# Difference between revisions of "Euler problems/71 to 80"

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problem_76 = partitions !! 100 - 1 |
problem_76 = partitions !! 100 - 1 |
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+ | </haskell> |
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+ | Here is a simpler solution: For each n, we create the list of the number of partitions of n |
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+ | whose lowest number is i, for i=1..n. We build up the list of these lists for n=0..100. |
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+ | <haskell> |
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+ | build x = (map sum (zipWith drop [0..] x) ++ [1]) : x |
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+ | problem_76' = (sum $ head $ iterate build [] !! 100) - 1 |
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</haskell> |
</haskell> |
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## Revision as of 10:36, 14 November 2007

## Contents

## Problem 71

Listing reduced proper fractions in ascending order of size.

Solution:

```
import Data.Ratio (Ratio, (%), numerator)
fractions :: [Ratio Integer]
fractions = [f | d <- [1..1000000], let n = (d * 3) `div` 7, let f = n%d, f /= 3%7]
problem_71 :: Integer
problem_71 = numerator $ maximum $ fractions
```

## Problem 72

How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?

Solution:

Using the Farey Sequence method, the solution is the sum of phi (n) from 1 to 1000000.

See problem 69 for phi function

```
problem_72 = sum [phi x|x <- [1..1000000]]
```

## Problem 73

How many fractions lie between 1/3 and 1/2 in a sorted set of reduced proper fractions?

Solution:

```
import Data.Ratio (Ratio, (%), numerator, denominator)
median :: Ratio Int -> Ratio Int -> Ratio Int
median a b = ((numerator a) + (numerator b)) % ((denominator a) + (denominator b))
count :: Ratio Int -> Ratio Int -> Int
count a b
| d > 10000 = 1
| otherwise = count a m + count m b
where
m = median a b
d = denominator m
problem_73 :: Int
problem_73 = (count (1%3) (1%2)) - 1
```

## Problem 74

Determine the number of factorial chains that contain exactly sixty non-repeating terms.

Solution:

```
import Data.Array (Array, array, (!), elems)
import Data.Char (ord)
import Data.List (foldl1')
import Prelude hiding (cycle)
fact :: Integer -> Integer
fact 0 = 1
fact n = foldl1' (*) [1..n]
factorDigits :: Array Integer Integer
factorDigits = array (0,2177281) [(x,n)|x <- [0..2177281], let n = sum $ map (\y -> fact (toInteger $ ord y - 48)) $ show x]
cycle :: Integer -> Integer
cycle 145 = 1
cycle 169 = 3
cycle 363601 = 3
cycle 1454 = 3
cycle 871 = 2
cycle 45361 = 2
cycle 872 = 2
cycle 45362 = 2
cycle _ = 0
isChainLength :: Integer -> Integer -> Bool
isChainLength len n
| len < 0 = False
| t = isChainLength (len-1) n'
| otherwise = (len - c) == 0
where
c = cycle n
t = c == 0
n' = factorDigits ! n
-- | strict version of the maximum function
maximum' :: (Ord a) => [a] -> a
maximum' [] = undefined
maximum' [x] = x
maximum' (a:b:xs) = let m = max a b in m `seq` maximum' (m : xs)
problem_74 :: Int
problem_74 = length $ filter (\(_,b) -> isChainLength 59 b) $ zip ([0..] :: [Integer]) $ take 1000000 $ elems factorDigits
```

Slightly faster solution :

```
{-# OPTIONS_GHC -fbang-patterns #-}
import Data.List
import Data.Array
explode 0 = []
explode n = n `mod` 10 : explode (n `quot` 10)
count :: (a -> Bool) -> [a] -> Int
count pred xs = lgo 0 xs
where lgo !c [] = c
lgo !c (y:ys) | pred y = lgo (c + 1) ys
| otherwise = lgo c ys
known = [([1,2,145,40585],1),([871,45361,872,45362],2),([169,363601,1454],3)]
mChain = array (1,1000000) $ (concat $ expand known)
++ [(x, n)|x<-[3..1000000]
, not $ x `elem` concat (map fst known)
, let n = 1 + chain (sumFactDigits x)]
where expand [] = []
expand ((xs,len):xxs) = map (flip (,) len) xs : expand xxs
chain x | x <= 1000000 = mChain ! x
| otherwise = 1 + chain (sumFactDigits x)
sumFactDigits = foldl' (\a b -> a + facts !! b) 0 . explode
facts = scanl (*) 1 [1..9]
problem_74 = count (== 60) $ elems mChain
```

## Problem 75

Find the number of different lengths of wire can that can form a right angle triangle in only one way.

