 One last thing that I want to do here before I have Fred come on up and welcome you is that I want to obviously think so Ben Taylor and Lauren For helping put this event together and also point them out because if you have questions They're the ones to go see or or myself here So thank you all again for coming on out here, and we'll have our Dean and should come on up here and welcome us Officially start us off Thanks Good morning It's great to have you Wow Full of house and to Have a conference here at Grand Valley State University of graph theorists Some mighty ones and some mighty ones in training That's a real gift to us We thank you. We really do want to welcome you Michael has been trying to tell me what Exactly a graph theorist does because I'm not a mathematician You will no doubt Figure that out as as I speak but I Understand a few things about you understand that you're kind of a network of people who think deeply about networks You come here at a time when our college has been paying special attention to how data and narrative Support one another and certainly visual representations of information and Interrelationships is critical to understanding Well, this coming week at Grand Valley is Also in the wake of having you also a Week of celebrating the classics So I thought it would be appropriate as a way of describing How the world is in need of specifically you To share with you mathematicians the famous passage in book four of the Aeneid Where Virgil describes the way? rumor when not pent By truth moves in the world a world desperate for ways to understand He was writing about our need for you When he wrote rumor Compared with whom no other is as swift She flourishes by speed and gains strength as she goes First limited by fear rumor soon reaches into the sky walks on the ground and hides her heads and head in the clouds earth inclined to anger against the gods so they say bore her last rumor a monster vast and terrible fleet winged and swift-footed Sister to coious and then sell it us who for every feather on her body Has as many watchful eyes below as Many tongues speaking as many listening ears She flies screeching by night Through the shadows between earth and sky Never closing her eyelids in the sweet sleep By day rumor Sits on guard on tall rooftops or high towers and scares great cities Tenacious of lies and evil as she is a messenger of some part of truth Well, whether you use a standard model of daily and Candle or the Maki Thompson model or your own model you too can tell us stories But true were stories of the spread of information through social media of our time My Virgilian example is meant to suggest that this has been a Concern of incredibly enduring interest to human beings and these days Maybe it's not too bold to say it's crucial your work is crucial to understanding our new civic reality in short your contributions to the understanding of complexity and in particular To ways we can get our head around it Have for a long time been powerful, but have never been more vital So we got you here at a good time on behalf of our college and the university Thank you for bringing your conversations here to us and our students are Really pumped up to hear you. Thank you so much. I Wish you a Continuously productive meeting and Ask you please disseminate your findings wide widely to a world in need of specifically you and In desperate need in wild search of ways to understand To which about which we look to you Thank you for coming. It is great to have you and now The real need of the matter Okay, so we are very happy to have here as our first plenary speaker Doug West Probably most of you already know him. So I have most of what I'll say is not news He has graduated 38 PhD students now has 59 academic descendants His publication list is quickly reaching towards the 250 mark and that's with over 200 co-authors already Aside from the refereed articles he's written some books depending on how you count them they are between two and seven Looking very much forward to it being seven someday because I would like a bound version And the last thing I want to do is take away any of our time to hear Doug talk to us about reconstruction So I'll just keep it there and please help me join Doug to GVSU Thank you, Ben Nice introduction Okay, I hope you can all see this big enough I'd like to oh, yeah, so So I've been coming to mighty meetings for a long time I'm not sure when the first was but it might have been 35 years ago So I'm I'm happy that mighty is still here. It's been a wonderful Series of meetings I think and it's great for Graduate students to come and be able to talk in a friendly environment And get themselves started onto bigger and better things So mighty has I think really been a great thing for the development of the subject Okay So I'm going to talk to you about Reconstruction This is joint work with the next to last student of mine Hannah Spinoza And you can find these slides at the top of my preprint page So you can review them later at your leisure or follow along now if you like Okay, so Let's talk first about what the classical problem of reconstruction is When we have a graph G We say that a card of G is an induced subgraph that we get by its leading one vertex And so when you have cards you have a deck of cards the deck is the multi-set of all the cards So if you have in vertices you will have in cards and some of them may be repeated So for example here This graph has these four cards