So in this video and the next, we're going to study a very cool divide and conquer
algorithm for the closest pair problem. this is a problem where you're given n
points in the plane and you want to figure out which pair of points are closest to
each other. So this would be the first taste we get of an application in
computational geometry, which is the part of algorithms which studies how to reason
and manipulate geometric objects. So those algorithms are important in, among other
areas robotics, computer vision and computer graphics. So this is relatively
advanced material, it's a bit more difficult than the other applications of
divide and conquer that we've seen. The algorithm's a little bit tricky and it has
a quite nontrivial proof of correctness, so just be ready for that and also be
warned that because it's more advanced I'm going to talk about the material in at a
slightly faster pace tha I do in most of the other videos. So let's begin now by
defining the problem formally, so we're given as imput endpoints in the plane, so
each one just define by its x coordinate and ist y coordinate. And when we talk
about the distance between two points in this problem, we're going to focus on
Euclidean distance. So, let me remind you what that is briefly, but we're going to
introduce some simple notation for that, which we'll use for the rest of the
lecture. So we're just going to note the Euclidean distance between two points, pi
and pj, by d of pi pj. So in terms of the x and y coordinates of these two points,
we just look at the squared differences in each coordinate, sum them up, and take the
square root. And now as the name of the problem would suggest, the goal is to
identify among all pairs of points that pair which has the smallest distance
between them. Next, let's start getting a feel for the
problem by making some preliminary observations. First, I want to make an
assumption purely for convenience that there's no ties. So that is I'm going to
assume all endpoints have distinct x coordinat es, and also all endpoints have
distinct y coordinates. It's not difficult to extend the algorithm
to accommodate ties. I'll leave it to you to think about how to do that. So next,
let's draw some parallels with the problem of counting inversions, which was a
earlier application of divide and conquer that we saw. The first parallel I want,
want to out is that, if we're comfortable with the quadratic time algorithm, then
this is not a hard problem, we can simply solve this by brute-force search. And
again, by brute-force search, I just mean we set up a double for loop, which
iterates over all distinct pairs of points. We compute the distance for each
such pair and we remember the smallest. That's clearly a correct algorithm, it has
to iterate over a quadratic number of pairs, so its running time is going to be
theta of n squared. And, as always, the question is can we apply some algorithmic
ingenuity to do better? Can we have a better algorithm than this naive one which
iterates over all pairs of points? You might have a, an initial instinct that
because the problem asks about a quadratic number of different objects, perhaps we
fundamentally need to do quadratic work. But again, recall back in counting
inversions, using divide and conquer, we were able to get an n log n algorithm
despite the fact that there might be as many as a quadratic number of inversions
in an array. So the question is, can we do something similar here for the closest
pair problem? Now, one of the keys to getting an n log n time algorithm for
counting inversions was to leverage a sorting subroutine. Recall that we
piggybacked on merge sort to count the number of inversions in n log n time. So
the question is, here, with the closest pair problem, perhaps, sorting again can
be useful in some way to beat the quadratic barrier. So, to develop some
evidence that sorting will indeed help us compute the closest pair of points
embedded in quadratic time, let's look at a special case of the problem, really, an
easier version of t he problem, which is when the points are just in one dimension,
so on the line rather that in two dimensions in the plane. So in the 1D
version, all the points just lie on a line like this one, and we're given the points
in some arbitrary order not necessarily in sorted order. So, a way to solve the
closest pair problem in one dimension, is to simply sort the points, and then of
course, the closest pair better be adjacent in this ordering, so you just
iterate through the n minus 1 consecutive pairs and see which one is closest to each
other So, more formally, here's how you solve the one-dimensional version of the
problem. You sort the points according to their only coordinate, because you're
going to remember, this is one dimension. So as we've seen, using merge sort, we can
sort the points in n log n time and then we just do a scan through the points, so
this takes linear time. And for each consecutive pair, we compute their
distance and we remember the smallest of those consecutive pairs and we return
that. That's gotta be the closest pair. So, in this picture here on the right, I'm
just going to circle here in green the closest pair of points. So this is
something we discover by sorting and then doing a linear scan. Now, needless to say,
this isn't directly useful, this is not the problem I started out with. We wanted
to find out the closest pair among of points in the plane not points in the
line. But, I want to point out that, this, even in the line, there are a quadratic
number of different pairs, so brute-force search is still a quadratic time algorythm
even in the 1D case. So at least, with one dimension, we can use sorting, piggyback
on it, to beat the naive brute-force search bound and solve the problem in n
log n time. So our goal for this lecture is going to be to devise an equally good
algorithm for the two-dimensional case, so we want to solve closest pair of points in
the plane, again, in n log n, n time. So we will succeed in this goal. I'm going
to show you an n log n time algo rithm for 2D closest pair. It's going to take us a
couple steps. So let me begin with a high level approach. Alright. So the first I
need to try is just to copy what worked for us in the one-dimensional case. So the
one-dimensional case, we first sorted the points by their coordinate and that was
really useful. Now, in the 2D case, points have two coordinates, x coordinates and y
coordinates, so there's two ways to sort them. So let's just sort them both ways,
that is, the first step of our algorithm, which you should really think of as a
preprocessing step. We're going to take the input points. We invoke merge sort
once to sort them according to x coordinate, that's one copy of the points.
