Answer:

sequence of partitions that result in a worst-case performance of the RANDOMIZED-SELECT algorithm for the given array A. Here is the sequence:

Start with the array A={3,2,9,0,7,5,4,8,6,1}.

The first call to RANDOMIZED-SELECT will choose a random pivot element, let's say it selects 9 as the pivot.

Partition the array based on the pivot, resulting in two subarrays:

Left subarray: {3,2,0,7,5,4,8,6,1}

Right subarray: {9}

Now, the algorithm needs to find the minimum element in the left subarray. It calls RANDOMIZED-SELECT recursively on the left subarray with p=1, r=9, and i=1.

In this recursive call, RANDOMIZED-SELECT again chooses a random pivot. Let's say it selects 8 as the pivot.

Partition the left subarray based on the pivot, resulting in two subarrays:

Left subarray: {3,2,0,7,5,4,6,1}

Right subarray: {8}

Continue this process recursively until the algorithm finally reaches the base case, where it's looking for the minimum element in a subarray of size 1.

This sequence of partitions ensures that the algorithm always chooses a pivot element that doesn't lead to a balanced partition. In the worst-case scenario, the pivot selected in each step ends up being the largest element in the current subarray, resulting in unbalanced partitions with only one element in the left subarray and all other elements in the right subarray.

Verified answer

Answer:

To describe a sequence of partitions that results in a worst-case performance of RANDOMIZED-SELECT for finding the minimum element in the array A={3,2,9,0,7,5,4,8,6,1}, we need to consider the specific behavior of the algorithm.

The RANDOMIZED-SELECT algorithm is based on the quicksort partitioning scheme. It randomly selects a pivot element and partitions the array around it, placing the pivot in its correct position. It then recursively applies the partitioning process to the subarrays on either side of the pivot until the desired element is found.

In the worst-case scenario for finding the minimum element, we want to ensure that the pivot is always selected as the maximum element in the current subarray. This will result in the maximum number of recursive calls, leading to poor performance.

One possible sequence of partitions that achieves this worst-case scenario is as follows:

1. Initial array: A={3,2,9,0,7,5,4,8,6,1}

2. First partition: Pivot = 9

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

3. Second partition: Pivot = 9 (maximum element in the subarray)

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

4. Third partition: Pivot = 9 (maximum element in the subarray)

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

5. Fourth partition: Pivot = 9 (maximum element in the subarray)

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

6. Fifth partition: Pivot = 9 (maximum element in the subarray)

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

7. Sixth partition: Pivot = 9 (maximum element in the subarray)

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

8. Seventh partition: Pivot = 9 (maximum element in the subarray)

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

9. Eighth partition: Pivot = 9 (maximum element in the subarray)

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

10. Ninth partition: Pivot = 9 (maximum element in the subarray)

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

11. Tenth partition: Pivot = 9 (maximum element in the subarray)

Partitioned array: A={0,2,1,3,7,5,4,8,6,9}

In this sequence of partitions, the pivot is always selected as the maximum element in the subarray. As a result, the algorithm will make ten recursive calls, each time reducing the size of the subarray by only one element. This worst-case scenario leads to inefficient performance.

It is important to note that this specific worst-case scenario is just one possibility, and there could be other sequences of partitions that also result in poor performance for finding the minimum element using RANDOMIZED-SELECT.