and the famous reconstruction conjecture By a published by Kelly and by Ulam is that any graph with three vertices is determined by its deck So you see I've sent the arrow the other way also here now If we have these four cards, then we know we have a triangle in our graph and The fourth vertex there must be four vertices because there are four cards and the cards all have three vertices The fourth vertex must be put back in adjacent to some number of the triangles And so you have the cases zero one two or three and the only one that will yield this deck Is when you put the fourth vertex back in adjacent to two of the vertices So the deck determines this graph So actually the reconstruction conjecture was mentioned earlier in Kelly's thesis Which was in 1942? So it's appropriate to talk about reconstruction now because this is the 75th anniversary of the reconstruction conjecture and So it's a hard problem. It's been around a long time So what do we do when we have a hard problem? Oh, yeah, well we write lots of surveys about it. Okay Okay, what else do we do? Let's see We say well Notice that actually This graph is determined by just three of its cards again. We have to have a triangle and If we put the fourth vertex back in adjacent to all three we'd have many triangles so that wouldn't be right If we made it adjacent to only one or two we would only have one triangle so because we have so we must Again with just knowing these three cards the only graph that has these three cards in its deck is K4 minus So on the other hand Which three cards if we take these three cards There's another graph that has these in their deck Namely if we take k13 and just add one edge we have a triangle and we have these two three vertex paths as induced subgraphs, so It's determined by three cards three of its cards, but you have to be careful about which cards so As I said, what do we do when we have a very hard problem? We define a more general problem, right? That's harder in some sense But it's more detailed so Harari and plant plant halt Define the reconstruction number of a graph to be the minimum number of its cards that Do not appear in anybody else's deck, so they determine the graph, okay? so In that the reconstruction number of this graph, I think is three You have another concept that you could consider which was introduced by Mirvold Which is the adversary reconstruction number saying, okay? What's the minimum K such that no matter what K cards I give you You would still Have this graph determined so that's going to be greater than or equal to the reconstruction number Okay, so you have those concepts and so just as a little bit of background for The idea of making a more refined version of the reconstruction problem Let me tell you a little bit of the results about reconstruction numbers so Mirvold's Showed that the reconstruction of any Disconnected graph with at least two non isomorphic components is three. So that's quite small And it's also three for all trees with at least five vertices Bolabash and earlier Mueller had proved that for The fraction of graphs as n tends to infinity that have reconstruction number three tends to one That's what we mean when we say this is true almost always On the other hand the reconstruction number can be large if you take a disjoint union of Complete graphs with our vertices then the reconstruction number will be r plus two and The reconstruction number of any graph is equal to the reconstruction number of its complement okay, because the deck of the complement is You get by taking the compliments of everything in the deck Okay So in particular if you take the complete bipartite graph with n over two vertices in each part It shares n over two plus one cards with The slightly unbalanced complete bipartite graph So that shows you that you need at least n over two plus two to determine this complete bipartite graph And Harari and plantel actually conjectured that this is the largest the reconstruction number can be for an n vertex graph and That equality would hold only for this graph the complete bipartite graph and Also for its complementing Okay Trying to avoid a cold here Okay So now in fact Later Not very long ago It was discovered that there are some other graphs that have reconstruction number this big this n over two plus two so This only This proves the uniqueness part of the conjecture. So they still did not have larger reconstruction number Okay, so we're going to consider a different direction of Making a more refined version of the reconstruction conjecture. So actually Kelly 60 years ago Had the following notion Reconstruction conjecture we delete one vertex and we have the deck What if we consider the deck of all the cards that we can get by deleting L vertices from our graph Then he conjectured that If you have enough vertices if n is sufficiently large Then every n vertex graph would be determined by the deck of cards obtained by deleting L vertices and the original And so we'll have this notion the graph would be L reconstructible so In a sense the larger that L is that you can make L as I'll Make precise in a moment The more easy the graph is for you easy to reconstruct So the original reconstruction is just that m sub one is three and if you have at least three vertices You can reconstruct from the deck So McMullen and Rod Satovsky actually asked Whether m sub two is equal to six when I say they asked that the computer search they proved that Every graph with between six