 Okay, so thanks everybody for making it out especially with the the time change. And thanks Andrew for for coming to speak. So this week we've got Andrew shorts from the Southeast Missouri State University who will be talking about zero forcing sets and each matchable grass. Go ahead and take us away. All right, well thank you. Oh, sorry, go ahead. Was somebody. Okay, I think it was made an accident. Go ahead Andrew, sorry. Oh, no worries. All right, well, thank you, Drew for inviting me to speak today at your seminar. As he indicated in one of his emails, you know, just seminars around the country, you know, unique opportunities given our current situation with zooming and whatnot to, you know, see some speakers that are well outside our normal our normal driving distance for instance I guess. And again thank you very much for having me today I greatly appreciate it. So today I'd like to talk about zero forcing sets in each matchable grass. Okay. There we go. All right. Now, just kind of go back over the abstract that we had here for the talk so in this talk. We have a graph G to be, you know, ordered pair of a vertex that has no vertices to be finite simple and undirected. And if we fix a non-trivial connected graph H, we consider a perfect H matching of a graph G is a set H sub one through H sub N of vertex and do sub graphs of G. In other words, if we isolate G down to its individual vertex sets H sub I then it's going to be equal to the H sub I so each in vertex induced copy right. So, and it also is going to partition all the vertices here G right and each sub graph H sub I is going to be isomorphic to this graph H that we have in mind. So two perfect H matchings are of G we would consider equal if they're equal as sets of graphs, and a perfect matching of G is then a perfect piece of two matching of G. All right, so we're kind of seeing that this is basically kind of one way of generalizing the whole concept of perfect matchings of graphs right just instead of little copies of piece of two. We consider a general graph H that we have inside of multiple copies of this H inside of G here. Okay, and we say that G is H matchable or matchable right so borrowing that terminology again. If and only if G has a perfect H matching. In other words, so again barring the terminology from perfect matchings. All right. So if G has a perfect H matching, then the order of the, the number of vertices of G right has to be congruent to zero mod, the number of vertices of age, right, noble assume that throughout. All right, so let's kind of just start out with a quick visual example. So this for instance, would be an example of a claw matchable graph. I highlighted just the show in green all these little different copies of the claw right so this is kind of that a visualization of this idea. And this is not a claw matchable graph. All right, so we can see that we do have one issue here. Did anybody want to point out exactly where where it's at. Absolutely right that's correct. Yeah, the upper left corner right and so I've highlighted that one on this slide in orange to show you know that it sticks out right one of these is not like the other right so it fails to be a perfect claw matchable graph right So this little one up here on the left. I have a piece of four there instead. All right. So, now we have, we want to give an algorithm for for looking at these so it generates all each matchable graphs with vertex sets are in and in this algorithm, this sum here from ghost and K to equals to the M of us of K V sub K. Now it's empty for whenever M is less than or equal to one. And here the order of H right the number of vertices of H is just going to be equal to our. So our algorithm. H matchable tree is going to start with the input of a tree T. Alright, let's just start as the empty set and eight and the output is it become an H matchable tree with vertex set our end. Alright, and then here's our algorithm. So I don't want to read off all of this pseudo code here, but we can see that we are basically going through and we're updating as as we move along. Alright, so we're using a copy H sub M sub or H sub M plus one, excuse me of H with its vertex set as a subset of that set of numbers one through our end right bracket, our end in other words. And it's a, of course, actually it's an iterative, we're iterative process right and so while M is less than and we're just going to keep repeating this step to here. Alright, where we're just kind of basically appending these edges right and that step to a BB. Alright, and then when M finally hits M we stop and then we output this graph this graph. Alright, so I kind of want to just go through just one quick example of how this algorithm is utilized. Alright. So, but before we do that, I just was going to state that our quick observation is that this does indeed outputs a tree. T is equal to h sub zero h sub one through h sub n plus sigma k equals two to end of u sub k b sub k with a perfect perfect h matching h sub one through h sub n specifically so we're looking to have a specific each matching that we actually have in mind that this is generating right a very specific example. And here's our ideas behind it right so suppose S is the tree of order and V is a vertex of that of S and let T sub S of V denote the number of distinct sequences of trees as sub one as sub two