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>> Alexandra Kolla: It's my pleasure to have Sebastian Cioaba, is that how you pronounce your name?
>> Sebastian Cioaba: That's good.
>> Alexandra Kolla: From the University of Delaware. He was brave enough to come on a holiday weekend as we all found out after we planned his visit. But he's going to talk to us about strongly deregular graphs.
>> Sebastian Cioaba: That's correct.
>> Alexandra Kolla: And give a blackboard talk.
>> Sebastian Cioaba: Thank you, Alexandra. So thank you for being here on a holiday weekend. I'll try not to make it too bad, the presentation. So I will talk about a problem with connectivity of strongly regular graphs. So just a few introductory facts.
So strongly regular graphs. So there are many strongly regular graphs. The definition is the following: We call G, there are four parameters, VK lambda, mu, strongly regular graph.
And I'll just abbreviate that SRG, if the following conditions are satisfied so 1G has V vertices to G's K regular, any two adjacent vertices are going to have exactly lambda comma neighbors. So for every U adjacent to V the cardinality of the intersection of the neighborhood of here with the neighborhood of V is lambda, and four, the same thing.
So for any two nonadjacent vertices, any two nonadjacent vertices will have exactly mu common neighbors. And so for any U and V are not adjacent to V and U and V are not the same. The cardinality of the intersection of these neighborhoods equals mu.
Okay. So these are strongly regular graphs. Obviously there is some kind of -- there are some relation between these parameters. You cannot just pick four numbers, VK lambda mu and plug them in and see if a graph of these parameters works.
Some simple examples of such graphs. So I guess the most famous one is the Paley graph. So Paley graphs, here the vertices. So you take Q, a power of a prime, which is congruent to one mod four. And the vertex set of the Paley graph is formed by all elements of the finite field with Q elements. And you make X adjacent to Y if and only if

X minus Y is a square. If and only if X minus Y can be written as A squared for some A in the field.
So, in other words, like the Paley graphs are, the Paley graphs with respect to the set of generated for this is the set of squares, if you know anything about Paley graphs and so on.
And the fact that you take Q congruent to one mod 4 ensures this is an undirected graph.
It ensures if X is adjacent to Y then Y is adjacent to X. And it's basically equivalent minus
1 is squared. Known as minus 1 is square if and only F is congruent to mode four. But this is an example of strongly deregular graphs. You have to prove that. So the parameters of the Paley. So Paley of Q has the parameters of the vertex, the number of vertices obviously is Q. Then you have as many squares as you have nonsquares, nonzero nonsquares. Nonzero squares and nonzero nonsquares so the regularity is Q minus 1 over 2, takes a bit more work. But it's not hard to prove that lambda is Q minus 5 over 4 and mu is Q minus 1 over 4. This is strongly regular graph of these parameters.
I mean this is a well-studied graph. Has all the pseudorandom properties. The eigenvalues are small. So the edges are very nicely distributed throughout the graph.
So I'll give you more examples about, of strongly regular graphs as we move along.
I mean, I can draw -- I think the smallest is maybe four strongly regular graph I can draw pictures of them. So C4. This is a 4, 2, 0, 1, strongly regular graph. C5. It's a 5, 2, 0, 1 strongly regular graph. And then you have this three-by-three grid where everything in the same row same column is adjacent to each other. So 9, 4, 1, 2, I think strongly regular graph.
And Peterson graph is also strongly regular graph. So this is a 10, 3, 0, 1, strongly regular graph. Again, there are many more examples.
Okay. So now the problem that I'll talk about is it's about connectivity of these graphs.
So the connectivity -- I mean, there are many measures of connectivity for graphs. The one I'll talk about is related to the vertex connectivity.
So K of G, which is the vertex connectivity of a graph of a connected graph G is the minimum number of vertices in a set S. So S is a subset of vertices of the graph. So G minus S is disconnected or a single vertex. And the last condition is put -- this or is to handle the connectivity of, to give a meaning to the vertex connectivity of the complete graph.

