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7.
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We got a little lucky.
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Let's pop you out and insert you right where you belong.
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And then 5, you belong a few shifts over.
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So let's shift three of you.
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And then, done.
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All right.
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So let's give all of you a stress ball.
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And maybe a round of applause.
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And let's see if we can't tease apart what just happened here.
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All right.
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And then you can exit stage left-- right if you'd like.
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All right, thank you.
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Derek, you're really racking them up here.
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All right.
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AUDIENCE: I'll [INAUDIBLE].
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SPEAKER 1: All right.
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Thank you.
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All right.
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Sure.
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So what was the point of this exercise?
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And how do we actually now get to a decision point where one of these
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is the way to go?
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So in bubble sort, recall that the key was to do something
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again and again until we did no swaps.
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At which point we could conclude that the list was sorted.
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So let's do exactly that, repeat until no swaps.
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And then, let's give ourselves a counter,
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just to make things a little more methodical,
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and say for i from 0 to n minus 1-- rather n minus 2.
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Why?
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Well, the end of the list recall is it n minus 1.
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Because if you have a list of size n, the beginning is at 0.
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The end is at n minus 1.
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And so one shy of the end is going to be n minus 2.
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So we want to do this up through the second to last element.
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If the i-th and the i-th plus 1 elements are out of order.
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So if left hand and right hand are out of order, swap them.
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And repeat this again and again and again.
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And thanks to that outer loop, we're going
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to do it again and again and again until we have no such swaps.
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And now to be clear, why this n minus 2?
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We want to make sure as we're sort of walking down our row of volunteers
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that my left hand does not actually point at the last human on stage
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because then, where would my right hand point?
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There's no person to beyond that point.
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So this then is bubble sort.
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Now, let's propose pseudocode for our other algorithm, selection sort,
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wherein we walked through the list iteratively
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trying to select on each pass the smallest element
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and putting it into its place.
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So for i from 0 to n minus 1, we're going
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to do this from beginning through the end,
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find the smallest element between i-th and n-th minus 1 element.
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Swap the smallest with that i-th element.
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In other words, walk through the list.
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Find the smallest, put it on the end.
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Walk through the list, swap it with the second to the end.
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Walk through the list, do the same, do the same, do the same,
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each time grabbing the then smallest element.
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And then lastly, we had insertion sort, whereby we iterated through the list,
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but we pretty much dealt with every element as we encountered it.
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So let me propose this pseudocode code here.
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For i from 1 to n minus 1.
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So let's just assume that the very beginning of the list
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is sort of trivially sorted, no matter what number he or she is.
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Because if that left side of the list is of size 1, it is, indeed, sorted.
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And indeed, let's call the 0-th element through the i minus [? 1-th ?] element,
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wherever we are, the sorted side.
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So essentially, call the left the sorted side.
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And you can think of the right-hand side as the unsorted side.
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Remove the i-th element, whatever element you're dealing with,
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and insert it into the sorted side in order.
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In other words, just walk through the list, one at a time,
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plucking out elements, and then forcibly insert them
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into the sorted side of the list, making room,
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as our humans did, for that list by shuffling everyone over.
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This then was bubble sort, selection sort, and insertion sort.
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But which to choose?
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So let's introduce now an answer to why.
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Not just what these things are, but why you might use one over the other.
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So the running time of an algorithm might
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be the number of seconds an algorithm takes to run,
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or the number of minutes, or the number of steps, or the number of comparisons.
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It doesn't really matter what your unit of measure
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is, so long as you're consistent, whatever feels the most
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expensive or useful to express.
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So let's consider an example, like bubble sort.
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If we want to consider the efficiency of bubble sort
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and make a claim as to whether we should or shouldn't
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use it because it's good or bad, let's consider it a little formulaically.
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So a little mathy, but pretty generically.
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So if there are n elements to store, and a computer scientist,
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recall from week 0, will generally just call n the size of his or her problem.
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n is the number of numbers to sort, or humans to sort.
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How many steps is it going to take to sort that many people?
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Well, when I had 8 music stands and 8 people here before,
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I considered a pair, a pair, a pair, a pair, a pair.
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And so if there were n humans up here.
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Or let's speak concretely, 8 people initially.
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