View source for Euler problems/91 to 100

== [http://projecteuler.net/index.php?section=view&id=91 Problem 91] ==
Find the number of right angle triangles in the quadrant.

Solution:
<haskell>
reduce x y = (quot x d, quot y d)
  where d = gcd x y

problem_91 n = 
    3*n*n + 2* sum others
    where
    others =[min xc yc|
        x1 <- [1..n],
        y1 <- [1..n],
        let (yi,xi) = reduce x1 y1,
        let yc = quot (n-y1) yi,
        let xc = quot x1 xi
        ]
</haskell>

== [http://projecteuler.net/index.php?section=view&id=92 Problem 92] ==
Investigating a square digits number chain with a surprising property.

Solution:
<haskell>
import Data.Array
import Data.Char
import Data.List
makeIncreas 1 minnum  = [[a]|a<-[minnum..9]]
makeIncreas digits minnum  = [a:b|a<-[minnum ..9],b<-makeIncreas (digits-1) a]
squares :: Array Char Int
squares = array ('0','9') [ (intToDigit x,x^2) | x <- [0..9] ]
 
next :: Int -> Int
next = sum . map (squares !) . show
factorial n = if n == 0 then 1 else n * factorial (n - 1)  
countNum xs=ys
    where
    ys=product$map (factorial.length)$group xs
yield :: Int -> Int
yield = until (\x -> x == 89 || x == 1) next
problem_92=sum[div p7 $countNum a|a<-tail$makeIncreas  7 0,let k=sum $map (^2) a,yield k==89]
    where
    p7=factorial 7
</haskell>

== [http://projecteuler.net/index.php?section=view&id=93 Problem 93] ==
Using four distinct digits and the rules of arithmetic, find the longest sequence of target numbers.

Solution:
<haskell>
import Data.List
import Control.Monad
 
solve [] [x] = [x]
solve ns stack = 
    pushes ++ ops
    where
    pushes = do
        x <- ns
        solve (x `delete` ns) (x:stack)
    ops = do
        guard (length stack > 1)
        x <- opResults (stack!!0) (stack!!1)
        solve ns (x : drop 2 stack)
 
opResults a b = 
    [a*b,a+b,a-b] ++ (if b /= 0 then [a / b] else [])

results xs = fun 1 ys
    where
    ys = nub $ sort $ map truncate $ 
        filter (\x -> x > 0 && floor x == ceiling x) $ solve xs [] 
    fun n (x:xs) 
        |n == x =fun (n+1) xs 
        |otherwise=n-1
 
cmp a b = results a `compare` results b
 
main = 
    appendFile "p93.log" $ show $ 
    maximumBy cmp $ [[a,b,c,d] | 
    a <- [1..10], 
    b <- [a+1..10], 
    c <- [b+1..10], 
    d <- [c+1..10]
    ]
problem_93 = main
</haskell>

== [http://projecteuler.net/index.php?section=view&id=94 Problem 94] ==
Investigating almost equilateral triangles with integral sides and area.

Solution:
<haskell>
import List
findmin d = d:head [[n,m]|m<-[1..10],n<-[1..10],n*n==d*m*m+1]
pow 1 x=x
pow n x =mult x $pow (n-1) x 
    where
    mult [d,a, b] [_,a1, b1]=d:[a*a1+d*b*b1,a*b1+b*a1]
--find it looks like (5-5-6)
f556 =takeWhile (<10^9) 
    [n2|i<-[1..],
        let [_,m,_]=pow i$findmin 12,
        let n=div (m-1) 6,
        let n1=4*n+1,       -- sides 
        let n2=3*n1+1       -- perimeter
    ]
--find it looks like (5-6-6)
f665 =takeWhile (<10^9)
    [n2|i<-[1..],
        let [_,m,_]=pow i$findmin 3,
        mod (m-2) 3==0,
        let n=div (m-2) 3,
        let n1=2*n,
        let n2=3*n1+2
    ]
problem_94=sum f556+sum f665-2
</haskell>

== [http://projecteuler.net/index.php?section=view&id=95 Problem 95] ==
Find the smallest member of the longest amicable chain with no element exceeding one million.