Complete graph you cannot break it into two components. So that's the reason for the last part. But in general, so if G is connected, then this is how you define the vertex connectivity of the graph.
So there are -- there is one family of strongly regular graphs, which is not connected.
And that consists of union of clicks. You can take a union of clicks of the same size. So if you take a union of clicks of the size, let's say, K plus one and so on, so take as many as you want, this is going to be -- it has some multiple of K plus one vertices is K regular.
Any two adjacent vertices they're going to leave in the same two clicks any nonadjacent so mu is zero. So basically this is the only graph of mu with zero. It's disconnected. So throughout this talk, when I say strongly regular graph, I'm not going to mean these graphs which are disconnected and their complement. Complement is multi-partite graph with K plus 1 parts on each side. These are called primitive strongly regular graphs and there's not much to talk there about connectivity and so on. Here this is disconnected anyway.
So in 1986, Brouwer and Messner -- so for all strongly regular graphs except for union of clicks and complement, so if G is VK lambda mu strongly regular graph, then the vertex connectivity of the graph equals its degree, it's obvious that this is the largest value of the vertex connectivity, as you can always delete the neighborhood of a vertex, and they also prove that the only disconnecting sets of this size of size K are of the form neighborhood of a vertex for X and V of G.
So disconnecting set is a set S whose removal disconnects your graph, breaks your graph into two pieces. So no matter what parameters you give me, you know, give me a strongly regular graph, which is not a union of clicks or its complement, then the vertex connectivity is maximum K the only disconnecting set of that size are the neighborhoods.
So that's what they proved. And the thing that I like about the result is that so this is a -- you see it's a purely combinatorial result. Their proof makes nontrivial use of algebraic techniques. So what they use in their paper, they use Seidel's classification, so Seidel, they classified all strongly regular graphs with eigenvalues great or equal to minus two.
So I haven't talked about what eigenvalues I mean here, but the eigenvalues that I'll talk here are eigenvalues of the adjacency matrix. So what Seidel did for the -- he said if you give me a strongly regular graph for which all the eigenvalues of the adjacency matrix are at least minus two then your graph is one of this, there are two infinite families and there's some kind of few other examples. So he basically classified all of them.

So that's what brewer and Messner. If you want details about their proof I can give it to you after the talk. But anyway, so it's a combinatorial result which proof makes nontrivial use of spectral result of algebraic result.
Okay. So in 90-96, Brouwer asked the question -- so what if I want the next basically disconnecting set? So all the ones of size K are neighborhoods. Now I want -- tell me what's the size of the next disconnecting set?
So before I tell you what he conjectured so the picture with disconnect sets is the following. You have a set S with vertices and you're going to have here a subset of vertices A in here B, here it could be disconnected. I don't know whether it can be more components but you have no edges between these parts. So that's kind of the picture in a vertex connectivity problem.
So let me denote by K-2 of G. So this is going to be the minimum cardinality of a set S such that G minus S is disconnected and has -- and every component of G minus S is not a singleton. So every component of G minus S is not a singleton. So basically here I mean I don't know how many -- I'm going to have at least two components. But you're going to have at least one edge in each of the components. So you cannot have just a single vertex. I mean, here if you look at this picture, this picture you have a vertex X, the disconnecting set is going to be formed by like S is the neighborhood of X and then here you might have something else. You're going to have something else.
So that's kind of the picture how he looks for -- this would be an A and this would be a B.
So that's how it looks in this situation.
All clear so far? I mean, you remove these vertices you separate X from everybody else.
So it's disconnecting. What he asked is what if I don't want to leave just one vertex on one side, I want to have at least one edge in every component. Okay. So what he conjectured, he said show -- he said show that this K-two of G equals 2 K minus lambda minus 2. Now where does this number come from? Well, in a strongly regular graph, if I take an edge -- I mean, he's kind of looking at the natural candidate for this disconnecting set.
The natural candidate is the following. You take an edge and you look at all the vertices adjacent to at least one of the endpoints. Because the graph is strongly regular, you know that these two guys have exactly lambda comma neighbors by the definition, and therefore let's say this vertex is X, this vertex is Y, X is going to have K minus lambda, and minus one edge that is here. Neighbors of its own. So X doesn't have anything there.

