# Euler problems/31 to 40

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problem_31 = length $ combinations coins !! 200 |
problem_31 = length $ combinations coins !! 200 |
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</haskell> |
</haskell> |
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+ | 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 |
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+ | <haskell> |
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+ | coins = [1,2,5,10,20,50,100,200] |
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+ | |||

+ | withcoins 1 x = [[x]] |
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+ | withcoins n x = concatMap addCoin [0 .. x `div` coins!!(n-1)] |
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+ | where addCoin k = map (++[k]) (withcoins (n-1) (x - k*coins!!(n-1)) ) |
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+ | problem_31 = length $ withcoins (length coins) 200 |
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+ | </haskell> |
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+ | |||

== [http://projecteuler.net/index.php?section=problems&id=32 Problem 32] == |
== [http://projecteuler.net/index.php?section=problems&id=32 Problem 32] == |

## Revision as of 16:38, 22 February 2008

## 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]

## 4 Problem 34

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

Solution:

--http://www.research.att.com/~njas/sequences/A014080 problem_34 = sum [145, 40585]

## 5 Problem 35

How many circular primes are there below one million?

Solution: millerRabinPrimality on the Prime_numbers page

--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]