Graph_Theory_Notes3.pdf

# Proof let g be a tree since g is connected there is

• Notes
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Proof. ( ) Let G be a tree. Since G is connected, there is at least one path between any two vertices. We need to prove that G has exactly one path between any two vertices. Suppose to the contrary that, for some pair of distinct vertices u, v V ( G ), there exist two distinct ( u, v )-paths P and Q in G . Let P = ( u = x 0 , x 1 , . . . , x k - 1 , x k = v ) Q = ( u = y 0 , y 1 , . . . , y l - 1 , y l = v ) . Since P 6 = Q , we have { x 1 , . . . , x k - 1 } 6 = { y 1 , . . . , y l - 1 } .

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Proof (continued). Since x 0 = y 0 = u , we can define i to be the largest subscript such that x 0 = y 0 , . . . , x i = y i . Then x i +1 6 = y i +1 . Since x k = y l = v , there are common vertices of P and Q after x i , and among them we choose the first one, say, w . Then the subpath of P from x i to w and the subpath of Q from y i to w form a cycle, which contradicts the assumption that G is a tree. Hence G has exactly one path between any two vertices. ( ) Suppose that there is exactly one path in G between any two ver- tices. Then G is connected. G cannot contain any cycle for otherwise there are two paths between any two vertices on the cycle, which is a contradic- tion. Hence G is a tree. Third characterisation of trees Theorem 5. A graph G is a tree if and only if it has no cycle and | E ( G ) | = | V ( G ) | - 1. Proof. If G is a tree, then it has no cycle by the definition of a tree, and | E ( G ) | = | V ( G ) | - 1 by the previous theorem. Conversely, suppose G has no cycle and | E ( G ) | = | V ( G ) | - 1. It suf- fices to prove that G is connected. Suppose G has k 1 components, say G 1 , . . . , G k . Since G has no cycles, each G i has no cycles. Since G i is connected, it must be a tree, and | E ( G i ) | = | V ( G i ) | - 1 for each i . So | V ( G ) | - 1 = | E ( G ) | = k i =1 | E ( G i ) | = k i =1 ( | V ( G i ) | - 1) = ( k i =1 | V ( G i ) | ) - k = | V ( G ) | - k , implying k = 1. Hence G is connected and therefore is a tree. Summary: properties of trees A tree T with order n and size m has the following properties: (1) connected; (2) no cycle; (3) exactly one path between any two vertices; (4) every edge is a bridge; (5) m = n - 1. Equivalent definitions of a tree: (1) + (2) (3) (1) + (4) (1) + (5) (see the 4th characterisation below) (2) + (5) Trees are “minimal” connected graphs A tree G is connected and has | E ( G ) | = | V ( G ) | - 1 edges. Every connected graph G satisfies (Exercise!) | V ( G ) | - 1 ≤ | E ( G ) | ≤ | V ( G ) | 2 .
Among all connected graphs of a given order, trees have the mini- mum number of edges. In other words, trees are “minimal” connected graphs. Trees are also the most “vulnerable” connected graphs since deleting any edge destroys connectedness. 2 Spanning trees Spanning trees Definition 3. A spanning tree of a graph is a spanning subgraph that is a tree. Theorem 6. A graph G has a spanning tree if and only if it is connected. Proof. (‘Only if’ part) Suppose that G has a spanning tree T . Then V ( G ) = V ( T ) and in T any two vertices are connected by a unique path. Since a path in T is a path in G , there is at least one path in G between any two vertices. Hence G is connected. Proof (continued). (‘If’ part) Suppose that G is connected. If G contains a cycle, then delete one edge in some cycle. We obtain a spanning subgraph G 1 of G .

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