Difference between revisions of "Euler problems/21 to 30"
(→Problem 21: Clarify problem and add a solution.) |
|||
(26 intermediate revisions by 14 users not shown) | |||
Line 1: | Line 1: | ||
== [http://projecteuler.net/index.php?section=problems&id=21 Problem 21] == |
== [http://projecteuler.net/index.php?section=problems&id=21 Problem 21] == |
||
− | Evaluate the sum of all amicable |
+ | Evaluate the sum of all amicable numbers (including those with a pair number over the limit) under 10000. |
− | Solution: |
+ | Solution: |
+ | (http://www.research.att.com/~njas/sequences/A063990) |
||
+ | |||
+ | This is a little slow because of the naive method used to compute the divisors. |
||
<haskell> |
<haskell> |
||
+ | problem_21 = sum [m+n | m <- [2..9999], let n = divisorsSum ! m, amicable m n] |
||
− | --http://www.research.att.com/~njas/sequences/A063990 |
||
+ | where amicable m n = m < n && n < 10000 && divisorsSum ! n == m |
||
− | problem_21 = sum [220, 284, 1184, 1210, 2620, 2924, 5020, 5564, 6232, 6368] |
||
+ | divisorsSum = array (1,9999) |
||
+ | [(i, sum (divisors i)) | i <- [1..9999]] |
||
+ | divisors n = [j | j <- [1..n `div` 2], n `mod` j == 0] |
||
+ | </haskell> |
||
+ | |||
+ | Here is an alternative using a faster way of computing the sum of divisors. |
||
+ | <haskell> |
||
+ | problem_21_v2 = sum [n | n <- [2..9999], let m = d n, |
||
+ | m > 1, m < 10000, n == d m, d m /= d (d m)] |
||
+ | d n = product [(p * product g - 1) `div` (p - 1) | |
||
+ | g <- group $ primeFactors n, let p = head g |
||
+ | ] - n |
||
+ | primeFactors = pf primes |
||
+ | where |
||
+ | pf ps@(p:ps') n |
||
+ | | p * p > n = [n] |
||
+ | | r == 0 = p : pf ps q |
||
+ | | otherwise = pf ps' n |
||
+ | where (q, r) = n `divMod` p |
||
+ | primes = 2 : filter (null . tail . primeFactors) [3,5..] |
||
+ | </haskell> |
||
+ | |||
+ | Here is another alternative solution that computes the sum-of-divisors for the numbers by iterating over products of their factors (very fast): |
||
+ | |||
+ | <haskell> |
||
+ | import Data.Array |
||
+ | |||
+ | max_ = 100000 |
||
+ | |||
+ | gen 100001 = [] |
||
+ | gen n = [(i*n,n)|i <- [2 .. max_ `div` n]] ++ (gen (n+1)) |
||
+ | |||
+ | arr = accumArray (+) 0 (0,max_) (gen 1) |
||
+ | |||
+ | problem_21_v3 = sum $ filter (\a -> let b = (arr!a) in b /= a && (arr!b) == a) [1 .. (10000 - 1)] |
||
+ | |||
</haskell> |
</haskell> |
||
Line 15: | Line 54: | ||
import Data.List |
import Data.List |
||
import Data.Char |
import Data.Char |
||
− | problem_22 = |
+ | problem_22 = |
− | input <- readFile "names.txt" |
+ | do input <- readFile "names.txt" |
− | let names = sort $ read$"["++ input++"]" |
+ | let names = sort $ read$"["++ input++"]" |
− | let scores = zipWith score names [1..] |
+ | let scores = zipWith score names [1..] |
− | + | print . sum $ scores |
|
+ | where score w i = (i *) . sum . map (\c -> ord c - ord 'A' + 1) $ w |
||
− | where |
||
− | score w i = (i *) $ sum $ map (\c -> ord c - ord 'A' + 1) w |
||
</haskell> |
</haskell> |
||
Line 39: | Line 77: | ||
isSum = any (abunds_array !) . rests |
isSum = any (abunds_array !) . rests |
||
− | problem_23 = |
+ | problem_23 = print . sum . filter (not . isSum) $ [1..n] |
