3 Euler Tours, Hamilton Cycles, and Their Applications

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1 3 Euler Tours, Hamilton Cycles, and Their Applications 3.1 Euler Tours and Applications Euler tours Carefully review the definition of (closed) walks, trails, and paths from Section 1... Definition 3.1 An Euler trail in a graph is a trail that contains every edge of the graph. An Euler tour is a closed Euler trail. A graph is called eulerian is it has an Euler tour. The origins of the problem of eulerian graphs and, in fact, the origins of graph theory, go back to the eighteenth century puzzle called the Bridges of Königsberg. In the town of Königsberg in what was then East Prussia, the two branches of the river Pregel converge and flow through to the Baltic Sea. The town is divided into four parts, which are connected to each other by the famous seven bridges (see Figure 11, left). The townspeople want to know if it is possible to take a walk that crosses each of the bridges exactly once before returning to the starting point. The Prussian emperor, Frederick the Great, brought the problem to the attention of the famous Swiss mathematician Leonhard Euler, and Euler proved in 13 that no such walk is possible. A A D B D B C C Figure 11: The bridges of Königsberg. Euler used the graph in Figure 11, right, to model the problem, and showed that this graph has no Euler tour. Explain. A generalization of this result a complete characterization of eulerian graphs is given by our next theorem. Theorem 3. Let G be a connected graph. Then G is eulerian if and only if the degree of every vertex in G is even. Proof. First notice that if G has no edges, then it must be isomorphic to K 1 (since it is connected), and the statement of the theorem is trivially true.. Hence we may assume G is not edgeless. ( ) Assume G has an Euler tour T starting at a vertex v 0. Each time an internal vertex u of the tour is encountered during a traversal of T, two edges incident with u are traversed. Since each edge of the graph is traversed exactly once, u must have even degree. For the initial vertex v 0 of T, the first and the last edge of T each add one to the count, and each 31

2 encounter of v 0 as an internal vertex adds two to the count of edges incident with v 0. Hence v 0 also has even degree. ( ) We must prove that every connected graph with all vertices of even degree is eulerian. We shall prove this by contradiction. Suppose there exists a connected graph with all vertices of even degree that is not eulerian. Among such graphs, let G be one with the smallest number of edges. Let T be a longest trail in G. Since every vertex in G has even degree and T cannot be extended, T must be a closed trail, and since G contains an edge, T has at least one edge. If T uses all the edges of G, then T is an Euler tour of G a contradiction. Hence T does not contain all the edges of G. Consider the graph G = G E(T ). This graph has all vertices of even degree and contains at least one edge. Let H be a non-trivial connected component of G. Hence H is connected, has at least one edge, and has all vertices of even degree. Moreover, H has fewer edges than G, so by the assumption on G, H has an Euler tour T. Now, since G is connected, T and T necessarily have a vertex w in common. We can now join T and T at this vertex w to obtain a closed trail T. But since T has at least one edge, T is longer than T, contradicting the choice of T. We conclude that every connected graph with all vertices of even degree must have an Euler tour. Figure 1: Exercise 3.3. Exercise 3.3 For each of the graphs in Figure 1, determine whether the graph has an Euler tour. Algorithm 3.4 Constructing an Euler Tour procedure Euler(G: eulerian graph) x := first vertex of G T := closed trail in G starting at x E := E(G) E(T ) while E begin G := G[E ] x := vertex on T incident with an edge in G T := closed trail in G starting at x T := trail T enlarged by inserting T into T at vertex x E := E E(T ) end {T is an Euler tour in G} 3

3 Exercise 3. Use Algorithm 3.4 to construct an Euler tour of the graph in Figure 13. Figure 13: Exercise 3.. Exercise 3. Let G be an eulerian graph. (a) Explain in detail how to construct a closed trail starting at a given vertex x of G. Prove that such a trail exists for every x V (G). (b) Prove that if T is a closed tour in G that is not an Euler tour, then there exists a vertex on T that is incident with an edge of G not on T. (c) Use (a) and (b), together with some ideas from the proof of Theorem 3., to show that Algorithm 3.4 correctly constructs an Euler tour of an eulerian graph. Exercise 3. Estimate the time complexity of Algorithm 3.4. Exercise 3.8 Prove that a graph with at least one edge and all vertices of even degree must contain a cycle. Exercise 3. Prove that every closed trail is an edge-disjoint union of cycles. (Hint: First show that every nontrivial closed trail contains a cycle. Then use strong induction on the number of edges to prove the statement.) Exercise 3.10 An open Euler trail in a graph is an Euler trail that is not a tour (i.e. it is not closed). Prove that a connected graph has an open Euler trail if an only if it has exactly two vertices of odd degree. (Hint: Use Theorem 3. for the graph G + uv, where u and v are two appropriately chosen vertices.) Exercise 3.11 For each of the graphs in Figure 1, determine whether the graph has an open Euler trail. Exercise 3.1 Describe an algorithm that determines whether or not a connected graph has an Euler tour or an open Euler trail, and if so, constructs one. 33

