# 99 questions/90 to 94

### From HaskellWiki

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 lisp>,<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 92 > take 1 queens [[4,2,7,3,6,8,5,1]]

Solution:

queens = queens' 8 where queens' 0 = [[]] queens' n = [ try:qs | qs <- queens' (n-1), try <- [1..8], isSafe try 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. "try `elem` alreadySet" checks for a queen in the same row, "abs(try - q) == col" checks for a queen in the same diagonal.

This is a modification of a function I wrote when I was just learning haskell, so there's certainly much to improve here! For one thing there is speedup potential in caching "blocked" rows, columns and diagonals.

Otherwise a smarter representation in memory might simplify the whole thing drastically as well.

## 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:

<example in lisp>

Example in Haskell:

<example in Haskell>

Solution:

<solution in haskell>

<description of implementation>

## 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:

<example in lisp>

Example in Haskell:

<example in Haskell>

Solution:

<solution in haskell>

<description of implementation>

## 7 Problem 95

(**) English number words

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.

Example in Haskell:

> fullWords 175 one-seven-five

Solution:

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

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

A minor variant of the above solution:

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

## 8 Problem 96

(**) Syntax checker

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

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.

Example in Prolog:

% identifier(Str) :- Str is a legal identifier

Example in Haskell:

> identifier "this-is-a-long-identifier" True > identifier "this-ends-in-" False > identifier "two--hyphens" False

Solution:

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

Simple functional transcription of the diagram.

Another direct transcription of the diagram:

identifier :: String -> Bool identifier (c:cs) = isLetter c && hyphen cs where hyphen [] = True hyphen ('-':cs) = alphas cs hyphen cs = alphas cs alphas [] = False alphas (c:cs) = isAlphaNum c && hyphen cs

The functions `hyphen` and `alphas` correspond to states in the automaton at the start of the loop and before a compulsory alphanumeric, respectively.

## 9 Problem 97

(**) Sudoku

Sudoku puzzles go like this:

Problem statement Solution . . 4 | 8 . . | . 1 7 9 3 4 | 8 2 5 | 6 1 7 | | | | 6 7 . | 9 . . | . . . 6 7 2 | 9 1 4 | 8 5 3 | | | | 5 . 8 | . 3 . | . . 4 5 1 8 | 6 3 7 | 9 2 4 --------+---------+-------- --------+---------+-------- 3 . . | 7 4 . | 1 . . 3 2 5 | 7 4 8 | 1 6 9 | | | | . 6 9 | . . . | 7 8 . 4 6 9 | 1 5 3 | 7 8 2 | | | | . . 1 | . 6 9 | . . 5 7 8 1 | 2 6 9 | 4 3 5 --------+---------+-------- --------+---------+-------- 1 . . | . 8 . | 3 . 6 1 9 7 | 5 8 2 | 3 4 6 | | | | . . . | . . 6 | . 9 1 8 5 3 | 4 7 6 | 2 9 1 | | | | 2 4 . | . . 1 | 5 . . 2 4 6 | 3 9 1 | 5 7 8

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.

Solutions: see Sudoku

## 10 Problem 98

<Problem description> (***) Nonograms

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 ;-).

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.

Problem statement: Solution:

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

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.

Example in Haskell:

|X|X|_|X| 2 1 |_|X|X|_| 2 |X|_|X|_| 1 1 |_|_|X|X| 2 1 2 3 1 1 1 > let horz = [[2,1], [2], [1,1], [2]] :: [[Int]] > let vert = [[1,1], [2], [3], [1,1]] :: [[Int]] > fst . showSolution . head $ solver (4,4) horz vert [[1,1,0,1] ,[0,1,1,0] ,[1,0,1,0] ,[0,0,1,1]]

For second solution:

