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Euler problems/1 to 10

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Do them on your own!
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== [http://projecteuler.net/index.php?section=view&id=1 Problem 1] ==
 +
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 04:53, 30 January 2008

Contents

1 Problem 1

Add all the natural numbers below 1000 that are multiples of 3 or 5.

Solution: 

problem_1 = 
    sum [ x |
    x <- [1..999],
    (x `mod` 3 == 0) ||  (x `mod` 5 == 0)
    ]
problem_1_v2 = 
    sum $ filter (\x -> ( x `mod` 3 == 0 || x `mod` 5 == 0 ) ) [1..999]
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)

2 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)

3 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)

This can be improved by using

null . tail

instead of

(== 1) . length
.

4 Problem 4

Find the largest palindrome made from the product of two 3-digit numbers.

Solution: 

problem_4 = 
    foldr max 0 [ x |
    y <- [100..999],
    z <- [100..999], 
    let x = y * z, 
    let s = show x,
    s == reverse s
    ]

An alternative to avoid evaluating twice the same pair of numbers: 

problem_4' = 
    foldr1 max [ x |
    y <- [100..999],
    z <- [y..999],
    let x = y * z,
    let s = show x,
    s == reverse s
    ]

5 Problem 5

What is the smallest number divisible by each of the numbers 1 to 20?

Solution: 

problem_5 = 
    head [ x |
    x <- [2520,5040..],
    all (\y -> x `mod` y == 0) [1..20]
    ]

An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom: 

problem_5' = foldr1 lcm [1..20]

6 Problem 6

What is the difference between the sum of the squares and the square of the sums?

Solution: 

problem_6 = 
    sum [ x^2 | x <- [1..100]] - (sum [1..100])^2

7 Problem 7

Find the 10001st prime.

Solution: 

--primes in problem_3
problem_7 = 
    head $ drop 10000 primes

As above, this can be improved by using

null . tail

instead of

(== 1) . length
.

Here is an alternative that uses a sieve of Eratosthenes: 

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

8 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

9 Problem 9

There is only one Pythagorean triplet, {a, b, c}, for which a + b + c = 1000. Find the product abc.

Solution: 

problem_9 = 
    head [a*b*c |
    a <- [1..500], 
    b <- [a..500], 
    let c = 1000-a-b, 
    a^2 + b^2 == c^2
    ]

Another solution using Pythagorean Triplets generation: 

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

10 Problem 10

Calculate the sum of all the primes below one million.

Solution: 

problem_10 = 
    sum (takeWhile (< 1000000) primes)
Retrieved from "http://www.haskell.org/haskellwiki/Euler_problems/1_to_10"