4 3.1. The Chinese Postman Problem A letter carrier wishes to deliver mail along every street in a city. There is a cost (travel time) associated with the traversal of every street. How can the carrier deliver the mail and then return to the starting point so that the total cost of her traversal is as small as possible? This problem became known as the Chinese Postman Problem. It asks to find a minimum closed walk in weighted graph that traverses each edge at least once, that is, a minimum postman tour. Definition 3.13 A postman tour in a graph G is a closed walk that traverses each edge of G at least once. A minimum postman tour in a weighted graph G is a postman tour of G of smallest weight. If the graph G is eulerian, then every Euler tour is in fact a minimum postman tour of the graph. If the graph is not eulerian, however, then some of the edges will have to be traversed more than once. In other words, we will have to duplicate some of the edges in G to obtain an eulerian graph from G. It makes sense to choose the edges to be duplicated so that the total weight is as small as possible, however, we have to make sure that the resulting graph is eulerian. Example 3.14 Find a minimum postman tour in the graph in Figure 14, extreme left Figure 14: But what edges should be duplicated if the vertices of odd degree in our graph are not adjacent? The idea of the algorithm of Edmonds and Johnson (Algorithm 3.18) is to pair up the vertices of odd degree so that duplicating the edges along a shortest path between the vertices in each pair gives the minimum total weight of duplicated edges. The algorithm makes use of a procedure that constructs a minimum perfect matching in a complete graph, to be studied in a later chapter. Definition 3.1 A matching in a graph G is a subset M of E(G) such that no two edges of M have an endpoint in common. A perfect matching of G is a matching M of G such that every vertex of G is an endpoint of an edge in M. In other words, a perfect matching of G is a partitioning of V (G) into pairs of adjacent vertices. Exercise 3.1 Show that if a graph has a perfect matching, then it must have an even number of vertices. Does the converse hold? 34

5 Exercise 3.1 For each of the graphs in Figure 14, find all perfect matchings of the graph. Which of the perfect matchings is minimum (i.e. has minimum total weight)? The Edmonds and Johnson algorithm starts by constructing a complete graph K with vertex set consisting of all vertices of odd degree in G. To each edge uv in K it then assigns the weight dist G (u, v), that is, the weight of a shortest path from u to v in the weighted graph G. Next, it constructs a minimum perfect matching M in K. Note that to each edge uv in M corresponds a (u, v)-path P (u, v) in G; the edges of P (u, v) (for all uv M) are then duplicated in G to obtain an eulerian graph G. Finally, an Euler tour of G is constructed, giving a minimum postman tour of G. Algorithm 3.18 Minimum Postman Tour (Edmonds and Johnson) procedure P ostman(g: connected graph with edge-weight function w) S := the set of odd-degree vertices of G for all u S {e.g. use Dijkstra s algorithm} for all v S {u} P (u, v) := a shortest (u, v)-path K := complete graph with V (K) = S and weight function c, where c(uv) = w(p (u, v)) M := minimum perfect matching in K {algorithm to be developed later} E := E(G) for all uv M E := E E(P (u, v)) {duplicate the edges of the path P (u, v) in G} G := eulerian graph with vertex set V (G), edge set E, and weight function w T := Euler tour of G {use Algorithm 3.4} {T is an minimum postman tour in G} Exercise 3.1 Apply the Edmonds and Johnson algorithm to each of the graphs in Figure 14 to find a minimum postman tour. Exercise 3.0 In this exercise you will prove correctness of Algorithm Let G be a connected graph and S the set of odd-degree vertices of G. 1. Let K be the complete graph on the vertex set S. Prove that K has a perfect matching.. Let M be a perfect matching in K, and for every edge uv M, let P (u, v) be a (u, v)- path in G. Obtain the graph G from G by duplicating each edge of P (u, v), for all uv M. Prove that G is an eulerian graph, and any Euler tour of G is a postman tour of G. 3. Recall that in Algorithm 3.18, P (u, v) was chosen to be a shortest (u, v)-path for every uv M, and M a minimum matching in K, where the weights of the edges in K are the corresponding distances between the endpoints in G. Prove that it follows that the Euler tour of G is indeed a minimum postman tour of G. 3

