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([http://projecteuler.net/index.php?section=problems&id=40 Problem 40]: a solution)
(Problem 31)
 
(48 intermediate revisions by 18 users not shown)
Line 4: Line 4:
 
Solution:
 
Solution:
   
This is the naive doubly recursive solution. Speed would be greatly improved by use of memoization, dynamic programming, or the closed form.
+
This is the naive doubly recursive solution. Speed would be greatly improved by use of [[memoization]], dynamic programming, or the closed form.
 
<haskell>
 
<haskell>
problem_31 = pence 200 [1,2,5,10,20,50,100,200]
+
problem_31 = ways [1,2,5,10,20,50,100,200] !!200
where pence 0 _ = 1
+
where ways [] = 1 : repeat 0
pence n [] = 0
+
ways (coin:coins) =n
pence n denominations@(d:ds)
+
where n = zipWith (+) (ways coins) (replicate coin 0 ++ n)
| n < d = 0
+
</haskell>
| otherwise = pence (n - d) denominations
+
+ pence n ds
+
A beautiful solution, making usage of laziness and recursion to implement a dynamic programming scheme, blazingly fast despite actually generating the combinations and not only counting them :
  +
<haskell>
  +
coins = [1,2,5,10,20,50,100,200]
  +
  +
combinations = foldl (\without p ->
  +
let (poor,rich) = splitAt p without
  +
with = poor ++ zipWith (++) (map (map (p:)) with)
  +
rich
  +
in with
  +
) ([[]] : repeat [])
  +
  +
problem_31 = length $ combinations coins !! 200
  +
</haskell>
  +
  +
The above may be ''a beautiful solution'', but I couldn't understand it without major mental gymnastics. I would like to offer the following, which I hope will be easier to follow for ordinary ''mentats'' -- HenryLaxen 2008-02-22
  +
<haskell>
  +
coins = [1,2,5,10,20,50,100,200]
  +
  +
withcoins 1 x = [[x]]
  +
withcoins n x = concatMap addCoin [0 .. x `div` coins!!(n-1)]
  +
where addCoin k = map (++[k]) (withcoins (n-1) (x - k*coins!!(n-1)) )
  +
  +
problem_31 = length $ withcoins (length coins) 200
  +
</haskell>
  +
  +
The program above can be slightly modified as shown below so it just counts the combinations without generating them.
  +
<haskell>
  +
coins = [1,2,5,10,20,50,100,200]
  +
  +
countCoins 1 _ = 1
  +
countCoins n x = sum $ map addCoin [0 .. x `div` coins !! pred n]
  +
where addCoin k = countCoins (pred n) (x - k * coins !! pred n)
  +
  +
problem_31 = countCoins (length coins) 200
 
</haskell>
 
</haskell>
   
Line 20: Line 20:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_32 = undefined
+
import Control.Monad
  +
  +
combs 0 xs = [([],xs)]
  +
combs n xs = [(y:ys,rest) | y <- xs, (ys,rest) <- combs (n-1) (delete y xs)]
  +
  +
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)
  +
  +
pandigiticals =
  +
nub $ do (beg,end) <- combs 5 [1..9]
  +
n <- [1,2]
  +
let (a,b) = splitAt n beg
  +
res = l2n a * l2n b
  +
guard $ sort (explode res) == end
  +
return res
  +
  +
problem_32 = sum pandigiticals
 
