Respuesta :

The value of full positive integer k such that  [tex]O(n^k)[/tex] is the most restrictive polynomial-time upper bound of f(n) are as follows : (A) 9 (B) no integer exists (C)5/3(not an integer) (D)no integer exists (E) [tex]3e^{99}[/tex]

Using the Lower and Upper Bound Theory, it is possible to identify the algorithm with the lowest level of complexity. Let's quickly review what Lower and Upper bounds are before we can understand the theory.

  • If there are two constants C and N such that U(n) = C*g(n) for n > N, then g(n) is the Upper Bound of A. Let U(n) be the running time of an algorithm A(say). An algorithm's upper bound is displayed using the asymptotic notation Big Oh(O).
  • If there are two constants C and N such that L(n) >= C*g(n) for n > N, then g(n) is the Lower Bound of A. Let L(n) be the running time of an algorithm A(say). The asymptotic notation known as Big Omega displays an algorithm's lower bound.

A)[tex]2^{lg(3n+4n+5)}+lg {\spaceh} n \inO(n^k)[/tex]

or, [tex]O(2^{lgn^9}+lgn)\inO(n^k)[/tex]    since   [tex]3n^9+4n+5=O(n^9)[/tex]  

or, [tex]O(2^{lgn^9})\in O(n^k)[/tex] as [tex]2^m+lg m=O(2^m)[/tex]

Applying lg 2 on both sides we get:

[tex]lg_22^{lgn^9}=lg_2n^k[/tex]

solving we get:

k=9 × lg2 × lg n

At n=1/2

k=9

Hence the minimum integer for K is 9.

B) [tex]7lgn+13lg^5n\in O(n^k)[/tex]

Solving for k we get:

k=lgₙ(lg⁵n)

C) [tex]\sqrt[3]{7n^5+2n^3-4}\in O(n^k)[/tex]

Solving by exponent rule we get:

[tex]O(n^{5/3})\in O(n^k)[/tex]

k=5/3

D)[tex]13lg2^{3^n\inO(n^k)}[/tex]

or, [tex]k=lg_n(lg2^{3^n})[/tex]

E)[tex]7^{20!+1} \inO(n^k)[/tex]

As we don't have any term of n on the Left Hand Side, therefore no most restrictive polynomial-time upper bound exist for [tex]O(n^k)[/tex]

[tex]k=e^{99}[/tex]

To learn more about upper bound of a polynomial visit:

https://brainly.com/question/22965427

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