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__NOTOC__
 
__NOTOC__
   
This is part of [[H-99:_Ninety-Nine_Haskell_Problems|Ninety-Nine Haskell Problems]], based on [http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/ Ninety-Nine Prolog 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].
   
 
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.
 
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.
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Example in Haskell:
 
Example in Haskell:
<pre>
+
  +
<haskell>
 
> length (queens 8)
 
> length (queens 8)
 
92
 
92
 
> head (queens 8)
 
> head (queens 8)
[4,2,7,3,6,8,5,1]
+
[1,5,8,6,3,7,2,4]
</pre>
 
 
Solution:
 
The simplest solution is a composition of separate functions to generate the list of candidates and to test each candidate:
 
<haskell>
 
queens :: Int -> [[Int]]
 
queens n = filter test (generate n)
 
where generate 0 = [[]]
 
generate k = [q : qs | qs <- generate (k-1), q <- [1..n]]
 
test [] = True
 
test (q:qs) = isSafe q qs && test qs
 
isSafe try qs = not (try `elem` qs || sameDiag try qs)
 
sameDiag try qs = any (\(colDist,q) -> abs (try - q) == colDist) $ zip [1..] qs
 
 
</haskell>
 
</haskell>
By definition/data representation no two queens can occupy the same column.
 
<hask>try `elem` alreadySet</hask> checks for a queen in the same row, <hask>abs (try - q) == col</hask> 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.
+
[[99 questions/Solutions/90 | Solutions]]
The key to speeding it up is to fuse the composition <hask>filter test . generate</hask> into a semantically equivalent function <hask>queens'</hask> 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.
+
This yields the following version, which is much faster:
 
<haskell>
 
queens :: Int -> [[Int]]
 
queens n = queens' n
 
where queens' 0 = [[]]
 
queens' k = [q:qs | qs <- queens' (k-1), q <- [1..n], isSafe q qs]
 
isSafe try qs = not (try `elem` qs || sameDiag try qs)
 
sameDiag try qs = any (\(colDist,q) -> abs (try - q) == colDist) $ zip [1..] qs
 
</haskell>
 
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.
 
 
 
== Problem 91 ==
 
== Problem 91 ==
   
 
(**) Knight's tour
 
(**) 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?
+
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).
 
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).
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Example in Haskell:
 
Example in Haskell:
<pre>
+
  +
<haskell>
 
Knights> head $ knightsTo 8 (1,1)
 
Knights> head $ knightsTo 8 (1,1)
 
[(2,7),(3,5),(5,6),(4,8),(3,6),(4,4),(6,5),(4,6),
 
[(2,7),(3,5),(5,6),(4,8),(3,6),(4,4),(6,5),(4,6),
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(6,2),(8,1),(7,3),(5,2),(7,1),(8,3),(7,5),(8,7),
 
(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)]
 
(6,8),(4,7),(2,8),(1,6),(2,4),(1,2),(3,1),(2,3)]
</pre>
 
 
Solution:
 
<haskell>
 
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)
 
 
</haskell>
 
</haskell>
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.
 
   
  +
[[99 questions/Solutions/91 | Solutions]]
   
Solution 2:
 
 
<haskell>
 
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)
 
</haskell>
 
 
This is just the naive backtracking approach. I tried a speedup using Data.Map, but the code got too verbose to post.
 
   
 
== Problem 92 ==
 
== Problem 92 ==
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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.
 
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.
   
http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p92a.gif
+
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.
 
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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Write a predicate that calculates a numbering scheme for a given tree. What is the solution for the larger tree pictured below?
 
Write a predicate that calculates a numbering scheme for a given tree. What is the solution for the larger tree pictured below?
   
http://www.hta-bi.bfh.ch/~hew/informatik3/prolog/p-99/p92b.gif
+
https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/p92b.gif
   
 
Example in Haskell:
 
Example in Haskell:
<pre>
+
  +
<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)]
 
> 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]
 
[6,7,8,9,3,4,10,11,5,12,2,13,14,1]
</pre>
 
 
Solution:
 
<haskell>
 
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
 
 
</haskell>
 
</haskell>
   
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.
+
[[99 questions/Solutions/92 | Solutions]]
  +
 
 
 
== Problem 93 ==
 
== Problem 93 ==
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Example in Haskell:
 
Example in Haskell:
<pre>
+
P93> putStr $ unlines $ puzzle [2,3,5,7,11]
+
<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)
Line 188: Line 108:
 
2-(3-5)+7 = 11
 
2-(3-5)+7 = 11
 
2-3+5+7 = 11
 
2-3+5+7 = 11
</pre>
+
</haskell>
  +
 
The other two solutions alluded to in the problem description are dropped by the Haskell solution as trivial variants:
 
The other two solutions alluded to in the problem description are dropped by the Haskell solution as trivial variants:
  +
 
<pre>
 
<pre>
 
2 = 3-(5+(7-11))
 
2 = 3-(5+(7-11))
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</pre>
 
</pre>
   
Solution:
+
[[99 questions/Solutions/93 | Solutions]]
<haskell>
 
module P93 where
 
   
import Control.Monad
 
import Data.List
 
import Data.Maybe
 
   
type Equation = (Expr, Expr)
+
== Problem 94 ==
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
+
(***) Generate K-regular simple graphs with N nodes
-- numbers, as pretty strings.
 
puzzle :: [Integer] -> [String]
 
puzzle ns = map (flip showsEquation "") (equations ns)
 
   
-- generate all correct equations from the list of numbers
+
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?
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
+
[https://prof.ti.bfh.ch/hew1/informatik3/prolog/p-99/p94.txt Sample results]
-- 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 = "/"
 
</haskell>
 
Unlike the Prolog solution, I've eliminated solutions like
 
<tt>"1+(2+3) = 6"</tt> as a trivial variant of <tt>"1+2+3 = 6"</tt> (cf the function <tt>right_associative</tt>).
 
Apart from that, the Prolog solution is shorter because it uses built-in evaluation and printing of expressions.
 
 
== Problem 94 ==
 
 
<Problem description>
 
 
Example in Haskell:
 
<pre>
 
<example in Haskell>
 
</pre>
 
   
 
Solution:
 
Solution:

Revision as of 17:08, 15 July 2010


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 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]

Solutions


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:

  1. find a tour ending at a particular square
  2. 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)]

Solutions


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.

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.

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?

p92b.gif

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]

Solutions


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

Solutions


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?

Sample results

Solution:

<solution in haskell>

<description of implementation>