The question is: "What is the minimum number of leaves in the decision tree for a comparison sort? Use this observation to derive a lower bound on the number of comparisons performed by a comparison sort in the worst case."
In a comparison sort, the decision tree represents all the possible comparisons made between elements during the sorting process. Each leaf node in the decision tree represents a possible permutation of the input elements after sorting.
For 𝑛n distinct elements, there are 𝑛! possible permutations. Each permutation corresponds to a unique leaf in the decision tree. The minimum number of leaves in the decision tree for a comparison sort is equal to the number of permutations of 𝑛n distinct elements, which is 𝑛!.
Now, let's derive a lower bound on the number of comparisons performed by a comparison sort in the worst case based on this observation.
Suppose a comparison sort algorithm constructs a decision tree to sort 𝑛n distinct elements. In the worst-case scenario, the algorithm must traverse the entire decision tree to find the correct permutation (i.e., the sorted order). Since each leaf in the decision tree represents a unique permutation of the input elements, the worst-case number of comparisons performed by the algorithm is at least equal to the depth of the decision tree.
In a complete binary tree, the number of leaves is 2^d, where d is the depth of the tree. So, for n! leaves, we have:
Taking the logarithm base 2 of both sides:
Using Stirling's approximation, we have,
