# cs313 unit 4.3

## 01 s Size of Input

Vertex cover is so much easier to solve for this graph over here. It would actually be solved by pre-processing alone, although both graphs have the same number of vertices. It seems like the size of the input is actually not the best way to measure how hard it is to solve an NP complete problem for that input. It seems kind of unfair to just say to this graph here--well, you're just as large as this one over here, so we're going to assume that you're just as hard to solve.

## 02 l Hardness

And that is why you could think of hardness, meaning how hard is a specific input to solve as a combination of 2 things. One is, I think, the size of the input, and we always measure this as a number N. But the other thing is the structure of the input. In a way, it's not larger instances are harder to solve, but in a way you could say, harder instances are harder to solve. This of course might seem like a bit nitpicking around, so the question is, is this actually useful, and the answer is yes. For some problems, you can actually express the structure of the input, and I'm soon going to show you how to do that, as a number K. And while in classical NP-completeness, so to say, you would measure hardness as some function of N. You can then measure hardness as a function of 2 parameters. One parameter is still the size of the input, and the other parameter is going to specify something about the structure of the input. Now this might sound a bit abstract, so what could this parameter be? And I'm going to show you 2 possibilities for this parameter, and we're going to discuss them in detail. One is size of the solution, and the other one is something I would call distance from triviality. So for example, for the network from the last quiz, you could actually see that both parameters could tell us that finding a solution here is easy, because first of all, the solution here and optimum vertex cover is very small, you only require 1 vertex. And also, you could argue that this graph here is very trivial because all you need is a pre-processing rule to actually find the best possible solution. But again, we'll get into more detail. Let's start out with the size of the solution.

## 03 q Measuring the Hardness of Vertex Cover

And now we already know that we can solve vertex cover using a search tree. So we started with the original graph, and I'm now going to use again the search tree where we branch into only 3 different possibilities. We know it's not the best possible one, but it will be easier to understand the rest this way, but it also works for the more complicated search trees. So then on the next level, we again branched into And we said that the search tree had a size of N half and the total number of solutions that we looked at in the search tree was 3 to the power of N half, which as you remember was about 1.734 to the power of N. Now I would like you to quickly think about the following: So as you'll remember, at each level of the search tree, we're looking at an edge where the 2 end points have not yet had an assignment, and we're branching into 3 cases, this one, this one, and this one. So now it's a very easy quiz actually. So how many vertices are added each time we go one level deeper into the tree? At least 1, at least 2, exactly 1, exactly 2, or is this something that actually depends on the structure of the input graph?

## 03 s Measuring the Hardness of Vertex Cover

And the answer here of course is that at least one vertex is added to the potential vertex cover each time we go down here in this tree or we branched because there's only three possibilities and in this possibility here, we add one vertex, and this possibility here, we also add one vertex, and here we will add too. And of course, there then comes data reduction and all that jazz, but we're always adding at least one.

## 04 q Decision Version of Vertex Cover

So what happens as we go down here, at least one vertex added and so on. So what does this have to do with measuring the hardness of vertex cover? Well, we'll get there in a minute. What I would like you to think about now is the following: Let's say we were looking at the decision variant of vertex cover where as input we were given a graph and an integer K, and our output was an answer to the question, does the graph have a vertex cover of size at most K? So now here comes my next question to you, what is the maximum height of the search tree if we are not using it to solve the optimization version of vertex cover but the decision version of vertex cover? Would that height be in half, would that height be K, would that height be N, or something we can't say? And please choose the best possible one here.

## 04 s Decision Version of Vertex Cover

And now this is the interesting insight here. The maximum height of the search tree is K Why is that? Well, we just noticed that each time we go 1 level deeper in the search tree, we add at least 1 vertex to the vertex cover. If the graph is supposed to have a vertex cover of size at most K, then the maximum number of levels that we can go down here is actually K levels, so for the decision variant, what we have here is that the height of the search tree is no longer in half, but it's actually bounded by K, and K is the size of the vertex cover that we're looking for.

