Difference between revisions of "Euler problems/31 to 40"
(→[http://projecteuler.net/index.php?section=problems&id=34 Problem 34]: restore erased solution) 
(→[http://projecteuler.net/index.php?section=problems&id=35 Problem 35]: rm premature "optimization") 

Line 150:  Line 150:  
Solution: 
Solution: 

−  millerRabinPrimality on the [[Prime_numbers]] page 

−  
−  <i>Note: Miller Rabin for primes less than 1000000? 

−  Why not use a primesieve?</i> 

−  
<haskell> 
<haskell> 

−  http://www.research.att.com/~njas/sequences/A068652 

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

−  isPrime x 

+  
−   x==1 = False 

+  primes :: [Integer] 

−   x==2 = True 

+  primes = 2 : filter ((==1) . length . primeFactors) [3,5..] 

−   x==3 = True 

+  
−   otherwise = millerRabinPrimality x 2 

+  primeFactors :: Integer > [Integer] 

−  
+  primeFactors n = factor n primes 

−  permutations n = take l 

+  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 
l = length s 

−  
+  
+  circular_primes :: [Integer] > [Integer] 

circular_primes [] = [] 
circular_primes [] = [] 

circular_primes (x:xs) 
circular_primes (x:xs) 

 all isPrime p = x : circular_primes xs 
 all isPrime p = x : circular_primes xs 

 otherwise = circular_primes xs 
 otherwise = circular_primes xs 

−  where p = permutations x 

+  where 

−  
+  p = permutations x 

−  x = [1,3,7,9] 

+  
−  
+  problem_35 :: Int 

−  dmm = foldl (\x y>x*10+y) 0 

+  problem_35 = length $ circular_primes $ takeWhile (<1000000) primes 

−  
−  xx n = map dmm (replicateM n x) 

−  
−  problem_35 = (+13) . length . circular_primes 

−  $ [a  a < concat [xx 3,xx 4,xx 5,xx 6], isPrime a] 

</haskell> 
</haskell> 

Revision as of 17:40, 25 February 2008
Contents
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) (take coin (repeat 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 20080222
coins = [1,2,5,10,20,50,100,200]
withcoins 1 x = [[x]]
withcoins n x = concatMap addCoin [0 .. x `div` coins!!(n1)]
where addCoin k = map (++[k]) (withcoins (n1) (x  k*coins!!(n1)) )
problem_31 = length $ withcoins (length coins) 200
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 (n1) (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
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 20080234
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)
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 (k1) (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 = foldr (\x y > (fac x) + y) 0 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)
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
Problem 36
Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2.
Solution:
http://www.research.att.com/~njas/sequences/A007632
problem_36 = sum [0, 1, 3, 5, 7, 9, 33, 99, 313, 585, 717,
7447, 9009, 15351, 32223, 39993, 53235,
53835, 73737, 585585]
Problem 37
Find the sum of all eleven primes that are both truncatable from left to right and right to left.
Solution:
 isPrime in p35
 http://www.research.att.com/~njas/sequences/A020994
problem_37 = sum [23, 37, 53, 73, 313, 317, 373, 797, 3137, 3797, 739397]
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 = maximum . map read . filter ((['1'..'9'] ==) .sort)
$ [mult n 1 []  n < [2..9999]]
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.
http://www.research.att.com/~njas/sequences/A046079
problem_39 = let t = 3*5*7
in floor(2^floor(log(1000/t)/log 2)*t)
Problem 40
Finding the nth digit of the fractional part of the irrational number.
Solution:
http://www.research.att.com/~njas/sequences/A023103
problem_40 = product [1, 1, 5, 3, 7, 2, 1]