Tutorial4-sol - Problem 1: Indexing on Sequential Files...

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Problem 1: Indexing on Sequential Files Solution 1. 100 / 10 = 10 blocks. Explanation: A block can hold 10 key-pointer pairs, and we need a pair for every tuple. 100 tuples / 10 key-points pairs per block = 10 blocks. 2. (100 / 2) / 10 = 5 blocks. Explanation: A block can hold 10 key-pointer pairs, and we need a pair for every tuple that is the first in its block. Every block can hold 2 tuples. Therefore there are 100 / 2 = 50 tuples that need key-pointer pairs. Now, it's as if we're building a dense index on these 50 tuples, and 50 / 10 = 5 blocks. 3. Clustered index reorders the data, unclustered does not. So with unclustered index, can't have ordered tuples within blocks (because nothing is sorted). Therefore sparse index is impossible.
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Problem 2: B+ Tree Solution 1. First valid answer (recommended): The following changes will be done to the original B-tree: Step 1. The frst node in the leaF level split to two nodes: (1,2) and (4,5,6). Step 2. The frst node in the second level split to two nodes
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This note was uploaded on 02/17/2012 for the course CS 411 taught by Professor Winslett during the Fall '07 term at University of Illinois at Urbana–Champaign.

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Tutorial4-sol - Problem 1: Indexing on Sequential Files...

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