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

Solution:

data Separator a = Sep Int a (Separator a) | EndSep Int deriving Show -- calculate all possible start positions a description like [3,2] can yield using shifts -- i.e. for size 7: [Sep 0 "3" (Sep 1 "2" (End 1)), ...] positions :: Int -> [Int] -> [Separator Int] positions n xs = map (merge xs) combis where merge [] [sep] = EndSep sep merge (e:es) (s:ss) = Sep s e (merge es ss) combis = fill (length xs - 1) (n - sum xs) fill seps space = filter ((space==) . sum) . sequence $ ([0..space-seps] : replicate seps [1..space-seps+1] ++ [[0..space-seps]]) -- check whether row number "row" (horz) and column number "col" (vert) fit together doesFit :: (Int, Separator Int) -> (Int, Separator Int) -> Bool doesFit (row, horz) (col, vert) = horz'!!col == vert'!!row where (horz',vert') = (expand horz, expand vert) expand (Sep n e rest) = replicate n 0 ++ replicate e 1 ++ expand rest expand (EndSep n) = replicate n 0 -- build all combinations of all possible shifts and check each row combination against each column -- not too astonishingly this is more than slow (i.e. unreasonably slow for size 9 with 270000*270000 combinations to check) solver (nx,ny) horz vert = [(r,c) | r <- sh, c <- sv, fits r c] where (sh, sv) = (sequence . map (positions nx) $ horz, sequence . map (positions ny) $ vert) fits rs cs = and [doesFit (i,r) (j,c) | (i,r) <- zip [0..] rs, (j,c) <- zip [0..] cs] showSolution (x,y) = (map expand x, map expand y)

__This is not a real solution__. It's FAR too slow for that.

It builds all combinations of rows and columns which can be achieved by shifting the allowed blocks arbitrarily. Then it builds all combinations of possible shifts and checks these against each other. NO WONDER IT'S SLOW!

I tried to write a version using two-dimensional recursion, but I simply not smart and patient enough for that. One nice idea would be to use dons' Sudoku Backtracking Monad or alternatively to solve it similarly to my version but with a few simplistic "heuristics" added which reduce the complexity early on.

A different representation (at least for checking collisions) might do wonders as well.

A faster solution:

module Nonogram where import Data.List data Square = Filled | Blank | Unknown deriving (Eq, Show) unify :: Square -> Square -> Square unify Filled Filled = Filled unify Blank Blank = Blank unify _ _ = Unknown match :: Square -> Square -> Bool match Filled Filled = True match Blank Blank = True match Unknown _ = True match _ _ = False name :: Square -> Char name Filled = 'X' name Blank = '_' name Unknown = '?' -- rows n ks = all possible ways of placing blocks of length ks -- in a row of length n. rows :: Int -> [Int] -> [[Square]] rows n [] = [replicate n Blank] rows n (k:ks) | n < k = [] rows n (k:ks) = [Blank:row | row <- rows (n-1) (k:ks)] ++ if null ks then [replicate k Filled ++ row | row <- rows (n-k) ks] else [replicate k Filled ++ Blank : row | row <- rows (n-k-1) ks] -- Does r2 match all the known squares of r1? matchRow :: [Square] -> [Square] -> Bool matchRow r1 r2 = and (zipWith match r1 r2) -- common n ks partial = commonality between all possible ways of -- placing blocks of length ks in a row of length n that match partial. common :: Int -> [Int] -> [Square] -> [Square] common n ks partial = foldr1 (zipWith unify) $ filter (matchRow partial) $ rows n ks solve :: [[Int]] -> [[Int]] -> [[Square]] solve rs cs = converge step init where nr = length rs nc = length cs step = transpose . improve nr cs . transpose . improve nc rs init = replicate nr (replicate nc Unknown) improve n = zipWith (common n) -- repeatedly apply f until a fixed point is reached converge :: Eq a => (a -> a) -> a -> a converge f s | s' == s = s | otherwise = converge f s' where s' = f s showBoard :: [[Int]] -> [[Int]] -> [[Square]] -> String showBoard rs cs ss = unlines (zipWith showRow rs ss ++ showCols cs) where showRow rs ss = concat [['|', name s] | s <- ss] ++ "| " ++ unwords (map show rs) showCols cs | all null cs = [] | otherwise = concatMap showCol cs : showCols (map advance cs) showCol (k:_) | k < 10 = ' ':show k | otherwise = show k showCol [] = " " advance [] = [] advance (x:xs) = xs nonogram :: [[Int]] -> [[Int]] -> String nonogram rs cs = showBoard rs cs (solve rs cs)

