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What is the time complexity \(T(n)\) of the nested loops below? For simplicity, you may assume that \(n\) is a power of \(2 .\) That is, \(n=2^{k}\) for some positive integer \(k\) \(:\) \\[ \begin{array}{} \text { for }(i=1 ; i<=n, i++)\\} \\ \ \begin{array}{} j=n \\ \text { while }(j>=1)\\{ \end{array} \end{array} \\] < body of the while loop> \(\quad\) I/ Needs \(\Theta(1)\) \\[ j=\lfloor j / 2\rfloor \\] } }

Short Answer

Expert verified
The time complexity for the nested loops is \(O(n\log_2{n})\), where \(n\) is a power of 2.

Step by step solution

01

Recognize iteration process

Firstly, SEE that the outer loop runs from \(i = 1\) to \(n\) which gives it n iterations, hence its time complexity would be \(O(n)\). However, the inner loop halves the value of 'j' with each iteration, where 'j' starts at n. So, it runs until \(j >= 1\), which happens \(k+1\) times, where \(j = n = 2^k\), hence its time complexity would be \(O(\log_2{n})\) or \(O(k+1)\).
02

Calculate total time complexity

Understanding that the time complexity of the outer loop is \(O(n)\), and the time complexity of the inner loop is \(O(\log_2{n})\). Since the loops are nested, the time complexities multiply: \(O(n \cdot \log_2{n})\). As per question assumption, \(n = 2^k\), so \(O(n \cdot \log_2{n})\) is same as \(O(2^k \cdot k)\).
03

Final simplification

Apply logarithmic properties \(2^k = 2^{(\log_2{2^k})} = 2^{(k \cdot \log_2{2})} = n^{\log_2{2}} = n^1 = n\). So, overall time complexity simplifies to \(O(nk)\). However, as the function is represented in terms of \(n\), so writing \(k\) in terms of \(n\), we have \(k = \log_2{n} \). Hence, the time complexity of the given nested loop is \(O(n\log_2{n})\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nested Loops
Understanding nested loops is fundamental in analyzing how certain algorithms behave, especially when it comes to their running time. In essence, a nested loop occurs when one loop runs inside another. The outer loop triggers a new iteration of the inner loop each time through. As a simple example, if the outer loop runs 3 times, and the inner loop runs 3 times for each outer loop iteration, the overall number of iterations becomes 3 multiplied by 3, which is 9.

In the provided exercise, we have an outer 'for' loop and inside it, a 'while' loop. The particularity of the inner loop is its halving mechanism, which will cause it to have a logarithmic number of iterations relative to its starting value. This is different from a typical inner loop that might have a linear number of iterations. As a result, the time complexity will not be as simple as multiplying the two numbers of iterations, but will instead involve understanding the effects of operations inside the loop.
Algorithm Analysis
Algorithm analysis allows us to predict the resources that an algorithm requires. Mostly, we look at time (how long does it take to run?) and space (how much memory does it use?). However, time complexity tends to be the focus for many algorithms since efficient time usage often translates to improved performance.

Time complexity is usually expressed as a function of the size of the input, denoted as 'n'. When we analyze the time complexity of an algorithm, we're often interested in its growth rate as 'n' increases rather than its exact running time, which is why we use Big O notation to articulate our findings. In the nested loops from our exercise, we dissect each loop's performance and then consider how they interact together to understand the overall time complexity of the algorithm.
Big O Notation
Big O notation is the language we use to describe the complexity of an algorithm. It gives a high-level understanding of the running time or space requirement in relation to the size of the input data. It doesn't care about precision — only about what ultimately dominates the growth of the resource requirement.

Terms like 'O(n)', 'O(log n)', and 'O(n log n)' describe linear, logarithmic, and linearithmic complexities respectively. A linear complexity, denoted as 'O(n)', increases proportionally with the size of the input. Logarithmic complexity, 'O(log n)', suggests that the running time increases logarithmically relative to the input size—meaning it grows much slower compared to linear complexity. Linearithmic, 'O(n log n)', combines both linear and logarithmic growth, which suggests a multiplicative relationship between them, as seen in our exercise's nested loop analysis, where the inner loop's halving process introduces logarithmic behavior.

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Most popular questions from this chapter

Write an algorithm that finds the greatest common divisor of two integers.

Show the correctness of the following statements. (a) \(\lg n \in O(n)\) (b) \(n \in O(n \lg n)\) (c) \(n \lg n \in O\left(n^{2}\right)\) (d) \(2^{n} \in \Omega\left(5^{\ln n}\right)\) (c) \(\lg ^{3} n \in o\left(n^{a \cdot 5}\right)\)

Give an algorithm for the following problem. Given a list of \(n\) distinct positive integers, partition the list into two sublists, each of size \(n / 2,\) such that the difference between the sums of the integers in the two sublists is minimized. Determine the time complexity of your algorithm. You may assume that \(n\) is a multiple of 2

Algorithm 1.7 (nth Fibonacci Term, Iterative) is clearly linear in \(n,\) but is it a linear-time algorithm? In Section 1.3 .1 we defined the input size as the size of the input. In the case of the \(nth\) Fibonacci term, \(n\) is the input, and the number of bits it takes to encode \(n\) could be used as the input size. Using this measure the size of 64 is \(\lg 64=6,\) and the size of 1024 is \(\lg 1024=\) 10\. Show that Algorithm 1.7 is exponential-time in terms of its input size. Show further that any algorithm for computing the \(nth\) Fibonacci term must be an exponential-time algorithm because the size of the output is exponential in the input size. See Section 9.2 for a related discussion of the input size.

Explain in English what functions are in the following sets. (a) \(n^{C(1)}\) (b) \(O\left(n^{O(1)}\right)\) (c) \(O\left(O\left(n^{O(1)}\right)\right)\)

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