kanishka/counterfactual-babylm-new_regex_aanns_removal

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So, that's roughly n operations for each of the n partial products .
So, that's roughly n squared operations.
It could be all of the partial products.
Then, we have to add it all up at the end.
But that takes just an extra number of constant times n primitive operations to do all of those additions.
So summarizing, overall, the number of primitive operations required to multiply to n digit numbers using this procedure grows quadratically with the length N. So, for example, if you double the length of two numbers, you expect to do roughly four times as many primitive operations if you use this iterative algorithm for multiplying two numbers.
Now, depending on what type of third grader you were, you might well have accepted this procedure as being the unique or at least the optimal way of multiplying two n digit numbers together.
And if you want to be an expert algorithm designer, that kind of obedient attitude is something you're going to have to learn to discard.
Here's a favorite quote of mine.
It's from a quite early textbook in the field The Design and Analysis of Computer Algorithms by Aho,
Hopcroft, and Ullman, and after highlighting a number of algorithmic design techniques, they conclude by saying, perhaps the most important principle for the good algorithm designer is to refuse to be content."
So, that's a beautifully accurate quote.
I might rephrase it more succinctly by just saying, if you're, want to be a good algorithm designer, you should adopt the following mantra.
You should always be asking, can we do better?
This is particularly appropriate question to ask when you're faced with some kind of naive or obvious algorithm for solving a problem, like the third grade algorithm for energy multiplication.
So, this will come up over and over again.
We'll see an algorithm.
There'll be an obvious solution but with some extra algorithmic ingenuity by detecting subtle structure in the problem, we'll be able to do signifigantly, qualitatively better than the naive or the obvious solution to the problem.
So, let's apply this cocky attitude to the problem of multiplying two in tegers.
Let's just take, suppose as a working hypothesis that there is some procedure which is better than what we learned back in elementary school.
What could it possibly look like?
So, as someone with some programming experience, you know that they are not only iterative algorithms, iterative programs, like the one we just outlined for multiplying two integers.
But they're also recursive programs.
These are programs that call themselves, typically, on an input of a similar form, but with smaller size.
So, as a working hypothesis,
let's imagine that there's some interesting recursive approach to multiplying two integers.
What must such an algorithm look like?
Well, it must somehow fundamentally involve calling itself on smaller problems.
That is, on smaller numbers, numbers with fewer digits.
So, what kind of decomposition on numbers could be used to enable this kind of recursive approach?
Well, if you think about it, there's a pretty natural way to break a number with a bunch of digits into a couple numbers with fewer digits.
Namely, just take the first half of the digits, the first n over two digits, regard that as a number in its own right.
And the second half of the digits, regard that as a number in its own right.
So, perhaps this decomposition will have some use, in enabling a recursive attack on how to multiply two images.
So, we're going to investigate that in more detail on this next slide.
Given x and y, each with n over two digits, we're going to represent x as terms of its first half of the digits a and its second half of the digit to b.
Similarly, we will write y, the second input in terms of its first half and second half of digits, c and d.
And in this expansion, a and b, c and d are all n over two digit numbers.
I'm assuming for convenience here, that n is even.
This will extend to the case where n is odd in the natural way, where you break it into n over two rounded up and n over two rounded down.
So, in our previous example, where x was 5,678 and y was 1,234, we would have a being equal to 56 .
The first two digits of the four in x.
And then b would be the other two digits.
And then similarly for c and d.
They have decomposition of y.
So now, in a purely mechanical way, we can expand the product of x times y in terms of this representation.
In terms of these smaller numbers, a, b, c, and d.
So, x times y, by a representation, is just the same thing as ten n over two times a plus b times quantity ten to the n over two c plus d.
And now, if we expand this in grouplike terms, we get a ten to the n times ac plus ten to the n over two times ad plus bc.
So, that's combining two terms from the expansion, plus bd.
And I'm going to call this expression circled in green, star.
