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== [http://projecteuler.net/index.php?section=view&id=1 Problem 1] ==
+
Do them on your own!
Add all the natural numbers below 1000 that are multiples of 3 or 5.
 
 
Solution:
 
<haskell>
 
problem_1 =
 
sum [ x |
 
x <- [1..999],
 
(x `mod` 3 == 0) || (x `mod` 5 == 0)
 
]
 
</haskell>
 
 
<haskell>
 
problem_1_v2 =
 
sum $ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999]
 
</haskell>
 
----
 
<haskell>
 
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)
 
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=view&id=2 Problem 2] ==
 
Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million.
 
 
Solution:
 
<haskell>
 
problem_2 =
 
sum [ x |
 
x <- takeWhile (<= 1000000) fibs,
 
x `mod` 2 == 0
 
]
 
where
 
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
 
</haskell>
 
 
The following two solutions use the fact that the even-valued terms in
 
the Fibonacci sequence themselves form a Fibonacci-like sequence
 
that satisfies
 
<hask>evenFib 0 = 0, evenFib 1 = 2, evenFib (n+2) = evenFib n + 4 * evenFib (n+1)</hask>.
 
<haskell>
 
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)
 
</haskell>
 
 
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):
 
<haskell>
 
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)
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=view&id=3 Problem 3] ==
 
Find the largest prime factor of 317584931803.
 
 
Solution:
 
<haskell>
 
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)
 
</haskell>
 
 
This can be improved by using
 
<hask>null . tail</hask>
 
instead of
 
<hask>(== 1) . length</hask>.
 
 
== [http://projecteuler.net/index.php?section=view&id=4 Problem 4] ==
 
Find the largest palindrome made from the product of two 3-digit numbers.
 
 
Solution:
 
<haskell>
 
problem_4 =
 
foldr max 0 [ x |
 
y <- [100..999],
 
z <- [100..999],
 
let x = y * z,
 
let s = show x,
 
s == reverse s
 
]
 
</haskell>
 
An alternative to avoid evaluating twice the same pair of numbers:
 
<haskell>
 
problem_4' =
 
foldr1 max [ x |
 
y <- [100..999],
 
z <- [y..999],
 
let x = y * z,
 
let s = show x,
 
s == reverse s
 
]
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=view&id=5 Problem 5] ==
 
What is the smallest number divisible by each of the numbers 1 to 20?
 
 
Solution:
 
<haskell>
 
problem_5 =
 
head [ x |
 
x <- [2520,5040..],
 
all (\y -> x `mod` y == 0) [1..20]
 
]
 
</haskell>
 
An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom:
 
<haskell>
 
problem_5' = foldr1 lcm [1..20]
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=view&id=6 Problem 6] ==
 
What is the difference between the sum of the squares and the square of the sums?
 
 
Solution:
 
<haskell>
 
problem_6 =
 
sum [ x^2 | x <- [1..100]] - (sum [1..100])^2
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=view&id=7 Problem 7] ==
 
Find the 10001st prime.
 
 
Solution:
 
<haskell>
 
--primes in problem_3
 
problem_7 =
 
head $ drop 10000 primes
 
</haskell>
 
 
As above, this can be improved by using
 
<hask>null . tail</hask>
 
instead of
 
<hask>(== 1) . length</hask>.
 
 
Here is an alternative that uses a
 
[http://www.haskell.org/pipermail/haskell-cafe/2007-February/022854.html sieve of Eratosthenes]:
 
 
<haskell>
 
primes' =
 
2 : 3 : sieve (tail primes') [5,7..]
 
where
 
sieve (p:ps) x =
 
h ++ sieve ps (filter (\q -> q `mod` p /= 0) t
 
where
 
(h, _:t) = span (p*p <) x
 
problem_7_v2 = primes' !! 10000
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=view&id=8 Problem 8] ==
 
Discover the largest product of five consecutive digits in the 1000-digit number.
 
 
Solution:
 
<haskell>
 
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
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=view&id=9 Problem 9] ==
 
There is only one Pythagorean triplet, {''a'', ''b'', ''c''}, for which ''a'' + ''b'' + ''c'' = 1000. Find the product ''abc''.
 
 
Solution:
 
<haskell>
 
problem_9 =
 
head [a*b*c |
 
a <- [1..500],
 
b <- [a..500],
 
let c = 1000-a-b,
 
a^2 + b^2 == c^2
 
]
 
</haskell>
 
 
Another solution using Pythagorean Triplets generation:
 
<haskell>
 
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
 
</haskell>
 
 
== [http://projecteuler.net/index.php?section=view&id=10 Problem 10] ==
 
Calculate the sum of all the primes below one million.
 
 
Solution:
 
<haskell>
 
problem_10 =
 
sum (takeWhile (< 1000000) primes)
 
</haskell>
 

Revision as of 21:39, 29 January 2008

Do them on your own!