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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.
in a garden 1/8 of the flowers are tulips. 1/4 of the tulips are rd. what fraction of the flowers in the garden are red tulips
This time we are going to take a look at an application of second order differential equations. It's now time take a look at mechanical vibrations. In exactly we are going to look
ABCD is a trapezium AB parallel to DC prove square of AC - square of BCC= AB*
4 1/2 ----2----1/3=3
you need to cut the proper to cut 2''*4*8 long studs to the proper length to make a finished wall 8" in height underneath the studs there will be a double plate made up of two piec
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Under this section we're going to go back and revisit the concept of modeling only now we're going to look at this in light of the fact as we now understand how to solve systems of
For queries Q 1 and Q 2 , we say Q 1 is contained in Q 2 , denoted Q 1 ⊆ Q 2 , iff Q 1 (D) ⊆ Q 2 (D) for every database D. The container problem for a fixed Query Q 0 i
48 more than the quotientvof a number and 64
10p=100
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