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Euler problems/81 to 90

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([http://projecteuler.net/index.php?section=problems&id=84 Problem 84]: rm c code - probable copyvio)
([http://projecteuler.net/index.php?section=problems&id=87 Problem 87]: restore simpler solution)
Line 257: Line 257:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
diag ((a:as):bss) = a:merge as (diag bss)
+
import List
map2 f as bs = [map (f a) bs | a<-as]
 
 
 
ordsums as bs = diag $ map2 (+) as bs
+
problem_87 = length expressible
sums = foldr1 ordsums $ map2 (flip (^)) [4,3,2] primes
+
where limit = 50000000
+
squares = takeWhile (<limit) (map (^2) primes)
problem_87 =print$ length $ takeWhile (<50000000) sums
+
cubes = takeWhile (<limit) (map (^3) primes)
merge xs@(x:xt) ys@(y:yt) = case compare x y of
+
fourths = takeWhile (<limit) (map (^4) primes)
LT -> x : (merge xt ys)
+
choices = [[s,c,f] | s <- squares, c <- cubes, f <- fourths]
EQ -> x : (merge xt yt)
+
unique = map head . group . sort
GT -> y : (merge xs yt)
+
expressible = filter (<limit) . unique . map sum $ choices
 
diff xs@(x:xt) ys@(y:yt) = case compare x y of
 
LT -> x : (diff xt ys)
 
EQ -> diff xt yt
 
GT -> diff xs yt
 
 
primes, nonprimes :: [Int]
 
primes = [2,3,5] ++ (diff [7,9..] nonprimes)
 
nonprimes = foldr1 f . map g $ tail primes
 
where f (x:xt) ys = x : (merge xt ys)
 
g p = [ n*p | n <- [p,p+2..]]
 
 
</haskell>
 
</haskell>
   

Revision as of 19:07, 25 February 2008

Contents

1 Problem 81

Find the minimal path sum from the top left to the bottom right by moving right and down.

Solution:

import Data.List (unfoldr)
 
columns s = 
    unfoldr f s
    where
    f [] = Nothing
    f xs = Just $ (\(a,b) -> (read a, drop 1 b)) $ break (==',') xs
 
firstLine ls = scanl1 (+) ls
 
nextLine pl [] = pl
nextLine pl (n:nl) = 
    nextLine p' nl
    where
    p' = nextCell (head pl) pl n
    nextCell _ [] [] = []
    nextCell pc (p:pl) (n:nl) = 
        pc' : nextCell pc' pl nl
        where pc' = n + min p pc
 
minSum (p:nl) = 
    last $ nextLine p' nl
    where
    p' = firstLine p
 
problem_81 c = minSum $ map columns $ lines c
main=do
    f<-readFile "matrix.txt"
    print$problem_81 f

2 Problem 82

Find the minimal path sum from the left column to the right column.

Solution:

import Data.List
import qualified Data.Map as M
import Data.Array
 
minPathSum xs t= 
    stepPath M.empty $ M.singleton t $ arr ! t
    where 
    len = genericLength $ head xs
    ys = concat $ transpose xs
    arr = listArray ((1, 1), (len, len)) ys
    nil = ((0,0),0)
    stepPath ds as 
        |fs2 p1==len =snd p1 
        |fs2 p2==len =snd p2 
        |fs2 p3==len =snd p3 
        |otherwise=stepPath ds' as3
        where
        fs2=fst.fst
        ((i, j), cost) = 
            minimumBy (\(_,a) (_,b) -> compare a b) $ M.assocs as
        tas = M.delete (i,j) as
        (p1, as1) = if i == len then (nil, tas) else check (i+1, j) tas
        (p2, as2) = if j == len then (nil, as1) else check (i, j+1) as1
        (p3, as3) = if j == 1   then (nil, as2) else check (i, j-1) as2
        check pos zs =
            if pos `M.member` tas || pos `M.member` ds 
            then (nil, zs)
            else (entry, uncurry M.insert entry $ zs)
            where
            entry = (pos, cost + arr ! pos)  
        ds' = M.insert (i, j) cost ds
 
main=do
    let parse = map (read . ("["++) . (++"]")) . words
    a<-readFile "matrix.txt"
    let s=parse a
    let m=minimum[p|a<-[1..80],let p=minPathSum s (1,a)]
    appendFile "p82.log"$show m
 
problem_82 = main

3 Problem 83

Find the minimal path sum from the top left to the bottom right by moving left, right, up, and down.

