The overarching problem of this project was trying to find the maximal number of disjoint subgraphs of a certain type we can pack into any graph. These disjoint graphs could be of any type in the original problem. However, they were limited to be T2 trees for my research (T2 trees are defined in section 2.1 of the paper). In addition, most of my work was focused on packing these T2 trees into constrained bipartite graphs (also defined in section 2.1 of the paper).
Even with these specific constraints applied to the overall problem, the project still branched into different subproblems such as packing trees into complete bipartite graphs and finding minimal and maximal bounds for packing these graphs

Wong, Collin Christopher

English

DigitalCommons@CalPoly

Bipartite Graph Packing ProblemsCollin WongJune 9, 20181 IntroductionThe overarching problem of this project was trying to find the maximal number of disjoint sub-graphs of a certain type we can pack into any graph. These disjoint graphs could be of any typein the original problem. However, they were limited to be T2 trees for my research (T2 trees aredefined in section 2.1 of the paper). In addition, most of my work was focused on packing theseT2 trees into constrained bipartite graphs (also defined in section 2.1 of the paper).Even with these specific constraints applied to the overall problem, the project still branched intodifferent subproblems such as packing trees into complete bipartite graphs and finding minimaland maximal bounds for packing these graphs.2 The Initial Problem2.1 Packing T2 Trees into Undirected Bipartite GraphsSuppose we have an undirected, bipartite graph G, where the vertices of G are partitioned intovertex sets S and T, and |S| = |T| + 1. Assume that G is a finite graph with no loops or multipleedges. We will define a T2 tree to be a spanning tree in which every vertex in T has a degree ofexactly 2. What is the maximum number of disjoint T2 trees we can form on graph G?2.2 Initial Notes on the ProblemFrom this problem, it is important to highlight important facts about these T2 trees. Because thetrees are disjoint, none of the T2 trees drawn on a graph G can have any similar edges (in otherwords, each edge in G can only be used in one T2 tree). In addition, each tree must be spanning,so every vertex in S and T must be touched by every tree. Lastly, these must be trees, so no cyclescan be formed in any one T2 tree.2.3 Drawing TreesInitially, it was easiest to draw complete bipartite graphs as it allowed more T2 trees to fit withinthe graph. Therefore, the upper three drawings in Figure 1 show the maximum number of treeson the first three complete bipartite graphs: K2,1,K3,2, and K4,3.After, I attempted to draw T2 trees that were not paths on some graphs. An example of thisapproach is Tree 1 in the bottom right graph of Figure 1. This showed to be much harder asforming cycles was much easier when not trying to find T2 paths. In addition, these trees usuallyresulted in failures to create the maximum number of T2 trees on the graphs.1Figure 1: Initial Drawings of Packing T2 Trees into Bipartite Graphs.2.4 Observation from Initial DrawingsDrawing T2 trees on different bipartite graphs gave some initial insights on certain bounds of thisproblem:● By definition, Graph G must have at least (2 x |T|) edges in order to have one T2 tree● Because the T2 trees must be spanning, the maximum number of T2 trees on G ≤ min degree ofall vertices in S● Also by the definition of the T2 tree, the maximum number of T2 trees on G ≤ ⌊min degree of all vertices in T2 ⌋● From studying the complete graphs, the maximum number of T2 trees on G ≤ ⌊ ∣S∣2 ⌋2.5 Theorem I: Minimum Edges to Form a T2 TreeFrom the initial drawings, we can see that Graph G must have at at least (2 x |T|) edges in orderto form one T2 tree.Theorem: In an undirected, bipartite graph G, where the vertices of G are partitioned into vertexsets S and T, and |S| = |T| + 1, G must have at at least (2 x |T|) edges in order to form one T2tree.2Proof. Suppose we have a Bipartite Graph G = (S ∪ T, E), where |S| = |T| + 1.By definition, every vertex in T must have a degree of exactly 2 to form one T2 tree on G.Therefore, there must be at least (2 x |T|) edges to form one T2 tree on G.2.6 Theorem II: Maximum Edges Without Forming a T2 TreeAnother interesting case was to see how many edges can be added to graph G until the additionof one more would surely create a T2 tree. For graph G, the max number of edges resulted fromconnecting every vertex minus one in S to every vertex in T.Theorem: In an undirected, bipartite graph G, where the vertices of G are partitioned into vertexsets S and T, and |S| = |T| + 1, the maximum number of edges in G without