Euler problems/91 to 100: Difference between revisions
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== [http://projecteuler.net/index.php?section=view&id=91 Problem 91] == | == [http://projecteuler.net/index.php?section=view&id=91 Problem 91] == | ||
Find the number of right angle triangles in the quadrant. | Find the number of right angle triangles in the quadrant. | ||
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problem_100 = undefined | problem_100 = undefined | ||
</haskell> | </haskell> | ||
Revision as of 12:13, 30 September 2007
Problem 91
Find the number of right angle triangles in the quadrant.
Solution:
reduce x y = (quot x d, quot y d)
where d = gcd x y
problem_91 n = 3*n*n + 2* sum others
where
others = do
x1 <- [1..n]
y1 <- [1..n]
let (yi,xi) = reduce x1 y1
let yc = quot (n-y1) yi
let xc = quot x1 xi
return (min xc yc)
Problem 92
Investigating a square digits number chain with a surprising property.
Solution:
problem_92 = undefined
Problem 93
Using four distinct digits and the rules of arithmetic, find the longest sequence of target numbers.
Solution:
problem_93 = undefined
Problem 94
Investigating almost equilateral triangles with integral sides and area.
Solution:
problem_94 = undefined
Problem 95
Find the smallest member of the longest amicable chain with no element exceeding one million.
Solution which avoid visiting a number more than one time :
import Data.Array.Unboxed
import qualified Data.IntSet as S
import Data.List
takeUntil _ [] = []
takeUntil pred (x:xs) = x : if pred x then takeUntil pred xs else []
chain n s = lgo [n] $ properDivisorsSum ! n
where lgo xs x | x > 1000000 || S.notMember x s = (xs,[])
| x `elem` xs = (xs,x : takeUntil (/= x) xs)
| otherwise = lgo (x:xs) $ properDivisorsSum ! x
properDivisorsSum :: UArray Int Int
properDivisorsSum = accumArray (+) 1 (0,1000000)
$ (0,-1):[(k,factor)|
factor<-[2..1000000 `div` 2]
, k<-[2*factor,2*factor+factor..1000000]
]
base = S.fromList [1..1000000]
problem_95 = fst $ until (S.null . snd) f ((0,0),base)
where
f (p@(n,m), s) = (p', s')
where
setMin = head $ S.toAscList s
(explored, chn) = chain setMin s
len = length chn
p' = if len > m then (minimum chn, len) else p
s' = foldl' (flip S.delete) s explored
Here is a more straightforward solution, without optimization. Yet it solves the problem in a few seconds when compiled with GHC 6.6.1 with the -O2 flag. I like to let the compiler do the optimization, without cluttering my code.
This solution avoids using unboxed arrays, which many consider to be somewhat of an imperitive-style hack. In fact, no memoization at all is required.
import Data.List (foldl1', group)
-- The sum of all proper divisors of n.
d n = product [(p * product g - 1) `div` (p - 1) |
g <- group $ primeFactors n, let p = head g
] - n
primeFactors = pf primes
where
pf ps@(p:ps') n
| p * p > n = [n]
| r == 0 = p : pf ps q
| otherwise = pf ps' n
where
(q, r) = n `divMod` p
primes = 2 : filter (null . tail . primeFactors) [3,5..]
-- The longest chain of numbers is (n, k), where
-- n is the smallest number in the chain, and k is the length
-- of the chain. We limit the search to chains whose
-- smallest number is no more than m and, optionally, whose
-- largest number is no more than m'.
longestChain m m' = (n, k)
where
(n, Just k) = foldl1' cmpChain [(n, findChain n) | n <- [2..m]]
findChain n = f [] n $ d n
f s n n'
| n' == n = Just $ 1 + length s
| n' < n = Nothing
| maybe False (< n') m' = Nothing
| n' `elem` s = Nothing
| otherwise = f (n' : s) n $ d n'
cmpChain p@(n, k) q@(n', k')
| (k, negate n) < (k', negate n') = q
| otherwise = p
problem_95_v2 = longestChain 1000000 (Just 1000000)
Problem 96
Devise an algorithm for solving Su Doku puzzles.
See numerous solutions on the Sudoku page.
Problem 97
Find the last ten digits of the non-Mersenne prime: 28433 × 27830457 + 1.
Solution:
problem_97 = (28433 * 2^7830457 + 1) `mod` (10^10)
Problem 98
Investigating words, and their anagrams, which can represent square numbers.
Solution:
problem_98 = undefined
Problem 99
Which base/exponent pair in the file has the greatest numerical value?
Solution:
problem_99 = undefined
Problem 100
Finding the number of blue discs for which there is 50% chance of taking two blue.
Solution:
problem_100 = undefined