I'm partitioning a partitioned unit.

This is called fuel partitioning in the lingo.

Spectral partitioning, L1 minimization. This comes from the optimization community.

It's sort of cool that Top K via partitioning as just described works.

The time it takes to run Top K via partitioning on one element.

It's kind of a partially sorted list, and we call this partitioning around V.

So, lets imagine there is some problem called graph partitioning which is a real problem.

Now, the partitioning algorithm is going to split this list L into a left and a right.

Do we know that graph partitioning is in NP, and therefore an exponential time solution would exist?

Let's go over now how we can actually solve the top K problem using this idea of partitioning.

When we observe the world, what we're doing is we're taking the current belief state and partitioning it up into pieces.

Right. Let's do an analysis of the time it takes for a Top K to do its work via this partitioning approach.

This is very similar to the K partitioning algorithm from the lectures, but returns a single value only, instead of a list.

There's our value v, and now, we're going to run the partitioning algorithm on this, and there is a couple of different cases that can happen.

It doesn't solve the TopK problem better than the randomized algorithm we talked about the partitioning algorithm but it does it pretty fast and there's other uses for this.

We're going to start off remember we know how to do a partitioning is giving some v, well I don't know which value to use, so let's just pick one at random.

And these includ I mean leading approaches from, from our book where we have bottom up detectors, which try to find boundaries based on brightness, color, and texture differences, as well as some globalization through the use of normalized cups into the spectral partitioning technique.

First, it does the partitioning operation, which runs through all the elements of the list that takes time n and then it's going to recurse assuming it hasn't gotten really lucky and since this is a less than or equal to let's assume it didn't get lucky.

But now, we have a fourth approach, which uses this partitioning idea recursively to get an expected running time Î (n), which is better and the worst case than sorting and better than selection insertion for any k other than a constant n for constant they come up the same.

Even if you don't know what it is, there are some things that you do know by the virtue of the fact that is is NP complete, assuming that you believe me, you'd know the following which of these, check all that apply, which of these do you know? if I tell you that graph partitioning is NP complete does that mean its

What happens when we run partitioning on that value v is it's going to separate all the values of L into the ones smaller than v, equal to v, and greater than v, and that happens to be position K, so the ones on the first K positioned here are exactly the top K for the list, the smallest K elements that we were looking for.