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我有两个疑问:-
1) 是 (log* n)^n = O((logn)!) 吗?
2) log(log* n) 或 log*(logn) 哪个更大?
对于 2),您有log*(log n) = log*(n)-1. 然后让m = log*(n). 你有m-1 > log(m)足够大的m.
log*(log n) = log*(n)-1
m = log*(n)
m-1 > log(m)
m
提示:
对于 1),让m = log*(n). 那么LHS是m^n,而RHS是指数塔高度的对数的m阶乘,即指数塔高度的阶乘m-1。
m^n
m-1
即使不考虑阶乘,指数塔的增长速度也应该比幂快得多。