And then we make a second copy of the points where they're sorted by y
coordinates. So we're going to call those copies of points px, that's an array of
the points sorted by x coordinate, and py for them sorted by y coordinate. Now, we
know merge short takes n log n times, so this preprocessing step only takes o of n
log n time. And again, given that we're shooting for an algorithm with running
time big O of n log n, why not sort the points? We don't even know how we're going
to use this fact right now, but it's sort of harmless. It's not going to effect our
goal of getting a big of O n log n time algorithm. And indeed, this illustrates a
broader point, which is one of the themes of this course. So recall, I hope one of
the things you take away from this course is a sense for what are the four free
primitives, what are manipulations or operations you can do on data which
basically are costless. Meaning that if your data set fits in the main memory of
your computer, you can basically invoke the primitive and it's just going to run
blazingly fast, and you can just do it even if you don't know why. And again,
sorting is the canonical for free primitive, although, we'll see some more
later in the course and so, here, we're using exactly that principle.
So we don't even understand why yet we might wa nt the points to be sorted. It
just seems like it's probably going to be useful, motivated by the 1D case, so let's
go ahead and make assorted copies of the points by x and y coordinate upfront. So
reasoning by analogy with the 1D suggests that sorting the points might be useful,
but we can't carry this analogy too far. So in particular, we're not going to be
able to get away with just a simple linear scan through these arrays to identify the
closest pair of points. So, to see that, consider the following example.
So we're going to look at a point set which has six points. There's going to be
two points, which I'll put in blue which are very close in x coordinate, but very
far away in y coordinate. And then there's going to be another pair of points which
I'll do in green, which are very close in y coordinate, but very far away in x
coordinate. And then there's going to be a red pair of points, which are not too far
away in either the x coordinate or the y coordinate. So in this set of six points,
the closest pair is the pair of red points. They're not even going to show up
consecutively on either of the two arrays, right? So in the array that's sorted by x
coordinate, this blue point here is going to be wedged in between the two red
points, they won't be consecutive. And similarly in the, in py, which is sort of
by y coordinate, this green coordinate is going to be wedged between the two red
points. So you won't even notice these red point if you just do a linear scan if your
px and py, or py look at the consecutive pairs of points. So, following our
preprocessing step where we just invert, invoke merge sort twice we're going to do
a quite nontrivial divide and conquer algorithm to compute the closest pair. So
really, in this algorithm, we're applying the divide and conquer algorithm twice.
First, internal to the sorting subroutine, assuming that we use the merge sort
algorithm to sort. Divide and conquer is being used there to get an n log n running
time in this preprocessing step, and the n, we're going to use it again on sorted
arrays in a new way and that's what I'm going to tell you about next. So let's
just briefly review the divide and conquer algorithm design paradigm before we apply
it to the closest pair problem. So, as usual, the first step is to figure out a
way to divide your problem into smaller subproblems. Sometimes this has a
reasonable amount of ingenuity, but it's not going to. Here in the closest pair
problem, we're going to proceed exactly as we did in the merge sort and counting
inversions problems, where we took the array and broke it into its left and right
half. So here, we're going to take the input point set, and again, just recurse
on the left half of the points, and recurse on the right half of the points.
