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Euler problems/31 to 40

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Line 6: Line 6:
This is the naive doubly recursive solution. Speed would be greatly improved by use of [[memoization]], dynamic programming, or the closed form.
This is the naive doubly recursive solution. Speed would be greatly improved by use of [[memoization]], dynamic programming, or the closed form.
<haskell>
<haskell>
-
problem_31 = pence 200 [1,2,5,10,20,50,100,200]
+
problem_31 =
-
where pence 0 _ = 1
+
pence 200 [1,2,5,10,20,50,100,200]
-
pence n [] = 0
+
where
-
pence n denominations@(d:ds)
+
pence 0 _ = 1
-
| n < d = 0
+
pence n [] = 0
-
| otherwise = pence (n - d) denominations
+
pence n denominations@(d:ds)
-
+ pence n ds
+
| n < d = 0
 +
| otherwise = pence (n - d) denominations + pence n ds
</haskell>
</haskell>
Line 27: Line 28:
) ([[]] : repeat [])
) ([[]] : repeat [])
-
problem_31 = length $ combinations coins !! 200
+
problem_31 =
 +
length $ combinations coins !! 200
</haskell>
</haskell>
Line 35: Line 37:
Solution:
Solution:
<haskell>
<haskell>
-
problem_32 = sum $ nub $ map (\(a, b) -> a * b) multiplicands
+
problem_32 =
 +
sum $ nub $ map (\(a, b) -> a * b) multiplicands
where
where
-
multiplicands =
+
multiplicands =
-
[(a,b)| a <- [2..5000], b <- [a..(9999 `div` a)], check a b]
+
[(a,b)| a <- [2..5000], b <- [a..(9999 `div` a)], check a b]
-
check a b =
+
check a b =
-
no_zero s
+
no_zero s
-
&& (length ss) == 9
+
&& (length ss) == 9
-
&& foldr (\x y -> length x == 1 && y) True ss
+
&& foldr (\x y -> length x == 1 && y) True ss
-
where
+
where
-
s = show a ++ show b ++ show (a*b)
+
s = show a ++ show b ++ show (a*b)
-
ss = group $ sort s
+
ss = group $ sort s
-
no_zero (x:xs)
+
no_zero (x:xs)
-
| x == '0' = False
+
| x == '0' = False
-
| null xs = True
+
| null xs = True
-
| otherwise = no_zero xs
+
| otherwise = no_zero xs
</haskell>
</haskell>
Line 72: Line 75:
]
]
-
rs' = filter (<1) $ map (\x -> denominator x % numerator x) rs
+
rs' =
 +
filter (<1) $ map (\x -> denominator x % numerator x) rs
</haskell>
</haskell>
Line 81: Line 85:
<haskell>
<haskell>
import Data.Char
import Data.Char
-
problem_34 = sum [ x | x <- [3..100000], x == facsum x ]
+
problem_34 =
-
where facsum = sum . map (product . enumFromTo 1 . digitToInt) . show
+
sum [ x | x <- [3..100000], x == facsum x ]
 +
where
 +
facsum = sum . map (product . enumFromTo 1 . digitToInt) . show
</haskell>
</haskell>
Line 92: Line 98:
import Data.List (tails, (\\))
import Data.List (tails, (\\))
-
primes :: [Integer]
+
primes =
-
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
+
2 : filter ((==1) . length . primeFactors) [3,5..]
-
 
+
primeFactors n =
-
primeFactors :: Integer -> [Integer]
+
factor n primes
-
primeFactors n = factor n primes
+
where
where
-
factor _ [] = []
+
factor n (p:ps)
-
factor m (p:ps) | p*p > m = [m]
+
| p*p > n = [n]
-
| m `mod` p == 0 = p : factor (m `div` p) (p:ps)
+
| n `mod` p == 0 = p : factor (n `div` p) (p:ps)
-
| otherwise = factor m ps
+
| otherwise = factor n ps
-
 
+
isPrime
-
isPrime :: Integer -> Bool
+
|1 = False
-
isPrime 1 = False
+
|otherwise=
-
isPrime n = case (primeFactors n) of
+
case (primeFactors n) of
-
(_:_:_) -> False
+
(_:_:_) -> False
-
_ -> True
+
_ -> True
-
 
