Difference between revisions of "Euler problems/1 to 10"
CaleGibbard (talk | contribs) (Fix bizarre layout.) |
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<haskell> |
<haskell> |
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sumOnetoN n = n * (n+1) `div` 2 |
sumOnetoN n = n * (n+1) `div` 2 |
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− | problem_1 = |
+ | problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999 |
⚫ | |||
− | sumStep 3 999 + sumStep 5 999 - sumStep 15 999 |
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⚫ | |||
sumStep s n = s * sumOnetoN (n `div` s) |
sumStep s n = s * sumOnetoN (n `div` s) |
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</haskell> |
</haskell> |
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<haskell> |
<haskell> |
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problem_2 = |
problem_2 = |
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− | sum [ x | |
+ | sum [ x | x <- takeWhile (<= 1000000) fibs, |
⚫ | |||
− | x <- takeWhile (<= 1000000) fibs, |
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⚫ | |||
− | ] |
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where |
where |
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fibs = 1 : 1 : zipWith (+) fibs (tail fibs) |
fibs = 1 : 1 : zipWith (+) fibs (tail fibs) |
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<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>. |
<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>. |
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<haskell> |
<haskell> |
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− | problem_2_v2 = |
+ | problem_2_v2 = sumEvenFibs $ numEvenFibsLessThan 1000000 |
⚫ | |||
− | sumEvenFibs $ numEvenFibsLessThan 1000000 |
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⚫ | |||
− | sumEvenFibs n = |
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⚫ | |||
− | evenFib n = |
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⚫ | |||
numEvenFibsLessThan n = |
numEvenFibsLessThan n = |
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− | + | floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5) |
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</haskell> |
</haskell> |
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problem_2 = sumEvenFibsLessThan 1000000 |
problem_2 = sumEvenFibsLessThan 1000000 |
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− | sumEvenFibsLessThan n = |
+ | sumEvenFibsLessThan n = (a + b - 1) `div` 2 |
⚫ | |||
− | (a + b - 1) `div` 2 |
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⚫ | |||
n2 = n `div` 2 |
n2 = n `div` 2 |
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− | (a, b) = |
+ | (a, b) = foldr f (0,1) |
− | + | . takeWhile ((<= n2) . fst) |
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− | + | . iterate times2E $ (1, 4) |
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− | + | f x y | fst z <= n2 = z |
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− | + | | otherwise = y |
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− | + | where z = x `addE` y |
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− | + | addE (a, b) (c, d) = (a*d + b*c - 4*ac, ac + b*d) |
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− | + | where ac=a*c |
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+ | |||
− | addE (a, b) (c, d) = |
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− | + | times2E (a, b) = addE (a, b) (a, b) |
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⚫ | |||
− | ac=a*c |
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− | times2E (a, b) = |
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− | addE (a, b) (a, b) |
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</haskell> |
</haskell> |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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⚫ | |||
− | primes = |
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+ | |||
⚫ | |||
− | primeFactors n = |
+ | primeFactors n = factor n primes |
⚫ | |||
− | factor n primes |
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− | where |
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factor n (p:ps) |
factor n (p:ps) |
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| p*p > n = [n] |
| p*p > n = [n] |
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| otherwise = factor n ps |
| otherwise = factor n ps |
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− | problem_3 = |
+ | problem_3 = last (primeFactors 317584931803) |
− | last (primeFactors 317584931803) |
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</haskell> |
</haskell> |
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== [http://projecteuler.net/index.php?section=problems&id=4 Problem 4] == |
== [http://projecteuler.net/index.php?section=problems&id=4 Problem 4] == |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | problem_4 = |
+ | problem_4 = maximum [ x | y <- [100..999], |
⚫ | |||
− | foldr1 max [ x | |
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+ | let x = y * z, |
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− | y <- [100..999], |
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⚫ | |||
⚫ | |||
− | + | s == reverse s ] |
|
⚫ | |||
− | s == reverse s |
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− | ] |
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</haskell> |
</haskell> |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | fun n= |
+ | fun n = a - b |
− | + | where |
|
− | where |
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a=div (n^2 * (n+1)^2) 4 |
a=div (n^2 * (n+1)^2) 4 |
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b=div (n * (n+1) * (2*n+1)) 6 |
b=div (n * (n+1) * (2*n+1)) 6 |
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+ | |||
− | problem_6=fun 100 |
+ | problem_6 = fun 100 |
</haskell> |
</haskell> |
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<haskell> |
<haskell> |
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--primes in problem_3 |
--primes in problem_3 |
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− | problem_7 = |
+ | problem_7 = head $ drop 10000 primes |
− | head $ drop 10000 primes |
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</haskell> |
</haskell> |
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== [http://projecteuler.net/index.php?section=problems&id=8 Problem 8] == |
== [http://projecteuler.net/index.php?section=problems&id=8 Problem 8] == |
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take n xs : groupsOf n ( tail xs ) |
take n xs : groupsOf n ( tail xs ) |
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− | problem_8 x= |
+ | problem_8 x = maximum . map product . groupsOf 5 $ x |
⚫ | |||
− | maximum . map product . groupsOf 5 $ x |
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⚫ | |||
− | main=do |
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⚫ | |||
⚫ | |||
⚫ | |||
⚫ | |||
</haskell> |
</haskell> |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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− | triplets l = |
+ | triplets l = [[a,b,c] | m <- [2..limit], |
− | + | n <- [1..(m-1)], |
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− | + | let a = m^2 - n^2, |
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− | let |
+ | let b = 2*m*n, |
− | let |
+ | let c = m^2 + n^2, |
− | + | a+b+c==l] |
|
⚫ | |||
− | a+b+c==l |
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+ | |||
− | ] |
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⚫ | |||
⚫ | |||
⚫ | |||
</haskell> |
</haskell> |
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<haskell> |
<haskell> |
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--http://www.research.att.com/~njas/sequences/A046731 |
--http://www.research.att.com/~njas/sequences/A046731 |
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− | problem_10 = |
+ | problem_10 = sum (takeWhile (< 1000000) primes) |
− | sum (takeWhile (< 1000000) primes) |
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</haskell> |
</haskell> |
Revision as of 19:12, 19 February 2008
Problem 1
Add all the natural numbers below 1000 that are multiples of 3 or 5.