So here, by left and right, I mean with respect to the points x coordinates.
There's pretty much never any ingenuity in the conquer step, that just means you take
the sub-problems you identified in the first step, and you solve them
recursively. That's what we'll do here, we'll recursively complete the closest
pair in the left half of the points, and the closest pair in the right half of the
points. So where all the creativity in divide and conquer algorithms is in the
combined step. Given the solutions to your sub problems, how do you somehow recover a
solution to the original problem? The one that you actually care about. So for
closest pair, the questionis going to be, given that you've computed the closest
pair on the left half of the points, and the closest pair on the right half of the
points, how do you then quickly recover the closest pair for the whole point set?
That's a tricky problem, that's what we're going to spend most of our time on. So
let's make this divide and conquer approach for closest pair a little bit
more precise, so let's now actually start spelling out our closest pair algorithm.
The input we're given, it's, this follows the preprocessing steps or recall that we
invoke, merge sort, we get our two sorted copies of the poin t set Px, sorted by x
coordinate, and py sorted by y coordinate. So the first dividend is the division
step. So given that we have a copy of the points px sorted by x coordinate, it's
easy to identify the leftmost half of the points, those with the, those n over two
smallest x coordinates and in the right half, those were the n over two largest x
coordinates. We're going to call those Q and R respectively. One thing I'm skipping
over is the base case. I'm not going to bother writing that down, so base case
omitted, but it's what you would think it would be. So basically once you have a
small number point, say two points or three points, then you can just solve the
problem in constant time by a brute-force search. You just look at all the pairs and
you return the closest pair. So think of it being at least four points in the
input. Now, in order to recurse, to call clo pair again, in the left and right
halves, we need sorted version of Q and R, both by x coordinate and by y coordinate,
so we're just going to form those by doing suitable linear scans through px and py.
And so one thing I encourage you to think through carefully or maybe even code up
after the video is how would you form Qx, Qy, Rx and Ry given that you already have
Px and Py. And if you think about it, because Px and Py are already sorted just
producing these sorted sublists takes linear time. It's in some sense the
opposite of the merge subroutine used in merge sort. Here, we're sort of splitting
rather than merging. But again, this can be done in linear time, that's something
you should think through carefully later. So that's the division step, now we just
conquer, meaning we recursively call closest pair line on each of the two
subproblems, so when we invoke closest pair on the left half of the points on Q
we're going to get back what are indeed, the closest pair of points amongst those
in Q. So we're going to call those P1 and Pq, So among all pairs of points that both
lie in Q, P1 and Q1 minimize the distance between them. Similarly, we're going to
call Q2Q2 the results of the second recursive call, that is, P2 and Q2 are
amongst all pairs of points that both lie in R, the pair that has the minimum
Euclidean distance. Now, conceptually, there's two cases. There's a lucky case
and there's an unlucky case. In the original point set P, if we're lucky, the
closest pair of points in all of P, actually, both of them lie in Q or both of
them lie in R. In this lucky case, we'd already be done if the closest pair in the
entire point set they happen to both lie in Q, then this first recursive call is
going to recover them and we just have them in our hands P1Q1. Similarly, if both
of the closest pair of points in all of P lies on the right side in R, then they get
handed to us on a silver platter by the second recursive call that just operate on
R. So in the unlucky case, the closest pair of point in P happens to be split.
That is, one of the points lies in the left half, in Q, and the other point lies
in the right half, in R. Notice, if the closest pair of points in all of P is
split, is half in Q and half in R, neither recursive call is going to find it. Okay?
The pair of points is not passed to either of the two recursive calls, so there's no
way it's going to be returned to us. Okay? So we have not identified the closest pair
after these two recursive calls, if the closest pair happens to be split.
