# Difference between revisions of "Euler problems/31 to 40"

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(9*x*z) + (y*z) == (10*x*y)] |
(9*x*z) + (y*z) == (10*x*y)] |
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where t = [1..9] |
where t = [1..9] |
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+ | </haskell> |
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+ | |||

+ | 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 |
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+ | <haskell> |
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+ | import Data.Ratio |
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+ | problem_33 = denominator $ product |
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+ | [ a%c | a<-[1..9], b<-[1..9], c<-[1..9], |
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+ | isCurious a b c, a /= b && a/= c] |
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+ | where isCurious a b c = ((10*a+b)%(10*b+c)) == (a%c) |
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</haskell> |
</haskell> |
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## Revision as of 18:09, 23 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 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
```

## 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
```

## 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)
```

## 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]
```

## 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]
```

## 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]
```