Difference between revisions of "Euler problems/121 to 130"

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Solution:
 
Solution:
 
<haskell>
 
<haskell>
  +
import Data.List
problem_121 = undefined
 
  +
problem_121 = possibleGames `div` winningGames
 
where
  +
possibleGames = product [1..16]
  +
winningGames =
  +
(1+) $ sum $ map product $ chooseUpTo 7 [1..15]
  +
chooseUpTo 0 _ = []
  +
chooseUpTo (n+1) x =
  +
[y:z |
  +
(y:ys) <- tails x,
  +
z <- []: chooseUpTo n ys
 
]
 
</haskell>
 
</haskell>
   
Line 154: Line 165:
 
<haskell>
 
<haskell>
 
import Data.List
 
import Data.List
import Data.Map(fromList,(!))
+
import Data.Array.IArray
  +
import Data.Array.Unboxed
rad x= product[a|(a,_)<-fstfac x]
 
  +
radMap=fromList[(a,fromList [(b,rad(a*1000+b))|b<-[0..999]])|a<-[0..100]]
 
 
main = appendFile "p127.log" $show$ solve 99999
fastrad x=
 
 
radMap!a!b
 
  +
rad x = fromIntegral $ product $ map fst $ primePowerFactors $ fromIntegral x
 
primePowerFactors x = [(head a ,length a)|a<-group$primeFactors x]
  +
solve :: Int -> Int
  +
solve n = sum [ c | (rc,c) <- invrads
 
, 2 * rc < c
  +
, (ra, a) <- takeWhile (\(a,_)->(c > 2*rc*a)) invrads
  +
, a < c `div` 2
  +
, gcd ra rc == 1
  +
, ra * rads ! (c - a) < c `div` rc]
 
where
 
where
  +
rads :: UArray Int Int
(a,b)=divMod x 1000
 
  +
rads = listArray (1,n) $ map rad [1..n]
maxSingFac x
 
  +
invrads = sort $ map (\(a,b) -> (b, a)) $ assocs rads
|not$null ar=last ar
 
|otherwise=0
 
where
 
ar=[a|(a,b)<-x,b==1]
 
testPrime x p divx= [swap(x,a,b)|
 
a<-[p2*a1|a1<-[1..n]],
 
gcd a x==1,
 
let b=x-a,
 
gcd b x==1,
 
gcd b a==1,
 
fastrad a*fastrad b<divx
 
]
 
where
 
p2=p^2
 
(n,m)=divMod x p2
 
swap (a,b,c)
 
|b<c=[a,b]
 
|otherwise=[a,c]
 
test1 x divx
 
|fastrad (x-1)<divx=[[x,1]]
 
|otherwise=[]
 
test x
 
|(length ff>4)=[]
 
|(maxSingFac ff>1200)=[]
 
|otherwise= nub d
 
where
 
ra=fastrad x
 
ff=fstfac x
 
divx=div x ra
 
ba=div divx 2
 
c1=takeWhile (<=ba) primes
 
d=test1 x divx++[b|p<-c1,b<-testPrime x p divx]
 
 
groups=1000
 
p127 k=[m|a<-[1+k*groups..groups*(k+1)],let t=test a,not$null t,m<-t]
 
show1 x=foldl (++) "" $map ((++" \n").show2) x
 
show2 [a,b]=show a++" "++show b
 
google num
 
-- write file to change bignum to small num
 
=if (num>99)
 
then return()
 
else do appendFile "p127.log" $(show1$p127 num)
 
google (num+1)
 
main=google 0
 
 
fstfac x = [(head a ,length a)|a<-group$primeFactors x]
 
 
primeFactors :: Integer -> [Integer]
 
primeFactors :: Integer -> [Integer]
 
primeFactors n = factor n primes
 
primeFactors n = factor n primes
Line 230: Line 205:
 
where f (x:xt) ys = x : (merge xt ys)
 
where f (x:xt) ys = x : (merge xt ys)
 
g p = [ n*p | n <- [p,p+2..]]
 
g p = [ n*p | n <- [p,p+2..]]
 
problem_127 = main
split :: Char -> String -> [String]
 
split = unfoldr . split'
 
 
split' :: Char -> String -> Maybe (String, String)
 
split' c l
 
| null l = Nothing
 
| otherwise = Just (h, drop 1 t)
 
where (h, t) = span (/=c) l
 
 
sToInt x=((+0).read) $head$split ' ' x
 
 
problem_127=do
 
x<-readFile "p127.log"
 
let y=sum$map sToInt $lines x
 
print (y-1-100000)
 
 
</haskell>
 
</haskell>
   

Revision as of 02:38, 14 January 2008

Problem 121

Investigate the game of chance involving coloured discs.