and nine vertices is Reconstructible from the deck of two vertex deleted sub graphs But you know that was the limit of their computers And the what about on the other end so if you take a four cycle plus one isolated vertex That has the same Deck of three vertex sub graphs deleting two as The graph that you get by taking the claw and subdividing one edge Okay, so you can check that out. So those graphs are not too reconstructible They have the same deck of three vertex sub graphs so I'll say that the k-deck of a graph is this multi-set of Induced sub graphs having k vertices then the reason why this L notion of L Reconstructibility makes sense is that the k-deck of a graph determines the k-1 deck It's very easy observation because each graph in the k-1 deck would occur and Minus k plus one times by deleting one vertex from some graph in the k-deck So that just means you know by counting the occurrences in your k-deck of a particular k-1 vertex and do sub graph We just divide by this amount and you and you find out how many times it appears in the k-1 deck Okay, so that means that the smaller you can make k and still be able to reconstruct the Graph the better off you are so we're interested in finding the least k so that the graph is reconstructible from the k-deck Okay, and this will be the have the same meaning as L Reconstructible when k plus L is the number of vertices so sometimes We're interested in talking about how many vertices do we need and other times how many vertices can we delete? so for example among our results The least in Such that even telling whether the graph is connected or not can always be Reconstructed from The deck you get by deleting L vertices is going to be more than two L okay that's not so surprising, but That's an example of what we can show so let me list for you some of the results that we have and So in particular what I just said follows from this result Which says that if you so for example if you have a cycle with n vertices and Your other graph is two cycles with n over two vertices You can't tell them apart By looking at the k-deck if k has less than n over two vertices Okay, that's maybe not too surprising, but yes, even you break up the cycle into two pieces. Well you You have the same number of each Induced sub graph with k vertices if k is less than n over two and similarly if you have a path with n vertices You can compare it with a cycle and a path that where each of those has about n over two vertices Let me not be too precise about that This statement is also sharp That if you have one more than that many vertices you will be able to tell them apart Okay, in fact you'll be able to reconstruct the cycle or the path So More generally for any graph with maximum degree to We can determine the least k such the graph is k-deck reconstructible okay, and and Well a corollary of these results as I mentioned with the floor of n over two vertices in The cards in your deck you may not be able to tell whether the graph is connected Now Manvel showed that it's quite easy to see that You can tell whether a graph is connected from the deck obtained by deleting one vertex And he showed that you can tell whether your graph is connected from the deck obtained by deleting two vertices So we strengthen this Yeah, so He showed that that's true when it ends at least six and again the same example of the cycle with an isolated vertex or The claw with one edge subdivided shows that you need and at least six because one of these is connected And the other one is not and they have the same N-2 deck Okay So what do we show? We show in particular that for three reconstructible If you have at least 25 vertices You will be able to tell whether the graph is connected or not from the deck obtained by deleting three vertices And for L in general for the deck obtained by deleting L vertices if N is kind of big It's bigger than something like L to the L plus one squared Then you can tell from the deck obtained by deleting L vertices whether the graph is connected or not And now and this one is kind of a generalization of that result of Bollabash and Mueller that again for a fixed L well as long as N is somewhat bigger than to L then there's a small set of Subgraphs obtained by deleting L vertices that will determine whether the graph is connected or not Okay So I want to tell you some about all these results We'll see how far we get But before we go into them, let me do a couple of things that are easier just to warm up Let's think about K deck reconstruction from the deck of K vertex induced subgraphs when K is small Well K is 2 Then all you know about the graph is the number of vertices and the number of edges So there aren't too many graphs that are determined by the number of vertices and the number of edges just complete graph Delete one edge. That's okay, and the complements of those and that's all that's determined by knowing the number of vertices and number of edges so So we want to use this result, which is also pretty easy Manvel shown that if you have the deck of K vertex induced subgraphs where K is at least the maximum degree plus two Then you can determine the degree list of the graph Okay, it's pretty easy if you have subgraphs that are that big Then you're going to see any vertex with all of its neighbors right and the fact that you don't see a Subgraph that has a spanning star Tells you what the maximum degree is and then you can continue on and find the whole degree list. It's not very hard