Difference between revisions of "Euler problems/31 to 40"
Line 74:  Line 74:  
primeFactors :: Integer > [Integer] 
primeFactors :: Integer > [Integer] 

primeFactors n = factor n primes 
primeFactors n = factor n primes 

−  where factor n (p:ps)  p*p > n = [n] 

+  where 

−   n `mod` p == 0 = p : factor (n `div` p) (p:ps) 

⚫  
−  +  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 :: Integer > Bool 

Line 97:  Line 97:  
where 
where 

p = permutations x 
p = permutations x 

⚫  
problem_35 :: Int 
problem_35 :: Int 
Revision as of 06:57, 7 August 2007
Contents
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
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
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
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
Problem 35
How many circular primes are there below one million?
Solution:
import Data.List (tails, (\\))
primes :: [Integer]
primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
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
otherwise > True
permutations :: Integer > [Integer]
permutations n = take l $ map (read . take l) $ tails $ take (2*l 1) $ cycle s
where
s = show n
l = length s
circular_primes :: [Integer] > [Integer]
circular_primes [] = []
circular_primes (x:xs)
 all isPrime p = x : circular_primes xs
 otherwise = circular_primes xs
where
p = permutations x
problem_35 :: Int
problem_35 = length $ circular_primes $ takeWhile (<1000000) primes
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)]
Problem 37
Find the sum of all eleven primes that are both truncatable from left to right and right to left.
Solution:
problem_37 = undefined
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 :: Int
problem_38 = maximum $ map read $ filter ((['1'..'9'] ==) .sort) $ [ mult n 1 []  n < [2..9999] ]
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]
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 !! (j1))