Difference between revisions of "Euler problems/71 to 80"
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− | [[Category:Programming exercise spoilers]] |
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== [http://projecteuler.net/index.php?section=view&id=71 Problem 71] == |
== [http://projecteuler.net/index.php?section=view&id=71 Problem 71] == |
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Listing reduced proper fractions in ascending order of size. |
Listing reduced proper fractions in ascending order of size. |
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Line 5: | Line 4: | ||
Solution: |
Solution: |
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<haskell> |
<haskell> |
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+ | -- http://mathworld.wolfram.com/FareySequence.html |
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− | import Data.Ratio (Ratio, (%), numerator) |
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+ | import Data.Ratio ((%), numerator,denominator) |
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− | |||
+ | fareySeq a b |
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− | fractions :: [Ratio Integer] |
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+ | |da2<=10^6=fareySeq a1 b |
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− | fractions = [f | d <- [1..1000000], let n = (d * 3) `div` 7, let f = n%d, f /= 3%7] |
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+ | |otherwise=na |
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− | |||
+ | where |
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− | problem_71 :: Integer |
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+ | na=numerator a |
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− | problem_71 = numerator $ maximum $ fractions |
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+ | nb=numerator b |
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+ | da=denominator a |
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+ | db=denominator b |
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+ | a1=(na+nb)%(da+db) |
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+ | da2=denominator a1 |
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+ | problem_71=fareySeq (0%1) (3%7) |
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</haskell> |
</haskell> |
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Line 18: | Line 23: | ||
Solution: |
Solution: |
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+ | |||
+ | Using the [http://mathworld.wolfram.com/FareySequence.html Farey Sequence] method, the solution is the sum of phi (n) from 1 to 1000000. |
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<haskell> |
<haskell> |
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+ | groups=1000 |
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− | problem_72 = undefined |
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+ | eulerTotient n = product (map (\(p,i) -> p^(i-1) * (p-1)) factors) |
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+ | where factors = fstfac n |
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+ | fstfac x = [(head a ,length a)|a<-group$primeFactors x] |
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+ | p72 n= sum [eulerTotient x|x <- [groups*n+1..groups*(n+1)]] |
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+ | problem_72 = sum [p72 x|x <- [0..999]] |
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</haskell> |
</haskell> |
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Line 26: | Line 38: | ||
Solution: |
Solution: |
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+ | |||
+ | If you haven't done so already, read about Farey sequences in Wikipedia |
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+ | http://en.wikipedia.org/wiki/Farey_sequence, where you will learn about |
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+ | mediants. Then divide and conquer. The number of Farey ratios between |
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+ | (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> |
<haskell> |
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+ | import Data.Ratio |
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− | problem_73 = undefined |
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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)) |
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+ | |||
+ | problem_73 :: Integer |
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+ | problem_73 = fareyCount 10000 (1%3,1%2) |
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</haskell> |
</haskell> |
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+ | |||
== [http://projecteuler.net/index.php?section=view&id=74 Problem 74] == |
== [http://projecteuler.net/index.php?section=view&id=74 Problem 74] == |
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Line 35: | Line 67: | ||
Solution: |
Solution: |
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<haskell> |
<haskell> |
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+ | import Data.List |
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− | problem_74 = undefined |
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+ | explode 0 = [] |
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+ | explode n = n `mod` 10 : explode (n `quot` 10) |
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+ | |||
+ | chain 2 = 1 |
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+ | chain 1 = 1 |
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+ | chain 145 = 1 |
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+ | chain 40585 = 1 |
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+ | chain 169 = 3 |
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+ | chain 363601 = 3 |
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+ | chain 1454 = 3 |
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+ | chain 871 = 2 |
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+ | chain 45361 = 2 |
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+ | chain 872 = 2 |
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+ | chain 45362 = 2 |
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+ | chain x = 1 + chain (sumFactDigits x) |
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+ | makeIncreas 1 minnum = [[a]|a<-[minnum..9]] |
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+ | makeIncreas digits minnum = [a:b|a<-[minnum ..9],b<-makeIncreas (digits-1) a] |