</haskell>
 
</haskell>
   
Line 28: Line 28:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_33 = undefined
+
import Data.Ratio
  +
problem_33 = denominator . product $ rs
  +
{-
  +
xy/yz = x/z
  +
(10x + y)/(10y+z) = x/z
  +
9xz + yz = 10xy
  +
-}
  +
rs = [(10*x+y)%(10*y+z) | x <- t,
  +
y <- t,
  +
z <- t,
  +
x /= y ,
  +
(9*x*z) + (y*z) == (10*x*y)]
  +
where t = [1..9]
  +
</haskell>
  +
  +
That is okay, but why not let the computer do the ''thinking'' for you? Isn't this a little more directly expressive of the problem? -- HenryLaxen 2008-02-34
  +
<haskell>
  +
import Data.Ratio
  +
problem_33 = denominator $ product
  +
[ a%c | a<-[1..9], b<-[1..9], c<-[1..9],
  +
isCurious a b c, a /= b && a/= c]
  +
where isCurious a b c = ((10*a+b)%(10*b+c)) == (a%c)
 
</haskell>
 
</haskell>
   
Line 36: Line 36:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_34 = undefined
+
import Data.Char
  +
problem_34 = sum [ x | x <- [3..100000], x == facsum x ]
  +
where facsum = sum . map (product . enumFromTo 1 . digitToInt) . show
  +
 
</haskell>
 
</haskell>
  +
  +
Another way:
  +
  +
<haskell>
  +
import Data.Array
  +
import Data.List
  +
  +
{-
  +
  +
The key comes in realizing that N*9! < 10^N when N >= 9, so we
  +
only have to check up to 9 digit integers. The other key is
  +
that addition is commutative, so we only need to generate
  +
combinations (with duplicates) of the sums of the various
  +
factorials. These sums are the only potential "curious" sums.
  +
  +
-}
  +
  +
fac n = a!n
  +
where a = listArray (0,9) (1:(scanl1 (*) [1..9]))
  +
  +
-- subsets of size k, including duplicates
  +
combinationsOf 0 _ = [[]]
  +
combinationsOf _ [] = []
  +
combinationsOf k (x:xs) = map (x:)
  +
(combinationsOf (k-1) (x:xs)) ++ combinationsOf k xs
  +
  +
intToList n = reverse $ unfoldr
  +
(\x -> if x == 0 then Nothing else Just (x `mod` 10, x `div` 10)) n
  +
  +
isCurious (n,l) = sort (intToList n) == l
  +
  +
-- Turn a list into the sum of the factorials of the digits
  +
factorialSum l = sum $ map fac l
  +
  +
possiblyCurious = map (\z -> (factorialSum z,z))
  +
curious n = filter isCurious $ possiblyCurious $ combinationsOf n [0..9]
  +
problem_34 = sum $ (fst . unzip) $ concatMap curious [2..9]
  +
</haskell>
  +
(The wiki formatting is messing up the unzip"&gt;unzip line above, it is correct in the version I typed in. It should of course just be fst . unzip)
   
 
== [http://projecteuler.net/index.php?section=problems&id=35 Problem 35] ==
 
== [http://projecteuler.net/index.php?section=problems&id=35 Problem 35] ==
Line 44: Line 83:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_35 = undefined
+
import Data.List (tails, (\\))
  +
  +
primes :: [Integer]
  +
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
  +
  +
primeFactors :: Integer -> [Integer]
  +
primeFactors n = factor n primes
  +
where
  +
factor _ [] = []
  +
factor m (p:ps) | p*p > m = [m]
  +
| m `mod` p == 0 = p : factor (m `div` p) (p:ps)
  +
| otherwise = factor m ps
  +
  +
isPrime :: Integer -> Bool
  +
isPrime 1 = False
  +
isPrime n = case (primeFactors n) of
  +
(_:_:_) -> False
  +
_ -> True
  +
  +
permutations :: Integer -> [Integer]
  +
permutations n = take l $ map (read . take l) $ tails $ take (2*l -1) $ cycle s
  +
where
  +
s = show n
  +
l = length s
  +
  +
circular_primes :: [Integer] -> [Integer]
  +
circular_primes [] = []
  +
circular_primes (x:xs)
  +
| all isPrime p = x : circular_primes xs
  +
| otherwise = circular_primes xs
  +
where
  +
p = permutations x
  +
  +
problem_35 :: Int
  +
problem_35 = length $ circular_primes $ takeWhile (<1000000) primes
 
