HW3SOL_111027

HW3SOL_111027 - MC 302 GRAPH THEORY SOLUTIONS TO HW#3 42...

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Page 1 of 5 MC 302 – GRAPH THEORY –SOLUTIONS TO HW #3 – 10/27/11 42 points plus 4 extra credit points 1. pp. 56-57, #2.2.3 (be sure to justify your answer!) G has 4 connected components and 24 edges – what’s the maximum number of vertices that G can have? SOLUTION TO #1 (4 pts): By a theorem (Thm. 2.9 in the text), q n - (G), where q, n and (G) are the numbers of edges, vertices and components, resp., of G. Thus n q + (G) = 24 + 4 = 28, i.e., G can have at most 28 vertices. 2. pp. 56-57, #2.2.4. Prove that a connected graph G with n vertices and e edges is unicyclic if and only if n = e . Hint: For sufficiency, you might want to argue the contrapositive. SOLUTION TO #2 (10 pts): Necessity ( ): We assume that the connected graph G is unicyclic and prove that n = e. So suppose G is connected and has a unique cycle C. Choose any edge f = u-v on C. Since the remaining section of C is a path from u to v , it can replace f in any walk from one vertex of G to another. Thus G\ f is still connected and is now acyclic. Thus G\ f is a tree, so by a theorem it has e = n -1 edges. Therefore G has e =n edges, proving necessity. Sufficiency ( ): (Proof by contrapositive) We prove the contrapositive, namely, we show that if G is not unicyclic, then n e. If G is not unicyclic, then G has either no cycles or more than one cycle. If G has no cycles, then since it’s connected, G is a tree. Thus n = e – 1 e. If G has more than one cycle, say it has distinct cycles C 1 and C 2 , choose an edge e 1 in C 1 but not in C 2 . Then G\e 1 is still connected, since any path using e 1 could instead use the rest of the cycle C 1 . But G\e 1 still has the cycle C 2 , thus G\e 1 is not a tree and so has more than n-1 edges, which implies that G has more than n edges. Sufficiency ( ): (Direct proof) Assume G is connected with e = n edges. Then G cannot be a tree, by the theorem that says a tree has n- 1 edges. So, since G is connected, it must have a cycle C. We must show that C is unique, i.e., if C and C’ are both cycles in G, then C = C’. So suppose C and C’ are both cycles in G , and let f be an edge of C . Then G\f is still connected, and it has n-1 edges, so G\f is a tree. Thus, removing f must have disconnected C’ as well, so the edge f is in both cycles. If the endpoints of f are u and v , then C\f and C’\f are both paths between u and v in the tree G\f . By the unique path property of trees, it follows that C\f = C’\f , and thus C = C’ , proving that G has a unique cycle. 3. For the graph shown at the right, a. Use Prim’s Algorithm starting at vertex A to build a minimum-weight spanning tree. Be sure to specify the order in which you add the edges to your tree, and to give the total weight of the tree.
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Page 2 of 5 b. Use Kruskal’s Algorithm to find a maximum -weight spanning tree. Again, be sure to specify the order in which you add the edges to your tree, and to give the total weight of the tree. SOLUTION TO #3 (8 pts):
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This note was uploaded on 01/15/2012 for the course ECON 101 taught by Professor N/a during the Fall '11 term at Middlesex CC.

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HW3SOL_111027 - MC 302 GRAPH THEORY SOLUTIONS TO HW#3 42...

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