of placing words onto sites. |
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− | |||

− | Hints: (1) The problem is not easy. You will need some time to thoroughly understand it. So, don't give up too early! And remember that the objective is a clean solution, not just a quick-and-dirty hack! |
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− | |||

− | (2) Reading the data file is a tricky problem for which a solution is provided in the file [http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p99-readfile.pl p99-readfile.pl]. See the predicate read_lines/2. |
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− | |||

− | (3) For efficiency reasons it is important, at least for larger puzzles, to sort the words and the sites in a particular order. For this part of the problem, the solution of P28 may be very helpful. |
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− | |||

− | Example in Haskell: |
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− | <pre> |
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− | ALPHA |
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− | ARES |
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− | POPPY |
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− | |||

− | . |
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− | . |
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− | ..... |
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− | . . |
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− | . . |
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− | . |
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− | |||

− | > solve $ readCrossword "ALPHA\nARES\nPOPPY\n\n . \n . \n.....\n . .\n . .\n .\n" |
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− | |||

− | [[((3,1),'A'),((3,2),'L'),((3,3),'P'),((3,4),'H'),((3,5),'A'),((1,3),'P'),((2,3) |
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− | ,'O'),((3,3),'P'),((4,3),'P'),((5,3),'Y'),((3,5),'A'),((4,5),'R'),((5,5),'E'),(( |
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− | 6,5),'S')]] |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | -- import Control.Monad |
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− | -- import Data.List |
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− | |||

− | type Coord = (Int,Int) |
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− | type Word = String |
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− | data Site = Site {siteCoords :: [Coord], siteLen :: Int} deriving (Show,Eq) |
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− | data Crossword = Crossword {cwWords :: [Word], cwSites :: [Site]} deriving (Show,Eq) |
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− | |||

− | comparing f = \a b -> f a `compare` f b |
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− | equaling f = \a b -> f a == f b |
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− | |||

− | -- convert the text lines from the file to the "Site" datatype, |
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− | -- which contain the adjacent coordinates of the site and its length |
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− | toSites :: [String] -> [Site] |
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− | toSites lines = find (index_it lines) ++ find (transpose . index_it $ lines) |
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− | where find = map makePos . concat . map extractor |
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− | extractor = filter ((>1) . length) . map (filter (\(_,x) -> x=='.')) . groupBy (equaling snd) |
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− | index_it = map (\(row,e) -> zip [(col,row) | col <- [1..]] e) . zip [1..] |
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− | makePos xs = Site {siteCoords = map fst xs, siteLen = length xs} |
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− | |||

− | -- test whether there exist no two different letters at the same coordinate |
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− | noCollision :: [(String, Site)] -> Bool |
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− | noCollision xs = all allEqual groupedByCoord |
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− | where groupedByCoord = map (map snd) . groupBy (equaling fst) . sortBy (comparing fst) . concatMap together $ xs |
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− | allEqual [] = True |
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− | allEqual (x:xs) = all (x==) xs |
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− | |||

− | -- merge a word and a site by assigning each letter to its respective coordinate |
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− | together :: (Word, Site) -> [(Coord, Char)] |
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− | together (w,s) = zip (siteCoords s) w |
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− | |||

− | -- returns all solutions for the crossword as lists of occupied coordinates and their respective letters |
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− | solve :: Crossword -> [[(Coord, Char)]] |
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− | solve cw = map (concatMap together) solution |
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− | where solution = solve' (cwWords cw) (cwSites cw) |
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− | |||

− | solve' :: [Word] -> [Site] -> [[(Word, Site)]] |
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− | solve' _ [] = [[]] |
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− | solve' words (s:ss) = if null possWords |
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− | then error ("too few words of length " ++ show (siteLen s)) |
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− | else do try <- possWords |
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− | let restWords = Data.List.delete try words |
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− | more <- solve' restWords ss |
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− | let attempt = (try,s):more |
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− | Control.Monad.guard $ noCollision attempt |
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− | return attempt |
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− | where possWords = filter (\w -> siteLen s == length w) words |
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− | |||

− | -- read the content of a file into the "Crossword" datatype |
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− | readCrossword :: String -> Crossword |
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− | readCrossword = (\(ws,ss) -> Crossword ws (toSites (drop 1 ss))) . break (""==) . lines |
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− | </haskell> |
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− | |||

− | |||

− | This is a simplistic solution with no consideration for speed. Especially sites and words aren't ordered as propesed in (3) of the problem. Words of the correct length are naively tried for all blanks (without heuristics) and the possible solutions are then backtracked. |
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− | |||

− | To test for collisions, all (Word, Site) pairs are merged to result in a list of (Coord, Char) elements which represent all letters placed so far. If all (two) characters of the same coordinate are identical, there exist no collisions between words. |
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[[Category:Tutorials]] |
[[Category:Tutorials]] |

## Revision as of 00:08, 16 December 2006

These are Haskell translations of Ninety-Nine Prolog Problems.

