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Euler problems/111 to 120

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Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_118 = undefined
+
find2km :: Integral a => a -> (a,a)
  +
find2km n = f 0 n
  +
where
  +
f k m
  +
| r == 1 = (k,m)
  +
| otherwise = f (k+1) q
  +
where (q,r) = quotRem m 2
  +
  +
millerRabinPrimality :: Integer -> Integer -> Bool
  +
millerRabinPrimality n a
  +
| a <= 1 || a >= n-1 =
  +
error $ "millerRabinPrimality: a out of range ("
  +
++ show a ++ " for "++ show n ++ ")"
  +
| n < 2 = False
  +
| even n = False
  +
| b0 == 1 || b0 == n' = True
  +
| otherwise = iter (tail b)
  +
where
  +
n' = n-1
  +
(k,m) = find2km n'
  +
b0 = powMod n a m
  +
b = take (fromIntegral k) $ iterate (squareMod n) b0
  +
iter [] = False
  +
iter (x:xs)
  +
| x == 1 = False
  +
| x == n' = True
  +
| otherwise = iter xs
  +
  +
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
  +
  +
mulMod :: Integral a => a -> a -> a -> a
  +
mulMod a b c = (b * c) `mod` a
  +
squareMod :: Integral a => a -> a -> a
  +
squareMod a b = (b * b) `rem` a
  +
powMod :: Integral a => a -> a -> a -> a
  +
powMod m = pow' (mulMod m) (squareMod m)
  +
--isPrime x=millerRabinPrimality x 2
  +
isPrime x
  +
|x<100=isPrime' x
  +
|otherwise=foldl (&& )True [millerRabinPrimality x y|y<-[2,3,7,61]]
  +
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
  +
  +
getprimes ""= [[]]
  +
getprimes s1=
  +
[n:f|
  +
let len=length s1,
  +
a<-[1..len],
  +
let b=take a s1,
  +
let n=read b::Integer,
  +
isPrime n,
  +
let k=getprimes $drop a s1,
  +
f<-k,
  +
a==len|| n<head f
  +
]
  +
perms :: [a] -> [[a]]
  +
perms [] = [ [] ]
  +
perms (x:xs) =
  +
concat (map (between x) (perms xs))
  +
where
  +
between e [] = [ [e] ]
  +
between e (y:ys) = (e:y:ys) : map (y:) (between e ys)
  +
fun x=do
  +
let cs=length$getprimes x
  +
if (cs/=0) then
  +
appendFile "p118.log"$(++"\n")$show cs
  +
else
  +
return ()
  +
sToInt =(+0).read
  +
problem_118a=do
  +
s<-readFile "p118.log"
  +
print$sum$map sToInt$lines s
  +
main=do
  +
mapM_ fun $perms ['1'..'9']
  +
problem_118a
  +
problem_118 = main
 
</haskell>
 
</haskell>
   

Revision as of 14:37, 15 January 2008

Contents

1 Problem 111

Search for 10-digit primes containing the maximum number of repeated digits.

Solution:

import Control.Monad (replicateM)
 
-- All ways of interspersing n copies of x into a list
intr :: Int -> a -> [a] -> [[a]]
intr 0 _ y      = [y]
intr n x (y:ys) = concat
                  [map ((replicate i x ++) . (y :)) $ intr (n-i) x ys
                       | i <- [0..n]]
intr n x _      = [replicate n x]
 
-- All 10-digit primes containing the maximal number of the digit d
maxDigits :: Char -> [Integer]
maxDigits d = head $ dropWhile null
              [filter isPrime $ map read $ filter ((/='0') . head) $
               concatMap (intr (10-n) d) $
               replicateM n $ delete d "0123456789"
                   | n <- [1..9]]
 
problem_111 = sum $ concatMap maxDigits "0123456789"

2 Problem 112

Investigating the density of "bouncy" numbers.

Solution:

isIncreasing' n p
    | n == 0 = True
    | p >= p1 = isIncreasing' (n `div` 10) p1
    | otherwise = False
    where
    p1 = n `mod` 10
 
isIncreasing :: Int -> Bool
isIncreasing n = isIncreasing' (n `div` 10) (n `mod` 10)
 
isDecreasing' n p
    | n == 0 = True
    | p <= p1 = isDecreasing' (n `div` 10) p1
    | otherwise = False
    where
    p1 = n `mod` 10
 
isDecreasing :: Int -> Bool
isDecreasing n = isDecreasing' (n `div` 10) (n `mod` 10)
 
isBouncy n = not (isIncreasing n) && not (isDecreasing n)
nnn=1500000
num150 =length [x|x<-[1..nnn],isBouncy x]
p112 n nb
    | fromIntegral nnb / fromIntegral n >= 0.99 = n
    | otherwise = prob112' (n+1) nnb
    where 
    nnb = if isBouncy n then nb + 1 else nb
 
problem_112=p112 (nnn+1) num150

3 Problem 113

How many numbers below a googol (10100) are not "bouncy"?