## 05 q Number of Assignments

Now my next question to you is what does this mean for the number of assignments that we're looking for in this new search tree? Is that still 3^n/2, is it k^n/2, is it 3^k/2, or is it 3^k?

## 05 s Number of Assignments

What you can easily see is that the size of this search tree here is actually 3^k because we branched into three possibilities each time we go a little deeper, and this can happen at most k times. The size of the search tree is 3^k.

## 06 l Search Tree Size

Of course, this is only a very coarse analysis actually so you can have much better search trees and even for this tree here, you can do a better analysis because actually, this tree doesn't always have height K for the 3 cases that you branch in. In one of the cases, as you'll remember, you put 2 vertices into the vertex cover, so this one here, this one here, and here you put 2. Actually, that already decreases the size of the tree, so doing this type of analysis is out of the scope of this course actually. If you were to do this, then you find out that the size of the tree is approximately 2.4 to the power of K. And again, there are even better search trees for this. The main point however is this, look at what we have just done. We have just discovered a search tree for a vertex cover for which the size does not depend on the size of the input at all. The size of this search tree only depends on the integer K. It doesn't matter if we do it for a graph with 10 vertices, 100 vertices, or 1,000 vertices, the size of the search tree only depends on K, and it gets better if we implement this as an algorithm to solve vertex cover, so the decision version of vertex cover. Now look at this running time here and what that means. The running time here will be depending on how you implement it, but into the power of 3 will do here.

## 07 q Time to Solve Decision Version

The decision variant of vertex cover can be solved in 2.4^kn³ time and this is by far not the best algorithm for vertex cover there is and I'm soon going to tell you a little bit about that. Even with this, think about what that means and I'm actually going to quiz you on this one. Does it mean that for any fixed k, vertex cover can be solved in polynomial time? Does it mean that the running time is independent of the size of the input graph? Or have I been tricking you all along and it's actually the case that P=NP? Please check all of those that are correct.

## 07 s Time to Solve Decision Version

Of course, this one here is correct. So if we say we want to find out if a graph has a vertex cover of size 10, then all we need is polynomial time because then this decision problem can be solved in O of 2.4 to the power of 10 times N cubes time, and the exponential factor here is no longer dependent on the input size. So for a fixed value of K, vertex cover can indeed be solved in polynomial time. Of course, the running time is not independent of the size of the input graph because you still have this polynomial factor here. Does this statement here mean that actually P=NP? Unfortunately, it doesn't because K can take any value, so K can also depend on N in the general case, but what it does mean is that if we consider any fixed value of K, then vertex cover becomes easy. In other words, small vertex covers are easy to find.

## 08 l Fixed Parameter Tractability

Now, so far, we've only worked with the decision variant of vertex cover. So, what about the optimization problem? Actually, that's not that difficult. We just run through all values of K. So we will first start off assuming K=1. Then our algorithm requires O(2.4¹n³), then we'll increase K=2, algorithm requires o(2.4²n³). So each time we're solving the decision problem here, and we increase K even further to K=3 and so on and so on and so on. Until we hit some value k’ which is the size of the smallest vertex cover of the input graph. Of course, we don't know what that size will be in advanced? But once we have increased K enough, solving the decision variant will give us the answer, yes. Now, how much running time is this in total? Well, you need to know a bit of discrete mathematics for this, but running time O(2.4¹+2.4²+)...(2.4^k')n²). Now if you have a discrete math course and remember what you learned there, you will recognize this part here as a geometric series and this evaluates to (2.4^k'?¹-1/2.4-1) which is the same as 0(2.4^k'), and k' is the size of the smallest vertex cover n³. We forgot the n³ here which is right here. And this is amazing because now it means that not only the decision variant of vertex cover can be solved in (2.4^k' n³) time, but it means that the optimization variant of vertex cover can also be solved in O(2.4^k'n³ )time, where k is the minimum size of a vertex cover for the input graph. And this again emphasizes the fact, small vertex covers are easy to find even if we don't know that a graph has a small vertex cover as long as it is small, it will be easy to find and for any fixed k, so either in the decision variant, we fixed k or we say we look at all graphs that say have a vertex cover of size 10 or 20 or 30, we can solve this NP complete problem in polynomial time. And this is a concept known as Fixed Parameter Tractability. Well, that means is that as long as the parameter that measures the hardness and in this case, it's the size of a vertex cover for your graph, as long as that parameter is small or fixed, then you can solve that problem in polynomial time and that is what this here says, so as long as the parameter is fixed, the problem is tractable, if the parameter is not fixed, so for example, there's a function of n, then the problem is not tractable. And this of course is a super neat technique for solving vertex cover and actually 2.4^k is actually a very bad exponent here. We only did that because of what's easier to explain the problem that way. The currently best algorithms have an exponential factor of about 1.3^k. Actually, a current state of the art is that you can solve vertex cover in Now, the polynomial here that depends on certain analysis factors and so on but the important thing here is to see, first of all, vertex cover is fixed parameter tractable and second of all, the base of the exponent is actually really good. It's only 1.3^k.

