# 99 questions/90 to 94

### From HaskellWiki

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__NOTOC__ |
__NOTOC__ |
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− | These are Haskell translations of [http://www.ic.unicamp.br/~meidanis/courses/mc336/2006s2/funcional/L-99_Ninety-Nine_Lisp_Problems.html Ninety Nine Lisp Problems], |
+ | This is part of [[H-99:_Ninety-Nine_Haskell_Problems|Ninety-Nine Haskell Problems]], based on [https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/ Ninety-Nine Prolog Problems]. |

− | which are themselves translations of [http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/ Ninety-Nine Prolog Problems]. |
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− | 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. |
+ | 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. |

== Miscellaneous problems == |
== Miscellaneous problems == |
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== Problem 90 == |
== Problem 90 == |
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− | Eight queens problem |
+ | (**) 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. |
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. |
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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. |
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. |
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− | <pre> |
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Example in Haskell: |
Example in Haskell: |
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− | > length queens |
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− | 92 |
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− | > take 1 queens |
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− | [[4,2,7,3,6,8,5,1]] |
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− | </pre> |
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− | Solution: |
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<haskell> |
<haskell> |
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− | queens = queens' 8 |
+ | > length (queens 8) |

+ | 92 |
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+ | > head (queens 8) |
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+ | [1,5,8,6,3,7,2,4] |
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+ | </haskell> |
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− | queens' 0 = [[]] |
+ | [[99 questions/Solutions/90 | Solutions]] |

− | queens' (n+1) = [ try:alreadySet | alreadySet <- queens' n, try <- [1..8], isSafe try alreadySet] |
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− | where isSafe try alreadySet = not (sameRow try alreadySet || sameDiagonal try alreadySet) |
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− | sameRow try alreadySet = try `elem` alreadySet |
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− | sameDiagonal try alreadySet = any (\(col,q) -> abs(try - q) == col) $ zip [1..] alreadySet |
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− | </haskell> |
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− | 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. |
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− | 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. |
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− | |||

== Problem 91 == |
== Problem 91 == |
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− | <Problem description> |
+ | (**) Knight's tour |

− | <pre> |
+ | 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? A set of solutions is given on the [[The_Knights_Tour]] page. |

− | Example: |
+ | |

− | <example in lisp> |
+ | 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: |
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+ | # find a tour ending at a particular square |
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+ | # find a circular tour, ending a knight's jump from the start (clearly it doesn't matter where you start, so choose (1,1)) |
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Example in Haskell: |
Example in Haskell: |
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− | <example in Haskell> |
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− | </pre> |
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− | Solution: |
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<haskell> |
<haskell> |
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− | <solution in haskell> |
+ | Knights> head $ knightsTo 8 (1,1) |

+ | [(2,7),(3,5),(5,6),(4,8),(3,6),(4,4),(6,5),(4,6), |
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+ | (5,4),(7,5),(6,3),(5,5),(4,3),(2,4),(1,6),(2,8), |
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+ | (4,7),(6,8),(8,7),(6,6),(4,5),(6,4),(5,2),(7,1), |
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+ | (8,3),(6,2),(8,1),(7,3),(8,5),(7,7),(5,8),(3,7), |
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+ | (1,8),(2,6),(3,4),(1,5),(2,3),(3,1),(1,2),(3,3), |
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+ | (1,4),(2,2),(4,1),(5,3),(7,4),(8,2),(6,1),(4,2), |
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+ | (2,1),(1,3),(2,5),(1,7),(3,8),(5,7),(7,8),(8,6), |
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+ | (6,7),(8,8),(7,6),(8,4),(7,2),(5,1),(3,2),(1,1)] |
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+ | Knights> head $ closedKnights 8 |
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+ | [(1,1),(3,2),(1,3),(2,1),(3,3),(5,4),(6,6),(4,5), |
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+ | (2,6),(1,8),(3,7),(5,8),(4,6),(2,5),(4,4),(5,6), |
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+ | (6,4),(8,5),(7,7),(6,5),(5,3),(6,1),(4,2),(6,3), |
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+ | (8,2),(7,4),(5,5),(3,4),(1,5),(2,7),(4,8),(3,6), |
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+ | (1,7),(3,8),(5,7),(7,8),(8,6),(6,7),(8,8),(7,6), |
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+ | (8,4),(7,2),(5,1),(4,3),(3,5),(1,4),(2,2),(4,1), |
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+ | (6,2),(8,1),(7,3),(5,2),(7,1),(8,3),(7,5),(8,7), |
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+ | (6,8),(4,7),(2,8),(1,6),(2,4),(1,2),(3,1),(2,3)] |
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</haskell> |
</haskell> |
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− | <description of implementation> |
+ | [[99 questions/Solutions/91 | Solutions]] |

