# Chapter 4 Index Structures

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1 Chapter 4 Index Structures Having seen the options available for representing records, we must now consider how whole relations, or the extents of classes, are represented. It is not sufficient

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3 4.1. INDEXES ON SEQUENTIAL FILES 125 Keys and More Keys The term "key" has several meanings, and this book uses "key" in each

4 126 CHAPTER 4. INDEX STRUCTURES Figure 4.2: A sequential file key every 10 integers, although in practice we would not expect to find such a regular pattern of keys. We have also assumed that index blocks can hold only four key-pointer pairs. Again, in practice we would find typically that there were many more pairs per block, perhaps hundreds.

5 4.1. INDEXES ON SEQUENTIAL FILES 127

6 128 CHAPTER 4. INDEX STRUCTURES Locating Index Blocks Wo have assumed that some mechanism exists for locating the index blocks, from which the individual tuples (if the index is dense) or blocks of the data file (if the index is sparse) can be found. Many ways of locating

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8 130 CHAPTER 4. INDEX STRUCTURES

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10 132 CHAPTER 4. INDEX STRUCTURES

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12 134 CHAPTER 4. INDEX STRUCTURES

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14 136 CHAPTER 4. INDEX STRUCTURES Preparing for Evolution of Data Since it is common for relations or class extents to grow with time, it is often wise to distribute extra space among blocks both data and index blocks. If blocks are, say, 75% full to begin with, then we can run for some time before having to create overflow blocks or slide records between

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16 138 CHAPTER 4. INDEX STRUCTURES block at the front. Since 40 is now the first key on the second data block, we need to update the index record for that block. We see in Fig that the key

17 4.1. INDEXES ON SEQUENTIAL FILES 139 Figure 4.13: Insertion into a file with a sparse index, using immediate reorganization Our last step is to modify the index entries of the changed blocks. Wo might have

18 140 CHAPTER 4. INDEX STRUCTURES Figure 4.14: Insertion into a file with a sparse index, using overflow blocks

19 4.1. INDEXES ON SEQUENTIAL FILES 141

20 142 CHAPTER 4. INDEX STRUCTURES 4.2 Secondary Indexes

21 4.2. SECONDARY INDEXES 143 Example 4.13: Figure 4.15 shows a typical secondary index. The data file

22 144 CHAPTER 4. INDEX STRUCTURES Applications of Secondary Indexes Besides suppoiting

23 4.2. SECONDARY INDEXES 145

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25 4.2. SECONDARY INDEXES 147 Example 4.16 : Consider the relation Movie(title, year, length, studioname)

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27 4.2. SECONDARY INDEXES 149

28 150 CHAPTER 4. INDEX STRUCTURES More About Information Retrieval There are a number of techniques for improving the effectiveness of retrieval of documents given keywords. While a complete treatment is be-

29 4.2. SECONDARY INDEXES 151 Insertion and Deletion From Buckets We show buckets in figures such as Fig as compacted an ays of appro-

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31 4.2. SECONDARY INDEXES 153

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33 1.3. B-TREES 155 At a leaf, the last pointer points to the next leaf block to the right, i.e., to the block with the next higher keys. Among the other n pointers in a leaf block, at least L 2^] ^tnese pointers are used and point to data records; unused pointers may be thought of as null and do not point anywhere. The zth pointer, if it is used, points to a record with the iih key. At an interior node, all n + 1 pointers can be used to point to B-tree

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35 4,3. B-TREES 157

36 158 CHAPTER 4. INDEX STRUCTURES

37 4.3. B-TREES , and 31 have all been replaced by 23. As a result, the key at the root is 17,

38 160 CHAPTER 4. INDEX STRUCTURES At that node, we find 31 < 40 < 43, so we follow the third pointer. We are

39 4.3 B-TREES 161 Example 4.24 : Suppose we have the B-tree of Fig. 4.23, and we are given the range (10,25) to search for. We look for key 10, which leads us to the second leaf. The first key is less than 10, but the second, 11, is at least 10. We follow

40 162 CHAPTER 4. INDEX STRUCTURES

41 4.3. B-TREES 163

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43 4.3. B-TREES 165

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45 4.3. B-TREES 167 Figure 4.29: Completing the deletion of key 11 disk I/O's needed foi manipulation of the record itself. We must thus ask: how many levels does a B-tree have? For the typical sizes of keys, pointers,

46 168 CHAPTER 4. INDEX STRUCTURES

47 4.3. B-TREES 169 Exercise 4.3.4: What are the minimum numbers of keys and pointers in B-

48 170 CHAPTER 4. INDEX STRUCTURES!! *! a) 6 records.!! b) 10 records. He) 15 records.

49 4.4. HASH TABLES 171

50 172 CHAPTER 4. INDEX STRUCTURES Choice of Hash Function The hash function should "hash" the key so the resulting integer is a seemingly random function of the key. Thus, buckets will tend to have

51 4.4. HASH TABLES 173

52 174 CHAPTER 4. INDEX STRUCTURES

53 4.4. HASH TABLES 175 smaller number of bits, say i bits, from the beginning of this sequence. That is, the bucket array will have 2* entries when i is the number of bits used. Example 4.31: Figure 4.33 shows a small extensible hash table. We suppose, for simplicity of the example, that k 4; i.e., the hash function produces a sequence of only four bits. At the moment, only one of these bits is used, as indicated by i = 1 in the box above the bucket array. The bucket array therefore has only two entries, one for 0 and one for 1. Buckets Data blocks Figure 4.33: An extensible hash table The bucket array entries point to two blocks. The first holds all the current records whose search keys hash to a bit sequence that begins with 0, and the second holds all those whose search keys hash to a sequence beginning with 1. For convenience, we show the keys of records as if they were the entire bit sequence that the hash function converts them to. Thus, the first block holds a record whose key hashes to 0001, and the second holds records whose keys hash to 1001 and D We should notice the number 1 appearing in the "nub" of each of the blocks in Fig This number, which would actually appear in the block header, indicates how many bits of the hash function's sequence is used to determine membership of records in this block. In the situation of Example 4.31, there

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55 4.4. HASH TABLES 177 Figure 4.34: Now, two bits of the hash function are used partition its records into those beginning 10 and those beginning 11. A 2 in each of these blocks indicates that two bits are used to determine membership. Fortunately, the split is successful; since each of the two new blocks gets at least one record, we do not have to split recursively. Now suppose we insert records whose keys hash to 0000 and These

56 178 CHAPTER 4. INDEX STRUCTURES Figure 4.35: The hash table now uses three bits of the hash function

57 4.4. HASH TABLES 179 one block. Since blocks cannot always be split, overflow blocks are permitted, although the average number of overflow blocks pei bucket will be much less than 1. The number of bits used to number the entries of the bucket array is [Iog 2 n~\, where n is the current number of buckets. These bits are always taken from the right (low-order) end of the bit sequence that is produced by the hash function. Suppose i bits of the hash function are being used to number array entries, and a record with key K is intended for bucket 0102 a z ; that is, aitt2

58 180 CHAPTER 4. INDEX STRUCTURES Insertion Into Linear Hash Tables

59 4.4. HASH TABLES 181 Figure 4.38: Overflow blocks are used if necessary

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61 4.4. HASH TABLES 183

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