forming a T2 tree is|T|2 edges.Proof. Suppose we want to find the maximum number of edges without forming a T2 tree on agraph where n = |S| and m = |T|.We can create a complete bipartite graph Km,m = (S ∪ T, E) and add a vertex into S, sd, sothat each vertex in T is connected to each vertex in S\sd. We will call this graph G.In graph G, there is no possible way to form a T2 tree because T2 trees must be spanning andthere are no edges incident to sd.In addition, we cannot add any more edges to S\sd, because it was created by the completegraph Km,m.Therefore, the only way to add an edge to G is with an edge from sd to a vertex in T.Claim: Adding one more edge to the G will always create a T2 tree.Proof:Label the vertices in S: s1, s2, s3, . . . , snLabel the vertices in T: t1, t2, t3, . . . , tn−1We will say there are edges from every vertex in T to every vertex in S except sn (In this proof ofthe claim, sn refers to the vertex sd in the encasing proof). Therefore, we can only add an edgefrom sn to any vertex in T.If we add an edge from sn to tx, where x is any integer from 1 to (n-1), we can use this formula tofind a T2 tree: sntxs1tx+1s2 . . . sn−2tx − 1sn−1Therefore, there will always be a T2 tree in the graph after adding one more edge to G. ∎Therefore, there can be no more edges added to G without forming a T2 tree on G.Therefore, the maximum number of edges that we can have in a graph without forming a T2tree is |T|2.Figure 2: Graph with the max number of edgeswithout forming a T2 Tree.3 The Complete Bipartite Graph: Kn,m3.1 Investigation of Kn,mThe Complete Bipartite Graph Kn,m is the graph G = (S ∪ T, E), where n = |S| and m = |T|, n =m + 1, and E contains an edge from every vertex in S to every vertex in T. The complete bipartite3Figure 3: T2 trees on K2,1, K3,2, and K4,3.graph Kn,m is an important special case because the graph has every possible edge connectingits vertices, while keeping the bipartite and no multiple edges or loops conditions. Investigationof Kn,m, showed that it was easiest to form T2 trees on the complete bipartite graph with paths.However, it is important to note that is may not be the case for every possible bipartite graph.In addition, the complete graph Kn,m was useful to research because it helped form an upperbound on the maximum number of T2 trees on all bipartite graphs, where n = m + 1. In Figure 3 ,I have drawn the maximum number of T2 trees on K2,1, K3,2, and K4,3. However, after continuingto draw K5,4, K6,5, and K7,6, it became obvious that each graph had ⌊n2 ⌋ disjoint T2 trees on it.Therefore, I conjectured that this special graph Kn,m must have ⌊n2 ⌋ disjoint T2 paths on the graph(paths are trees by definition).3.2 Theorem III: Packing Maximal T2 trees on Kn,mTheorem: A complete bipartite graph G = (S ∪ T, E), where |S| = |T| + 1, must have ⌊ ∣S∣2⌋disjoint T2 paths on the graph.Input: A complete bipartite graph G = (S ∪ T, E), where |S| = n, |T| = n - 1, and n > 1. Goal:⌊n2⌋ disjoint T2 pathsProof. Pink with rgbLabel the vertices in S: s1, s2, s3, . . . , snLabel the vertices in T: t1, t2, t3, . . . , tn−1Paths:First : s1t1s2t2s3t3 . . . sn−1tn−1snSecond : s3t1s4t2s5t3 . . . sntn−2s1tn−1s2...General: skt1sk+1t2sk+2t3 . . . sk−2tn−1sk−1Case 1: n is evenLast : sn−1t1snt2s1t3 . . . sn−3tn−1sn−2These are n2disjoint T2 paths on G. Because all edges are used, there cannot be more T2trees/paths.Case 2: n is oddLast : sn−2t1sn−1t2snt3 . . . sn−4tn−1sn−3These are n−12or ⌊n2⌋ disjoint T2 paths on G.The graph G initially has |S| x |T|, or n(n - 1), edges.We know each disjoint T2 tree uses 2|T|, or 2(n - 1), edges of G.We have created n−12disjoint T2 paths on G, so we have used:4[n − 12trees] ∗ [2(n − 1) edgestree] = (n − 1)2 edges (1)We can use this to determine how many edges remaining in graph G.n(n − 1) − (n − 1)2 = (n − 1) edges remaining (2)We are left with (n - 1), or |T|, edges left in G, and we know we need at least 2(n - 1), or 2|T|,edges to form another T2 path/tree.For all n > 1, 2(n - 1) > (n - 1).Therefore, there cannot be more T2 trees/paths.3.3 Failures on packing the Maximal Number of T2 trees on Kn,mWhile attempting to draw T2 trees on Kn,m, there were many failures, where failures are definedas drawing less than ⌊ ∣S∣2⌋ disjoint T2 trees on the graph. These failures resulted from startingon a non-minimal degree vertex of S when drawing the T2 paths, avoiding paths altogether, andcreating cycles inside the T2 trees.If S contains an odd number of vertices, then we can still create ⌊ ∣S∣2⌋ disjoint T2 paths on thegraph while starting on non-minimal edges of the graph. However, when S has an even number ofvertices, each T2 path must start on minimal degree vertex of S.In addition, attempting to use non-paths in the Kn,m