Difference between revisions of "Euler problems/81 to 90"
(→Problem 82) 
(→Problem 81: Same algorithm, smaller and cleaner code) 

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Solution: 
Solution: 

<haskell> 
<haskell> 

−  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 

−  
⚫  
−  main=do 

⚫  
−  print$problem_81 f 

−  </haskell> 

−  Another solution that uses Data.Array: 

⚫  
−  <haskell> 

−  import Data.Array 

−  +  parse :: String > [Int] 

−  +  parse = read . ('[':) . (++ "]") 

−  where 

−  a = listArray ((0,0), (79,79)) . concatMap read . map (\xs > '[' : xs ++ "]") . lines $ txt 

−  b = array ((0,0), (80,80)) $ [((80, i), 0)  i < [0..80]] 

−  ++ [((i, 80), 0)  i < [0..80]] 

−  ++ [((79, i), a ! (79, i) + b ! (79, i+1))  i < [79,78..0]] 

−  ++ [((i, 79), a ! (i, 79) + b ! (i+1, 79))  i < [79,78..0]] 

−  ++ [((i, j), a ! (i,j) + min (b ! (i+1, j)) (b ! (i, j+1)))  i < [78,77..0], j < [78,77..0]] 

−  main :: IO () 

+  minSum :: [[Int]] > Int 

⚫  
+  minSum (x:xs) = last $ (foldl nextLine) (scanl1 (+) x) xs 

−  txt < readFile "matrix.txt" 

+  
⚫  
+  nextLine :: [Int] > [Int] > [Int] 

+  nextLine (p:pl) (n:nl) = scanl nextCell (p+n) (zip pl nl) 

