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So we've been talking about insertion sorting and selection sorting, and
we've been analyzing their performance.
And in this video we're going to talk about yet another algorithm
to accomplish the exact same task, which is sorting a collection of data.
And so by the end of this video, you'll be able to describe this algorithm,
the merge sort algorithm.
And the reason we're talking about it is because it's a very
different kind of algorithm from the ones that we've looked at so far.
And in particular, it uses recursion.
So you'll be able to explain how this particular algorithm uses recursion, and
then discuss its performance.
So let's start with a description of the algorithm and
how it goes about sorting a collection of data.
So we're presented with some input data and we think of it as a list.
And let's just think about the easiest case we might have.
And in the easiest case that list will only have one element.
And so if our list has just one element then really we don't need to do anything.
We can return.
Because just one element is already sorted.
There's nothing that could be out of place so
it's already all organized in ascending order.
Okay, so if we're in that special base case, the small case.
Then we're done, but if not then we actually have to do some work and
the merge sort algorithm says well,
I've got a big collection of data that I need to organize from smallest to biggest.
And that seems really hard so why don't I make my life a little bit easier and
just cut the list right in half.
And then I'll focus on say the first have of the collections of data and
say can I organize that piece.
Okay, maybe that's a slightly easier task because I've got fewer elements there.
And so I'll think about how to do that in a minute.
But, suppose I was able to organize that first half of the list.
Then I go ahead and say, well what about the second half of the list?
Can I organize that?
Can I sort it in order from smallest to biggest?
And again, that's a smaller collection than what I started with so
maybe it's a little bit easier.
And so once I do these two steps,
then I've got two smaller lists that are each sorted, and now what
remains is to somehow combine those two lists while maintaining the order.
And so I'd wanna merge the two sorted lists and then output the result.
And so we really haven't done any comparison of elements yet,
in this description.
But what we've done is described a high level strategy of how we might sort
of big list by dividing the problem into something a little bit smaller.
Sort the first half, sort the second half, and then merge the results
while preserving the order of the two smaller lists that are sorted.
Okay, so that seems like a plausible strategy, when we have a big problem,
break it down, divide and conquer.
But we've really kind of punted, we've postponed our solution for
how do we actually go ahead and sort these two smaller sublists?
We haven't said how we're going to do that, and
this is when something really powerful comes into play.
What we're doing here is using the power of recursion.
And what recursion is all about is breaking a big problem down
into a smaller subproblem and
then a manageable amount of work to incrementally change what we need to do.
So we had a big list to begin with, and we're going to
think about solving the problem on the big list as solving smaller,
similar subproblems and then combining the results.
And what's really helping us out here is that once was broken down the problem
into smaller and smaller and smaller list, eventually we know what to do because
eventually, if we get down to just a single element, then that list has sorted.
Okay, so, this all kinda feels a little bit abstract.
So, let's work through an example and see how it all plays out.
So let's think about a five element list.
And think about the numbers, the integers between one and
five that we've messed around a bit, and so they're not in order anymore.
And so we'd like to use the merge sort algorithm to rearrange the numbers in this
list and put them in order.
So, according to our merge sort algorithm, we first of all observe that this list has
more than one element, so we're not in the base case.
We have to do something, and
so we're going to divide the list in half and then sort the two halves.
So we start by dividing the list in half and so we've got the first half.
Well, half of 5 is two and a half and so we round down to two elements, and
then the rest of the list is in the second half.
And now we need to sort that first half, but sorting means using merge sort, again,
using our overall strategy, and so that means we start at the beginning and
we check, does the list that contained 5 and 3 have just one element?
No, it's got two.
So we need to divide that list in half again.
And so the list that contains 5 and 3 gets divided down into a list that
just contains 5, and a list that just contains 3.
And similarly the list that contained 2, 4, and 1, gets divided down into halves.
One that contains just the element 2, and then the second part of that list is
the list that contains the elements 4 and 1.
Okay and now what we want to do is
sort each of those halves that we got out of dividing our lists.
And so we look at that first half of the list, and that just has the element 5.
And so when we reach down to this base case of a list with just one element,
then that one list is already sorted so we can return.
And so when we've sorted the list five, well we don't need to do anything.
So the yellow shading indicates that that list is sorted.
