[英]Why does the time complexity of the Master theorem differ in comparison to other recurrence relation solving methods?
[英]Solving a recurrence with exponential rule where master theorem does not apply
當您擴展公式時,我們將有:
T(n) = 3 log(n^{2/3}) + 3^2 log(n^((2/3)^2)) + ... + 3^k log(n^((2/3)^k)) + log(n)
在上面的等式中, k
是樹的高度。 如果我們假設n = 2 ^ ((3/2)^k)
,最后我們將在n^((2/3)^k)
有2
。 因此, k = log_{3/2)(log(n))
。 此外,我們知道log(n^a) = a log(n)
:
T(n) = 2 log(n) + 2^2 log(n) + ... + 2^k log(n) + log(n) =
log(n) (1 + 2 + 2^2 + ... + 2^k) =
(2^(k+1) - 1) log(n)
因此,作為2^k = O(log^2(n))
, T(n) = O(log^2(n) * log(n)) = \\O(log^3(n))
。
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