We build up knowledge of which squares must be filled and which must be blank, until we can't make any more deductions.

The method of trying all placements of blocks in a row (`rows`) to determine commonality (`common`) could be more sophisticated, but seems fast enough.

## 11 Problem 99

(***) Crossword puzzle

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.

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 p99a.dat, other examples are p99b.dat and p99d.dat. There is also an example of a puzzle (p99c.dat) which does not have a solution.

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.

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!

(2) Reading the data file is a tricky problem for which a solution is provided in the file p99-readfile.pl. See the predicate read_lines/2.

(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.

Example in Haskell:

ALPHA ARES POPPY . . ..... . . . . . > solve $ readCrossword "ALPHA\nARES\nPOPPY\n\n . \n . \n.....\n . .\n . .\n .\n" [[((3,1),'A'),((3,2),'L'),((3,3),'P'),((3,4),'H'),((3,5),'A'),((1,3),'P'),((2,3) ,'O'),((3,3),'P'),((4,3),'P'),((5,3),'Y'),((3,5),'A'),((4,5),'R'),((5,5),'E'),(( 6,5),'S')]]

Solution:

-- import Control.Monad -- import Data.List type Coord = (Int,Int) type Word = String data Site = Site {siteCoords :: [Coord], siteLen :: Int} deriving (Show,Eq) data Crossword = Crossword {cwWords :: [Word], cwSites :: [Site]} deriving (Show,Eq) comparing f = \a b -> f a `compare` f b equaling f = \a b -> f a == f b -- convert the text lines from the file to the "Site" datatype, -- which contain the adjacent coordinates of the site and its length toSites :: [String] -> [Site] toSites lines = find (index_it lines) ++ find (transpose . index_it $ lines) where find = map makePos . concat . map extractor extractor = filter ((>1) . length) . map (filter (\(_,x) -> x=='.')) . groupBy (equaling snd) index_it = map (\(row,e) -> zip [(col,row) | col <- [1..]] e) . zip [1..] makePos xs = Site {siteCoords = map fst xs, siteLen = length xs} -- test whether there exist no two different letters at the same coordinate noCollision :: [(String, Site)] -> Bool noCollision xs = all allEqual groupedByCoord where groupedByCoord = map (map snd) . groupBy (equaling fst) . sortBy (comparing fst) . concatMap together $ xs allEqual [] = True allEqual (x:xs) = all (x==) xs -- merge a word and a site by assigning each letter to its respective coordinate together :: (Word, Site) -> [(Coord, Char)] together (w,s) = zip (siteCoords s) w -- returns all solutions for the crossword as lists of occupied coordinates and their respective letters solve :: Crossword -> [[(Coord, Char)]] solve cw = map (concatMap together) solution where solution = solve' (cwWords cw) (cwSites cw) solve' :: [Word] -> [Site] -> [[(Word, Site)]] solve' _ [] = [[]] solve' words (s:ss) = if null possWords then error ("too few words of length " ++ show (siteLen s)) else do try <- possWords let restWords = Data.List.delete try words more <- solve' restWords ss let attempt = (try,s):more Control.Monad.guard $ noCollision attempt return attempt where possWords = filter (\w -> siteLen s == length w) words -- read the content of a file into the "Crossword" datatype readCrossword :: String -> Crossword readCrossword = (\(ws,ss) -> Crossword ws (toSites (drop 1 ss))) . break (""==) . lines

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.

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.