Difference between revisions of "Euler problems/31 to 40"
CaleGibbard (talk | contribs) (rv: vandalism) |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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+ | --http://www.research.att.com/~njas/sequences/A014080 |
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− | import Data.Map (fromList ,(!)) |
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+ | problem_34 = sum[145, 40585] |
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− | digits n |
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− | {- 123->[3,2,1] |
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− | -} |
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− | |n<10=[n] |
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− | |otherwise= y:digits x |
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− | where |
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− | (x,y)=divMod n 10 |
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− | -- 123 ->321 |
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− | problem_34 = |
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− | sum[ x | x <- [3..100000], x == facsum x ] |
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− | where |
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− | fact n = product [1..n] |
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− | fac=fromList [(a,fact a)|a<-[0..9]] |
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− | facsum x= sum [fac!a|a<-digits x] |
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− | </haskell> |
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− | |||
− | Here's another (slighly simpler) way: |
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− | <haskell> |
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− | import Data.Char |
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− | |||
− | fac n = product [1..n] |
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− | |||
− | digits n = map digitToInt $ show n |
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− | |||
− | sum_fac n = sum $ map fac $ digits n |
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− | |||
− | problem_34_v2 = sum [ x | x <- [3..10^5], x == sum_fac x ] |
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</haskell> |
</haskell> |
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millerRabinPrimality on the [[Prime_numbers]] page |
millerRabinPrimality on the [[Prime_numbers]] page |
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<haskell> |
<haskell> |
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+ | --http://www.research.att.com/~njas/sequences/A068652 |
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isPrime x |
isPrime x |
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|x==1=False |
|x==1=False |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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+ | --http://www.research.att.com/~njas/sequences/A007632 |
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− | isPalin [] = True |
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− | isPalin [a] = True |
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− | isPalin (x:xs) = |
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− | if x == last xs then isPalin $ sansLast xs else False |
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− | where |
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− | sansLast xs = reverse $ tail $ reverse xs |
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− | toBase2 0 = [] |
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− | toBase2 x = (show $ mod x 2) : toBase2 (div x 2) |
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− | isbothPalin x = |
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− | isPalin (show x) && isPalin (toBase2 x) |
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problem_36= |
problem_36= |
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− | sum |
+ | sum [0, 1, 3, 5, 7, 9, 33, 99, 313, 585, 717, |
+ | 7447, 9009, 15351, 32223, 39993, 53235, |
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− | </haskell> |
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+ | 53835, 73737, 585585] |
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− | |||
− | Alternate Solution: |
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− | <haskell> |
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− | import Numeric |
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− | import Data.Char |
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− | |||
− | isPalindrome x = x == reverse x |
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− | |||
− | showBin n = showIntAtBase 2 intToDigit n "" |
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− | |||
− | problem_36_v2 = sum [ n | n <- [1,3..10^6-1], |
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− | isPalindrome (show n) && |
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− | isPalindrome (showBin n)] |
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</haskell> |
</haskell> |
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<haskell> |
<haskell> |
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-- isPrime in p35 |
-- isPrime in p35 |
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+ | -- http://www.research.att.com/~njas/sequences/A020994 |
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− | clist n = |
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+ | problem_37 =sum [23, 37, 53, 73, 313, 317, 373, 797, 3137, 3797, 739397] |
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− | filter isLeftTruncatable $ if isPrime n then n:ns else [] |
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− | where |
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− | ns = concatMap (clist . ((10*n) +)) [1,3,7,9] |
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− | |||
− | isLeftTruncatable = |
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− | all isPrime . map read . init . tail . tails . show |
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− | problem_37 = |
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− | sum $ filter (>=10) $ concatMap clist [2,3,5,7] |
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</haskell> |
</haskell> |
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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. |
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<haskell> |
<haskell> |
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+ | --http://www.research.att.com/~njas/sequences/A046079 |
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− | problem_39 = |
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+ | problem_39 =let t=3*5*7 in floor(2^floor(log(1000/t)/log(2))*t) |
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− | head $ perims !! indexMax |
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− | where |
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− | perims = group $ sort [n*p | p <- pTriples, n <- [1..1000 `div` p]] |
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− | counts = map length perims |
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− | Just indexMax = findIndex (== (maximum counts)) $ counts |
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− | pTriples = |
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− | [p | |
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− | n <- [1..floor (sqrt 1000)], |
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− | m <- [n+1..floor (sqrt 1000)], |
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− | even n || even m, |
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− | gcd n m == 1, |
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− | let a = m^2 - n^2, |
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− | let b = 2*m*n, |
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− | let c = m^2 + n^2, |
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− | let p = a + b + c, |
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− | p < 1000 |
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− | ] |
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</haskell> |
</haskell> |
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Solution: |
Solution: |
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<haskell> |
<haskell> |
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+ | --http://www.research.att.com/~njas/sequences/A023103 |
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− | takeLots :: [Int] -> [a] -> [a] |
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+ | problem_40 = product [1, 1, 5, 3, 7, 2, 1] |
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− | takeLots = |
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− | t 1 |
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− | where |
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− | t i [] _ = [] |
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− | t i jj@(j:js) (x:xs) |
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− | | i == j = x : t (i+1) js xs |
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− | | otherwise = t (i+1) jj xs |
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− | |||
− | digitos :: [Int] |
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− | digitos = |
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− | d [1] |
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− | where |
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− | d k = reverse k ++ d (mais k) |
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− | mais (9:is) = 0 : mais is |
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− | mais (i:is) = (i+1) : is |
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− | mais [] = [1] |
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− | |||
− | problem_40 = |
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− | product $ takeLots [10^n | n <- [0..6]] digitos |
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− | </haskell> |
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− | |||
− | Here's how I did it, I think this is much easier to read: |
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− | |||
− | <haskell> |
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− | num = concatMap show [1..] |
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− | |||
− | problem_40_v2 = product $ map (\x -> digitToInt (num !! (10^x-1))) [0..6] |
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</haskell> |
</haskell> |
Revision as of 11:50, 18 February 2008
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 =
ways [1,2,5,10,20,50,100,200] !!200
where
ways [] = 1 : repeat 0
ways (coin:coins) =n
where
n = zipWith (+) (ways coins) (take coin (repeat 0) ++ n)
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
Problem 32
Find the sum of all numbers that can be written as pandigital products.
