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Euler problems/151 to 160

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1 Problem 151

Paper sheets of standard sizes: an expected-value problem.

Solution:

problem_151 = undefined

2 Problem 152

Writing 1/2 as a sum of inverse squares

Note that if p is an odd prime, the sum of inverse squares of all terms divisible by p must have reduced denominator not divisible by p.

Solution:

import Data.Ratio
import Data.List
 
invSq n = 1 % (n * n)
sumInvSq = sum . map invSq
 
subsets (x:xs) = let s = subsets xs in s ++ map (x :) s
subsets _      = [[]]
 
primes = 2 : 3 : 7 : [p | p <- [11, 13..79],
                          all (\q -> p `mod` q /= 0) [3, 5, 7]]
 
-- All subsets whose sum of inverse squares,
-- when added to x, does not contain a factor of p
pfree s x p = [(y, t) | t <- subsets s, let y =  x + sumInvSq t,
                        denominator y `mod` p /= 0]
 
-- Verify that we need not consider terms divisible by 11, or by any
-- prime greater than 13. Nor need we consider any term divisible
-- by 25, 27, 32, or 49.
verify = all (\p -> null $ tail $ pfree [p, 2*p..85] 0 p) $
         11 : dropWhile (< 17) primes ++ [25, 27, 32, 49]
 
-- All pairs (x, s) where x is a rational number whose reduced
-- denominator is not divisible by any prime greater than 3;
-- and s is all sets of numbers up to 80 divisible
-- by a prime greater than 3, whose sum of inverse squares is x.
only23 = foldl f [(0, [[]])] [13, 7, 5]
  where
    f a p = collect $ [(y, u ++ v) | (x, s) <- a,
                                     (y, v) <- pfree (terms p) x p,
                                     u <- s]
    terms p = [n * p | n <- [1..80`div`p],
                       all (\q -> n `mod` q /= 0) $
                           11 : takeWhile (>= p) [13, 7, 5]
              ]
    collect = map (\z -> (fst $ head z, map snd z)) .
              groupBy fstEq . sortBy cmpFst
    fstEq  (x, _) (y, _) = x == y
    cmpFst (x, _) (y, _) = compare x y
 
-- All subsets (of an ordered set) whose sum of inverse squares is x
findInvSq x y = f x $ zip3 y (map invSq y) (map sumInvSq $ init $ tails y)
  where
    f 0 _        = [[]]
    f x ((n, r, s):ns)
     | r > x     = f x ns
     | s < x     = []
     | otherwise = map (n :) (f (x - r) ns) ++ f x ns
    f _ _        = []
 
-- All numbers up to 80 that are divisible only by the primes
-- 2 and 3 and are not divisible by 32 or 27.
all23 = [n | a <- [0..4], b <- [0..2], let n = 2^a * 3^b, n <= 80]
 
solutions = if verify
              then [sort $ u ++ v | (x, s) <- only23,
                                    u <- findInvSq (1%2 - x) all23,
                                    v <- s]
              else undefined
 
problem_152 = length solutions

3 Problem 153

Investigating Gaussian Integers

Solution:

problem_153 = undefined

4 Problem 154

Exploring Pascal's pyramid.

Solution:

problem_154 = undefined

5 Problem 155

Counting Capacitor Circuits.

Solution:

problem_155 = undefined

6 Problem 156

Counting Digits

Solution:

digits =reverse.digits' 
    where
    digits' n 
        |n<10=[n]
        |otherwise= y:digits' x 
        where
        (x,y)=divMod n 10
digitsToNum n=foldl dmm 0  n
    where
    dmm=(\x y->x*10+y)
countA :: Int -> Integer
countA 0 = 0
countA k = fromIntegral k * (10^(k-1))
 
countFun :: Integer -> Integer -> Integer
countFun _ 0 = 0
countFun d n = countL ds k
      where
        ds = digits n
        k = length ds - 1
        countL [a] _
            | a < d     = 0
            | otherwise = 1
        countL (a:tl) m
            | a < d     = a*countA m + countL tl (m-1)
            | a == d    = a*countA m + digitsToNum tl + 1 + countL tl (m-1)
            | otherwise = a*countA m + 10^m + countL tl (m-1)
 
