It doesn't have to be a top level definition, it works anyway. -- Lennart Bruno Abdon wrote:
'hamming', in your code, is a top-level definition. When used three times inside its own definition, it's the same variable being used three times. You don't recompute a variable value in order to reuse it.
As an example, if you do
foo :: [Integer] foo = [1,2,3] + [4,5]
bar = foo ++ foo ++ foo
the concatenation used to produce foo will not be done three times in order to calculate the value of bar. That would be true for any function would foo be defined upon, not only concatenation.
Bruno Abdon
On Mon, 24 Jan 2005 10:38:35 +0100, Francis Girard <francis.girard@free.fr> wrote:
Hi,
The classical Hamming problem have the following solution in Haskell :
*** BEGIN SNAP -- hamming.hs
-- Merges two infinite lists merge :: (Ord a) => [a] -> [a] -> [a] merge (x:xs)(y:ys) | x == y = x : merge xs ys | x < y = x : merge xs (y:ys) | otherwise = y : merge (x:xs) ys
-- Lazily produce the hamming sequence hamming :: [Integer] hamming = 1 : merge (map (2*) hamming) (merge (map (3*) hamming) (map (5*) hamming)) *** END SNAP
I just love these algorithms that run after their tail (they make my brain melt) but I don't know how is it that they are efficient.
Here, the hamming recursively calls itself three times. For this algorithm to be efficient, the Haskell system, somehow, has to "remember" the already generated sequence THROUGH RECURSION (i.e. not only intermediate "local" results) otherwise it would end up regenerating the beginning of the sequence over and over again.
Obviously, Haskell does remember what had already been generated THROUGH RECURSION since executing the program with GHCI runs quite smoothly and responsively.
That Haskell manages to do that is for me "magnifique". But I need to know (if only a little) about how it achieves this in order to know what I, as a lambda programmer, can do, and how to compute the Big-Oh complexity of the algorithm.
Thank you,
Francis Girard FRANCE
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