Solution: This is only slightly harder than problem 39. The search condition is simpler but the search space is larger.

```
problem_75 = length . filter ((== 1) . length) $ group perims
where perims = sort [scale*p | p <- pTriples, scale <- [1..10^6 `div` p]]
pTriples = [p |
n <- [1..1000],
m <- [n+1..1000],
even n || even m,
gcd n m == 1,
let a = m^2 - n^2,
let b = 2*m*n,
let c = m^2 + n^2,
let p = a + b + c,
p <= 10^6]
```

## Problem 76

How many different ways can one hundred be written as a sum of at least two positive integers?

Solution:

Calculated using Euler's pentagonal formula and a list for memoization.

```
partitions = 1 : [sum [s * partitions !! p| (s,p) <- zip signs $ parts n]| n <- [1..]]
where
signs = cycle [1,1,(-1),(-1)]
suite = map penta $ concat [[n,(-n)]|n <- [1..]]
penta n = n*(3*n - 1) `div` 2
parts n = takeWhile (>= 0) [n-x| x <- suite]
problem_76 = partitions !! 100 - 1
```

Here is a simpler solution: For each n, we create the list of the number of partitions of n whose lowest number is i, for i=1..n. We build up the list of these lists for n=0..100.

```
build x = (map sum (zipWith drop [0..] x) ++ [1]) : x
problem_76' = (sum $ head $ iterate build [] !! 100) - 1
```

## Problem 77

What is the first value which can be written as the sum of primes in over five thousand different ways?

Solution:

Brute force but still finds the solution in less than one second.

```
combinations acc 0 _ = [acc]
combinations acc _ [] = []
combinations acc value prim@(x:xs) = combinations (acc ++ [x]) value' prim' ++ combinations acc value xs
where
value' = value - x
prim' = dropWhile (>value') prim
problem_77 :: Integer
problem_77 = head $ filter f [1..]
where
f n = (length $ combinations [] n $ takeWhile (<n) primes) > 5000
```

## Problem 78

Investigating the number of ways in which coins can be separated into piles.

Solution:

Same as problem 76 but using array instead of lists to speedup things.

```
import Data.Array
partitions :: Array Int Integer
partitions = array (0,1000000) $ (0,1) : [(n,sum [s * partitions ! p| (s,p) <- zip signs $ parts n])| n <- [1..1000000]]
where
signs = cycle [1,1,(-1),(-1)]
suite = map penta $ concat [[n,(-n)]|n <- [1..]]
penta n = n*(3*n - 1) `div` 2
parts n = takeWhile (>= 0) [n-x| x <- suite]
problem_78 :: Int
problem_78 = head $ filter (\x -> (partitions ! x) `mod` 1000000 == 0) [1..]
```

## Problem 79

By analysing a user's login attempts, can you determine the secret numeric passcode?

Solution:

A bit ugly but works fine

```
import Data.List
problem_79 :: String -> String
problem_79 file = map fst $ sortBy (\(_,a) (_,b) -> compare (length b) (length a)) $ zip digs order
where
nums = lines file
digs = map head $ group $ sort $ filter (\c -> c >= '0' && c <= '9') file
prec = concatMap (\(x:y:z:_) -> [[x,y],[y,z],[x,z]]) nums
order = map (\n -> map head $ group $ sort $ map (\(_:x:_) -> x) $ filter (\(x:_) -> x == n) prec) digs
```

## Problem 80

Calculating the digital sum of the decimal digits of irrational square roots.

Solution:

```
import Data.List ((\\))
hundreds :: Integer -> [Integer]
hundreds n = hundreds' [] n
where
hundreds' acc 0 = acc
hundreds' acc n = hundreds' (m : acc) d
where
(d,m) = divMod n 100
squareDigs :: Integer -> [Integer]
squareDigs n = p : squareDigs' p r xs
where
(x:xs) = hundreds n ++ repeat 0
p = floor $ sqrt $ fromInteger x
r = x - (p^2)
squareDigs' :: Integer -> Integer -> [Integer] -> [Integer]
squareDigs' p r (x:xs) = x' : squareDigs' (p*10 + x') r' xs
where
n = 100*r + x
(x',r') = last $ takeWhile (\(_,a) -> a >= 0) $ scanl (\(_,b) (a',b') -> (a',b-b')) (0,n) rs
rs = [y|y <- zip [1..] [(20*p+1),(20*p+3)..]]
sumDigits n = sum $ take 100 $ squareDigs n
problem_80 :: Integer
problem_80 = sum $ map sumDigits [x|x <- [1..100] \\ [n^2|n<-[1..10]]]
```