6 Directed Graphs 6.1 Digraphs In some problems the relation between the objects is not symmetric. For these cases we need directed graphs, where the edges are oriented from one vertex to another. As an example consider a map of a small town. Can you make the streets one-way, and still be able to drive from one house to another (or exit the town)? Definitions DEFINITION. A digraph (or a directed graph) consists of the vertices and (directed) edges (without loops ). We still write for , but note that now . For each pair define the inverse of as . Note that does not imply . DEFINITION. Let be a digraph. Then is its subdigraph, if and , induced subdigraph, , if and . The underlying graph of a digraph is the graph on such that if , then the undirected edge with the same ends is in . A digraph is an orientation of a graph , if and implies . In this case, is said to be an oriented graph. DEFINITION. Let be a digraph. A walk of is a directed walk, if for all . Similarly, we define directed paths and directed cycles as directed walks and closed directed walks without repetitions of vertices. The digraph is di-connected, if, for all , there exist directed paths and . The maximal induced di-connected subdigraphs are the di-components of . 6.1 Digraphs 84 Note that a graph might be connected, although the digraph is not di- connected. DEFINITION. The indegree and the outdegree of a vertex are defined as follows We have the following handshaking lemma. (You offer and accept a handshake.) Lemma 6.1. Let be a digraph. Then Directed paths The relationship between paths and directed paths is in gen- eral rather complicated. This digraph has a path of length five, but its directed paths are of length one. There is a nice connection between the lengths of directed paths and the chromatic number . Theorem 6.1 (ROY (1967),GALLAI (1968)). A digraph has a directed path of length . Proof. Let be a minimal set of edges such that the subdigraph contains no directed cycles. Let be the length of the longest directed path in . For each vertex , assign a colour , if a longest directed path from has length in . Here . First we observe that if ( ) is any directed path in , then . Indeed, if , then there exists a directed path of length , and is a directed path, since does not contain directed cycles. Since , . In particular, if , then . Consider then an edge . By the minimality of , contains a directed cycle , where the part is a directed path in , and hence . This shows that is a proper colouring of , and therefore , that is, . The bound is the best possible in the following sense: Theorem 6.2. Every graph has an orientation , where the longest directed paths have lengths . 6.1 Digraphs 85 Proof. Let and let be a proper -colouring of . As usual the set of colours is . We orient each edge by setting , if . Clearly, the so obtained orientation has no directed paths of length . DEFINITION. An orientation of an undirected graph is acyclic, if it has no directed cycles. Let be the number of acyclic orientations of . The next result is charming, since measures the number of proper colourings of using colours! Theorem 6.3 (STANLEY (1973)). Let be a graph of order . Then the number of the acyclic orientations of is where is the chromatic polynomial of . Proof. The proof is by induction on . First, if is discrete, then , and as required. Now is a polynomial that satisfies the recurrence . To prove the claim, we show that satisfies the same recurrence. Indeed, if (6.1) then, by the induction hypothesis, For (6.1), we observe that every acyclic orientation of gives an acyclic orientation of . On the other hand, if is an acyclic orientation of for , it extends to an acyclic orientation of by putting or . Indeed, if has no directed path , we choose , and if has no directed path , we choose . Note that since is acyclic, it cannot have both ways and . We conclude that , where is the number of acyclic orientations of that extend in both ways and . The acyclic orientations that extend in both ways are exactly those that contain neither nor as a directed path. (6.2) Each acyclic orientation of corresponds in a natural way to an acyclic orientation of that satisfies (6.2). Therefore , and the proof is completed. One-way traffic Every graph can be oriented, but the result may not be di-connected. In the one-way traffic problem the resulting orientation should be di-connected, for otherwise someone is not able to drive home. ROBBINS’ theorem solves this problem. 