</haskell>
 
</haskell>
  +
  +
Using isPrime from above, and observing that one that can greatly reduce the search space because no circular prime can contain an even number, nor a 5, since eventually such a digit will be at the end of the number, and
  +
hence composite, we get: (HenryLaxen 2008-02-27)
  +
  +
<haskell>
  +
import Control.Monad (replicateM)
  +
  +
canBeCircularPrimeList = [1,3,7,9]
  +
  +
listToInt n = foldl (\x y -> 10*x+y) 0 n
  +
rot n l = y ++ x where (x,y) = splitAt n l
  +
allrots l = map (\x -> rot x l) [0..(length l)-1]
  +
isCircular l = all (isPrime . listToInt) $ allrots l
  +
circular 1 = [[2],[3],[5],[7]] -- a slightly special case
  +
circular n = filter isCircular $ replicateM n canBeCircularPrimeList
  +
  +
problem_35 = length $ concatMap circular [1..6]
  +
</haskell>
  +
   
 
== [http://projecteuler.net/index.php?section=problems&id=36 Problem 36] ==
 
== [http://projecteuler.net/index.php?section=problems&id=36 Problem 36] ==
Line 52: Line 110:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_36 = undefined
+
import Numeric
  +
import Data.Char
  +
  +
showBin = flip (showIntAtBase 2 intToDigit) ""
  +
  +
isPalindrome x = x == reverse x
  +
  +
problem_36 = sum [x | x <- [1,3..1000000], isPalindrome (show x), isPalindrome (showBin x)]
 
</haskell>
 
</haskell>
   
Line 60: Line 118:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_37 = undefined
+
import Data.List (tails, inits, nub)
  +
  +
primes :: [Integer]
  +
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
  +
  +
primeFactors :: Integer -> [Integer]
  +
primeFactors n = factor n primes
  +
where
  +
factor _ [] = []
  +
factor m (p:ps) | p*p > m = [m]
  +
| m `mod` p == 0 = p : factor (m `div` p) (p:ps)
  +
| otherwise = factor m ps
  +
  +
isPrime :: Integer -> Bool
  +
isPrime 1 = False
  +
isPrime n = case (primeFactors n) of
  +
(_:_:_) -> False
  +
_ -> True
  +
  +
truncs :: Integer -> [Integer]
  +
truncs n = nub . map read $ (take l . tail . tails) s ++ (take l . tail . inits) s
  +
where
  +
l = length s - 1
  +
s = show n
  +
  +
problem_37 = sum $ take 11 [x | x <- dropWhile (<=9) primes, all isPrime (truncs x)]
 
</haskell>
 
</haskell>
  +
  +
Or, more cleanly:
  +
  +
<haskell>
  +
import Data.Numbers.Primes (primes, isPrime)
  +
  +
test' :: Int -> Int -> (Int -> Int -> Int) -> Bool
  +
test' n d f
  +
| d > n = True
  +
| otherwise = isPrime (f n d) && test' n (10*d) f
  +
  +
test :: Int -> Bool
  +
test n = test' n 10 (mod) && test' n 10 (div)
  +
  +
problem_37 = sum $ take 11 $ filter test $ filter (>7) primes
  +
</haskell>
  +
   
 
== [http://projecteuler.net/index.php?section=problems&id=38 Problem 38] ==
 
== [http://projecteuler.net/index.php?section=problems&id=38 Problem 38] ==
Line 68: Line 143:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_38 = maximum $ catMaybes [result | j <- [1..9999],
+
import Data.List
let p2 = show j ++ show (2*j),
+
let p3 = p2 ++ show (3*j),
+
mult n i vs
let p4 = p3 ++ show (4*j),
+
| length (concat vs) >= 9 = concat vs
let p5 = p4 ++ show (5*j),
+
| otherwise = mult n (i+1) (vs ++ [show (n * i)])
let result
+
| isPan p2 = Just p2
+
problem_38 :: Int
| isPan p3 = Just p3
+
problem_38 = maximum . map read . filter ((['1'..'9'] ==) . sort)
| isPan p4 = Just p4
+
$ [mult n 1 [] | n <- [2..9999]]
| isPan p5 = Just p5
 
| otherwise = Nothing]
 
where isPan s = sort s == "123456789"
 
