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2 Level Indexes

2 Level Indexes. Indexed Files - Part Two Portions of this lecture stolen from Foster's 325 Lecture Notes. Where we left off last class. The primary purpose of using indexes is to speed searching. In a single layer indexed file, the index-to-data relationship is 1:1.

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2 Level Indexes

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  1. 2 Level Indexes Indexed Files - Part Two Portions of this lecture stolen from Foster's 325 Lecture Notes

  2. Where we left off last class The primary purpose of using indexes is to speed searching. In a single layer indexed file, the index-to-data relationship is 1:1. The index resides in main memory for fast access. What if that 1:1 index is too big for memory?

  3. 2 Levels of Indexes First Level Index • resides in memory • entries point to the Second Level • entries are ordered for fast searching • entries contain • key • an IRRN - Index Relative Record Number (pointer into the Second Level) • size is TBD Second Level Index • entirety stays in a file • 1:1 ratio of entries to records in data file • entries are ordered for fast searching • entries contain • key • DRRN - Data RRN (pointer into the data file)

  4. An Overly Simple Example

  5. Search Algorithm Can this be a binary search? Data Size = N records Level One Size = K1 entries Preconditions : K2 = N / K1 level one index is already in an array in memory (arrary1) i = 0; while (Target > array1[i].Key) && (i < K1) i++; i = i-1; SeekG (secondaryfile, array1[i].IRRN*sizeof(index records)) Read (secondaryfile, K2 records, into array2) binary search array2 for Target SeekG (datafile, array2[location].DRRN*sizeof(data records)) read record from datafile

  6. A Better Example

  7. Add Algorithm YIKES! Sorting a File takes a long time! Append new record to end of datafile add entry (Key and DRRN) to end of secondary file sort secondary key K2 = N / K1 for (i=0; i<K1; i++) array1[i].key = secondarykey (i * K2) array1[i].IRRN = i * K2

  8. Better Structure when Additions are Frequent • Instead of filling the secondary index, leave room for expansion. • Example • between Adams and Foster, put 15 names instead of 20 • that leaves 5 growth spots before an adjustment is needed • when adding "Baker", only need to sort (move) Adams to Foster-1

  9. Theoretical Best Size of Index 1 • Remember: • Level 1 index stays in memory • only a portion of Level 2 goes into memory • To minimize search times of those two arrays, optimal size of Index 1 is sqrt(N) • Example • Assume N = 100 • Size of Level 1 = sqrt(100) = 10 • each of those level 1 entries points to 10 level 2 entries • so we end up searching two arrays of 10 elements each

  10. Real Best Size of Index 1 "To minimize search times of those two arrays, optimal size of Index 1 is sqrt(N)" But array2 must be read from a file over and over and over. So, the smaller array2 the better! Hence, optimal size of Index 1 = as big as main memory allows

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