+
permutations n =
-
permutations :: Integer -> [Integer]
+
take l $ map (read . take l) $
-
permutations n = take l $ map (read . take l) $ tails $ take (2*l -1) $ cycle s
+
tails $ take (2*l -1) $ cycle s
where
where
-
s = show n
+
s = show n
-
l = length s
+
l = length s
-
 
+
-
circular_primes :: [Integer] -> [Integer]
+
circular_primes [] = []
circular_primes [] = []
circular_primes (x:xs)
circular_primes (x:xs)
Line 121: Line 124:
| otherwise = circular_primes xs
| otherwise = circular_primes xs
where
where
-
p = permutations x
+
p = permutations x
-
problem_35 :: Int
+
problem_35 =
-
problem_35 = length $ circular_primes $ takeWhile (<1000000) primes
+
length $ circular_primes $ takeWhile (<1000000) primes
</haskell>
</haskell>
Line 135: Line 138:
import Data.Char
import Data.Char
-
showBin = flip (showIntAtBase 2 intToDigit) ""
+
showBin =
 +
flip (showIntAtBase 2 intToDigit) ""
-
isPalindrome x = x == reverse x
+
isPalindrome x =
 +
x == reverse x
-
problem_36 = sum [x |
+
problem_36 =
 +
sum [x |
x <- [1,3..1000000],
x <- [1,3..1000000],
isPalindrome (show x),
isPalindrome (show x),
Line 153: Line 159:
import Data.List (tails, inits, nub)
import Data.List (tails, inits, nub)
-
primes :: [Integer]
+
truncs n =
-
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
+
nub . map read $
-
 
+
-
primeFactors :: Integer -> [Integer]
+
-
primeFactors n = factor n primes
+
-
where
+
-
factor _ [] = []
+
-
factor m (p:ps) | p*p > m = [m]
+
-
| m `mod` p == 0 = p : factor (m `div` p) (p:ps)
+
-
| otherwise = factor m ps
+
-
 
+
-
isPrime :: Integer -> Bool
+
-
isPrime 1 = False
+
-
isPrime n = case (primeFactors n) of
+
-
(_:_:_) -> False
+
-
_ -> True
+
-
 