6 3. Hamilton Cycles and Applications 3..1 Hamilton cycles In 18, the Irish mathematician Sir William Rowan Hamilton described in a letter to a friend the following game played on the dodecahedron (Figure 1, left): One person sticks five pins into any five consecutive vertices, thus creating a path of length four, while the other person is requires to complete this path to a spanning cycle of the dodecahedron. This game became known as the Icosian Game. In honour of Hamilton, a spanning cycle of a graph is called a Hamilton cycle. Figure 1: The dodecahedron and the Herschel graph. Definition 3.1 A Hamilton cycle (path) is a cycle (path) that contains all vertices of the graph. A graph is called hamiltonian if it contains a Hamilton cycle. Example 3. Try playing the Icosian Game with a friend. You should now be convinced that the dodecahedron is a hamiltonian graph. Example 3.3 Show that the Herschel graph (Figure 1, right) is not hamiltonian. Example 3.4 Show that the two graphs in Figure 1 are not hamiltonian. Figure 1: Example 3.4. Although Hamilton cycles are in some sense analogous to Euler tours, there is no characterization of hamiltonian graphs as simple as the even-degree characterization of eulerian graphs. In fact, the decision problem Is a graph G hamiltonian? is NP-complete. However, some simple sufficient conditions for a graph to be hamiltonian are known. For example, one would expect that the more edges in a graph, the more likely the graph is hamiltonian. The following theorem confirms such a hypothesis and makes it precise. 3

7 Theorem 3. (Dirac) Let G be a simple graph with n vertices and let δ = min{deg(u) : u V (G)}. If n 3 and δ n, then G is hamiltonian. Proof. By contradiction. Suppose G is a maximal simple graph with n 3 vertices and δ n that is not hamiltonian. This means that for any two non-adjacent vertices u and v, the graph G + uv is hamiltonian. Moreover, since G is not hamiltonian, every Hamilton cycle of G + uv must contain the edge uv. Hence G contains a Hamilton path u 1 u... u n with u 1 = u and u n = v. Let Since u n S T, we have Furthermore, S = {u i : u 1 u i+1 } and T = {u i : u i u n }. S T < n. (i) S T = 0 (ii) since if there existed u k S T, then u 1 u... u k u n u n 1... u k+1 u 1 would be a Hamilton cycle in G, a contradiction (see Figure 1). 1 3 k k+1 n-1 n From (i) and (ii) we obtain Figure 1: Theorem 3.. deg(u) + deg(v) = S + T = S T + S T < n, contradicting the hypothesis that deg(u) + deg(v) δ + δ n. Hence every simple graph with n 3 vertices and δ n is hamiltonian. Exercise 3. Let G be a simple graph with n vertices and u, v two non-adjacent vertices in G such that deg(u) + deg(v) n. Use Dirac s Theorem to prove that G is hamiltonian if and only if G + uv is hamiltonian. Exercise 3. A mouse eats his way through a cube of cheese by tunnelling through all of the twenty-seven subcubes. If he starts at one corner and always moves on to an uneaten subcube, can he finish at the centre of the cube? (Hint: represent the cube of cheese by a graph. What are we trying to find in this graph? What properties does this graph have?) Exercise 3.8 For each of the graphs in Figure 18, find a Hamilton cycle in the graph or else, prove that the graph is not hamiltonian. References for Section 3..1: Bondy, Gross 3

8 Figure 18: Exercise The Travelling Salesman Problem A travelling salesman wishes to make a tour of n cities, visiting each city exactly once before returning home, so that the cost of travel is as small as possible (see Figure 1). This problem has become known as the Travelling Salesman Problem (TSP). It asks to find a minimum Hamilton cycle in a weighted graph. We shall be assuming that the weight function is non-negative Figure 1: Weighted graph for a TSP. Since the TSP is, like the Hamilton cycle problem, NP-complete, there is a trade-off between algorithms that guarantee an optimal solution and those that run quickly at the expense of a formal performance guarantee; the latter are called heuristic. Definition 3. A heuristic is a guideline for a decision step in an algorithm that helps to choose from several possible alternatives. A heuristic algorithm is an algorithm whose steps are guided by heuristics. Note that a heuristic may in fact miss the best move, and hence a heuristic algorithm sacrifices the guarantee to find the best solution so that it can terminate quickly. We shall consider two heuristic algorithms for the TSP, the Nearest Neighbour Algorithm and the Tree-Doubling Algorithm. As we shall see, the Tree-Doubling Algorithm in fact performs as a true approximation algorithm (with performance guarantee) for weighted graphs satisfying the triangle inequality. The Nearest Neighbour Algorithm, on the other hand, though very fast and easy to implement, comes with no such guarantee even for these special cases. Definition 3.30 An approximation algorithm is an algorithm that does not necessarily produce an optimal solution, however, the solution is guaranteed to be a fixed percentage away from the optimum. 38