This is exactly analagous to what happened when we were counting inversions. The
recursive call on the left half of the array counted the left inversions. The
recursive call on the right half of the array counted the right inversions. But we
still had to count the split inversions, so in this closest pair algorithm, we
still need a special purpose subroutine that computes the closest pair for the
case in which it is split, in which there is one point in Q and one point in R. So
just like in counting inversions, I'm going to write down that subroutine and
I'm going to leave it unimplemented for now, we'll figur e out how to implement it
quickly in the rest of the lecture. Now, if we have a correct implementation of
closest split pair, so that takes us input the original point set sort of the x and y
coordinate, and returns the smallest pair that's split or one points in Q and one
points in R, then we're done. So then, the split, then the closest pair has to either
be on the lef or onm the right or it has to be split. Steps two through four
compute the closest pair in each of those categories, so those are the only possible
candidates for the closest pair and we just returned the best of them. So that's
an argument for y, if we have a correct implementation of the closest split para
subroutine, then that implies a correct implementation of closest pair. Now, what
about the running time? So the running time of the closest para algorithm is
going to be in part determined by the running time of closest split pair. So in
the next quiz, I want you to think about what kind of running time we should be
shooting for with a closest split pair subroutine. So the correct response of
this quiz is the second one, and the reasoning is just by analogy with our
previous algorithms for merge sort and for counting inversions. So, what is all of
the work that we would do in this algorithm or we do have this preprocessing
step we call merge sort twice, we know that's n log n, so we're not going to have
a running time better than n log n cause we sort at the beginning. And then, we
have a recursive algorithm with the following flavor, it makes two recursive
calls. Each recursive call is on a problem of
exactly half the size with half the points of the original one. And outside of the
recursive calls, by assumption, by, in the problem, we do a linear amount of work in
computing the closest split pair. So we, the exact same recursion tree which proves
an n log n bound for merge sort, proves an n log n bound for how much work we do
after the preprocessing step, so that gives us an overall running time bound of
n log n. Remem ber, that's what we were shooting for. We were working n log n
already to solve the one-dimensional version of closest pair and the goal of
these lectures is to have an n log n algorithm for the 2D versions. So this
would be great. So in other words, the goal should be to have a correct linear
time implementation of the closest split pair subroutine. If we can do that, we're
home-free, we get the desired n log algorithm. Now, I'm going to proceed in a
little bit to show you how to implement closest split pair, but before I do that,
I want to point out one subtle, the key idea, which is going to allow us to get
this linear time correct implementation. So, let me just put that on the slide. So,
the key idea is that we don't actually need a full-blown correct implementation
of the closets split pair subroutine. So, I'm not actually going to show you a
linear time subroutine that always correctly computes the closets split pair
of a point set. The reason I'm going to do that is that's actually a strictly harder
problem than what we need to have a correct recursive algorithm. We do not
actually need a subroutine that, for every point sets, always correctly computes the
closest split pair of points. Remember, there's a lucky case and there's an
unlucky case. The lucky case is where the closest pair in the whole point set P
happens to lie entirely in the left half of the points Q or in the right half of
the points R In that lucky case, we, one of our recursive calls will identify this
closest pair and hand it over to us on a silver platter. We could care less about
the split pairs in that case. We get the right answer without even looking at the
split pair, pairs. Now, there's this unlucky case where the split pairs happens
to be the closest pair of points. That is when we need this linear time subroutine,
and only. then, only in the unlucky case where the closest pair of points happens
to be split. Now, that's in some sense, a fairly trivial observation, but, there's a
lot of ingenuity here i n figuring out how to use that observation. The fact that we
only need to solve a strictly easier problem and that will enable the linear
time implementation that I'm going to show you next. So now, let's rewrite the high
level recursive algorithm slightly to make use of this observation that the closest
split pair subroutine only has to operate correctly in the regime of the unlucky
case, when in fact, the closest split pair is closer than the result of either
recursive call. So I've erased the previous steps 4 and 5, that, but we're
going to rewrite them in a second. So, before we invoke close split pair, what
we're going to do is we're going to see how well did our recursive calls do. That
is, we're going to define a parameter little delta, which is going to be the
closest pair that we found or the distance of the closest pair we found by either
recursive call. So the minimum of the distance between P1 and Q1, the closest