So what are we going to do with that? We're going to show that if you have a graph That is every component is complete okay, then as long as the components all have at most m vertices then you can Reconstruct from the deck of m plus one Vertex subgraphs So why is this? Well m plus one is at least three This we We're given the m plus one deck so we know the three deck So we know whether or not the three vertex path is an induced subgraph With three perfect vertex path is an induced subgraph if and only if You are not a disjoint union of complete graphs right So we find out that G has the form is disjoint union of complete graphs and all we care about now is the sizes of those complete graphs and That's going to be determined by the degree list of the graph Okay, so that's fairly easy that we can reconstruct all of those so we can also reconstruct their complements which are complete M-partite graphs with at most in m. Sorry We don't care how many parts there are the complement just says it's a complete multi-partite graph where all the Parts have size at most now So a little bit more interesting is to say okay Let's allow large parts But we have an r-partite complete graph Okay So that's what we can we show that any complete r-partite graph Will be reconstructable from its r plus one deck So again since we know the three deck We know since there's Since we don't have The complement of p3 as a card. We know that we have a complete Multi-partite graph and since we're given the r plus one deck We know we do not have a complete graph with r plus one vertices So we know it is a complete r-partite graph and now the question is just what the sizes of the parts are So we do something a little more interesting to determine that If the part sizes are q1 through qr Let's form this polynomial The product of x minus qi x minus the part sizes Okay Well if you take this polynomial and multiply it out then When you want the coefficient of x to the r minus i you have the product of i Of those terms that look like minus q sub i. I have too many i's there, but that's okay Well, what is the product of i choices from q1 through qr? That's the number of i vertex complete subgraphs that you have Okay, so since you know the r plus one deck You know the decks for all these smaller things the d sub i with i vertex induced subgraphs As I said s sub i is the number of cards in d sub i that are complete So we know that so we know all the coefficients of this polynomial So we can find these integer roots and that tells us the sizes of the parts Okay, so That's a little bit interesting not too hard okay, so I'm going to go kind of in reverse order to the of the results that I told you I would tell you about So let's start by thinking about almost all graphs and the the key Lama here As I said Miller did this in 1976 Bollabache came along and did it again in 1990 If you have a bit more than half of the vertices of the graph then those induced subgraphs almost always are Paralyzed isomorphic and have no non-trivial automorphisms So random graphs are really random in other words. They have there's no symmetry Even when you look at these subgraphs with slightly more than half of the vertices So that's the the key thing that we use and the theorem that we prove So I'll say that the subgraphs with a certain number of vertices are good if they are paralyzed non-osomorphic and have no non-trivial automorphisms, okay, so our theorem is that if the Subgraphs that we can get by deleting L plus one vertices are good Then we will be able to reconstruct the graph from some set of Subgraphs obtained by deleting L vertices and not very many of them I mean there are in choose L of these subgraphs. We only need L plus two choose two of them, okay, so and As I said that what we will get as a corollary from this lemma About random subgraphs Random graphs is that the fraction of graphs that are reconstructable from the substrafts obtained by deleting Slightly less than half of the vertices will tend to one as n tends to infinity Okay Okay, so the first thing I need to tell you is which subgraphs I'm going to use, okay So what we're going to And and this is a short version of what I just said the theorem is So let n be the number of vertices in our graph and let's fix L plus one vertices Okay, a particular set of L plus one vertices Let H be the subgraph that we get by deleting those L plus one Remember we really want to Reconstruct from the deck obtained by deleting L vertices, but I have this special graph Where I deleted L plus one vertices And let H be the number of vertices in it so what I'm going to do is Give you some Subgraphs with H plus one vertices. That's where I've deleted L So if I have this X sub I in my special set Consider the subgraph Induced by all the vertices down here plus X sub I okay So when I do that For all I then I will get L plus one subgraphs. I call that collection of subgraphs C Okay, C sub I when I leave X sub I in Now I need some more Besides these I need some subgraphs where I leave two of the special vertices in Okay, and if I'm going to do that then I need to leave one of the vertices down here out So that again, I have the same number of vertices Okay, so that's going to give me L plus one subgraphs like this and L plus one choose two subgraphs like that and this where I get my collection of L plus two choose two subgraphs and the idea is Because the H vertex subgraphs are good That I'm going to be able to tell which are the special