through S of n where we have this nested sequence right when we finally took till we finally obtain S of n which is equal to S and each S tree S sub k has ordered k and for a specific matching h one through h and in the total number of matching n of p sub two, we find that fs of m goes from a of mp sub two to tm where fs of mt is a graph with vertex set m obtained from all trees in that set of a mp sub two by contracting each tree right h sub i to the vertex h sub i. So we're taking each one of those copies of h and we want to think about just shrinking it down to an individual vertex. Um, also another observation is and let's see I'm kind of respect and move this off my slide here a little bit there we go. All right. The tree T in a and h is generated by this. Some V goes in all the matchings M of T sub S of the times in a algorithm H matchable tree, where T is in a of mh and now this is for you again a unique M in the set of all matchings M script M. And h and s is this piece of M of T so you want to think about a piece of M as this contraction map. All right, you know, I get that out of the way. All right, now I'll go to. Sorry, I got to kind of move this little screen sharing thing around so it's not covering up pieces of my slide. Now, here's our here's the example that I promised you if we take M to be the matching 123456 right in the set of all matchings on on three vertices for piece of two and tree T is in the set of all trees with this perfect piece of two matching right with vertex set bracket six and the tree set 123456 then the tree S is equal to a fee sub M of T has three vertices 123456 and where 12 is adjacent to 34 is adjacent to five. All right, so then the way this kind of looks is we have T sub at T sub S of 12 is one right we have 12 where 12 is adjacent to 34 and then 12 is adjacent to 34 is adjacent to 56 and then T sub S of 34 is equal to two where we have these two items here 3034 where 34 is adjacent to 1256 is adjacent to 34 is adjacent to 12 and 34 were 34 is adjacent to 56 and 1212 is adjacent to 34 is adjacent to 56. And then T sub S of 56 is equal to one where we have the order triple 5656 is adjacent to 34 and 56 is adjacent to 34 is adjacent to 12. So, hence our our sum that we were looking at a piece of S of V is equal to four. All right, an algorithm H matchable tree generates precisely four distinct tuples, giving our tree T that were that's of interest. So we have fee of 123456 and then add in 13 and 35. And we also have fee of 341256 and then add in 13 and 35. And then we have fee of 345612 and then add in 35 and 13. And then we also have fee of 563412, and we add in 3513. All right. Oops, let me see here. There we go. All right, so this is the tea. This is the tree T that we generated here in the in the in the example given right so there's the one to that gets contracted to a single edge underneath the contraction map 34. All right, to a single vertex, excuse me, one to get contracted to a single vertex, 34, of course, as well, and 56. Right. So that's just a real simple example of the implementation number. All right. Now, this algorithm is, I kind of summarize some, some results here in table one of, of, of previous results, right to kind of show that I, as far as finding a perfect H matching in general, when the graph H has a component of order greater than three. That was actually known with instance being a graph G that was actually known to be complete and a publication by hell. And if the instance and set is a graph G with a perfect K three matching or a tripart. It's a triangle matching that's known to be NP complete in a publication that I've labeled a GT 11 that's listed later on in the references. And if instead G is a bipartite graph and whether it has a perfect matching or not. That's actually complexity is known to be a book horse polynomial due to Edmonds. So the instance switches instead to a tree G, and we have a sub graph isomorphic to a force, force H, that's known to be NP complete and that was proven by mature in the GMG chief of 48, and forced G. All right that sub graph or some more for to a tree H that's known to actually be polynomial by mature in that same publication. Now, if G is planar, and that's the instance instead and we're looking for covered by very text just joint copies of H with connected planar graph H of order greater than equal three that's known to be NP complete by So here's a little bit of background some of the history behind what motivated some of the study of looking at these H matchable graphs. Now, we'll move on to the graph theoretic portion of the. I'm sorry that's zero forcing portion of the talk so just kind of a reminder of those for those of you who aren't familiar with zero forcing. And this whole idea is that we let each vertex of a graph G, right be given one or two colors. Let's say, red and blue, and let's let's denote the initial set of red vertices of G. Now the color change rule converts the color of a vertex from blue to red, if the blue vertex is the only blue neighbor of a red vertex. The set Z is said to be a zero forcing set of G, if all vertices of G will be turned. Excuse me, if all vertices of G will be turned red after finally many applications of the color change. And the zero forcing number of G, Z of G is a minimum of all Z's overall zero forcing set C that are a subset of the