pair that lies entirely on the left, and P2Q2, the closest pair that lies entirely
on the right. Now, we're going to pass this delta information as a parameter into
our closest split pair subroutine. We're going to have to see why on earth that
would be useful and I still owe you that information, but, for now, we're just
going to pass delta as a parameter for use in the closest split pair. And then, as
before we just do a comparison between the three candidate closest pairs and return
the best of the, of the trio. And so, just so we're all clear on, on where things
stand, so what remains is to describe the implementation of closest split pair, and
before I describe it, let me just be crystal clear on what it is that we're
going to demand of the subroutine. What do we need to have a correct in o of n log n
time closest pair algorithm. Well, as you saw on the quiz, we want the running time
to be o of n always, and for correctness, what do we need? Again, we don't need it
to always compute the closest split pair, but we need it to compute the closest
split pair in the events that there is a split pair of distance strictly less than
delta, strictly better than the outcome of either recursive call. So now that we're
clear on what we want, let's go ahead and go through the pseudocode for this closest
split pair subroutine. And I'm going to tell you upfront, iIt's going to be fairly
straightforward to figure out that the subroutine runs in linear time, o of n
time. The correctness requirement of closest split pair will be highly
non-obvious. In fact, after I show you this pseudo you're not going to believe
me. You're going to look at the pseudocode and you'd be like, what are you talking
about? But in the second video, on the closest pair lecture, we will in fact show
that this is a correct sub-routine. So, how does it work? Well, let's look at a
point set. So, the first thing we're going to do is a filtering step. We're going to
prune a bunch of the points away and so to zoom in on a subset of the points. And the
subset of the points we're going to look at is those that lie in a vertical strip,
which is roughly centered in the middle of the point set. So, here's what I mean. By
center dot, we're going to look at the middle x coordinate. So, let x bar be the
biggest x coordinate in the left half, so that is in the sorted version of the
points by x coordinate, we look at the n over two smallest ex-coordinate. So, in
this example where we have six points, all this means is we draw, we imagine drawing
a line between the third points, so that's going to be x bar, the x coordinate of the
third point from the left. Now, since we're passed as input, a copy of the
points sorted by x coordinate, we can figure out what x bar is in constant time.
Just by accessing the relevant entry of the array, px. Now, the way we're going to
use this parameter delta that we're passed, so remember what delta is. So
before we invoke the closest split pair subroutine in the recursive algorithm, we
make our two recursive calls, we find the closest pair on the left, the closest pair
on the right, and delta is whatever the smaller of those two distances are. So
delta is the parameter that controls whether or not we actually care about the
closest split pair or not, we care if and only if there is a split pair at distance
less than delta. So, how do we use delta? Well, that's going to determine the width
of our strip, so the strip's going to have width 2 delta, and it's going to be
centered around x. And the first thing we're going to do is we're going to
ignore, forevermore, points which do not line in this vertical strip.
So the rest of the algorithm will operate only on the subset of p, the subset of the
points that lie on the strip, and we're going to keep track of them sorted by y
coordinate. So the formal way to say that they line the strip, is that they have x
coordinate in the interval with lower endpoint x bar minus delta and upper
endpoint x bar plus delta. Now, how long does it take to construct this set Sy
sorted by y coordinate? Well fortunately, we've been passed as input a sorted
version of the points Py So to extract Sy from Py, all we need to do is a simple
linear scan through p y checking for each point where its x coordinate is. So this
can be done in linear time. Now, I haven't yet shown you why it's useful to have this
sorted set as y, but if you take it on faith that it's useful to have the points
in this vertical strip sorted by y coordinate. You now see why it was useful
that we did this merge sort all the way at the beginning of the algorithm before we
even underwent any recurssion. Remember, what is our running time goal for closest
split pair? We want this to run in linear time, that means we cannot sort inside the
closest split pair subroutine. That would take too long. We want this to be in
linear time. Fortunately, since we sorted once and for all at the beginning of the
closest pair algorithm, extracting sorted sublists from those sorted lists of points
can be done, done in linear time, which is within our goals here. Now, it's the rest
of t he subroutine where you're never going to believe me that it does anything
useful. So, I claim that essentially with a linear scan through Sy, we're going to
be able to identify the closest split pair of points in the interesting, unlucky case
where there is such a split pair with distance less than delta. So here's what I
mean by that linear scan through Sy. So as we do the scan, we're, we're going to keep
track of the closest pair of points of a particular type that we've seen so far.