+  where nextCell acc (prev, new) = new + min prev acc 

</haskell> 
</haskell> 

Line 164:  Line 137:  
</haskell> 
</haskell> 

−  
−  
== [http://projecteuler.net/index.php?section=problems&id=82 Problem 82] == 
== [http://projecteuler.net/index.php?section=problems&id=82 Problem 82] == 
Latest revision as of 03:07, 8 December 2011
Contents
Problem 81
Find the minimal path sum from the top left to the bottom right by moving right and down.
Solution:
main = do
file < readFile "matrix.txt"
print $ problem_82 file
problem_82 = minSum . map parse . lines
parse :: String > [Int]
parse = read . ('[':) . (++ "]")
minSum :: [[Int]] > Int
minSum (x:xs) = last $ (foldl nextLine) (scanl1 (+) x) xs
nextLine :: [Int] > [Int] > [Int]
nextLine (p:pl) (n:nl) = scanl nextCell (p+n) (zip pl nl)
where nextCell acc (prev, new) = new + min prev acc
I am offering this solution not because it is particularly brilliant, but because it introduces the wonderful fgl Graph Library written by Martin Erwig. Martin's Data.Graph.Inductive library allows you to solve problems 81,82, and 83 with exactly the same code, and best of all, little or no thinking. The idea is to convert the n by n matrix into an n^2 by n^2 graph whose edges depend on the allowed paths. Fortunately these graphs are very sparse, averaging only 4 edges per node. This allows us to use the Dijkstra algorithm to find the shortest path in a graph.
The only slightly dodgy bit is problem 82, where we must find the shortest path from the first column to the last column. In order to avoid recomputing the Dijkstra algorithm over and over again, you have to be a little careful in the order of evaluation. I used spTree function from Data.Graph.Inductive.Query.SP which generated the shortest path tree from a given initial node to all other nodes. I then map over this tree with the nodes of the graph that are in the last column. The tree only needs to be calculated once for each element in the first column, rather than for every pair (i,j). This reduces the running time by a factor of n. Henry Laxen  Apr. 27, 2008
Note that problem 82 may also be solved using a straightforward Dijkstra by adding an initial node A connected to all the nodes in the first column, and a final node B that all the nodes of the last column connect to, and then searching for a path from A to B.
import Data.Graph.Inductive
import Data.Graph.Inductive.Graph
import Data.Graph.Inductive.Query.SP
import Data.Graph.Inductive.Internal.RootPath
import Data.List (unfoldr, minimumBy)
import Data.Ord (comparing)
type Matrix = [[Int]]
type IJ = (Int, Int)
connect81, connect82, connect83 :: [IJ]
connect81 = [(1,0),(0,1)]
connect82 = [(1,0),(1,0),(0,1)]
connect83 = [(1,0),(0,1),(1,0),(0,1)]
dimensions :: Matrix > IJ
dimensions matrix = (length matrix, length (matrix!!0))
ijToindex :: Matrix > IJ > Int
ijToindex matrix (i,j) = i*rows + j
where (rows,cols) = dimensions matrix
indexToij :: Matrix > Int > IJ
indexToij matrix index = divMod index rows
where (rows,cols) = dimensions matrix
ijValid :: Matrix > [IJ] > [IJ]
ijValid matrix ijs = filter f ijs
where (rows,cols) = dimensions matrix
f (i,j) = i >= 0 && i < rows && j >= 0 && j < cols
ijPlus :: IJ > IJ > IJ
ijPlus (i1,j1) (i2,j2) = ((i1+i2),(j1+j2))
mEdges :: Matrix > [IJ] > IJ > [(Int, Int, Int)]
mEdges matrix connectL (i,j) =
let ijs = ijValid matrix $ map (ijPlus (i,j)) connectL
in map (\(x,y) > (ijToindex matrix (i,j),
ijToindex matrix (x,y),
matrix!!x!!y)) ijs
mGraph :: Matrix > [IJ] > Gr IJ Int
mGraph matrix connectL =
let (rows,cols) = dimensions matrix
ijs = [(i,j)  i<[0..(rows1)], j<[0..(cols1)]]
mnodes = map (\(x,y) > (ijToindex matrix (x,y) ,(x,y))) ijs
medges = concatMap (mEdges matrix connectL) ijs
 Everything written above is leading up to this line,
 namely transforming an m x n matrix into an mn x mn graph
in mkGraph mnodes medges
mSPlen :: Matrix > [IJ] > [IJ] > [IJ] > ((IJ, IJ), Int)
mSPlen matrix connectL from to =
let (rows,cols) = dimensions matrix
mx (i,j) = matrix!!i!!j
ijI = ijToindex matrix
gr = mGraph matrix connectL
spTrees = [(x,spTree (ijI x) gr)  x < from]
distance (i,j) = getDistance (ijI (i,j))
distances = [((a,y), distance y b + mx a)  (a,b) < spTrees, y < to]
in minimumBy (comparing snd) distances
debug = False
mName = if debug then "small_matrix.txt" else "matrix.txt"
columns :: [Char] > [Int]
columns s =
unfoldr f s
where
f [] = Nothing
f xs = Just $ (\(a,b) > (read a, drop 1 b)) $ break (==',') xs
main = do
f<readFile mName
let matrix = map columns $ lines f
(rows,cols) = dimensions matrix
firstColumn = [(i,0)  i<[0..(rows1)]]
lastColumn = [(i,(rows1))  i<[0..(rows1)]]
topLeft = [(0,0)]
bottomRight = [(rows1,cols1)]
putStrLn $ "Problem 81: " ++
(show $ mSPlen matrix connect81 topLeft bottomRight)
putStrLn $ "Problem 82: " ++
(show $ mSPlen matrix connect82 firstColumn lastColumn)
putStrLn $ "Problem 83: " ++
(show $ mSPlen matrix connect83 topLeft bottomRight)
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
import Data.Ord (comparing)
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 (comparing snd) $ 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, j1) 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[minPathSum s (1,a)a<[1..80]]
appendFile "p82.log"$show m
problem_82 = main
Another concise approach:
import Data.List
main = do
s < readFile "matrix.txt"
let a = transpose . map (\x > read ("["++x++"]")) . lines $ s
print $ minimum $ foldl1 (\u v >
let l1 = (head u + head v) : zipWith3 (\x y z > x + min y z) (tail v) l1 (tail u)
v' = reverse v
l1' = reverse l1
l2 = head l1' : zipWith3 (\x y z > min x (y+z)) (tail l1') l2 (tail v')
in reverse l2) a
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 (unless)
import Control.Monad.State (State, execState, get, put)