Now, what I want you to think about and make sure you understand is that, this expansion that I've circled and called star, immediately suggests a recursive algorithm for multiplying two integers.
Now, it's not clear whether that algorithm is going to be fast or slow, whether this is a good idea or a crackpot idea.
But certainly, it's a legitimate recursive approach to multiplying two integers.
Specifically, the relevant product in star namely ac, ad, bc, and bd all involve smaller numbers than what we stared with.
Those are all n over two digit numbers.
Those all original inputs at n digits so we can legitimately solve those recursively.
We can invoke our same algorithm to compute those in a recursive way.
Now, once we've invoked our multiplication algorithm recursively four times to compute these four relevant products, now we can just compute the expression in star in the obvious way.
We take ad and bc.
We add them together, using the just, standard first grade iterative addition algorithm.
Then, we add both ac and the result of that addition by a bunch of zeros, n over two zeros or n zeros as appropriate.
Now, we have these three constituent terms and we just add them up again using the grade school algorithms to compute the overall expression.
So, to keep this intro lecture brisk, I haven't discussed the base case of this recursive al gorithm.
As, hopefully, you all know, recursive algorithms do need base cases.
At some point, the recursion has got to stop.
Otherwise, your algorithm is never going to terminate.
So, in multiplication, all you'd do is, if your past as input two single digit numbers, you'd multiply them in one primitive operation, and return the result.
If the numbers have two or more digits, you would recurse in the way we have described here.
If there's one digit, you just compute them and you're done.
So, who knows whether this recursive algorithm is a good idea or not, it's totally not obvious, you should not even necessarily have any intuition about how this compares to the iterative algorithm you learned back in elementary school.
So, in lieu of answering that question, although we will answer it several lectures hence, I want to show you a second, still more clever, recursive algorithm, for integer multiplication, so this goes by the name of Karatsuba multiplication after the founder, although really it's sort of the key point, there's a trick pointed out by the mathematician, Gauss, in the early nineteenth century.
So, to explain this algorithm, let me write once again our expansion in terms of the n over two digit numbers.
So, we have the product x times y, which we wrote as ten to the n ac plus ten to the n over two ad plus bc plus bd.
Now, the previous approach was in recognition of the four products that we see in this expression.
We made four recursive calls.
So, here's the trick, and this is really what Gauss figured out over 200 years ago which is that it seems there are four products.
But fundamentally, in this expression, there's really only three quantities that we care about.
There's the ac, the coefficient of this first term.
There's ad plus bc, the coefficient of this middle term, and there's bd.
So, we care about ad and bc as quantities individually, only in as much as we care about their sum.
We really don't care about them individually.
So, that motivates the question, perhaps we can somehow uncover these three d ifferent quantities using only three recursive calls rather than four.
In fact, we can't, and that's Karatsuba multiplication.
So, to be more specific, our first two recursive calls are shared with the previous recursive algorithm.
We go ahead and compute ac and bd recursively.
The third step is the, is the clever one where we recursively compute the product of quantity a plus b with c plus d.
Now, expanding what are we going to get when we recursively compute this product, we're going to get ac plus bd plus ad plus bc.
And here is Gauss's trick for observation which is that the result of the third recursive call minus each of the first two, what happens?
Well, the ac is going to cancel the ac, the bd is going to cancel the bd and we will be left with ad + bc, which is exactly that middle coefficience that we cared about.
So, we can indeed compute the sum of ad and bc without separately computing each of them, and that's exactly what we wanted.
So, what is this bias?
This gives us another second recursive algorithm from multiplying two integers that makes use of fewer recursive calls.
Only three recursive calls rather than four.
And as before, there's a little bit of work to do beyond the recursive calls but not much.
You just have to do some padding by zeroes and adding up the results.
So, there is two points I want you to appreciate about all these energy multiplication algorithms.
The first point is that the algorithm design space is incredibly rich.
Much richer than you might have had initially had intuition for.
Multiply two integers.

