Euler problems/41 to 50: Difference between revisions
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doub xs = nub xs /= xs | doub xs = nub xs /= xs | ||
</haskell> | |||
An arguably cleaner, alternate solution uses nondeterminism + state to create a backtracking monad particularly suited to this problem: | |||
<haskell> | |||
import Control.Monad.State | |||
import Data.Set | |||
type Select elem a = StateT (Set elem) [] a | |||
assert :: Bool -> Select elem () | |||
assert True = return () | |||
assert False = lift mzero | |||
select :: (Ord elem) => [elem] -> Select elem elem | |||
select as = do | |||
set <- get | |||
a <- lift as | |||
assert (not (member a set)) | |||
put (insert a set) | |||
return a | |||
runSelect :: Select elem a -> [a] | |||
runSelect m = Prelude.map fst (runStateT m empty) | |||
fromDigits = foldl (\tot d -> 10 * tot + d) 0 | |||
ds = runSelect $ do | |||
d6 <- select [0,5] | |||
d4 <- select [0,2..8] | |||
d3 <- select [0..9] | |||
d5 <- select [0..9] | |||
assert ((d3 + d4 + d5) `mod` 3 == 0) | |||
d7 <- select [0..9] | |||
assert ((2 * d5 + 3 * d6 + d7) `mod` 7 == 0) | |||
d8 <- select [0..9] | |||
assert ((d6 - d7 + d8) `mod` 11 == 0) | |||
d9 <- select [0..9] | |||
assert ((9 * d7 + 10 * d8 + d9) `mod` 13 == 0) | |||
d10 <- select [0..9] | |||
assert ((-2 * d8 + 10 * d9 + d10) `mod` 17 == 0) | |||
d2 <- select [0..9] | |||
d1 <- select [0..9] | |||
return (fromDigits [d1, d2, d3, d4, d5, d6, d7, d8, d9, d10]) | |||
answer = sum ds | |||
main = do | |||
print ds | |||
print answer | |||
</haskell> | </haskell> | ||
Revision as of 02:41, 10 September 2008
Problem 41
What is the largest n-digit pandigital prime that exists?
Solution:
import Data.List
isprime a = isprimehelper a primes
isprimehelper a (p:ps)
| a == 1 = False
| p*p > a = True
| a `mod` p == 0 = False
| otherwise = isprimehelper a ps
primes = 2 : filter isprime [3,5..]
problem_41 =
head . filter isprime . filter fun $ [7654321,7654320..]
where
fun = (=="1234567") . sort . show
Problem 42
How many triangle words can you make using the list of common English words?
Solution:
import Data.Char
trilist = takeWhile (<300) (scanl1 (+) [1..])
wordscore xs = sum $ map (subtract 64 . ord) xs
problem_42 megalist =
length [ wordscore a | a <- megalist,
elem (wordscore a) trilist ]
main = do f <- readFile "words.txt"
let words = read $"["++f++"]"
print $ problem_42 words
Problem 43
Find the sum of all pandigital numbers with an unusual sub-string divisibility property.
Solution:
import Data.List
l2n :: (Integral a) => [a] -> a
l2n = foldl' (\a b -> 10*a+b) 0
swap (a,b) = (b,a)
explode :: (Integral a) => a -> [a]
explode =
unfoldr (\a -> if a==0 then Nothing else Just $ swap $ quotRem a 10)
problem_43 = sum . map l2n . map (\s -> head ([0..9] \\ s):s)
. filter (elem 0) . genSeq [] $ [17,13,11,7,5,3,2]
mults mi ma n = takeWhile (< ma) . dropWhile (<mi) . iterate (+n) $ n
sequ xs ys = tail xs == init ys
addZ n xs = replicate (n - length xs) 0 ++ xs
genSeq [] (x:xs) = genSeq (filter (not . doub)
. map (addZ 3 . reverse . explode)
$ mults 9 1000 x)
xs
genSeq ys (x:xs) =
genSeq (do m <- mults 9 1000 x
let s = addZ 3 . reverse . explode $ m
y <- filter (sequ s . take 3) $ filter (not . elem (head s)) ys
return (head s:y))
xs
genSeq ys [] = ys
doub xs = nub xs /= xs
An arguably cleaner, alternate solution uses nondeterminism + state to create a backtracking monad particularly suited to this problem:
import Control.Monad.State
import Data.Set
type Select elem a = StateT (Set elem) [] a
assert :: Bool -> Select elem ()
assert True = return ()
assert False = lift mzero
select :: (Ord elem) => [elem] -> Select elem elem
select as = do
set <- get
a <- lift as
assert (not (member a set))
put (insert a set)
return a
runSelect :: Select elem a -> [a]
runSelect m = Prelude.map fst (runStateT m empty)
fromDigits = foldl (\tot d -> 10 * tot + d) 0
ds = runSelect $ do
d6 <- select [0,5]
d4 <- select [0,2..8]
d3 <- select [0..9]
d5 <- select [0..9]
assert ((d3 + d4 + d5) `mod` 3 == 0)
d7 <- select [0..9]
assert ((2 * d5 + 3 * d6 + d7) `mod` 7 == 0)
d8 <- select [0..9]
assert ((d6 - d7 + d8) `mod` 11 == 0)
d9 <- select [0..9]
assert ((9 * d7 + 10 * d8 + d9) `mod` 13 == 0)
d10 <- select [0..9]
assert ((-2 * d8 + 10 * d9 + d10) `mod` 17 == 0)
d2 <- select [0..9]
d1 <- select [0..9]
return (fromDigits [d1, d2, d3, d4, d5, d6, d7, d8, d9, d10])
answer = sum ds
main = do
print ds
print answer
Problem 44
Find the smallest pair of pentagonal numbers whose sum and difference is pentagonal.