graph resulted in failures. For example,creating a T2 tree that contains a star tree (with one vertex in S as the center vertex and all othervertices in T) will result in a failure (proved in Conjecture II). This example is shown in Figure 5.The most common error was creating a cycle while trying to create a T2 tree when not usingthe Construction in Conjecture I for finding T2 trees on Kn,m. It could also be possible to useDepth First Search in order to find a path to fix this problem.Figure 4: Example of a graph where you cannot create a T2 tree.3.4 Theorem IV: Disjoint Star TreesTheorem: A graph can never have two disjoint spanning trees, one of which is a star.Proof. Suppose we have a graph G = (V, E) that has no multiple edges.Suppose graph G contains a star tree, S, that has a center vertex vc and edges ES .5Since vc is adjacent to every other vertex in G, G\ES leaves vc as an isolated vertex.Therefore, there cannot be another spanning tree in G\ES as vc is an isolated vertex.Therefore, a graph with a star spanning tree cannot have another disjoint spanning tree.Figure 5: An graphical representation of Conjecture II.4 Research on Related Problems4.1 Lovász on Kneser’s ConjectureIn the paper Kneser’s Conjecture, Chromatic Number, and Homotopy, Lovász attempts to gener-alize Kneser’s Conjecture. "If the simplicial complex formed by the neighborhoods of points of agraph is (k-2)-connected, then the graph is not k-colorable" [1]. This conjecture was used to provethis theorem: "If we split the n-subsets of a (2n+k) element subset, one of the classes will containtwo disjoint n-subsets" [1]. In these definitions, it is still unclear what a simplicial complex is, butan n-subset is a subset of exactly n distinct elements of the set.As an example of this, suppose n = 2 and k = 1, and the set is 1, 2, 3, 4, 5. Therefore, the2-subsets are: (1,2), (1,3), (1,4), (1,5), (2,3), (2,4), (2,5), (3,4), (3,5), (4,5). We will separateclasses into subsets that do not contain the element 1 and subsets that do contain the element 1.The two classes are:Class 1 (that contains 1) : (1,2), (1,3), (1,4), (1,5)Class 2 (does not contain 1): (2,3), (2,4), (2,5), (3,4), (3,5), (4,5)As shown, there are actually three pairs of two disjoint n-subsets in Class 2: (2,3)/(4,5), (2,4)/(3,5),and (3,4)/(2,5). Moreover, Lovász includes another theorem that can be deduced from the firsttheorem: "The chromatic number of Kneser’s graph KGn,k is k+1" [Lov77].In Figure 6, we can see the previous example in graph form. This shows the 2 classes from theearlier example. In addition, it shows the minimum coloring of the graph is 3 (red, yellow, andblue vertices). Therefore, the chromatic number of KG2,1 is 3.Unfortunately, after reading this paper, it was decided that it had no applications to the graphpacking problems I was working on.6Figure 6: Kneser’s graph KG2,1 (n = 2, k = 1).4.2 Lovász on Konig’s TheoremIn his other paper, A Generalization of Konig’s Theorem, Lovász gives a generalization of Konig’sTheorem which was originally stated for bipartite graphs. The theorem states "the maximumnumber of (pairwise) independent edges of a bipartite graph G equals to the minimum number ofvertices covering all the edges of G" [Lov70].Lovász uses many variables and functions without clear explanations in this paper, so it wasdifficult to understand this paper. Nevertheless, it was also decided that this edge/vertex covertheorem was also not helpful for bipartite graph packing.4.3 Lovász on Covering GraphsIn the paper On Covering of Graphs, Lovász discusses how we can minimally cover graphs givencertain constraints. In this paper, he considers undirected graphs without multiple edges. Theinvestigation begins with covering graphs with a minimal number of paths and circuits. He givesa theorem that says "A graph of n vertices can be covered by ≤ ⌊n2⌋ disjoint paths and [cycles]"[Lov68]. He also goes on to give another theorem about complete graphs that says "a graph of nvertices can be covered by ⌊n22⌋ complete graphs" [Lov68].Although this paper initially seemed related to the problem, it only talked about minimal cov-erings instead of maximal coverings which would relate to our problem of packing the maximalamount of subgraphs into a graph.75 The Chinese Postman Problem5.1 DefinitionsIn his paper Matching, Euler Tours and the Chinese Postman, Edmonds discusses the interestingproblem of the Chinese postman, as well as its subproblems of finding matchings and Euler tours.In order to understand the problem, it is important to define these five terms:• "A simple tour of G is a [cycle] which contains every edge e at least once" [Edm73].• "A postman tour of G is a [cycle] which contains every edge at least once" [Edm73].