 
</haskell>
 
</haskell>
   
Line 89: Line 164:
 
$ sort [n*p | p <- pTriples, n <- [1..1000 `div` p]]
 
$ sort [n*p | p <- pTriples, n <- [1..1000 `div` p]]
 
counts = map length perims
 
counts = map length perims
Just indexMax = findIndex (== (maximum counts)) $ counts
+
Just indexMax = elemIndex (maximum counts) $ counts
 
pTriples = [p |
 
pTriples = [p |
 
n <- [1..floor (sqrt 1000)],
 
n <- [1..floor (sqrt 1000)],
Line 111: Line 186:
 
d j = Data.Char.digitToInt (n !! (j-1))
 
d j = Data.Char.digitToInt (n !! (j-1))
 
</haskell>
 
</haskell>
 
[[Category:Tutorials]]
 
[[Category:Code]]
 

Latest revision as of 00:16, 27 September 2012

Contents

[edit] 1 Problem 31

Investigating combinations of English currency denominations.

Solution:

This is the naive doubly recursive solution. Speed would be greatly improved by use of memoization, dynamic programming, or the closed form.

problem_31 = ways [1,2,5,10,20,50,100,200] !!200
  where ways [] = 1 : repeat 0
        ways (coin:coins) =n 
          where n = zipWith (+) (ways coins) (replicate coin 0 ++ n)

A beautiful solution, making usage of laziness and recursion to implement a dynamic programming scheme, blazingly fast despite actually generating the combinations and not only counting them :

coins = [1,2,5,10,20,50,100,200]
 
combinations = foldl (\without p ->
                          let (poor,rich) = splitAt p without
                              with = poor ++ zipWith (++) (map (map (p:)) with)
                                                          rich
                          in with
                     ) ([[]] : repeat [])
 
problem_31 = length $ combinations coins !! 200

The above may be a beautiful solution, but I couldn't understand it without major mental gymnastics. I would like to offer the following, which I hope will be easier to follow for ordinary mentats -- HenryLaxen 2008-02-22

coins = [1,2,5,10,20,50,100,200]
 
withcoins 1 x = [[x]]
withcoins n x = concatMap addCoin [0 .. x `div` coins!!(n-1)]
  where addCoin k = map (++[k]) (withcoins (n-1) (x - k*coins!!(n-1)) )
 
problem_31 = length $ withcoins (length coins) 200

The program above can be slightly modified as shown below so it just counts the combinations without generating them.

coins = [1,2,5,10,20,50,100,200]
 
countCoins 1 _ = 1
countCoins n x = sum $ map addCoin [0 .. x `div` coins !! pred n]
  where addCoin k = countCoins (pred n) (x - k * coins !! pred n)
 
problem_31 = countCoins (length coins) 200

[edit] 2 Problem 32

Find the sum of all numbers that can be written as pandigital products.

Solution:

import Control.Monad
 
combs 0 xs = [([],xs)]
combs n xs = [(y:ys,rest) | y <- xs, (ys,rest) <- combs (n-1) (delete y xs)]
 
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)
 
pandigiticals =
  nub $ do (beg,end) <- combs 5 [1..9]
           n <- [1,2]
           let (a,b) = splitAt n beg
               res = l2n a * l2n b
           guard $ sort (explode res) == end
           return res
 
problem_32 = sum pandigiticals

[edit] 3 Problem 33

Discover all the fractions with an unorthodox cancelling method.