6.1 Digraphs 86 DEFINITION. A graph is di-orientable, if there is a di-connected oriented graph such that . Theorem 6.4 (ROBBINS (1939)). A connected graph is di-orientable if and only if has no bridges. Proof. If has a bridge , then any orientation of has at least two di-components (both sides of the bridge). Suppose then that has no bridges. Hence has a cycle , and a cycle is always di- orientable. Let then be maximal such that it has a di-orientation . If , then we are done. Otherwise, there exists an edge such that but (because is connected). The edge is not a bridge and thus there exists a cycle in , where is the last vertex inside . In the di-orientation of there is a directed path . Now, we orient and the edges of in the direction to obtain a directed cycle . In conclusion, has a di-orientation, which contradicts the maximality assumption on . This proves the claim. Example 6.1. Let be a digraph. A directed Euler tour of is a directed closed walk that uses each edge exactly once. A directed Euler trail of is a directed walk that uses each edge exactly once. The following two results are left as exercises. (1) Let be a digraph such that is connected. Then has a directed Euler tour if and only if for all vertices . (2) Let be a digraph such that is connected. Then has a directed Euler trail if and only if for all vertices with possibly excepting two vertices for which . The above results hold equally well for multidigraphs, that is, for directed graphs, where we allow parallel directed edges between the vertices. Example 6.2. The following problem was first studied by HUTCHINSON AND WILF (1975) with a motivation from DNA sequencing. Consider words over an alphabet of letters, that is, each word is a sequence of letters. In the case of DNA, the let- ters are . In a problem instance, we are given nonnegative integers and for , and the question is: does there exist a word in which each letter occurs exactly times, and is followed by exactly times. For instance, if , , and , , , , then the word is a solution to the problem. 6.1 Digraphs 87 Consider a multidigraph with for which there are edges . It is rather obvious that a directed Euler trail of gives a solution to the sequencing problem. Tournaments DEFINITION. A tournament is an orientation of a complete graph. Example 6.3. There are four tournaments of four vertices that are not isomorphic with each other. (Isomorphism of directed graphs is defined in the obvious way.) Theorem 6.5 (RÉDEI (1934)). Every tournament has a directed Hamilton path. Proof. The chromatic number of is , and hence by Theorem 6.1, a tournament of order has a directed path of length . This is then a directed Hamilton path visiting each vertex once. The vertices of a tournament can be easily reached from one vertex (sometimes called the king). Theorem 6.6 (LAUDAU (1953)). Let be a vertex of a tournament of maximum outdegree. Then for all , there is a directed path of length at most two. Proof. Let be an orientation of , and let be the maximum outdegree in . Suppose that there exists an , for which the directed distance from to is at least three. It follows that and for all with . But there are vertices in , and thus vertices in . It follows that the outdegree of is , which contradicts the maximality assumption made for . Problem. Ádám’s conjecture states that in every digraph with a directed cycle there exists an edge the reversal of which decreases the number of directed cycles. Here the new digraph has the edge instead of . Example 6.4. Consider a tournament of teams that play once against each other, and sup- pose that each game has a winner. The situation can be presented as a tournament, where the vertices correspond to the teams , and there is an edge , if won in their mutual game. DEFINITION. A team is a winner (there may be more than one winner), if comes out with the most victories in the tournament. 6.1 Digraphs 88 Theorem 6.6 states that a winner either defeated a team or defeated a team that defeated . A ranking of a tournament is a linear ordering of the teams that should reflect the scoring of the teams. One way of ranking a tournament could be by a Hamilton path: the ordering can be obtained from a directed Hamilton path . However, a tournament may have several directed Hamilton paths, and some of these may do unjust for the ‘real’ winner. Example 6.5. Consider a tournament of six teams , and let be the scoring digraph as in the figure. Here is a di- rected Hamilton path, but this extends to a directed Hamilton cycle (by adding )! So for every team there is a Hamilton path, where it is a winner, and in another, it is a looser. Let be the winning number of the team (the number of teams beaten by ). In the above tournament, So, is team 1 the winner? If so, is 2 or 3 next? Define the second-level scoring for each team by This tells us how good teams beat. In our example, we have Now, it seems that 3 is the winner,but 4 and 6 have the same score. We continue by defining inductively the th-level scoring by It can be proved (using matrix methods) that for a di-connected tournament with at least four teams, the level scorings will eventually stabilize in a ranking of the tournament: there exits an for which the th-level scoring gives the same ordering as do the th-level scorings for all . If is not di-connected, then the level scoring should be carried out with respect to the di-components. In our example the level scoring gives as the ranking of the tournament. 