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Then t is a spanning tree for g since g was an

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Pick such a vertex, say w, and let T = (V(G), E(T 0 ) {vw}). Then T is a spanning tree for G. Since G was an arbitrary connected graph of order k+1, S(k+1) follows. Thus, S(k) S(k+1). Since k was arbitrary, we have ( k
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TEST2/MAD3305 Page 3 of 4 _________________________________________________________________ 5. (10 pts.) Apply Kruskal’s algorithm to find a minimum spanning tree in the weighted graph below. When you do this, list the edges in the order that you select them from left to right. What is the weight w(T) of your minimum spanning tree T? There are several different correct solutions. First, in some order you will take two of the three edges from the "1" 3- cycle: de, ef, fd. Then you will need both "2" edges in some order: ae and ac. Then exactly one of the two "3" edges: ab or bd. For any of the 6 possible trees T, w(T) = 9. _________________________________________________________________ 6. (15 pts.) (a) Suppose G 1 and G 2 are nontrivial graphs. What does it mean mathematically to say that G 1 and G 2 are isomorphic?? [This is really a request for the definition!] Two graphs G 1 and G 2 are isomorphic if there is a bijection φ : V(G 1 ) V(G 2 ) such that uv ε E(G 1 ) if, and only if φ (u) φ (v) ε E(G 2 ). (b) Sketch two graphs G and H that have the degree sequence s: 2, 2, 2, 2, 2, 2 and have the same order and size, but are not isomorphic. Explain briefly how one can readily see that the graphs are not isomorphic. Since G is not connected and H is connected, G and H cannot be isomorphic. Why? Any graph isomorphic to H must be connected. (c) Explicitly realize C 5 and its complement below. [You may provide carefully labelled sketches.] Next, explicitly define an isomorphism from C 5 to its complement that reveals that C 5 is self-complementary. Define φ from C 5 to its complement by φ ( a )=a , φ ( b )=c , φ ( c )=e , φ ( d )=b , and φ ( e )=d . φ
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Then T is a spanning tree for G Since G was an arbitrary...

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