# Difference between revisions of "Euler problems/181 to 190"

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Aren't you glad you didn't skip that math class now? HenryLaxen Apr 18,2008 |
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## Latest revision as of 01:29, 26 April 2021

## Contents

## Problem 181

Investigating in how many ways objects of two different colours can be grouped.

## Problem 182

RSA encryption.

Solution:

```
fun a1 b1 = sum [ e | e <- [2..a*b-1],
gcd e (a*b) == 1,
gcd (e-1) a == 2,
gcd (e-1) b == 2 ]
where a = a1-1
b = b1-1
problem_182 = fun 1009 3643
```

## Problem 183

Maximum product of parts.

Solution:

```
-- Does the decimal expansion of p/q terminate?
terminating p q = 1 == reduce [2,5] (q `div` gcd p q)
where reduce [] n = n
reduce (x:xs) n | n `mod` x == 0 = reduce (x:xs) (n `div` x)
| otherwise = reduce xs n
-- The expression (round $ fromIntegral n / e) computes the integer k
-- for which (n/k)^k is at a maximum. Also note that, given a rational number
-- r and a natural number k, the decimal expansion of r^k terminates if
-- and only if the decimal expansion of r does.
answer = sum [if terminating n (round $ fromIntegral n / e) then -n else n
| n <- [5 .. 10^4]]
where e = exp 1
main = print answer
```

## Problem 184

Triangles containing the origin.

## Problem 185

Number Mind

Solution:

This approach does **NOT** solve the problem in under a minute,
unless of course you are extremely lucky. The best time I've
seen so far has been about 76 seconds. Before I came up with
this code, I tried to search for the solution by generating a
list of all possible solutions based on the information given in
the guesses. This was feasible with the 5 digit problem, but
completely intractable with the 16 digit problem. The approach
here, simple yet effective, is to make a random guess, and then
vary each digit in the guess from [0..9], generating a score of
how well the guess matched the given numbermind clues. You then
*improve* the guess by selecting those digits that had a unique
low score. It turns out this approach converges rather quickly,
but can often be stuck in cycles, so we test for this and try a
differenct random first guess if a cycle is detected. Once you
run the program, you might have time for a cup of coffee, or
maybe even a dinner. HenryLaxen 2008-03-12

```
import Data.List
import Control.Monad
import Data.Char
import System.Random
type Mind = [([Char],Int)]
values :: [Char]
values = "0123456789"
score :: [Char] -> [Char] -> Int
score guess answer = foldr (\(a,b) y -> if a == b then y+1 else y) 0
(zip guess answer)
scores :: Mind -> [Char] -> [Int]
scores m g = map (\x -> abs ((snd x) - score (fst x) g)) m
scoreMind :: Mind -> [Char] -> Int
scoreMind m g = sum $ scores m g
ex1 :: Mind
ex1 =
[("90342",2),
("39458",2),
("51545",2),
("34109",1),
("12531",1),
("70794",0)]
ex2 :: Mind
ex2 =
[
("5616185650518293",2),
("3847439647293047",1),
("5855462940810587",3),
("9742855507068353",3),
("4296849643607543",3),
("3174248439465858",1),
("4513559094146117",2),
("7890971548908067",3),
("8157356344118483",1),
("2615250744386899",2),
("8690095851526254",3),
("6375711915077050",1),
("6913859173121360",1),
("6442889055042768",2),
("2321386104303845",0),
("2326509471271448",2),
("5251583379644322",2),
("1748270476758276",3),
("4895722652190306",1),
("3041631117224635",3),
("1841236454324589",3),
("2659862637316867",2)]
guesses :: [Char] -> Int -> [[Char]]
guesses str pos = [ left ++ n:(tail right) | n<-values]
where (left,right) = splitAt pos str
bestGuess :: Mind -> [[Char]] -> [Int]
bestGuess mind guesses = bestGuesses
where scores = map (scoreMind mind) guesses
bestScore = minimum scores
bestGuesses = elemIndices bestScore scores
iterateGuesses :: Mind -> [Char] -> [Char]
iterateGuesses mind value = nextguess value mins
where allguesses = map (guesses value) [0..(length value)-1]
mins = map (bestGuess mind) allguesses
nextguess :: [Char] -> [[Int]] -> [Char]
nextguess = zipWith choose
where choose x [y] = intToDigit y
choose x _ = x
iterateMind :: Mind -> [Char] -> [([Char], Int)]
iterateMind mind n = zip b c
where a = drop 2 $ inits $ iterate (iterateGuesses mind) n
b = last $ takeWhile (\x -> (last x) `notElem` (init x)) a
c = map (scoreMind mind) b
randomStart :: Int -> IO [Char]
randomStart n = replicateM n (getStdRandom (randomR ('0','9')))
main :: IO ()
main = do
let ex = ex1
x <- randomStart (length (fst (head ex)))
let y = iterateMind ex x
done = snd (last y) == 0
if done then (putStrLn $ (fst.last) y)
else main
```