And the same way Y is going to have K minus lambda minus one neighbors. And maybe there's something else, something left here. So now if you look at this, if you look at this set, the set of -- the set of vertices which are adjacent to at least X and Y and are not in X and Y, this is kind of like an A, this is S was I guess you don't see here.
So this would be S, whose size is actually two K minus lambda minus two. And, I don't know, there's some edges here. And I don't know what could be. So he conjectured, okay if, I want to break the graph into nonsingletons this is the best you can do.
Everything clear so far. If you have any questions, just stop me.
Okay. So what we've proven, this is joint work with -- so this is joint work with my name,
Sebastian with Jacques Quilan [phonetic] and Qujin Kim [phonetic], they're at POSTECH in South Korea. So what we showed is that the conjecture is false in general. So they're actually graphs for which you can break them into nonsingletons by removing less than what he conjectured.
So I'm going to give you first -- we have four infinite families of contra examples. I'll give you the easiest one, which might make you think, you know, make you how long did
Brouwer think about making such conjecture because the first contra example is simple.
So the first contra example for the minimum size of disconnecting set whose removal leaves you with nonsingleton components is the following.
So you take M -- it's a triangular graph. So the triangular graph, so M is an integer, I don't know, greater than let's say eight or something like that. The triangular graph is just a line graph of the complete graph, KM. Now, if you're familiar with these notations so the vertex set of the triangular graph is just formed by pairs, XY.
X and Y between one and M. And A is adjacent to B. These two subsets are adjacent if and only if their intersection is one. You think of this as the edges of your graph of the complete graph and two edges are adjacent if and only if they share one vertex. Okay.
Now, why is this strongly regular graph? Well, you can go through the, calculate what V is and so on.
So the triangular graph, the number of vertices is obviously M choose two. Now it's easy to see, I think, that the number of the degrees two times M minus two. And then lambda it's N minus two. And I think mu is four.
Something like this. Let me just double-check it. I mean, you can kind of prove this stuff.
It's not very complicated. Anyway, when you plug this number into what Brouwer said, so

two K minus lambda minus two, it's going to be 3M minus eight. I'm not going to go through the calculation. That's what the value is.
So that's the size of the neighborhood of one edge of any edge pretty much in the triangular graph. So what we showed is that the actual value of the connectivity is one less. So K-two of TM is actually 3M minus nine. And we showed the only disconnecting sets of size 3M minus nine are of the form the neighborhood of modular permutation of your elements one, two, three up to M of one, two, one, three, and two, three. So I'm just abusing the notation instead of making all the brackets I'm just going to write the subset one, two, one, two. And so on.
So, first of all, so why -- here's the picture. So you have the vertex one, two, one, three, two, three, and so here you have a triangle because any two of them share one part.
And now you're going to look at -- so this is kind of my set A. And there I'm going to look at the neighborhood of A. And here it's going to be something that's left, I'll call it B. And obviously by the definition of the neighborhood, I don't have any edges here. So how many vertices do I have here? Well, if here I will have vertices of the form let's say UV.
Well, one of this U has to be one the same as one of the numbers in here. So U has to be between one and three and V has to be between four and M.
That's the neighborhood. And that tells you the cardinality of NA is three N minus nine.
And we actually -- in order to show that -- so what I'm proving here, I'm proving that this
K-two TM is less than or equal than 3M minus nine. In order to show that this is actually a disconnecting set of these properties you have to show when I remove this set from here, I don't end up with singletons here. I don't know. It might happen. But here what do you have? In B you have all the vertices of let's say XY with four less than X, less than Y, less than M, and this part is basically the same, it's isomorphic to the graph, the triangle graph TN minus three, which is connected.
Okay. Now, so, again, I gave this talk about a month or so ago at a workshop at Banff and Godsell kind of pointed out that he didn't know how much he doubted that maybe
Brouwer thought much about this problem.
Anyway, this example is one of the first strongly regular graphs that you think of to try it.
It's pretty simple.
So this gives the upper bound. How do you know this is basically the exact value? So here, again, when you're trying to prove a lower bound on this parameter, the picture for us is always the same. You have a set A such that what you have here is a connected subgraph. On, let's say, at least two vertices.

