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Midterm2

# Midterm2 - nodes is 2 The example is the tree(1 2(2 3 n-1,n...

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Solutions to Midterm 2 Peng Du University of California, San Diego 1 (15 points) 1.1 Part (a) Exponential time. Consider a complete graph G with n nodes and an edge of weight one between every pair of nodes. For any permutation σ of 1 , 2 , . . . , n , the path ( σ (1) , σ (2)) , ( σ (2) , σ (3)) , . . . , ( σ ( n - 1) , σ ( n )) is a minimum spanning tree. Therefore, G has at least n ! / 2 number of minimum spanning trees. 1.2 Part (b) False. Consider a tree with weight one for all its edges. 1.3 Part (c) True. Suppose G has k connected components with n 1 , n 2 , . . . , n k nodes, respec- tively. Since a connected graph with n nodes has at least n - 1 edges, we have | E | ≥ ( n 1 - 1) + ( n 2 - 1) + · · · + ( n k - 1) = ( n 1 + n 2 + · · · + n k ) - k = | V | - k , so k ≥ | V | - | E | . 1.4 Part (d) True. Use the same proof as the cut property. 1.5 Part (e) The largest possible number of leaves for a tree with n nodes is n - 1. The example is the tree (1 , 2) , (1 , 3) , . . . , (1 , n ). The smallest possible number of leaves for a tree with

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Unformatted text preview: nodes is 2. The example is the tree (1 , 2) , (2 , 3) ,..., ( n-1 ,n ). 2 (6 points) 2.1 Part (a) The recurrence relation is T ( n ) = T (3 n/ 4) + O ( n ) and T ( n ) = O ( n ) by master theorem. 2 2.2 Part (b) The recurrence relation is T ( n ) = 2 T ( n-1)+ O (1) and T ( n ) = 2 n by induction. 2.3 Part (c) The recurrence relation is T ( n ) = 4 T ( n/ 2) + O ( n 2 ) and T ( n ) = n 2 log n by master theorem. 3 (9 points) 3.1 Part (a) S ( n ). 3.2 Part (b) S ( i ) = 10 , 8 , 18 , 33 , 26 , 36 , 56 , 66 , 60 , 70. 3.3 Part (c) S ( i ) = min { S ( j ) + h i : j < i and d i-d j ≤ M } . 3.4 Part (d) S (0) = 0; for i = 1 ...n S ( i ) = ∞ ; for i = 1 ...n for j = 0 ...i-1 if d i-d j ≤ M and S ( j ) + h i < S ( i ) S ( i ) = S ( j ) + h i ; return S ( n ); 3.5 Part (e) O ( n 2 )....
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