Difference between revisions of "Euler problems/71 to 80"
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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 = |
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− | + | |otherwise=na |
|
+ | where |
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− | d <- [1..1000000], |
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+ | na=numerator a |
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− | let n = (d * 3) `div` 7, |
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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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− | ] |
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+ | a1=(na+nb)%(da+db) |
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− | |||
+ | da2=denominator a1 |
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− | problem_71 :: Integer |
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− | problem_71 |
+ | problem_71=fareySeq (0%1) (3%7) |
</haskell> |
</haskell> |
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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. |
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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− | |||
− | See problem 69 for phi function |
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− | |||
<haskell> |
<haskell> |
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+ | groups=1000 |
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− | problem_72 = sum [phi x|x <- [1..1000000]] |
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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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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | import Data. |
+ | import Data.Array |
+ | twix k = crude k - fd2 - sum [ar!(k `div` m) | m <- [3 .. k `div` 5], odd m] |
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− | |||
− | median :: Ratio Int -> Ratio Int -> Ratio Int |
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− | median a b = ((numerator a) + (numerator b)) % ((denominator a) + (denominator b)) |
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− | |||
− | count :: Ratio Int -> Ratio Int -> Int |
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− | count a b |
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− | | d > 10000 = 1 |
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− | | otherwise = count a m + count m b |
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where |
where |
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− | + | fd2 = crude (k `div` 2) |
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− | + | ar = array (5,k `div` 3) $ |
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+ | ((5,1):[(j, crude j - sum [ar!(j `div` m) | m <- [2 .. j `div` 5]]) |
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− | |||
+ | | j <- [6 .. k `div` 3]]) |
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− | problem_73 :: Int |
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+ | crude j = |
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− | problem_73 = (count (1%3) (1%2)) - 1 |
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+ | m*(3*m+r-2) + s |
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+ | where |
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+ | (m,r) = j `divMod` 6 |
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+ | s = case r of |
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+ | 5 -> 1 |
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+ | _ -> 0 |
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+ | |||
+ | problem_73 = twix 10000 |
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</haskell> |
</haskell> |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | import Data.Array (Array, array, (!), elems) |
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− | import Data.Char (ord) |
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− | import Data.List (foldl1') |
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− | import Prelude hiding (cycle) |
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− | |||
− | fact :: Integer -> Integer |
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− | fact 0 = 1 |
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− | fact n = foldl1' (*) [1..n] |
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− | |||
− | factorDigits :: Array Integer Integer |
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− | factorDigits = |
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− | array (0,2177281) |
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− | [(x,n)| |
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− | x <- [0..2177281], |
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− | let n = sum $ |
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− | map (\y -> fact (toInteger $ ord y - 48)) $ |
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− | show x |
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− | ] |
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− | |||
− | cycle :: Integer -> Integer |
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− | cycle 145 = 1 |
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− | cycle 169 = 3 |
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− | cycle 363601 = 3 |
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− | cycle 1454 = 3 |
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− | cycle 871 = 2 |
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− | cycle 45361 = 2 |
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− | cycle 872 = 2 |
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− | cycle 45362 = 2 |
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− | cycle _ = 0 |
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− | |||
− | isChainLength :: Integer -> Integer -> Bool |
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− | isChainLength len n |
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− | | len < 0 = False |
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− | | t = isChainLength (len-1) n' |
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− | | otherwise = (len - c) == 0 |
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− | where |
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− | c = cycle n |
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− | t = c == 0 |
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− | n' = factorDigits ! n |
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− | |||
− | -- | strict version of the maximum function |
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− | maximum' :: (Ord a) => [a] -> a |
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− | maximum' [] = undefined |
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− | maximum' [x] = x |
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− | maximum' (a:b:xs) = let m = max a b in m `seq` maximum' (m : xs) |
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− | |||
− | problem_74 :: Int |
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− | problem_74 = |
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− | length $ filter |
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− | (\(_,b) -> isChainLength 59 b) $ |
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− | zip ([0..] :: [Integer]) $ |
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− | take 1000000 $ elems factorDigits |
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− | </haskell> |
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− | |||
− | Slightly faster solution : |
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− | <haskell>{-# OPTIONS_GHC -fbang-patterns #-} |
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import Data.List |
import Data.List |
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− | import Data.Array |
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− | |||
explode 0 = [] |
explode 0 = [] |
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explode n = n `mod` 10 : explode (n `quot` 10) |
explode n = n `mod` 10 : explode (n `quot` 10) |
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+ | |||