9 Definition 3.31 A graph G with a weight function w is said to satisfy the triangle inequality if w(xz) w(xy) + w(yz) for every triple of pairwise adjacent vertices x, y, z. The Nearest Neighbour Algorithm is an example of a greedy algorithm: wherever it happens to be, it picks the cheapest way to get somewhere else as long as this move does not create a short cycle. Note that by assigning a very large weight to each missing edge, we may assume that our weighted graph is complete. Algorithm 3.3 Nearest Neighbour Algorithm procedure N earestn eighbour(g: weighted complete graph with n vertices) x := first vertex of G y := x C := trivial path with vertex set {y} {C is a path to become a Hamilton cycle; y is the last vertex on the path C} for i := to n begin z := vertex not in V (C) such that w(yz) is as small as possible C := C + yz y := z end C := C + xy {C is a Hamilton cycle in G} Example 3.33 Use the Nearest Neighbour Algorithm to find a Hamilton cycle in the graph in Figure 1. Is this Hamilton cycle minimum? The Nearest Neighbour Algorithm sometimes performs quite well, however, it can produce arbitrarily bad Hamilton cycles. See the graph in Figure 0 and explain. M Figure 0: A bad example for the Nearest Neighbour Algorithm. The following result shows that performance guarantees for the general TSP are highly unlikely. Theorem 3.34 If there exists a polynomial-time algorithm whose solution to every instance of the general TSP is never worse than some constant times the optimum, then P=NP. 3

10 As mentioned above, algorithms with such guarantees do exist for TSPs satisfying the triangle inequality. However, the Nearest Neighbour Algorithm is not one of them. Theorem 3.3 For every r > 1 there exists a graph G satisfying the triangle inequality such that the solution for the TSP on G obtained by the Nearest Neighbour Algorithm is at least r times the optimum. The Tree-Doubling Algorithm starts by finding a minimum spanning tree T in a weighted complete graph G satisfying the triangle inequality. By doubling each edge in T, it obtains an eulerian graph H, and in H, it constructs an Euler tour T. Finally, it converts T into a Hamilton cycle by taking shortcuts whenever vertex doubling would occur while traversing T. Note that the resulting Hamilton cycle depends on which was the first vertex of the Euler tour T. Algorithm 3.3 Tree-Doubling Algorithm procedure T reedoubling(g: weighted complete graph with n vertices) {assuming the weight function w is non-negative and satisfies the triangle inequality} T := minimum spanning tree in G H := eulerian graph obtained by taking two copies of each edge in T T := Euler tour of H x := first vertex of T y := x {y will be the first endpoint of a new edge on C} C := trivial path with vertex set {y} {C is a path to become a Hamilton cycle of G} z := y {z is the last visited vertex on T } for i := to (n 1) begin u := first vertex on T following z if u V (C) then begin C := C + yu y := u end z := u end C := C + yx {C is a Hamilton cycle in G} Example 3.3 Suppose Figure 1 shows a minimum spanning tree in a complete graph. Using the Tree-Doubling Algorithm, find the corresponding Hamilton cycle starting at (a) vertex 1, (b) vertex. Exercise 3.38 Use the Tree-Doubling Algorithm to find a Hamilton cycle in the graph in Figure 1. Is this Hamilton cycle minimum? 40

