The recurrence relation
的递归关系
T(n) = 2T(n/2) + n lg lg n
T(n) = 2T(n/2) + nlgn。
(where lg is logarithm to base 2) can be solved using the master theorem but I am not very sure about the answer. I have found my answer but am not mentioning it here in order to prevent information cascades. Please help me find the big O and Ω for above.
(lg是log以2为底的对数)可以用主定理求解，但我不太确定答案。我已经找到了我的答案，但我并没有在这里提到它，以防止信息级联。请帮我找到上面的大O和。
1 个解决方案
#1
3
None of the 3 cases in the master theorem apply for
主定理中的3个案例都不适用。
T(n)=2 T(n/2) + n log(log n)
(With arbitrary base, it doesn't really matter)
(带有任意的碱基，这并不重要)
Case 1: f(n)=n log(log n) is 'bigger' than nlog2 2=n1
案例1:f(n)=n log(log n)大于nlog2 =n1。
Case 2: f(n) does not fit n logk(n)
情形2:f(n)不符合n logk(n)
Case 3: f(n) is smaller than n1+e
情形3:f(n)小于n1+e。
U(n)=2 U(n/2) + n log n
L(n)=2 L(n/2) + n
You can show that: U(n) >= T(n) and L(n) <= T(n). So U gives a upper bound, and L a lower bound for T.
你可以证明:U(n) >= T(n)和L(n) <= T(n)。所以U给出了上界，而L是T的下界。
Applying the master theorem for U(n), gives
为U(n)应用主定理，给出。
Case 2: f(n)=n log n=Θ(n1 log1 n) thus U(n)=Θ(n log2 n)
案例2:f(n)= n o(log n)=Θ(n1 log1 n)因此U(n)=Θ(n log2 n)
Applying the master theorem for L(n), gives
给出了L(n)的主定理。
Case 2: f(n)=n =Θ(n1 log0 n) thus L(n)=Θ(n log n)
案例2:f(n)= n =Θ(n1 log0 n)因此L(n)=Θ(n o(log n))
Because L(n)<=T(n)<=U(n) it follows that T(n)=O(n log2 n) and T(n)=Ω(n log n)
因为L(n)< = T(n)< = U(n),T(n)= O(n log2 n)和T(n)=Ω(n O(log n))
Also, note that O(log2n)=O((log n)/log 2)=O((log n) * c)=O(log n).
另外，注意O(log2n)=O((log n)/log 2)=O((log n) * c)=O(log n)。
#1
3
None of the 3 cases in the master theorem apply for
主定理中的3个案例都不适用。
T(n)=2 T(n/2) + n log(log n)
(With arbitrary base, it doesn't really matter)
(带有任意的碱基，这并不重要)
Case 1: f(n)=n log(log n) is 'bigger' than nlog2 2=n1
案例1:f(n)=n log(log n)大于nlog2 =n1。
Case 2: f(n) does not fit n logk(n)
情形2:f(n)不符合n logk(n)
Case 3: f(n) is smaller than n1+e
情形3:f(n)小于n1+e。
U(n)=2 U(n/2) + n log n
L(n)=2 L(n/2) + n
You can show that: U(n) >= T(n) and L(n) <= T(n). So U gives a upper bound, and L a lower bound for T.
你可以证明:U(n) >= T(n)和L(n) <= T(n)。所以U给出了上界，而L是T的下界。
Applying the master theorem for U(n), gives
为U(n)应用主定理，给出。
Case 2: f(n)=n log n=Θ(n1 log1 n) thus U(n)=Θ(n log2 n)
案例2:f(n)= n o(log n)=Θ(n1 log1 n)因此U(n)=Θ(n log2 n)
Applying the master theorem for L(n), gives
给出了L(n)的主定理。
Case 2: f(n)=n =Θ(n1 log0 n) thus L(n)=Θ(n log n)
案例2:f(n)= n =Θ(n1 log0 n)因此L(n)=Θ(n o(log n))
Because L(n)<=T(n)<=U(n) it follows that T(n)=O(n log2 n) and T(n)=Ω(n log n)
因为L(n)< = T(n)< = U(n),T(n)= O(n log2 n)和T(n)=Ω(n O(log n))
Also, note that O(log2n)=O((log n)/log 2)=O((log n) * c)=O(log n).
另外，注意O(log2n)=O((log n)/log 2)=O((log n) * c)=O(log n)。