Solution which avoid visiting a number more than one time :
<haskell>
import Data.Array.Unboxed
import qualified Data.IntSet as S
import Data.List

takeUntil _ [] = []
takeUntil pred (x:xs) = x : if pred x then takeUntil pred xs else []

chain n s =  lgo [n] $ properDivisorsSum ! n
    where lgo xs x | x > 1000000 || S.notMember x s = (xs,[])
                   | x `elem` xs = (xs,x : takeUntil (/= x) xs)
                   | otherwise = lgo (x:xs) $ properDivisorsSum ! x

properDivisorsSum :: UArray Int Int
properDivisorsSum = accumArray (+) 1 (0,1000000) 
                    $ (0,-1):[(k,factor)| 
                               factor<-[2..1000000 `div` 2]
                             , k<-[2*factor,2*factor+factor..1000000]
                             ]

base = S.fromList [1..1000000]

problem_95 = fst $ until (S.null . snd) f ((0,0),base)
    where 
      f (p@(n,m), s) = (p', s')
          where 
            setMin = head $ S.toAscList s
            (explored, chn) = chain setMin s
            len = length chn
            p' = if len > m then (minimum chn, len) else p
            s' = foldl' (flip S.delete) s explored
</haskell>
----
Here is a more straightforward solution, without optimization.
Yet it solves the problem in a few seconds when
compiled with GHC 6.6.1 with the -O2 flag. I like to let
the compiler do the optimization, without cluttering my code.

This solution avoids using unboxed arrays, which many consider to be
somewhat of an imperitive-style hack. In fact, no memoization
at all is required.

<haskell>
import Data.List (foldl1', group)
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    = [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..]] 
fstfac x = [(head a ,length a)|a<-group$primeFactors x] 
-- The sum of all proper divisors of n.
sumDivi m =
    product [div (a^(n+1)-1) (a-1)|
    (a,n)<-fstfac m
    ]-m
 
primeFactors = pf primes
 where
   pf ps@(p:ps') n
    | p * p > n = [n]
    | r == 0    = p : pf ps q
    | otherwise = pf ps' n
    where
      (q, r) = n `divMod` p
 
 
-- The longest chain of numbers is (n, k), where
-- n is the smallest number in the chain, and k is the length
-- of the chain. We limit the search to chains whose
-- smallest number is no more than m and, optionally, whose
-- largest number is no more than m'.
chain s n n'
    | n' == n               = s
    | n' < n                = []
    | (< n') 1000000 = []
    | n' `elem` s           = []
    | otherwise             = chain(n' : s) n $ sumDivi n'
findChain n = length$chain [] n $ sumDivi n
longestChain =
    foldl1' cmpChain [(n, findChain n) | n <- [12496..15000]]
    where
    cmpChain p@(n, k) q@(n', k')
        | (k, negate n) < (k', negate n') = q
        | otherwise                       = p
problem_95 = fst$longestChain  
</haskell>

== [http://projecteuler.net/index.php?section=view&id=96 Problem 96] ==
Devise an algorithm for solving Su Doku puzzles.

See numerous solutions on the [[Sudoku]] page.
<haskell>
import Data.List
import Char
 
top3 :: Grid -> Int
top3 g =
    read . take 3 $ (g !! 0)

type Grid = [String]
type Row  = String
type Col  = String
type Cell = String
type Pos = Int
 
row :: Grid -> Pos -> Row
row [] _ = []
row g  p = filter (/='0') (g !! (p `div` 9))
 
col :: Grid -> Pos -> Col
col [] _ = []
col g  p = filter (/='0') ((transpose g) !! (p `mod` 9))
 
cell :: Grid -> Pos -> Cell
cell [] _ = []
cell g  p =
    concat rows
    where
    r = p `div` 9 `div` 3 * 3
    c = p `mod` 9 `div` 3 * 3
    rows =
        map (take 3 . drop c) . map (g !!) $ [r, r+1, r+2]
 
groupsOf _ [] = []
groupsOf n xs = 
    front : groupsOf n back
    where
    (front,back) = splitAt n xs

extrapolate :: Grid -> [Grid]
extrapolate [] = []
extrapolate g  =
    if null zeroes
    then [] -- no more zeroes, must have solved it
    else map mkGrid possibilities
    where
    flat = concat g
    numbered = zip [0..] flat
    zeroes = filter ((=='0') . snd) numbered
    p = fst . head $ zeroes
    possibilities =
        ['1'..'9'] \\ (row g p ++ col g p ++ cell g p)
    (front,_:back) = splitAt p flat
    mkGrid new = groupsOf 9 (front ++ [new] ++ back)
 
loop :: [Grid] -> [Grid]
loop [] = []
loop xs = concat . map extrapolate $ xs
 
solve :: Grid -> Grid
solve g =
    head .
    last .
    takeWhile (not . null) .
    iterate loop $ [g]

main = do
    contents <- readFile "sudoku.txt"
    let
        grids :: [Grid]
        grids =
            groupsOf 9 .
            filter ((/='G') . head) .
            lines $ contents
    let rgrids=map (concat.map words) grids
    writeFile "p96.log"$show$  sum $ map (top3 . solve) $ rgrids
problem_96 =main
</haskell>
== [http://projecteuler.net/index.php?section=view&id=97 Problem 97] ==
Find the last ten digits of the non-Mersenne prime: 28433 × 2<sup>7830457</sup> + 1.