</haskell> |
</haskell> |
||
Line 52: | Line 90: | ||
fac n = n * fac (n - 1) |
fac n = n * fac (n - 1) |
||
perms [] _= [] |
perms [] _= [] |
||
− | perms xs n= |
+ | perms xs n= x : perms (delete x xs) (mod n m) |
− | + | where m = fac $ length xs - 1 |
|
− | + | y = div n m |
|
+ | x = xs!!y |
||
− | m=fac$(length(xs) -1) |
||
− | y=div n m |
||
− | x = xs!!y |
||
− | problem_24 = |
+ | problem_24 = perms "0123456789" 999999 |
+ | </haskell> |
||
+ | |||
+ | Or, using Data.List.permutations, |
||
+ | <haskell> |
||
+ | import Data.List |
||
+ | problem_24 = (!! 999999) . sort $ permutations ['0'..'9'] |
||
+ | </haskell> |
||
+ | |||
+ | Casey Hawthorne |
||
+ | |||
+ | For Project Euler #24 you don't need to generate all the lexicographic permutations by Knuth's method or any other. |
||
+ | |||
+ | You're only looking for the millionth lexicographic permutation of "0123456789" |
||
+ | |||
+ | <haskell> |
||
+ | |||
+ | -- Plan of attack. |
||
+ | |||
+ | -- The "x"s are different numbers |
||
+ | -- 0xxxxxxxxx represents 9! = 362880 permutations/numbers |
||
+ | -- 1xxxxxxxxx represents 9! = 362880 permutations/numbers |
||
+ | -- 2xxxxxxxxx represents 9! = 362880 permutations/numbers |
||
+ | |||
+ | |||
+ | -- 20xxxxxxxx represents 8! = 40320 |
||
+ | -- 21xxxxxxxx represents 8! = 40320 |
||
+ | |||
+ | -- 23xxxxxxxx represents 8! = 40320 |
||
+ | -- 24xxxxxxxx represents 8! = 40320 |
||
+ | -- 25xxxxxxxx represents 8! = 40320 |
||
+ | -- 26xxxxxxxx represents 8! = 40320 |
||
+ | -- 27xxxxxxxx represents 8! = 40320 |
||
+ | |||
+ | |||
+ | module Euler where |
||
+ | |||
+ | import Data.List |
||
+ | |||
+ | factorial n = product [1..n] |
||
+ | |||
+ | -- lexOrder "0123456789" 1000000 "" |
||
+ | |||
+ | lexOrder digits left s |
||
+ | | len == 0 = s ++ digits |
||
+ | | quot > 0 && rem == 0 = lexOrder (digits\\(show (digits!!(quot-1)))) rem (s ++ [(digits!!(quot-1))]) |
||
+ | | quot == 0 && rem == 0 = lexOrder (digits\\(show (digits!!len))) rem (s ++ [(digits!!len)]) |
||
+ | | rem == 0 = lexOrder (digits\\(show (digits!!(quot+1)))) rem (s ++ [(digits!!(quot+1))]) |
||
+ | | otherwise = lexOrder (digits\\(show (digits!!(quot)))) rem (s ++ [(digits!!(quot))]) |
||
+ | where |
||
+ | len = (length digits) - 1 |
||
+ | (quot,rem) = quotRem left (factorial len) |
||
+ | |||
</haskell> |
</haskell> |
||
Line 67: | Line 155: | ||
Solution: |
Solution: |
||
<haskell> |
<haskell> |
||
+ | fibs = 0:1:(zipWith (+) fibs (tail fibs)) |
||
− | import Data.List |
||
+ | t = 10^999 |
||
− | fib x |
||
− | |x==0=0 |
||
− | |x==1=1 |
||
− | |x==2=1 |
||
− | |odd x=(fib (d+1))^2+(fib d)^2 |
||
− | |otherwise=(fib (d+1))^2-(fib (d-1))^2 |
||
− | where |
||
− | d=div x 2 |
||
+ | problem_25 = length w |
||
− | phi=(1+sqrt 5)/2 |
||
− | dig x=floor( (fromInteger x-1) * log 10 /log phi) |
||
− | problem_25 = |
||
− | head[a|a<-[dig num..],(>=limit)$fib a] |
||
where |
where |
||
+ | w = takeWhile (< t) fibs |
||
− | num=1000 |
||
+ | </haskell> |
||
− | limit=10^(num-1) |
||
+ | |||
+ | |||
+ | Casey Hawthorne |
||
+ | |||
+ | I believe you mean the following: |
||
+ | |||
+ | <haskell> |
||
+ | |||
+ | fibs = 0:1:(zipWith (+) fibs (tail fibs)) |
||
+ | |||
+ | last (takeWhile (<10^1000) fibs) |