Solution:
import Data.Set
problem_44 = head solutions
where solutions = [a-b | a <- penta,
b <- takeWhile (<a) penta,
isPenta (a-b),
isPenta (b+a) ]
isPenta = (`member` fromList penta)
penta = [(n * (3*n-1)) `div` 2 | n <- [1..5000]]
Problem 45
After 40755, what is the next triangle number that is also pentagonal and hexagonal?
Solution:
isPent n = (af == 0) && ai `mod` 6 == 5
where (ai, af) = properFraction . sqrt $ 1 + 24 * (fromInteger n)
problem_45 = head [x | x <- scanl (+) 1 [5,9..], x > 40755, isPent x]
Problem 46
What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
Solution:
This solution is inspired by exercise 3.70 in Structure and Interpretation of Computer Programs, (2nd ed.).
millerRabinPrimality on the Prime_numbers page
import Data.List
isPrime x | x==3 = True
| otherwise = millerRabinPrimality x 2
problem_46 = find (\x -> not (isPrime x) && check x) [3,5..]
where
check x = not . any isPrime
. takeWhile (>0)
. map (\y -> x - 2 * y * y) $ [1..]
Alternate Solution:
Considering that the answer is less than 6000, there's no need for fancy solutions. The following is as fast as most C++ solutions.
primes :: [Int]
primes = 2 : filter isPrime [3, 5..]
isPrime :: Int -> Bool
isPrime n = all (not . divides n) $ takeWhile (\p -> p^2 <= n) primes
where
divides n p = n `mod` p == 0
compOdds :: [Int]
compOdds = filter (not . isPrime) [3, 5..]
verifConj :: Int -> Bool
verifConj n = any isPrime (takeWhile (>0) $ map (\i -> n - 2*i*i) [1..])
problem_46 :: Int
problem_46 = head $ filter (not . verifConj) compOdds
Problem 47
Find the first four consecutive integers to have four distinct primes factors.
Solution:
import Data.List
problem_47 = find (all ((==4).snd)) . map (take 4) . tails
. zip [1..] . map (length . factors) $ [1..]
fstfac x = [(head a ,length a) | a <- group $ primeFactors x]
fac [(x,y)] = [x^a | a <- [0..y]]
fac (x:xs) = [a*b | a <- fac [x], b <- fac xs]
factors x = fac $ fstfac x
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
primeFactors n = factor n primes
where factor _ [] = []
factor m (p:ps) | p*p > m = [m]
| m `mod` p == 0 = p : [m `div` p]
| otherwise = factor m ps
Problem 48
Find the last ten digits of 11 + 22 + ... + 10001000.
Solution: If the problem were more computationally intensive, modular exponentiation might be appropriate. With this problem size the naive approach is sufficient.
powMod on the Prime_numbers page
problem_48 = flip mod limit $ sum [powMod limit n n | n <- [1..1000]]
where limit=10^10
Problem 49
Find arithmetic sequences, made of prime terms, whose four digits are permutations of each other.
Solution: millerRabinPrimality on the Prime_numbers page
import Control.Monad
import Data.List
isPrime x
| x==3 = True
| otherwise = millerRabinPrimality x 2
primes4 = takeWhile (<10000) $ dropWhile (<1000) primes
problem_49 = do a <- primes4
b <- dropWhile (<= a) primes4
guard (sort $ show a == sort $ show b)
let c = 2 * b - a
guard (c < 10000)
guard (sort $ show a == sort $ show c)
guard $ isPrime c
return (a, b, c)
primes = 2 : filter (\x -> isPrime x ) [3..]
Problem 50
Which prime, below one-million, can be written as the sum of the most consecutive primes?
Solution: (prime and isPrime not included)
import Control.Monad
findPrimeSum ps
| isPrime sumps = Just sumps
| otherwise = findPrimeSum (tail ps) `mplus` findPrimeSum (init ps)
where
sumps = sum ps
problem_50 = findPrimeSum $ take 546 primes