CSL 451 Introduction to Database Systems Indexing and Hashing Department of Computer Science and Engineering Indian Institute of Technology Ropar Narayanan (CK) Chatapuram Krishnan! Summary • • • • • • • • Multilevel indices • B+-Tree Index Files Access types Access time Insertion time Deletion time Space overhead Search key Clustering index – Balanced tree – ceil(n/2) – n number of children for a nonleaf node. • • • • – primary indices • Nonclustering indices – secondary indices • index-sequential files • Dense and Sparse indices 30/03/15! structure of a B+ tree queries insertion deletion • B+-Tree Extensions – file organization – indexing strings • prefix compression – B tree Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 2! 11.1 Indices speed query processing, but it is usually a bad idea to create indices on every attribute, and every combination of attributes, that is a potential search key. Explain why? 30/03/15! Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 3! 11.2 Is it possible in general to have two clustering indices on the same relation for different search keys? Explain? 30/03/15! Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 4! 11.15 When is it preferable to use a dense index rather than a sparse index? Explain your answer. 30/03/15! Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 5! 11.16 What is the difference between a clustering index and a secondary index? 30/03/15! Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 6! 11.3.a Construct a B+ -tree for the following set of key values: (2, 3, 5, 7, 11, 17, 19, 23, 29, 31) Assume that the tree is initially empty and values are added in ascending order. Let the number of pointers that will fit in one node to be 4. 2 3 5 5 2 2 3 3 5 11 5 7 5 5 5 11 7 2 17 2 5 2 3 5 11 7 3 17 19 7 11 5 11 5 7 11 17 19 17 19 23 29 23 5 30/03/15! 5 19 11 2 3 3 5 7 11 19 11 Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 7! 11.3.a Construct a B+ -tree for the following set of key values: (2, 3, 5, 7, 11, 17, 19, 23, 29, 31) Assume that the tree is initially empty and values are added in ascending order. Let the number of pointers that will fit in one node to be 4. 5 2 3 5 7 11 19 11 17 19 23 29 19 23 29 30/03/15! Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 31 8! 11.3.a Construct a B+ -tree for the following set of key values: (2, 3, 5, 7, 11, 17, 19, 23, 29, 31) Assume that the tree is initially empty and values are added in ascending order. Let the number of pointers that will fit in one node to be 4. 5 2 3 5 11 7 19 11 17 19 23 29 19 5 2 30/03/15! 3 5 11 7 29 11 17 19 23 Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 29 31 9! 11.3.b Construct a B+ -tree for the following set of key values: (2, 3, 5, 7, 11, 17, 19, 23, 29, 31) Assume that the tree is initially empty and values are added in ascending order. Let the number of pointers that will fit in one node to be 6. 2 30/03/15! 3 5 7 19 7 11 17 19 Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 23 29 31 10! 11.17 For the B+-tree show the steps involved in the following queries a. find records with a search-key value of 11 b. find records with a search-key value between 7 and 17, inclusive. 19 5 2 30/03/15! 3 5 11 7 29 11 17 19 23 Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 29 31 11! 11.4 Construct a B+ -tree for the following set of key values: (2, 3, 5, 7, 11, 17, 19, 23, 29, 31) Assume that the tree is initially empty and values are added in ascending order. Let the number of points that will fit in one node to be 4. Show the form of the tree after a. insert 9 19 5 2 3 5 11 29 7 11 17 19 23 29 31 19 5 2 30/03/15! 