## 09 l FPT Vertex Cover Clique Independent Set Shortest Tour

So now coming back to this picture here, you see what I meant when I said that hardness of a problem can be measured as both a function of the size of the imput and the structure of the imput, so in the case of the vertex covered, it turns out that what makes solving vertex cover hard is actually the size of the vertex cover itself. So if there's a small vertex cover, then the problem is easy and if it's large, then the problem becomes hard. Now of course, this immediately raises a question for the book of the laws of NP-completeness and that question is, is every problem fixed-parameter tractable? That depends, it depends on what you consider as a parameter, and one parameter I will soon introduce to you as well is a parameter called distance from triviality, but unfortunately in general, not every NP-complete problem seems to be fixed-parameter tractable. So for vertex cover, we have just shown that if you use the size of the solution as a parameter, then yes, this problem is fix-parameter tractable using K, and by K I mean the size of the solution of course. Now what about clique and independent set? And again, we can consider these 2 problems together because they're so similar. And here unfortunately, I have a bit of bad news for you because using K, by which I mean the size of the solution, that's highly unlikely to be fixed-parameter tractable. So by highly unlikely, what I mean is it would not be something as similar as P=NP or something like that, but still there is evidence in a way that is similar to P versus NP that clique and independent set are not fixed-parameter tractable at least if you consider the parameter to be the size of an optimal solution. Now what about shortest tour? Well, one possible parameter here would be using D, the size of a shortest tour, and here again, it seems somewhat unlikely that the problem is fixed-parameter tractable, but for shortest tour, another parameter has been investigated that actually seems rather promising, and that parameter is called using distance from triviality, and there it turns out shortest tour actually could be fixed-parameter tractable. Now what is distance from triviality?

## 10 q Shortest Tour

Of course, we'll do a little quiz to have you figure that out. What I would like you think about is the following: I'm going to give you 3 graphs, and I'm going to ask you for which of these graphs is shortest toward the hardest and the easiest to solve. So here you have 3 graphs, and each of these graphs has the same number of vertices in case you're wondering, or at least I hope it does, I counted twice, but you never know. And what I want I would like you to tell me now is for which of these graphs you think shortest tour is the hardest to solve and for which of these graphs you think is the easiest to solve and I've not written any distances on the edges, but you can basically assume that I would be able to write any distances that I want on these edges. It's more of a question about the structure of these graphs, and for the one that you think is hardest, I want you to enter the number 3 in one of these boxes, and for the one that you think is easiest, I want you to enter the number 1, and guess what number I want you to enter for the one that is in between. So please, go ahead.