6.2 Network Flows 89 6.2 Network Flows Various transportation networks or water pipelines are conveniently represented by weighted directed graphs. These networks usually possess also some additional requirements. Goods are transported from specific places (warehouses) to final locations (marketing places) through a network of roads. In modeling a transportation network by a digraph, we must make sure that the number of goods remains the same at each crossing of the roads. The problem setting for such networks was proposed by T.E. Harris in the 1950s. The connection to Kirchhoff’s Cur- rent Law (1847) is immediate. According to this law, in every electrical network the amount of current flowing in a vertex equals the amount flowing out that vertex. Flows DEFINITION. A network consists of an underlying digraph , two distinct vertices and , called the source and the sink of , and a capacity function (nonnegative real numbers), for which , if . Denote and . Let be a set of vertices, and any function such that , if . We adopt the following notations: and In particular, and DEFINITION. A flow in a network is a function such that for all and for all Example 6.6. The value can be taught of as the rate at which transportation actually happens along the channel which has the maximum capacity . The second condition states that there should be no loss. 6.2 Network Flows 90 If is a network of water pipes, then the value gives the capacity ( ) of the pipe . The previous network has a flow that is indicated on the right. A flow in is something that the network can handle. E.g., in the above figure the source should not try to feed the network the full capacity ( ) of its pipes, because the junctions cannot handle this much water. DEFINITION. Every network has a zero flow defined by for all . For a flow and each subset , define the resultant flow from and the value of as the numbers and A flow of a network is a maximum flow, if there does not exist any flow such that . The value of a flow is the overall number of goods that are (to be) transported through the network from the source to the sink. In the above example, . Lemma 6.2. Let be a network with a flow . (i) If , then . (ii) . Proof. Let . Then where the third equality holds since the values of the edges with cancel each out. The second claim is also clear. Improvable flows Let be a flow in a network , and let be an undirected path in where an edge is along , if , and against , if . We define a nonnegative number for as follows: where if is along if is against . DEFINITION. Let be a flow in a network . A path is ( -)improvable, if . On the right, the bold path has value , and therefore this path is improvable. 6.2 Network Flows 91 Lemma 6.3. Let be a network. If is a maximum flow of , then it has no improvable paths. Proof. Define if is along P if is against P if is not in P Then is a flow, since at each intermediate ver- tex , we have , and the capacities of the edges are not exceeded. Now , since has exactly one edge for the source . Hence, if , then we can improve the flow. Max-Flow Min-Cut Theorem DEFINITION. Let be a network. For a subset with and , let the cut by be The capacity of the cut is the sum A cut is a minimum cut, if there is no cut with . Example 6.7. In our original network the capac- ity of the cut for the indicated vertices is equal to . Lemma 6.4. For a flow and a cut of , Proof. Let . Now (since ), and . Hence 6.2 Network Flows 92 Theorem 6.7. For a flow and any cut of , . Furthermore, equality holds if and only if for each and , (i) if , then , (ii) if , then . Proof. By the definition of a flow, and . By Lemma 6.4, , and hence , as required. Also, the equality holds if and only if (1) and (2) . This holds if and only if for all (since ), and (2) for all with , . This proves the claim. In particular, if is a maximum flow and a minimum cut, then Corollary 6.1. If is a flow and a cut such that , then is a maximum flow and a minimum cut. The following main result of network flows was proved independently by ELIAS, FEIN- STEIN, SHANNON, by FORD AND FULKERSON, and by ROBACKER in 1955 – 56. The present approach is due to Ford and Fulkerson. Theorem 6.8. A flow of a network is maximum if and only if there are no -improvable paths in . Proof. By Lemma 6.3, a maximum flow cannot have improvable paths. Conversely, assume that contains no -improvable paths, and let for some path Set . Consider an edge , where and . Since , there exists a path with . Moreover, since , for the path . Therefore , and so . By the same argument, for an edge with and , . By Theorem 6.7, we have . Corollary 6.1 implies now that is a maximum flow (and is a minimum cut). [...]