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Silberschatz Korth Sudarshan: Database System Concepts, Fourth Edition
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bucket 3
General extendable hash structure
We do not create a bucket for each hash value Indeed, 232 is over 4 billion, and that many buckets is unreasonable for all but the largest databases Instead, we create buckets on demand, as records are inserted into the le We do not use the entire b bits of the hash value initially At any point, we use i bits, where 0 i b These i bits are used as an offset into an additional table of bucket addresses The value of i grows and shrinks with the size of the database Figure 1224 shows a general extendable hash structure The i appearing above the bucket address table in the gure indicates that i bits of the hash value h(K) are required to determine the correct bucket for K This number will, of course, change as the le grows Although i bits are required to nd the correct entry in the bucket address table, several consecutive table entries may point to the same bucket All such entries will have a common hash pre x, but the length of this pre x may be less than i Therefore, we associate with each bucket an integer giving the length of the common hash pre x In Figure 1224 the integer associated with bucket j is shown as ij The number of bucket-address-table entries that point to bucket j is 2(i ij )
1262 Queries and Updates
We now see how to perform lookup, insertion, and deletion on an extendable hash structure To locate the bucket containing search-key value Kl , the system takes the rst i high-order bits of h(Kl ), looks at the corresponding table entry for this bit string, and follows the bucket pointer in the table entry To insert a record with search-key value Kl , the system follows the same procedure for lookup as before, ending up in some bucket say, j If there is room in the bucket,
Silberschatz Korth Sudarshan: Database System Concepts, Fourth Edition
IV Data Storage and Querying
12 Indexing and Hashing
The McGraw Hill Companies, 2001
Dynamic Hashing
the system inserts the record in the bucket If, on the other hand, the bucket is full, it must split the bucket and redistribute the current records, plus the new one To split the bucket, the system must rst determine from the hash value whether it needs to increase the number of bits that it uses If i = ij , only one entry in the bucket address table points to bucket j Therefore, the system needs to increase the size of the bucket address table so that it can include pointers to the two buckets that result from splitting bucket j It does so by considering an additional bit of the hash value It increments the value of i by 1, thus doubling the size of the bucket address table It replaces each entry by two entries, both of which contain the same pointer as the original entry Now two entries in the bucket address table point to bucket j The system allocates a new bucket (bucket z), and sets the second entry to point to the new bucket It sets ij and iz to i Next, it rehashes each record in bucket j and, depending on the rst i bits (remember the system has added 1 to i), either keeps it in bucket j or allocates it to the newly created bucket The system now reattempts the insertion of the new record Usually, the attempt will succeed However, if all the records in bucket j, as well as the new record, have the same hash-value pre x, it will be necessary to split a bucket again, since all the records in bucket j and the new record are assigned to the same bucket If the hash function has been chosen carefully, it is unlikely that a single insertion will require that a bucket be split more than once, unless there are a large number of records with the same search key If all the records in bucket j have the same search-key value, no amount of splitting will help In such cases, over ow buckets are used to store the records, as in static hashing If i > ij , then more than one entry in the bucket address table points to bucket j Thus, the system can split bucket j without increasing the size of the bucket address table Observe that all the entries that point to bucket j correspond to hash pre xes that have the same value on the leftmost ij bits The system allocates a new bucket (bucket z), and set ij and iz to the value that results from adding 1 to the original ij value Next, the system needs to adjust the entries in the bucket address table that previously pointed to bucket j (Note that with the new value for ij , not all the entries correspond to hash pre xes that have the same value on the leftmost ij bits) The system leaves the rst half of the entries as they were (pointing to bucket j), and sets all the remaining entries to point to the newly created bucket (bucket z) Next, as in the previous case, the system rehashes each record in bucket j, and allocates it either to bucket j or to the newly created bucket z The system then reattempts the insert In the unlikely case that it again fails, it applies one of the two cases, i = ij or i > ij , as appropriate Note that, in both cases, the system needs to recompute the hash function on only the records in bucket j To delete a record with search-key value Kl , the system follows the same procedure for lookup as before, ending up in some bucket say, j It removes both the
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