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Algorithms and Complexity Homework #5 Solution

1    Minimum spanning tree

Do problem  4.9 on page  192 of the  textbook.  For  your  convenience,  here  is the  problem.    Let G = (V, E)  be a connected  (undirected) graph  with  n vertices,  m edges and  positive  edge costs (assume  edge costs are distinct). Let T = (V, E0) be a spanning  tree of G.  The bottleneck  edge of T is the edge of T with the greatest cost.  A spanning  tree T of G is called a minimum  bottleneck spanning  tree where there  exists no spanning  tree of G with a cheaper  bottleneck  edge.

Questions:  (a)  Is every minimum  bottleneck  tree  of G a minimum  spanning  tree  of G?  Prove or give a counter-example.  (b)  Is every minimum  spanning  tree  of G a minimum  bottleneck  tree of G?  Prove  or give a counter-example.

 

 

2    MST

 

Let G be a connected,  undirected graph,  where the edge weights are all distinct. You are also given a specific edge e in G.  You want  to find out  whether  e is contained  in some minimum  spanning tree.

 

1.  First  prove that e = (u, v) does not  belong to any MST iff there  is a path  between  u and  v

with edges all cheaper  than  e.

 

2.  Give an O(|V | + |E|) algorithm  for this problem.

 

3    Divide and conquer algorithm

 

You are given a rooted  tree  T with  n nodes.  The  tree  may not  be binary.   You want  to find the depth  of T . Here, we define the depth  of a node v in the tree to be the maximum  number  of nodes along a path  from v to a leaf under  v (going downwards).  The depth  of T is the depth  of the root of T . Now give a divide and conquer algorithm  for finding the depth  of a tree.  You need to analyze its running  time.

 

 

4    Recurrence

 

Solve the following recurrence.  Show the steps.

(12T (n/4) + n1.5,    if x ≥ 4

1.  T (n) =

 

 

 

2.  T (n) =


 

1,                             otherwise

(T (√n) + logn,    if x ≥ 4

1,                          otherwise



3.  T (n) = T (n/2) + Θ(1),  which is the running  time of the binary  search.

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