vertices and So I'll be able to find from I'll be able to identify C So I'll be able to find from C this subgraph and all the edges between S and here and then I go and look at these and Use them to determine for each pair of special vertices whether they're adjacent or not That's the basic idea Okay So to make this so this is my claim right that I've be able to reconstruct the graph from these vertices these subgraphs that I've picked Okay, and let me point out that there are many such families Because I could have chosen my special set of vertices as any L plus one vertices out of the end so that's n choose L plus one right there and then for each pair here when I make this subgraph I could pick almost any There's a little bit of constraint. We'll see of these vertices down here to delete So that is an additional something like H to the L plus one if so many many collections that will work Okay, so now to do the reconstruction The cards have H plus one vertices and The vertices with that many vertices are good Sorry the subgraphs with that many vertices are good meaning they're pairwise nine isomorphic and so let's ask which H vertex subgraphs can show up in these cards by deleting one vertex well if you If your H vertex subgraph has three of the special vertices then it can never show up Right because the cards we picked never used more than two of the special vertices and the H vertex subgraphs are pairwise non-isomorphic so those Subgraphs H vertex subgraphs can't show up at all as subgraphs of cards in the deck If we have two of the special vertices then it can show up only if in In the card d sub ij that we had okay If we have one of the special vertices Well that occurs as a one of the C's right one special vertex and all the others down in H and It also can occur sometimes in the d ij if you keep this one special vertex And you've thrown away the other guy it matches and you have just the right set down there okay Well, we want it to show up a total of it at most L times and we can ensure that but just making sure that When the cards that d ij Overall j right new you keeping x i then when you put in some xj the w ij that you chose to Delete down here is not the same vertex all the time Okay, so that's all you need to make sure that one of these guys is going to show up in at most L cards But the special vertex X The special subgraph H Sorry is One that doesn't share any of the special vertices So he's specifically that graph H. He will show up in all of the L plus one cards that are in C So he's the only graph With H vertices that can show up in the deck L plus one times Aha, so we can tell who he is once we know who he is we look at all the places where he occurs that C So we know and you know and we know which subgraph he is because Again he shows up there because the guys are good So we determine each of the special vertices From C we have all those edges and then as I said we can go back now that we know who X i and X j Are we can go over to the DIJ and find out whether X i and X j are adjacent So this is the summary of those things I just said, okay Okay, so we've proved that one So now let's think about connectedness Okay So we're going to fix L the number of vertices that we delete to make our deck and n is going to be large So let me let C of D when I have some deck D be the number of connected cards in the deck So here's the easy thing about deleting one vertex when you a graph is connected One of the first things we learn in graph theory If a graph is connected it has at least two vertices that are not cut vertices Right leaves of a spanning tree So that means a connected graph has at least two connected cards deleting one vertex But a disconnected graph has at most one connected card so from the deck Deleting one vertex you can easily determine whether a graph is connected or not so Man Vell As I said did it with n minus two vertices as long as you have it six vertices in the graph so Let's try to get a bound on how many vertices we can have if we have a connected graph and a Disconnected graph that have the same n minus L deck Okay, so let's suppose we have G and H G is connected H is disconnected and They have the same n minus L deck so We must have at least one connected card in the deck right because I mean when you have a connected graph You're looking at G if you just peel away leaves from a spanning tree. You're always going to have a connected sub graph Okay, so our disconnected graph must also have a card that's connected right because they have the same deck So that means it has to have a component with at least n minus L vertices As long as L as long as n is bigger than 2L. There's only going to be one such component okay, and Let it actually have n minus p vertices. Okay, so p is less than or equal to L so When we look at H The number of connected cards Again with n minus L vertices We must throw away the vertices that are not in the big component and Some more vertices and L minus p of them from the connected component So to obtain connected cards in the deck that comes from H the disconnected graph there will be At most n minus p choose L minus p of those right? We throw away the cards outside the vertices outside the big component and some more some L minus p more on the other hand an Upper bound would be n minus one choose L minus one on the number of connected cards Well, just because that's an upper bound on n minus p choose L minus p. Okay on The other hand if we let ah, here's the