Solution:

import Array
 
mkArray b f = listArray b $ map f (range b)
 
digits = 100
 
inc = mkArray ((1, 0), (digits, 9)) ninc
dec = mkArray ((1, 0), (digits, 9)) ndec
 
ninc (1, _) = 1
ninc (l, d) = sum [inc ! (l-1, i) | i <- [d..9]]
 
ndec (1, _) = 1
ndec (l, d) = sum [dec ! (l-1, i) | i <- [0..d]]
 
problem_113 = sum [inc ! i | i <- range ((digits, 0), (digits, 9))]
               + sum [dec ! i | i <- range ((1, 1), (digits, 9))]
               - digits*9 -- numbers like 11111 are counted in both inc and dec 
               - 1 -- 0 is included in the increasing numbers

Note: inc and dec contain the same data, but it seems clearer to duplicate them.

it is another way to solution this problem:

binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
prodxy x y=product[x..y]
problem_113=sum[binomial (8+a) a+binomial (9+a) a-10|a<-[1..100]]

4 Problem 114

Investigating the number of ways to fill a row with separated blocks that are at least three units long.

Solution:

-- fun in p115
problem_114=fun 3 50

5 Problem 115

Finding a generalisation for the number of ways to fill a row with separated blocks.

Solution:

binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
prodxy x y=product[x..y]
fun m n=sum[binomial (k+a) (k-a)|a<-[0..div (n+1) (m+1)],let k=1-a*m+n]
problem_115 = (+1)$length$takeWhile (<10^6) [fun 50 i|i<-[1..]]

6 Problem 116

Investigating the number of ways of replacing square tiles with one of three coloured tiles.

Solution:

binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
prodxy x y=product[x..y]
f116 n x=sum[binomial (a+b) a|a<-[1..div n x],let b=n-a*x]
p116 x=sum[f116 x a|a<-[2..4]]
problem_116 = p116 50

7 Problem 117

Investigating the number of ways of tiling a row using different-sized tiles.

Solution:

fibs5 = 0 : 0 :1: 1:zipWith4 (\a b c d->a+b+c+d) fibs5 a1 a2 a3 
    where
    a1=tail fibs5
    a2=tail a1
    a3=tail a2
p117 x=fibs5!!(x+2)
problem_117 = p117 50

8 Problem 118

Exploring the number of ways in which sets containing prime elements can be made.

Solution:

find2km :: Integral a => a -> (a,a)
find2km n = f 0 n
    where 
        f k m
            | r == 1 = (k,m)
            | otherwise = f (k+1) q
            where (q,r) = quotRem m 2        
 
millerRabinPrimality :: Integer -> Integer -> Bool
millerRabinPrimality n a
    | a <= 1 || a >= n-1 = 
        error $ "millerRabinPrimality: a out of range (" 
              ++ show a ++ " for "++ show n ++ ")" 
    | n < 2 = False
    | even n = False
    | b0 == 1 || b0 == n' = True
    | otherwise = iter (tail b)
    where
        n' = n-1
        (k,m) = find2km n'
        b0 = powMod n a m
        b = take (fromIntegral k) $ iterate (squareMod n) b0
        iter [] = False
        iter (x:xs)
            | x == 1 = False
            | x == n' = True
            | otherwise = iter xs
 
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
 
mulMod :: Integral a => a -> a -> a -> a
mulMod a b c = (b * c) `mod` a
squareMod :: Integral a => a -> a -> a
squareMod a b = (b * b) `rem` a
powMod :: Integral a => a -> a -> a -> a
powMod m = pow' (mulMod m) (squareMod m)
--isPrime x=millerRabinPrimality x 2
isPrime x
    |x<100=isPrime' x
    |otherwise=foldl   (&& )True [millerRabinPrimality x y|y<-[2,3,7,61]]
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
 
getprimes ""= [[]]
getprimes s1= 
    [n:f|
    let len=length s1,
    a<-[1..len],
    let b=take a s1,
    let n=read b::Integer,
    isPrime n,
    let k=getprimes $drop a s1,
    f<-k,
    a==len|| n<head f
    ]
perms :: [a] -> [[a]]
perms [] = [ [] ]
perms (x:xs) = 
    concat (map (between x) (perms xs))
    where 
    between e [] = [ [e] ]
    between e (y:ys) = (e:y:ys) : map (y:) (between e ys)  
fun x=do
    let cs=length$getprimes x
    if (cs/=0) then
        appendFile "p118.log"$(++"\n")$show cs
        else
        return ()
sToInt =(+0).read 
problem_118a=do
    s<-readFile "p118.log"
    print$sum$map sToInt$lines s
main=do
    mapM_ fun $perms ['1'..'9']
    problem_118a
problem_118 = main

9 Problem 119

Investigating the numbers which are equal to sum of their digits raised to some power.

Solution:

import Data.List
digits n 
{-  123->[3,2,1]
 -}
    |n<10=[n]
    |otherwise= y:digits x 
    where
    (x,y)=divMod n 10
problem_119 =sort [(a^b)|
    a<-[2..200],
    b<-[2..9],
    let m=a^b,
    let n=sum$digits m,
    n==a]!!29

10 Problem 120

Finding the maximum remainder when (a − 1)n + (a + 1)n is divided by a2.

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
fun x
    |even x=x*(x-2)
    |not$null$funb x=head$funb x
    |odd e=x*(x-1)
    |otherwise=2*x*(e-1)
    where
    e=div x 2
 
funb x=take 1 [nn*x|
    a<-[1,3..x],
    let n=div (x-1) 2,
    let p=x*a+n,
    isPrime p,
    let nn=mod (2*(x*a+n)) x
    ]
 
problem_120 = sum [fun a|a<-[3..1000]]