3 5 11 7 29 9 11 17 19 23 Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 29 31 12! 11.4 Construct a B+ -tree for the following set of key values: (2, 3, 5, 7, 11, 17, 19, 23, 29, 31) Assume that the tree is initially empty and values are added in ascending order. Let the number of points that will fit in one node to be 4. Show the form of the tree after b. insert 10 19 5 2 3 11 5 7 29 9 11 17 19 23 29 31 19 5 2 3 5 9 11 29 7 11 9 30/03/15! 17 19 23 29 31 10 Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 13! 11.4 Construct a B+ -tree for the following set of key values: (2, 3, 5, 7, 11, 17, 19, 23, 29, 31) Assume that the tree is initially empty and values are added in ascending order. Let the number of points that will fit in one node to be 4. Show the form of the tree after c. insert 8 19 5 2 3 5 9 11 29 7 11 9 17 19 23 29 31 19 23 29 31 10 19 5 2 3 5 9 7 11 29 8 11 9 30/03/15! 17 10 Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 14! 11.4 Construct a B+ -tree for the following set of key values: (2, 3, 5, 7, 11, 17, 19, 23, 29, 31) Assume that the tree is initially empty and values are added in ascending order. Let the number of points that will fit in one node to be 4. Show the form of the tree after d. delete 23 19 5 2 3 5 9 7 11 29 8 11 9 17 19 29 23 31 10 11 5 2 30/03/15! 3 9 5 19 7 8 9 10 11 17 Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 19 29 31 15! 11.12 What would the occupancy of each leaf node of a B+-tree be, if index entries are inserted in sorted order? Explain why? 30/03/15! Department of Computer Science and Engineering ! Indian Institute of Technology Ropar! 16! Static Hashing ■ A bucket is a unit of storage containing one or more records (a bucket is typically a disk block). ■ In a hash file organization we obtain the bucket of a record directly from its search-key value using a hash function. ■ Hash function h is a function from the set of all search-key values K to the set of all bucket addresses B. ■ Hash function is used to locate records for access, insertion as well as deletion. ■ Records with different search-key values may be mapped to the same bucket; thus entire bucket has to be searched sequentially to locate a record. Database System Concepts - 6th Edition 11.17 ©Silberschatz, Korth and Sudarshan Example of Hash File Organization Hash file organization of instructor file, using dept_name as key (See figure in next slide.) ■ There are 10 buckets, ■ The binary representation of the ith character is assumed to be the integer i. ■ The hash function returns the sum of the binary representations of the characters modulo 10 ● E.g. h(Music) = 1 h(History) = 2 h(Physics) = 3 h(Elec. Eng.) = 3 Database System Concepts - 6th Edition 11.18 ©Silberschatz, Korth and Sudarshan Example of Hash File Organization bucket 0 bucket 4 12121 Wu 76543 Singh bucket 1 15151 Mozart Music 40000 History 80000 58583 Califieri History 60000 98345 Kim Finance 90000 80000 Biology 72000 bucket 5 bucket 2 32343 El Said bucket 3 22222 Einstein 33456 Gold Finance 76766 Crick bucket 6 10101 Srinivasan Comp. Sci. 65000 45565 Katz Comp. Sci. 75000 83821 Brandt Comp. Sci. 92000 bucket 7 Physics 95000 Physics 87000 Elec. Eng. 80000 Hash file organization of instructor file, using dept_name as key (see previous slide for details). Database System Concepts - 6th Edition 11.19 ©Silberschatz, Korth and Sudarshan