Difference between revisions of "Euler problems/41 to 50"
CaleGibbard (talk  contribs) 
m (Added code and nondeterminism categories) 

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Line 4:  Line 4:  
Solution: 
Solution: 

<haskell> 
<haskell> 

⚫  
+   Assuming isPrime has been implemented 

−  isprime a = isprimehelper a primes 

+  import Data.Char (intToDigit) 

−  isprimehelper a (p:ps) 

+  problem_41 = maximum [ n'  d < [3..9], n < permute ['1'..intToDigit d], 

−  +  let n' = read n, isPrime n'] 

−   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 
where 

−  fun = (=="1234567") . sort . show 

+  permute "" = [""] 

+  permute str = [(x:xs) x < str, xs < permute (delete x str)] 

</haskell> 
</haskell> 

Line 58:  Line 59:  
genSeq (do m < mults 9 1000 x 
genSeq (do m < mults 9 1000 x 

let s = addZ 3 . reverse . explode $ m 
let s = addZ 3 . reverse . explode $ m 

−  y < filter (sequ s . take 3) $ filter ( 
+  y < filter (sequ s . take 3) $ filter (notElem (head s)) ys 
return (head s:y)) 
return (head s:y)) 

xs 
xs 

Line 64:  Line 65:  
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 

+  
+  select :: (Ord elem) => [elem] > Select elem elem 

+  select as = do 

+  set < get 

+  a < lift as 

+  guard (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 

+  d4 < select [0,2..8] 

+  d3 < select [0..9] 

+  d5 < select [0..9] 

+  guard ((d3 + d4 + d5) `mod` 3 == 0) 

+  d6 < select [0,5] 

+  d7 < select [0..9] 

+  guard ((100 * d5 + 10 * d6 + d7) `mod` 7 == 0) 

+  d8 < select [0..9] 

+  guard ((d6  d7 + d8) `mod` 11 == 0) 

+  d9 < select [0..9] 

+  guard ((100 * d7 + 10 * d8 + d9) `mod` 13 == 0) 

+  d10 < select [0..9] 

+  guard ((100 * 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> 

+  
+  An almost instant answer can be generated by only creating permutations which fulfil the requirement of particular digits being multiples of certain numbers. 

+  
+  <haskell> 

+  import Data.List ((\\), nub) 

+  
+  main = print q43 

+  
+  q43 = sum [ read n  (d7d8d9, remDigits) < permMults digits 17, 

+  (d4d5d6, remDigits') < permMults remDigits 7, 

+  d4d5d6 !! 1 == '0'  d4d5d6 !! 1 == '5', 

+  (d1d2d3, remDigit) < permMults remDigits' 2, 

+  let n = remDigit ++ d1d2d3 ++ d4d5d6 ++ d7d8d9, 

⚫  
+  where 

+  digits = "0123456789" 

+  primes = [2,3,5,7,11,13,17] 

+  hasProperty _ [] = True 

+  hasProperty c (p:ps) = (read $ take 3 c) `mod` p == 0 

+  && hasProperty (tail c) ps 

+  permMults cs p = [ (ds, cs \\ ds)  n < [p,2*p..987], 

+  let ds = leadingZero n, 

+  ds == nub ds, 

+  all (flip elem cs) ds] 

+  where 

+  leadingZero n 

⚫  
+   n < 100 = "0" ++ show n 

+   otherwise = show n 

</haskell> 
</haskell> 

Line 80:  Line 158:  
penta = [(n * (3*n1)) `div` 2  n < [1..5000]] 
penta = [(n * (3*n1)) `div` 2  n < [1..5000]] 

</haskell> 
</haskell> 

+  
+  The above solution finds the correct answer but searches the pairs in the wrong order. Lengthier and slower but perhaps more correct solution [https://gist.github.com/2079968 here]. 

== [http://projecteuler.net/index.php?section=problems&id=45 Problem 45] == 
== [http://projecteuler.net/index.php?section=problems&id=45 Problem 45] == 

Line 110:  Line 190:  
. takeWhile (>0) 
. takeWhile (>0) 

. map (\y > x  2 * y * y) $ [1..] 
. map (\y > x  2 * y * y) $ [1..] 

+  </haskell> 

+  
+  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. 

+  
+  <haskell> 

+  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 

</haskell> 
</haskell> 

Line 129:  Line 232:  
where factor _ [] = [] 
where factor _ [] = [] 

factor m (p:ps)  p*p > m = [m] 
factor m (p:ps)  p*p > m = [m] 

−   m `mod` p == 0 = p 
+   m `mod` p == 0 = [p, m `div` p] 
 otherwise = factor m ps 
 otherwise = factor m ps 

+  </haskell> 

+  
+  
+  Alternate Solution: 

+  The previous solution actually didn't give the correct answer for me. The following method did. 