Solution:

A very verbose solution based on the Dijkstra algorithm. Infinity could be represented by any large value instead of the data type Distance. Also, some equality and ordering tests are not really correct. To be semantically correct, I think infinity == infinity should not be True and infinity > infinity should fail. But for this script's purpose it works like this.

import Array (Array, listArray, bounds, inRange, assocs, (!))
import qualified Data.Map as M 
    (fromList, Map, foldWithKey, 
    lookup, null, delete, insert, empty, update)
import Data.List (unfoldr)
import Control.Monad.State (State, execState, get, put)
import Data.Maybe (fromJust, fromMaybe)
 
type Weight  = Integer
 
data Distance = D Weight | Infinity
    deriving (Show)
 
instance Eq Distance where
    (==) Infinity Infinity = True
    (==) (D a) (D b) = a == b
    (==) _ _ = False
 
instance Ord Distance where
    compare Infinity Infinity = EQ
    compare Infinity (D _) = GT
    compare (D _) Infinity = LT
    compare (D a) (D b) = compare a b
 
data (Eq n, Num w) => Arc n w = A {node :: n, weight :: w}
    deriving (Show)
 
type Index   = (Int, Int)
type NodeMap = M.Map Index Distance
type Matrix  = Array Index Weight
type Path    = Arc Index Weight
type PathMap  = M.Map Index [Path]
 
data Queues = Q {input :: NodeMap, output :: NodeMap, pathMap :: PathMap}
    deriving (Show)
 
listToMatrix :: [[Weight]] -> Matrix
listToMatrix xs = listArray ((1,1),(cols,rows)) $ concat $ xs
    where
        cols = length $ head xs
        rows = length xs
 
directions :: [Index]
directions = [(0,-1), (0,1), (-1,0), (1,0)]
 
add :: (Num a) => (a, a) -> (a, a) -> (a, a)
add (a,b) (a', b') = (a+a',b+b')
 
arcs :: Matrix -> Index -> [Path]
arcs a idx = do
    d <- directions
    let n = add idx d
    if (inRange (bounds a) n) then
        return $ A n (a ! n)
        else
            fail "out of bounds"
 
paths :: Matrix -> PathMap
paths a = M.fromList $ map (\(idx,_) -> (idx, arcs a idx)) $ assocs a
 
nodes :: Matrix -> NodeMap
nodes a = 
    M.fromList $ (\((i,_):xs) -> (i, D (a ! (1,1))):xs) $ 
    map (\(idx,_) -> (idx, Infinity)) $ assocs a
 
extractMin :: NodeMap -> (NodeMap, (Index, Distance))
extractMin m = (M.delete (fst minNode) m, minNode)
    where
        minNode = M.foldWithKey mini ((0,0), Infinity) m
        mini i' v' (i,v)
            | v' < v    = (i', v')
            | otherwise = (i,v)
 
dijkstra :: State Queues ()
dijkstra = do
    Q i o am <- get
    let (i', n) = extractMin i
    let o' = M.insert (fst n) (snd n) o
    let i'' = updateNodes n am i'
    put $ Q i'' o' am
    if M.null i'' then return () else dijkstra
 
updateNodes :: (Index, Distance) -> PathMap -> NodeMap -> NodeMap
updateNodes (i, D d) am nm = foldr f nm ds
    where
        ds = fromJust $ M.lookup i am
        f :: Path -> NodeMap -> NodeMap
        f (A i' w) m = fromMaybe m val
            where
                val = do
                    v <- M.lookup i' m
                    if (D $ d+w) < v then
                        return $ M.update (const $ Just $ D (d+w)) i' m
                        else return m
 
shortestPaths :: Matrix -> NodeMap
shortestPaths xs = output $ dijkstra `execState` (Q n M.empty a)
    where
        n = nodes xs
        a = paths xs
 
problem_83 :: [[Weight]] -> Weight
problem_83 xs = jd $ M.lookup idx $ shortestPaths matrix
    where
        matrix = listToMatrix xs
        idx = snd $ bounds matrix
        jd (Just (D d)) = d
main=do
    f<-readFile "matrix.txt"
    let m=map sToInt $lines f
    print $problem_83 m
split :: Char -> String -> [String]
split = unfoldr . split'
 
split' :: Char -> String -> Maybe (String, String)
split' c l
    | null l = Nothing
    | otherwise = Just (h, drop 1 t)
    where (h, t) = span (/=c) l
sToInt x=map ((+0).read) $split ',' x

4 Problem 84

In the game, Monopoly, find the three most popular squares when using two 4-sided dice.

5 Problem 85

Investigating the number of rectangles in a rectangular grid.

Solution:

import List
problem_85 = snd$head$sort 
    [(k,a*b)|
    a<-[1..100],
    b<-[1..100],
    let k=abs (a*(a+1)*(b+1)*b-8000000)
    ]

6 Problem 86

Exploring the shortest path from one corner of a cuboid to another.