Hash Functions ■ Worst hash function maps all search-key values to the same bucket; this makes access time proportional to the number of search-key values in the file. ■ An ideal hash function is uniform, i.e., each bucket is assigned the same number of search-key values from the set of all possible values. ■ Ideal hash function is random, so each bucket will have the same number of records assigned to it irrespective of the actual distribution of search-key values in the file. ■ Typical hash functions perform computation on the internal binary representation of the search-key. ● For example, for a string search-key, the binary representations of all the characters in the string could be added and the sum modulo the number of buckets could be returned. . Database System Concepts - 6th Edition 11.20 ©Silberschatz, Korth and Sudarshan Handling of Bucket Overflows ■ Bucket overflow can occur because of ● Insufficient buckets ● Skew in distribution of records. This can occur due to two reasons: ! multiple records have same search-key value ! chosen hash function produces non-uniform distribution of key values ■ Although the probability of bucket overflow can be reduced, it cannot be eliminated; it is handled by using overflow buckets. Database System Concepts - 6th Edition 11.21 ©Silberschatz, Korth and Sudarshan Handling of Bucket Overflows (Cont.) ■ Overflow chaining – the overflow buckets of a given bucket are chained together in a linked list. ■ Above scheme is called closed hashing. ● An alternative, called open hashing, which does not use overflow buckets, is not suitable for database applications. bucket 0 bucket 1 overflow buckets for bucket 1 bucket 2 bucket 3 Database System Concepts - 6th Edition 11.22 ©Silberschatz, Korth and Sudarshan Hash Indices ■ Hashing can be used not only for file organization, but also for index- structure creation. ■ A hash index organizes the search keys, with their associated record pointers, into a hash file structure. ■ Strictly speaking, hash indices are always secondary indices ● if the file itself is organized using hashing, a separate primary hash index on it using the same search-key is unnecessary. ● However, we use the term hash index to refer to both secondary index structures and hash organized files. Database System Concepts - 6th Edition 11.23 ©Silberschatz, Korth and Sudarshan Example of Hash Index bucket 0 76766 bucket 1 45565 76543 bucket 2 22222 bucket 3 10101 bucket 4 bucket 5 15151 33456 58583 98345 76766 10101 45565 83821 98345 12121 76543 32343 58583 15151 22222 33465 Crick Srinivasan Katz Brandt Kim Wu Singh El Said Califieri Mozart Einstein Gold Biology Comp. Sci. Comp. Sci. Comp. Sci. Elec. Eng. Finance Finance History History Music Physics Physics 72000 65000 75000 92000 80000 90000 80000 60000 62000 40000 95000 87000 bucket 6 83821 hash index on instructor, on attribute ID bucket 7 12121 32343 Database System Concepts - 6th Edition 11.24 ©Silberschatz, Korth and Sudarshan Deficiencies of Static Hashing ■ In static hashing, function h maps search-key values to a fixed set of B of bucket addresses. Databases grow or shrink with time. ● If initial number of buckets is too small, and file grows, performance will degrade due to too much overflows. ● If space is allocated for anticipated growth, a significant amount of space will be wasted initially (and buckets will be underfull). ● If database shrinks, again space will be wasted. ■ One solution: periodic re-organization of the file with a new hash