[英]Find complexity of the following recurrence: T(n) = T(n/2) + log(n)
我如何估計兩個循環的運行時間,每個循環以對數時間運行,如下所示:
for(int i=n; i>=0; i /= 2)
{
for( int j=i; j>=0; j /= 2)
{
count++;
}
}
我的分析:
i = n , 1 => j = lg n | 1+lg (n)
i = n/2, 1 => j = lg n/2 | 1+lg (n/2)
i = 1 , 1 => j = 0 | 1
在這種情況下如何總結對數項?
i = n. | Inner loop runs log(n) iterations. | O(log(n))
i = n/2. | Inner loop runs log(n/2) iterations. | O(log(n))
i = n/4. | Inner loop runs log(n/4) iterations. | O(log(n))
. | |
. | |
i = log(n). | Inner loop runs log(log(n)) iterations. | O(log(log(n)))
我們注意到了什么? 我們每個n
都添加O(log(n))
這種重現是T(n) = T(n/2) + log(n) + 1
可以通過以下方式擴展:
T(n) = log(n) + log(n/2) + log(n/4) + ... + 1
= log(n) + [log(n)-1] + [log(n)-2] + [log(n) - 3] + .... + 1
= log(n)*log(n) - (2 + 3 + .... log(n))
= log(n)*log(n) - ([(2+log(n))*log(n)]/2)
= log(n)*log(n) - ~0.5log(n)*log(n)
= ~0.5log(n)*log(n)
T(n) = O(log(n)*log(n))
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