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Euler problems/51 to 60

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Line 3: Line 3:
   
 
Solution:
 
Solution:
  +
  +
millerRabinPrimality on the [[Prime_numbers]] page
  +
 
<haskell>
 
<haskell>
problem_51 = undefined
+
isPrime x
  +
|x==3=True
  +
|otherwise=millerRabinPrimality x 2
  +
ch='1'
  +
numChar n= sum [1|x<-show(n),x==ch]
  +
replace d c|c==ch=d
  +
|otherwise=c
  +
nextN repl n= (+0)$read $map repl $show n
  +
same n= [if isPrime$nextN (replace a) n then 1 else 0|a<-['1'..'9']]
  +
problem_51=head [n|
  +
n<-[100003,100005..999999],
  +
numChar n==3,
  +
(sum $same n)==8
  +
]
 
</haskell>
 
</haskell>
   
Line 12: Line 15:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_52 = undefined
+
import List
  +
  +
has_same_digits a b = (show a) \\ (show b) == []
  +
  +
check n = all (has_same_digits n) (map (n*) [2..6])
  +
  +
problem_52 = head $ filter check [1..]
 
</haskell>
 
</haskell>
   
Line 20: Line 23:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_53 = undefined
+
facs = scanl (*) 1 [1..100]
  +
comb (r,n) = facs!!n `div` (facs!!r * facs!!(n-r))
  +
perms = [(n,x) | x<-[1..100], n<-[1..x]]
  +
problem_53 = length $ filter (>1000000) $ map comb $ perms
 
</haskell>
 
</haskell>
   
 
== [http://projecteuler.net/index.php?section=problems&id=54 Problem 54] ==
 
== [http://projecteuler.net/index.php?section=problems&id=54 Problem 54] ==
How many hands did player one win in the game of poker?
+
How many hands did player one win in the [http://www.pokerroom.com poker games]?
   
 
Solution:
 
Solution:
  +
  +
probably not the most straight forward way to do it.
  +
 
<haskell>
 
<haskell>
problem_54 = undefined
+
import Data.List
  +
import Data.Maybe
  +
import Control.Monad
  +
  +
readCard [r,s] = (parseRank r, parseSuit s)
  +
where parseSuit = translate "SHDC"
  +
parseRank = translate "23456789TJQKA"
  +
translate from x = fromJust $ elemIndex x from
  +
  +
solveHand hand = (handRank,tiebreak)
  +
where
  +
handRank
  +
| flush && straight = 9
  +
| hasKinds 4 = 8
  +
| all hasKinds [2,3] = 7
  +
| flush = 6
  +
| straight = 5
  +
| hasKinds 3 = 4
  +
| 1 < length (kind 2) = 3
  +
| hasKinds 2 = 2
  +
| otherwise = 1
  +
tiebreak = kind =<< [4,3,2,1]
  +
hasKinds = not . null . kind
  +
kind n = map head $ filter ((n==).length) $ group ranks
  +
ranks = sortBy (flip compare) $ map fst hand
  +
flush = 1 == length (nub (map snd hand))
  +
straight = length (kind 1) == 5 && 4 == head ranks - last ranks
  +
  +
gameLineToHands = splitAt 5 . map readCard . words
  +
p1won (a,b) = solveHand a > solveHand b
  +
  +
problem_54 = do
  +
f <- readFile "poker.txt"
  +
let games = map gameLineToHands $ lines f
  +
wins = filter p1won games
  +
print $ length wins
 
</haskell>
 
</haskell>
   
Line 36: Line 42:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_55 = undefined
+
reverseNum = read . reverse . show
  +
  +
palindrome x =
  +
sx == reverse sx
  +
where
  +
sx = show x
  +
  +
lychrel =
  +
not . any palindrome . take 50 . tail . iterate next
  +
where
  +
next x = x + reverseNum x
  +
  +
problem_55 = length $ filter lychrel [1..10000]
 
</haskell>
 
</haskell>
   
 
== [http://projecteuler.net/index.php?section=problems&id=56 Problem 56] ==
 
== [http://projecteuler.net/index.php?section=problems&id=56 Problem 56] ==
Considering natural numbers of the form, ab, finding the maximum digital sum.
+
Considering natural numbers of the form, a<sup>b</sup>, finding the maximum digital sum.
   
