Euler problems/131 to 140

From HaskellWiki
< Euler problems
Revision as of 02:19, 20 December 2007 by Lisp (talk | contribs)
Jump to navigation Jump to search
The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

Problem 131

Determining primes, p, for which n3 + n2p is a perfect cube.

Solution:

primes=sieve [2..]
sieve (x:xs)=x:sieve [y|y<-xs,mod y x>0]
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 n = case (primeFactors n) of
                (_:_:_)   -> False
                _         -> True
problem_131 =
    length $ takeWhile (<1000000) 
    [x|
    a<-[1 .. ],
    let x=(3*a*(a+1)+1),
    isPrime x]

Problem 132

Determining the first forty prime factors of a very large repunit.

Solution:

-- primes powMod in problem_133
fun x = (powMod x 10 n)==1
    where
    n=10^9
--add 3
p132 =sum$take 41 [a|a<-primes,fun a]
problem_132 =p132-3

Problem 133

Investigating which primes will never divide a repunit containing 10n digits.

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)

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..]]
fact25 m  | m `mod` 2 == 0 = 2 : fact25 (m `div` 2) 
          | m `mod` 5 == 0 = 5 : fact25 (m `div` 5)
          | otherwise      = []
fun x
    |n==x-1=True
    |otherwise= (powMod x 10 n)==1
    where
    n=product$fact25 (x-1)
--miss 2 3 5
test =sum$takeWhile (<100)[a|a<-primes,not$fun a]
p133 =sum$takeWhile (<100000)[a|a<-primes,not$fun a]
problem_133 = p133+2+3+5

Problem 134

Finding the smallest positive integer related to any pair of consecutive primes.

Solution:

import List

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..]]

dign x=(+1)$floor$logBase 10$fromInteger x  

euler x1 x2 x3 y1 y2 1=(x3,y2) 
euler x1 x2 x3 y1 y2 y3 
    =euler y1 y2 y3 t1 t2 t3
    where
    (k,t3)=divMod x3 y3
    t1=x1-k*y1
    t2=x2-k*y2
-- find a ,b
-- mod (x*a) y = 1
-- mod (y*b) x = 1
congrue x y |x>y=euler 1 0 x 0 1 y
    |otherwise =(a,b)
    where
    (b,a)=congrue y x
--fastfun 7=1219
fastfun x
    |x==1=0
    |p1>1000000=0
    |otherwise= a*d+p1 
    where
    p1=primes!!x
    p2=primes!!(x+1)
    dp=p2-p1
    d=10^dign p1
    dmod=mod d p2
    eu=(+p2)$fst$congrue dmod p2
    a=mod (eu*dp) p2

groups=1000

funsum k=sum[fastfun a|a<-[1+k*groups..groups*(k+1)]]

google num
-- write file to change bignum to small num
  =if (num>79)
      then return()
      else do appendFile "file.log" $(show$funsum num)  ++"  "++(show num) ++"\n"
              google (num+1)
-- first use main to make file.log
-- then run problem_134
main=google 0


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_134=do
    x<-readFile "file.log"
    let y=sum$map sToInt $lines x
    print  y

Problem 135

Determining the number of solutions of the equation x2 − y2 − z2 = n.

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
 
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
fastfun x
    |mod x 4==3=[a|a<-factors x,a*a<3*x]
    |mod x 16==4=[a|let n=div x 4,a<-factors n,a*a<3*n]
    |mod x 16==12=[a|let n=div x 4,a<-factors n,a*a<3*n]
    |mod x 16==0=[a|let n=div x 16,a<-factors n,a*a<3*n]
    |otherwise=[]

slowfun x =[a|a<-factors x,a*a<3*x,let b=div x a,mod (a+b) 4==0]

problem_135 =[a|a<-[1..groups],(length$fastfun a)==10]

Problem 136

Discover when the equation x2 − y2 − z2 = n has a unique solution.

Solution:

-- fastfun in the problem 135
groups=1000000
pfast=[a|a<-[1..5000],(length$fastfun a)==1]
pslow=[a|a<-[1..5000],(length$slowfun a)==1]
-- find len pfast=len pslow+2
-- so sum file.log and +2
problem_136 b=[a|a<-[1+b*groups..groups*(b+1)],(length$fastfun a)==1]
google num
-- write file to change bignum to small num
  =if (num>49)
      then return()
      else do appendFile "file.log" ((show$length$problem_136 num)  ++ "\n")
              google (num+1)
main=google 0

Problem 137

Determining the value of infinite polynomial series for which the coefficients are Fibonacci numbers.

Solution:

-- afx=x/(1-x-x^2)=n
--   ->5*n^2+2*n+1=d^2
--   ->let k=10*n+2
--   ->20*d^2=k^2+16
--   ->5*d^2=k^2+4
--   ->let d k is even
--   ->5*d^2=k^2+1
--   ->let d k is odd
--   ->5*d^2=k^2+4

import Data.List
findmin d = d:head [[n,m]|m<-[1..10],n<-[1..10],n*n==d*m*m+1]
findmin_s d = d:head [[n,m]|m<-[1..10],n<-[1..10],n*n==d*m*m-1]
findmu d y= d:head [[n,m]|m<-[1..10],n<-[1..10],n*n==d-y*m]
mux2 [d,a, b]=[d,a,-b]
mult [d,a, b] [_,a1, b1]=d:[a*a1+d*b*b1,a*b1+b*a1]
pow 1 x=x
pow n x =mult x $pow (n-1) x 
    where
    mult [d,a, b] [_,a1, b1]=d:[a*a1+d*b*b1,a*b1+b*a1]
fun =[c|a<-[1..20],[_,b,_]<-powmu a,let bb=abs(b),mod bb 5==1,let c=div bb 5]
powmu n =
    [a,b]
    where
    c=pow n $findmin 5
    x1=findmu 5 4
    x2=mux2 x1
    a=mult c x1 
    b=mult c x2
fun2=[c|a<-[1..20],let[_,b,_]=pow a $findmin_s 5,let bb=b*2,mod bb 5==1,let c=div bb 5]
problem_137 =(!!14)$sort $(++fun)$fun2

Problem 138

Investigating isosceles triangle for which the height and base length differ by one.