Solution:

import Data.List
problem_121 = possibleGames `div` winningGames
   where
   possibleGames = product [1..16]
   winningGames = 
       (1+) $ sum $ map product $ chooseUpTo 7 [1..15]
   chooseUpTo 0     _ = []
   chooseUpTo (n+1) x = 
       [y:z | 
       (y:ys) <- tails x,
       z <- []: chooseUpTo n ys
       ]

Problem 122

Finding the most efficient exponentiation method.

Solution using a depth first search, pretty fast :

import Data.List
import Data.Array.Diff
import Control.Monad

depthAddChain 12 branch mins = mins
depthAddChain d branch mins = foldl' step mins $ nub $ filter (> head branch)
                               $ liftM2 (+) branch branch
    where
      step da e | e > 200 = da
                | otherwise =
                    case compare (da ! e) d of
                      GT -> depthAddChain (d+1) (e:branch) $ da // [(e,d)]
                      EQ -> depthAddChain (d+1) (e:branch) da
                      LT -> da

baseBranch = [2,1]

baseMins :: DiffUArray Int Int
baseMins = listArray (1,200) $ 0:1: repeat maxBound

problem_122 = sum . elems $ depthAddChain 2 baseBranch baseMins

Problem 123

Determining the remainder when (pn − 1)n + (pn + 1)n is divided by pn2.

Solution:

primes = 2 : filter ((==1) . length . primeFactors) [3,5..]
 
primeFactors n = factor n primes
    where
        factor _ [] = []
        factor m (p:ps) | p*p > m        = [m]
                        | m `mod` p == 0 = p : [m `div` p]
                        | otherwise      = factor m ps
 
isPrime :: Integer -> Bool
isPrime 1 = False
isPrime n = case (primeFactors n) of
                (_:_:_)   -> False
                _         -> True
 
problem_123 = 
    head[a+1|a<-[20000,20002..22000],
    let n=2*(a+1)*primes!!(fromInteger a),
    n>10^10
    ]

Problem 124

Determining the kth element of the sorted radical function.

Solution:

import List
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
problem_124=snd$(!!9999)$sort[(product$nub$primeFactors x,x)|x<-[1..100000]]

Problem 125

Finding square sums that are palindromic.

Solution:

import Data.List 
import Data.Map(fromList,(!))

toFloat = (flip encodeFloat 0) 
digits n 
{-  123->[3,2,1]
 -}
    |n<10=[n]
    |otherwise= y:digits x 
    where
    (x,y)=divMod n 10

palind n=foldl dmm 0 (digits n) 
-- 123 ->321
dmm=(\x y->x*10+y)

makepalind n=(n*d+p):[c+b*d|b<-[0..9]]
    where
    a=(+1)$floor$logBase 10$fromInteger n
    d=10^a
    p=palind n
    c=n*10*d+p

twomakep n=(n*d+p)
    where
    a=(+1)$floor$logBase 10$fromInteger n
    d=10^a
    p=palind n

p125=sum[b|a<-[1..999], b<-makepalind a,not$null$ funa b]
p125a=sum[b|a<-[1000..9999], let b=twomakep a,not$null$ funa b]
p125b=sum[a|a<-[1..9], not$null$ funa a]

findmap=fromList[(a,2*fill_map a)|a<-[0..737]]
fill_map x
    |odd x=fastsum $div (x-1) 2
    |otherwise=fastsumodd (x-1)
    where
    fastsum  y=div (y*(y+1)*(2*y+1)) 6
    fastsumodd  y=let n=div (y+1) 2 in div (n*(4*n*n-1)) 3

funa x=[(a,x)|a<-takeWhile (\a->a*a*a<4*x) [2..],funb a x]
funb x n
    |odd x=d2==0 && 4*d1>=(x+1)^2 && isSq d1
    |otherwise=d4==0 && odd d3 && d3>=(x+1)^2 && isSq d3
    where
    x1=fromInteger x
    (d1,d2)=divMod ((n-findmap! x1)) (x)
    (d3,d4)=divMod ((4*n-findmap!x1)) (x)
isSq x=(floor$sqrt$toFloat x)^2==x
problem_125 = (p125+p125a+p125b)