So, that's roughly n operations for each of the n partial products .

So, that's roughly n squared operations.

It could be all of the partial products.

Then, we have to add it all up at the end.

But that takes just an extra number of constant times n primitive operations to do all of those additions.

So summarizing, overall, the number of primitive operations required to multiply to n digit numbers using this procedure grows quadratically with the length N. So, for example, if you double the length of two numbers, you expect to do roughly four times as many primitive operations if you use this iterative algorithm for multiplying two numbers.

Now, depending on what type of third grader you were, you might well have accepted this procedure as being the unique or at least the optimal way of multiplying two n digit numbers together.

And if you want to be an expert algorithm designer, that kind of obedient attitude is something you're going to have to learn to discard.

Here's a favorite quote of mine.

It's from a quite early textbook in the field The Design and Analysis of Computer Algorithms by Aho,

Hopcroft, and Ullman, and after highlighting a number of algorithmic design techniques, they conclude by saying, perhaps the most important principle for the good algorithm designer is to refuse to be content."

So, that's a beautifully accurate quote.

I might rephrase it more succinctly by just saying, if you're, want to be a good algorithm designer, you should adopt the following mantra.

You should always be asking, can we do better?

This is particularly appropriate question to ask when you're faced with some kind of naive or obvious algorithm for solving a problem, like the third grade algorithm for energy multiplication.

So, this will come up over and over again.

We'll see an algorithm.

There'll be an obvious solution but with some extra algorithmic ingenuity by detecting subtle structure in the problem, we'll be able to do signifigantly, qualitatively better than the naive or the obvious solution to the problem.

So, let's apply this cocky attitude to the problem of multiplying two in tegers.

Let's just take, suppose as a working hypothesis that there is some procedure which is better than what we learned back in elementary school.

What could it possibly look like?

So, as someone with some programming experience, you know that they are not only iterative algorithms, iterative programs, like the one we just outlined for multiplying two integers.

But they're also recursive programs.

These are programs that call themselves, typically, on an input of a similar form, but with smaller size.

So, as a working hypothesis,

let's imagine that there's some interesting recursive approach to multiplying two integers.

What must such an algorithm look like?

Well, it must somehow fundamentally involve calling itself on smaller problems.

That is, on smaller numbers, numbers with fewer digits.

So, what kind of decomposition on numbers could be used to enable this kind of recursive approach?

Well, if you think about it, there's a pretty natural way to break a number with a bunch of digits into a couple numbers with fewer digits.

Namely, just take the first half of the digits, the first n over two digits, regard that as a number in its own right.

And the second half of the digits, regard that as a number in its own right.

So, perhaps this decomposition will have some use, in enabling a recursive attack on how to multiply two images.

So, we're going to investigate that in more detail on this next slide.

Given x and y, each with n over two digits, we're going to represent x as terms of its first half of the digits a and its second half of the digit to b.

Similarly, we will write y, the second input in terms of its first half and second half of digits, c and d.

And in this expansion, a and b, c and d are all n over two digit numbers.

I'm assuming for convenience here, that n is even.

This will extend to the case where n is odd in the natural way, where you break it into n over two rounded up and n over two rounded down.

So, in our previous example, where x was 5,678 and y was 1,234, we would have a being equal to 56 .

The first two digits of the four in x.

And then b would be the other two digits.

And then similarly for c and d.

They have decomposition of y.

So now, in a purely mechanical way, we can expand the product of x times y in terms of this representation.

In terms of these smaller numbers, a, b, c, and d.

So, x times y, by a representation, is just the same thing as ten n over two times a plus b times quantity ten to the n over two c plus d.

And now, if we expand this in grouplike terms, we get a ten to the n times ac plus ten to the n over two times ad plus bc.

So, that's combining two terms from the expansion, plus bd.

And I'm going to call this expression circled in green, star.

