# Euler problems/31 to 40

### From HaskellWiki

## Contents |

## 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) (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 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

## 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

## 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)

## 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 = 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)

## 5 Problem 35

How many circular primes are there below one million?

Solution: millerRabinPrimality on the Prime_numbers page

*Note: Miller Rabin for primes less than 1000000?
Why not use a primesieve?*

--http://www.research.att.com/~njas/sequences/A068652 isPrime x | x==1 = False | x==2 = True | x==3 = True | otherwise = millerRabinPrimality x 2 permutations n = take l . map (read . take l) . tails . take (2*l-1) . cycle $ s where s = show n l = length s circular_primes [] = [] circular_primes (x:xs) | all isPrime p = x : circular_primes xs | otherwise = circular_primes xs where p = permutations x x = [1,3,7,9] dmm = foldl (\x y->x*10+y) 0 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]

## 6 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]

## 7 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]

## 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 = maximum . map read . filter ((['1'..'9'] ==) .sort) $ [mult n 1 [] | n <- [2..9999]]

## 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.

--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)

## 10 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]