So, let me introduce some variables to keep track of the best candidate we've
seen so far. There's going to be a vary, variable best which will initialize to be
delta. Remember, we're uninterested in split pairs unless they have distance
strictly less than delta. So, and then we're going to keep track of the points
themselves, so we'll initialize the best pair to be null. Now, here is the linear
scan. So we go through the points of Sy in order y coordinate. Okay, well, not quite
all the points of Sy. We stop at the eighth to last point and you'll see why in
a second. And then, for each position I of the array Sy, we investigate the seven
subsequent points of the same array Sy. So for j going from one to seven, we look at
the Ith, and I plus jth entry of Sy. So if sy looks something like this array here,
in any given point in this double for loop, we're generally looking at an index
I, a point in this, in this of the array, and then some really quite nearby point in
the array I plus j, because j here's going to be at most seven. Okay? So we're
constantly looking at pairs in this array, but we're not looking at all pairs of all.
We're only looking at pairs that are very close to each other, within seven
positions of each other. And what do we do for each choice of i and j? Well, we just
look at those points, we compute the distance, we see if it's better than all
of the pairs of points of this form that we've looked at in the past and if it is
better, then we remember it. So we just remember the best, ie c losest pair of
points, of this particular type for choices of i and j of this form. So in
more detail, if the distance between the current pair of points of p and q is
better than the best we've seen so far, we reset the best pair of points to be equal
to p and q, and we reset the best distance, the closest distance seemed so
far to be the distance between p and q and that's it. Then, once this double for loop
terminates, we just return it the best pair. So one possible execution of closest
split pair is that it never finds a pair of points, p and q, at distance less than
delta. In that case, this is going to return null and then in the outer call. In
the closet pair, obviously, you interpret a null pair of points to have an infinite
distance. So if you call closest split pair, and it doesn't return any points,
then the interpretation is that there's no interesting split pair of points and you
just return the better of the results of the two recursive calls p1Q1 or P2Q2. Now,
as far as the running time of the subroutine, what happens here? Well, we do
constant work just initializing the variables. Then notice that the number of
points in Sy, well in the worst case, you have all of the points of P. So, it's
going to be the most endpoints, and so, you do a linear number of iterations in
the outer for loop. But here is the key point, in the inner for loop, right,
normally double for loops give rise to quadratic running time, but in this inner
for loop we only look at a constant number of other positions. We only look at seven
other positions and for each of those seven positions, we only do a constant
number of work. Right? We just, we want to compare distance and make a couple other
comparisons, and reset some variables. So for each of the linear number of outer
iterations, we do a constant amount of work, so that gives us a running time of o
of n for this part of the algorithm. So as I promised, analyzing the running time of
this closest split pair subroutine was not challenging. We just , in a
straightforward way, looked at all the operations. Again, because in the key
linear scan, we only do constant work per index, the overall running time is big O
of n, just as we wanted. So this does mean that our overall recursive algorithm will
have running time o of n log n. What is totally not obvious and perhaps even
unbelievable, is that this subroutine satifies the correctness requirements that
we wanted. Remember, what we needed, we needed that whenever we're in the unlucky
case, whenever, in fact, the closest pair of points in the whole point set is split,
this subroutine better find it. So, but it does, and that's being precise in the
following correctness claim. So let me rephrase the claim in terms of an
arbitrary split pair, which has distance less than delta, not necessarily the
closest such pair. So suppose, there exists, a p on the left, a point on the
left side and a point on the right side so that is a split pair and suppose the
distance of this pair is less than Q. Now, there may or may not be such a pair of
points, PQ.. Don't forget what this parameter delta means. What delta is, by
definition, is the minimum of d of p1q1, for p1q1 is the closest pair of points
that lie entirely in the left half of the point set Q and d of p2q2, or similarly,
p2Q2 is the closest pair of points that entirely on the right inside of R. So, if
there's a split pair with distance less than delta, this is exactly the unlucky
case of the algorithm. This is exactly where neither recursive call successfully
identifies the closest pair of points, instead that closest pair is a split pair.