− | + | ||

+ | |||

== Problem 92 == |
== Problem 92 == |
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− | <Problem description> |
+ | (***) Von Koch's conjecture |

− | <pre> |
+ | 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. |

− | Example: |
+ | |

− | <example in lisp> |
+ | https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/p92a.gif |

+ | |||

+ | 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. |
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+ | |||

+ | 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! |
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+ | |||

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

+ | https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/p92b.gif |
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Example in Haskell: |
Example in Haskell: |
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− | <example in Haskell> |
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− | </pre> |
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− | Solution: |
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<haskell> |
<haskell> |
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− | <solution 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] |
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</haskell> |
</haskell> |
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− | <description of implementation> |
+ | [[99 questions/Solutions/92 | Solutions]] |

+ | |||

== Problem 93 == |
== Problem 93 == |
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− | An arithmetic puzzle |
+ | (***) 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!). |
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!). |
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+ | |||

+ | Division should be interpreted as operating on rationals, and division by zero should be avoided. |
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Example in Haskell: |
Example in Haskell: |
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− | <pre> |
+ | |

− | P93> putStr $ unlines $ puzzle [2,3,5,7,11] |
+ | <haskell> |

+ | P93> mapM_ putStrLn $ puzzle [2,3,5,7,11] |
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2 = 3-(5+7-11) |
2 = 3-(5+7-11) |
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2 = 3-5-(7-11) |
2 = 3-5-(7-11) |
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2-(3-5)+7 = 11 |
2-(3-5)+7 = 11 |
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2-3+5+7 = 11 |
2-3+5+7 = 11 |
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− | </pre> |
+ | </haskell> |

− | The other two solutions alluded to in the problem description are dropped by solution 2 as trivial variants: |
+ | |

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

<pre> |
<pre> |
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2 = 3-(5+(7-11)) |
2 = 3-(5+(7-11)) |
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Line 97: | Line 83: | ||

</pre> |
</pre> |
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− | Solution 1: |
+ | [[99 questions/Solutions/93 | Solutions]] |

− | <haskell> |
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− | import Text.ParserCombinators.Parsec |
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− | import Text.ParserCombinators.Parsec.Expr |
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− | puzzle :: [Integer] -> [String] |
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− | puzzle xs = filter ((\(a,b) -> calcExp a == calcExp b) . split) equations |
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− | where equations = map concat . filter isValid . sequence . intersperse ops . map ((:[]) . show) $ xs |
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− | split = (\(a,b) -> (a, drop 1 b)) . break ('='==) |
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− | ops = ["+", "-", "/", "*", "="] |
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− | isValid = (1==) . length . filter ("="==) |
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− | |||

− | |||

− | expr :: Parser Integer |
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− | expr = buildExpressionParser optable factor |
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− | |||

− | optable = [[op "*" (*) AssocLeft, op "/" div AssocLeft] |
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− | ,[op "+" (+) AssocLeft, op "-" (-) AssocLeft] |
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− | ] |
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− | where op s f assoc = Infix (do{ string s; return f}) assoc |
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− | |||

− | factor = do{ char '(' |
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− | ; x <- expr |
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− | ; char ')' |
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− | ; return x |
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− | } |
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− | <|> number |
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− | |||

− | number :: Parser Integer |
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− | number = do{ ds <- many1 digit |
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− | ; return (read ds) |
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− | } |
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− | |||

− | calcExp e = either (error . show) id (parse expr "source" e) |
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− | </haskell> |
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− | |||

− | I chose using parsec to evaluate the arithmetic expressions. Parentheses are not yet supported. Integer division is used (i.e. 7/11 == 0). |
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− | |||

− | Using Template Haskell or hs-plugins would likely yield a more concise solution. |
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− | |||

− | Solution 2: |
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− | <haskell> |
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− | module P93 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 Equation = (Expr, Expr) |
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− | -- cache value in binary expression |
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− | data Expr = Const Integer | Binary Value Expr Op Expr |
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− | deriving (Eq, Show) |
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− | data Op = Plus | Minus | Multiply | Divide |
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− | deriving (Bounded, Eq, Enum, Show) |
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− | type Value = Rational |
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− | |||

− | -- top-level function: all correct equations generated from the list of |
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− | -- numbers, as pretty strings. |
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− | puzzle :: [Integer] -> [String] |
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− | puzzle ns = map (flip showsEquation "") (equations ns) |
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− | |||