Here's another solution, and this one squeaks by in just under a minute on my machine. The basic idea is to just do a back-tracking search, but with some semi-smart pruning and guess ordering. The code is in pretty much the order I wrote it, so most prefixes of this code should also compile. This also means you should be able to figure out what each function does one at a time.

```
import Control.Monad
import Data.List
import qualified Data.Set as S
ensure p x = guard (p x) >> return x
selectDistinct 0 _ = [[]]
selectDistinct n [] = []
selectDistinct n (x:xs) = map (x:) (selectDistinct (n - 1) xs) ++ selectDistinct n xs
data Environment a = Environment { guesses :: [(Int, [a])]
, restrictions :: [S.Set a]
, assignmentsLeft :: Int
} deriving (Eq, Ord, Show)
reorder e = e { guesses = sort . guesses $ e }
domain = S.fromList "0123456789"
initial = Environment gs (replicate a S.empty) a where
a = length . snd . head $ gs
gs = [(2, "5616185650518293"), (1, "3847439647293047"), (3, "5855462940810587"), (3, "9742855507068353"), (3, "4296849643607543"), (1, "3174248439465858"), (2, "4513559094146117"), (3, "7890971548908067"), (1, "8157356344118483"), (2, "2615250744386899"), (3, "8690095851526254"), (1, "6375711915077050"), (1, "6913859173121360"), (2, "6442889055042768"), (0, "2321386104303845"), (2, "2326509471271448"), (2, "5251583379644322"), (3, "1748270476758276"), (1, "4895722652190306"), (3, "3041631117224635"), (3, "1841236454324589"), (2, "2659862637316867")]
acceptableCounts e = small >= 0 && big <= assignmentsLeft e where
ns = (0:) . map fst . guesses $ e
small = minimum ns
big = maximum ns
positions s = map fst . filter (not . snd) . zip [0..] . zipWith S.member s
acceptableRestriction r (n, s) = length (positions s r) >= n
acceptableRestrictions e = all (acceptableRestriction (restrictions e)) (guesses e)
firstGuess = head . guesses
sanityCheck e = acceptableRestrictions e && acceptableCounts e
solve e@(Environment _ _ 0) = [[]]
solve e@(Environment [] r _) = mapM (S.toList . (domain S.\\)) r
solve e' = do
is <- m
newE <- f is
rest <- solve newE
return $ interleaveAscIndices is (l is) rest
where
f = ensure sanityCheck . update e
m = selectDistinct n (positions g (restrictions e))
e = reorder e'
l = fst . flip splitAscIndices g
(n, g) = firstGuess e
splitAscIndices = indices 0 where
indices _ [] xs = ([], xs)
indices n (i:is) (x:xs)
| i == n = let (b, e) = indices (n + 1) is xs in (x:b, e)
| True = let (b, e) = indices (n + 1) (i:is) xs in (b, x:e)
interleaveAscIndices = indices 0 where
indices _ [] [] ys = ys
indices n (i:is) (x:xs) ys
| i == n = x : indices (n + 1) is xs ys
| True = head ys : indices (n + 1) (i:is) (x:xs) (tail ys)
update (Environment ((_, a):gs) r l) is = Environment newGs restriction (l - length is) where
(assignment, newRestriction) = splitAscIndices is a
(_, oldRestriction) = splitAscIndices is r
restriction = zipWith S.insert newRestriction oldRestriction
newGs = map updateEntry gs
updateEntry (n', a') = (newN, newA) where
(dropped, newA) = splitAscIndices is a'
newN = n' - length (filter id $ zipWith (==) assignment dropped)
problem_185 = head . solve $ initial
```

## Problem 186

Connectedness of a network.

This implements the known tree structure for imperative union/find. On my machine it runs in about 3 seconds.

```
module ProjectEuler186 where
import qualified Control.Monad.ST.Strict as StrictST
import qualified Control.Monad.ST.Lazy as LazyST
import Data.Array.ST (STUArray, newArray, newListArray, readArray, writeArray, )
import Control.Monad (liftM2, when, )
data Network s =
Network
(STUArray s Telephone Telephone)
(STUArray s Telephone Size)
type Telephone = Int
type Size = Int
initialize :: StrictST.ST s (Network s)
initialize =
let bnds = (0,999999)
in liftM2 Network
(newListArray bnds [0..])
(newArray bnds 1)
union :: Network s -> Telephone -> Telephone -> StrictST.ST s ()
union net@(Network repr size) m n = do
mroot <- find net m
nroot <- find net n
when (mroot/=nroot) $ do
msize <- readArray size mroot
nsize <- readArray size nroot
let totalSize = msize+nsize
if msize < nsize
then do
writeArray repr mroot nroot
writeArray size nroot totalSize
else do
writeArray repr nroot mroot
writeArray size mroot totalSize
find :: Network s -> Telephone -> StrictST.ST s Telephone
find (Network repr _size) =
let go n = do
m <- readArray repr n
if n==m
then return m
else go m
in go
callersStart :: [Telephone]
callersStart =
map (\k -> fromInteger $ mod (100003 - 200003*k + 300007*k^(3::Int)) 1000000) [1..]
callers :: [Telephone]
callers =
take 55 callersStart ++
zipWith (\s24 s55 -> mod (s24+s55) 1000000) (drop 31 callers) callers
pairs :: [a] -> [(a,a)]
pairs (x0:x1:xs) = (x0,x1) : pairs xs
pairs _ = error "list of callers must be infinite"
answer :: Int
answer =
(1+) . length . takeWhile (<990000) $
LazyST.runST (do
net@(Network _repr size) <- LazyST.strictToLazyST initialize
mapM
(\(from,to) ->
LazyST.strictToLazyST $
union net from to >>
find net 524287 >>=
readArray size) $
filter (uncurry (/=)) $
pairs $ callers)
main :: IO ()
main = print answer
```