function ● Expensive, disrupts normal operations ■ Better solution: allow the number of buckets to be modified dynamically. Database System Concepts - 6th Edition 11.25 ©Silberschatz, Korth and Sudarshan Dynamic Hashing ■ Good for database that grows and shrinks in size ■ Allows the hash function to be modified dynamically ■ Extendable hashing – one form of dynamic hashing ● ● ● Hash function generates values over a large range — typically b-bit integers, with b = 32. At any time use only a prefix of the hash function to index into a table of bucket addresses. Let the length of the prefix be i bits, 0 ≤ i ≤ 32. ! Bucket address table size = 2i. Initially i = 0 ! Value ● ● of i grows and shrinks as the size of the database grows and shrinks. Multiple entries in the bucket address table may point to a bucket Thus, actual number of buckets is < 2i ! The number of buckets also changes dynamically due to coalescing and splitting of buckets. Database System Concepts - 6th Edition 11.26 ©Silberschatz, Korth and Sudarshan General Extendable Hash Structure hash prefix i1 i 00.. 01.. bucket 1 10.. i2 11.. … bucket 2 i3 bucket 3 bucket address table … In this structure, i2 = i3 = i, whereas i1 = i – 1 (see next slide for details) Database System Concepts - 6th Edition 11.27 ©Silberschatz, Korth and Sudarshan Use of Extendable Hash Structure ■ Each bucket j stores a value ij ● All the entries that point to the same bucket have the same values on the first ij bits. ■ To locate the bucket containing search-key Kj: 1. Compute h(Kj) = X 2. Use the first i high order bits of X as a displacement into bucket address table, and follow the pointer to appropriate bucket ■ To insert a record with search-key value Kj ● follow same procedure as look-up and locate the bucket, say j. ● If there is room in the bucket j insert record in the bucket. ● Else the bucket must be split and insertion re-attempted (next slide.) ! Overflow Database System Concepts - 6th Edition buckets used instead in some cases (will see shortly) 11.28 ©Silberschatz, Korth and Sudarshan Insertion in Extendable Hash Structure (Cont) To split a bucket j when inserting record with search-key value Kj: ■ If i > ij (more than one pointer to bucket j) ● ● ● ● allocate a new bucket z, and set ij = iz = (ij + 1) Update the second half of the bucket address table entries originally pointing to j, to point to z remove each record in bucket j and reinsert (in j or z) recompute new bucket for Kj and insert record in the bucket (further splitting is required if the bucket is still full) ■ If i = ij (only one pointer to bucket j) ● ● If i reaches some limit b, or too many splits have happened in this insertion, create an overflow bucket Else ! increment i and double the size of the bucket address table. ! replace each entry in the table by two entries that point to the same bucket. ! recompute new bucket address table entry for Kj Now i > ij so use the first case above. Database System Concepts - 6th Edition 11.29 ©Silberschatz, Korth and Sudarshan Deletion in Extendable Hash Structure ■ To delete a key value, ● locate it in its bucket and remove it. ● The bucket itself can be removed if it becomes empty (with appropriate updates to the bucket address table). ● Coalescing of buckets can be done (can coalesce only with a “buddy” bucket having same value of ij and same ij –1 prefix, if it is present) ● Decreasing bucket address table size is also possible ! Note: decreasing bucket address table size is an expensive operation and should be done only if number of buckets becomes much smaller than the size of the table Database System Concepts - 6th Edition 11.30 ©Silberschatz, Korth and Sudarshan Use of Extendable Hash Structure: Example !"