Euler problems/71 to 80
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)
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]]
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)
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]
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
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
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
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..]
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
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