## Problem 187

Semiprimes

Solution:

The solution to this problem isn't terribly difficult, once you know that the numbers the problem is referring to are called semiprimes. In fact there is an excellent write-up at: http://mathworld.wolfram.com/Semiprime.html which provides an explicit formula for the number of semiprimes less than a given number.

The problem with this formula is the use of pi(n) where pi is the number of primes less than n. For Mathematica users this is no problem since the function is built in, but for us poor Haskeller's we have to build it ourselves. This is what took the most time for me, to come up with a pi(n) function that can compute pi(50000000) in less than the lifetime of the universe. Thus I embarked on a long and circuitous journey that eventually led me to the PI module below, which does little more than read a 26MB file of prime numbers into memory and computes a map from which we can calculate pi(n). I am sure there must be a better way of doing this, and I look forward to this entry being amended (or replaced) with a more reasonable solution. HenryLaxen Mar 24, 2008

```
module PI (prime,primes,pI) where
import Control.Monad.ST
import System.IO.Unsafe
import qualified Data.Map as M
import Data.ByteString.Char8 (readFile,lines,readInt)
import Data.Maybe
import Data.Array.ST
import Data.Array.IArray
r = runSTUArray
(do a <- newArray (0,3001134) 0 :: ST s (STUArray s Int Int)
writeArray a 0 1
zipWithM_ (writeArray a) [0..] primes
return a)
{-# NOINLINE s #-}
s = M.fromDistinctAscList $ map (\(x,y) -> (y,x)) (assocs r)
prime n = r!n
pI :: Int -> Int
pI n =
case M.splitLookup n s of
(_,Just x,_) -> (x+1)
(_,_,b) -> snd (head (M.assocs b))
{-# NOINLINE primes #-}
primes :: [Int]
primes = unsafePerformIO $ do
l <- Data.ByteString.Char8.readFile "primes_to_5M.txt"
let p = Data.ByteString.Char8.lines l
r = map (fst . fromJust . readInt) p::[Int]
return r
import PI
import Data.List
s n k = ( n `div` (prime (k-1))) - k + 1
semiPrimeCount n =
let last = pI $ floor $ sqrt (fromIntegral n)
s k = pI ( n `div` (prime (k-1))) - k + 1
pI2 = foldl' (\x y -> s y + x) 0 [1..last]
in pI2
main = print (semiPrimeCount 100000000)
```

The file primes_to_5M.txt is: 2 3 5 .. 49999991 50000017

## Problem 188

hyperexponentiation

The idea here is actually very simple. Euler's theorem tells us that a^phi(n) mod n = 1, so the exponent never need grow above 40000000 in this case, which is phi(10^8). Henry Laxen April 6, 2008

```
fastPower x 1 modulus = x `mod` modulus
fastPower x n modulus
| even n = fastPower ((x*x) `mod` modulus) (n `div` 2) modulus
| otherwise = (x * (fastPower x (n-1) modulus)) `mod` modulus
modulus = 10^8
phi = 40000000 -- eulerTotient of 10^8
a :: Int -> Int -> Int
a n 1 = n
-- a n k = n^(a n (k-1))
a n k = fastPower n (a n (k-1) `mod` phi) modulus
problem_188 = a 1777 1855
```

## Problem 190

Maximising a weighted product

For those of us who can reach all the back into our sophomore calculus class memories, we will find the dreaded Lagrange Multipliers, which say that if you want to max(min)imize f subject to a constraint g, then the gradients of f and g have to be parallel, or grad f = L * grad g.

Happy happy, for in our case f = x * y^2 * .. z^n and even better g = x + y + .. z - n, so grad g = (1,1,..1) and grad f = (f/x, 2f/y, .. nf/z) That leads us to:

x = f/L, y = 2f/L .. z = nf/L and sum (1..n) x == n so x = 2/(n+1)

Thus the Haskell program is nothing more than:

```
f l = product $ zipWith (^) l [1..]
fmin n = floor $ f [ i*2/(n+1) | i<-[1..n]]
problem_190 = sum $ map fmin [2..15]
```

Aren't you glad you didn't skip that math class now? HenryLaxen Apr 18,2008