# A tree can be defined in a variety of ways as is shown in the following theorem: 2. There exists a unique path between every two vertices of G.

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1 7 Basic Properties 24 TREES 7 Basic Properties Definition 7.1: A connected graph G is called a tree if the removal of any of its edges makes G disconnected. A tree can be defined in a variety of ways as is shown in the following theorem: Theorem 7.1: The following statements are equivalent: 1. G is a tree. 2. There exists a unique path between every two vertices of G. 3. G does not hold any closed paths and for n > 2 every additional edge creates a closed path in G. 4. G is connected and contains no closed path. 5. G does not contain a closed path and e = n G is connected and e = n 1. Proof: ) There exists at least one path between u and v as G is connected. If there was a second path between u and v, any edge on this path (not belonging to the first path) could be removed without making G disconnected ) G cannot contain a closed path that holds the vertices u and v as this results in two paths between u and v. Let u v E(G), i.e., u v is an edge that is not a part of G, then adding u v would create two paths between u and v ) If G is not connected, we could take u and v such that they do not reside in the same component. Adding the edge uv would not create a

2 7 Basic Properties 25 closed path ) Apply induction on n (the case n = 1 is trivial). Let H = G uv, where uv is an arbitrary edge in G. u and v have to belong to a different component of H (say, H u and H v ), otherwise G would hold a closed path containing u and v. For each of these components criteria 4. is met, thus e = e Hu + e Hv + 1 = n Hu + n Hv 1 = n ) Assume G has k components (say, H 1, H 2,..., H k ). For each component H i we have e Hi = n Hi 1, meaning that e = i e H i = i (n H i 1) = n k. Hence, k = ) Removing any edge makes G disconnected, because a graph with n vertices clearly needs at least n 1 edges to be connected. Definition 7.2: A tree T is called a subtree of the graph G if T G. A spanning tree T of G is defined as a maximum subtree of G. It should be clear that any spanning tree of G contains all the vertices of G. Moreover, for any edge e, there exists at least one spanning tree that contains e [Proof: Take an arbitrary tree T and assume e T. When we add the edge e to T, the graph T + e holds a closed path. Removing any edge e e from this path provides us with a spanning tree T that contains e]. Theorem 7.2 (Cayley (1889)): Let V = n, there are n n 2 spanning trees on V, that is, K n has n n 2 spanning trees. Proof: Let v be any vertex in K n and define X(k) as the number of spanning trees where v has a degree equal to k. Assume k > 1 and n > 2 (the cases n = 1, 2 are trivial). Let T be a spanning tree such that d T (v) = k. (A) Choose an edge uu that is not incident to v (i.e., v u, u ). T uu is disconnected and u and u belong to a different component. Without loss of generality assume v belongs to the component of u. Then T = T uu +vu is a spanning tree. The number of couples (T, T ) equals X(k 1)(n k) as there where n 1 (k 1) edges in T not incident to v. (B) Let T be a spanning tree with d T (v) = k. If we remove the i-th edge vu i incident to v (i = 1,..., k), we obtain a disconnected graph with two components T i and T T i, where T i is said to hold u i. If we now connect u i with any vertex v v in T T i, we obtain a spanning tree T with d T(v) = k 1. There are X(k)( i (n T T i 1)) = X(k)(nk (n 1) k) = X(k)(n 1)(k 1) couples ( T, T ).

3 8 The Connector Problem 26 (C) Clearly, the number of couples (T, T ) and ( T, T ) are identical. Thus, X(k 1) = (n 1)(k 1)X(k)/(n k). Replacing k by n j gives X(n j 1)j/[(n j 1)(n 1)] = X(n j). Moreover, X(n 1) = 1, therefore, it is easy to show by induction on j that X(n j) = (n 1) j 1( n 2 j 1). The total number of spanning trees equals n 1 j=1 X(n j) = n 1 j=1 ((n 1) + 1) n 2 = n n 2 by Newton s binomial theorem. Exercises 7.1: On spanning trees: ( n 2 j 1) (n 1) j 1 = 1. We state that two graphs G and G are isomorphic if there exists a bijection α : V (G) V (G ) such that uv E(G) α(u)α(v) E(G ). Thus, α simply renames the vertices of G. Determine the number of spanning trees in K n up to isomorphisms for n = 1, The Connector Problem To build a network connecting n nodes (towns, computers, chips in a computer) it is desirable to decrease the cost of construction of the links to the minimum. This is the connector problem. In graph theoretical terms we wish to find an optimal spanning subgraph of a weighted graph. Such an optimal subgraph is clearly a spanning tree, for, otherwise a deletion of any edge that is part of a cycle will reduce the total weight of the subgraph. Let then G α be a graph G together with a weight function α : E(G) R + (positive reals) on the edges. Kruskal s algorithm (also known as the greedy algorithm) provides a solution to the connector problem. Algorithm 8.1 (Kruskal s algorithm (1956)): For a connected and weighted graph G α of order n: 1. Let e 1 be an edge of smallest weight, and set E 1 = {e 1 }. 2. For each i = 2, 3,..., n 1 in this order, choose an edge e i E i of smallest possible weight such that e i does not produce a cycle when added to G[E i 1 ], and let E i = E i 1 {e i }. Notice, G[E i ] need not to be connected. The final outcome is T = (V (G), E n 1 ) is a spanning tree with a minimal weight.

