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Euler problems/71 to 80

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(solution to problem 73)
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Solution:
 
Solution:
This was my code, published here without my permission nor any attribution, shame on whoever put it here. [[User:Daniel.is.fischer|Daniel.is.fischer]]
+
  +
If you haven't done so already, read about Farey sequences in Wikipedia
  +
http://en.wikipedia.org/wiki/Farey_sequence, where you will learn about
  +
mediants. Then divide and conquer. The number of Farey ratios between
  +
(a, b) is 1 + the number between (a, mediant a b) + the number between
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(mediant a b, b). Henrylaxen 2008-03-04
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  +
<haskell>
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import Data.Ratio
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  +
mediant :: (Integral a) => Ratio a -> Ratio a -> Ratio a
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mediant f1 f2 = (numerator f1 + numerator f2) %
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(denominator f1 + denominator f2)
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fareyCount :: (Integral a, Num t) => a -> (Ratio a, Ratio a) -> t
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fareyCount n (a,b) =
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let c = mediant a b
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in if (denominator c > n) then 0 else
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1 + (fareyCount n (a,c)) + (fareyCount n (c,b))
  +
  +
problem_73 :: Integer
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problem_73 = areyCount 10000 (1%3,1%2)
  +
</haskell>
  +
   
 
== [http://projecteuler.net/index.php?section=view&id=74 Problem 74] ==
 
== [http://projecteuler.net/index.php?section=view&id=74 Problem 74] ==

Revision as of 05:59, 5 March 2008

Contents

1 Problem 71

Listing reduced proper fractions in ascending order of size.

Solution:

-- http://mathworld.wolfram.com/FareySequence.html 
import Data.Ratio ((%), numerator,denominator)
fareySeq a b
    |da2<=10^6=fareySeq a1 b
    |otherwise=na
    where
    na=numerator a
    nb=numerator b
    da=denominator a
    db=denominator b
    a1=(na+nb)%(da+db)
    da2=denominator a1
problem_71=fareySeq (0%1) (3%7)

2 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.

groups=1000
eulerTotient n = product (map (\(p,i) -> p^(i-1) * (p-1)) factors)
    where factors = fstfac n
fstfac x = [(head a ,length a)|a<-group$primeFactors x] 
p72 n= sum [eulerTotient x|x <- [groups*n+1..groups*(n+1)]]
problem_72 = sum [p72 x|x <- [0..999]]

3 Problem 73

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

Solution:

If you haven't done so already, read about Farey sequences in Wikipedia http://en.wikipedia.org/wiki/Farey_sequence, where you will learn about mediants. Then divide and conquer. The number of Farey ratios between (a, b) is 1 + the number between (a, mediant a b) + the number between (mediant a b, b). Henrylaxen 2008-03-04

import Data.Ratio
 
mediant :: (Integral a) => Ratio a -> Ratio a -> Ratio a
mediant f1 f2 = (numerator f1 + numerator f2) % 
                (denominator f1 + denominator f2)
fareyCount :: (Integral a, Num t) => a -> (Ratio a, Ratio a) -> t
fareyCount n (a,b) =
  let c = mediant a b
  in  if (denominator c > n) then 0 else 
         1 + (fareyCount n (a,c)) + (fareyCount n (c,b))
 
problem_73 :: Integer
problem_73 =  areyCount 10000   (1%3,1%2)


4 Problem 74

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

Solution:

import Data.List
explode 0 = []
explode n = n `mod` 10 : explode (n `quot` 10)
 
chain 2    = 1
chain 1    = 1
chain 145    = 1
chain 40585    = 1
chain 169    = 3
chain 363601 = 3
chain 1454   = 3
chain 871    = 2
chain 45361  = 2
chain 872    = 2
chain 45362  = 2
chain x = 1 + chain (sumFactDigits x)
makeIncreas 1 minnum  = [[a]|a<-[minnum..9]]
makeIncreas digits minnum  = [a:b|a<-[minnum ..9],b<-makeIncreas (digits-1) a]
p74=
    sum[div p6 $countNum a|
    a<-tail$makeIncreas  6 1,
    let k=digitToN a,
    chain k==60
    ]
    where
    p6=facts!! 6
sumFactDigits = foldl' (\a b -> a + facts !! b) 0 . explode
factorial n = if n == 0 then 1 else n * factorial (n - 1)
digitToN = foldl' (\a b -> 10*a + b) 0 .dropWhile (==0)
facts = scanl (*) 1 [1..9]
countNum xs=ys
    where
    ys=product$map (factorial.length)$group xs 
problem_74= length[k|k<-[1..9999],chain k==60]+p74
test = print $ [a|a<-tail$makeIncreas 6 0,let k=digitToN a,chain k==60]

5 Problem 75

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

Solution:

import Data.Array
 
triplets = 
    [p | 
    n <- [2..706],
    m <- [1..n-1],
    gcd m n == 1, 
    let p = 2 * (n^2 + m*n),
    odd (m + n),
    p <= 10^6
    ]
 
hist bnds ns = 
    accumArray (+) 0 bnds [(n, 1) |
        n <- ns,
        inRange bnds n
        ]
 
problem_75 = 
    length $ filter (\(_,b) -> b == 1) $ assocs arr
    where
    arr = hist (12,10^6) $ concatMap multiples triplets
    multiples n = takeWhile (<=10^6) [n, 2*n..]

6 Problem 76

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

Solution:

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

7 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.

counter = foldl (\without p ->
                     let (poor,rich) = splitAt p without
                         with = poor ++ 
                                zipWith (+) with rich
                     in with
                ) (1 : repeat 0)
 
problem_77 =  
    find ((>5000) . (ways !!)) $ [1..]
    where
    ways = counter $ take 100 primes

8 Problem 78

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

Solution:

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..]

9 Problem 79

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

Solution:

import Data.Char (digitToInt, intToDigit)
import Data.Graph (buildG, topSort)
import Data.List (intersect)
 
p79 file= 
    (+0)$read . intersect graphWalk $ usedDigits
    where
    usedDigits = intersect "0123456789" $ file
    edges = concat . map (edgePair . map digitToInt) . words $ file
    graphWalk = map intToDigit . topSort . buildG (0, 9) $ edges
    edgePair [x, y, z] = [(x, y), (y, z)]
    edgePair _         = undefined
 
problem_79 = do
    f<-readFile  "keylog.txt"
    print $p79 f

10 Problem 80

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

Solution:

import Data.Char
problem_80=
    sum [f x |
    a <- [1..100],
    x <- [intSqrt $ a * t],
    x * x /= a * t
    ]
    where
    t=10^202
    f = (sum . take 100 . map (flip (-) (ord '0') .ord) . show)