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Euler problems/71 to 80

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Do them on your own!
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== [http://projecteuler.net/index.php?section=view&id=71 Problem 71] ==
 +
Listing reduced proper fractions in ascending order of size.
 +
 
 +
Solution:
 +
<haskell>
 +
-- 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)
 +
</haskell>
 +
 
 +
== [http://projecteuler.net/index.php?section=view&id=72 Problem 72] ==
 +
How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?
 +
 
 +
Solution:
 +
 
 +
Using the [http://mathworld.wolfram.com/FareySequence.html Farey Sequence] method, the solution is the sum of phi (n) from 1 to 1000000.
 +
<haskell>
 +
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]]
 +
</haskell>
 +
 
 +
== [http://projecteuler.net/index.php?section=view&id=73 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
 +
 
 +
<haskell>
 +
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 =  fareyCount 10000  (1%3,1%2)   
 +
</haskell>
 +
 
 +
 
 +
== [http://projecteuler.net/index.php?section=view&id=74 Problem 74] ==
 +
Determine the number of factorial chains that contain exactly sixty non-repeating terms.
 +
 
 +
Solution:
 +
<haskell>
 +
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]
 +
</haskell>
 +
== [http://projecteuler.net/index.php?section=view&id=75 Problem 75] ==
 +
Find the number of different lengths of wire can that can form a right angle triangle in only one way.
 +
 
 +
Solution:
 +
<haskell>
 +
import Data.Array
 +
 
 +
triangs :: [Int]
 +
triangs = [p | n <- [2..1000],
 +
              m <- [1..n-1],
 +
              gcd m n == 1,
 +
              odd (m+n),
 +
              let p = 2 * (n^2 + m*n),
 +
              p <= 2*10^6]
 +
 
 +
problem_75 :: Int
 +
problem_75 = length
 +
      $ filter (\(_, c) -> c == 1)
 +
      $ assocs
 +
      $ (\ns -> accumArray (+) 0 (1, 2*10^6) [(n, 1) | n <- ns, inRange (1, 2*10^6) n])
 +
      $ concatMap (\n -> takeWhile (<=2*10^6) [n,2*n..]) triangs
 +
</haskell>
 +
 
 +
== [http://projecteuler.net/index.php?section=view&id=76 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.
 +
<haskell>
 +
build x = (map sum (zipWith drop [0..] x) ++ [1]) : x
 +
problem_76 = (sum $ head $ iterate build [] !! 100) - 1
 +
</haskell>
 +
 
 +
== [http://projecteuler.net/index.php?section=view&id=77 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.
 +
<haskell>
 +
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
 +
</haskell>
 +
 
 +
== [http://projecteuler.net/index.php?section=view&id=78 Problem 78] ==
 +
Investigating the number of ways in which coins can be separated into piles.
 +
 
 +
Solution:
 +
<haskell>
 +
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..]
 +
</haskell>
 +
 
 +
== [http://projecteuler.net/index.php?section=view&id=79 Problem 79] ==
 +
By analysing a user's login attempts, can you determine the secret numeric passcode?
 +
 
 +
Solution:
 +
<haskell>
 +
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 = concatMap (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
 +
</haskell>
 +
 
 +
== [http://projecteuler.net/index.php?section=view&id=80 Problem 80] ==
 +
Calculating the digital sum of the decimal digits of irrational square roots.
 +
 
 +
This solution uses binary search to find the square root of a large Integer:
 +
<haskell>
 +
import Data.Char (digitToInt)
 +
 
 +
intSqrt :: Integer -> Integer
 +
intSqrt n = bsearch 1 n
 +
    where
 +
      bsearch l u = let m = (l+u) `div` 2
 +
                        m2 = m^2
 +
                    in if u <= l
 +
                      then m
 +
                      else if m2 < n
 +
                            then bsearch (m+1) u
 +
                            else bsearch l m
 +
 
 +
problem_80 :: Int
 +
problem_80 = sum [f r | a <- [1..100],
 +
                        let x = a * e,
 +
                        let r = intSqrt x,
 +
                        r*r /= x]
 +
    where
 +
      e = 10^202
 +
      f = sum . take 100 . map digitToInt . show
 +
</haskell>

Revision as of 10:37, 13 December 2009

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 =  fareyCount 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
 
triangs :: [Int]
triangs = [p | n <- [2..1000],
               m <- [1..n-1],
               gcd m n == 1,
               odd (m+n),
               let p = 2 * (n^2 + m*n),
               p <= 2*10^6]
 
problem_75 :: Int
problem_75 = length
       $ filter (\(_, c) -> c == 1)
       $ assocs
       $ (\ns -> accumArray (+) 0 (1, 2*10^6) [(n, 1) | n <- ns, inRange (1, 2*10^6) n])
       $ concatMap (\n -> takeWhile (<=2*10^6) [n,2*n..]) triangs

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 = concatMap (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.

This solution uses binary search to find the square root of a large Integer:

import Data.Char (digitToInt)
 
intSqrt :: Integer -> Integer
intSqrt n = bsearch 1 n
    where
      bsearch l u = let m = (l+u) `div` 2
                        m2 = m^2
                    in if u <= l
                       then m
                       else if m2 < n
                            then bsearch (m+1) u
                            else bsearch l m
 
problem_80 :: Int
problem_80 = sum [f r | a <- [1..100],
                        let x = a * e,
                        let r = intSqrt x,
                        r*r /= x]
    where
      e = 10^202
      f = sum . take 100 . map digitToInt . show