... (circles), 68 Kuratowski graph, 61 latin rectangle, 37 latin square, 37 leaf, 7 line segment graph, 69 linked cycles, 77 list chromatic number, 67 list colouring, 67 loop, 4 Mửbius band, 71 map, 65 matching, 34 maximal planar graph, 60 maximum degree à, 7 ow, 85 matching, 34 minimum cut, 87 degree à, 7 weighted distance, 12 minor, 69 monochromatic, 46 multigraph, 4 Ă ặ near-triangulation, 67 neighbour,... orientation of Then necessarily the Theorem 6. 12 A connected graph has a ow number degrees of the vertices are even, and so is eulerian ỉ à ô à ắ ĩ í ĩí ĩí ĩí à , and if ề , then Theorem 6. 13 We have à ắ ềà if ề is odd if ề is even Example 6. 9 For each -regular bipartite graph , we have Indeed, let be -biparte By Corollary 3.1, a -regular graph has a perfect matching Orient the edges... values A -ow is for all ắ nowhere zero, if ô à ẳ ô à ô à In the -ows we do not have any source or sink For convenience, let for all ắ in the orientation of so that the condition (6. 3) becomes ắ ô à ẳ ẵ (6. 4) ẵ ẵ ắ Example 6. 8 A graph with a nowhere zero -ow ẵ The condition (6. 4) generalizes to the subsets ắ since the values of the edges inside Lemma 6. 6 If ô à ẳ ẻ ắ ắ in a natural way, (6. 5) cancel... edge connectivity Corollary 6. 4 A graph is -edge connected if and only if any two distinct vertices of connected by at least independent paths are ỉ Proof The claim follows immediately from Corollary 6. 3 Seymours 6- owsÊ ôà of an undirected graph is an orientation of ô ẳ ẵ such that for all vertices ắ ẻ , ô à ô à D EFINITION A -ow with an edge colouring ắ ắ together (6. 3) that is, the sum of... spanning subgraph, 10 spanning tree, 19 spatial embedding, 77 sphere, 73 sphere with a handle, 74 stable matching, 39 stable set, 15 subdigraph, 79 subdivision, 57 subgraph, 10 surface, 71 symmetric difference, 3 topologically equivalent, 71 torus, 74 tournament, 83 trail, 28 transversal, 36 tree, 17 triangle, 71 triangle-free, 51 triangulation, 71 trivial graph, 9 trivial path, 11 2-cell, 76 2-switch,... connectivity number à, 23 contracted vertex, 55 critical, 51 crossing number, 64 cube, 9 cut (in a network), 87 cut vertex, 22 cycle, 11 degree à, 7 di-connected di-component, 80 di-orientable, 82 digraph, 79 directed Euler tour, trail, 82 walk, path, cycle, 80 directed graphs (digraph), 4 disconnected, 12 disconnecting set, 25 discrete graph, 9 disjoint walks, 11 distance, 12 distance function, 5 edge,... 58 face, 58 fan, 26 ow, 85 ow number, 90 forest, 17 genus, 75 graph, 4 graphical sequence, 8 Hamilton path, cycle, 30 hamiltonian, 30 Hamming distance, 21 homeomorphic, 71 improvable (path), 86 Index 97 improvement (colouring), 42 incident colours, 41 indegree, 80 independent paths, 11 induced subdigraph, 79 induced subgraph, 10 interior: face, vertex, edge, 58 intersection graph, 6 inverse pair, 79... graphs The proof of this uses the -Colour Theorem In order to fully appreciate Seymours result, Theorem 6. 11, we mention that it was proved as late as 19 76 (by JAEGER) that every bridgeless has a nowhere zero -ow for some integer S EYMOURs remarkable result reads as follows: 6. 2 Network Flows 95 Theorem 6. 11 (S EYMOURs (1981)) Every bridgeless graph has a nowhere zero -ow ỉ Proof Omitted D EFINITION... other In particular, has a nowhere zero -ow for some , then has no bridges Tuttes Problem It was conjectured by T UTTE (1954) that every bridgeless graph has a nowhere zero -ow The Petersen graph has a nowhere zero -ow but does not have any nowhere -ows, and so is the best one can think of Tuttes conjecture resembles the Colour Theorem, and indeed, the conjecture is known to hold for the planar graphs... nontrivial graph, 9 NP-complete problems, 3 ặ odd component, 37 odd cycle, 11 optimal colouring, 42 order , 4 orientable surface, 71 orientation, 79 oriented, 71 oriented graph, 79 outdegree, 80 parallel edges, 4 partition, 3 path, 11 perfect matching, 34 Petersen graph, 9, 31 planar graph, 57 plane embedding, 57 plane model, 72 proper colouring, 41, 50 Ramsey number, 47 ranking, 83 regular graph, 9 . vertices. The digraph is di-connected, if, for all , there exist directed paths and . The maximal induced di-connected subdigraphs are the di-components of . 6. 1 Digraphs 84 Note that a graph might. problem. 6. 1 Digraphs 86 DEFINITION. A graph is di-orientable, if there is a di-connected oriented graph such that . Theorem 6. 4 (ROBBINS (1939)). A connected graph is di-orientable if and only. (circles), 68 Kuratowski graph, 61 latin rectangle, 37 latin square, 37 leaf, 7 line segment graph, 69 linked cycles, 77 list chromatic number, 67 list colouring, 67 loop, 4 Möbius band, 71 map, 65 matching,