||
</haskell> |
</haskell> |
||
Line 91: | Line 180: | ||
Solution: |
Solution: |
||
<haskell> |
<haskell> |
||
− | problem_26 = |
+ | problem_26 = fst $ maximumBy (comparing snd) |
+ | [(n,recurringCycle n) | n <- [1..999]] |
||
+ | where recurringCycle d = remainders d 10 [] |
||
+ | remainders d 0 rs = 0 |
||
+ | remainders d r rs = let r' = r `mod` d |
||
+ | in case elemIndex r' rs of |
||
+ | Just i -> i + 1 |
||
+ | Nothing -> remainders d (10*r') (r':rs) |
||
</haskell> |
</haskell> |
||
Line 99: | Line 195: | ||
Solution: |
Solution: |
||
<haskell> |
<haskell> |
||
− | problem_27= |
+ | problem_27 = -(2*a-1)*(a^2-a+41) |
+ | where n = 1000 |
||
− | negate (2*a-1)*(a^2-a+41) |
||
+ | m = head $ filter (\x->x^2-x+41>n) [1..] |
||
− | where |
||
− | + | a = m-1 |
|
− | m=head $filter (\x->x^2-x+41>n)[1..] |
||
− | a=m-1 |
||
</haskell> |
</haskell> |
||
Line 113: | Line 207: | ||
<haskell> |
<haskell> |
||
problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1 |
problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1 |
||
+ | </haskell> |
||
+ | |||
+ | Alternatively, one can use the fact that the distance between the diagonal numbers increases by 2 in every concentric square. Each square contains four gaps, so the following <hask>scanl</hask> does the trick: |
||
+ | |||
+ | <haskell> |
||
+ | euler28 n = sum $ scanl (+) 0 |
||
+ | (1:(concatMap (replicate 4) [2,4..(n-1)])) |
||
</haskell> |
</haskell> |
||
Line 122: | Line 223: | ||
import Control.Monad |
import Control.Monad |
||
problem_29 = length . group . sort $ liftM2 (^) [2..100] [2..100] |
problem_29 = length . group . sort $ liftM2 (^) [2..100] [2..100] |
||
+ | </haskell> |
||
+ | |||
+ | We can also solve it in a more naive way, without using Monads, like this: |
||
+ | <haskell> |
||
+ | import List |
||
+ | problem_29 = length $ nub pr29_help |
||
+ | where pr29_help = [z | y <- [2..100], |
||
+ | z <- lift y] |
||
+ | lift y = map (\x -> x^y) [2..100] |
||
+ | </haskell> |
||
+ | |||
+ | Simpler: |
||
+ | |||
+ | <haskell> |
||
+ | import List |
||
+ | problem_29 = length $ nub [x^y | x <- [2..100], y <- [2..100]] |
||
+ | </haskell> |
||
+ | |||
+ | Instead of using lists, the Set data structure can be used for a significant speed increase: |
||
+ | |||
+ | <haskell> |
||
+ | import Set |
||
+ | problem_29 = size $ fromList [x^y | x <- [2..100], y <- [2..100]] |
||
</haskell> |
</haskell> |
||
Line 129: | Line 253: | ||
Solution: |
Solution: |
||
<haskell> |
<haskell> |
||
+ | import Data.Char (digitToInt) |
||
− | --http://www.research.att.com/~njas/sequences/A052464 |
||
+ | |||
− | problem_30 = sum [4150, 4151, 54748, 92727, 93084, 194979] |
||
+ | limit :: Integer |
||
+ | limit = snd $ head $ dropWhile (\(a,b) -> a > b) $ zip (map (9^5*) [1..]) (map (10^) [1..]) |
||
+ | |||
+ | fifth :: Integer -> Integer |
||
+ | fifth = sum . map ((^5) . toInteger . digitToInt) . show |
||
+ | |||
+ | problem_30 :: Integer |
||
+ | problem_30 = sum $ filter (\n -> n == fifth n) [2..limit] |
||
</haskell> |
</haskell> |
Latest revision as of 15:53, 11 October 2015
Problem 21
Evaluate the sum of all amicable numbers (including those with a pair number over the limit) under 10000.