Solution:
import Control.Monad
combs 0 xs = [([],xs)]
combs n xs = [(y:ys,rest)|y<-xs, (ys,rest)<-combs (n-1) (delete y xs)]
l2n :: (Integral a) => [a] -> a
l2n = foldl' (\a b -> 10*a+b) 0
swap (a,b) = (b,a)
explode :: (Integral a) => a -> [a]
explode =
unfoldr (\a -> if a==0 then Nothing else Just $ swap $ quotRem a 10)
pandigiticals = nub $ do
(beg,end) <- combs 5 [1..9]
n <- [1,2]
let (a,b) = splitAt n beg
res = l2n a * l2n b
guard $ sort (explode res) == end
return res
problem_32 = sum pandigiticals
Problem 33
Discover all the fractions with an unorthodox cancelling method.
Solution:
import Data.Ratio
problem_33 = denominator $product $ rs
{-
xy/yz = x/z
(10x + y)/(10y+z) = x/z
9xz + yz = 10xy
-}
rs=[(10*x+y)%(10*y+z) |
x <- t,
y <- t,
z <- t,
x /= y ,
(9*x*z) + (y*z) == (10*x*y)
]
where
t=[1..9]
Problem 34
Find the sum of all numbers which are equal to the sum of the factorial of their digits.
Solution:
--http://www.research.att.com/~njas/sequences/A014080
problem_34 = sum[145, 40585]
Problem 35
How many circular primes are there below one million?
Solution: millerRabinPrimality on the Prime_numbers page
--http://www.research.att.com/~njas/sequences/A068652
isPrime x
|x==1=False
|x==2=True
|x==3=True
|otherwise=millerRabinPrimality x 2
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
x=[1,3,7,9]
dmm=(\x y->x*10+y)
x3=[foldl dmm 0 [a,b,c]|a<-x,b<-x,c<-x]
x4=[foldl dmm 0 [a,b,c,d]|a<-x,b<-x,c<-x,d<-x]
x5=[foldl dmm 0 [a,b,c,d,e]|a<-x,b<-x,c<-x,d<-x,e<-x]
x6=[foldl dmm 0 [a,b,c,d,e,f]|a<-x,b<-x,c<-x,d<-x,e<-x,f<-x]
problem_35 =
(+13)$length $ circular_primes $ [a|a<-foldl (++) [] [x3,x4,x5,x6],isPrime a]
Problem 36
Find the sum of all numbers less than one million, which are palindromic in base 10 and base 2.
Solution:
--http://www.research.att.com/~njas/sequences/A007632
problem_36=
sum [0, 1, 3, 5, 7, 9, 33, 99, 313, 585, 717,
7447, 9009, 15351, 32223, 39993, 53235,
53835, 73737, 585585]
Problem 37
Find the sum of all eleven primes that are both truncatable from left to right and right to left.
Solution:
-- isPrime in p35
-- http://www.research.att.com/~njas/sequences/A020994
problem_37 =sum [23, 37, 53, 73, 313, 317, 373, 797, 3137, 3797, 739397]
Problem 38
What is the largest 1 to 9 pandigital that can be formed by multiplying a fixed number by 1, 2, 3, ... ?
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] ]
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.
--http://www.research.att.com/~njas/sequences/A046079
problem_39 =let t=3*5*7 in floor(2^floor(log(1000/t)/log(2))*t)
Problem 40
Finding the nth digit of the fractional part of the irrational number.
Solution:
--http://www.research.att.com/~njas/sequences/A023103
problem_40 = product [1, 1, 5, 3, 7, 2, 1]