Problem 126

Exploring the number of cubes required to cover every visible face on a cuboid.

Solution:

problem_126 = undefined

Problem 127

Investigating the number of abc-hits below a given limit.

Solution:

import Data.List
import Data.Array.IArray
import Data.Array.Unboxed

main = appendFile "p127.log" $show$ solve 99999
 
rad x = fromIntegral $ product $ map fst $ primePowerFactors $ fromIntegral x
primePowerFactors x = [(head a ,length a)|a<-group$primeFactors x] 
solve :: Int -> Int
solve n = sum [ c | (rc,c) <- invrads
                  , 2 * rc < c
                  , (ra, a) <- takeWhile (\(a,_)->(c > 2*rc*a)) invrads
                  , a < c `div` 2
                  , gcd ra rc == 1
                  , ra * rads ! (c - a) < c `div` rc]
    where
     rads :: UArray Int Int
     rads = listArray (1,n) $ map rad [1..n]
     invrads = sort $ map (\(a,b) -> (b, a)) $ assocs rads
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
merge xs@(x:xt) ys@(y:yt) = case compare x y of
    LT -> x : (merge xt ys)
    EQ -> x : (merge xt yt)
    GT -> y : (merge xs yt)
    
diff  xs@(x:xt) ys@(y:yt) = case compare x y of
    LT -> x : (diff xt ys)
    EQ -> diff xt yt
    GT -> diff xs yt
 
primes, nonprimes :: [Integer]
primes    = [2,3,5] ++ (diff [7,9..] nonprimes) 
nonprimes = foldr1 f . map g $ tail primes
    where f (x:xt) ys = x : (merge xt ys)
          g p = [ n*p | n <- [p,p+2..]]
problem_127 = main

Problem 128

Which tiles in the hexagonal arrangement have prime differences with neighbours?

Solution:

problem_128 = undefined

Problem 129

Investigating minimal repunits that divide by n.

Solution:

problem_129 = undefined

Problem 130

Finding composite values, n, for which n−1 is divisible by the length of the smallest repunits that divide it.

Solution:

import Data.List
mulMod :: Integral a => a -> a -> a -> a
mulMod a b c= (b * c) `rem` a
squareMod :: Integral a => a -> a -> a
squareMod a b = (b * b) `rem` a
pow' :: (Num a, Integral b) => (a -> a -> a) -> (a -> a) -> a -> b -> a
pow' _ _ _ 0 = 1
pow' mul sq x' n' = f x' n' 1
    where
    f x n y
        | n == 1 = x `mul` y
        | r == 0 = f x2 q y
        | otherwise = f x2 q (x `mul` y)
        where
            (q,r) = quotRem n 2
            x2 = sq x
powMod :: Integral a => a -> a -> a -> a
powMod m = pow' (mulMod m) (squareMod m)
 
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
 
primes=2:[a|a<-[2..],isPrime a]
isPrime :: Integer -> Bool
isPrime 1 = False
isPrime n = case (primeFactors n) of
                (_:_:_)   -> False
                _         -> True
fstfac x = [(head a ,length a)|a<-group$primeFactors x]
fac [(x,y)]=[x^a|a<-[0..y]]
fac (x:xs)=[a*b|a<-fac [x],b<-fac xs]
factors x=fac$fstfac x
fun x |(not$null a)=head a 
    |otherwise=0
    where 
    a=take 1 [n|n<-sort$factors (x-1),(powMod x 10 n)==1]
problem_130 =sum$take 25[a|a<-[1..],
    not$isPrime a,
    let b=fun a, 
    b/=0,
    mod (a-1) b==0,
    mod a 3 /=0]