Question

Show that if \(d(n)\) is \(O(f(n))\) and \(e(n)\) is \(O(g(n)),\) then \(d(n)-e(n)\) is not necessarily \(O(f(n)-g(n))\)

Short answer

d(n) and e(n) can individually be O(f(n)) and O(g(n)), but d(n) - e(n) is not necessarily O(f(n) - g(n)).

Step-by-step solution

Understanding Big O Notation

Big O notation provides an upper bound on the growth rate of a function. If a function d(n) is O(f(n)), it means there exist positive constants C and n0 such that for all n ≥ n0, |d(n)| ≤ C|f(n)|. Similarly, if e(n) is O(g(n)), there exist positive constants C' and n0' such that for all n ≥ n0', |e(n)| ≤ C'|g(n)|.

Define the Problem

We need to show that even if d(n) = O(f(n)) and e(n) = O(g(n)), it does not necessarily follow that d(n) - e(n) is O(f(n) - g(n)).

Provide Counterexample

To show this, we can find a counterexample. Consider d(n) = n, e(n) = n, f(n) = n+1, and g(n) = 1. Clearly, d(n) and e(n) are both O(n) since |n| ≤ C|n| for some constant C. Additionally, e(n) is O(1) since |n| can be bounded by any sufficiently large constant multiple of 1. Similarly, f(n) = n + 1 can be considered O(n) since |n+1| ≤ C|n| for large n.

Examine the Difference

Now consider d(n) - e(n) and f(n) - g(n). We have d(n) - e(n) = n - n = 0 and f(n) - g(n) = (n + 1) - 1 = n. Clearly, 0 is not O(n) because it doesn't grow as n grows.

Conclusion

From the counterexample provided, we see that despite d(n) = O(f(n)) and e(n) = O(g(n)), it does not follow that d(n) - e(n) = O(f(n) - g(n)). Thus, the proposition is false.

Key concepts

Growth Rate

Understanding the growth rate of functions is crucial in analyzing algorithms. The growth rate tells us how fast a function increases as its input gets larger. When discussing the efficiency of algorithms, we commonly use Big O notation to describe the upper bound of this growth rate.

Functions can grow at different rates, such as constant (1\text{, which does not change), linear (n\text{, which grows proportionally to the input), quadratic (n^2\text{, which grows faster as the square of the input), and so on. Knowing these differences is key for comparing algorithms in terms of their efficiency.

In Big O notation, saying that a function d(n) is O(f(n)) means we're providing an upper bound on the growth rate of d(n). Specifically, there exist positive constants C and n0 such that for all n ≥ n0:

\(\big|d(n)\big| ot\text{ is less than or equal to} \)\text{C code.}}

This tells us that the function d(n) will not grow faster than a constant multiple of f(n) beyond some point n0.

Counterexample

A counterexample is a specific case that disproves a general statement or proposition. In our exercise, we need a counterexample to show that if d(n) = O(f(n)) and e(n) = O(g(n)), it does not imply that d(n) - e(n) is O(f(n) - g(n)).

Let's look at the provided counterexample:

1. Define d(n) = n
2. Define e(n) = n
3. Define f(n) = n + 1
4. Define g(n) = 1

Here, both d(n) and e(n) are O(n), as |n| is less than or equal to a constant multiple of |n|. Similarly, e(n) is also O(1) because |n| can be bounded by a constant. f(n) = n + 1 is O(n) since |n + 1| is less than or equal to a constant multiple of |n| for large n.

When we consider the difference, d(n) - e(n) results in 0, and f(n) - g(n) equals n. Clearly, 0 is not O(n), which shows that d(n) - e(n) is not necessarily O(f(n) - g(n)). This counterexample effectively disproves the proposition.

Upper Bound

In the context of Big O notation, the term 'upper bound' is used to describe the maximum growth rate of a function. It provides a way to categorize functions according to their growth rates, which is fundamental in analyzing the efficiency of algorithms.

When we say a function \(d(n)\) is O(f(n)), we are setting an upper bound on the growth rate of \(d(n)\). This means we are stating \(d(n)\) will not grow faster than a constant multiplied by \(f(n)\) beyond a certain point n0.

This is written mathematically as:

\(\big|d(n)\big| ≤ C\big|f(n)\big| ot\text{for all} n≥n0\)

Here, \(C\) and \(n0\) are positive constants. The upper bound helps us to anticipate the worst-case scenario for the given function as the input size \(n\) becomes large.

Providing an upper bound is crucial in computer science as it ensures we understand the limits of our algorithms’ performance. It helps in comparing algorithm efficiency and selecting the most appropriate one for a given problem, especially when dealing with large inputs.