You take here the neighborhood of A and here you take what is left, which we call B. So now what I'm trying to show is actually something -- yeah, what I'm trying to show is that for any A with at least two vertices and at most let's say half the number of vertices which induces a connected subgraph, then the size of the neighborhood of A has to be great or equal than whatever value you want these 3M minus nine.
So to prove something like this in general it's not -- we didn't find it that easy. So for specific graphs like this one, you can prove it using the definition, the structure of the graph.
So you split this proof into two cases. So the first case is if A is a click. Now, if A is a click in this triangular graph, then you have two situations. It either looks something like
A looks like one, two, one, three, and two, three, that's one type of a click. Another click would be if A is an edge, in which case you know exactly what the neighborhood is.
And the other kind of a click will be of the form one, two, one, three, one -- let's say A plus one or something like that. So all the vertices share one endpoint.
So in each case here you know what the cardinality of what NA is going to be. It's already done. 3M minus 9. In this situation the cardinality of the neighborhood is -- it's going to be bigger. It's much bigger than 3 M minus 9. You can do this combinatorially too -- if your A is like this then you know exactly what you're going to have here.
It's a vertex which pairs which contain at least one, two, three, A up to one. There are many of them.
The second part of the proof is if both A and B are not clicks, so in this situation if A and
B are not clicks, it means that A is going to contain a claw, something like this, which, again, with all this loss of cardinality you can assume it's 1, 2, 1, 3, and 2, 4; and then B it's going to contain, again, a claw here. Again assume it's 5, 6, 5, 7, and 6, 8.
And now to show that I want to show that in the neighborhood of A has at least 3 M minus 9 vertices so here the approach is similar to if, you're familiar with the Mengers' theorem for vertex connectivity, so vertex connectivity of a graph, you know, to disconnect these guys from these guys, what is an obvious lower bound for that?
Well, it's the maximum number of interior vertex disjoint paths that start in this set and end up in this set. So here basically at this point it's combinatorial proof. So we look for

many paths that go like this or we're going to find many vertex, interior vertex disjoint paths that start with one of these vertices.
This is very easy to -- I mean, you have to play a bit with them, but you can have paths of this form one, two, goes to let's say 1 X which goes to 5 X and here this X is anything between 9 and M.
This is a path, an example of a path. And then you can do something similar. 1, 2, 2x, and here you put 6X, the same thing. And you can find more paths.
At the end you end up with at least I think it was 4 M minus 16 interior vertex disjoint paths. From this set of three vertices and this set of three vertices.
So if there's at least this many, it means if NA is going to break that part from this part, it has to contain at least one vertex from each of these paths. So this tells me that the size of NA is at least 4 M minus 16, which beats 3M minus 9 or something like that.
So here it's just combinatorial, looking for paths. So I'll leave that there. And let me write the second contra example.
So the second contra example is the following: It's called simplectic [phonetic] graph. So there's two parameters, R and Q. R is an integer great or equal to Q and Q is a prime power.
And there is a matrix involved. So M is going to be a two R by two R block diagonal matrix with a diagonal block is this one, 0 minus 1, 1 and 0.
So basically for R equals 2, this is 0 minus 1, one 0, 0 minus 1, 0, 0 and 0. That's in that case.
So how do you define a graph from here? The vertex set of this graph is going to be formed by the one dimensional sub spaces of two R dimensional space, vector space over FQ.
And so this one dimensional, if I refer to a one dimensional sub space by X this is one of the elements of the basis of that sub space.
So I say X is adjacent to Y if and only if the value of the following expression X transpose
MY is non-zero.