+  
+  <haskell> 

+  
⚫  
+  import Data.Numbers 

+  import Data.Numbers.Primes 

+  import qualified Data.Set as Set 

+  
+  dPrimeFactors n = Set.fromList $ primeFactors n 

+  
+  dPFList n = [(k, dPrimeFactors k) 

+   k < filter (\z > (not $ isPrime z)) [1..n]] 

+  
+  nConsec n s = 

+  let dpf = dPFList s 

+  fltrd = filter (\z > Set.size (snd z) == n) dpf 

+  gps = [take (fromIntegral n) (drop (fromIntegral k) fltrd) 

+   k < [0..(length fltrd  n)] ] 

+  gps2 = filter (\z > isConsec (map fst z)) gps 

+  in filter (\zz > Set.empty == 

+  foldl (\acc z > Set.intersection acc (snd z)) 

+  (snd (head zz)) 

+  zz) gps2 

+  
+  isConsec xs = (sort xs) == [(minimum xs)..(maximum xs)] 

+  
+  problem_47 = (fst . head . head) $ nConsec 4 20000 

+  
</haskell> 
</haskell> 

Line 142:  Line 278:  
<haskell> 
<haskell> 

−  problem_48 = 
+  problem_48 = (`mod` limit) $ sum [powMod limit n n  n < [1..1000]] 
where limit=10^10 
where limit=10^10 

+  </haskell> 

+  
+  Another oneliner for this problem, with no use of other functions is the following: 

+  <haskell> 

+  problem_48 = reverse $ take 10 

+  $ reverse $ show $ sum $ map (\x > x^x) [1..1000] 

</haskell> 
</haskell> 

Line 153:  Line 295:  
<haskell> 
<haskell> 

⚫  
import Data.List 
import Data.List 

+  
isPrime x 
isPrime x 

 x==3 = True 
 x==3 = True 

 otherwise = millerRabinPrimality x 2 
 otherwise = millerRabinPrimality x 2 

−  
−  primes4 = takeWhile (<10000) $ dropWhile (<1000) primes 

⚫  
+  primes4 = filter isPrime [1000..9999] 

⚫  
+  
⚫  
⚫  
−  +  b < dropWhile (<= a) primes4, 

−  +  sort (show a) == sort (show b), 

−  +  let c = 2 * b  a, 

−  +  c `elem` primes4, 

−  +  sort (show a) == sort (show c) ] 

−  
−  primes = 2 : filter (\x > isPrime x ) [3..] 

</haskell> 
</haskell> 

Line 187:  Line 327:  
problem_50 = findPrimeSum $ take 546 primes 
problem_50 = findPrimeSum $ take 546 primes 

</haskell> 
</haskell> 

+  
+  * This code is wrong: if you switch ''init'' and ''tail'', it produces wrong result, so it is by sheer luck that it produces the right answer at all. As written, it will produce the longest chain of primes ending at 546th prime, summing up to a prime  using 546 as it is the longest prefix of primes summing up to less than a million. That's not what the problem asks for. 

+  
+  :What's to guarantee us there's no longer chain ending at 545th prime? 544th? For instance, for 1,100,000 the longest sequence ends at 568th instead of 571st prime which is what the above code would use. 

+  
+  :Moreover, cutting the search short at just first 546 primes is wrong too. What if the longest chain was really short, like 10 or 20 primes? Then we'd have to go much higher into the primes. We have no way of knowing that length in advance. 

+  
+  * Here's my solution, it's not the fastest but is correct, feel free to criticise (isPrime and primes not included): 

+  <haskell> 

+  import Euler.Helpers 

+  import qualified Data.List as L 

+  
+  prime n = takeWhileSum n primes 

+  takeWhileSum n = takeWhileArr (\x > sum x <= n) 

+  takeWhileArr f xs = takeWhileF f [] xs 

+  where 

+  takeWhileF f rs [] = reverse rs 

+  takeWhileF f rs (x:xs) 

+   f (x:rs) = takeWhileF f (x:rs) xs 

+   otherwise = reverse rs 

+  
+  primeSums n = map (map (\x > (isPrime x,x) ) . takeWhile (<n) . scanl1 (+)) (L.tails (prime n)) 

+  main = print . maximum $ map index (primeSums 100000) 

+  where index x = if null $ ind x 

+  then (0,0) 

+  else (last $ ind x, snd (x !! (last $ ind x))) 

+  ind = L.findIndices (fst) 

+  </haskell> 

+  
+  
+  [[Category:Code]] 

