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Unformatted text preview: 6.896 Sublinear Time Algorithms September 21, 2001 Lecture 3: Estimating the Number of Connected Components Lecturer: Ronitt Rubinfeld Scribe: Kevin Matulef 1 Testing for Connectedness Reviewing from last class, we had the following tester for connectedness: 1. Choose m = O ( 1 d ) nodes. 2. For every chosen node s , perform a BFS until (a) ≥ 8 d new nodes are found. (b) s is found to be in a component of size ≤ 8 d . 3. If (b) occurs, reject, otherwise accept. Note that this algorithm requires O ( 1 2 d ) total queries. For a BFS to discover up to 8 d new nodes, it requires up to O ( 8 d · d ) = O ( 1 ) queries. Since we do O ( 1 d ) BFSs, the total number of queries is O ( 1 2 d ). To analyze the validity of the algorithm, we had the following lemma: Lemma 1 G is-far from P ( d ) n ⇒ G has ≥ dn 4 components From this, we derive the following corollary: Corollary 2 G is-far from P ( d ) n ⇒ G has ≥ dn 8 “small” components of size ≤ 8 d . Proof Let = number of components of size ≤ 8 d and = number of components of size > 8 d . From the lemma, we know that + ≥ dn 4 . We can bound the number of large components with the equation 8 d ≤ n . This simply comes from the fact that there can’t be more nodes in large components than the total number of nodes. We can rearrange this to say ≤ dn 8 . Combining our lower bound on the total number of components with our upper bound on the number of large components, we get the lower bound ≥ dn 4 − dn 8 = dn 8 on the number of small components....
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This note was uploaded on 04/02/2010 for the course CS 6.896 taught by Professor Ronittrubinfeld during the Spring '04 term at MIT.
- Spring '04