− | |||
+ | chain 2 = 1 |
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− | count :: (a -> Bool) -> [a] -> Int |
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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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− | lgo !c (y:ys) | pred y = lgo (c + 1) ys |
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+ | chain 169 = 3 |
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− | | otherwise = lgo c ys |
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+ | chain 363601 = 3 |
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− | |||
+ | chain 1454 = 3 |
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− | known = [([1,2,145,40585],1),([871,45361,872,45362],2),([169,363601,1454],3)] |
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+ | chain 871 = 2 |
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− | mChain = array (1,1000000) $ (concat $ expand known) |
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+ | chain 45361 = 2 |
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− | ++ [(x, n)|x<-[3..1000000] |
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+ | chain 872 = 2 |
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− | , not $ x `elem` concat (map fst known) |
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+ | chain 45362 = 2 |
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− | , let n = 1 + chain (sumFactDigits x)] |
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+ | chain x = 1 + chain (sumFactDigits x) |
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− | where expand [] = [] |
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+ | makeIncreas 1 minnum = [[a]|a<-[minnum..9]] |
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− | expand ((xs,len):xxs) = map (flip (,) len) xs : expand xxs |
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+ | makeIncreas digits minnum = [a:b|a<-[minnum ..9],b<-makeIncreas (digits-1) a] |
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− | chain x | x <= 1000000 = mChain ! x |
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+ | p74= |
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− | | otherwise = 1 + chain (sumFactDigits x) |
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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 |
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] |
facts = scanl (*) 1 [1..9] |
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+ | countNum xs=ys |
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− | |||
+ | where |
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− | problem_74 = count (== 60) $ elems mChain</haskell> |
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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> |
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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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− | |||
− | Calculated using Euler's pentagonal formula and a list for memoization. |
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− | <haskell> |
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− | partitions = |
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− | 1 : [sum |
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− | [s * partitions !! p| |
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− | (s,p) <- zip signs $ parts n |
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− | ]| |
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− | n <- [1..]] |
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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_76 = partitions !! 100 - 1 |
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− | </haskell> |
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Here is a simpler solution: For each n, we create the list of the number of partitions of n |
Here is a simpler solution: For each n, we create the list of the number of partitions of n |
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<haskell> |
<haskell> |
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build x = (map sum (zipWith drop [0..] x) ++ [1]) : x |
build x = (map sum (zipWith drop [0..] x) ++ [1]) : x |
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− | problem_76 |
+ | problem_76 = (sum $ head $ iterate build [] !! 100) - 1 |
</haskell> |
</haskell> |
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Brute force but still finds the solution in less than one second. |
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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− | combinations acc 0 _ = [acc] |
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+ | let (poor,rich) = splitAt p without |
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− | combinations acc _ [] = [] |
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+ | with = poor ++ |
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− | combinations acc value prim@(x:xs) = |
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+ | zipWith (+) with rich |
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− | combinations (acc ++ [x]) value' prim' ++ combinations acc value xs |
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+ | in with |
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− | where |
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− | + | ) (1 : repeat 0) |
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+ | |||
− | prim' = dropWhile (>value') prim |
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+ | problem_77 = |
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− | |||
+ | find ((>5000) . (ways !!)) $ [1..] |
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− | problem_77 :: Integer |
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− | problem_77 = head $ filter f [1..] |
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where |
where |
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− | + | ways = counter $ take 100 primes |
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</haskell> |
</haskell> |
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Revision as of 07:51, 20 January 2008
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:
import Data.Array
twix k = crude k - fd2 - sum [ar!(k `div` m) | m <- [3 .. k `div` 5], odd m]
where
fd2 = crude (k `div` 2)
ar = array (5,k `div` 3) $
((5,1):[(j, crude j - sum [ar!(j `div` m) | m <- [2 .. j `div` 5]])
| j <- [6 .. k `div` 3]])
crude j =
m*(3*m+r-2) + s
where
(m,r) = j `divMod` 6
s = case r of
5 -> 1
_ -> 0
problem_73 = twix 10000
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: This is only slightly harder than problem 39. The search condition is simpler but the search space is larger.
problem_75 =
length . filter ((== 1) . length) $ group perims
where perims = sort [scale*p | p <- pTriples, scale <- [1..10^6 `div` p]]
pTriples = [p |
n <- [1..1000],
m <- [n+1..1000],
even n || even m,
gcd n m == 1,
let a = m^2 - n^2,
let b = 2*m*n,
let c = m^2 + n^2,
let p = a + b + c,
p <= 10^6]
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
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:
Same as problem 76 but using array instead of lists to speedup things.
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:
A bit ugly but works fine
import Data.List
problem_79 :: String -> String
problem_79 file =
map fst $
sortBy (\(_,a) (_,b) ->
compare (length b) (length a)) $
zip digs order
where
nums = lines file
digs =
map head $ group $
sort $ filter (\c -> c >= '0' && c <= '9') file
prec = concatMap (\(x:y:z:_) -> [[x,y],[y,z],[x,z]]) nums
order =
map (\n -> map head $
group $ sort $ map (\(_:x:_) -> x) $
filter (\(x:_) -> x == n) prec) digs
main=do
f<-readFile "keylog.txt"
print$problem_79 f
Problem 80
Calculating the digital sum of the decimal digits of irrational square roots.
Solution:
import Data.List ((\\))
hundreds :: Integer -> [Integer]
hundreds n = hundreds' [] n
where
hundreds' acc 0 = acc
hundreds' acc n = hundreds' (m : acc) d
where
(d,m) = divMod n 100
squareDigs :: Integer -> [Integer]
squareDigs n = p : squareDigs' p r xs
where
(x:xs) = hundreds n ++ repeat 0
p = floor $ sqrt $ fromInteger x
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]]]