+  [[Category:Nondeterminism]] 
Latest revision as of 05:45, 9 March 2021
Contents
Problem 41
What is the largest ndigit pandigital prime that exists?
Solution:
 Assuming isPrime has been implemented
import Data.Char (intToDigit)
problem_41 = maximum [ n'  d < [3..9], n < permute ['1'..intToDigit d],
let n' = read n, isPrime n']
where
permute "" = [""]
permute str = [(x:xs) x < str, xs < permute (delete x str)]
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 substring 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 (notElem (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
import Control.Monad.State
import Data.Set
type Select elem a = StateT (Set elem) [] a
select :: (Ord elem) => [elem] > Select elem elem
select as = do
set < get
a < lift as
guard (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
d4 < select [0,2..8]
d3 < select [0..9]
d5 < select [0..9]
guard ((d3 + d4 + d5) `mod` 3 == 0)
d6 < select [0,5]
d7 < select [0..9]
guard ((100 * d5 + 10 * d6 + d7) `mod` 7 == 0)
d8 < select [0..9]
guard ((d6  d7 + d8) `mod` 11 == 0)
d9 < select [0..9]
guard ((100 * d7 + 10 * d8 + d9) `mod` 13 == 0)
d10 < select [0..9]
guard ((100 * 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
An almost instant answer can be generated by only creating permutations which fulfil the requirement of particular digits being multiples of certain numbers.
import Data.List ((\\), nub)
main = print q43
q43 = sum [ read n  (d7d8d9, remDigits) < permMults digits 17,
(d4d5d6, remDigits') < permMults remDigits 7,
d4d5d6 !! 1 == '0'  d4d5d6 !! 1 == '5',
(d1d2d3, remDigit) < permMults remDigits' 2,
let n = remDigit ++ d1d2d3 ++ d4d5d6 ++ d7d8d9,
hasProperty (tail n) primes]
where
digits = "0123456789"
primes = [2,3,5,7,11,13,17]
hasProperty _ [] = True
hasProperty c (p:ps) = (read $ take 3 c) `mod` p == 0
&& hasProperty (tail c) ps
permMults cs p = [ (ds, cs \\ ds)  n < [p,2*p..987],
let ds = leadingZero n,
ds == nub ds,
all (flip elem cs) ds]
where
leadingZero n
 n < 10 = "00" ++ show n
 n < 100 = "0" ++ show n
 otherwise = show n
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 = [ab  a < penta,
b < takeWhile (<a) penta,
isPenta (ab),
isPenta (b+a) ]
isPenta = (`member` fromList penta)
penta = [(n * (3*n1)) `div` 2  n < [1..5000]]
The above solution finds the correct answer but searches the pairs in the wrong order. Lengthier and slower but perhaps more correct solution here.
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
Alternate Solution:
The previous solution actually didn't give the correct answer for me. The following method did.
import Data.List
import Data.Numbers
import Data.Numbers.Primes
import qualified Data.Set as Set
dPrimeFactors n = Set.fromList $ primeFactors n
dPFList n = [(k, dPrimeFactors k)
 k < filter (\z > (not $ isPrime z)) [1..n]]
nConsec n s =
let dpf = dPFList s
fltrd = filter (\z > Set.size (snd z) == n) dpf
gps = [take (fromIntegral n) (drop (fromIntegral k) fltrd)
 k < [0..(length fltrd  n)] ]
gps2 = filter (\z > isConsec (map fst z)) gps
in filter (\zz > Set.empty ==
foldl (\acc z > Set.intersection acc (snd z))
(snd (head zz))
zz) gps2
isConsec xs = (sort xs) == [(minimum xs)..(maximum xs)]
problem_47 = (fst . head . head) $ nConsec 4 20000
Problem 48
Find the last ten digits of 1^{1} + 2^{2} + ... + 1000^{1000}.
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 = (`mod` limit) $ sum [powMod limit n n  n < [1..1000]]
where limit=10^10
Another oneliner for this problem, with no use of other functions is the following:
problem_48 = reverse $ take 10
$ reverse $ show $ sum $ map (\x > x^x) [1..1000]
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 Data.List
isPrime x
 x==3 = True
 otherwise = millerRabinPrimality x 2
primes4 = filter isPrime [1000..9999]
problem_49 = [ (a,b,c)  a < primes4,
b < dropWhile (<= a) primes4,
sort (show a) == sort (show b),
let c = 2 * b  a,
c `elem` primes4,
sort (show a) == sort (show c) ]
Problem 50
Which prime, below onemillion, 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
 This code is wrong: if you switch init and tail, it produces wrong result, so it is by sheer luck that it produces the right answer at all. As written, it will produce the longest chain of primes ending at 546th prime, summing up to a prime  using 546 as it is the longest prefix of primes summing up to less than a million. That's not what the problem asks for.
 What's to guarantee us there's no longer chain ending at 545th prime? 544th? For instance, for 1,100,000 the longest sequence ends at 568th instead of 571st prime which is what the above code would use.
 Moreover, cutting the search short at just first 546 primes is wrong too. What if the longest chain was really short, like 10 or 20 primes? Then we'd have to go much higher into the primes. We have no way of knowing that length in advance.
 Here's my solution, it's not the fastest but is correct, feel free to criticise (isPrime and primes not included):
import Euler.Helpers
import qualified Data.List as L
prime n = takeWhileSum n primes
takeWhileSum n = takeWhileArr (\x > sum x <= n)
takeWhileArr f xs = takeWhileF f [] xs
where
takeWhileF f rs [] = reverse rs
takeWhileF f rs (x:xs)
 f (x:rs) = takeWhileF f (x:rs) xs
 otherwise = reverse rs
primeSums n = map (map (\x > (isPrime x,x) ) . takeWhile (<n) . scanl1 (+)) (L.tails (prime n))
main = print . maximum $ map index (primeSums 100000)
where index x = if null $ ind x
then (0,0)
else (last $ ind x, snd (x !! (last $ ind x)))
ind = L.findIndices (fst)