And similarly the list that just contains 3 and the list that just contains 2.
When we look at the list it contains 4 and 1 and we try to apply merge sort to that
list in order to sort that half so then we can merge all the way back.
We notice that it has more than one element so
we need to divide it in half and then we'll be sorting each of it's halves,
but they're just one element list so they're already sorted.
Okay, so what we've done at this point is just broken down our problem over and
over again to narrow into the smallest sub problems that we can actually manage.
And so we've started with a big list and we've divided, and divided, and
divided in half until we get a whole bunch of lists that each have one element.
And so those one element lists are sorted.
But now we have to do that crucial step of merging the sorted lists
while preserving the order.
So let's think about what that happens.
We want to merge the lists 5 and 3,
and that means we're going to combine their elements.
The resulting list is gonna have two elements, but we want to do it in order,
so we choose to pick off the elements from the top of the list, or
the head of the list.
Based on which one is smaller.
And so we're going to start with the 3 because it's smaller, and
afterwards put in the 5.
And notice that what we got as a result is a sorted list, but it's a little bit
bigger than what we had before, and so it's bringing us one step closer to our
result, which is that sorted, five element array that we're looking for.
Okay, so we've combined two halves of a list that were part of our
recursive descent into the smallest sub problems, and now we keep on going.
So the 1 and the 4 get combined to a two element list and
then we add in that list two.
And notice that when we merge the list two with the list one four,
what's going to happen is that the 2 will get inserted between the 1 and the 4.
Because we're comparing the heads of the list at each point, and
making sure to merge those lists so that they preserve the order.
And so then what we have is a two element list on one side that's sorted.
And a three element list on the other side, also sorted.
And we're going to merge those.
And what we'll get is exactly the sorted array 1, 2, 3, 4, 5 that we want.
Okay, so this is a bit of a quick and
dirty high level description of the merge sort algorithm.
I encourage you to try implementing it.
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It's very, it feels really good to see it in action in front of you.
But, it is a bit tricky because it is using this principle of recursion.
And so, what we're focusing on in this module of the course isn't so
much an in-depth analysis of recursion.
But rather an illustration of the different possible
algorithms that are out there.
For solving the same problem,
and then thinking about how that translates to different performance.
So, let's think about that performance.
Let's think about how we might analyze how long it takes merged sort to run.
How many operations are required.
And so if we think about this algorithm,
really at the beginning all we were doing is saying I don't want to sort yet.
I don't want to sort yet,
I'm just going to define my problem and smaller subproblems.
And so we were organizing our data until the step where we had to merge.
That's where we starting doing comparisons between list elements and
really putting them together.
And at each level of the recursion once we started merging the sorted lists,
what we had to do was compare, pair-wise, basically all the elements.
But only once, each of them, were we comparing it with another element.
And so if we count how many comparisons are required there, it's O(n).
Where n, remember, is our input size, the size of the list that we started with.
So really where we were doing a lot of work was at a single level, merging
all those smaller lists together and figuring out how to maintain the order.
Okay. So then the question is, how many times
did we have to do this big merge step, and the number of times that we have to do
the merge step depended upon how many times we had to divide the lists in half.
And to get them into smaller and smaller chunks so that we could sort them.
Okay, so each time that we divide, we got two smaller lists.
So if we started with a list of size five, when we divided we got
lists of size two and three and then smaller and smaller and smaller.
And we got to stop.
When all our lists were of size 1.
So we had to keep dividing until all of our lists were of size 1.
And so we're asking ourselves how many times do we have to keep dividing by two.
If we start at n, I want to get all the way to 1.
But that's exactly the logarithm base, too.
So when you start at a number and you wanna keep dividing by 2, dividing by 2,
dividing by 2, and count how many times we have to do that until we get to 1,
that's log base 2.
And so what we're seeing here is that we have log base 2 of
n many times that we had to break down our problem into sub-problems, and
for each one of those, we had to do a merge step.
So, each one of those recursion levels
we had about O(n) work to merge all of the little pieces of lists.
And that repeated log n times.
And so our total performance is n times log n, or O(n log n).
Christine AlvaradoAssociate Teaching Professor
Mia MinnesAssistant Teaching Professor
Leo PorterAssistant Teaching Professor