+
-
truncs :: Integer -> [Integer]
+
-
truncs n = nub . map read $
+
(take l . tail . tails) s ++ (take l . tail . inits) s
(take l . tail . tails) s ++ (take l . tail . inits) s
where
where
Line 177: Line 166:
s = show n
s = show n
-
problem_37 = sum $ take 11 [x |
+
problem_37 =
 +
sum $ take 11 [x |
x <- dropWhile (<=9) primes,
x <- dropWhile (<=9) primes,
all isPrime (truncs x)
all isPrime (truncs x)
Line 189: Line 179:
Solution:
Solution:
<haskell>
<haskell>
-
problem_38 = maximum $ catMaybes [result | j <- [1..9999],
+
problem_38 =
-
let p2 = show j ++ show (2*j),
+
maximum $ catMaybes [result | j <- [1..9999],
-
let p3 = p2 ++ show (3*j),
+
let p2 = show j ++ show (2*j),
-
let p4 = p3 ++ show (4*j),
+
let p3 = p2 ++ show (3*j),
-
let p5 = p4 ++ show (5*j),
+
let p4 = p3 ++ show (4*j),
-
let result
+
let p5 = p4 ++ show (5*j),
-
| isPan p2 = Just p2
+
let result
-
| isPan p3 = Just p3
+
| isPan p2 = Just p2
-
| isPan p4 = Just p4
+
| isPan p3 = Just p3
-
| isPan p5 = Just p5
+
| isPan p4 = Just p4
-
| otherwise = Nothing]
+
| isPan p5 = Just p5
-
where isPan s = sort s == "123456789"
+
| otherwise = Nothing
 +
]
 +
where
 +
isPan s = sort s == "123456789"
</haskell>
</haskell>
Line 207: Line 200:
import Data.List
import Data.List
-
mult n i vs | length (concat vs) >= 9 = concat vs
+
mult n i vs
-
| otherwise = mult n (i+1) (vs ++ [show (n * i)])
+
| length (concat vs) >= 9 = concat vs
 +
| otherwise = mult n (i+1) (vs ++ [show (n * i)])
-
problem_38 :: Int
+
problem_38 =
-
problem_38 = maximum $ map read $ filter
+
maximum $ map read $ filter
((['1'..'9'] ==) .sort) $
((['1'..'9'] ==) .sort) $
[ mult n 1 [] | n <- [2..9999] ]
[ mult n 1 [] | n <- [2..9999] ]
Line 222: Line 216:
We use the well known formula to generate primitive Pythagorean triples. All we need are the perimeters, and they have to be scaled to produce all triples in the problem space.
We use the well known formula to generate primitive Pythagorean triples. All we need are the perimeters, and they have to be scaled to produce all triples in the problem space.
<haskell>
<haskell>
-
problem_39 = head $ perims !! indexMax
+
problem_39 =
-
where perims = group
+
head $ perims !! indexMax
-
$ sort [n*p | p <- pTriples, n <- [1..1000 `div` p]]
+
where
-
counts = map length perims
+
perims = group $ sort [n*p | p <- pTriples, n <- [1..1000 `div` p]]
-
Just indexMax = findIndex (== (maximum counts)) $ counts
+
counts = map length perims
-
pTriples = [p |
+
Just indexMax = findIndex (== (maximum counts)) $ counts
-
n <- [1..floor (sqrt 1000)],
+
pTriples =
-
m <- [n+1..floor (sqrt 1000)],
+
[p |
-
even n || even m,
+
n <- [1..floor (sqrt 1000)],
-
gcd n m == 1,
+
m <- [n+1..floor (sqrt 1000)],
-
let a = m^2 - n^2,
+
even n || even m,
-
let b = 2*m*n,
+
gcd n m == 1,
-
let c = m^2 + n^2,
+
let a = m^2 - n^2,
-
let p = a + b + c,
+
let b = 2*m*n,
-
p < 1000]
+
let c = m^2 + n^2,
 +
let p = a + b + c,
 +
p < 1000
 +
]
</haskell>
</haskell>

Revision as of 00:44, 17 January 2008

Contents

1 Problem 31

Investigating combinations of English currency denominations.

Solution:

This is the naive doubly recursive solution. Speed would be greatly improved by use of memoization, dynamic programming, or the closed form. 

problem_31 = 
    pence 200 [1,2,5,10,20,50,100,200]
    where 
    pence 0 _  = 1
    pence n [] = 0
    pence n denominations@(d:ds)
        | n < d     = 0
        | otherwise = pence (n - d) denominations + pence n ds

A beautiful solution, making usage of laziness and recursion to implement a dynamic programming scheme, blazingly fast despite actually generating the combinations and not only counting them : 

coins = [1,2,5,10,20,50,100,200]
 
combinations = foldl (\without p ->
                          let (poor,rich) = splitAt p without
                              with = poor ++ 
                                     zipWith (++) (map (map (p:)) with)
                                                  rich
                          in with
                     ) ([[]] : repeat [])
 
problem_31 = 
    length $ combinations coins !! 200

2 Problem 32

Find the sum of all numbers that can be written as pandigital products.

Solution: 

problem_32 = 
    sum $ nub $  map (\(a, b) -> a * b) multiplicands
    where
    multiplicands =
        [(a,b)| a <- [2..5000], b <- [a..(9999 `div` a)], check a b]
    check a b =
        no_zero s
        && (length ss) == 9
        && foldr (\x y -> length x == 1 && y) True ss
        where
            s = show a ++ show b ++ show (a*b)
            ss = group $ sort s
    no_zero (x:xs)
        | x == '0'  = False
        | null xs   = True
        | otherwise = no_zero xs

3 Problem 33

Discover all the fractions with an unorthodox cancelling method.

