The edges represent various routes that we can send the raw materials along, with the capacities being how much material we can ship along those routes. In a strong component of such a graph, if we give a message to any one person, that message could then get to any other person in that component. The graph we obtain through this process is called the closure. They idea is that eventually, once the number of vertices gets large enough, exponential algorithms are far worse than polynomial algorithms. A weighted graph is a graph where we label each edge with a number, called its weight, like shown below. To do this, we look for what's called an unavoidable set, a set of graphs such that every maximal planar graph with at least five vertices must contain at least one graph in that set. In general, it's possible to find graphs with arbitrarily high chromatic number that have no triangles. But some graphs are more connected than others. Since it is planar, by Theorem 29 it has a vertex v of degree 5 or less. Suppose I give you a program that finds the optimal solution to the Traveling salesman problem. There he had to push bricks around in a wagon, and crossing over wagon ruts in the ground was difficult. In particular, suppose man p and woman q prefer each other to their current partners p' and q'. This graph is isomorphic to C5, which we've drawn at right. It is not possible to have both of these cycles in the graph without them crossing over each other, either at a vertex or at an edge. Here is a little introduction to how running times of algorithms are typically measured. In fact, we have the following theorem. Not only can we use the König-Egerváry theorem to prove Hall's theorem, but it is also possible to go the other way—namely, we can prove Hall's theorem from scratch and then use Hall's theorem to prove the König-Egerváry theorem. The proof of this theorem is the same as for the previous theorem except that if there are no triangles, then every face has at least 4 edges around its boundary. In every simple graph with at least two vertices, there are at least two vertices with the same degree. It is possible for every edge of a graph on 5 vertices to be a cut edge. The next vertex to visit would be e, as its total cost is currently 2, cheaper than any other labeled vertex. If there is a transversal after doing this, then stop. After doing that, a now has degree 5, so now we have deg(a)+deg(c) = 7, so we can connect a and c. However, this is as far as we can go. Suppose we want to place ATMs around a city so that no one ever has to walk more than 1 block to get to an ATM. It does this by looking for paths from the source to the sink along which to improve the flow. On the left is a graph and on the right is its closure. Circle all that apply. Instead, in our induction proof below we work in the other direction, thinking about removing things from a plane graph instead of adding things. To find the closure, first note that there are 8 vertices, so we are looking to add edges between nonadjacent vertices whose degrees add up to at least 8. Most mathematicians and computer scientists think that P is not equal to NP, but not everyone agrees, and no one has come close to solving the problem. For instance, if a graph is regular and each vertex is adjacent to at least one third of the vertices in the graph, then the graph has a Hamiltonian cycle. This is a very simple augmenting path. This means that K5 is not planar. We then seek a minimum counterexample, the smallest planar graph that can't be colored with 4 colors or less, and try to show that no such counterexample can exist. It takes a graph with no triangles and chromatic number k and produces a graph with no triangles and chromatic number k+1. What is interesting is that the converse is also true: if a connected graph has no odd vertices, then the graph has an Eulerian circuit. Is it possible? Prim's and Kruskal's algorithms work fine with negative weights, but Dijkstra's algorithm doesn't. [Hint: the second answer has 8 vertices.]. A few years later Øystein Ore noticed that it isn't necessary that every single vertex have degree at least n/2. This would be a cut vertex. The trickiest part is checking to see if adding an edge creates a cycle. Strong components are useful in a number of contexts. The problem is they don't all get along, so some of them can't travel together in the same car. Each person is only available for certain time slots, and we want to allow as many people to give their presentations. This is shown in the figure below. It is not to hard to show that all complete graphs with an even number of vertices are 1-factorable. To see why the chromatic number goes up, note that we need as many colors to properly color the copy vertices as we do to properly color G. This is because each copy vertex w' is adjacent to all of the neighbors of some vertex w in G. So any color that we use on w' we could also use on w, meaning if we could get away with less than χ(G) colors on the copy vertices, then we could do the same on G. Finally, since x is adjacent to every copy vertex, x will need a color different from all of them, forcing χ(G') = χ(G)+1. What does each of them have to say about the chromatic number of this graph? The standard way to show this is something called Mycielski's construction. After a few more steps we get to the point that g is the cheapest labeled vertex. Thus the minimum number of vertices needed to separate s from t is at least as large as the maximum number of disjoint paths from s to t. Going the other direction, showing it is no larger, is quite a bit more involved, and we will skip it. See below where a hypothetical portion of S is shown along with d and e. Consider removing d from S and replacing it with e (this is the tree S-d+e). Note: A common mistake is to try to use the same vertex to subdivide multiple edges. So the running time of this algorithm on a particular computer might be something like 2.3n + .2 microseconds. For instance, graphs are often used to visualize data, and they are easier for people to follow when there are few crossing edges. Note that some planar graphs, like K4, need four colors, while others can be colored with fewer. Recall that for graph coloring we have the result χ ≥ n/α, where n is the number of vertices in the graph and