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_56 = undefined
+
digitalSum 0 = 0
  +
digitalSum n =
  +
let (d,m) = quotRem n 10 in m + digitalSum d
  +
  +
problem_56 =
  +
maximum [digitalSum (a^b) | a <- [99], b <- [90..99]]
  +
</haskell>
  +
  +
Alternate solution:
  +
<haskell>
  +
import Data.Char (digitToInt)
  +
  +
digiSum :: Integer -> Int
  +
digiSum = sum . map digitToInt . show
  +
  +
problem_56 :: Int
  +
problem_56 = maximum $ map digiSum [a^b | a <- [1..100], b <- [1..100]]
 
</haskell>
 
</haskell>
   
Line 52: Line 58:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_57 = undefined
+
twoex = zip ns ds
  +
where
  +
ns = 3 : zipWith (\x y -> x + 2 * y) ns ds
  +
ds = 2 : zipWith (+) ns ds
  +
  +
len = length . show
  +
  +
problem_57 =
  +
length $ filter (\(n,d) -> len n > len d) $ take 1000 twoex
  +
</haskell>
  +
  +
The following solution is based on the observation that the fractions needed appear regularly in the repeating pattern _______$____$ where underscores are ignored and dollars are interesting fractions.
  +
  +
<haskell>
  +
calc :: Int -> Int
  +
calc n = nd13 * 2 + ((n-nd13*13) `div` 8)
  +
where
  +
nd13 = n `div` 13
  +
  +
problem_57 :: Int
  +
problem_57 = calc 1000
 
</haskell>
 
</haskell>
   
Line 60: Line 66:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_58 = undefined
+
isPrime x
  +
|x==3=True
  +
|otherwise=and [millerRabinPrimality x n|n<-[2,3]]
  +
diag = 1:3:5:7:zipWith (+) diag [8,10..]
  +
problem_58 =
  +
result $ dropWhile tooBig $ drop 2 $ scanl primeRatio (0,0) diag
  +
where
  +
primeRatio (n,d) num = (if d `mod` 4 /= 0 && isPrime num then n+1 else n,d+1)
  +
tooBig (n,d) = n*10 >= d
  +
result ((_,d):_) = (d+2) `div` 4 * 2 + 1
 
</haskell>
 
</haskell>
   
Line 68: Line 74:
 
Solution:
 
Solution:
 
<haskell>
 
<haskell>
problem_59 = undefined
+
import Data.Bits
  +
import Data.Char
  +
import Data.List
  +
import Data.Ord (comparing)
  +
  +
keys = [ [a,b,c] | a <- [97..122], b <- [97..122], c <- [97..122] ]
  +
allAlpha = all (\k -> let a = ord k in (a >= 32 && a <= 122))
  +
howManySpaces = length . filter (==' ')
  +
  +
problem_59 = do
  +
s <- readFile "cipher1.txt"
  +
let
  +
cipher = (read ("[" ++ s ++ "]") :: [Int])
  +
decrypts = [ map chr (zipWith xor (cycle key) cipher) | key <- keys ]
  +
alphaDecrypts = filter allAlpha decrypts
  +
message = maximumBy (comparing howManySpaces) alphaDecrypts
  +
asciisum = sum (map ord message)
  +
print asciisum
 
</haskell>
 
</haskell>
   
Line 75: Line 81:
   
 
Solution:
 
Solution:
  +
  +
Breadth first search that works on infinite lists. Breaks the 60 secs rule. This program finds the solution in 185 sec on my Dell D620 Laptop.
 
