4.1 Eulerian Tours. 4. Eulerian Tours and Trails. Königsberg. Seven Bridges. Try. Königsberg bridge problem

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1 4. Eulerian Tours and Trails 4.1 Eulerian Tours FMONG NIE 1 FMONG NIE 2 Königsberg Königsberg was a city in eastern Prussia in 18 th century. river flowed through this city and separated it into four pieces. River Island ank ank Island FMONG NIE 3 Seven ridges Seven bridges connected the two islands and the two opposite banks. River Island ank ank Island FMONG NIE 4 Königsberg bridge problem Is there a tour that starts from any one of the places,, and, passes each bridge eactly once and returns to the starting place? Try an you find a tour that starts from any one of the places,, and, passes each bridge eactly once and returns to the starting place? River River FMONG NIE 5 FMONG NIE 6

2 River Transfer to a graph problem Is there a closed trail in the following graph which passes each edge eactly once? onverted into a graph FMONG NIE 7 FMONG NIE 8 Königsberg bridge problem This problem has no solutions. It was first proved by Euler, a Swiss mathematician in 18 century. This problem is also called the seven bridge problem. Generalization Euler generalized the Königsberg bridge problem to a general graph problem: What graphs have a closed trail which passes each edge eactly once? Erler characterized all such graphs. FMONG NIE 9 FMONG NIE 10 Eulerian Tours closed trail is a closed walk in a graph which does not repeat any edge. Eulerian tour in a graph G is a closed walk which goes through each edge in G eactly once. graph G is called an Euler graph if G has a Eulerian tour. Is the following graph an Euler graph? Yes FMONG NIE 11 FMONG NIE 12

3 Is the following graph an Euler graph? No. Is the following graph an Euler graph? Yes. FMONG NIE 13 FMONG NIE 14 Property If G is an Euler graph, then every verte of G has an even degree. Proof: Let T be an Eulerian tour of G and be any verte in G. If T passes eactly k times, then T goes into eactly k times and leaves also eactly k time. Thus d() = 2k. Hence every verte in G has an even degree. go in k times leave k times go in k times leave k times FMONG NIE 15 FMONG NIE 16 Sufficient condition The condition that every verte has an even degree is also sufficient for a connected graph to be Eulerian. FMONG NIE 17 Lemma If G has no verte of degree 1, then either G is empty or G contains a cycle. Proof. ssume that G is not empty. Then G has a component H which is not K 1. Since H has no verte of degree 1, H is not a tree. Thus H contains a cycle, which is also a cycle in G. FMONG NIE 18

4 Lemma If two edge-disjoint closed trails T 1 and T 2 in G have some vertices in common, then they can be merged into one closed trail. Let be a verte in botht 1 and T 2. ssume that the two closed trails are T e e Le e : s s s+ 1 T2 e' 1 y1e' 2 y2 Le' yte' : t t+ 1 T 1 T 2 FMONG NIE 19 where e i and e j are edges in G and i and y j are vertices in G. Then they can be merged into one closed trail: T : e 1e 22 Les ses + 1e' 1 y1e' 2 y2 Le' t yte' t+ 1 1 FMONG NIE 20. Theorem For any connected graph G, the following statements are equivalent: (i) G is an Euler graph; (ii) all vertices in G are of even degrees; (iii) there eists a collection of cycles in G such that each edge of G belongs to eactly one cycle of this collection. The following graph is Eulerian; Every verte has even degree; There is a collection of cycles such that every edge belongs to eactly one cycle. or FMONG NIE 21 FMONG NIE 22 Proof (i) (ii) (iii) (i). (i) (ii) follows from Lemma (ii) (iii) Suppose that all vertices in G are even degrees. Let k be the maimum integer such that G contains a collection of k edge-disjoint cycles 1, 2,, k. Let H be the graph obtained from G by deleting all edges in these cycles. FMONG NIE 23 Whenever all edges in a cycle are deleted from a graph, a verte in new graph has either the same degree or 2 less than its degree in the original graph. Since all vertices in G are of even degrees, all vertices in H are also even degrees. If H is not empty, by Lemma 4.1.2, H contains a cycle k+1, implying that 1, 2,, k, k+1 form a collection of edge-disjoint cycles in G, a contradiction. Hence H is empty, implying that (iii) holds. FMONG NIE 24