definition of C hat The number of cards having a small component a component with fewer than L vertices Well if you keep one of the vertices that's outside the big component and Then just delete some other L minus one vertices Then you're going to have a small component So you get a lower bound on the number of cards of H that have a component That's small a lower bound which is n choose And my L as I said you pick L vertices to throw away besides the one that you've kept That's outside the big component So the idea here is we have looked at H and In looking at H. We obtained an upper bound on the number of Connected cards in the deck and a lower bound on the number of cards in the deck that have a small component okay so But G and H have the same deck That was the argument I gave for this upper bound as I said these bounds come from H And we look at G. We're going to get a lower bound on The number of connected cards and an upper bound on the number of cards with a small component And when we put all this together we get a contradiction So we cannot have a connected G and a disconnected H that have the same deck okay Okay, so At the other key idea that makes this work fairly nicely is we look at a spanning tree of our connected graph And spanning tree and and look at its deck Where we throw away L vertices Okay, well when you have this spanning tree Any connected card here the same vertices Would give you a connected card in G because G just adds more edges so The number of connected cards in D prime will be a lower bound on the thing we were looking for on the other hand if you have something that Some card of G that has some small component When you look at the spanning tree the same car fonding vertices will also give you a card with a small component so that Means the number of cards with a small component from the tree is going to be at least as big as the number of cards with a small component from G So we have these two inequalities Which means that to get our lower bound on C of D and our upper bound on C hat of D We can bound these quantities coming from the spanning tree instead Okay, the spanning tree is better behaved as a connected graph. We can do some counting a little better So Okay So let's look at cards in this deck that comes from the spanning tree Let T be the number of leaves When you delete leaves you don't disconnect the graph So we're going to have at least T choose L connected cards when they throw away L vertices, okay? and so This gives us on the other hand remember our upper bound was L and choose L Blowing this one on the number of connected cards So we get some sort of inequality here that Says basically that T cannot be too big the number of leaves and T cannot be big So here's an upper bound on the number of leaves and T so Now if we take a corn of our spanning tree and It has a small component okay Well that component is some subtree and It is cut off from the rest of the tree by deleting at most L vertices Okay, because we threw away L vertices So what we're going to want to do is Get a bound on how many sub trees can be cut off by at most L vertices Well each particular sub tree like this that's cut off by J vertices We can show is a component in fewer than this many cards Fewer than N choose L minus J cards And so this is where our upper bound on the number of cards with a small component is going to come from This And choose L minus J cards in which a particular small component can particular subtree can show up as a small component and An upper bound B sub J on the number of such sub trees so And we will show an upper bound on that product and So when we put it all together This looks fairly horrible, but but the point is here is the lower bound that we had from H on the number of cards with a small component and We compare that with the upper bound that comes here and Basically that gives you an inequality on N that says N has to be less than this something like L to the L plus 1 squared So that's the idea of it and when am I supposed to stop then I'm okay, so This part is well, okay, I'll show you a little more of this Okay, so here's a tree At least I have some pictures here, so Here's a tree and it has eight leaves and suppose L is 11 and we're looking for Subtrees cut off by four vertices Okay, so Why did that move Anyway, oh this is the statement of the zero Okay This is this is I mean that we were looking for this upper bound on a certain number of certain kind of subtrees, right? Okay, so So let me let me just talk about when J is at least three So we're talking about subtrees with at most L vertices cut off by J vertices from the rest of the graph okay, so So here I have some subtree and these red things Are the set of outside vertices that have neighbors in my sub tree, okay, then Our subtree will be the component of Take my spanning to my treaty throw away those vertices It's going to be the component that has vertices that are on paths between vertices of s all the rest This stuff is outside s right Okay, so that's the component. We're looking at and When you look at a path from this subtree through a vertex of s and you continue on out You will eventually get to a leaf of tea Okay, so in that way we can pick J vertices remember J is the number of vertices cutting off the tree. We can pick J leaves of T So for each such choice Of J leaves of T so their teachers J of them. That's where this factor teaches day J is coming from We want to think about how many Subtrees like