Solution:

import Data.Ratio
problem_33 = denominator . product $ rs
{-
 xy/yz = x/z
(10x + y)/(10y+z) = x/z
9xz + yz = 10xy
 -}
rs = [(10*x+y)%(10*y+z) | x <- t, 
                          y <- t, 
                          z <- t,
                          x /= y ,
                          (9*x*z) + (y*z) == (10*x*y)]
  where t = [1..9]

That is okay, but why not let the computer do the thinking for you? Isn't this a little more directly expressive of the problem? -- HenryLaxen 2008-02-34

import Data.Ratio
problem_33 = denominator $ product 
             [ a%c | a<-[1..9], b<-[1..9], c<-[1..9],
                     isCurious a b c, a /= b && a/= c]
   where isCurious a b c = ((10*a+b)%(10*b+c)) == (a%c)

[edit] 4 Problem 34

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

Solution:

import Data.Char
problem_34 = sum [ x | x <- [3..100000], x == facsum x ]
    where facsum = sum . map (product . enumFromTo 1 . digitToInt) . show

Another way:

import Data.Array
import Data.List
 
{-
 
The key comes in realizing that N*9! < 10^N when N >= 9, so we
only have to check up to 9 digit integers.  The other key is
that addition is commutative, so we only need to generate
combinations (with duplicates) of the sums of the various
factorials.  These sums are the only potential "curious" sums.
 
-}
 
fac n = a!n
  where a = listArray (0,9)  (1:(scanl1 (*) [1..9]))
 
-- subsets of size k, including duplicates
combinationsOf 0 _ = [[]]
combinationsOf _ [] = []
combinationsOf k (x:xs) = map (x:) 
  (combinationsOf (k-1) (x:xs)) ++ combinationsOf k xs
 
intToList n = reverse $ unfoldr 
  (\x -> if x == 0 then Nothing else Just (x `mod` 10, x `div` 10)) n
 
isCurious (n,l) =  sort (intToList n) == l
 
-- Turn a list into the sum of the factorials of the digits
factorialSum l = sum $ map fac l
 
possiblyCurious = map (\z -> (factorialSum z,z)) 
curious n = filter isCurious $ possiblyCurious $ combinationsOf n [0..9]
problem_34 = sum $ (fst . unzip) $ concatMap curious [2..9]

(The wiki formatting is messing up the unzip">unzip line above, it is correct in the version I typed in. It should of course just be fst . unzip)

[edit] 5 Problem 35

How many circular primes are there below one million?

Solution:

import Data.List (tails, (\\))
 
primes :: [Integer]
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
 
primeFactors :: Integer -> [Integer]
primeFactors n = factor n primes
    where
        factor _ [] = []
        factor m (p:ps) | p*p > m        = [m]
                        | m `mod` p == 0 = p : factor (m `div` p) (p:ps)
                        | otherwise      = factor m ps
 
isPrime :: Integer -> Bool
isPrime 1 = False
isPrime n = case (primeFactors n) of
                (_:_:_)   -> False
                _         -> True
 
permutations :: Integer -> [Integer]
permutations n = take l $ map (read . take l) $ tails $ take (2*l -1) $ cycle s
    where
        s = show n
        l = length s
 
circular_primes :: [Integer] -> [Integer]
circular_primes []     = []
circular_primes (x:xs)
    | all isPrime p = x :  circular_primes xs
    | otherwise     = circular_primes xs
    where
        p = permutations x
 
problem_35 :: Int
problem_35 = length $ circular_primes $ takeWhile (<1000000) primes

Using isPrime from above, and observing that one that can greatly reduce the search space because no circular prime can contain an even number, nor a 5, since eventually such a digit will be at the end of the number, and hence composite, we get: (HenryLaxen 2008-02-27)

import Control.Monad (replicateM)
 
canBeCircularPrimeList = [1,3,7,9]
 
listToInt n = foldl (\x y -> 10*x+y) 0 n
rot n l = y ++ x where (x,y) = splitAt n l
allrots l = map (\x -> rot x l) [0..(length l)-1]
isCircular l =  all (isPrime . listToInt) $ allrots l
circular 1 = [[2],[3],[5],[7]]  -- a slightly special case
circular n = filter isCircular $ replicateM n canBeCircularPrimeList
 
problem_35 = length $ concatMap circular [1..6]


[edit] 6 Problem 36

Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2.