Now, what I want you to think about and make sure you understand is that, this expansion that I've circled and called star, immediately suggests a recursive algorithm for multiplying two integers.

Now, it's not clear whether that algorithm is going to be fast or slow, whether this is a good idea or a crackpot idea.

But certainly, it's a legitimate recursive approach to multiplying two integers.

Specifically, the relevant product in star namely ac, ad, bc, and bd all involve smaller numbers than what we stared with.

Those are all n over two digit numbers.

Those all original inputs at n digits so we can legitimately solve those recursively.

We can invoke our same algorithm to compute those in a recursive way.

Now, once we've invoked our multiplication algorithm recursively four times to compute these four relevant products, now we can just compute the expression in star in the obvious way.

We take ad and bc.

We add them together, using the just, standard first grade iterative addition algorithm.

Then, we add both ac and the result of that addition by a bunch of zeros, n over two zeros or n zeros as appropriate.

Now, we have these three constituent terms and we just add them up again using the grade school algorithms to compute the overall expression.

So, to keep this intro lecture brisk, I haven't discussed the base case of this recursive al gorithm.

As, hopefully, you all know, recursive algorithms do need base cases.

At some point, the recursion has got to stop.

Otherwise, your algorithm is never going to terminate.

So, in multiplication, all you'd do is, if your past as input two single digit numbers, you'd multiply them in one primitive operation, and return the result.

If the numbers have two or more digits, you would recurse in the way we have described here.

If there's one digit, you just compute them and you're done.

So, who knows whether this recursive algorithm is a good idea or not, it's totally not obvious, you should not even necessarily have any intuition about how this compares to the iterative algorithm you learned back in elementary school.

So, in lieu of answering that question, although we will answer it several lectures hence, I want to show you a second, still more clever, recursive algorithm, for integer multiplication, so this goes by the name of Karatsuba multiplication after the founder, although really it's sort of the key point, there's a trick pointed out by the mathematician, Gauss, in the early nineteenth century.

So, to explain this algorithm, let me write once again our expansion in terms of the n over two digit numbers.

So, we have the product x times y, which we wrote as ten to the n ac plus ten to the n over two ad plus bc plus bd.

Now, the previous approach was in recognition of the four products that we see in this expression.

We made four recursive calls.

So, here's the trick, and this is really what Gauss figured out over 200 years ago which is that it seems there are four products.

But fundamentally, in this expression, there's really only three quantities that we care about.

There's the ac, the coefficient of this first term.

There's ad plus bc, the coefficient of this middle term, and there's bd.

So, we care about ad and bc as quantities individually, only in as much as we care about their sum.

We really don't care about them individually.

So, that motivates the question, perhaps we can somehow uncover these three d ifferent quantities using only three recursive calls rather than four.

In fact, we can't, and that's Karatsuba multiplication.

So, to be more specific, our first two recursive calls are shared with the previous recursive algorithm.

We go ahead and compute ac and bd recursively.

The third step is the, is the clever one where we recursively compute the product of quantity a plus b with c plus d.

Now, expanding what are we going to get when we recursively compute this product, we're going to get ac plus bd plus ad plus bc.

And here is Gauss's trick for observation which is that the result of the third recursive call minus each of the first two, what happens?

Well, the ac is going to cancel the ac, the bd is going to cancel the bd and we will be left with ad + bc, which is exactly that middle coefficience that we cared about.

So, we can indeed compute the sum of ad and bc without separately computing each of them, and that's exactly what we wanted.

So, what is this bias?

This gives us another second recursive algorithm from multiplying two integers that makes use of fewer recursive calls.

Only three recursive calls rather than four.

And as before, there's a little bit of work to do beyond the recursive calls but not much.

You just have to do some padding by zeroes and adding up the results.

So, there is two points I want you to appreciate about all these energy multiplication algorithms.

The first point is that the algorithm design space is incredibly rich.

Much richer than you might have had initially had intuition for.

Multiply two integers.