So some people call -- this is called simplectic graph over FQ.
Some people call the simplectic graph the complement of this one where X transpose MY equals 0.
So this expression, this form it's called simplected form, I guess it appears in geometry and so on.
Now, why is this -- why is this a strongly regular graph? I'm not going to give you the whole proof, but it's not very complicated. So anyway let me just write down the parameters.
So the parameters of this simplectic graph 2 RQ, the parameters are the following. So V, the vertices are the one dimensional sub spaces, so that's going to be Q to the 2R minus
1 divided by Q minus one divided by Q minus 1, and you can calculate K lambda and mu and they're very close.
K is Q to the 2R. Lambda and mu are the same. And they're Q to the 2 R minus Q to the
2 R minus 1.
And this is not -- this is just basically inclusion/exclusion, something very simple using dimensional arguments. It's not very complicated. So here's a very dense graph here if you think about it you have 2 QR to something a bit smaller and it's very, very dense graph.
So the value that Brouwer conjectured, 2 K minus 2 lambda minus 2, if you plug it in it's going to be 2 QR to the minus 1 plus Q to the 2R the minus 2 and what we proved is that the actual value of this connectivity for this graph is Q to the 2 R minus 1 plus two Q to the 2 minus 2 minus 1.
>>: The first parameter is that vertices?
>>: Yes.
>>: The second one is the number of neighbors? Thank you.
>> Sebastian Cioaba: So somebody is paying attention. That's good. Thank you.
Yeah. So it's Q to the Q R minus 1. Thanks.

So but these values are the same. So when you plug these numbers into here, if you plug this -- let me see, did I mess up another one? 2 R minus 1. Two R minus 2. Sorry about this.
Also lambda and mu have to be smaller than K. Yeah, those are the exponents. You put
2 K minus lambda minus one, yeah, that's correct.
So that's what we prove. So it's basically the difference, if you let your Q be big, the difference between what Brouwer conjecture and what the actual value can be, can be as large as possible.
And so how do you prove something like that? The easy, to prove an upper bound you have to come up with again a set A whose neighborhood has that many vertices and whose removal of the neighborhood is going to leave you with nonsingleton so the A we took is the following. So you take X and Y to adjacent vertices in your graph. So XY in the vertex set of this graph. And then you take A to be formed the set from X, Y, and everything that looks like X plus alpha Y, where alpha is an element of FQ. Say nonzero.
So X and Y and everything that looks like X plus Y. So this A is going to have size Q plus one. 2 here and Q minus 1 there. Also another important property this is a click. You can show in this graph any two -- because X and Y are adjacent, X and Y are adjacent, this is going to mean that X transpose MY equals 1 and then you can use that relation to prove if I take X, I don't know, let's say X plus alpha Y transpose N times X plus beta Y, well, what's going to happen with this form, with this matrix M, every vector is perpendicular on itself. If I took X transpose MXI get 0. So at the end when you compute this -- here, this means nonzero because it's adjacent. When you do this calculation, what is it going to be? Anyway, you show it's always nonzero so they're always adjacent.
This is a click of size Q plus one. In geometry they call it hyperbolic line.
And now I want to figure out -- this is kind of the set A, and you want to figure out how many vertices you have in here. Well, so the size of A is going to be Q plus 1. So you can figure out what's in here by doing some inclusion/exclusion argument which takes about two pages, and you can calculate the size of NA is actually going to be what I wrote here.
Or if you have friends or colleagues who are finite geometries you can tell them about this problem they can tell you again A is hyperbolic line it's called inflicted power space, and then the vertices in B are the ones which are orthogonal on your A with respect to this M. And now to be orthogonal here, it's enough to be orthogonal in X and Y because then you're going to be orthogonal on every X plus Y. So the size of B is going to be

