Question

We determined that the actual number of visits in the selection sort algorithm is $$ T(n)=\frac{1}{2} n^{2}+\frac{5}{2} n-3 $$ We characterized this function as having \(O\left(n^{2}\right)\) growth. Compute the actual ratios $$ \begin{aligned} &T(2,000) / T(1,000) \\ &T(4,000) / T(1,000) \\ &T(10,000) / T(1,000) \end{aligned} $$ and compare them with $$ \begin{aligned} &f(2,000) / f(1,000) \\ &f(4,000) / f(1,000) \\ &f(10,000) / f(1,000) \end{aligned} $$ where \(f(n)=n^{2}\)

Short answer

The ratios for \( T(n) \) are approximately 3.989, 15.94, and 99.57, showing \( T(n) \) is \( O(n^2) \).

Step-by-step solution

Calculate T(1,000)

First, substitute \( n = 1000 \) into the formula \( T(n) = \frac{1}{2} n^2 + \frac{5}{2} n - 3 \). Calculate:\[T(1000) = \frac{1}{2} (1000)^2 + \frac{5}{2} (1000) - 3\]Simplify:\[ T(1000) = 500,000 + 2,500 - 3 = 502,497\]

Calculate T(2,000)

Substitute \( n = 2000 \) into the function:\[T(2000) = \frac{1}{2} (2000)^2 + \frac{5}{2} (2000) - 3\]Simplify:\[T(2000) = 2,000,000 + 5,000 - 3 = 2,004,997\]

Calculate T(4,000)

Substitute \( n = 4000 \) into the function:\[T(4000) = \frac{1}{2} (4000)^2 + \frac{5}{2} (4000) - 3\]Simplify:\[T(4000) = 8,000,000 + 10,000 - 3 = 8,009,997\]

Calculate T(10,000)

Substitute \( n = 10000 \) into the function:\[T(10000) = \frac{1}{2} (10000)^2 + \frac{5}{2} (10000) - 3\]Simplify:\[T(10000) = 50,000,000 + 25,000 - 3 = 50,024,997\]

Calculate Ratio T(2,000) / T(1,000)

Using the computed values, evaluate the ratio:\[\frac{T(2000)}{T(1000)} = \frac{2,004,997}{502,497} \approx 3.989\]

Calculate Ratio T(4,000) / T(1,000)

Evaluate the following ratio:\[\frac{T(4000)}{T(1000)} = \frac{8,009,997}{502,497} \approx 15.94\]

Calculate Ratio T(10,000) / T(1,000)

Evaluate the following ratio:\[\frac{T(10000)}{T(1000)} = \frac{50,024,997}{502,497} \approx 99.57\]

Calculate f(n) at key points

Given \( f(n) = n^2 \), find: \[ f(1000) = 1000^2 = 1,000,000 \]\[ f(2000) = 2000^2 = 4,000,000 \]\[ f(4000) = 4000^2 = 16,000,000 \]\[ f(10000) = 10000^2 = 100,000,000 \]

Calculate Ratios f(n) / f(1,000)

Evaluate each target ratio:\[\frac{f(2000)}{f(1000)} = \frac{4,000,000}{1,000,000} = 4\]\[\frac{f(4000)}{f(1000)} = \frac{16,000,000}{1,000,000} = 16\]\[\frac{f(10000)}{f(1000)} = \frac{100,000,000}{1,000,000} = 100\]

Compare Ratios T(n) and f(n)

The ratios for \( T(n) \) are approximately 3.989, 15.94, and 99.57. These are close to the ratios for \( f(n) \), which are 4, 16, and 100, respectively, reflecting the \( O(n^2) \) complexity of \( T(n) \).

Key concepts

Selection Sort

Selection Sort is a simple comparison-based sorting algorithm. It works by repeatedly selecting the smallest (or largest, depending on sorting order) element from the unsorted portion of the list and moving it to the front or back of the sorted portion. This process is continued until the entire list is sorted.

Here's how Selection Sort operates in simple terms:

• The algorithm maintains two subarrays: one already sorted and one unsorted.
• Initially, the entire array is unsorted.
• The smallest element is selected from the unsorted section and is swapped with the first unsorted element, thereby growing the sorted subarray.
• This process is repeated, moving from the front to the end until the whole list is sorted.

Due to its straightforward process, Selection Sort is easy to understand and implement. However, it might not be the most efficient for large datasets due to its quadratic time complexity.

Time Complexity

Time Complexity is a concept that helps us understand how the performance of an algorithm changes as the size of the input data grows. It describes the speed of an algorithm in terms of the input size, usually denoted by the variable \( n \).

Time Complexity is categorized into different classes like:

• Constant Time \( O(1) \)
• Logarithmic Time \( O(\log n) \)
• Linear Time \( O(n) \)
• Quadratic Time \( O(n^2) \)
• Exponential Time \( O(2^n) \)

For example, when discussing Selection Sort, its time complexity is \( O(n^2) \), meaning that as the input size \( n \) increases, the processing time grows quadratically. This is due to the nested loops in the algorithm where each element is compared with every other element.

Big O Notation

Big O Notation provides a high-level understanding of the worst-case scenario for an algorithm's runtime or space requirement. It is a mathematical notation that describes the upper bound of an algorithm's complexity.

When we say a sorting algorithm has a complexity of \( O(n^2) \), we're indicating that the maximum time it might take to run is proportional to the square of the number of items being worked with. This upper bound offers a valuable benchmark for comparing the efficiency of different algorithms, especially as datasets grow large.

Examples of Big O complexities are:

• \( O(1) \): constant time regardless of input size.
• \( O(n) \): linear time in relation to input size.
• \( O(n^2) \): quadratic time, where time taken grows as a square function of input size.

Using Big O, you can predict scalability and help in optimizing your code.

Quadratic Growth

Quadratic Growth happens when the time taken by an algorithm is proportional to the square of the input size. This is often observed in algorithms with nested loops where each pass itself is another loop through the input data.

Consider Selection Sort: each element needs to be compared with every other element, leading to a time complexity of \( O(n^2) \). As the input size doubles, the required processing power can quadruple, making this approach less efficient for large data volumes.

To illustrate this, let's look at the growth comparison in our problem, where we calculated:

• The ratios for Selection Sort, \( T(n) \) as 3.989, 15.94, and 99.57.
• For quadratic function \( f(n) = n^2 \), with ratios of 4, 16, and 100.

Both comparisons grew near these ratios as expected, affirming the quadratic nature of Selection Sort. This demonstrates how quadratic growth might be acceptable for smaller inputs but can drastically increase processing time for larger datasets.