# Problem 8.6

## Lower bound on merging sorted lists

The problem of merging two sorted lists arises frequently. We have seen a procedure for it as the subroutine

`MERGE`

in Section 2.3.1. In this problem, we will prove a lower bound of $2n - 1$ on the worst-case number of comparisons required to merge two sorted lists, each containing $n$ items.First we will show a lower bound of $2n - o(n)$ comparisons by using a decision tree.

- Given $2n$ numbers, compute the number of possible ways to divide them into two sorted lists, each with $n$ numbers.
- Using a decision tree and your answer to part (a), show that any algorithm that correctly merges two sorted lists must perform $2n - o(n)$ comparisons.
Now we will show a slightly tighter $2n - 1$ bound.

- Show that if two elements are consecutive in the sorted order and from different lists, then they must be compared.
- Use your answers to the previous part to show a lower bound of $2n - 1$ comparisons for merging two sorted lists.

### The first bound

This can be reduced to a counting problem: in how many ways can we pick $n$ items out of $2n$ items? That way we pick the first sorted list and each such pick determines a single, unique second list. In exercise C.1-13 we already proved that:

$$ \binom{2n}{n} = \frac{2^{2n}}{\sqrt{\pi n}}(1 + \O(1/n)) $$

Using the familiar reasoning, if there are $l$ leaves in the decision tree and its height is $h$, the following inequality must hold:

$$ \frac{2^{2n}}{\sqrt{\pi n}}(1 + \O(1/n)) \le l \le 2^h $$

Taking the logarithm of the leftmost and rightmost parts (and inverting their positions) we get;

$$ \begin{aligned} h & \ge \lg\bigg(\frac{2^{2n}}{\sqrt{\pi n}}(1 + \O(1/n))\bigg) \\ & = \lg{2^{2n}} - \lg\sqrt{\pi n} + \lg(1 + O(1/n)) \\ & = 2n - \o(n) \end{aligned} $$

### The tighter bound

If we don't compare two consecutive elements, we don't know which one comes first. They are completely indistinguishable when comparing to the other elements. We have no way in determining which one should come first. (Note that if they were in the same sorted list, we don't need to compare them, since we already have that information).

If the sorted order of the elements is $\langle a_1, b_2, a_2, b_2, \ldots, a_n, b_n \rangle$ and we have the two lists $\langle a_1, a_2, \ldots, a_n \rangle$ and $\langle b_1, b_2, \ldots, b_n \rangle$, then there are $2n - 1$ pairs of elements that need to be compared. Any algorithm that handles this case must perform $2n - 1$ comparisons in the worst case.