If you want to work on one of these, put your name in the block so we know someone's working on it. Then, change n in your block to the appropriate problem number, and fill in the <Problem description>,<example in Haskell>,<solution in haskell> and <description of implementation> fields.

## 1 Miscellaneous problems

## 2 Problem 90

(**) Eight queens problem

This is a classical problem in computer science. The objective is to place eight queens on a chessboard so that no two queens are attacking each other; i.e., no two queens are in the same row, the same column, or on the same diagonal.

Hint: Represent the positions of the queens as a list of numbers 1..N. Example: [4,2,7,3,6,8,5,1] means that the queen in the first column is in row 4, the queen in the second column is in row 2, etc. Use the generate-and-test paradigm.

Example in Haskell:

> length (queens 8) 92 > head (queens 8) [4,2,7,3,6,8,5,1]

Solution: The simplest solution is a composition of separate functions to generate the list of candidates and to test each candidate:

queens :: Int -> [[Int]] queens n = filter test (generate n) where generate 0 = [[]] generate k = [q : qs | qs <- generate (k-1), q <- [1..n]] test [] = True test (q:qs) = isSafe q qs && test qs isSafe try qs = not (try `elem` qs || sameDiag try qs) sameDiag try qs = any (\(colDist,q) -> abs (try - q) == colDist) $ zip [1..] qs

By definition/data representation no two queens can occupy the same column.

This is easy to understand, but it's also quite slow, as it generates and tests N^N possible N-queen configurations.

The key to speeding it up is to fuse the compositionIf a list already contains two queens in a line, there's no point in considering all the possible ways of adding more queens. This yields the following version, which is much faster:

queens :: Int -> [[Int]] queens n = queens' n where queens' 0 = [[]] queens' k = [q:qs | qs <- queens' (k-1), q <- [1..n], isSafe q qs] isSafe try qs = not (try `elem` qs || sameDiag try qs) sameDiag try qs = any (\(colDist,q) -> abs (try - q) == colDist) $ zip [1..] qs

If you approach this problem with an imperative mindset, you might be tempted to use an accumulating parameter for the list of candidates. This would make the function harder to understand, and would not help much (if at all): the important thing here is the breadth of the search tree, not its depth.

## 3 Problem 91

(**) Knight's tour

Another famous problem is this one: How can a knight jump on an NxN chessboard in such a way that it visits every square exactly once?

Hints: Represent the squares by pairs of their coordinates of the form X/Y, where both X and Y are integers between 1 and N. (Note that '/' is just a convenient functor, not division!) Define the relation jump(N,X/Y,U/V) to express the fact that a knight can jump from X/Y to U/V on a NxN chessboard. And finally, represent the solution of our problem as a list of N*N knight positions (the knight's tour).

There are two variants of this problem:

- find a tour ending at a particular square
- find a circular tour, ending a knight's jump from the start (clearly it doesn't matter where you start, so choose (1,1))

Example in Haskell:

Knights> head $ knightsTo 8 (1,1) [(2,7),(3,5),(5,6),(4,8),(3,6),(4,4),(6,5),(4,6), (5,4),(7,5),(6,3),(5,5),(4,3),(2,4),(1,6),(2,8), (4,7),(6,8),(8,7),(6,6),(4,5),(6,4),(5,2),(7,1), (8,3),(6,2),(8,1),(7,3),(8,5),(7,7),(5,8),(3,7), (1,8),(2,6),(3,4),(1,5),(2,3),(3,1),(1,2),(3,3), (1,4),(2,2),(4,1),(5,3),(7,4),(8,2),(6,1),(4,2), (2,1),(1,3),(2,5),(1,7),(3,8),(5,7),(7,8),(8,6), (6,7),(8,8),(7,6),(8,4),(7,2),(5,1),(3,2),(1,1)] Knights> head $ closedKnights 8 [(1,1),(3,2),(1,3),(2,1),(3,3),(5,4),(6,6),(4,5), (2,6),(1,8),(3,7),(5,8),(4,6),(2,5),(4,4),(5,6), (6,4),(8,5),(7,7),(6,5),(5,3),(6,1),(4,2),(6,3), (8,2),(7,4),(5,5),(3,4),(1,5),(2,7),(4,8),(3,6), (1,7),(3,8),(5,7),(7,8),(8,6),(6,7),(8,8),(7,6), (8,4),(7,2),(5,1),(4,3),(3,5),(1,4),(2,2),(4,1), (6,2),(8,1),(7,3),(5,2),(7,1),(8,3),(7,5),(8,7), (6,8),(4,7),(2,8),(1,6),(2,4),(1,2),(3,1),(2,3)]