this can give us that set of J leaves Okay, so we're going to get a bound on the number of subgraphs that can generate that set of J leaves Well, if I'm given a particular blue set of leaves Which could be chosen in teacher's J ways Consider the tree the smallest subtree containing those leaves Okay, that's the tree generated by this. So that's this this brown stuff plus the orange stuff But not this or that So the vertex you that's in s It's on the path from the leaf to f and The point is that because I want subtrees that have fewer than L vertices. I can't be farther from this leaf than L so The number of ways to place these break vertices and define this Subtree is it most the number of solutions to this Elementary combinatorial problem when you're looking for non negative integers that add up to something. That's it most L that you know because you have At most L vertices in your tree and the answer to that elementary combinatorial problem is is just this binomial coefficient So that's why we get the teacher's J times this other ball binomial coefficient okay, so that's the idea in and The cases J equal to J equal one or a little bit different you do your computation. Yes, so when L is three If you follow these computations that implies the connectedness is three reconstructible when n is least something like 59,000 or so but for any L any L equals three we looked in more detail and Was able to show that when n is at least 25 That you can win Okay Again, we think probably you win when n is bigger than two L or so But so there's plenty of work left to do but there's some interesting things so far Okay. Oh, yes. Yes, if you want to know how to do it when L is L is three There's an appendix at the end of this file that I'm not going to show you Okay, but you can find it on the web if you like and you want to see how You can prove this statement when L equals three that when n is at least 25 you can reconstruct connectedness Okay, so let's talk a little bit in the few remaining minutes about Maximum degree two the graphs with maximum degree two So it turned out that the sort of thing That we needed or at least was relevant to what we were doing Was proposed as a problem in the monthly by Richard Stanley around the time that we were thinking about this and he said Suppose you have an n-vertex graph Whose components are all cycles of lengths bigger than k Then the number of independent sets of size k Would depend only on k and the number of vertices not how you arrange them into cycles of different lengths Okay Well, that's just a very small piece of saying that the deck k-decks are the same The k-decks the cards that are independent sets There would be the same number of those no matter how you arrange the lengths of the cycles So we proved the more general theorem We extended that and said if you have two n-vertex graphs with maximum degree two and the same number of edges The only reason why we're saying that is because maybe you have some paths in there Which are number of edges is one less than the number of vertices if every component in each of those graphs is Either a cycle with more than k vertices or a path with at least k-1 vertices Then the two graphs have the same deck Okay, so this is where all our lower bounds For Graphs with maximum degree two are coming from the lower bounds on how many vertices we need in the deck to be reconstructible in In the graphs in the deck so in particular a special case of this is if you have two cycles and Each has at least k plus one vertices They will have the same deck as the graph where you make one big cycle out of those vertices Okay, and similarly if you have a path and a cycle you can combine them into a long path as long as they're each long enough and The important thing for proving this thing is that if you have two paths Each At least k vertices roughly each at least k-1 vertices then You take the same number of vertices and it's not going to matter how you distribute them into the two paths As long as both paths remain big enough. You'll have the same deck so A lemma that says that You know if you have two graphs and they have the same deck that you can add any other Component to them and again the two graphs will still have the same deck That actually enables you to reduce the theorem to just proving these three statements, okay? so then so then one two and three suffice to prove the theorem and Okay This is the statement of actually it's almost the statement of actually what the The minimum k that you need to reconstruct their graph with maximum degree two it has a bunch of paths and cycles and The answer just depends on the order of the biggest component and the order of the next biggest component Okay, so that's enough said about that So here's this basic idea that When you have these two graphs like this each consists of two paths G1 is those two g2 is these two We want to show that they're going to have the same deck and the idea is going to be to do this by sticking in three vertices To join the two paths and talk about the number of induced copies of any particular Subgraph that includes this vertex as an isolated vertex Okay, so it just be The induced subgraph is just add p1 as a component to the induced subgraph you were interested in over here and then what we show inductively is that The answer here is not going to depend on which guy You specify as the special guy as long as he's not too close to the ends Okay, so that's the biggest basic idea What I