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+ | p74= |
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+ | sum[div p6 $countNum a| |
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+ | a<-tail$makeIncreas 6 1, |
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+ | let k=digitToN a, |
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+ | chain k==60 |
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+ | ] |
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+ | where |
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+ | p6=facts!! 6 |
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+ | sumFactDigits = foldl' (\a b -> a + facts !! b) 0 . explode |
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+ | factorial n = if n == 0 then 1 else n * factorial (n - 1) |
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+ | digitToN = foldl' (\a b -> 10*a + b) 0 .dropWhile (==0) |
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+ | facts = scanl (*) 1 [1..9] |
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+ | countNum xs=ys |
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+ | where |
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+ | ys=product$map (factorial.length)$group xs |
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+ | problem_74= length[k|k<-[1..9999],chain k==60]+p74 |
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+ | test = print $ [a|a<-tail$makeIncreas 6 0,let k=digitToN a,chain k==60] |
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</haskell> |
</haskell> |
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− | |||
== [http://projecteuler.net/index.php?section=view&id=75 Problem 75] == |
== [http://projecteuler.net/index.php?section=view&id=75 Problem 75] == |
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Find the number of different lengths of wire can that can form a right angle triangle in only one way. |
Find the number of different lengths of wire can that can form a right angle triangle in only one way. |
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Solution: |
Solution: |
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− | 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. |
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<haskell> |
<haskell> |
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+ | module Main where |
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− | problem_75 = length . filter ((== 1) . length) $ group perims |
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+ | |||
− | where perims = sort [scale*p | p <- pTriples, scale <- [1..10^6 `div` p]] |
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+ | import Data.Array.Unboxed (UArray, accumArray, elems) |
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− | pTriples = [p | |
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+ | |||
− | n <- [1..1000], |
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+ | main :: IO () |
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− | m <- [n+1..1000], |
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+ | main = print problem_75 |
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− | even n || even m, |
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+ | |||
− | gcd n m == 1, |
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+ | limit :: Int |
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− | let a = m^2 - n^2, |
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+ | limit = 2 * 10 ^ 6 |
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− | let b = 2*m*n, |
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+ | |||
− | let c = m^2 + n^2, |
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+ | triangs :: [Int] |
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− | let p = a + b + c, |
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− | + | triangs = [p | n <- [2 .. 1000], m <- [1 .. n - 1], odd (m + n), |
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+ | m `gcd` n == 1, let p = 2 * (n ^ 2 + m * n), p <= limit] |
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+ | |||
+ | problem_75 :: Int |
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+ | problem_75 = length $ filter (== 1) $ elems $ |
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+ | (\ns -> accumArray (+) 0 (1, limit) [(n, 1) | n <- ns] :: UArray Int Int) $ |
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+ | take limit $ concatMap (\m -> takeWhile (<= limit) [m, 2 * m .. ]) triangs |
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</haskell> |
</haskell> |
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Solution: |
Solution: |
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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> |
<haskell> |
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+ | build x = (map sum (zipWith drop [0..] x) ++ [1]) : x |
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− | problem_76 = undefined |
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+ | problem_76 = (sum $ head $ iterate build [] !! 100) - 1 |
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</haskell> |
</haskell> |
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Line 70: | Line 144: | ||
Solution: |
Solution: |
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+ | |||
+ | Brute force but still finds the solution in less than one second. |
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<haskell> |
<haskell> |
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+ | counter = foldl (\without p -> |
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− | problem_77 = undefined |
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+ | let (poor,rich) = splitAt p without |
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+ | with = poor ++ |
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+ | zipWith (+) with rich |
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+ | in with |
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+ | ) (1 : repeat 0) |
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+ | |||
+ | problem_77 = |
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+ | find ((>5000) . (ways !!)) $ [1..] |
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+ | where |
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+ | ways = counter $ take 100 primes |
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</haskell> |