fixedPoints :: Integer -> [Integer]
fixedPoints d
    = [a*10^10+b | a <- [0 .. d-1], b <- findFrom 0 (10^10-1)]
      where
        fun = countFun d
        good r = r == fun r
        findFrom lo hi
            | hi < lo   = []
            | good lo   = lgs ++ findFrom (last lgs + 2) hi
            | good hi   = findFrom lo (last hgs - 2) ++ reverse hgs
            | h1 < l1   = []
            | l1 == h1  = if good l1 then [l1] else []
            | m0 == m1  = findFrom l1 (head mgs - 2) ++ mgs
                             ++ findFrom (last mgs + 2) h1
            | m0 < m1   = findFrom l1 (m0-1) ++ findFrom (goUp h1 m1) h1
            | otherwise = findFrom l1 (goDown l1 m1) ++ findFrom (m0+1) h1
              where
                l1 = goUp hi lo
                h1 = goDown l1 hi
                goUp bd k
                    | k < k1 && k < bd  = goUp bd k1
                    | otherwise         = k
                      where
                        k1 = fun k
                goDown bd k
                    | k1 < k && bd < k  = goDown bd k1
                    | otherwise         = k
                      where
                        k1 = fun k
                m0 = (l1 + h1) `div` 2
                m1 = fun m0
                lgs = takeWhile good [lo .. hi]
                hgs = takeWhile good [hi,hi-1 .. lo]
                mgs = reverse (takeWhile good [m0,m0-1 .. l1])
                        ++ takeWhile good [m0+1 .. h1]
problem_156=sum[sum $fixedPoints a|a<-[1..9]]

7 Problem 157

Solving the diophantine equation 1/a+1/b= p/10n

Solution:

problem_157 = undefined

8 Problem 158

Exploring strings for which only one character comes lexicographically after its neighbour to the left.

Solution:

problem_158 = undefined

9 Problem 159

Digital root sums of factorisations.

Solution:

problem_159 = undefined

10 Problem 160

Factorial trailing digits

We use the following two facts:

Fact 1:
(2^(d + 4*5^(d-1)) - 2^d) `mod` 10^d == 0
Fact 2:
product [n | n <- [0..10^d], gcd n 10 == 1] `mod` 10^d == 1

We really only need these two facts for the special case of

d == 5
, and we can verify that directly by

evaluating the above two Haskell expressions.

More generally:

Fact 1 follows from the fact that the group of invertible elements

of the ring of integers modulo
5^d
has
4*5^(d-1)
elements.

Fact 2 follows from the fact that the group of invertible elements

of the ring of integers modulo
10^d
is isomorphic to the product

of a cyclic group of order 2 and another cyclic group.

Solution:

problem_160 = trailingFactorialDigits 5 (10^12)
 
trailingFactorialDigits d n = twos `times` odds
  where
    base = 10 ^ d
    x `times` y = (x * y) `mod` base
    multiply = foldl' times 1
    x `toPower` k = multiply $ genericReplicate n x
    e = facFactors 2 n - facFactors 5 n
    twos
     | e <= d    = 2 `toPower` e
     | otherwise = 2 `toPower` (d + (e - d) `mod` (4 * 5 ^ (d - 1)))
    odds = multiply [odd | a <- takeWhile (<= n) $ iterate (* 2) 1,
                           b <- takeWhile (<= n) $ iterate (* 5) a,
                           odd <- [3, 5 .. n `div` b `mod` base],
                           odd `mod` 5 /= 0]
 
-- The number of factors of the prime p in n!
facFactors p = sum . zipWith (*) (iterate (\x -> p * x + 1) 1) .
               tail . radix p
 
-- The digits of n in base b representation
radix p = map snd . takeWhile (/= (0, 0)) .
          iterate ((`divMod` p) . fst) . (`divMod` p)

it have another fast way to do this .

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)
 
productMod =foldl (mulMod (10^5)) 1
hFacial 0=1
hFacial a
    |gcd a 5==1=mod (a*hFacial(a-1)) (5^5)
    |otherwise=hFacial(a-1)
fastFacial a= hFacial $mod a 6250
numPrime x p=takeWhile(>0) [div x (p^a)|a<-[1..]]
p160 x=mulMod t5 a b
    where
    t5=10^5
    lst=numPrime x 5
    a=powMod t5 1563 $mod c 2500
    b=productMod  c6 
    c=sum lst
    c6=map fastFacial $x:lst
problem_160 = p160 (10^12)