Naïve Divide and Conquer Algorithm

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En provenance du cours de University of California San Diego

Algorithmic Toolbox

3499 notes

University of California San Diego

Algorithmic Toolbox

3499 notes

Cours 1 sur 6 dans Specialization Data Structures and Algorithms

The course covers basic algorithmic techniques and ideas for computational problems arising frequently in practical applications: sorting and searching, divide and conquer, greedy algorithms, dynamic programming. We will learn a lot of theory: how to sort data and how it helps for searching; how to break a large problem into pieces and solve them recursively; when it makes sense to proceed greedily; how dynamic programming is used in genomic studies. You will practice solving computational problems, designing new algorithms, and implementing solutions efficiently (so that they run in less than a second).

À partir de la leçon

Divide-and-Conquer

In this module you will learn about a powerful algorithmic technique called Divide and Conquer. Based on this technique, you will see how to search huge databases millions of times faster than using naïve linear search. You will even learn that the standard way to multiply numbers (that you learned in the grade school) is far from the being the fastest! We will then apply the divide-and-conquer technique to design two efficient algorithms (merge sort and quick sort) for sorting huge lists, a problem that finds many applications in practice. Finally, we will show that these two algorithms are optimal, that is, no algorithm can sort faster!

Problem Overview and Naïve Solution6:21

Naïve Divide and Conquer Algorithm7:09

Faster Divide and Conquer Algorithm6:41

Rencontrer les enseignants

Alexander S. Kulikov
Visiting Professor
Department of Computer Science and Engineering
Michael Levin
Lecturer
Computer Science
Neil Rhodes
Adjunct Faculty
Computer Science and Engineering
Pavel Pevzner
Professor
Department of Computer Science and Engineering
Daniel M Kane
Assistant Professor
Department of Computer Science and Engineering / Department of Mathematics

So let's look at a naive divide and

conquer algorithm, to solve polynomial multiplication problem.

The idea is, we're going to take our long polynomial and

we're going to break it in two parts.

The upper half and the lower half.

So A(x) is going to be D sub one of X ,times x sub n over 2,

plus d sub 0 of x, the bottom half.

D sub 1 of x, since we've pulled out x sub n over 2 terms,

it's lowest term is actually, just a sub n over 2.

So we have two parallel sub polynomials, the high and the low.

We do the same thing for B.

So we break that into E sub 1 of x, and E sub 0 of x.

Again, where E sub 1 of x is the high terms, E sub 0 of x is the low terms.

When we do our multiplication, then, we just multiply together D 1,

x sub n over 2 plus D 0 and E 1 times x sub n over 2 plus E 0.

And that then yields for terms, D sub 1 E sub 1 times x sub n,

D sub 1 E sub 0 + D sub 0 E sub 1 times x sub n/2 + D sub 0 E sub 0.

The key here is that, we now just need to calculate D1 E1, D1 E0, D0 E1, and D0 E0.

Those are all polynomials of degree n over 2.

And so, now we can go ahead and use a recursive solution to solve this problem.

So it gives us a divide and conquer problem.

Its run time is T of n, equals 4 T of n over 2.

Why 4?

4, because we're breaking into 4 subproblems.

Each of them takes time T of n over 2

ecause the problem is broken in half.

Plus, then, in order to take the results and

do our addition that's going to take order n time.

So some constant k, times that.

Let's look at an example.

So we have, n is 4, so we have degree three polynomials.

And we're going to break up A of x into the top half, 4x plus 3,

and the bottom half, 2x plus 1.

Similarly, we're going to break up the top half of B of x.

X cubed plus 2 x squared just becomes x plus 2.

And 3x plus 4, stays at 3x plus 4.

Now, we compute D1 E1.

So multiplying together, 4x + 3, times x plus 2, gives us 4 x squared + 11x + 6.

Similarly, we calculate D1 E0, D0 E1, and D0 E0.

Now we've done all four of those computations, AB is just D1 E1,

4 x squared + 11x + 6 times x to the 4th, plus the sum of D1 E0 and

D0 E1, times x squared, plus finally D0 E0.

If we sum this all together, we get 4 x to the 6th, plus 11 x to the 5th,

plus 20 x to the 4th, plus 30 x cubed, plus 20 x squared, plus 11x plus 4.

Which is our solution.

Now, how long's this take to run?

We're going to look at that in a moment.

Let's look at the actual code for it.

So we're going to compute a resulting array, from 0 to 2n-2, so

is all the results coefficients.

And our base case is that if n of size 1,

we're going to multiply together A at a sub l, plus B at b sub l.

Let's look at those parameters again.

So A and B are our arrays of coefficients, n is the size of the problem,

a sub l is the first coefficient that we're interested in.

And b sub l is the coefficient in B, that we're interested in.

So we're going to be going from b sub l, b sub l plus one, b sub l plus two, etc.

And for n times.

First thing we'll do, is multiply together the D sub one and E sub one.

So basically what we're doing, I'm sorry, the D sub zero and E sub zero.

So, what we're doing is taking A and B, we're reducing the problem size by 2 and

we're starting with those same coefficients.

And we're going to assign those to the lower half of the elements in R.

Then we're going to do something similar,

where we take the upper halves of each of A and B.

So again, the problem size becomes n/2, but now we're moving the lower coefficient

we're interested in from a sub l to a sub l + n/2 and b sub l to b sub l + n/2.

And we're going to assign those to the high coefficients in our result.

Then, what we have to do is calculate D sub 0 E1, and D1 E0.

And then, sum those together.

When we sum those together,

we're going to assign those to the middle elements of the resulting array.

And we'll then return that result.

Now the question comes up, how long does it take?

So we have an original problem of size n,

we break it into four problems of size n over 2.

So, level 0 we have size n, level 1 we have size of n over 2,

at level i, our problems are of size n over 2 to the i.

And all the way down to the bottom of the tree is at log base 2 of n,

and each of the problems are of size 1.

How many problems do we have?

At level 0, we have 1 problem.

We have then 4 problems.

If we go to the i'th level, we have 4 to the i problems.

And at the very bottom, then we have 4 to the log base 2 of n problems.

How much work is there?

Well, we just need to multiply together the number of problems

times the amount of work, so we have kn here, and

4 times, 4 because there are 4 problems,

kn over 2, because the problem size is n over 2 and

the amount of work we're doing at each level is k times n over 2 per problem.

So 4 kn over 2 just equals k times 2n, At the ith level for

the i problems, each problem takes k times n over 2 to the i

to deal with, we multiply together, k 2 to the i times n.

And at the very bottom, we have k amount of work,

we have a problem size of 1, times 4 to the log base 2 of n.

Well four the log base two of n, is just n squared.

So we have k n squared. Our total

as we sum up all the work is going to be summation from i equals zero

to log base two of n of four to the i k times n over two to the i.

And that just gets dominated by the very bottom term which is

big theta of n squared.

So that's what our runtime takes. This is kind of weird.

We went through all this work to create a divide and conquer algorithm.

And yet,

the run time is the same run time as it was with our naive original algorithm.

We're going to see in the next video, a way to redo our divide and

conquer algorithm, so we have less work to do at each level, and so

we actually get a better final run time.

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