Chapter 5: Problem 16
Prove or find a counterexample. If \(\mathbf{O}(G)=\mathbf{O}(F)\) and
\(\mathbf{O}(H) \subset \mathbf{O}(F)\), then for all but finitely many \(n \in
\mathbb{N}, H(n)
Short Answer
Expert verified
The statement is true: \(H(n) < G(n)\) for all but finitely many \(n\).
Step by step solution
01
Understand the Given Condition
We are given that \(\mathbf{O}(G) = \mathbf{O}(F)\), which implies that the growth rate of the function \(G(n)\) is asymptotically equivalent to the growth rate of \(F(n)\). Essentially, for very large values of \(n\), G(n) behaves similarly to F(n). Additionally, \(\mathbf{O}(H) \subset \mathbf{O}(F)\) means that the growth rate of \(H(n)\) is strictly less than the growth rate of \(F(n)\), and hence \(G(n)\) as well.
02
Interpret Asymptotic Notations
If \(H(n)\) is \(\mathbf{O}(F)\), it implies that \(H(n) = o(F(n))\) which means \(\lim_{{n \to \infty}} \frac{H(n)}{F(n)} = 0\). Similarly, since \(G(n)\) is \(\mathbf{O}(F)\), we know it grows at the same rate as \(F(n)\).
03
Compare Growth Rates for Large n
Because \(H(n) = o(F(n))\), for all but finitely many values of \(n\), \(H(n) < cF(n)\) for any \(c > 0\). Since \(G(n)\) behaves as \(F(n)\), we can write \(G(n) = dF(n)\) where \(d > 0\). Hence for large \(n\), \(H(n) < G(n)\).
04
Formulate Conclusion
Given the asymptotic behaviors derived, it follows that for all but finitely many \(n\), the inequality \(H(n) < G(n)\) must hold true due to the faster growth rate of \(G(n)\) compared to \(H(n)\). Therefore, the initial statement is a valid conclusion.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Growth Rate Comparison
Understanding growth rates is central to analyzing algorithms. Growth rates help us compare how functions behave as inputs become large.
This comparison is crucially useful in determining the efficiency and performance of algorithms. In practice:
Essentially, understanding these growth rates and their comparisons assists in predicting how efficiently algorithms can solve problems as the size of the input grows.
This comparison is crucially useful in determining the efficiency and performance of algorithms. In practice:
- If two functions, say \(f(n)\) and \(g(n)\), have the same growth rate, their complexity is similar for very large inputs.
- When functions grow at different rates, the one with a higher growth rate will eventually overpower the other as inputs become sufficiently large.
Essentially, understanding these growth rates and their comparisons assists in predicting how efficiently algorithms can solve problems as the size of the input grows.
Big O Notation
Big O notation provides a way to describe the upper bound of a function's growth rate. It becomes especially valuable for understanding and managing the time and space complexity of algorithms.
These relationships are pivotal in making informed decisions in computer science, particularly concerning which algorithm to implement.
- When we express a function \(G(n)\) to be \(\mathbf{O}(F(n))\), we are stating that for sufficiently large \(n\), \(G(n)\) does not grow faster than a constant multiple of \(F(n)\).
- If \(\mathbf{O}(G) = \mathbf{O}(F)\), it implies both functions, for all practical purposes, have the same asymptotic behavior as \(n\) becomes large.
These relationships are pivotal in making informed decisions in computer science, particularly concerning which algorithm to implement.
Limit of Functions
Limits help us understand the behavior of functions as the input value approaches a particular point, often infinity. In asymptotic notation, limits can clarify the relative growth of different functions.
- For example, when \(\lim_{{n \to \infty}} \frac{H(n)}{F(n)} = 0\), it indicates that \(H(n)\) grows much slower than \(F(n)\) as \(n\) becomes very large.
- Similarly, a situation where \(\lim_{{n \to \infty}} \frac{G(n)}{F(n)} = d\), with \(d > 0\), suggests \(G(n)\) grows steadily at the pace of \(F(n)\).