</haskell> |
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Line 79: | Line 165: | ||
Solution: |
Solution: |
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<haskell> |
<haskell> |
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+ | import Data.Array |
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− | problem_78 = undefined |
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+ | |||
+ | partitions :: Array Int Integer |
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+ | partitions = |
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+ | array (0,1000000) $ |
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+ | (0,1) : |
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+ | [(n,sum [s * partitions ! p| |
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+ | (s,p) <- zip signs $ parts n])| |
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+ | n <- [1..1000000]] |
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+ | where |
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+ | signs = cycle [1,1,(-1),(-1)] |
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+ | suite = map penta $ concat [[n,(-n)]|n <- [1..]] |
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+ | penta n = n*(3*n - 1) `div` 2 |
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+ | parts n = takeWhile (>= 0) [n-x| x <- suite] |
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+ | |||
+ | problem_78 :: Int |
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+ | problem_78 = |
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+ | head $ filter (\x -> (partitions ! x) `mod` 1000000 == 0) [1..] |
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</haskell> |
</haskell> |
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Line 87: | Line 190: | ||
Solution: |
Solution: |
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<haskell> |
<haskell> |
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+ | import Data.Char (digitToInt, intToDigit) |
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− | problem_79 = undefined |
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+ | import Data.Graph (buildG, topSort) |
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+ | import Data.List (intersect) |
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+ | |||
+ | p79 file= |
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+ | (+0)$read . intersect graphWalk $ usedDigits |
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+ | where |
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+ | usedDigits = intersect "0123456789" $ file |
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+ | edges = concatMap (edgePair . map digitToInt) . words $ file |
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+ | graphWalk = map intToDigit . topSort . buildG (0, 9) $ edges |
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+ | edgePair [x, y, z] = [(x, y), (y, z)] |
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+ | edgePair _ = undefined |
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+ | |||
+ | problem_79 = do |
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+ | f<-readFile "keylog.txt" |
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+ | print $p79 f |
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</haskell> |
</haskell> |
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Line 93: | Line 211: | ||
Calculating the digital sum of the decimal digits of irrational square roots. |
Calculating the digital sum of the decimal digits of irrational square roots. |
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+ | This solution uses binary search to find the square root of a large Integer: |
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− | Solution: |
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<haskell> |
<haskell> |
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+ | import Data.Char (digitToInt) |
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− | problem_80 = undefined |
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− | </haskell> |
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+ | intSqrt :: Integer -> Integer |
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− | [[Category:Tutorials]] |
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+ | intSqrt n = bsearch 1 n |
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− | [[Category:Code]] |
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+ | where |
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+ | bsearch l u = let m = (l+u) `div` 2 |
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+ | m2 = m^2 |
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+ | in if u <= l |
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+ | then m |
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+ | else if m2 < n |
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+ | then bsearch (m+1) u |
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+ | else bsearch l m |
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+ | |||
+ | problem_80 :: Int |
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+ | problem_80 = sum [f r | a <- [1..100], |
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+ | let x = a * e, |
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+ | let r = intSqrt x, |
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+ | r*r /= x] |
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+ | where |
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+ | e = 10^202 |
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+ | f = sum . take 100 . map digitToInt . show |
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+ | </haskell> |
Latest revision as of 02:57, 3 May 2015
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:
module Main where
import Data.Array.Unboxed (UArray, accumArray, elems)
main :: IO ()
main = print problem_75
limit :: Int
limit = 2 * 10 ^ 6
triangs :: [Int]
triangs = [p | n <- [2 .. 1000], m <- [1 .. n - 1], odd (m + n),
m `gcd` n == 1, let p = 2 * (n ^ 2 + m * n), p <= limit]
problem_75 :: Int
problem_75 = length $ filter (== 1) $ elems $
(\ns -> accumArray (+) 0 (1, limit) [(n, 1) | n <- ns] :: UArray Int Int) $
take limit $ concatMap (\m -> takeWhile (<= limit) [m, 2 * m .. ]) 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 = 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
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