<haskell>
 
<haskell>
problem_60 = undefined
+
problem_60 = print$sum $head solve
  +
isPrime x = x==3 || millerRabinPrimality x 2
  +
  +
solve = do
  +
a <- primesTo10000
  +
let m = f a $ dropWhile (<= a) primesTo10000
  +
b <- m
  +
let n = f b $ dropWhile (<= b) m
  +
c <- n
  +
let o = f c $ dropWhile (<= c) n
  +
d <- o
  +
let p = f d $ dropWhile (<= d) o
  +
e <- p
  +
return [a,b,c,d,e]
  +
where
  +
f x = filter (\y -> and [isPrime $read $shows x $show y,
  +
isPrime $read $shows y $show x])
  +
primesTo10000 = 2:filter isPrime [3,5..9999]
 
</haskell>
 
</haskell>
 
[[Category:Tutorials]]
 
[[Category:Code]]
 

Latest revision as of 00:17, 17 February 2010

Contents

[edit] 1 Problem 51

Find the smallest prime which, by changing the same part of the number, can form eight different primes.

Solution:

millerRabinPrimality on the Prime_numbers page

isPrime x
    |x==3=True
    |otherwise=millerRabinPrimality x 2
ch='1'
numChar n= sum [1|x<-show(n),x==ch]
replace d c|c==ch=d
           |otherwise=c
nextN repl n= (+0)$read $map repl $show n  
same n= [if isPrime$nextN (replace a) n then 1 else 0|a<-['1'..'9']]
problem_51=head [n|
    n<-[100003,100005..999999],
    numChar n==3,
    (sum $same n)==8
    ]

[edit] 2 Problem 52

Find the smallest positive integer, x, such that 2x, 3x, 4x, 5x, and 6x, contain the same digits in some order.

Solution:

import List
 
has_same_digits a b = (show a) \\ (show b) == []
 
check n = all (has_same_digits n) (map (n*) [2..6])
 
problem_52 = head $ filter check [1..]

[edit] 3 Problem 53

How many values of C(n,r), for 1 ≤ n ≤ 100, exceed one-million?

Solution:

facs = scanl (*) 1 [1..100]
comb (r,n) = facs!!n `div` (facs!!r * facs!!(n-r))
perms = [(n,x) | x<-[1..100], n<-[1..x]]
problem_53 = length $ filter (>1000000) $ map comb $ perms

[edit] 4 Problem 54

How many hands did player one win in the poker games?

Solution:

probably not the most straight forward way to do it.

import Data.List
import Data.Maybe
import Control.Monad
 
readCard [r,s] = (parseRank r, parseSuit s)
 where parseSuit = translate "SHDC"
       parseRank = translate "23456789TJQKA"
       translate from x = fromJust $ elemIndex x from
 
solveHand hand = (handRank,tiebreak)
 where
 handRank
  | flush && straight   = 9
  | hasKinds 4          = 8
  | all hasKinds [2,3]  = 7
  | flush               = 6
  | straight            = 5
  | hasKinds 3          = 4
  | 1 < length (kind 2) = 3
  | hasKinds 2          = 2
  | otherwise           = 1
 tiebreak = kind =<< [4,3,2,1]
 hasKinds = not . null . kind
 kind n = map head $ filter ((n==).length) $ group ranks
 ranks  = sortBy (flip compare) $ map fst hand
 flush  = 1 == length (nub (map snd hand))
 straight = length (kind 1) == 5 && 4 == head ranks - last ranks
 
gameLineToHands = splitAt 5 . map readCard . words
p1won (a,b) = solveHand a > solveHand b
 
problem_54 = do
    f <- readFile "poker.txt"
    let games = map gameLineToHands $ lines f
        wins = filter p1won games
    print $ length wins

[edit] 5 Problem 55

How many Lychrel numbers are there below ten-thousand?