## 10 s Shortest Tour

Now again, of course, this is a bit debatable, but I think you will all agree that this one here is actually super easy to solve shortest path on, because let's say you start here, this is your mail station where you start. You go all the way here and then you have no other choice but to go here, because you have to visit all those vertices, and then you go back. so there's not really difficult choices that you have to make when you solve shortest path for this one here. Now this one here is a little bit harder. Let's say it doesn't really matter where you start out because for a tour you can start out at any point because you have to come back to it anyway. And let's say you start out here and then you go up here and then you already have your first choice to make, so do you go this way or do you go that way? Well, probably you want to go that way, but actually there are certain distances where you might want to go this way, so you have to make a choice here. Let's say we go this way, and here we have to make a choice again, so do we go this way and back here or do we go this way and here? So a little bit harder to solve because we have to make certain choices, but still not quite as hard as this one over here, which I would think is the hardest because say we start out here, oh, first choice to make, this way or that way, well, that way. Oh, again, new choice, this way or that way, let's go that way. So all the time we have to make choices, which city to visit next, and this is really tough, so what does this have to do with what I just call distance from triviality? So what does with what I just call distance from triviality? If I mark the vertices in each of the networks, if I mark the vertices in each of the graphs that has more than 2 neighbors, then here we have just 1, here we already have a couple of them, and this one also although this one is fairly easy, but let's be consistent here, and here I have a lot of them, and I'll give you one additional graph, which looks something like this. Now this has a bit fewer vertices, but that's not the point here. The point is that this graph down here doesn't have any vertices that I would color because all vertices have exactly 2 neighbors, and this here is what I would call a trivial instance for shortest path, because you don't really have to make any choices where you want to go. Here you also don't have to make any choices, but what you see here is that the number of vertices that have more than 2 neighbors actually is an indicator of how hard it is to solve shortest tour on that respective graph. If you have a few of these blue vertices here that have more than 2 neighbors, then it's easier to solve shortest tour. Now I cannot give you the precise algorithm for this here because actually that is a bit more complicated to figure out and doesn't only have to do with these vertices here, but I think you get the intuition that we could try to use the number of these blue vertices or some additional structural information as a parameter, and it turns out that if you do this the right way, then shortest tour is somewhat fixed-parameter tractable.

## 11 q Closest to 2 SAT

Now let's do a final quiz here using again distance from triviality so that you have a chance to understand this concept a bit better. Let's consider 3-SAT this time. You already know that 3-SAT is NP-complete, and you 2-SAT is contained in P, so 2-SAT is polynomial time solvable. So let's consider 2 Boolean formulas here. This is the first one here, x1 or not x2 or not x3 and x1 or not x5 or x6 and x2 or x5 or x7 and x3 or not x6 or not x7. Now it's not really difficult to see if either of these 2 formulas has a satisfying assignment or not, and that's not really the point. What I would like you to think about is the following: Considering x1, x2, and x3, which of these Boolean formulas is closer to triviality, and by triviality I mean which one of these is closer to a 2-SAT formula. So look at x1, x2, and x3 and consider what would happen if you were to assign them true or false values. And then you just check the appropriate box.

## 11 s Closest to 2 SAT

And I think the correct answer here is this one above. Again, it's a bit subjective using these small examples, but otherwise it's just too complicated for you to see and explain. So my point here is basically this, if you have decided what you're going to do with x1, x2, and x3, then the following thing happens. You've already decided what you're going to do with x1, you've already decided what you're going to do with x2, and you've already decided what you're going to do with x3, and in that case, what remains is an instance of 2-SAT, and that is solvable in polynomial time. Of course this is a very simple example, so you could have already figured out how to solve it. So there's many different ways to solve it, but I just wanted to show you what I mean by distance of a 3-SAT instance to a 2-SAT instance, because over here you decide what to do with these values. Well, you have it figured out for this clause here, you have it figured out for this one here, but you have not figured it out for that one here. So in a way, the above formula is closer to a polynomial-time solvable instance of SAT than the one down here. But of course you would need a really, really large and tedious Boolean formula to actually take advantage of this. But in fact, many professional SAT solvers or SAT solving packages actually make use of this fact. There are a number of variants of SAT that are known to be solvable in polynomial time, and what these solvers do is they will try to figure out if the Boolean formula that they are given is actually close to one of those polynomial-time solvable instances. Actually, I don't even know if somebody has considered this as a parameter for fixed-parameter tractability, so there might actually be an open research problem for you here if you're ready to get into it.