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Euler problems/71 to 80

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Solution:
 
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>
 
<haskell>
import Data.Array
+
import Data.Ratio
twix k = crude k - fd2 - sum [ar!(k `div` m) | m <- [3 .. k `div` 5], odd m]
+
 
    where
+
mediant :: (Integral a) => Ratio a -> Ratio a -> Ratio a
    fd2 = crude (k `div` 2)
+
mediant f1 f2 = (numerator f1 + numerator f2) %
    ar = array (5,k `div` 3) $
+
                (denominator f1 + denominator f2)
          ((5,1):[(j, crude j - sum [ar!(j `div` m) | m <- [2 .. j `div` 5]])
+
fareyCount :: (Integral a, Num t) => a -> (Ratio a, Ratio a) -> t
                      | j <- [6 .. k `div` 3]])
+
fareyCount n (a,b) =
    crude j =  
+
  let c = mediant a b
        m*(3*m+r-2) + s
+
  in  if (denominator c > n) then 0 else
        where
+
        1 + (fareyCount n (a,c)) + (fareyCount n (c,b))
            (m,r) = j `divMod` 6
+
       
            s = case r of
+
problem_73 :: Integer
                  5 -> 1
+
problem_73 =  fareyCount 10000   (1%3,1%2)   
                  _ -> 0
+
+
problem_73 =  twix 10000
+
 
</haskell>
 
</haskell>
 +
  
 
== [http://projecteuler.net/index.php?section=view&id=74 Problem 74] ==
 
== [http://projecteuler.net/index.php?section=view&id=74 Problem 74] ==
Line 102: Line 107:
  
 
Solution:
 
Solution:
This is only slightly harder than [[Euler problems/31 to 40#39|problem 39]].  The search condition is simpler but the search space is larger.
 
 
<haskell>
 
<haskell>
problem_75 =
+
import Data.Array
    length . filter ((== 1) . length) $ group perims
+
 
    where  perims = sort [scale*p | p <- pTriples, scale <- [1..10^6 `div` p]]
+
triangs :: [Int]
          pTriples = [p |
+
triangs = [p | n <- [2..1000],
                      n <- [1..1000],
+
              m <- [1..n-1],
                      m <- [n+1..1000],
+
              gcd m n == 1,
                      even n || even m,
+
              odd (m+n),
                      gcd n m == 1,
+
              let p = 2 * (n^2 + m*n),
                      let a = m^2 - n^2,
+
              p <= 2*10^6]
                      let b = 2*m*n,
+
 
                      let c = m^2 + n^2,
+
problem_75 :: Int
                      let p = a + b + c,
+
problem_75 = length
                      p <= 10^6]
+
      $ 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>
 
</haskell>
  
Line 155: Line 162:
  
 
Solution:
 
Solution:
 
Same as problem 76 but using array instead of lists to speedup things.
 
 
<haskell>
 
<haskell>
 
import Data.Array
 
import Data.Array
Line 182: Line 187:
  
 
Solution:
 
Solution:
 
A bit ugly but works fine
 
 
<haskell>
 
<haskell>
import Data.List
+
import Data.Char (digitToInt, intToDigit)
 +
import Data.Graph (buildG, topSort)
 +
import Data.List (intersect)
 
   
 
   
problem_79 :: String -> String
+
p79 file=  
problem_79 file =  
+
     (+0)$read . intersect graphWalk $ usedDigits
     map fst $
+
    sortBy (\(_,a) (_,b) ->
+
        compare (length b) (length a)) $
+
    zip digs order
+
 
     where
 
     where
     nums = lines file
+
     usedDigits = intersect "0123456789" $ file
     digs =  
+
     edges = concatMap (edgePair . map digitToInt) . words $ file
        map head $ group $
+
    graphWalk = map intToDigit . topSort . buildG (0, 9) $ edges
        sort $ filter (\c -> c >= '0' && c <= '9') file
+
     edgePair [x, y, z] = [(x, y), (y, z)]
     prec = concatMap (\(x:y:z:_) -> [[x,y],[y,z],[x,z]]) nums
+
     edgePair _        = undefined
     order =
+
        map (\n -> map head $
+
problem_79 = do
            group $ sort $ map (\(_:x:_) -> x) $
+
     f<-readFile "keylog.txt"
         filter (\(x:_) -> x == n) prec) digs
+
     print $p79 f
main=do  
+
     f<-readFile "keylog.txt"
+
     print$problem_79 f
+
 
</haskell>
 
</haskell>
  
Line 211: Line 209:
 
Calculating the digital sum of the decimal digits of irrational square roots.
 
Calculating the digital sum of the decimal digits of irrational square roots.
  
Solution:
+
This solution uses binary search to find the square root of a large Integer:
 
<haskell>
 
<haskell>
import Data.List ((\\))
+
import Data.Char (digitToInt)
  
hundreds :: Integer -> [Integer]
+
intSqrt :: Integer -> Integer
hundreds n = hundreds' [] n
+
intSqrt n = bsearch 1 n
 
     where
 
     where
        hundreds' acc 0 = acc
+
      bsearch l u = let m = (l+u) `div` 2
        hundreds' acc n = hundreds' (m : acc) d
+
                        m2 = m^2
            where
+
                    in if u <= l
                (d,m) = divMod n 100
+
                      then m
 +
                      else if m2 < n
 +
                            then bsearch (m+1) u
 +
                            else bsearch l m
  
squareDigs :: Integer -> [Integer]
+
problem_80 :: Int
squareDigs n = p : squareDigs' p r xs
+
problem_80 = sum [f r | a <- [1..100],
 +
                        let x = a * e,
 +
                        let r = intSqrt x,
 +
                        r*r /= x]
 
     where
 
     where
        (x:xs) = hundreds n ++ repeat 0
+
      e = 10^202
        p = floor $ sqrt $ fromInteger x
+
      f = sum . take 100 . map digitToInt . show
        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]]]
+
 
</haskell>
 
</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