− | -- generate all correct equations from the list of numbers |
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− | equations :: [Integer] -> [Equation] |
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− | equations [] = error "empty list of numbers" |
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− | equations [n] = error "only one number" |
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− | equations ns = [(e1, e2) | |
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− | (ns1, ns2) <- splits ns, |
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− | e1 <- exprs ns1, e2 <- exprs ns2, |
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− | value e1 == value e2] |
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− | |||

− | -- generate all expressions from the numbers, except those containing |
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− | -- a division by zero, or redundant right-associativity. |
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− | exprs :: [Integer] -> [Expr] |
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− | exprs [n] = [Const n] |
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− | exprs ns = [Binary n e1 op e2 | (ns1, ns2) <- splits ns, |
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− | e1 <- exprs ns1, e2 <- exprs ns2, op <- [minBound..maxBound], |
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− | not (right_associative op e2), |
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− | n <- maybeToList (apply op (value e1) (value e2))] |
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− | |||

− | -- splittings of a list into two non-empty lists |
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− | splits :: [a] -> [([a],[a])] |
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− | splits xs = tail (init (zip (inits xs) (tails xs))) |
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− | |||

− | -- the cached values of an expression |
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− | value :: Expr -> Value |
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− | value (Const n) = fromInteger n |
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− | value (Binary n _ _ _) = n |
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− | |||

− | -- applying an operator to arguments may fail (division by zero) |
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− | apply :: Op -> Value -> Value -> Maybe Value |
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− | apply Plus x y = Just (x + y) |
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− | apply Minus x y = Just (x - y) |
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− | apply Multiply x y = Just (x * y) |
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− | apply Divide x 0 = Nothing |
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− | apply Divide x y = Just (x / y) |
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− | |||

− | -- e1 op (e2 op' e3) == (e1 op e2) op' e3 |
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− | right_associative :: Op -> Expr -> Bool |
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− | right_associative Plus (Binary _ _ Plus _) = True |
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− | right_associative Plus (Binary _ _ Minus _) = True |
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− | right_associative Multiply (Binary _ _ Multiply _) = True |
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− | right_associative Multiply (Binary _ _ Divide _) = True |
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− | right_associative _ _ = False |
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− | |||

− | -- Printing of equations and expressions |
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− | |||

− | showsEquation :: Equation -> ShowS |
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− | showsEquation (l, r) = showsExprPrec 0 l . showString " = " . showsExprPrec 0 r |
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− | |||

− | -- all operations are left associative |
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− | showsExprPrec :: Int -> Expr -> ShowS |
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− | showsExprPrec _ (Const n) = shows n |
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− | showsExprPrec p (Binary _ e1 op e2) = showParen (p > op_prec) $ |
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− | showsExprPrec op_prec e1 . showString (opName op) . |
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− | showsExprPrec (op_prec+1) e2 |
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− | where op_prec = precedence op |
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− | |||

− | precedence :: Op -> Int |
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− | precedence Plus = 6 |
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− | precedence Minus = 6 |
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− | precedence Multiply = 7 |
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− | precedence Divide = 7 |
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− | |||

− | opName :: Op -> String |
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− | opName Plus = "+" |
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− | opName Minus = "-" |
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− | opName Multiply = "*" |
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− | opName Divide = "/" |
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− | </haskell> |
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− | I've used rational division, because that's what the Prolog solution does. |
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− | Unlike the Prolog solution, I've eliminated solutions like |
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− | <tt>"1+(2+3) = 6"</tt> as a trivial variant of <tt>"1+2+3 = 6"</tt> (cf the function <tt>right_associative</tt>). |
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− | Apart from that, the Prolog solution is shorter because it uses built-in evaluation and printing of expressions. |
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== Problem 94 == |
== Problem 94 == |
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− | <Problem description> |
+ | (***) Generate K-regular simple graphs with N nodes |

− | <pre> |
+ | In a K-regular graph all nodes have a degree of K; i.e. the number of edges incident in each node is K. How many (non-isomorphic!) 3-regular graphs with 6 nodes are there? |

− | Example: |
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− | <example in lisp> |
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− | Example in Haskell: |
+ | [https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/p94.txt Sample results] |

− | <example in Haskell> |
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− | </pre> |
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− | Solution: |
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− | <haskell> |
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− | <solution in haskell> |
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− | </haskell> |
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− | |||

− | <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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− | |||

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

− | Solution: |
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<haskell> |
<haskell> |
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− | import Data.List |
+ | length $ regular 6 3 |

− | import Data.Maybe |
+ | 2 |

− | |||

− | numbers :: Integer -> String |
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− | numbers 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> |
</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. |
+ | [[99 questions/Solutions/94 | Solutions]] |

− | |||

− | == Problem 96 == |
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− | |||

− | Syntax checker |
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− | |||

− | In a certain programming language (Ada) identifiers are defined by the [http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p96.gif syntax diagram]. 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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− | |||

