Observe that g 1 is connected if g 1 contains a cycle

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Observe that G 1 is connected. If G 1 contains a cycle, then delete one edge on some cycle of G 1 to obtain another connected spanning subgraph G 2 of G . Repeat this process until we obtain a connected spanning subgraph with no cycles. This is a spanning tree of G . Fourth characterisation of trees Theorem 7. A graph G is a tree if and only if it is connected and | E ( G ) | = | V ( G ) | - 1. Proof. If G is a tree, then it is connected and | E ( G ) | = | V ( G ) |- 1 as proved earlier. Conversely, suppose G is connected and | E ( G ) | = | V ( G ) | - 1. By the previous theorem, G has a spanning tree T . Then | E ( T ) | = | V ( T ) | - 1 = | V ( G ) | - 1 = | E ( G ) | . Since E ( T ) E ( G ), it follows that E ( T ) = E ( G ). Hence T = G and so G is a tree. Fundamental cycles Lemma 3. Let T be a spanning tree of G . Let e E ( G ) \ E ( T ). Then T + e contains a unique cycle C ( T, e ); for each edge e 0 on C ( T, e ), T + e - e 0 is a spanning tree of G .
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Proof. Let e = uv . T + e is connected since T is connected. T + e contains a cycle since T + e has n edges (where n = | V ( G ) | ). Since T has no cycle, every cycle of T + e contains e and the unique ( u, v )-path in T . Hence T + e has a unique cycle C . Let e 0 be an edge of C . Then T + e - e 0 contains no cycle and has n vertices and n - 1 edges. By the third characterisation of trees, T + e - e 0 is a tree. Since T + e - e 0 has the same vertex set as G and is a subgraph of G , it is a spanning tree of G . Definition 4. C ( T, e ) in the lemma above is called a fundamental cycle . The change from T to T + e - e 0 is called a spanning tree transformation . Exercise 1. Prove that in any connected graph any two spanning trees can be obtained from each other by a sequence of spanning tree transformations. A characterisation of bipartite graphs (again) Lemma 4. Any nontrivial tree is bipartite. Proof. Let T be a nontrivial tree. Fix x V ( T ). Since T is a tree, for every w V ( T ) there is a unique path from x to w . Let A be the set of vertices w for which this path is of even length, and set B = V ( T ) - A . Since T is a tree, for any uv E ( T ), uv is the last edge of the unique path from x to u or the last edge of the unique path from x to v . In either case, u and v cannot be both in A or both in B . Therefore, T is bipartite with bipartition { A, B } . Theorem 8. A graph is bipartite iff it contains no odd cycles. Proof (second proof). A graph is bipartite iff each of its component is bi- partite, and contains an odd cycle iff one of its components contains an odd cycle. Thus, it suffices to prove the result for connected graphs. ( ) This was done in Part 2. ( ) Suppose that G is a connected graph without odd cycles. Then G has a spanning tree T . Fix a vertex x of T . By the previous lemma, T is bipartite with bipartition { A, B } as in the proof above. We claim that { A, B } is also a bipartition of G . Consider an arbitrary edge e = uv E ( G ) - E ( T ). Let P be the unique path in T between u and v . Then P + e is a cycle of G . Since G has no odd cycles, P + e is of even length and so P is of odd length. Thus, the end-vertices of P , namely u and v , must be in different parts of { A, B } . Hence G is bipartite with bipartition { A, B } .
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