If f(n) = O(g(n)) and g(n) = O(h(n)) then h(n) = Ω(f(n)) 这个表述正确吗
时间: 2023-11-18 15:02:21 浏览: 31
根据引用\[2\]中的定义,如果f(n) = O(g(n))和g(n) = O(h(n)),那么意味着存在正常数c1和c2,以及n0,使得对于所有的n > n0,有0 ≤ f(n) ≤ c1 * g(n)和0 ≤ g(n) ≤ c2 * h(n)。根据这些不等式,我们可以得出结论:0 ≤ f(n) ≤ c1 * g(n) ≤ c1 * (c2 * h(n)) = (c1 * c2) * h(n)。因此,我们可以说h(n) = Ω(f(n))是正确的。
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相关问题
f(N) = log10(2N)
The function f(N) returns the base-10 logarithm of 2N. This means that it tells us what power of 10 we need to raise 2N to in order to get the result. For example, if N = 3, then 2N = 8, and f(3) = log10(2N) = log10(8) = 0.903. This means that 2N needs to be raised to the power of 0.903 in order to get the result.
In general, as N increases, so does 2N, and therefore f(N) increases as well. This is because the logarithm of a number grows slowly at first, but then increases rapidly as the number gets larger. For example, if we compare f(10) and f(100), we get:
f(10) = log10(2^10) = log10(1024) = 3.010
f(100) = log10(2^100) = log10(1.2676506 x 10^30) = 30.104
As we can see, the value of f(N) increases rapidly as N gets larger. This means that the function f(N) grows exponentially, but at a slower rate than 2N itself.
prove of disprove f(n) = O(g(n)) implies g(n) = O(f(n))
This statement is false.
To disprove it, we can provide a counterexample. Let f(n) = n and g(n) = n^2.
Then, f(n) = O(g(n)) because n <= n^2 for all n >= 1. However, g(n) = n^2 is not O(f(n)) because there does not exist a positive constant c such that n^2 <= c*n for all n >= 1.
Therefore, f(n) = O(g(n)) does not imply g(n) = O(f(n)).
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