11 1 3 4 Figure 1: Example 3.3 In the following theorem we shall prove the performance guarantee for the Tree-Doubling Algorithm, namely, that it always produces a Hamilton cycle whose weight is at most twice the weight of a minimum Hamilton cycle. Theorem 3.3 Let G be a complete weighted graph satisfying the triangle inequality, and C a minimum Hamilton cycle in G. The Tree-Doubling Algorithm produces a Hamilton cycle C with w(c) w(c ). Proof. Let T be the minimum spanning tree obtained by the Tree-Doubling Algorithm. Since C e is a spanning tree for all edges e E(C ), we have w(t ) w(c e) w(c ) (note that we needed the assumption w(e) 0). Clearly, the weight of the Euler tour T is then at most w(c ). Now consider the Hamilton cycle C = u 0 u 1 u... u n 1 u 0 obtained from the Euler tour T by the Tree-Doubling Algorithm. We have w(c) = w(u 0 u 1 ) + w(u 1 u ) w(u n u n 1 ) + (u n 1 u 0 ). Each edge u i u i+1 (subscripts modulo n) of C is obtained from a subtrail T i = u i0 u i1... u ik(i) (where u i0 = u i and u ik(i) = u i+1 ) of the Euler tour T, and T = T 0 T 1... T n 1. Clearly, since G satisfies the triangle inequality, n 1 n 1 w(c) = w(u i u i+1 ) w(u i0 u i1 ) + w(u i1 u i ) w(u ik(i) 1 u ik(i) ) = w(t ) w(c ). i=0 i=0 Cristofides realized that not all of the edges of the minimum spanning tree need to be duplicated to obtain an eulerian graph. The result of his algorithm, a variation of the Tree- Doubling Algorithm using a minimum perfect matching in the subgraph of G induced by the vertices of odd degree in T, is a Hamilton cycle that is never worse than 3 times the optimal. Exercise 3.40 Estimate the order of complexity of the Nearest Neighbour and Tree-Doubling Algorithms. Exercise 3.41 Perform each of the Nearest Neighbour and Tree-Doubling Algorithms on each of the graphs in Figure, starting from vertex a and resolving ties alphabetically. Indicate the edges of the resulting approximation for a minimum Hamilton cycle, as well as its weight. 41

12 e d a 10 c b e d a c b f 3 e a d c b Figure : Exercise 3.4 Minimum Hamilton path.to find a minimum Hamilton path in a weighted complete graph G with V (G) = {u 1,..., u n } and non-negative weight function w, create a graph G by adding to G a new vertex u 0 and edges u 0 u i, for i = 1,..., n, with weights w(u 0 u i ) = 0. (a) Explain how a minimum Hamilton cycle in G corresponds to a minimum Hamilton path in G. (b) Using the Nearest Neighbour and Tree-Doubling Algorithms, find approximations for a minimum Hamilton path in each of the graphs in Figure. Exercise 3.43 Minimum Hamilton path with specified endpoints. (a) Let G be a weighted complete graph with V (G) = {u 1,..., u n } and non-negative weight function w. Explain how to convert the problem of finding a minimum Hamilton path from u 1 to u n to the problem of finding a minimum Hamilton cycle in an appropriate graph G. Describe this graph G. (See Exercise 3.4 for ideas.) (b) Using the Nearest Neighbour and Tree-Doubling Algorithms, find approximations for a minimum Hamilton path from vertex a to vertex d in each of the graphs in Figure. Exercise 3.44 Job Sequencing on a Single Machine. We have n jobs that need to be processed on a single machine. The time required to process job j immediately after job i is c ij. We would like to sequence the n jobs so that the total time is minimized. Describe the graph model that can be used to solve this problem Gray Codes Definition 3.4 A Gray code of order n is an ordering of the n binary strings of length n such that two consecutive binary strings in the sequence (viewed cyclically) differ in exactly one bit. Example 3.4 Construct Gray codes of orders 1,, and 3. Application 3.4 A spacecraft transmits a picture back to earth using long sequences of numbers, where each number represents a darkness value for one of the dots in the picture. Using a Gray code rather than usual binary representation of numbers to encode the picture has the following advantage: if an error from cosmic noise causes one binary digit in a sequence to be misread by the receiver, then the mistaken binary string will be interpreted as a darkness value that is almost the same as the true darkness number. 4

13 Example 3.48 How large can transmission error be when using the usual binary numbers to represent darkness values from 0 to? Compare this to the Gray code ordering. Give an ordering of binary strings of length 3 that yields the greatest possible error in the transmission of one faulty bit. Theorem 3.4 For every positive integer n there exists a Gray code of order n. Proof. A Gray code of order n corresponds to a Hamilton cycle in the n-dimensional cube Q n. Using induction, we prove that Q n is hamiltonian for all n. Q is a 4-cycle and therefore hamiltonian. This proves the basis of induction. Suppose Q n is hamiltonian for some n. Let C = b 1 b... b nb 1 be a Hamilton cycle in Q n. (Here, the b i are binary strings of length n.) Then b 1 0 b 0... b n 10 b n0 b n1 b n b 1 b 1 1 b 1 0 is a Hamilton cycle in Q n+1. By induction, Q n is hamiltonian for all n, and hence there exists a Gray code of order n for all n. Exercise 3.0 Construct Gray codes of orders 4 and from Gray codes of orders 3 and 4, respectively. References for Section 3.: Gross 43