basically Q to the 2 R minus 2 minus 1 divided by Q minus one. This is going to be like a
2 R minus 2 dimensional space.
And so you have this many, this many, so that will make N of A to be the number of vertices of your graph minus this and minus this.
Which when you do the calculation it gives you exactly that. So that's that. Now, so this tells you that the cardinality of this neighborhood is at most this value. Now, what you want to show is that in B you're not going to have Singletons. And that's a bit more work, which I'm not going to go into now. It's not complicated. You prove it by contradiction.
You the that you have a singleton here, and you use the fact here you have a click and now the one fact that you use, if you were to have a singleton here, all its neighborhoods would have to go here and now what would happen?
A vertex in your click and this guy in the original graph are not adjacent. So because they're not adjacent, you know exactly the number of common neighbors. So they would have here you know exactly the number of common neighbors. And then you can do some averaging argument to show that this cannot happen.
Anyway, so I'm not going to go into details of that. So that's -- so this is how you get an upper bound on this value of this connectivity. Now, how do you prove a lower bound?
So for lower bound, so if I have a set A, let's say in here such that this induces a connected subgraph, and let's say the cardinality of A is let's call it T, then it's actually easy to show that the neighborhood of A has to have at least 2 K minus lambda minus T vertices.
This is true in any strongly regular graph. So here's a picture how this works. So if you have your A here, it's connected. So you have at least -- actually you have at least one edge in there. And now you're looking at the neighborhood of A so you could have edges here. But you're going to have edges from here. And these edges could have -- these vertices could have neighbors here at well.
So if I look at these two vertices which are adjacent here. The cardinality of the neighborhood of that is going to be 2 K minus lambda minus 2 and in the worst situation, every vertex in here, there are T minus 2 of them, are going to be adjacent to at least one of these guys.
So here you're always going to have at least 2 K minus lambda minus two minus T minus
2, which is exactly this one.

So this is very simple. And you can analyze this, what happens in equality case. You get some constraints. So for this problem to show that this connectivity actually equals that value, if T is less than or equal than Q plus 1, then this is good enough, because you plug it in, the number, it gets you exactly what you want.
Now, if T is bigger -- so I'm just going to erase here. If T is great or equal than 2 plus Q then you have to do something else because this is no longer good enough. So you use eigenvalue bounds. So vertex connectivity, so there are large literature on spectral graph theory in connectivity, expansion constant and so on and eigenvalues.
Vertex connectivity sometimes the bounds you get are not the nicest, they're pretty messy and technical. So here's what is known, and this is the result of Hammers, and I think there's another paper by Mohar and some other co-authors on the bandwidth of graphs. So people proved the results of the following formation. You have this picture.
So this is S. This is A and this is B. And you have no edges between A and B. Then you can find an upper bound on something like this size of A times the size of BN minus the size of A times, N is the number of vertex minus B is less than or equal to do with something with the eigenvalue of the graph. You have to basically -- to get your hand on
S you have to impact this and try to see where S lies.
So for strongly regular graphs it turns out that things work out nicer for strongly regular graph if you impact this result, you get the size of S is at least something as follow four times AB mu divided by lambda minus mu square plus four times K minus mu. Where A and B little A is the size of A and the little b is the size of B. And this mu lambda K are the parameters of your strongly regular graph.
So that's pretty much what you get in general for any graph. And now how do you get this result? Well, if T, which was now in my case the size of A, if this is bigger than Q plus 2 and I the by contradiction that the size of S is less than what I proved, then what's going to happen? Well, I'm going to have A plus B is the number of vertices in my graph minus the size of S, but if I the that S is less than this, then this is going to be bigger than something. And now if A is bigger than Q plus 2 and A plus B is bigger than something then the product A times B is going to be bigger than Q plus 2 times something. And you plug the numbers in. It's simpler here because you see lambda and mu are the same.
This is 0. Simplifies. It's pretty technical and it goes on.
And that's how you pretty much prove it. Actually, so there are two other contra examples. I'm not going to talk about them now. There are two other contra examples in which you, the connectivity and the bound, the value of the actual value of this connectivity and what Brouwer conjecture differ by 1 and those two graphs are induced