#$%&'(" !"!"#$%&'("# $%&'&() **+*ȱ++*+ȱ++++ȱ+*++ȱ**+*ȱ++**ȱ**++ȱ**** ,&-./ȱ01%/ ++++ȱ***+ȱ**+*ȱ*+**ȱ+**+ȱ**++ȱ*++*ȱ++*+ 2'31/ȱ24(/ *+**ȱ**++ȱ+*+*ȱ++**ȱ++**ȱ*++*ȱ++*+ȱ++++ 5%46413 +*+*ȱ**++ȱ+*+*ȱ****ȱ++**ȱ*++*ȱ+**+ȱ++++ 7%89&:) ++**ȱ*+++ȱ+++*ȱ++*+ȱ+*++ȱ++++ȱ**++ȱ+*+* ;<8%1 **++ȱ*+*+ȱ+*+*ȱ*++*ȱ++**ȱ+**+ȱ+++*ȱ+*++ =!)8%18ȱȱȱȱȱȱȱȱȱȱȱȱ+**+ȱ+***ȱ**++ȱ++++ȱ+**+ȱ++**ȱ****ȱ***+ Database System Concepts - 6th Edition 11.31 ©Silberschatz, Korth and Sudarshan Example (Cont.) ■ Initial Hash structure; bucket size = 2 hash prefix 0 0 bucket address table Database System Concepts - 6th Edition bucket 1 11.32 ©Silberschatz, Korth and Sudarshan Example (Cont.) ■ Hash structure after insertion of “Mozart”, “Srinivasan”, and “Wu” records hash prefix 1 1 15151 Mozart bucket address table !"#$%&'(" 40000 1 !"!"#$%&'("# $%&'&() **+*ȱ++*+ȱ++++ȱ+*++ȱ**+*ȱ++**ȱ**++ȱ**** ,&-./ȱ01%/ ++++ȱ***+ȱ**+*ȱ*+**ȱ+**+ȱ**++ȱ*++*ȱ++*+ 2'31/ȱ24(/ *+**ȱ**++ȱ+*+*ȱ++**ȱ++**ȱ*++*ȱ++*+ȱ++++ 5%46413 +*+*ȱ**++ȱ+*+*ȱ****ȱ++**ȱ*++*ȱ+**+ȱ++++ 7%89&:) ++**ȱ*+++ȱ+++*ȱ++*+ȱ+*++ȱ++++ȱ**++ȱ+*+* ;<8%1 **++ȱ*+*+ȱ+*+*ȱ*++*ȱ++**ȱ+**+ȱ+++*ȱ+*++ =!)8%18ȱȱȱȱȱȱȱȱȱȱȱȱ+**+ȱ+***ȱ**++ȱ++++ȱ+**+ȱ++**ȱ****ȱ***+ Database System Concepts - 6th Edition Music 11.33 10101 Srinivasan Comp. Sci. 90000 12121 Wu Finance 90000 Add “Einstein” record ©Silberschatz, Korth and Sudarshan Example (Cont.) ■ Hash structure after insertion of Einstein record hash prefix 1 2 15151 Mozart Music 40000 12121 Wu Finance 90000 22222 Einstein Physics 95000 2 bucket address table !"#$%&'(" 2 !"!"#$%&'("# $%&'&() **+*ȱ++*+ȱ++++ȱ+*++ȱ**+*ȱ++**ȱ**++ȱ**** ,&-./ȱ01%/ ++++ȱ***+ȱ**+*ȱ*+**ȱ+**+ȱ**++ȱ*++*ȱ++*+ 2'31/ȱ24(/ *+**ȱ**++ȱ+*+*ȱ++**ȱ++**ȱ*++*ȱ++*+ȱ++++ 5%46413 +*+*ȱ**++ȱ+*+*ȱ****ȱ++**ȱ*++*ȱ+**+ȱ++++ 7%89&:) ++**ȱ*+++ȱ+++*ȱ++*+ȱ+*++ȱ++++ȱ**++ȱ+*+* ;<8%1 **++ȱ*+*+ȱ+*+*ȱ*++*ȱ++**ȱ+**+ȱ+++*ȱ+*++ =!)8%18ȱȱȱȱȱȱȱȱȱȱȱȱ+**+ȱ+***ȱ**++ȱ++++ȱ+**+ȱ++**ȱ****ȱ***+ Database System Concepts - 6th Edition 11.34 10101 Srinivasan Comp. Sci. 65000 Add “Gold” and “El Said” records – Physics and History ©Silberschatz, Korth and Sudarshan Example (Cont.) ■ Hash structure after insertion of Gold and El Said records 1 hash prefix 3 Mozart Music 40000 22222 Einstein Physics 33456 Physics 95000 87000 Finance 90000 15151 3 Gold 3 12121 Wu !"#$%&'(" !"!"#$%&'("# bucket address table $%&'&() **+*ȱ++*+ȱ++++ȱ+*++ȱ**+*ȱ++**ȱ**++ȱ**** ,&-./ȱ01%/ ++++ȱ***+ȱ**+*ȱ*+**ȱ+**+ȱ**++ȱ*++*ȱ++*+ 2'31/ȱ24(/ *+**ȱ**++ȱ+*+*ȱ++**ȱ++**ȱ*++*ȱ++*+ȱ++++ 5%46413 +*+*ȱ**++ȱ+*+*ȱ****ȱ++**ȱ*++*ȱ+**+ȱ++++ 7%89&:) ++**ȱ*+++ȱ+++*ȱ++*+ȱ+*++ȱ++++ȱ**++ȱ+*+* ;<8%1 **++ȱ*+*+ȱ+*+*ȱ*++*ȱ++**ȱ+**+ȱ+++*ȱ+*++ =!)8%18ȱȱȱȱȱȱȱȱȱȱȱȱ+**+ȱ+***ȱ**++ȱ++++ȱ+**+ȱ++**ȱ****ȱ***+ Database System Concepts - 6th Edition 11.35 2 10101 32343 Srinivasan Comp. Sci. 65000 60000 El Said History ©Silberschatz, Korth and Sudarshan Example (Cont.) 2 15151 Mozart 76766 Crick hash prefix 3 Music Biology 40000 72000 22222 Einstein Physics 33456 Gold Physics 95000 87000 Finance 90000 Finance 80000 History 60000 History 62000 3 And after insertion of eleven records 3 12121 Wu 76543 Singh 3 bucket address table 32343 El Said 58583 Califieri 3 10101 Srinivasan Comp. Sci. 45565 Katz Database System Concepts - 6th Edition 65000 Comp. Sci. 75000 11.36 83821 Brandt Comp. Sci. 92000 ©Silberschatz, Korth and Sudarshan
© Copyright 2024