Solution:

module Knights where import Data.List type Square = (Int, Int) -- Possible knight moves from a given square on an nxn board knightMoves :: Int -> Square -> [Square] knightMoves n (x, y) = filter (onBoard n) [(x+2, y+1), (x+2, y-1), (x+1, y+2), (x+1, y-2), (x-1, y+2), (x-1, y-2), (x-2, y+1), (x-2, y-1)] -- Is the square within an nxn board? onBoard :: Int -> Square -> Bool onBoard n (x, y) = 1 <= x && x <= n && 1 <= y && y <= n -- Knight's tours on an nxn board ending at the given square knightsTo :: Int -> Square -> [[Square]] knightsTo n finish = [pos:path | (pos, path) <- tour (n*n)] where tour 1 = [(finish, [])] tour k = [(pos', pos:path) | (pos, path) <- tour (k-1), pos' <- sortImage (entrances path) (filter (`notElem` path) (knightMoves n pos))] entrances path pos = length (filter (`notElem` path) (knightMoves n pos)) -- Closed knight's tours on an nxn board closedKnights :: Int -> [[Square]] closedKnights n = [pos:path | (pos, path) <- tour (n*n), pos == start] where tour 1 = [(finish, [])] tour k = [(pos', pos:path) | (pos, path) <- tour (k-1), pos' <- sortImage (entrances path) (filter (`notElem` path) (knightMoves n pos))] entrances path pos | pos == start = 100 -- don't visit start until there are no others | otherwise = length (filter (`notElem` path) (knightMoves n pos)) start = (1,1) finish = (2,3) -- Sort by comparing the image of list elements under a function f. -- These images are saved to avoid recomputation. sortImage :: Ord b => (a -> b) -> [a] -> [a] sortImage f xs = map snd (sortBy cmpFst [(f x, x) | x <- xs]) where cmpFst x y = compare (fst x) (fst y)

This has a similar structure to the 8 Queens problem, except that we apply a heuristic invented by Warnsdorff: when considering next possible moves, we prefer squares with fewer open entrances. This speeds things up enormously, and finds the first solution to boards smaller than 76x76 without backtracking.

Solution 2:

knights :: Int -> [[(Int,Int)]] knights n = loop (n*n) [[(1,1)]] where loop 1 = map reverse . id loop i = loop (i-1) . concatMap nextMoves nextMoves already@(x:xs) = [next:already | next <- possible] where possible = filter (\x -> on_board x && not (x `elem` already)) $ jumps x jumps (x,y) = [(x+a, y+b) | (a,b) <- [(1,2), (2,1), (2,-1), (1,-2), (-1,-2), (-2,-1), (-2,1), (-1,2)]] on_board (x,y) = (x >= 1) && (x <= n) && (y >= 1) && (y <= n)

This is just the naive backtracking approach. I tried a speedup using Data.Map, but the code got too verbose to post.

## 4 Problem 92

(***) Von Koch's conjecture

Several years ago I met a mathematician who was intrigued by a problem for which he didn't know a solution. His name was Von Koch, and I don't know whether the problem has been solved since.

Anyway the puzzle goes like this: Given a tree with N nodes (and hence N-1 edges). Find a way to enumerate the nodes from 1 to N and, accordingly, the edges from 1 to N-1 in such a way, that for each edge K the difference of its node numbers equals to K. The conjecture is that this is always possible.

For small trees the problem is easy to solve by hand. However, for larger trees, and 14 is already very large, it is extremely difficult to find a solution. And remember, we don't know for sure whether there is always a solution!

Write a predicate that calculates a numbering scheme for a given tree. What is the solution for the larger tree pictured below?

Example in Haskell:

> head $ vonKoch [(1,6),(2,6),(3,6),(4,6),(5,6),(5,7),(5,8),(8,9),(5,10),(10,11),(11,12),(11,13),(13,14)] [6,7,8,9,3,4,10,11,5,12,2,13,14,1]

Solution:

vonKoch edges = do let n = length edges + 1 nodes <- permutations [1..n] let nodeArray = listArray (1,n) nodes let dists = sort $ map (\(x,y) -> abs (nodeArray ! x - nodeArray ! y)) edges guard $ and $ zipWith (/=) dists (tail dists) return nodes

This is a simple brute-force solver. This function will permute all assignments of the different node numbers and will then verify that all of the edge differences are different. This code uses the List Monad.

