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Unformatted text preview: 4.15 Application: Gaussian Elimination 223 V ( D i ) = { v π ( n i +1) ,v π ( n i +2) ,...,v π ( n i +1 ) } . It is easy to see that B has ( n 1 ,...,n p )blocktriangular structure. This im plies that A has blocktriangular structure. The above observation suggests the following procedure to reveal hidden blocktriangular structure of A . 1. Replace every nonzero entry of A by 1 to obtain a (0 , 1)matrix B . 2. Construct a directed pseudograph D with vertex set { v 1 ,...,v n } such that B is the adjacency matrix of D . 3. Find the strong components of D . If D is strong, then B (and thus A ) does not have hidden blocktriangular structure 6 . If D is not strong, let D 1 ,...,D p be the strong components of D (in acyclic order). Find a permutation π on { 1 ,...,n } such that V ( D i ) = { v π ( n i +1) ,v π ( n i +2) ,...,v π ( n i +1 ) } . This permutation reveals hidden blocktriangular structure of B (and thus A ). Use π to permute rows and columns of A and coordinates of x and b . To perform Step 3 one may use Tarjan’s algorithm in Section 4.4. We will illustrate the procedure above by the following example. Suppose we wish to solve the system: x 1 + 3 x 3 + 8 x 4 = 2 , x 2 + 5 x 4 = 1 , 2 x 1 + 2 x 2 + 4 x 3 + 9 x 4 = 6 , 3 x 2 + 2 x 4 = 3 . We first construct the matrix B and the directed pseudograph D . We have V ( D ) = { v 1 ,v 2 ,v 2 ,v 4 } and A ( D ) = { v 1 v 3 ,v 1 v 4 ,v 2 v 4 ,v 3 v 1 ,v 3 v 2 ,v 3 v 4 ,v 4 v 2 } ∪ { v i v i : i = 1 , 2 , 3 , 4 } . The digraph D has strong components D (1) and D (2) , which are subdigraphs of D induced by { v 1 ,v 3 } and { v 2 ,v 4 } , respectively. These components suggest the following permutation π , π ( i ) = i for i = 1 , 4, π (2) = 3 and π (3) = 2, of rows and columns of A as well as elements of x and b , the righthand side. As a result, we obtain the following: x 1 + 3 x 2 + 8 x 4 = 2 , 2 x 1 + 4 x 2 + 2 x 3 9 x 4 = 6 , x 3 + 5 x 4 = 1 , 3 x 3 + 2 x 4 = 3 , 6 Provided we do not change the set of entries of the diagonal of A 224 4. Classes of Digraphs where x i = x i for i = 1 , 4, x 2 = x 3 and x 3 = x 2 . Solving the last two equations separately, we obtain x 3 = 1 , x 4 = 0 . Now solving the first two equations, we see that x 1 = 2 , x 2 = 0 . Hence, x 1 = 2 , x 2 = 1 , x 3 = x 4 = 0 . A discussion on practical experience with revealing and exploiting block triangular structures is given in [208]. 4.16 Exercises 4.1. Let φ ( u ) be the forefather of a vertex u as defined in Section 4.4. Combining (4.2) and (4.3), prove that φ ( φ ( u )) = φ ( u ). 4.2. Prove Proposition 4.3.1. 4.3. Prove Lemma 4.4.1. 4.4. In part (ii) ⇒ (i) of Theorem 4.5.1, prove that σ ( D ) = L ( Q )....
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 Spring '11
 Algorithms

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