The pseudocode provided is an implementation of Bubble Sort. Let’s analyze the number of swap operations in the case where the array is initially sorted in descending order:
Step 1: Understanding the Loop Execution
The outer loop runs from \( i = 0 \) to \( n-2 \), where \( n = 30 \). Therefore, the outer loop executes \( 29 \) times.
The inner loop runs from \( j = 0 \) to \( n-i-2 \), meaning the number of iterations decreases as \( i \) increases.
Step 2: Calculating Total Comparisons
The number of comparisons for each value of \( i \) is:
- For \( i = 0 \), the inner loop runs \( n-1 \) times.
- For \( i = 1 \), the inner loop runs \( n-2 \) times.
- For \( i = 2 \), the inner loop runs \( n-3 \) times.
- …
- For \( i = n-2 \), the inner loop runs 1 time.
Thus, the total number of comparisons is the sum of the first \( n-1 \) integers:
\[ \text{Total comparisons} = (n-1) + (n-2) + \ldots + 1 = \frac{n(n-1)}{2} \] For \( n = 30 \):
\[ \text{Total comparisons} = \frac{30 \times 29}{2} = 435 \] Step 3: Number of Swaps
In Bubble Sort, a swap occurs every time a comparison finds that \( A[j] > A[j+1] \). Since the array is initially sorted in descending order, every comparison will result in a swap.
Thus, the number of swaps is equal to the total number of comparisons, which is 435.
Final Answer: The number of swap operations performed is 435.
Consider game trees Tree-1 and Tree-2 as shown. The first level is a MAX agent and the second level is a MIN agent. The value in the square node is the output of the utility function.
For what ranges of \( x \) and \( y \), the right child of node B and the right child of node E will be pruned by the alpha-beta pruning algorithm?
A disk of size 512M bytes is divided into blocks of 64K bytes. A file is stored in the disk using linked allocation. In linked allocation, each data block reserves 4 bytes to store the pointer to the next data block. The link part of the last data block contains a NULL pointer (also of 4 bytes). Suppose a file of 1M bytes needs to be stored in the disk. Assume, 1K = \(2^{10}\) and 1M = \(2^{20}\). The amount of space in bytes that will be wasted due to internal fragmentation is ___________. (Answer in integer)
Three villages P, Q, and R are located in such a way that the distance PQ = 13 km, QR = 14 km, and RP = 15 km, as shown in the figure. A straight road joins Q and R. It is proposed to connect P to this road QR by constructing another road. What is the minimum possible length (in km) of this connecting road?
Note: The figure shown is representative.