− | |||

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

− | Example in Haskell: |
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− | > syntax_check "Str" |
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− | True |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | syntax_check [] = False |
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− | syntax_check (x:xs) = letter x && loop xs |
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− | where loop [] = True |
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− | loop (x':xs') | x' == '-' = (not . null) xs' && (letter (head xs') || digit (head xs')) && loop (tail xs') |
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− | | letter x' = loop xs' |
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− | | digit x' = loop xs' |
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− | | otherwise = False |
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− | letter x = x `elem` (['a'..'z'] ++ ['A'..'Z']) |
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− | digit x = x `elem` ['0'..'9'] |
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− | </haskell> |
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− | |||

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

− | <Problem description> |
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− | |||

− | <pre> |
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− | Example: |
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− | <example in lisp> |
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− | |||

− | Example in Haskell: |
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− | <example in Haskell> |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | <solution in haskell> |
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− | </haskell> |
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− | |||

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

− | 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 opposite 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. |
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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 p99-readfile.lisp. Use 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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− | |||

− | <pre> |
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− | Example in Haskell: |
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− | </pre> |
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− | |||

− | Solution: |
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− | <haskell> |
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− | type Word = String |
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− | data LayoutPos = LayoutPos {coords :: [(Int,Int)], len :: Int} deriving (Show,Eq) |
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− | data Desc = Desc {words :: [Word], layout :: [LayoutPos]} deriving (Show,Eq) |
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− | |||

− | word = do w <- many1 letter |
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− | newline |
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− | return w |
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− | <?> "word" |
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− | |||

− | emptyLine = char '\n' <?> "emptyLine" |
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− | |||

− | layoutLine = do line <- many1 (oneOf [' ', '.']) |
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− | newline |
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− | return line |
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− | <?> "layoutLine" |
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− | |||

− | description = do words <- manyTill word emptyLine |
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− | layout <- manyTill layoutLine eof |
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− | return $ Desc words (toLayout layout) |
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− | <?> "description" |
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− | |||

− | |||

− | --toLayout :: [String] -> Layout |
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− | toLayout 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 (('.'==) . snd)) . groupBy (\a b -> snd a == snd b) |
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− | index_it = map (\(n,e) -> zip [(n,x) | x <- [1..]] e) . zip [1..] |
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− | makePos xs = LayoutPos {coords = map fst xs, len = length xs} |
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− | |||

− | findCrossings lps = concat . nub . map commonSquares $ pairs |
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− | where pairs = [(a,b) | a <- map coords lps, b <- map coords lps, a /= b] |
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− | commonSquares (a,b) = [i | i <- a, j <- b, i == j] |
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− | |||

− | restrictToCrossings lps cross = map (\l@LayoutPos{coords = cs} -> l{coords = filter (`elem` cross) cs}) lps |
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− | |||

− | readDesc e = either (error . show) id (parse description "source" e) |
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− | |||

− | -- let lyt = layout . readDesc $ desc in restrictToCrossings lyt (findCrossings lyt) |
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− | --desc = "WORD\nWORD\n\n . \n . \n.....\n . .\n . .\n .\n" |
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− | </haskell> |
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− | |||

− | STILL FAR FROM FINISHED! feel free to radically alter/delete this. |
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− | |||

− | The parser parses the input string (desc) into the words and layout. Easier/shorter than parsec would probably be 'splitAt "\n" . lines' |
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− | |||

− | |||

− | toLayout puts the layout lines into the LayoutPos datatype which contains the coordinates and orientation of word-blanks. |
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− | |||

− | findCrossings find all overlapping coordinates between word-blanks. |
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− | |||

− | restrictToCrossings restricts the coordinates in LayoutPos to overlapping ones, since these are the only ones (apart from blank length) that have to be considered during filling. |
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− | |||

− | Now the real work begins and my code is already horridly bulky :( |
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− | |||

− | See the prolog version of the problems for [http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/ working links] |
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[[Category:Tutorials]] |
[[Category:Tutorials]] |

## Latest revision as of 20:38, 22 November 2013

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.

## [edit] 1 Miscellaneous problems

## [edit] 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) [1,5,8,6,3,7,2,4]

## [edit] 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? A set of solutions is given on the The_Knights_Tour page.

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)]

## [edit] 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]

## [edit] 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> mapM_ putStrLn $ 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

## [edit] 6 Problem 94

(***) Generate K-regular simple graphs with N nodes

In a K-regular graph all nodes have a degree of K; i.e. the number of edges incident in each node is K. How many (non-isomorphic!) 3-regular graphs with 6 nodes are there?

Example in Haskell:

length $ regular 6 3 2