Solution: (http://www.research.att.com/~njas/sequences/A063990)
This is a little slow because of the naive method used to compute the divisors.
problem_21 = sum [m+n | m <- [2..9999], let n = divisorsSum ! m, amicable m n]
where amicable m n = m < n && n < 10000 && divisorsSum ! n == m
divisorsSum = array (1,9999)
[(i, sum (divisors i)) | i <- [1..9999]]
divisors n = [j | j <- [1..n `div` 2], n `mod` j == 0]
Here is an alternative using a faster way of computing the sum of divisors.
problem_21_v2 = sum [n | n <- [2..9999], let m = d n,
m > 1, m < 10000, n == d m, d m /= d (d m)]
d n = product [(p * product g - 1) `div` (p - 1) |
g <- group $ primeFactors n, let p = head g
] - n
primeFactors = pf primes
where
pf ps@(p:ps') n
| p * p > n = [n]
| r == 0 = p : pf ps q
| otherwise = pf ps' n
where (q, r) = n `divMod` p
primes = 2 : filter (null . tail . primeFactors) [3,5..]
Here is another alternative solution that computes the sum-of-divisors for the numbers by iterating over products of their factors (very fast):
import Data.Array
max_ = 100000
gen 100001 = []
gen n = [(i*n,n)|i <- [2 .. max_ `div` n]] ++ (gen (n+1))
arr = accumArray (+) 0 (0,max_) (gen 1)
problem_21_v3 = sum $ filter (\a -> let b = (arr!a) in b /= a && (arr!b) == a) [1 .. (10000 - 1)]
Problem 22
What is the total of all the name scores in the file of first names?
Solution:
import Data.List
import Data.Char
problem_22 =
do input <- readFile "names.txt"
let names = sort $ read$"["++ input++"]"
let scores = zipWith score names [1..]
print . sum $ scores
where score w i = (i *) . sum . map (\c -> ord c - ord 'A' + 1) $ w
Problem 23
Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.
Solution:
--http://www.research.att.com/~njas/sequences/A048242
import Data.Array
n = 28124
abundant n = eulerTotient n - n > n
abunds_array = listArray (1,n) $ map abundant [1..n]
abunds = filter (abunds_array !) [1..n]
rests x = map (x-) $ takeWhile (<= x `div` 2) abunds
isSum = any (abunds_array !) . rests
problem_23 = print . sum . filter (not . isSum) $ [1..n]
Problem 24
What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
Solution:
import Data.List
fac 0 = 1
fac n = n * fac (n - 1)
perms [] _= []
perms xs n= x : perms (delete x xs) (mod n m)
where m = fac $ length xs - 1
y = div n m
x = xs!!y
problem_24 = perms "0123456789" 999999
Or, using Data.List.permutations,
import Data.List
problem_24 = (!! 999999) . sort $ permutations ['0'..'9']
Casey Hawthorne
For Project Euler #24 you don't need to generate all the lexicographic permutations by Knuth's method or any other.
You're only looking for the millionth lexicographic permutation of "0123456789"
-- Plan of attack.
