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Prove that a simple graph is connected if and only if it has a spanning tree.
Ans: First assume that a simple graph G has a spanning tree T. T consists of every node of G. By the definition of a tree, there is a path among any two nodes of T. As T is a subgraph of G, there is a path among each pair of nodes in G. Hence G is connected.
Here now let G is connected. If G is a tree then nothing to prove. If G is not a tree, it must consist of a simple circuit. Let G has n nodes. We can choose (n - 1) arcs from G in such type of a way that they not form a circuit. It results into a subgraph comprising all nodes and only (n - 1) arcs. So by definition this subgraph is a spanning tree.
It is totally possible that a or b could be zero and thus in 16 i the real part is zero. While the real part is zero we frequently will call the complex numbers a purely imaginar
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Evaluate following indefinite integrals. (a) ∫ 5t 3 -10t -6 + 4 dt (b) ∫ dy Solution (a) ∫ 5t 3 -10t -6 + 4 dt There's not whole lot to do here other than u
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