Solution:
sumOnetoN n = n * (n+1) `div` 2
problem_1 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
where
sumStep s n = s * sumOnetoN (n `div` s)
Problem 2
Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million.
Solution:
problem_2 =
sum [ x | x <- takeWhile (<= 1000000) fibs,
x `mod` 2 == 0]
where
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
The following two solutions use the fact that the even-valued terms in
the Fibonacci sequence themselves form a Fibonacci-like sequence
that satisfies
evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)
.
problem_2_v2 = sumEvenFibs $ numEvenFibsLessThan 1000000
sumEvenFibs n = (evenFib n + evenFib (n+1) - 2) `div` 4
evenFib n = round $ (2 + sqrt 5) ** (fromIntegral n) / sqrt 5
numEvenFibsLessThan n =
floor $ (log (fromIntegral n - 0.5) + 0.5*log 5) / log (2 + sqrt 5)
The first two solutions work because 10^6 is small. The following solution also works for much larger numbers (up to at least 10^1000000 on my computer):
problem_2 = sumEvenFibsLessThan 1000000
sumEvenFibsLessThan n = (a + b - 1) `div` 2
where
n2 = n `div` 2
(a, b) = foldr f (0,1)
. takeWhile ((<= n2) . fst)
. iterate times2E $ (1, 4)
f x y | fst z <= n2 = z
| otherwise = y
where z = x `addE` y
addE (a, b) (c, d) = (a*d + b*c - 4*ac, ac + b*d)
where ac=a*c
times2E (a, b) = addE (a, b) (a, b)
Problem 3
Find the largest prime factor of 317584931803.
Solution:
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
primeFactors n = factor n primes
where
factor n (p:ps)
| p*p > n = [n]
| n `mod` p == 0 = p : factor (n `div` p) (p:ps)
| otherwise = factor n ps
problem_3 = last (primeFactors 317584931803)
Problem 4
Find the largest palindrome made from the product of two 3-digit numbers.
Solution:
problem_4 = maximum [ x | y <- [100..999],
z <- [y..999],
let x = y * z,
let s = show x,
s == reverse s ]
Problem 5
What is the smallest number divisible by each of the numbers 1 to 20?
Solution:
--http://www.research.att.com/~njas/sequences/A003418
problem_5 = foldr1 lcm [1..20]
Problem 6
What is the difference between the sum of the squares and the square of the sums?
Solution:
fun n = a - b
where
a=div (n^2 * (n+1)^2) 4
b=div (n * (n+1) * (2*n+1)) 6
problem_6 = fun 100
Problem 7
Find the 10001st prime.
Solution:
--primes in problem_3
problem_7 = head $ drop 10000 primes
Problem 8
Discover the largest product of five consecutive digits in the 1000-digit number.
Solution:
import Data.Char
groupsOf _ [] = []
groupsOf n xs =
take n xs : groupsOf n ( tail xs )
problem_8 x = maximum . map product . groupsOf 5 $ x
main = do t <- readFile "p8.log"
let digits = map digitToInt $foldl (++) "" $ lines t
print $ problem_8 digits
Problem 9
There is only one Pythagorean triplet, {a, b, c}, for which a + b + c = 1000. Find the product abc.
Solution:
triplets l = [[a,b,c] | m <- [2..limit],
n <- [1..(m-1)],
let a = m^2 - n^2,
let b = 2*m*n,
let c = m^2 + n^2,
a+b+c==l]
where limit = floor . sqrt . fromIntegral $ l
problem_9 = product . head . triplets $ 1000
Problem 10
Calculate the sum of all the primes below one million.
Solution:
--http://www.research.att.com/~njas/sequences/A046731
problem_10 = sum (takeWhile (< 1000000) primes)