-- The "x"s are different numbers
-- 0xxxxxxxxx represents 9! = 362880 permutations/numbers
-- 1xxxxxxxxx represents 9! = 362880 permutations/numbers
-- 2xxxxxxxxx represents 9! = 362880 permutations/numbers
-- 20xxxxxxxx represents 8! = 40320
-- 21xxxxxxxx represents 8! = 40320
-- 23xxxxxxxx represents 8! = 40320
-- 24xxxxxxxx represents 8! = 40320
-- 25xxxxxxxx represents 8! = 40320
-- 26xxxxxxxx represents 8! = 40320
-- 27xxxxxxxx represents 8! = 40320
module Euler where
import Data.List
factorial n = product [1..n]
-- lexOrder "0123456789" 1000000 ""
lexOrder digits left s
| len == 0 = s ++ digits
| quot > 0 && rem == 0 = lexOrder (digits\\(show (digits!!(quot-1)))) rem (s ++ [(digits!!(quot-1))])
| quot == 0 && rem == 0 = lexOrder (digits\\(show (digits!!len))) rem (s ++ [(digits!!len)])
| rem == 0 = lexOrder (digits\\(show (digits!!(quot+1)))) rem (s ++ [(digits!!(quot+1))])
| otherwise = lexOrder (digits\\(show (digits!!(quot)))) rem (s ++ [(digits!!(quot))])
where
len = (length digits) - 1
(quot,rem) = quotRem left (factorial len)
Problem 25
What is the first term in the Fibonacci sequence to contain 1000 digits?
Solution:
fibs = 0:1:(zipWith (+) fibs (tail fibs))
t = 10^999
problem_25 = length w
where
w = takeWhile (< t) fibs
Casey Hawthorne
I believe you mean the following:
fibs = 0:1:(zipWith (+) fibs (tail fibs))
last (takeWhile (<10^1000) fibs)
Problem 26
Find the value of d < 1000 for which 1/d contains the longest recurring cycle.
Solution:
problem_26 = fst $ maximumBy (comparing snd)
[(n,recurringCycle n) | n <- [1..999]]
where recurringCycle d = remainders d 10 []
remainders d 0 rs = 0
remainders d r rs = let r' = r `mod` d
in case elemIndex r' rs of
Just i -> i + 1
Nothing -> remainders d (10*r') (r':rs)
Problem 27
Find a quadratic formula that produces the maximum number of primes for consecutive values of n.
Solution:
problem_27 = -(2*a-1)*(a^2-a+41)
where n = 1000
m = head $ filter (\x->x^2-x+41>n) [1..]
a = m-1
Problem 28
What is the sum of both diagonals in a 1001 by 1001 spiral?
Solution:
problem_28 = sum (map (\n -> 4*(n-2)^2+10*(n-1)) [3,5..1001]) + 1
Alternatively, one can use the fact that the distance between the diagonal numbers increases by 2 in every concentric square. Each square contains four gaps, so the following scanl
does the trick:
euler28 n = sum $ scanl (+) 0
(1:(concatMap (replicate 4) [2,4..(n-1)]))
Problem 29
How many distinct terms are in the sequence generated by ab for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?
Solution:
import Control.Monad
problem_29 = length . group . sort $ liftM2 (^) [2..100] [2..100]
We can also solve it in a more naive way, without using Monads, like this:
import List
problem_29 = length $ nub pr29_help
where pr29_help = [z | y <- [2..100],
z <- lift y]
lift y = map (\x -> x^y) [2..100]
Simpler:
import List
problem_29 = length $ nub [x^y | x <- [2..100], y <- [2..100]]
Instead of using lists, the Set data structure can be used for a significant speed increase:
import Set
problem_29 = size $ fromList [x^y | x <- [2..100], y <- [2..100]]
Problem 30
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.
Solution:
import Data.Char (digitToInt)
limit :: Integer
limit = snd $ head $ dropWhile (\(a,b) -> a > b) $ zip (map (9^5*) [1..]) (map (10^) [1..])
fifth :: Integer -> Integer
fifth = sum . map ((^5) . toInteger . digitToInt) . show
problem_30 :: Integer
problem_30 = sum $ filter (\n -> n == fifth n) [2..limit]