Solution:
<haskell>
mulMod :: Integral a => a -> a -> a -> a
mulMod a b c= (b * c) `rem` a
squareMod :: Integral a => a -> a -> a
squareMod a b = (b * b) `rem` a
pow' :: (Num a, Integral b) => (a -> a -> a) -> (a -> a) -> a -> b -> a
pow' _ _ _ 0 = 1
pow' mul sq x' n' = f x' n' 1
    where
    f x n y
        | n == 1 = x `mul` y
        | r == 0 = f x2 q y
        | otherwise = f x2 q (x `mul` y)
        where
            (q,r) = quotRem n 2
            x2 = sq x
powMod :: Integral a => a -> a -> a -> a
powMod m = pow' (mulMod m) (squareMod m)
problem_97 = 
    flip mod limit $ 28433 * powMod limit 2 7830457 + 1 
    where
    limit=10^10
</haskell>

== [http://projecteuler.net/index.php?section=view&id=98 Problem 98] ==
Investigating words, and their anagrams, which can represent square numbers.

Solution:
<haskell>
import Data.List
import Data.Maybe

-- Replace each letter of a word, or digit of a number, with
-- the index of where that letter or digit first appears
profile :: Ord a => [a] -> [Int]
profile x = map (fromJust . flip lookup (indices x)) x
  where
    indices = map head . groupBy fstEq . sort . flip zip [0..]

-- Check for equality on the first component of a tuple
fstEq :: Eq a => (a, b) -> (a, b) -> Bool
fstEq x y = (fst x) == (fst y)

-- The histogram of a small list
hist :: Ord a => [a] -> [(a, Int)]
hist = let item g = (head g, length g) in map item . group . sort

-- The list of anagram sets for a word list.
anagrams :: Ord a => [[a]] -> [[[a]]]
anagrams x = map (map snd) $ filter (not . null . drop 1) $
             groupBy fstEq $ sort $ zip (map hist x) x

-- Given two finite lists that are a permutation of one
-- another, return the permutation function
mkPermute :: Ord a => [a] -> [a] -> ([b] -> [b])
mkPermute x y = pairsToPermute $ concat $
                zipWith zip (occurs x) (occurs y)
  where
    pairsToPermute ps = flip map (map snd $ sort ps) . (!!)
    occurs = map (map snd) . groupBy fstEq . sort . flip zip [0..]

problem_98 :: [String] -> Int
problem_98 ws = read $ head
  [y | was <- sortBy longFirst $ anagrams ws,     -- word anagram sets
       w1:t <- tails was, w2 <- t,
       let permute = mkPermute w1 w2,
       nas <- sortBy longFirst $ anagrams $
              filter ((== profile w1) . profile) $
              dropWhile (flip longerThan w1) $
              takeWhile (not . longerThan w1) $
              map show $ map (\x -> x * x) [1..], -- number anagram sets
       x:t <- tails nas, y <- t,
       permute x == y || permute y == x
  ]

run_problem_98 :: IO Int
run_problem_98 = do
  words_file <- readFile "words.txt"
  let words = read $ '[' : words_file ++ "]"
  return $ problem_98 words

-- Sort on length of first element, from longest to shortest
longFirst :: [[a]] -> [[a]] -> Ordering
longFirst (x:_) (y:_) = compareLen y x

-- Is y longer than x?
longerThan :: [a] -> [a] -> Bool
longerThan x y = compareLen x y == LT

-- Compare the lengths of lists, with short-circuiting
compareLen :: [a] -> [a] -> Ordering
compareLen (_:xs) y  = case y of (_:ys) -> compareLen xs ys
                                 _      -> GT
compareLen _      [] = EQ
compareLen _      _  = LT
</haskell>

== [http://projecteuler.net/index.php?section=view&id=99 Problem 99] ==
Which base/exponent pair in the file has the greatest numerical value?

Solution:
<haskell>
import Data.List
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
lognum [a, b]=b*log a
logfun x=lognum$map ((+0).read)  $split ',' x 
problem_99 file = 
    head$map fst $ sortBy (\(_,a) (_,b) -> compare  b a) $ 
    zip [1..] $map logfun $lines file
main=do
    f<-readFile "base_exp.txt"
    print$problem_99 f
</haskell>

== [http://projecteuler.net/index.php?section=view&id=100 Problem 100] ==
Finding the number of blue discs for which there is 50% chance of taking two blue.

Solution:
<haskell>
nextAB a b 
    |a+b>10^12 =[a,b]
    |otherwise=nextAB (3*a+2*b+2) (4*a+3*b+3)
problem_100=(+1)$head$nextAB 14 20
</haskell>