35 lines
2.2 KiB
Haskell
35 lines
2.2 KiB
Haskell
-- directly find n-th Hamming number, in ~ O(n^{2/3}) time.
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-- based on "top band" idea by Louis Klauder, from the DDJ discussion.
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-- by Will Ness, original post: drdobbs.com/blogs/architecture-and-design/228700538
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import Data.List (sortBy, foldl') -- '
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import Data.Function (on)
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main = let (r,t) = nthHam 1000000 in print t >> print (trival t)
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trival (i,j,k) = 2^i * 3^j * 5^k
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nthHam :: Int -> (Double, (Int, Int, Int)) -- ( 64bit: use Int!!! NB! )
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nthHam n -- n: 1-based: 1,2,3...
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| n <= 0 = error $ "n is 1--based: must be n > 0: " ++ show n
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| n < 2 = ( 0.0, (0, 0, 0) ) -- trivial case so estimation works for rest
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| w >= 1 = error $ "Breach of contract: (w < 1): " ++ show w
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| m < 0 = error $ "Not enough triples generated: " ++ show ((c,n) :: (Int, Int))
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| m >= nb = error $ "Generated band is too narrow: " ++ show (m,nb)
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| otherwise = sortBy (flip compare `on` fst) b !! m -- m-th from top in sorted band
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where
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lb3 = logBase 2 3; lb5 = logBase 2 5; lb30_2 = logBase 2 30 / 2
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v = (6*lb3*lb5* fromIntegral n)**(1/3) - lb30_2 -- estimated logval, base 2
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estval n = (v + (1/v), 2/v) -- the space tweak! (thx, GBG!)
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(hi,w) = estval n -- hi > logval > hi-w
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m = fromIntegral (c - n) -- target index, from top
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nb = length (b :: [(Double, (Int, Int, Int))]) -- length of the band
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(c,b) = foldl_ (\(c,b) (i,t)-> let c2=c+i in c2 `seq` -- ( total count, the band )
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case t of []-> (c2,b);[v]->(c2,v:b) ) (0,[]) -- ( =~= mconcat )
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[ ( fromIntegral i+1, -- total triples w/ this (j,k)
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[ (r,(i,j,k)) | frac < w ] ) -- store it, if inside band
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| k <- [ 0 .. floor ( hi /lb5) ], let p = fromIntegral k*lb5,
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j <- [ 0 .. floor ((hi-p)/lb3) ], let q = fromIntegral j*lb3 + p,
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let (i,frac) = pr (hi-q) ; r = hi - frac -- r = i + q
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] where pr = properFraction -- pr 1.24 => (1,0.24)
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foldl_ = foldl'
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