4 8 The Connector Problem 27 Proof: By the construction, T = (V (G), E n 1 ) is a spanning tree of G, because it contains no cycles, it must be connected and has n 1 edges 6. We now show that T has the minimum total weight among the spanning trees of G. Suppose T 1 is any spanning tree of G. Let e k be the first edge produced by the algorithm that is not in T 1. If we add e k to T 1, then a cycle C containing e k is created. Also, C must contain an edge e that is not in T. When we replace e by e k in T 1, we still have a spanning tree, say T 2. However, by the construction, α(e k ) α(e), and therefore α(t 2 ) α(t 1 ). Note that T 2 has more edges in common with T than T 1. Repeating the above procedure, we can transform T 1 to T by replacing edges, one by one, such that the total weight does not increase. We deduce that α(t) α(t 1 ). The outcome of Kruskal s algorithm need not be unique. Indeed, there may exist several optimal spanning trees (with the same weight, of course) for a graph. The runtime of this algorithm equals O(e log e). Another algorithm that can be used for the same purpose is Prim s, it has a runtime complexity of O(n 2 ) [Prim s algorithm can be implemented in O(e + n log n) time using more advanced data structures (in particular, Fibonacci heaps), this will not be worth the trouble unless you have extremely large, fairly sparse graphs]. Thus, Prim s algorithm is faster on dense graphs, while Kruskal s is faster on sparse graphs. Prim s algorithm was also invented earlier by Jarnick (1930). Algorithm 8.2 (Prim s algorithm (1957)): For a connected and weighted graph G α of order n with V (G) = {v 1,...,v n }, define f(v 1 ) = 0 and f(v i ) = α(v 1 v i ) if v 1 v i E(G) and f(v i ) = otherwise (i > 1): 1. Let E = and set U = {v 1 }. 2. Choose w V (G) U with f(w) minimal. 3. Replace E by E {e}, where e is an edge incident to w and U for which α(e) = f(w) and set U = U {w}. If U = V (G) stop. 6 At each step i = 2,...,n 1 we can always find at least one edge that does not create a cycle. Indeed, at step i, the graph (V (G), E i 1 = {e 1,..., e i 1 ) has n nodes and less than n 1 edges, thus, it contains at least two components. As G is connected there has to exist a path P in G that connects these components. Any edge of this path can be added without creating a cycle.

5 8 The Connector Problem For each v U for which wv E(G): f(v) = min{f(v), α(wv)}. Return to Step 2. The final outcome is T = (U, E) is a spanning tree of G with minimal weight. Notice, f(v) equals the minimal weight of any edge connecting v with U (f(v) equals if there is no such edge). Thus, at each iteration we simply add a new edge e to T with a weight α(e) as small as possible. Proof: Let G α be a connected, weighted graph. By induction we show that T = (U, E) is a spanning tree on the vertices of U during each iteration. At step 1 we have n T = 1, e T = 0, hence, it is a spanning tree on {v 1 }. At every other iteration we augment n T and e T by one, meaning n T = e T + 1 and T is connected. Let T 1 be any spanning tree for G. Let e be the first edge not belonging to T 1 that was added when T = (U, E) was constructed. Then one endpoint of e was in U and the other was not. Since T 1 is a spanning tree of G, there is a path in T 1 joining the two endpoints of e. As one travels along the path, one must encounter an edge f joining a vertex in U to one that is not in U. Now, at the iteration when e was added to E, f could also have been added and it would be added instead of e if α(f) < α(e). Since f was not added, we conclude that α(f) α(e). Let T 2 = T 1 + e f, then α(t 2 ) α(t 1 ). Repeating the above procedure, we can transform T 1 to T by replacing edges, one by one, such that the total weight does not increase. We deduce that α(t) α(t 1 ). Many other algorithms exist to solve this problem, e.g., Boruvka s algorithm (1926) with runtime O(e log n). A faster algorithm due to Karger, Klein and Tarjan runs in O(n) time. Exercises 8.1: On Kruskal s and Prim s Algorithm: 1. Apply Kruskal s Algorithm to the graph G depicted in Figure Apply Prim s Algorithm to the graph G depicted in Figure 8. Minimum weight spanning trees are useful in a variety of disciplines. For instance, minimum spanning trees are useful in constructing networks, by describing the way to connect a set of sites using the smallest total amount of wire. Much of the work on minimum spanning (and related Steiner) trees

6 8 The Connector Problem 29 N1 2 N2 4 N3 1 7 N N4 3 8 N7 8 N6 1 N5 Fig. 8: Apply Kruskal s and Prim s algorithm on the graph G has been conducted by the phone company. They also provide a reasonable way for clustering points in space into natural groups. When the cities are points in the Euclidean plane, the minimum spanning tree provides a good heuristic for traveling salesman problems. The optimum traveling salesman tour is at most twice the length of the minimum spanning tree.

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