vertex. Now, the zero forcing number came about as a of a simple graph was introduced by the aim minimum rank special graphs workgroup. And this was to be able to bound the minimum rank for numerous families of graphs, zero forcing parameters have been investigated further and have been applied to the minimum rank problem and lots of recent literature. If we let Z be this initial set of red vertices, right, the color change rule changes the color of a vertex W from a blue to red, if the blue vertex is the only blue neighbor of a red vertex you in this case we may say that you forces W, and we write you right arrow W. Of course, there may be more than one red vertex capable of forcing W but we associate only one forcing vertex to W edit. And then applying the color change rule to all vertices of Z. We obtain an updated set of red vertices Z sub one, a super set of Z. Clearly not all sorry there should be not all vertices and Z need to be forcing vertices and if a vertex you and Z forces W, and you becomes inactive, in other words unable to force thereafter. And the vertex W replaces you as a potential forcing vertex and Z sub one, the Z sub one has at most the order of Z many potentially forcing vertices. So applying the color change rule to Z one results in another updated set Z sub two, which is a super set of Z sub one of now read vertices. And a chronological list of forces, right is a record of forcing actions in the order in which they are performed. A chronological list of forces a forcing chain is a sequence use of one use of two through use of T such that use of one forces use of one, use of or sorry use of I forces use of I plus one excuse me for I equals one to through T sub T minus one. Oops. And so what we've actually found right so then there's some canonical results that from zero forcing that I left out like for instance that the number of a path is just we want you just take one of the leaves and it can knock everything down all the rest of the way. Or for like a cycle for instance you take two that are adjacent to each other. And that's enough to force all the other vertices around the edge of the cycle. So, I did leave out some of those I wanted to make for sure present this result which we found with this was a. Most of the proof and then my one of my former advisees, Houston Houston sure who just recently learned earned is a PhD from University of North Texas for all matching graphs, G, such that the under the contraction of that piece of M of G is tree. We have that the zero forcing number of G has to be bounded in between the zero forcing number for H and N, where N is the number of copies of H in the graph G, so N times the zero forcing number of H right so here, and is the number of copies of H, as I said in G and fees of M is that function that we that we used to contract each copy of H and H magical graph to a vertex. And thus, piece of M of G is the graph resulting from replacing each copy of H in G by a single vertex. Alright, therefore, if there are n vertex induced and independent copies of H and G, then piece of M of G has and vertices, and whenever piece of M of G is a tree, we will call these tree like matchable graphs note that neither H nor G necessarily have to be trees themselves. Here's an example of an instance of G that realizes the lower bound so here we have obviously like a big path but we can think of being it being broke down into copies of piece of three right and the zero forcing number as I said earlier on pass right is just one. Alright, now, even though this past since it's evenly divisible by three. So we can still just use that one vertex as long as we're taking one of the leaves, and that's enough this red vertex would be enough to force all these other vertex vertices and sequence. Alright, and then turn every single vertex right. So here's an instance of G that realizes the lower bound not very interesting but it is an example not nonetheless. And here's a little bit more complicated that example that realizes the upper bound. Right. So in general, if we want, we could consider a actually more general than this specific example we could in general. So here's a PS box of P piece of T with fee of M being fee sub M excuse me of G being a path right, so we can't actually eliminate any of these red vertices in this instance and still be able to force everything in the graph right. So it's absolutely necessary that we actually take the number of vertices from each copy right of these to be able to force all of our vertices read. Okay. And these last couple slides are just a couple extra bonus slides that I threw in now this one right here actually is already as someone someone else's previous results and this is on pineapple graphs. So here we consider a complete graph case of N with key pendant vertices off of one vertex. And here we see that the zero forcing number of P and P is will copy P sub M comma P will call it is equal to N plus P minus two. So we actually have to take all of the vertices, except for the one that has the pendant vertices off of it and all but one vertex from the list dependence vertices in order to be able to force this crowd. This kind of borrow some of its ideas also from the zero forcing number on a complete graph as well. Right you have to take all but one vertices or otherwise you basically have always at least what I