import Data.Maybe (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', (x,y)) = extractMin i
let o' = M.insert x y o
let i'' = updateNodes n am i'
put $ Q i'' o' am
unless (M.null i'') dijkstra
updateNodes :: (Index, Distance) > PathMap > NodeMap > NodeMap
updateNodes (i, D d) am nm = foldr f nm ds
where
Just ds = 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
Problem 84
In the game, Monopoly, find the three most popular squares when using two 4sided dice.
This may not be the shortest or the fastest implementation, but I hope it is one of the clearest. I have one comment about the experience of solving this problem that I would like to share with you. At first I thought I would have to make use of the Control.Monad.State library, but being relatively new to Haskell, I quickly found myself in the slough of type checker despond. It was then that I remembered that foldl/foldr can used instead of "State," and now I found myself in the celestial city of type checker heaven, with Haskell preventing me from making silly mistakes at every turn. HenryLaxen May 7, 2008
import Data.Array.IArray
import Data.List
import Data.Ord
import System.Random
data Squares =
GO  A1  CC1  A2  T1  R1  B1  CH1  B2  B3  JAIL 
C1  U1  C2  C3  R2  D1  CC2  D2  D3  FP 
E1  CH2  E2  E3  R3  F1  F2  U2  F3  G2J 
G1  G2  CC3  G3  R4  CH3  H1  T2  H2
deriving (Eq,Ord,Enum,Read,Show,Ix)
type Roll = [Int]
data Cards = GoTo Squares  R  U  Back3  Other
deriving (Eq,Ord,Read,Show)
type Deck = [Cards]
data GameState = GameState
{ position :: Squares,
doublesCount :: Int,
chance :: [Cards],
communityChest :: [Cards],
history :: [Squares]
} deriving (Eq,Ord,Read,Show)
deckCommunityChest = [ GoTo JAIL, GoTo GO ] ++ replicate 14 Other
deckChance = [ GoTo GO, GoTo JAIL, GoTo C1,
GoTo E3, GoTo H2, GoTo R1] ++
[ R, U, Back3] ++
replicate 6 Other
doubles :: Roll > Bool
doubles r = r!!0 == r!!1
defaultGameState = GameState
{ position = GO,
doublesCount = 0,
chance = deckChance,
communityChest = deckCommunityChest,
history = [GO]
}
takeCard :: Deck > (Cards,Deck)
takeCard (c:cs) = (card,deck)
where card = c
deck = cs ++ [card]
nextR g = case position g of
CH1 > R2
CH2 > R3
CH3 > R1
nextU g = case position g of
CH1 > U1
CH2 > U2
CH3 > U1
doCommunityChest :: GameState > GameState
doCommunityChest g =
let (card,deck) = takeCard (communityChest g)
rotate g = g {communityChest = deck }
cases = case card of
GoTo sq > g { position = sq }
Other > g
in rotate cases
doChance :: GameState > GameState
doChance g =
let (card,deck) = takeCard (chance g)
rotate g = g {chance = deck }
cases = case card of
GoTo sq > g { position = sq}
R > g { position = nextR g }
U > g { position = nextU g }
 you might back up from CH3 to CC3 so checkForCards again
Back3 > checkForCards (g { position = position (newPosition g (3))})
Other > g
in rotate cases
newPosition :: GameState > Int > GameState
newPosition g n = g {position = toEnum $
(fromEnum (position g) + n) `mod` (fromEnum H2 + 1)}
checkForCards :: GameState > GameState
checkForCards g
 (position g) `elem` [CH1, CH2, CH3] = doChance g
 (position g) `elem` [CC1, CC2, CC3] = doCommunityChest g
 otherwise = g
travel :: GameState > [Int] > GameState
travel g roll =
let value = sum roll
checkDoubles
 doubles roll && doublesCount g == 2 =
g { position = JAIL,
doublesCount = 0 }
 doubles roll = move $ g { doublesCount = (doublesCount g) + 1}
 otherwise = move $ g { doublesCount = 0}
move g = newPosition g value
checkForJail g
 (position g) == G2J = g { position = JAIL }
 otherwise = g
saveHistory g = g { history = (position g) : (history g) }
in saveHistory $ checkForCards $ checkForJail $ checkDoubles
game :: GameState > [Roll] > GameState
 As an exercise in what a difference strictness can make
 compare the performance of this with replacing foldl' by foldl
game g rolls = foldl' (\x y > travel x y) g rolls
statistics :: [Squares] > [(Squares, Float)]
statistics history =
let a = accumArray (+) 0 (GO,H2) (zip history (repeat 1)) :: Array Squares Int
b = assocs a
c = reverse $ sortBy (comparing snd) b
(sq,cnt) = unzip c  wiki formatting bug, should be unzip c
total = sum cnt
stats = map (\x > ((fromIntegral x) / (fromIntegral total) * 100)) cnt
in take 3 $ zip sq stats
r = [[1,1],[2,2],[2,2],[4,4]]
t = game defaultGameState r  useful for debugging
pairs :: [a] > [[a]]
pairs [] = [[]]
pairs (x:y:xs) = [[x,y]] ++ (pairs xs)
dieSides :: (Int,Int)
 dieSides = (1,6)
dieSides = (1,4)
maxRolls = 100000
main = do
seed < newStdGen
let rolls = pairs (randomRs dieSides seed)
stats = statistics (history (game defaultGameState
(take maxRolls rolls)))
result = map (fromEnum . fst) stats
print (stats,result)
Problem 85
Investigating the number of rectangles in a rectangular grid.
Solution:
import List
problem_85 = snd$minimum
[(k,a*b)
a<[1..100],
b<[1..100],
let k=abs (a*(a+1)*(b+1)*b8000000)
]
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..]))
Problem 87
Investigating numbers that can be expressed as the sum of a prime square, cube, and fourth power?
Solution:
import Data.Array.Unboxed
takeMapPrimes :: Integer > (Integer > Integer) > [Integer]
takeMapPrimes u f = takeWhile (<u) . map f $ primes
squares = takeMapPrimes 50000000 (^2)
cubes = takeMapPrimes 50000000 (^3)
fourths = takeMapPrimes 50000000 (^4)
expressible :: UArray Integer Bool
expressible = accumArray () False (1, 50000000) [(t, True)  a < squares,
b < takeWhile (<(50000000a)) cubes,
c < takeWhile (<(50000000ab)) fourths,
let t = a + b + c]
problem_87 :: Int
problem_87 = length $ filter id $ elems expressible
Problem 88
Exploring minimal productsum 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
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")]
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