• "An Euler tour of G is a [cycle] which contains every edge exactly once" [Edm73].• "The length ce of an edge e, [the weight of the edge], is assumed to be a non-negative number"[Edm73].• "The length of a tour is ∑li=1 cei , where l is the number of edges in the tour" [Edm73].5.2 The ProblemThe Chinese postman problem is to find the shortest postman tour in a undirected, connectedgraph, where shortest does not mean the least number of edges, but rather the shortest length ofthe tour. While this definition may seem like it is really one problem, it is actually almost alwaysgoing to be two different problems.If the graph already contains an Euler tour, then we will just need to find the Euler tour. Thisis because if the graph contains an Euler tour, then that tour is the shortest postman tour as noedges will have to be repeated in the tour and each edge is included at least once. In order todetermine this, there is a theorem that says: if every vertex in graph G has an even degree, thenG for sure has a Euler Tour. If this is true, we will say the graph is graph G’ and move onto 5.4Finding The Euler Tour.However, if the graph does not contain a Euler tour, we will have to add edges to the graphin order to create this Euler Tour, as well as finding the Euler tour.5.3 Matching ProblemSuppose we have a graph G = (V,E), which does not contain an Euler tour. We will be addingedges to graph G to create graph G’. We will do this by implementing a Matching Algorithm:1. Add each odd-degree vertex in G to Gp.2. Find the shortest path between every odd-degree vertex in G (There are O(∣V ∣3) time algo-rithms for this), and add an edge in Gp between every edge. Each edge in Gp will have aweight of the length of the shortest path in G of the two adjacent vertices to the edge.3. Find an optimum 1-matching, M: a set of edges with the least possible total weight whereevery vertex only touches an edge once.4. For each edge in M, m, add an edge into G with the same weight as each edge in the shortestpath corresponding to the edge m. We will call this new graph G’.In this algorithm, there must be a 1-matching because of the theorem that says: in a connectedgraph, there must be an even number of odd degree vertices. Because G was connected and addedonly its odd degree vertices to Gp, Gp must have an even number of vertices. In addition, Gpwill always be the complete graph because each vertex has an edge to every other vertex in Gp.Therefore, there must always be a 1-matching in Gp.85.4 Finding the Euler TourOnce you have graph G’, a graph with an Euler tour, we will need an algorithm to actually finda series of edges to follow to find an Euler tour. Edmonds calls this an End-Pairing Algorithm,Algorithm 1. This algorithm will return the tour in the form of a list of edges of the tour. Keepin mind that this is not a unique Euler Tour, as there can be many Euler tours. However, we nowhave the shortest Postman tour of graph G’.5.5 Applications to the ProblemAlthough this problem does not pack graphs, the matching algorithm relates to our problem ofpacking because it is packing as many edges in a graph with the constraint that each vertex canonly touch one edge. However, this matching problem is different because we are adding in edgesinto a graph whereas our problem does not allow adding edges into the graph.Algorithm 1 End-Pairing AlgorithmInput: Graph G’ = (V’, E’) that contains an Euler TourOutput: A list of edges of a Euler Tour of G’1: function Pair-Ends(G’ = (V’, E’))2: tour = ∅3: mark all edges in E’ as unused4: do5: if tour == ∅ then6: vo = any vertex in V’ connected to an unused edge in E’7: else8: vo = any vertex in V’ connected to an unused edge in E’ and connected to an edgein the tour9: end if10: v = vo11: currentTour = ∅12: do13: e = an unused edge in E’ incident to v14: mark e as used15: append e to currentTour16: v = the other vertex incident to e that is not the current v17: while v is incident to an unused edge in E’18: if tour == ∅ then19: tour = currentTour20: else21: interject currentTour into tour between two consecutive edges in tour that are bothincident to v022: end if23: while There exists an unused edge in E’24: return tour25: end functionReferences[Edm73] Jack Edmonds. Matching, euler tours and the chinese postman. 1973. Waterloo, Canada.[Lov68] Lubromir Lovász. On covering of graphs. 1968. Budapest, Hungary.[Lov70] Lubromir Lovász. A generalization of konig’s theorm. 1970. Budapest, Hungary.[Lov77] Lubromir Lovász. Kneser’s conjecture, chromatic number, and homotopy. 1977. Szeged,Hungary.9

Bipartite Graph Packing Problems

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Bipartite Graph Packing Problems

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