# 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) |

## Revision as of 23:55, 23 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:

diag ((a:as):bss) = a:merge as (diag bss) map2 f as bs = [map (f a) bs | a<-as] ordsums as bs = diag $ map2 (+) as bs sums = foldr1 ordsums $ map2 (flip (^)) [4,3,2] primes problem_87 =print$ length $ takeWhile (<50000000) sums merge xs@(x:xt) ys@(y:yt) = case compare x y of LT -> x : (merge xt ys) EQ -> x : (merge xt yt) GT -> y : (merge xs yt) 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..]]

## 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