On the other hand, if we are in the lucky case, then there will not be any split
pairs with distance less than delta, because the closest pair lies either all
on the left or on the right, and it's not split. But remember, we're interested in
the case where there is a split pair that has a distance less than delta where there
is a split pair that is the closest pair. So the claim has two parts. The first
part, part A, says the following. It says that if there's a split pair p and, and q
of this type, then p and q are members of Sy. And let me just sort of redraw the
cartoon. So remember what Sy is. Sy is that vertical strip. And again, the way we
got that is we drew a line through a median x coordinate and then we fattened
it by delta on either side, and then, we focused only on points that lie in the
vertical strip. Now, notice our counts split pair subroutine, if it ever returns
a pair of points, it's going to return a pair of points pq that belong to Sy.
First, it filters down to Sy, then it does a linear search through Sy. So if we want
to believe that our subroutine identifies best split pairs of points, then, in
particular, such split pairs of points better show up in Sy, they better survive
the filtering step. So that's precisely what part A of the claim is. Here's part B
of the claim and this is the more remarkable part of the claim, which is
that p and q are almost next to each other in this sorted array, Sy. So they're not
necessarily adjacent, but they're very close, they're within seven positions away
from each other. So, this is really the remarkable part of the algorithm. This is
really what's surprising and what makes the whole algorithm work. So, just to make
sure that we're all clear on everything, let's show that if we prove this claim,
then we're done, then we have a correct fast implementation of a closest pair
algorithm. I certainly owe you the proof of the claim, that's what the next video
is going to be all about, but let's show that if the claim is true, then, we're
home-free. So if this claim is true, then so is the following corollary, which I'll
call corollaryl 1. So corollary 1 says, if we're in the unlucky case that we
discussed earlier, if we're in the case where the closest point and the whole
points of p does not lie both on the left, does not lie both on the right, but rather
has one point on the left and one on the right but as it's a split pair, th en in
fact, the count split pair subroutine will correctly identify the closest split pair
and therefore the closest pair overall. Why is this true? Well what does count
split pair do? Okay, so it has this double for loop, and thereby, explicitly examines
a bunch of pairs of points and it remembers the closest pair of all of the
pairs of points that it examines. What does this, so what are the criteria that
are necessary for count split pair to examine a pair point? Well, first of all,
the points p and q both have to survive the filtering step and make it into the
array Sy. Right? So count split pair only searches over the array Sy. Secondly, it
only searches over pairs of points that are almost adjacent in Sy, that are only
seven positions apart, but amongst pairs of points that satisfy those two criteria,
counts but pair will certainly compute the closest such pair, right? It just
explicitly remembers the best of them. Now, what's the content of the claim?
Well, the claim is guaranteeing that every potentially interesting split pair of
points and every split pair of points with distance less than delta meets both of the
criteria which are necessary to be examined by the count split pair
subroutine. So first of all, and this is the content of part A, if you have an
interesting split pair of points with distance less than delta, then they'll
both survive the filtering step. They'll both make it into the array Sy., part A
says that. Part B says they're almost adjacent in Sy. So if you have an
interesting split pair of points, meaning it has distance less than delta, then they
will, in fact, be at most seven positions apart. Therefore, count split pair will
examine all such split pairs, all split pairs with distance less than delta, and
just by construction, it will compute the closest pair of all of them. So again, in
the unlucky case where the best pair of points is a split pair, then this claim
guarantees that the count split pair will compute the closest pair of points.
Therefore, having h andled correctness, we can just combine that with our earlier
observations about running time and corollary 2 just says, if we can prove the
claim, then we have everything we wanted. We have a correct O of n log n
implementation for the closest pair of points. So with further work and a lot
more ingenuity, we've replicated the guarantee that we got just by sorting for
the one-dimensional case. Now again, these corrollaries hold only if this claim is,
in fact, true and I have given you no justification for this claim. And even the
statement of the claim, I think, is a little bit shocking. So if I were you I
would demand an explanation for why this claim is true, and that's what I'm going
to give you in the next video.