α is the size of the largest independent set. We'll take S = {a,b}. It has a natural analog for digraphs. For example, C12 fails every one of these conditions yet still has a Hamiltonian cycle. Here we have v = 6 and e = 9, and the inequality e ≤ 2v-4 becomes 9 ≤ 8, which does not hold. Both of the endpoints of that path are of degree 1 in the tree, with their only neighbors being the vertices immediately before them on the path. This process of adding vertices into edges is called subdivision. Here is the formal definition. It's not too hard to write some code implementing Kruskal's algorithm, though we won't do so here. Show that Menger's theorem works for the graphs below by finding a set of, Shown below are flows on digraphs. In a weighted graph, people are often interested in the cheapest path between two vertices. The genus g = 0 case is the Four color theorem. The problem is in NP because if someone gives us a Hamiltonian cycle in a graph, it is easy (polynomial-time) to check that it really is a Hamiltonian cycle. [Degree sum formula] In any graph, the sum of the degrees of all vertices is twice the number of edges. Vertices f and h are not adjacent and their degrees add up to 8, so we can add edge fh. This bound is sharp in that there are graphs (like K7) that require 7 colors. As mentioned in Section 5.5, the idea is to try to color maps so that adjacent regions (countries, states, etc.) Leave it out and we have a one-way edge. Consider any path of maximum length in the tree. It has one vertex of each of the degrees 1 through 7. The graphs K5 and K3,3 turn out to be extremely important in terms of telling which graphs are planar. The rest of the vertices go towards subdividing edges. We see that the graph that Prim's algorithm builds up is a tree at every step. Each of these labeled trees is equivalent to a spanning trees on Kn, so we have τ(Kn) = nn-2. Construct a bipartite graph with 8 vertices that has as many edges as possible. Corollary 3. Then at each stage throughout the process, we pick any man x who is not currently engaged. If a graph contains a vertex that is adjacent to every other vertex, then it is connected. Prove that the graph is not planar. Now suppose the graph has two vertices of odd degree. So in the end, the closure ends up being the complete graph K8, which clearly has a Hamiltonian cycle. Draw them. Graph theorists also study the crossing number of a graph, which is the minimum number of crossings needed to draw the graph in the plane. Nowadays, few mathematicians, if any, doubt the correctness of the proof, though many mathematicians would prefer a simpler, more aesthetically pleasing proof. The term “planar” comes from the fact that we are trying to draw graphs in the plane. Then delete any row and column and take the determinant. The connectivity is tells us the maximum k for which a graph is k-connected. We assume that a region is contiguous (i.e. Note that the flow is not optimal, as it would be possible to push more stuff through the network than the current total of 6 units. Prove that if a graph contains two vertices that are at a distance of, Prove if every vertex in a graph has degree at least. Draw the labeled tree that has Prüfer code (1,4,3,5,5,7,2). 2. Suppose we start with a graph G. Mycielski's construction says to make one copy of each vertex v of G. For each v, add edges from its copy to each of the neighbors of v in the original graph. Now each Prüfer code has n-2 entries, each of which can be any integer from 1 through n, so there are nn-2 possible Prüfer codes and hence that many labeled trees on n vertices. In other words, it requires an equal number of people as jobs. We are often interested in perfect matchings, matchings where every vertex is matched. That is, if we imagine a torus made of clay, it is possible to deform that torus by stretching and reshaping the clay until we have a sphere with a handle. Induction on Graphs Theorem: P(G) for every graph G. The absolute value of the determinant gives the number of spanning trees of the graph. The ordering is indicated by the circled numbers. For example, in the collection of sets below on the left, an SDR is highlighted. The König-Egerváry theorem provides another example of TONCAS, and there are many more examples throughout graph theory. We have k ≥ 3 since the faces of a polyhedron are polygons, which must have at least 3 sides. Prove that a DAG has a unique topological ordering if and only if it has a Hamiltonian path. Now every vertex in the graph has degree at least 6 except for e, which has degree 2. It does this by maintaining two sets, R and S. The set R is all the “reached” vertices and the set S is all the “searched” vertices. That man proposes to the first woman on his list and they become engaged. We can do this search by starting at an unmatched vertex of X, following an edge not in the matching, then an edge that is in the matching, then one not in the matching, etc. There are still others, such as Dillworth's theorem and the von Neumann minimax theorem, that we haven't covered. Planarity thus has applications in a variety of places. Essentially what happens is a and b act as a bottleneck. Next, T is connected, since if it weren't, we could add an edge between components of T without causing a cycle. We will start building the tree from the upper left. A graph with 10 vertices has 20 edges. We subtract that value from all the uncovered entries and add it to all the doubled covered entries (just the last two entries in row 2). Then by the theorem, we have χ ≥ n/α, so χ ≥ 7/2, which is 3.5. The claw is a star with a central vertex and three outer vertices. The crossing number of the complete graph Kn is also not known in general. People often confuse what this theorem says. See the figure below for three digraphs. A matching is shown in the graph below on the left. This might be a little surprising since there are so many more connections in the Cartesian product than in the union. Independent sets and coverings turn out to be closely related to each other. To see this, start by assuming the two partite sets are named, Relationship between the algorithm and the theorem. Remember that what limits an Eulerian circuit to having all even vertices is that whenever we enter a vertex via an edge, we need another edge to be able