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Home Objects of Vectors Mergesort  
See also: Analysis of Mergesort  
Mergesort
In Section 13.7, we saw a simple sorting algorithm that turns out not to be very efficient. In order to sort n items, it has to traverse the vector n times, and each traversal takes an amount of time that is proportional to n. The total time, therefore, is proportional to n^{2}. In this section I will sketch a more efficient algorithm called mergesort. To sort n items, mergesort takes time proportional to n log n. That may not seem impressive, but as n gets big, the difference between n^{2} and n log n can be enormous. Try out a few values of n and see. The basic idea behind mergesort is this: if you have two subdecks, each of which has been sorted, it is easy (and fast) to merge them into a single, sorted deck. Try this out with a deck of cards:
The result should be a single sorted deck. Here's what this looks like in pseudocode: Deck merge (const Deck& d1, const Deck& d2) {// create a new deck big enough for all the cards Deck result (d1.cards.length() + d2.cards.length()); // use the index i to keep track of where we are in // the first deck, and the index j for the second deck int i = 0; int j = 0; // the index k traverses the result deck for (int k = 0; k<result.cards.length(); k++) { // if d1 is empty, d2 wins; if d2 is empty, d1 wins; // otherwise, compare the two cards // add the winner to the new deck } return result; } I chose to make merge a nonmember function because the two arguments are symmetric. The best way to test merge is to build and shuffle a deck, use subdeck to form two (small) hands, and then use the sort routine from the previous chapter to sort the two halves. Then you can pass the two halves to merge to see if it works. If you can get that working, try a simple implementation of mergeSort: Deck Deck::mergeSort () const {// find the midpoint of the deck // divide the deck into two subdecks // sort the subdecks using sort // merge the two halves and return the result } Notice that the current object is declared const because mergeSort does not modify it. Instead, it creates and returns a new Deck object. If you get that version working, the real fun begins! The magical thing about mergesort is that it is recursive. At the point where you sort the subdecks, why should you invoke the old, slow version of sort? Why not invoke the spiffy new mergeSort you are in the process of writing? Not only is that a good idea, it is necessary in order to achieve the performance advantage I promised. In order to make it work, though, you have to add a base case so that it doesn't recurse forever. A simple base case is a subdeck with 0 or 1 cards. If mergesort receives such a small subdeck, it can return it unmodified, since it is already sorted. The recursive version of mergesort should look something like this: Deck Deck::mergeSort (Deck deck) const {// if the deck is 0 or 1 cards, return it // find the midpoint of the deck // divide the deck into two subdecks // sort the subdecks using mergesort // merge the two halves and return the result } As usual, there are two ways to think about recursive programs: you can think through the entire flow of execution, or you can make the "leap of faith." I have deliberately constructed this example to encourage you to make the leap of faith. When you were using sort to sort the subdecks, you didn't feel compelled to follow the flow of execution, right? You just assumed that the sort function would work because you already debugged it. Well, all you did to make mergeSort recursive was replace one sort algorithm with another. There is no reason to read the program differently. Well, actually you have to give some thought to getting the base case right and making sure that you reach it eventually, but other than that, writing the recursive version should be no problem. Good luck!


Home Objects of Vectors Mergesort 