Solution:

import Data.List
isSquare x = 
    (truncate $ sqrt $ fromIntegral x)^2 == x
 
cube m = 
    sum [ (a`div`2) - if a > m then (a - m -1) else 0|
    a <- [1..2*m],
    isSquare ((a)^2 + m2)
    ]
    where
    m2 = m * m
 
problem_86 =
    findIndex (>1000000) (scanl (+) 0 (map cube [1..]))

7 Problem 87

Investigating numbers that can be expressed as the sum of a prime square, cube, and fourth power?

Solution:

import List
 
problem_87 = length expressible
    where limit = 50000000
          squares = takeWhile (<limit) (map (^2) primes)
          cubes   = takeWhile (<limit) (map (^3) primes)
          fourths = takeWhile (<limit) (map (^4) primes)
          choices = [[s,c,f] | s <- squares, c <- cubes, f <- fourths]
          unique  = map head . group . sort
          expressible = filter (<limit) . unique . map sum $ choices

8 Problem 88

Exploring minimal product-sum numbers for sets of different sizes.

Solution:

import Data.List
import qualified Data.Set as S
import qualified Data.Map as M
 
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]  
primeFactors n = factors n primes
  where factors n (p:ps) | p*p > n        = [n]
                         | n `mod` p == 0 = p : factors (n `div` p) (p:ps)
                         | otherwise      = factors n ps
isPrime n | n > 1     = (==1) . length . primeFactors $ n
          | otherwise = False
 
facts = concat . takeWhile valid . iterate facts' . (:[])
  where valid xs = length (head xs) > 1
        facts' = nub' . concatMap factsnext
        nub' = S.toList . S.fromList
        factsnext xs = 
          let factsnext' [] = []
              factsnext' (y:ys) = map (form y) ys ++ factsnext' ys
              form a b = a*b : (delete b . delete a $ xs)
          in map sort . factsnext' $ xs        
 
problem_88 =  sum' . extract . scanl addks M.empty . filter (not . isPrime) $ [2..]
  where extract = head . dropWhile (\nm -> M.size nm < 11999)
        sum' = S.fold (+) 0 . S.fromList . M.elems
        addks nm n = foldl (addk n) nm . facts . primeFactors $ n
        addk n nm ps =
          let k = length ps + n - sum ps
              kGood = k > 1 && k < 12001 && k `M.notMember` nm
          in if kGood then M.insert k n nm else nm

9 Problem 89

Develop a method to express Roman numerals in minimal form.

Solution:

replace ([], _) zs = zs
replace _ [] = []
replace (xs, ys) zzs@(z:zs)
    | xs == lns = ys ++ rns
    | otherwise = z : replace (xs, ys) zs
    where
    (lns, rns) = splitAt (length xs) zzs
 
problem_89 = 
    print . difference . words =<< readFile "roman.txt"
    where
    difference xs = sum (map length xs) - sum (map (length . reduce) xs)
    reduce xs = foldl (flip replace) xs [("DCCCC","CM"), ("CCCC","CD"), 
                                         ("LXXXX","XC"), ("XXXX","XL"), 
                                         ("VIIII","IX"), ("IIII","IV")]

10 Problem 90

An unexpected way of using two cubes to make a square.

Solution:

Basic brute force: generate all possible die combinations and check each one to see if we can make all the necessary squares. Runs very fast even for brute force.

-- all lists consisting of n elements from the given list
choose 0 _  = [[]]
choose _ [] = []
choose n (x:xs) =
    ( map ( x : ) ( choose ( n - 1 ) xs ) ) ++ ( choose n xs )
 
-- cross product helper function
cross f xs ys = [ f x y | x <- xs, y <- ys ]
 
-- all dice combinations
-- substitute 'k' for both '6' and '9' to make comparisons easier
dice = cross (,) ( choose 6 "012345k78k" ) ( choose 6 "012345k78k" )
 
-- can we make all square numbers from the two dice
-- again, substitute 'k' for '6' and '9'
makeSquares dice =
    all ( makeSquare dice ) [ "01", "04", "0k", "1k", "25", "3k", "4k", "k4", "81" ]
 
-- can we make this square from the two dice
makeSquare ( xs, ys ) [ d1,  d2 ] =
    ( ( ( d1 `elem` xs ) && ( d2 `elem` ys ) ) || ( ( d2 `elem` xs ) && ( d1 `elem` ys ) ) )
 
problem_90 =
    ( `div` 2 ) . -- because each die combinations will appear twice
    length .
    filter makeSquares
    $ dice