Solution:

{-
 - 4*m^2-16*m*n-4*n^2+1=0
 - 4*m^2-16*m*n-4*n^2-1=0
 - (m-2*n)^2-5*n^2=1
 - (m-2*n)^2-5*n^2=-1
 -}
import Data.List
mult [d,a, b] [_,a1, b1]=d:[a*a1+d*b*b1,a*b1+b*a1]
pow 1 x=x
pow n x =mult x $pow (n-1) x 
    where
    mult [d,a, b] [_,a1, b1]=d:[a*a1+d*b*b1,a*b1+b*a1]
-- 2^2-5*1^2=-1
-- so [5,2,1]
fun =[d^2+c^2|a<-[1..20],let [_,b,c]=pow a [5,2,1],let d=2*c+b]
-- 9^2-5*4^2=1
-- so [5,9,4]
fun2 =[d^2+c^2|a<-[1..20],let [_,b,c]=pow a [5,9,4],let d=2*c+b]
problem_138 =sum$take 12 $nub$sort (fun++fun2)

Problem 139

Finding Pythagorean triangles which allow the square on the hypotenuse square to be tiled.

Solution:

{-
 -                              2                        2
 -                     (n - 1) y  - 2 n x y + (- n - 1) x  = 0
 --->
 -                                           2       2       2
 -                          ((n - 1) y - n x)  = (2 n  - 1) x
 --->
 -                       2        2
 -                    2 n  - 1 = k
 -
 -}
import Data.List
mult [d,a, b] [_,a1, b1]=d:[a*a1+d*b*b1,a*b1+b*a1]
pow 1 x=x
pow n x =mult x $pow (n-1) x 
    where
    mult [d,a, b] [_,a1, b1]=d:[a*a1+d*b*b1,a*b1+b*a1]
div2 [x,y,z]
    |mod x 2==0 && mod y 2==0 && mod z 2==0=
        [div x 2,div y 2,div z 2]
    |otherwise=[x,y,z]
-- 1^2-2*1^2=-1
-- so [2,1,1]
fun =map div2 [
    side
    |a<-[3,5..40],
    let [_,k,n]=pow a [2,1,1],
    let m=lcm (n+k) (n-1),
    let x=div m (n+k),
    let y=div m (n-1),
    let side=[y^2-x^2, 2*x*y, y^2+x^2]
    ]
limit=100000000
problem_139=sum [div limit$sum a|a<-fun,  sum a<limit]

Problem 140

Investigating the value of infinite polynomial series for which the coefficients are a linear second order recurrence relation.

Solution:

{-
                                      2
                                   3 x  + x
                            agx= ------------
                                    2
                                 - x  - x + 1
--->
                                   2             2
                                5 n  + 14 n + 1=d

--->                          k = 10 n + 14
--->                          20*d^2=k^2-176
--->                          k = 5 n + 2
--->                          5*d^2=k^2-44
-}
import Data.List
findmin d = d:head [[n,m]|m<-[1..10],n<-[1..10],n*n==d*m*m+1]
findmin_s d = d:head [[n,m]|m<-[1..10],n<-[1..10],n*n==d*m*m+1]
findmu d y= d:head [[n,m]|m<-[1..10],n<-[1..10],n*n==d+y*m]
mux2 [d,a, b]=[d,a,-b]
mult [d,a, b] [_,a1, b1]=d:[a*a1+d*b*b1,a*b1+b*a1]
div2 [d,a, b] =d:[div a 2,div b 2]
pow 1 x=x
pow n x =mult x $pow (n-1) x 
    where
    mult [d,a, b] [_,a1, b1]=d:[a*a1+d*b*b1,a*b1+b*a1]
fun =[c|a<-[1..20],[_,b,_]<-powmu a,let bb=abs(b),mod bb 5==2,let c=div bb 5]
fun2=[c|a<-[1..20],[_,b,_]<-powmu1 a ,let bb=(abs b)*2,mod bb 5==2,let c=div bb 5]
powmu n =
    [a,b,a1,a2,b1,b2]
    where
    c=pow n $findmin 5
    x1=findmu 5 44
    x2=mux2 x1
    a=mult c x1 
    b=mult c x2
    a1=div2$mult a [5,3, -1]
    a2=div2$mult a [5,3, 1]
    b1=div2$mult b [5,3, -1]
    b2=div2$mult b [5,3, 1]
powmu1 n =
    [a,b]
    where
    c=pow n $findmin_s 5
    x1=findmu 5 11
    x2=mux2 x1
    a=mult c x1 
    b=mult c x2
problem_140 =sum $take 30 [a-1|a<-nub$sort (fun++fun2)]