99 questions/90 to 94
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.
(**) 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]
Solution: The simplest solution is a composition of separate functions to generate the list of candidates and to test each candidate:
queens :: Int -> [[Int]] queens n = filter test (generate n) where generate 0 = [] generate k = [q : qs | q <- [1..n], qs <- generate (k-1)] test  = True test (q:qs) = isSafe q qs && test qs isSafe try qs = not (try `elem` qs || sameDiag try qs) sameDiag try qs = any (\(colDist,q) -> abs (try - q) == colDist) $ zip [1..] qs
By definition/data representation no two queens can occupy the same column.
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 easy to understand, but it's also quite slow, as it generates and tests N^N possible N-queen configurations.
The key to speeding it up is to fuse the composition
filter test . generate into a semantically equivalent function
queens' that does the tests as early as possible.
If a list already contains two queens in a line, there's no point in considering all the possible ways of adding more queens.
Now that the recursive call incorporates testing, we avoid recomputing it by interchanging the two generators, and reverse each answer at the end to obtain the original order.
This yields the following version, which is much faster:
queens :: Int -> [[Int]] queens n = map reverse $ queens' n where queens' 0 = [] queens' k = [q:qs | qs <- queens' (k-1), q <- [1..n], isSafe q qs] isSafe try qs = not (try `elem` qs || sameDiag try qs) sameDiag try qs = any (\(colDist,q) -> abs (try - q) == colDist) $ zip [1..] qs
If you approach this problem with an imperative mindset, you might be tempted to use an accumulating parameter for the list of candidates. This would make the function harder to understand, and would not help much (if at all): the important thing here is the breadth of the search tree, not its depth.
(**) 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)]
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.
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.
(***) 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]
vonKoch edges = do let n = length edges + 1 nodes <- permutations [1..n] let nodeArray = listArray (1,n) nodes let dists = sort $ map (\(x,y) -> abs (nodeArray ! x - nodeArray ! y)) edges guard $ and $ zipWith (/=) dists (tail dists) return nodes
This is a simple brute-force solver. This function will permute all assignments of the different node numbers and will then verify that all of the edge differences are different. This code uses the List Monad.
(***) 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
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.
Example in Haskell:
<example in Haskell>
<solution in haskell>
<description of implementation>