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Euler problems/1 to 10

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(Removing category tags. See Talk:Euler_problems)
Line 4: Line 4:
Solution:
Solution:
<haskell>
<haskell>
-
problem_1 = sum [ x | x <- [1..999], (x `mod` 3 == 0) || (x `mod` 5 == 0)]
+
problem_1 = sum [ x |
 +
x <- [1..999],
 +
(x `mod` 3 == 0) || (x `mod` 5 == 0)
 +
]
</haskell>
</haskell>
Line 23: Line 26:
Solution:
Solution:
<haskell>
<haskell>
-
problem_2 = sum [ x | x <- takeWhile (<= 1000000) fibs, x `mod` 2 == 0]
+
problem_2 = sum [ x |
-
where fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
+
x <- takeWhile (<= 1000000) fibs,
 +
x `mod` 2 == 0
 +
]
 +
where
 +
fibs = 1 : 1 : zipWith (+) fibs (tail fibs)
</haskell>
</haskell>
Line 79: Line 86:
Solution:
Solution:
<haskell>
<haskell>
-
problem_4 = foldr max 0 [ x | y <- [100..999], z <- [100..999], let x = y * z, let s = show x, s == reverse s]
+
problem_4 = foldr max 0 [ x |
 +
y <- [100..999],
 +
z <- [100..999],
 +
let x = y * z,
 +
let s = show x,
 +
s == reverse s
 +
]
</haskell>
</haskell>
An alternative to avoid evaluating twice the same pair of numbers:
An alternative to avoid evaluating twice the same pair of numbers:
<haskell>
<haskell>
-
problem_4' = foldr1 max [ x | y <- [100..999], z <- [y..999], let x = y * z, let s = show x, s == reverse s]
+
problem_4' = foldr1 max [ x |
 +
y <- [100..999],
 +
z <- [y..999],
 +
let x = y * z,
 +
let s = show x,
 +
s == reverse s
 +
]
</haskell>
</haskell>
Line 91: Line 110:
Solution:
Solution:
<haskell>
<haskell>
-
problem_5 = head [ x | x <- [2520,5040..], all (\y -> x `mod` y == 0) [1..20]]
+
problem_5 = head [ x |
 +
x <- [2520,5040..],
 +
all (\y -> x `mod` y == 0) [1..20]
 +
]
</haskell>
</haskell>
An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom:
An alternative solution that takes advantage of the Prelude to avoid use of the generate and test idiom:
Line 140: Line 162:
Solution:
Solution:
<haskell>
<haskell>
-
num = ... -- 1000 digit number as a string
+
import Data.Char
-
digits = map digitToInt num
+
-
 
+
groupsOf _ [] = []
groupsOf _ [] = []
groupsOf n xs = take n xs : groupsOf n ( tail xs )
groupsOf n xs = take n xs : groupsOf n ( tail xs )
-
 
+
-
problem_8 = maximum . map product . groupsOf 5 $ digits
+
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>
</haskell>
Line 159: Line 183:
Another solution using Pythagorean Triplets generation:
Another solution using Pythagorean Triplets generation:
<haskell>
<haskell>
-
triplets :: Int -> [(Int, Int, Int)]
+
triplets l = [[a,b,c]|
-
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]
+
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
where limit = floor $ sqrt $ fromIntegral l
-
 
+
problem_9 = product $ head $ triplets 1000
-
tripletWithLength :: Int -> [(Int, Int, Int)]
+
-
tripletWithLength n = filter ((==n) . f) $ triplets n
+
-
where
+
-
f (a,b,c) = a+b+c
+
-
 
+
-
problem_9 :: Int
+
-
problem_9 = prod3 $ head $ tripletWithLength 1000
+
-
where
+
-
prod3 (a,b,c) = a*b*c
+
</haskell>
</haskell>

Revision as of 10:57, 5 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]
sum1to n = n * (n+1) `div` 2
 
problem_1_v3 = sumStep 3 999 + sumStep 5 999 - sumStep 15 999
    where sumStep s n = s * sum1to (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_v3 = 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) = let ac = a*c in (a*d + b*c - 4*ac, ac + b*d)
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 = 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_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 = let (h, _:t) = span (p*p <) x
                     in h ++ sieve ps (filter (\q -> q `mod` p /= 0) t
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"