0:00

So as we said last time, about the running time of the brute force algorithm for

computing distances, is that you know, we got a very complicated formula.

But, at the end what we could say also, one of

the things that we could say without going through all these complications and

details, is that the running time of that algorithm.

Is at least 2 to the n minus 1 if you remember.

We said because the algorithm was going to try every possible subset, of size and

every possible subset of a set of size n minus 1.

And, it was going to, inspect each one of them and do a lot of work.

So even it's going to do definitely more than, 2 to n minus 1.

But, we can say that,

we can put a lower bound on the earning time that is 2 to the n minus 1.

Now why does that matter?

Because we have shown already, that when we talk about growth of functions, that 2

to the n is growing actually very, very fast, that even when n was 256.

It will take 10 to the 59 years, for

a computer that can handle 10 billion operations a second, okay.

2 to the n, 2 to the n minus 1, are basically the same,

it's just that 1 is divided by 2, okay.

So, using this kind of technique by putting lower bounds and

upper bounds on an algorithm, is very helpful because.

As I said from the beginning,

that we're not trying to get at the exact running time of the algorithm.

We're trying to get an approximation, that gives us a good idea about,

how long that our algorithm would take once we implement it.

So in the case of brute force algorithm.

Using that kind of technique by bounding the running time from below,

it told me that this algorithm is going to take at least exponential amount of

time 2 to the n.

Order of 2 to the n.

And that's enough for me to know, that this algorithm is not practical,

is not worth implementing and giving to the user.

1:53

In the same sense also we can usually put an upper bound on the running time, by, so

that we get an idea of, how much time is

this algorithm going to take in what worst case, in the worst case, okay?

So, how fast is this function going to go?

When we have a function like, 5n plus 3n squared, minus 17n cubed,

minus 19n to the 17, you know, that's a function that, you know,

we have to start plugging in values of n to evaluate it.

But, if we use this kind of lower bound and upper bound technique.

We can get the, an easier function that we can understand better,

it's much easier to say, 2 to the n grows fast and

look at a function like 2 to the n and decides that it grows very fast.

Than a very complicated function that has summation and combi, and

combinations and choose k and choose n infactorial and so on.

Okay.

So for that there is there is a set of mathematical tools that we use,

which we refer to collectively as asymptotic notations.

Asymptotic notations allows us, allow us to put upper and

lower bounds on functions, as well as the two bounds as the same time,

that will bound these functions and will give us an idea of.

How fast the function grows?

How bad can it be?

Or how good it can be, as well?

Okay? So, I will start with with the,

with the first one of the simp,

there are three main asymptotic notations that we will use.

The first one is, corresponds to putting an upper bound on the running time of a,

of a, of an algorithm.

3:26

So if I have the, if I have an algorithm, I have analyzed the running time of

the algorithm, and I get a function of the input size.

So we'll assume that the input size, that the parameter that corresponds to

input size is n, and I will assume that I have done all my work, and

I got that the running time is f of n.

4:48

Faster than f of n after some point, okay?

So in the asymptotic notations when I have a function f of n,

I am interested in getting a function g of n, and getting a function g of n.

That except for a few values of n in the beginning.

Except for a few values of n in the beginning, after a certain point

the function f of n is going to grow at most as fast,

as most as fast as some constant times that function g of n.

Okay, so, remember, again, in asymptotic notations, I have f of n.

It's a function that someone analyzed, or

the running time of an algorithm I obtained, okay.

Someone analyzed the running time of a certain algorithm,

got the function f of n.

I am interested in quickly getting some sort of upper bound on that function, so

that I can understand, how bad can it be?

Okay?

Now, in asymptotic notations, we are interested in getting another function,

g of n, that will bound f of n from above, that, when I draw them,

g of n is going to be on top of f of n, but, again, with the caveat that,

if multiplied by a constant, c.

Okay?

So, there's a constant, c, here,

that if I multiplied by g of n, is going to be higher than f of n.

The other issue, as I mentioned, is that.

In the beginning, I can tolerate a few values of n,

because remember we focus on the growth of function as n goes to infinity.

I am not much concerned with what happens when n, with n is very small, okay?

So when we talk about putting an upper bound,

we are interested basically in this function g of n.

And we have to think about two constants at the same time.

One constant that we are going to call n0, and

other constant that we are going to call c.

And we want that function g of n such that, except for

that these values here smaller than n0, or

for every value of n greater than or equal to n0, for every value here.

I want f of n to be at most c times g of n, okay.

Or in other words we want, we want a function

7:07

g of n.

Such that, there exist

two constants, c and

n0, where f of n,

is at most, c times g of n for

every n greater than n0.

Kay?

Again, if we look at this picture here, it shows clearly that,

if you ignore all the values of n that are smaller than n0, if you ignore them.

And just focus on values of n greater than or equal to n0.

You will notice that f of n is always smaller than c times g of n.

And this is what this definition is saying,

that we have a function g of n, two constants c and n0.

C has to be strictly greater than 0, n0 is greater than or equal to 0.

Such that f of n is smaller than or equal to c of g of n,

c times g of n, for every n greater than or equal to n0.

This upside down A is basically, in English, we say,

for all, or for every value of n that is greater than or equal to n0.

If we have this scenario.

If we have this scenario, we have a term for it.

We say that, in this case, in this case, f of

n is big O of g of n.

And this is the first asymptotic notation I'm introducing, the Big O notation.

f of n is O(g(n)) if we have this scenario or if we can,

if, if I can find two constants c and n0 such that for

every n greater than or equal to n0, f of n is going to be at most c times g of n.

Okay?

So this is the first notation that we can use.

And it tells us to put an upper bound, on the running time of an algorithm.

Again, in this notation, we will ignore finite set of values in the beginning,

and focus on the running time as the, as the input size grows.

Okay?

So this is the first, first asymptotic notation that we,

we learned, and again it's about putting an upper bound.

This is the definition.

If you want to put an upper bound on a function f of n, you find the g of n, and

the c and then n0, that will satisfy this definition.

Okay?

9:40

In the case of the brute force algorithm that we devi we devised for

the distance problem, we actually looked at the lower bound.

We said that that algorithm will take at least 2 to the n amount of time.

So what does it mean to put a lower bound?

We can look at the same thing, but, we can think about g now, c times g of n being.

A lower bound on this, and in this case,

we can think about it as, I can think about it as the following.

If I have f of n like this here.

10:24

And I can find, a g of n for

example c times g of n.

And I can focus on the n0.

So now I found a, a function g of n such that f of n is going to be greater than or

equal to c time g of n.

Once n is greater than n 0.

In this case here, again it's similar, definition, we need to find the g of n and

two constants, such that f of n is greater than or

equal to c times g of n for every n greater than or equal to n0.

In this case say that f of n equals omega of g of n, okay?

So omega is for lower bound, okay?

The third asymptotic notation is that, I can take this function, f of n.

11:28

Okay?

So we have now c1 times g of n and

we have c2 times g of n, and I say f of n is bounded from above and

below by g of n times two different constants.

In this case, what we have is that f of n is, we say it is theta(g(n)).

It is sandwiched between c1 times g of n and c2 times g of n.

So we have these three asymptotic notations f of n equals theta of g of n,

f of n equals big O of g of n.