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Euler problems/91 to 100

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<haskell>
 
<haskell>
 
import Data.List (foldl1', group)
 
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.
 
-- The sum of all proper divisors of n.
d n = product [(p * product g - 1) `div` (p - 1) |
+
sumDivi m =
g <- group $ primeFactors n, let p = head g
+
product [div (a^(n+1)-1) (a-1)|
] - n
+
(a,n)<-fstfac m
+
]-m
  +
 
primeFactors = pf primes
 
primeFactors = pf primes
 
where
 
where
Line 173: Line 173:
 
where
 
where
 
(q, r) = n `divMod` p
 
(q, r) = n `divMod` p
+
primes = 2 : filter (null . tail . primeFactors) [3,5..]
+
 
 
-- The longest chain of numbers is (n, k), where
 
-- The longest chain of numbers is (n, k), where
 
-- n is the smallest number in the chain, and k is the length
 
-- n is the smallest number in the chain, and k is the length
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-- smallest number is no more than m and, optionally, whose
 
-- smallest number is no more than m and, optionally, whose
 
-- largest number is no more than m'.
 
-- largest number is no more than m'.
longestChain m m' = (n, k)
+
chain s n n'
where
+
| n' == n = s
(n, Just k) = foldl1' cmpChain [(n, findChain n) | n <- [2..m]]
+
| n' < n = []
findChain n = f [] n $ d n
+
| (< n') 1000000 = []
f s n n'
+
| n' `elem` s = []
| n' == n = Just $ 1 + length s
+
| otherwise = chain(n' : s) n $ sumDivi n'
| n' < n = Nothing
+
findChain n = length$chain [] n $ sumDivi n
| maybe False (< n') m' = Nothing
+
longestChain =
| n' `elem` s = Nothing
+
foldl1' cmpChain [(n, findChain n) | n <- [12496..15000]]
| otherwise = f (n' : s) n $ d n'
+
where
 
cmpChain p@(n, k) q@(n', k')
 
cmpChain p@(n, k) q@(n', k')
| (k, negate n) < (k', negate n') = q
+
| (k, negate n) < (k', negate n') = q
| otherwise = p
+
| otherwise = p
+
problem_95 = fst$longestChain
problem_95_v2 = longestChain 1000000 (Just 1000000)
 
 
</haskell>
 
</haskell>
   

Revision as of 04:25, 8 January 2008

Contents

1 Problem 91

Find the number of right angle triangles in the quadrant.

Solution:

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
        ]

2 Problem 92

Investigating a square digits number chain with a surprising property.

Solution:

import Data.List
import Data.Map(fromList,(!))
pow2=[a*a|a<-[0..9]]
sum_pow 1=0
sum_pow 89=1
sum_pow x= sum_pow(sum [pow2 !! y |y<-digits x])
sumMap=fromList[(k,
    fromList[(i,sum_pow ki)
    |i<-[0..1000],
    let ki=k*1000+i
    ])|
    k<-[0..400]
    ]
fastsumpow x
    |x<400000=sumMap!d!m
    |otherwise=fastsumpow(sum [pow2 !! y |y<-digits x])
    where
    (d,m)=divMod x 1000
digits n 
{-  change 123 to [3,2,1]
 -}
    |n<10=[n]
    |otherwise= y:digits x 
    where
    (x,y)=divMod n 10
groups=100000
problem_92 b=sum [fastsumpow a|a<-[1+b*groups..groups+b*groups]]
google num
  =if (num>99)
      then return()
      else do let lst=[show$problem_92 num ,"   ",(show num),"\n"]
              appendFile "files.log" $foldl (++) "" lst
              google (num+1)
main=google 0
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=((+0).read) $head$split ' ' x
problem_92a=do
    x<-readFile "files.log"
    print $sum$map sToInt $lines x

3 Problem 93

Using four distinct digits and the rules of arithmetic, find the longest sequence of target numbers.

Solution:

problem_93 = undefined

4 Problem 94

Investigating almost equilateral triangles with integral sides and area.

Solution:

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

5 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 :

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

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.

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

6 Problem 96

Devise an algorithm for solving Su Doku puzzles.

See numerous solutions on the Sudoku page.

7 Problem 97

Find the last ten digits of the non-Mersenne prime: 28433 × 27830457 + 1.

Solution:

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

8 Problem 98

Investigating words, and their anagrams, which can represent square numbers.

Solution:

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

9 Problem 99

Which base/exponent pair in the file has the greatest numerical value?

Solution:

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

10 Problem 100

Finding the number of blue discs for which there is 50% chance of taking two blue.

Solution:

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