# Difference between revisions of "Euler problems/21 to 30"

(→[http://projecteuler.net/index.php?section=problems&id=26 Problem 26]: restore a correct solution that was replaced by a wrong one) |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | --http://www.research.att.com/~njas/sequences/A052464 |
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+ | import Data.Char (ord) |
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− | problem_30 = sum [4150, 4151, 54748, 92727, 93084, 194979] |
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− | </haskell> |
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− | I'm sorry, but I find the solution to problem 30 very unsatisfying. I'm using the Euler problems to learn Haskell, so looking up the answer and adding the terms isn't really that helpful. I would like to present the following as a clearer solution that perhaps gives a little more insight into the problem and programming in Haskell. -- Henry Laxen, Feb 20, 2008 |
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+ | limit :: Integer |
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+ | limit = snd $ head $ dropWhile (\(a,b) -> a > b) $ zip (map (9^5*) [1..]) (map (10^) [1..]) |
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+ | fifth :: Integer -> Integer |
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+ | fifth n = foldr (\a b -> (toInteger(ord a) - 48)^5 + b) 0 $ show n |
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− | <haskell> |
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+ | problem_30 :: Integer |
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− | problem_30 |
+ | problem_30 = sum $ filter (\n -> n == fifth n) [2..limit] |

− | -- we drop 2 because the first two members of the ans are 0 and 1, |
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− | -- which are considered "trivial" solutions and should not count in the sum |
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− | where maxFirstDigit = (6*9^5 `div` 10^5) + 1 |
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− | -- The largest number that can be the sum of fifth powers |
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− | -- is 6*9^5 = 354294, which has 6 digits |
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− | listToInt n = foldl (\x y -> 10*x+y) 0 n |
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− | isSumOfPowers p n = (sum $ map (\x -> x^p) n) == listToInt n |
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− | ans = filter (isSumOfPowers 5) [ [a,b,c,d,e,f] | |
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− | a <- [0..maxFirstDigit], |
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− | b <- [0..9], |
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− | c <- [0..9], |
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− | d <- [0..9], |
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− | e <- [0..9], |
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− | f <- [0..9] ] |
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</haskell> |
</haskell> |

## Revision as of 17:23, 25 February 2008

## Contents

## Problem 21

Evaluate the sum of all amicable pairs under 10000.

Solution: (http://www.research.att.com/~njas/sequences/A063990)

This is a little slow because of the naive method used to compute the divisors.

```
problem_21 = sum [m+n | m <- [2..9999], let n = divisorsSum ! m, amicable m n]
where amicable m n = m < n && n < 10000 && divisorsSum ! n == m
divisorsSum = array (1,9999)
[(i, sum (divisors i)) | i <- [1..9999]]
divisors n = [j | j <- [1..n `div` 2], n `mod` j == 0]
```

Here is an alternative using a faster way of computing the sum of divisors.

```
problem_21_v2 = sum [n | n <- [2..9999], let m = d n,
m > 1, m < 10000, n == d m]
d n = product [(p * product g - 1) `div` (p - 1) |
g <- group $ primeFactors n, let p = head g
] - n
primeFactors = pf primes
where
pf ps@(p:ps') n
| p * p > n = [n]
| r == 0 = p : pf ps q
| otherwise = pf ps' n
where (q, r) = n `divMod` p
primes = 2 : filter (null . tail . primeFactors) [3,5..]
```

## Problem 22

What is the total of all the name scores in the file of first names?

Solution:

```
import Data.List
import Data.Char
problem_22 =
do input <- readFile "names.txt"
let names = sort $ read$"["++ input++"]"
let scores = zipWith score names [1..]
print . show . sum $ scores
where score w i = (i *) . sum . map (\c -> ord c - ord 'A' + 1) $ w
```

## Problem 23

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Solution:

```
--http://www.research.att.com/~njas/sequences/A048242
import Data.Array
n = 28124
abundant n = eulerTotient n - n > n
abunds_array = listArray (1,n) $ map abundant [1..n]
abunds = filter (abunds_array !) [1..n]
rests x = map (x-) $ takeWhile (<= x `div` 2) abunds
isSum = any (abunds_array !) . rests
problem_23 = putStrLn . show . foldl1 (+) . filter (not . isSum) $ [1..n]
```

## Problem 24

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

Solution:

```
import Data.List
fac 0 = 1
fac n = n * fac (n - 1)
perms [] _= []
perms xs n= x : perms (delete x xs) (mod n m)
where m = fac $ length xs - 1
y = div n m
x = xs!!y
problem_24 = perms "0123456789" 999999
```

## Problem 25

What is the first term in the Fibonacci sequence to contain 1000 digits?

Solution:

```
valid ( i, n ) = length ( show n ) == 1000
problem_25 = fst . head . filter valid . zip [ 1 .. ] $ fibs
where fibs = 1 : 1 : 2 : zipWith (+) fibs ( tail fibs )
```

## Problem 26

Find the value of d < 1000 for which 1/d contains the longest recurring cycle.

Solution:

```
problem_26 = fst $ maximumBy (\a b -> snd a `compare` snd b)
[(n,recurringCycle n) | n <- [1..999]]
where recurringCycle d = remainders d 10 []
remainders d 0 rs = 0
remainders d r rs = let r' = r `mod` d
in case findIndex (== r') rs of
Just i -> i + 1
Nothing -> remainders d (10*r') (r':rs)
```

## Problem 27

Find a quadratic formula that produces the maximum number of primes for consecutive values of n.

Solution:

```
problem_27 = -(2*a-1)*(a^2-a+41)
where n = 1000
m = head $ filter (\x->x^2-x+41>n) [1..]
a = m-1
```

## Problem 28

What is the sum of both diagonals in a 1001 by 1001 spiral?

Solution:

```
problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1
```

## Problem 29

How many distinct terms are in the sequence generated by a^{b} for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

Solution:

```
import Control.Monad
problem_29 = length . group . sort $ liftM2 (^) [2..100] [2..100]
```

## Problem 30

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

Solution:

```
import Data.Char (ord)
limit :: Integer
limit = snd $ head $ dropWhile (\(a,b) -> a > b) $ zip (map (9^5*) [1..]) (map (10^) [1..])
fifth :: Integer -> Integer
fifth n = foldr (\a b -> (toInteger(ord a) - 48)^5 + b) 0 $ show n
problem_30 :: Integer
problem_30 = sum $ filter (\n -> n == fifth n) [2..limit]
```