## 5 Problem 93

(***) An arithmetic puzzle

Given a list of integer numbers, find a correct way of inserting arithmetic signs (operators) such that the result is a correct equation. Example: With the list of numbers [2,3,5,7,11] we can form the equations 2-3+5+7 = 11 or 2 = (3*5+7)/11 (and ten others!).

Division should be interpreted as operating on rationals, and division by zero should be avoided.

Example in Haskell:

P93> putStr $ unlines $ puzzle [2,3,5,7,11] 2 = 3-(5+7-11) 2 = 3-5-(7-11) 2 = 3-(5+7)+11 2 = 3-5-7+11 2 = (3*5+7)/11 2*(3-5) = 7-11 2-(3-(5+7)) = 11 2-(3-5-7) = 11 2-(3-5)+7 = 11 2-3+5+7 = 11

The other two solutions alluded to in the problem description are dropped by the Haskell solution as trivial variants:

2 = 3-(5+(7-11)) 2-3+(5+7) = 11

Solution:

module P93 where import Control.Monad import Data.List import Data.Maybe type Equation = (Expr, Expr) data Expr = Const Integer | Binary Expr Op Expr deriving (Eq, Show) data Op = Plus | Minus | Multiply | Divide deriving (Bounded, Eq, Enum, Show) type Value = Rational -- top-level function: all correct equations generated from the list of -- numbers, as pretty strings. puzzle :: [Integer] -> [String] puzzle ns = map (flip showsEquation "") (equations ns) -- generate all correct equations from the list of numbers equations :: [Integer] -> [Equation] equations [] = error "empty list of numbers" equations [n] = error "only one number" equations ns = [(e1, e2) | (ns1, ns2) <- splits ns, (e1, v1) <- exprs ns1, (e2, v2) <- exprs ns2, v1 == v2] -- generate all expressions from the numbers, except those containing -- a division by zero, or redundant right-associativity. exprs :: [Integer] -> [(Expr, Value)] exprs [n] = [(Const n, fromInteger n)] exprs ns = [(Binary e1 op e2, v) | (ns1, ns2) <- splits ns, (e1, v1) <- exprs ns1, (e2, v2) <- exprs ns2, op <- [minBound..maxBound], not (right_associative op e2), v <- maybeToList (apply op v1 v2)] -- splittings of a list into two non-empty lists splits :: [a] -> [([a],[a])] splits xs = tail (init (zip (inits xs) (tails xs))) -- applying an operator to arguments may fail (division by zero) apply :: Op -> Value -> Value -> Maybe Value apply Plus x y = Just (x + y) apply Minus x y = Just (x - y) apply Multiply x y = Just (x * y) apply Divide x 0 = Nothing apply Divide x y = Just (x / y) -- e1 op (e2 op' e3) == (e1 op e2) op' e3 right_associative :: Op -> Expr -> Bool right_associative Plus (Binary _ Plus _) = True right_associative Plus (Binary _ Minus _) = True right_associative Multiply (Binary _ Multiply _) = True right_associative Multiply (Binary _ Divide _) = True right_associative _ _ = False -- Printing of equations and expressions showsEquation :: Equation -> ShowS showsEquation (l, r) = showsExprPrec 0 l . showString " = " . showsExprPrec 0 r -- all operations are left associative showsExprPrec :: Int -> Expr -> ShowS showsExprPrec _ (Const n) = shows n showsExprPrec p (Binary e1 op e2) = showParen (p > op_prec) $ showsExprPrec op_prec e1 . showString (opName op) . showsExprPrec (op_prec+1) e2 where op_prec = precedence op precedence :: Op -> Int precedence Plus = 6 precedence Minus = 6 precedence Multiply = 7 precedence Divide = 7 opName :: Op -> String opName Plus = "+" opName Minus = "-" opName Multiply = "*" opName Divide = "/"

Unlike the Prolog solution, I've eliminated solutions like
`"1+(2+3) = 6"` as a trivial variant of `"1+2+3 = 6"` (cf the function `right_associative`).
Apart from that, the Prolog solution is shorter because it uses built-in evaluation and printing of expressions.

## 6 Problem 94

<Problem description>

Example in Haskell:

<example in Haskell>

Solution:

<solution in haskell>

<description of implementation>