to exit that vertex. However, Chvátal's condition gives us a way out. We can use Euler's formula to get the following inequalities that we can use to show certain graphs are not planar. Since the only 2-regular graphs are cycles, a 2-factor is a collection of cycles in the graph that collectively use all the vertices exactly once. A natural question arises: if we start with a matching that is not the largest possible, is there an augmenting path in the graph? In this case, the trick was to travel along a cycle and detour along the smaller Eulerian circuits. An edge between a person and job indicates that person can do that job. Notice also that these components are as large as possible. Kempe's 1879 proof was widely celebrated and helped get him elected to the Royal Society. Also, no player should play two matches in the same day. In everything that follows, assume the two partite sets are X and Y. Draw it. The answer is yes. Dijkstra's algorithm is quite famous and is used in a number of real applications, such as network packet routing. A strongly connected digraph has an Eulerian circuit if and only if the indegree of every vertex matches its outdegree. Then visit the vertices in that order, assigning each vertex the smallest color available that hasn't already been used on one of its neighbors. As with complete graphs, for bipartite graphs there is a close relationship between matchings and edge colorings. It comes from one of the earliest and most influential problems in graph theory, called the four color problem. The problem is vertices a, b, and c only have edges to A and B. We are interested in the sum of the capacities of the edges in cuts. In the graph below, vertices A and C have degree 4, since there are 4 edges leading into each vertex. In a DAG, there is a faster way to find a minimum weight path than using Dijkstra's algorithm. Notice the similarity between flows and cuts and what we were interested in with Menger's theorem—disjoint paths between vertices and sets that cut off the two vertices from each other. The connectivity of a graph G, denoted κ'(G), is the minimum number of edges that need to be deleted in order to disconnect the graph. Then we have what is called an Eulerian trail. For example, if we were to list out all the subsets of {1,2,…,n}, that would be an O(2n) algorithm, as there are 2n subsets. To see why, suppose there were no vertices of indegree 0. There exists a graph with no triangles and chromatic number 10. A matching would assign time slots to people. The relation to graph theory is that maps can be represented by graphs by making each region (county, state, etc.) The copy a' is made adjacent to a's neighbor, and the copy b' is made adjacent to b's neighbor. Prove that there exists a pair (v1;v2) of leaves that have a common neighbor.Solution: Suppose not, i.e. Here is the equivalent for a digraph. If we begin with just the vertices and no edges, every vertex has degree zero, so the sum of those degrees is zero, an even number. This generalizes to larger cliques (complete subgraphs). In that case, the cycle minus edge uv would be a path from u to v containing every vertex of G (a Hamiltonian path). The vertices x, y, and z form a triangle, so at most one of them can be part of an independent set. Therefore, the sum of the degrees of every vertex in a graphGis always even. But the converse isn't true as C5 has a chromatic number of 3 but no triangle. That is, they are independent sets of edges. Therefore, we can't have both the 1, 3 path and the 2, 4 path in the graph. A little later we will explain where the term “colors” comes from, but for now the colors we assign will be positive integers. A big part of this step consists of walking back and forth between X and Y along edges that alternately have flow 0 and flow 1 in a way that is exactly analogous to the bouncing back and forth between unmatched and matched edges that the augmenting path algorithm for matching does. What is the degree of the remaining vertex? They are named for William Rowan Hamilton, best known for his work in physics and quaternions, who once tried marketing a board game based on Hamiltonian cycles. In particular, on a torus we can embed K3,3 and K4,4 but not K5,5. If Y happens to be bigger than X, then we are interested in matchings in which every vertex of X is matched. ... Said another way, a graph is Eulerian if and only if every vertex has even degree. Every edge in the graph is incident on one of those vertices. In the first part of the proof, we add the vertex v adjacent to a and b to make the graph even. It also has a K5 minor, which can be seen by contracting each of the edges that lead from the outside cycle to the inside cycle. Prim's algorithm is another approach to finding a minimum spanning tree. Note that the other lower bound, χ ≥ n/α is not of much help here. Once our goal vertex has the cheapest label, no other route to that goal could possibly be better than what we have. Prüfer codes that have distinct values in all positions? One thing to notice right away is that if a vertex has degree k, then we will need at least k different colors for the edges incident on that vertex. The rightmost graph is strongly connected as we can follow directed edges to get from any vertex to any other. Prove that no 4-regular bipartite graph can be planar. With the König-Egerváry theorem it was that a minimum vertex cover led to a maximum matching of the same size. On the other hand, the graph on the right fails Ore's condition because the two degree 3 vertices are not adjacent, and their degrees sum to 3+3 = 6, which is not at least 8. Here is how it works: Start anywhere. Consider the following problem: We have three people, A, B, and C, and three jobs a, b, and c. There is a certain cost associated with each person doing a given job. It has indegree greater than 0, so there must be some edge v2 → v1. At the same time as it is doing this, it searches for cuts. This argument, while intuitively clear, is a little tricky to make rigorous because we would have to prove that every graph can be built up in this way by starting with a tree and adding edges and leaves. Explain why. [Degree sum formula] In any graph, the sum of the degrees of all vertices is twice the number of edges. 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