Personal tools

Euler problems/111 to 120

From HaskellWiki

(Difference between revisions)
Jump to: navigation, search
Line 30: Line 30:
Solution:
Solution:
<haskell>
<haskell>
-
import Data.List
+
isIncreasing' n p
-
digits n
+
| n == 0 = True
-
{- change 123 to [3,2,1]
+
| p >= p1 = isIncreasing' (n `div` 10) p1
-
-}
+
| otherwise = False
-
|n<10=[n]
+
-
|otherwise= y:digits x
+
where
where
-
(x,y)=divMod n 10
+
p1 = n `mod` 10
-
isdecr x=
+
 
-
null$filter (\(x, y)->x-y<0)$zip di k
+
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
where
-
di=digits x
+
p1 = n `mod` 10
-
k=0:di
+
 
-
isincr x=
+
isDecreasing :: Int -> Bool
-
null$filter (\(x, y)->x-y<0)$zip di k
+
isDecreasing n = isDecreasing' (n `div` 10) (n `mod` 10)
-
where
+
 
-
di=digits x
+
isBouncy n = not (isIncreasing n) && not (isDecreasing n)
-
k=tail$di++[0]
+
nnn=1500000
nnn=1500000
-
num150 =length [x|x<-[1..nnn],isdecr x||isincr x]
+
num150 =length [x|x<-[1..nnn],isBouncy x]
-
istwo x|isdecr x||isincr x=1
+
p112 n nb
-
|otherwise=0
+
| fromIntegral nnb / fromIntegral n >= 0.99 = n
-
problem_112 n1 n2=
+
| otherwise = prob112' (n+1) nnb
-
if (div n1 n2==100)
+
where
-
then do appendFile "file.log" ((show n1) ++" "++ (show n2)++"\n")
+
nnb = if isBouncy n then nb + 1 else nb
-
return()
+
-
else problem_112 (n1+1) (n2+istwo (n1+1))
+
-
main= problem_112 nnn num150
+
 +
problem_112=p112 (nnn+1) num150
</haskell>
</haskell>
Line 90: Line 91:
it is another way to solution this problem:
it is another way to solution this problem:
<haskell>
<haskell>
-
import List
+
binomial x y =div (prodxy (y+1) x) (prodxy 1 (x-y))
-
series 2 =replicate 10 1
+
prodxy x y=product[x..y]
-
series n=sumkey$map (\(x, y)->map (*y) x)$zip key (series (n-1))
+
problem_113=sum[binomial (8+a) a+binomial (9+a) a-10|a<-[1..100]]
-
key =[replicate (a+1) 1++replicate (9-a) 0|a<-[0..9]]
+
-
sumkey k=[sum [a!!m|a<-k]|m<-[0..9]]
+
-
fun x= sum [(sum$series i)-1|i<-[2..x]]-(x-1)*9-1+(sum$series x)
+
-
problem_113 =fun 101
+
</haskell>
</haskell>
== [http://projecteuler.net/index.php?section=view&id=114 Problem 114] ==
== [http://projecteuler.net/index.php?section=view&id=114 Problem 114] ==
Line 103: Line 100:
Solution:
Solution:
<haskell>
<haskell>
-
slowfibs n
+
-- fun in p115
-
|n<4=1
+
problem_114=fun 3 50
-
|otherwise=2*slowfibs (n-1)-slowfibs (n-2)+slowfibs(n-4)
+
-
fibs = 1 : 1: 1: 1: zipWith3 (\a b c->2*a-b+c) c b a
+
-
where
+
-
a=fibs
+
-
b=tail$tail fibs
+
-
c=tail$tail$tail fibs
+
-
fast=[fibs!! a|a<-[1..51]]
+
-
test=[slowfibs a|a<-[1..21]]
+
-
problem_114=fibs!!51
+
</haskell>
</haskell>
Line 124: Line 112:
prodxy x y=product[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]
fun m n=sum[binomial (k+a) (k-a)|a<-[0..div (n+1) (m+1)],let k=1-a*m+n]
-
p114=fun 3 50
 
problem_115 = (+1)$length$takeWhile (<10^6) [fun 50 i|i<-[1..]]
problem_115 = (+1)$length$takeWhile (<10^6) [fun 50 i|i<-[1..]]
</haskell>
</haskell>

Revision as of 05:45, 14 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: 

problem_116 = undefined

7 Problem 117

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

Solution: 

problem_117 = undefined

8 Problem 118

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

Solution: 

problem_118 = undefined

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]]
Retrieved from "http://www.haskell.org/haskellwiki/Euler_problems/111_to_120"