Solution:

reverseNum = read . reverse . show
 
palindrome x =
    sx == reverse sx
    where
    sx = show x 
 
lychrel = 
    not . any palindrome . take 50 . tail . iterate next
    where
    next x = x + reverseNum x
 
problem_55 = length $ filter lychrel [1..10000]

[edit] 6 Problem 56

Considering natural numbers of the form, ab, finding the maximum digital sum.

Solution:

digitalSum 0 = 0
digitalSum n = 
    let (d,m) = quotRem n 10 in m + digitalSum d 
 
problem_56 = 
    maximum [digitalSum (a^b) | a <- [99], b <- [90..99]]

Alternate solution:

import Data.Char (digitToInt)
 
digiSum :: Integer -> Int
digiSum = sum . map digitToInt . show
 
problem_56 :: Int
problem_56 = maximum $ map digiSum [a^b | a <- [1..100], b <- [1..100]]

[edit] 7 Problem 57

Investigate the expansion of the continued fraction for the square root of two.

Solution:

twoex = zip ns ds 
    where
    ns = 3 : zipWith (\x y -> x + 2 * y) ns ds
    ds = 2 : zipWith (+) ns ds
 
len = length . show
 
problem_57 = 
    length $ filter (\(n,d) -> len n > len d) $ take 1000 twoex

The following solution is based on the observation that the fractions needed appear regularly in the repeating pattern _______$____$ where underscores are ignored and dollars are interesting fractions.

calc :: Int -> Int
calc n = nd13 * 2 + ((n-nd13*13) `div` 8)
    where
      nd13 = n `div` 13
 
problem_57 :: Int
problem_57 = calc 1000

[edit] 8 Problem 58

Investigate the number of primes that lie on the diagonals of the spiral grid.

Solution:

isPrime x
    |x==3=True
    |otherwise=and [millerRabinPrimality x n|n<-[2,3]]
diag = 1:3:5:7:zipWith (+) diag [8,10..]
problem_58 = 
    result $ dropWhile tooBig $ drop 2 $ scanl primeRatio (0,0) diag
    where
    primeRatio (n,d) num = (if d `mod` 4 /= 0 && isPrime num then n+1 else n,d+1)
    tooBig (n,d) = n*10 >= d
    result ((_,d):_) = (d+2) `div` 4 * 2 + 1

[edit] 9 Problem 59

Using a brute force attack, can you decrypt the cipher using XOR encryption?

Solution:

import Data.Bits
import Data.Char
import Data.List
import Data.Ord (comparing)
 
keys = [ [a,b,c] | a <- [97..122], b <- [97..122], c <- [97..122] ]
allAlpha = all (\k -> let a = ord k in (a >= 32 && a <= 122))
howManySpaces = length . filter (==' ')
 
problem_59 = do
	s <- readFile "cipher1.txt"
	let 
		cipher = (read ("[" ++ s ++ "]") :: [Int])
		decrypts = [ map chr (zipWith xor (cycle key) cipher) | key <- keys ]
		alphaDecrypts = filter allAlpha decrypts
		message = maximumBy (comparing howManySpaces) alphaDecrypts
		asciisum = sum (map ord message)
	print asciisum

[edit] 10 Problem 60

Find a set of five primes for which any two primes concatenate to produce another prime.

Solution:

Breadth first search that works on infinite lists. Breaks the 60 secs rule. This program finds the solution in 185 sec on my Dell D620 Laptop.

problem_60 = print$sum $head solve 
isPrime x = x==3 || millerRabinPrimality x 2 
 
solve = do
 a <- primesTo10000
 let m = f a $ dropWhile (<= a) primesTo10000
 b <- m
 let n = f b $ dropWhile (<= b) m
 c <- n
 let o = f c $ dropWhile (<= c) n
 d <- o
 let p = f d $ dropWhile (<= d) o
 e <- p
 return [a,b,c,d,e]
 where
 f x = filter (\y -> and [isPrime $read $shows x $show y,
                 isPrime $read $shows y $show x])
primesTo10000 = 2:filter isPrime [3,5..9999]