# N 1 2 k i 2 i 2 n i test2mad3305 page 2 of 4 2 15 pts

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n 1 2 k i 2 ( i 2) n i

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TEST2/MAD3305 Page 2 of 4 _________________________________________________________________ 2. (15 pts.) For the graph G below, determine the cut-vertices, bridges, and blocks of G. List the cut-vertices and bridges in the appropriate places, and provide carefully labelled sketches of the blocks. Cut-vertices: c, d, f, and i Bridge(s): ci and fg Block(s): _________________________________________________________________ 3. (10 pts.) Below, provide a proof by induction on the order of the graph G that every nontrivial connected graph G has a spanning tree. [Hint: If the order of the graph G is at least 3, Theorem 1.10 implies that G has a vertex v with G - v connected.] // Let S(k) be the following assertion: "For any graph G, if G is connected with k vertices, then G has a spanning tree T." If G is a connected graph of order 2, then G is K 2 which is a tree and there is nothing to show. Thus S(2) is true, and we get a basis for the induction with no real work. To deal with the induction step, we must provide a proof of the proposition ( k 2)( S(k) S(k+1) ). To show this, we let k 2 be fixed and arbitrary. We need to show for this k that S(k) S(k+1). Suppose that S(k) is true. To show that S(k+1) follows, we may assume as true the hypothesis of S(k+1), namely that we are dealing with G, an arbitrary connected graph with k+1 vertices. We need to show that G has a spanning tree. Since G has at least 3 vertices, Theorem 1.10 implies that there are two vertices in G, say u and v, with G - u and G - v connected. We shall focus on G - v. G - v is a connected graph with k 2 vertices. From the induction hypothesis, S(k), with G - v replacing G, it follows that G - v has a spanning tree, T 0 say. Since G is connected, v is adjacent to at least one of the vertices of G - v. 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 2)( S(k) S(k+1) ). We may now apply the principle of induction. [Hello, modus ponens.]
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?

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• Summer '12
• Rittered
• Graph Theory, Vertex, Planar graph, κ

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