Solution: 

import Ratio
 
problem_33 = denominator (product $ rs ++ rs')
 
rs = [(x%y) | 
    a <- [0..9], 
    b <- [1..9], 
    c <- [1..9], 
    let x = 10*a + c, 
    let y = 10*c + b, 
    x /= y, 
    x%y < 1, 
    x%y == a%b
    ]
 
rs' = 
    filter (<1) $ map (\x -> denominator x % numerator x) rs

4 Problem 34

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

Solution: 

import Data.Char
problem_34 = 
    sum [ x | x <- [3..100000], x == facsum x ]
    where
    facsum = sum . map (product . enumFromTo 1 . digitToInt) . show

5 Problem 35

How many circular primes are there below one million?

Solution: 

import Data.List (tails, (\\))
 
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
isPrime 
    |1 = False
    |otherwise=
        case (primeFactors n) of
            (_:_:_)   -> False
            _         -> True
permutations n = 
    take l $ map (read . take l) $ 
    tails $ take (2*l -1) $ cycle s
    where
    s = show n
    l = length s
circular_primes []     = []
circular_primes (x:xs)
    | all isPrime p = x :  circular_primes xs
    | otherwise     = circular_primes xs
    where
    p = permutations x
 
problem_35 = 
    length $ circular_primes $ takeWhile (<1000000) primes

6 Problem 36

Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2.

Solution: 

import Numeric
import Data.Char
 
showBin = 
    flip (showIntAtBase 2 intToDigit) ""
 
isPalindrome x = 
    x == reverse x
 
problem_36 = 
    sum [x | 
    x <- [1,3..1000000], 
    isPalindrome (show x), 
    isPalindrome (showBin x)
    ]

7 Problem 37

Find the sum of all eleven primes that are both truncatable from left to right and right to left.

Solution: 

import Data.List (tails, inits, nub)
 
truncs n = 
    nub . map read $
    (take l . tail . tails) s ++ (take l . tail . inits) s
    where
    l = length s - 1
    s = show n
 
problem_37 = 
    sum $ take 11 [x |
    x <- dropWhile (<=9) primes,
    all isPrime (truncs x)
    ]

8 Problem 38

What is the largest 1 to 9 pandigital that can be formed by multiplying a fixed number by 1, 2, 3, ... ?

Solution: 

problem_38 = 
    maximum $ catMaybes [result | j <- [1..9999],
    let p2 = show j ++ show (2*j),
    let p3 = p2 ++ show (3*j),
    let p4 = p3 ++ show (4*j),
    let p5 = p4 ++ show (5*j),
    let result
        | isPan p2 = Just p2
        | isPan p3 = Just p3
        | isPan p4 = Just p4
        | isPan p5 = Just p5
        | otherwise = Nothing
    ]
    where 
    isPan s = sort s == "123456789"

Other solution: 

import Data.List
 
mult n i vs 
    | length (concat vs) >= 9 = concat vs
    | otherwise               = mult n (i+1) (vs ++ [show (n * i)])
 
problem_38 = 
    maximum $ map read $ filter
    ((['1'..'9'] ==) .sort) $
    [ mult n 1 [] | n <- [2..9999] ]

9 Problem 39

If p is the perimeter of a right angle triangle, {a, b, c}, which value, for p ≤ 1000, has the most solutions?

Solution: We use the well known formula to generate primitive Pythagorean triples. All we need are the perimeters, and they have to be scaled to produce all triples in the problem space. 

problem_39 = 
    head $ perims !! indexMax
    where
    perims = group $ sort [n*p | p <- pTriples, n <- [1..1000 `div` p]]
    counts = map length perims
    Just indexMax = findIndex (== (maximum counts)) $ counts
    pTriples = 
        [p |
        n <- [1..floor (sqrt 1000)],
        m <- [n+1..floor (sqrt 1000)],
        even n || even m,
        gcd n m == 1,
        let a = m^2 - n^2,
        let b = 2*m*n,
        let c = m^2 + n^2,
        let p = a + b + c,
        p < 1000
        ]

10 Problem 40

Finding the nth digit of the fractional part of the irrational number.

Solution: 

problem_40 =
    (d 1)*(d 10)*(d 100)*(d 1000)*(d 10000)*(d 100000)*(d 1000000)
    where 
    n = concat [show n | n <- [1..]]
    d j = Data.Char.digitToInt (n !! (j-1))
Retrieved from "http://www.haskell.org/haskellwiki/Euler_problems/31_to_40"