sub graphs of this guy when Q equals 2. They're called hyperbolic quadric and elliptic quadric graphs. And I'm not going to talk more about them.
I want to tell you about the method about proving that the conjecture is true for some graphs. So what we know about other graphs. So we can show that the conjecture is true for many graphs. And the only way, the only method that we have to prove that is through this eigenvalue thing. So if you gave me a strongly regular graph for which I don't know much about its structure, I just know the parameters. You tell me the structure. Then the following is the result that we have. So let me erase it here.
Something of the form if 4 times K minus 2 lambda times K minus mu is bigger than lambda minus mu square times 2 K minus 3, minus lambda minus 3 then the conjecture is true for that VK lambda mu.
Obviously you want your K to be strictly bigger than 2 lambda. And if lambda and mu are close enough, then that would work. So this kills many actually parameters. So this will kill Paley graphs and a bunch of other graphs. But, again, like I said for a general graph, if you give me just the parameters, without telling me how is this graph constructed and so on, this is the only result that we have. And we like to extend that to do something better. And the way you get this is following pretty much the same steps here through the eigenvalues. You follow the same step.
You say it's pretty much the same, the same idea. What else can I tell you? This is where the conjecture is true. It's also true with lattice graphs which are the graphs where it's basically an M by NI showed you a three by three but you can do M by N lattice graph
M is just the line graph of K and M and that's true and there's a generalization of, extension of this graph which comes from Latin squares, which they're called Latin square graphs. It's also true there. Anyways...but again you use the structure of the graph. You don't have some general argument that can -- okay -- prove it.
So what we like to do, we like to classify all the graphs that are contra examples, because it seems like they have this -- so in strongly regular graphs there are what are called geometric graphs come from finite geometry so that will be a step for us to do to figure out which graphs if there are more or contra examples.
Another thing to do would be to extend this problem to study this problem for so the generalization of strongly regular graph are called, it's one generalization is to look at distance regular graphs. So strongly regular graph have these regular property and they have diameter 2. If you look at the vertex, the neighborhood and what is left, you have something very regular.

But you could have similar graphs for which the diameter is bigger than 2. So those are distance regular graphs. And more general, there is a kind of unifying concept in algebraic graph theory which is called association schemes. Again, I'm not going to talk about that. But that would be an interesting place to study this problem.
So I think I'll stop here. I think I'll try to keep it short. Thank you very much.
[applause]
>>: Is there a classification for which parameters there exist strongly --
>> Sebastian Cioaba: That's very good. There are necessary conditions. So when I -- so you know like the simple thing you can get is if I have a graph, I take a vertex, the neighborhood has size K and then you can do some double counting of the edges in here, and you get something very easy like K times K minus lambda minus one equals mu times V minus K minus 1. So this is the simplest constraint. And then you can go a bit deeper into the theory and you can look what happens here, the adjacency matrix of the graph to guess I have them defined and identity they want a matrix they generate a algebra, which is called a Bows Messner algebra, the same Messner. You can look at what are called a principal item potence of that which are the projection on the eigen spaces and you can get some constraints, more constraints on this which are called the crime bounds in the literature you can --
>>: What are the matrices the adjacency and the --
>> Sebastian Cioaba: And the all one. The all one matrix. Actually, it's equivalent to -- so if you look at identity A and J, whatever they generate it's the same thing as identity A, and the adjacency matrix of the complement of the graph, which is -- and this -- so if you look at, for example, if I look at these three matrices, they commute. So what you can do, you can diagonal lies them at the same time so you have a common basis of eigenvectors and for strongly regular graph you have three eigen spaces correspond to K the all one vector pretty much and two which are orthogonal to that and you can take the matrices which are the projections these are principal importance, projections on the eigen spaces positive semidefinite matrices you can play a bit with it. And just from positive semidefinite constraint you're going to get some constraints about this. So these are called crime bounds and they kill off many parameters.
So if you go on an dress Brouwer Web page he has a table of kind of feasible parameters for strongly regular graphs up to a thousand and something. So they go in groups of 50.
And on that table it's very nicely organized because it tells you if the graph exists it gives you a list of the parameters. If the graph exists, they're always in green. It tells you the

construction. If the graph doesn't exist, it's in all red and it tells you what condition actually kills it.
So this is kind of a trivial condition. So parameters which do not satisfy this are not going to be there. But there are parameters which do not satisfy this thing that comes from this
Bosner Messner algebra there are bounds what are called crime bounds and for whatever parameters -- you know, for many parameters for which people have proved nonexistent it's many times like an ad hoc argument combines in computer search with some combinatorial things and so on.
So --
>>: Density of parameters?
>> Sebastian Cioaba: Yeah, not such thing. I mean, no. No. But that's a good question. Thanks.
Any other questions?
>> Alexandra Kolla: If you have any more questions [indiscernible] ask a few more questions later on.
[applause]