5 (iii) (i). Let T 1, T 2,, T k be a collection of cycles in G such that each edge of G belongs to eactly one cycle of this collection. Since G is connected, these cycles can be reordered such that for each i 2, V ( i j T ) V ( T ) φ holds for some 1 j<i. Suppose that T 1, T 2,, T s can be merged into a closed trail U s, where 1 s k-1. Since V(T s+1 ) V(T j ) for some j<s+1, T s+1 and U s has some vertices in common. y Lemma 4.1.2, T s+1 and U s can be merged into a closed trail U s+1. Hence T 1, T 2,, T k can be merged into a closed trail U k, which passes each edge eactly once. Therefore G is an Euler graph. FMONG NIE 25 FMONG NIE 26 Eulerian? To know whether a connected graph is an Euler graph, we just need to check its degrees. If some verte is not even degree, then this graph is not Eulerian. Verte has an odd degree and so the graph is not Eulerian. FMONG NIE 27 FMONG NIE 28 Eulerian? To know whether a connected graph is an Euler graph, we can also check it by (1) removing cycles one by one, until no cycles. (2) If there are no edges remaining, then this graph is Eulerian; otherwise, it is not. FMONG NIE 29 Not an Eulerian graph FMONG NIE 30

6 It is an Eulerian graph For what integers n, K n is Eulerian? Odd integers n. etermine integers m and n such that K m,n is Eulerian. oth m and n are even. FMONG NIE 31 FMONG NIE 32 Let n be an even positive integer. t least how many new edges should be added to K n so that the new graph (it is multigraph) is Eulerian? nswer: n/2 How do we add these edges? The n vertices in K n is partitioned into n/2 pairs, and add one edge to join each of these pairs of vertices. FMONG NIE 33 Let n be an even positive integer. t least how many edges should be removed from K n so that the new graph is Eulerian? n/2. FMONG NIE 34 Question Let G be any connected graph of order n, where n 2. t most how many new edges should be added to G so that the new graph is Eulerian? nswer: n / 2. t least how many edges should be added to this graph to change it into an Eulerian graph? FMONG NIE 35 FMONG NIE 36

7 4.2 Eulerian trails n Eulerian trail in G is a trail which passes each edge in G eactly once. oes the following has an Eulerian trail? yes. w y u z FMONG NIE 37 FMONG NIE 38 Eulerain graphs If G is an Euler graph, then G has a Eulerian trail. FMONG NIE 39 Theorem ssume that G has an Eulerian trail. Then we can get an Euler graph by adding at most one edge to G. Proof: Let T be an Eulerian trail of G. If the two ends of T are the same verte, then G is an Euler graph. If the two ends of T are different, then the new graph obtained from G by adding a new edge joining the two ends of T is an Euler graph. FMONG NIE 40 The following graph has an Euler trail. y adding one edge to it, it becomes an Eulerian graph. w y z FMONG NIE 41 u Theorem Let G be a connected graph. Then G has an Eulerian trail if and only if G has at most two vertices of odd degrees. Proof. (Necessity) Let G have an Eulerian trail. y Theorem 4.2.1, we can get an Eulerian graph by adding at most one edge to G. y Theorem 4.1.2, all vertices in this new graph are of even degrees. Thus G has at most two vertices of odd degrees. FMONG NIE 42

8 (Sufficiency) If G has no vertices of odd degree, then G has an Eulerian tour, which is also an Eulerian trail. G cannot have eactly one verte of odd degree. ssume that G has two vertices,y of odd degree. Let H be the graph obtained from G by adding a new edge e joining and y. Then all vertices in H are of even degree. y Theorem 4.1.2, H has an Eulerian tour T. Thus T-e is an Eulerian trail in G. FMONG NIE 43 FMONG NIE 44 Eercise Let m,n be positive integers, where m n. etermine the values of m and n such that K m,n has an Eulerian trail. nswer {(m,n): m n, K m,n has an Eulerian trail} = {(1,1),(1,2)} {(2,n):n 2} {(m,n): both m,n are even} FMONG NIE 45 FMONG NIE 46 Eercise What trees have an Eulerian trail? Why? nswer: Only path graphs have an Eulerian trail. Suppose that G has an Eulerian trail. s G is a tree, G has at least two vertices of degree 1. Since G has an Eulerian trail, G has at most two vertices of odd degrees. Thus G has only two vertices of degree 1. tree with eactly two vertices of degree 1 is a path graph, implying that G is a path graph. FMONG NIE 47 FMONG NIE 48