just said is there Okay, and so I'm basically out of time so So what I would just want to say is that that's what we do Okay, we show that that's independent So this is the inductive proof of that that's independent of where you place that special vertex the idea is you you You do your counting based on counting in smaller graphs and in the smaller graphs you show you're still far enough away from the ends okay, and then So here you get that first statement the third one and Then you use that to prove the second and you use the second to prove the first about Distributing that they into two cycles. Okay, again the same thing so two reduces to statement three and one reduces to statement two Okay, so that's how you prove those things and then As I said that gives you all your lower bounds for how big the cards in the deck need to be to reconstruct your graph with maximum degree two and the upper bounds are Involved this sort of stuff to show that you will be able to Reconstruct when you have that many vertices in each card again. They're counting arguments there Okay, and I won't have time to tell you about that, but let me just mention some open questions to end up here So yes, really this M. Sebel that Kelly mentioned if you are deleting L vertices How many vertices do you want to have in your graph to guarantee that you'll be able to reconstruct? Show that it exists Maybe it's just linear and then maybe it's two L plus one They maybe these these cycles and paths that we were talking about are the worst case They I think that's true Other kinds of questions you might be interested in it If you can't do Reconstructibility the entire thing just what's the minimum and you need so that when you delete L vertices You're always going to be able to determine whether the graph is connected What other parameters can you determine from the deck of k vertex subgraphs when n is large enough, you know We connectedness what about connectivity chromatic number matching number all kinds of things Kind of interesting thing about bipartite graphs For the I showed you that the complete bipartite graph is reconstructible from the deck of three vertex induced subgraphs But for a cycle with length 2m you need to go up to m vertex do subgraphs So is it true that as you add edges to a bipartite graph that the size of the deck that you need actually goes down That might be too much to hope for but it's perhaps interesting to investigate You know other classes of graphs like we did maximum degree to other classes of graphs maybe you can figure out what the least k is so that the graph would be Reconstructible and finally here's just a very small question In complete our part type graph we know is reconstructible from the deck of r plus 1 vertex induced subgraphs But is that sharp can you reconstruct it from the deck of our vertex induced subgraphs and We just have this one little example here to say that well not always because here is a complete tripartite graph Okay, and it has the same Three-deck as this complete for part type graph Okay, that's very easy to check and see that that's true. So but we don't have constructions for a larger r of Complete our part type graph that has the same our deck as some complete r plus 1 vertex graph So there's a little question for you. Okay. Yeah, and the next four slides are what I'm not going to show you about Okay, any questions for Doug Hello interwebs So I was curious is there a connection between the automorphism group of the graph and Sort of the decks that show up of the graph or sort of maybe like the multiplicities of graphs that show up in the deck Yeah, so if you have a vertex transitive graph Then there is only one graph in its deck by deleting one vertex, right? And it's just there with multiplicity n and you know similarly the Well there If all of the two vertex deleted subgraphs are pairwise isomorphic then I think you have to be the complete graph or its complement because You would change the number of edges depending on whether you delete two adjacent vertices or two non-adjacent vertices so that was that's fairly trivial, but nevertheless if you have a graph with a lot of symmetries the Number of distinct cards in the L vertex deleted subgraph will be Substantially smaller and okay, so yeah, and maybe you can make use of that Maybe you can right I mean when when you have the one vertex deleted deck and the graph is vertex transitive You can see that you have only one subgraph so you know the graph is vertex transitive, right? So, you know you're in that class. So so yeah, so it can give you more information Thank you. Yeah strongly regular maybe That's an interesting question. I don't know whether it has been shown that strongly regular graphs are reconstructible probably true Lowell do you know Yes, so that's an interesting question. Thank you Stanley's proof that he provided with his Proposal of the problem was by generating functions He of course it's Stanley Yeah, so he he showed that Made I can he made some generating function out of this number of independent sets and so forth and showed that for the two graphs It would have to be the same Yes, so our inductive proof I Initially gave an inductive proof that just gave that thing about independent sets for for the Two regular graph So but if you want to get the whole deck Then it turned out to be easiest to prove this more general result And it's it's well, yeah, it did look like it was several slides there, but But it's not that hard