Solution:

import Numeric
import Data.Char
 
showBin = flip (showIntAtBase 2 intToDigit) ""
 
isPalindrome x = x == reverse x
 
problem_36 = sum [x | x <- [1,3..1000000], isPalindrome (show x), isPalindrome (showBin x)]

[edit] 7 Problem 37

Find the sum of all eleven primes that are both truncatable from left to right and right to left.

Solution:

import Data.List (tails, inits, nub)
 
primes :: [Integer]
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
 
primeFactors :: Integer -> [Integer]
primeFactors n = factor n primes
    where
        factor _ [] = []
        factor m (p:ps) | p*p > m        = [m]
                        | m `mod` p == 0 = p : factor (m `div` p) (p:ps)
                        | otherwise      = factor m ps
 
isPrime :: Integer -> Bool
isPrime 1 = False
isPrime n = case (primeFactors n) of
                (_:_:_)   -> False
                _         -> True
 
truncs :: Integer -> [Integer]
truncs n = nub . map read $ (take l . tail . tails) s ++ (take l . tail . inits) s
    where
        l = length s - 1
        s = show n
 
problem_37 = sum $ take 11 [x | x <- dropWhile (<=9) primes, all isPrime (truncs x)]

Or, more cleanly:

import Data.Numbers.Primes (primes, isPrime)
 
test' :: Int -> Int -> (Int -> Int -> Int) -> Bool
test' n d f
    | d > n = True
    | otherwise = isPrime (f n d) && test' n (10*d) f
 
test :: Int -> Bool
test n = test' n 10 (mod) && test' n 10 (div)
 
problem_37 = sum $ take 11 $ filter test $ filter (>7) primes


[edit] 8 Problem 38

What is the largest 1 to 9 pandigital that can be formed by multiplying a fixed number by 1, 2, 3, ... ?

Solution:

import Data.List
 
mult n i vs 
    | length (concat vs) >= 9 = concat vs
    | otherwise               = mult n (i+1) (vs ++ [show (n * i)])
 
problem_38 :: Int
problem_38 = maximum . map read . filter ((['1'..'9'] ==) . sort) 
               $ [mult n 1 [] | n <- [2..9999]]

[edit] 9 Problem 39

If p is the perimeter of a right angle triangle, {a, b, c}, which value, for p ≤ 1000, has the most solutions?

Solution: We use the well known formula to generate primitive Pythagorean triples. All we need are the perimeters, and they have to be scaled to produce all triples in the problem space.

problem_39 = head $ perims !! indexMax
    where  perims = group
                    $ sort [n*p | p <- pTriples, n <- [1..1000 `div` p]]
           counts = map length perims
           Just indexMax = elemIndex (maximum counts) $ counts
           pTriples = [p |
                       n <- [1..floor (sqrt 1000)],
                       m <- [n+1..floor (sqrt 1000)],
                       even n || even m,
                       gcd n m == 1,
                       let a = m^2 - n^2,
                       let b = 2*m*n,
                       let c = m^2 + n^2,
                       let p = a + b + c,
                       p < 1000]

[edit] 10 Problem 40

Finding the nth digit of the fractional part of the irrational number.

Solution:

problem_40 = (d 1)*(d 10)*(d 100)*(d 1000)*(d 10000)*(d 100000)*(d 1000000)
    where n = concat [show n | n <- [1..]]
          d j = Data.Char.digitToInt (n !! (j-1))