9 Let G be a connected graph with eactly 2k vertices of odd degrees, where k>0. Is there a collection of k edge-disjoint trails in G such that each edge in G belongs to eactly one of these trails? Why? Yes. FMONG NIE 49 fter adding two edges, the new graph has an Eulerian tour. Removing the two edges breaks the Eulerian tour into two edge-disjoint trails. FMONG NIE 50 Let 1, 2,, k and y 1,y 2,,y k be 2k vertices in G which are of odd degrees. Let H be the new graph obtained from G by adding k edges e 1, e 2,, e k joining i to y i for i=1,2,,k. Then in H all vertices are of even degrees. Thus H has an Eulerian tour T. Then T-{e 1,e 2,,e k } is the union of k trails, which are trails in G. 4.3 Eulerian Tours and Trails in digraphs FMONG NIE 51 FMONG NIE 52 efinitions Let be a digraph. directed Eulerian tour in is a directed closed trail which passes each arc of eactly once. digraph with a directed Eulerian tour is called an Euler digraph. directed Eulerian trail in is a directed trail which passes each arc of eactly once. Is the following digraph an Euler digraph? Yes. FMONG NIE 53 FMONG NIE 54

10 Is the following digraph an Euler digraph? No. oes it have a directed Eulerian trail? Yes. Theorem Let be a weakly connected digraph. The following statements are equivalent: (i) is an Euler digraph; (ii) od() = id() for each verte in ; (iii) There eists a collection of directed cycles in such that each arc of belongs to eactly one cycle of this collection. FMONG NIE 55 FMONG NIE 56 The following digraph is Eulerian. Every verte has the same in-degree and outdegree. There is a collection of arc-disjoint directed cycles such that every arc belongs to one cycle. (ii) (iii) Let k be the maimum integer such that has a collection of k arc-disjoint cycles. Let be the new digraph obtained from by deleting all arcs in these cycles. Observe that every verte in has the same in-degree and out-degree. FMONG NIE 57 FMONG NIE 58 Suppose that is not empty. Let H be a component of such that H is not K 1. Observe that H has no any verte with outdegree 0, implying that H is not acyclic. So H contains a cycle, which is also a cycle in. Hence has a collection of more than k arcdisjoint cycles, a contradiction. Therefore is empty, implying that has a collection of k cycles such that each arc is on eactly one of these cycles. FMONG NIE 59 FMONG NIE 60

11 Theorem Let be a weakly connected digraph which is not an Euler digraph. Then has a directed Eulerian trail if and only if there eists two vertices,y in such that id()=od()+1, id(y)=od(y)-1 and id(z)=od(z) for all other vertices z in. The following digraph has a directed Eulerian trail, but is not an Erler digraph. It has two vertices,y such that id()=od()+1, id(y)=od(y)-1 and id(z)=od(z) for all other vertices z in. y FMONG NIE 61 FMONG NIE 62 Proof (Necessity) ssume that has a directed Eulerian trail T with distinct starting verte y and ending verte. Then id() = od()+1, id(y)=od(y)-1 and id(z)=od(z) for all other vertices z. (Sufficiency) Let be the new digraph obtained from by adding a new arc (,y). Then for every verte w in, id (w)=od (w). y Theorem 4.3.1, has a directed Eulerian tour T. Then T-(,y) is a directed Eulerian trial of. FMONG NIE 63 FMONG NIE 64 Let be a weakly connected digraph. Let k = id( ) od( ) V ( ) 1. Then k is even. Why? 2. If k>0, then has a collection of k/2 directed trails such that each arc of is on eactly one of these trails. Why? k = V ( ) V ( ) id( ) od( ) [ id( ) od( )] id( ) V ( ) V ( ) od( ) (mod 2) ( ) ( ) (mod 2) = 0. (mod 2) FMONG NIE 65 FMONG NIE 66

12 End FMONG NIE 67