would call like a co defender right helping that vertex from herring. Right. Right now, this however is a new result this is just kind of something after I had seen this result I kind of played around with so instead of pineapple graphs why now this have mutant pineapples right so people kind of think of a pineapple graph. And right as you had that little grid area that looks like the fruit part of the pineapple and then the little leafs coming off the top. But what if we had leaves, you know those little sets of tells of leaves coming off all over the place, right. And I kind of just jokingly coined the term, mutant pineapple graphs. Alright, so maybe some pineapples that were exposed to a little bit of radiation or something. And if this is the case and we have a piece of one through piece of K different pennant vertices all the way around right and instead our zero forcing number here becomes piece of NP, which is equal to zero zero forcing number of piece of NP becomes N plus the sum of piece of one through piece of K and then we subtract off the, the K K plus one. Okay, so we do actually have to go one over that. All right. Um, very good. All right, let's see I think I have. Oh yeah, have a nice little area here for direction for future research. We can invest other kick classes of graphs such that piece of M is a member of that class of grass such as cycles complete graphs and wheels. I also thought about maybe perhaps applying piece of M more than once to a graph right so maybe we have such a nice little neat structure that when we do the contraction we can still find a new age matching on the result and graph and maybe kind of go through an iterative process of taking piece of M now of course it's not necessarily going to be piece of M since M reference references a specific matching. Of course, it might be a piece of M and then a piece of M plot prime that we apply to it the sec for the second iteration. Right. So, maybe something where we could kind of iterate this process. And initial ideas that I had for direction for future result with combining these ideas of these age matchings with the zero forcing numbers. Now here's some of those references that I had promised earlier there's the aim minimum rank special work work group. That was one of the, the big ones that generated the whole idea of zero zero forcing sets to begin with. And then here's some of the references as well that I used for that table that was constructed earlier the different instances. And if you want to know a little bit more about the age matching graphs that was actually something that I come up with with my advisor lane, Dr Lane Clark. And there's we have a couple publications out on that and that's basically our generalized matches and force and freeze. And since then we've also done a paper on the distribution and we have a paper on the degree distribution of these age matchable graphs coming out shortly. And I think the remainder of these references are mostly from the table or some of the ideas that I've gotten where I got some of my ideas for the zero forcing numbers and how to combine them with. So, whoops. All right, and that's all I got for you today. So if we could all thank our speaker in some way and then we'll open it up for some questions. Looks like I get somebody here in chat. Oh, just just some thank yous I think. Okay. I'm so I'm so used to doing that automatically now from classes this semester's like some students will only ask them, you know, in the chat box and some will only ask them verbally. Yeah, you gotta be on top of. Yeah. Okay, do we have any questions for our speaker. Go on. So Andrew, you listed a number of empty complete problems where there was a component of size. So what if you want to match with an edge and then isolated vertex. Is it also empty complete or unknown. So, let me understand. So you want to match both. An edge in a isolated vertex. My three vertex graph is an edge and then isolated vertex. And I think you want to make induced copies of this to cover all vertices. Let me think so. So you're referencing back to the table like you have a isolated vertex and it's and two vertices with an edge right so you're looking at an age that's disconnected itself. Yeah, since then you don't have a component of order three. Right, right. Yeah, that's a good. That's a good question. Now, let's see. I mean, I guess that would be closest. So I guess that would be closest to, yeah, that's a tough question that because that would be really close to the instance being a tree. Well, you wouldn't know because the tree itself the trees, it wouldn't have a tree itself so you'd be looking more like a forest G right in a sub graph. Isomorphic to a tree age but you're really your subgrass really isomorphic to a forest, rather than a tree. Right. I honestly do not know off the top of my head. My initial intuition, though, would be that it would be it would be polynomial. My initial intuition that that would be I would cheat more towards the polynomial side than the, because if I remember correctly from it's been a while since I've constructed this table that first motivated some of this research, but I believe that the the stronger the connectivity of the graph the more difficult it was to find matchings in general a lot of the other literature. So the disconnectivity there, I think actually