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Efficiency of `times1p` for `[]` #9

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ppetr opened this Issue Dec 20, 2012 · 1 comment

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ppetr commented Dec 20, 2012

The instance Semigroup [a] uses the default implementation of times1p for replicating lists. I wonder, is there any advantage compared to the "standard" implementation

    times1p n x = let rep 0 = x
                      rep i = x ++ rep (i - 1)
                  in rep n

? Both have O(n) time overhead. But if the resulting list is consumed lazily, the default implementation of times1p has O(n * length x) memory overhead, while the "standard" variant has O(length x).

(Also head (times1p n x) takes O(1) time for the "standard" implementation but O(ln n) for the current default one. But I don't think this is important, since replicating a list to only read its head is probably something that never happens.)

@ekmett ekmett closed this in 194c8da Dec 20, 2012

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ekmett commented Dec 20, 2012

Done. Thanks!

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