# 99 questions/95 to 99

### From HaskellWiki

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− | Solutions: |
+ | Solutions: see [[Nonogram]] |

− | 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. If I were to check intermediate solutions against the blocks that should come out instead of sequencing everything, the List monad would fail much earlier... |
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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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− | |||

− | type Row s = [s] |
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− | type Grid s = [Row s] |
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− | |||

− | -- partial information about a square |
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− | type Square = Maybe Bool |
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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 Bool] |
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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 head . group |
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− | |||

− | -- A nonogram with all the values we can deduce |
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− | deduction :: [[Int]] -> [[Int]] -> Maybe (Grid Square) |
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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 Nothing) |
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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 Square -> Maybe (Row Square) |
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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) (map (map Just) rs)) |
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− | where unify :: Square -> Square -> Square |
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− | unify x y |
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− | | x == y = x |
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− | | otherwise = Nothing |
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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] -> Row Square -> [Row Bool] |
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− | rowsMatching n [] [] = [[]] |
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− | rowsMatching n ks [] = [] |
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− | rowsMatching n ks (Nothing:partial) = |
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− | rowsMatchingAux n ks True partial ++ |
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− | rowsMatchingAux n ks False partial |
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− | rowsMatching n ks (Just s:partial) = |
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− | rowsMatchingAux n ks s partial |
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− | |||

− | rowsMatchingAux :: Int -> [Int] -> Bool -> Row Square -> [Row Bool] |
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− | rowsMatchingAux n ks False partial = |
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− | [False : row | row <- rowsMatching (n-1) ks partial] |
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− | rowsMatchingAux n [k] True partial = |
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− | [replicate k True ++ replicate (n-k) False | |
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− | n >= k && all (/= Just False) front && all (/= Just True) back] |
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− | where (front, back) = splitAt (k-1) partial |
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− | rowsMatchingAux n (k:ks) True partial = |
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− | [replicate k True ++ False : row | |
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− | n > k+1 && all (/= Just False) front && blank /= Just True, |
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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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− | |||

− | showGrid :: [[Int]] -> [[Int]] -> Grid Bool -> 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 [['|', cellChar 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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− | cellChar True = 'X' |
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− | cellChar False = '_' |
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− | advance [] = [] |
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− | advance (x:xs) = xs |
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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 == |
== Problem 99 == |

## Revision as of 22:32, 14 January 2007

This is part of Ninety-Nine Haskell Problems, based on 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 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.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"]

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

## 3 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.

Here is a solution that parses the identifier using Parsec, a parser library that is commonly used in Haskell code:

identifier x = either (const False) (const True) $ parse parser "" x where parser = letter >> many (optional (char '-') >> alphaNum)

## 4 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

## 5 Problem 98

(***) 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:

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

Solutions: see Nonogram

## 6 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.