aids and finding and helping you would be able to find your matching and especially if you have your isolated vertices it seems like you could just kind of keep your isolated vertices almost off to the side if you will just find those copies of the edge inside right and just grab one of those isolated vertices well with those isolated vertices might be obviously isolated in the sense of the inside of the copy. Or whenever you look at the vertex induced copy of that age, but obviously they're not necessarily isolated whenever you are looking at the overall graph G before the contractive maps of map applies so. That's that's a good question I'm not that I'm not to think about that one a little bit more. Thank you. No problem. Yeah. It's equivalent to packing matching with P threes, the compliment. Because yeah because you would have. Yes, so you because you're looking at the so you'd be looking at the compliment yeah. I'm like searching right now I see some approximation algorithms for the maximum p three packing so maybe it's hard for the exact. You think it's MP hard. Just from the initial just from the initial search, of course, I mean, I don't know. But um, yeah. Yeah, because I mean yeah if because I did first I was I think I was thinking of it. When I first said polynomial I think I think I was thinking like okay well, you know, I wasn't thinking back to the originating graph G before you apply that map because yeah when you're back in that map I mean you have all these other edges that could potentially be in there. Right that could that you don't just get a, as I said earlier like try to simply just set them to the isolated vertices of the side and then grab them as you need them. But yeah because of the fact that you're looking at it in terms of the compliment being p three map then you would be well as far as the compliment is concerned then you're back to just a tree. Which is, which is, which is known. With that specific question and that specific instance. But let me, yeah. No, I said that's a, that's a, that's a, that's a good question. But that the algorithm you just described still is exhaustive though, right, I mean you can just take all of your possible however many isolated vertices you need. Check if there's a perfect matching on the rest and then check and see if there's edges across. Once you add a back in anyway, and that's still all polynomial. Yeah, exhausted. Yeah. No, that is. If we use, whoops. So, because you have an H matchable H matchable tree. Yeah, because you're sorry not sorry not this algorithm that the algorithm that you had just decided putting the isolated vertices away. And then check matching in the rest, and then just check if there's edges across and then just do that for all possible sets of vertices that you need to throw away. That would all be polynomial I think. Yeah, because it can. Yeah, because you would have the advantage of that you were looking for just a pat a piece of two and isolated vertex in your tree so you look at it as far as looking at it for underneath the map yes you're right after the map as already applied and you had like say a perfect matching in hand. Yeah, and then you're just checking for it. Yeah, I think I think I think you're, I think you're right there I think you're, I think you're good. No, I haven't honestly I haven't really explored a whole bunch of the, I'm really glad that you actually asked that question because I really honestly haven't explored a lot of the, like the disconnected cat cases of h yet like I my natural intuition on a lot of this stuff was just to look at like, you know connected graphs but I think those those disconnected graphs post measuring questions. Yeah, those are definitely worth looking, looking into for sure. It looks like Patrick and hell showed that partitioning the vertex set of an arbitrary graph into induced p threes induced I'm not sure. Is MP hard. Yeah, I think that that Kirk Patrick and hell that might be the same reference that I have here on the first on that first instance in the table actually that might be the same publication where they do that. The 83. Yeah. Let me see what's going on. Where is that. There's the home paper. There's the hall paper that I maybe, maybe I try to squeeze too much on a slide and it might have one either looks like it was one of. It looks like McCurry was the last one it looks like it went above it looks like I tried to squeeze too many references on a slide. But yeah but I believe that's the same paper that I was referring to in the reference but yeah that that was, that would go along with of course the first entry we have here in this table. You know, MP, or, you know, seem to agree with that at least anyway. No, it's always it's always interesting. I think we're, we're these things break, if you will, right where where they switch from polynomial and be hard it's just there's there's that there's that threshold where it just kind of, you know, where it occurs, which is always, I don't know I always find fascinating. Do we have any other questions for our speaker. If not, thanks again Andrew for the talk and thanks everybody for making it out and have a good Thanksgiving break we won't have a seminar next week obviously and then the last one of the semester will be the week after that so thanks everybody for making out and have a Thanksgiving.