# 119 answer let i denote the number of bits of the

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11.9 Answer: Let i denote the number of bits of the hash value used in the hash table. Let bsize denote the maximum capacity of each bucket. The pseudocode is shown in Figure 11.2. Note that we can only merge two buckets at a time. The common hash prefix of the resultant bucket will have length one less than the two buckets merged. Hence we look at the buddy bucket of bucket j differing from it only at the last bit. If the common hash prefix of this bucket is not i j , then this implies that the buddy bucket has been further split and merge is not possible. When merge is successful, further merging may be possible, which is handled by a recursive call to coalesce at the end of the function. 11.10 Answer: If the hash table is currently using i bits of the hash value, then maintain a count of buckets for which the length of common hash prefix is exactly i . Consider a bucket j with length of common hash prefix i j . If the bucket is being split, and i j is equal to i , then reset the count to 1. If the bucket is being split and i j is one less that i , then increase the count by 1. It the bucket if being coalesced, and i j is equal to i then decrease the count by 1. If the count becomes 0, then the bucket address table can be reduced in size at that point. However, note that if the bucket address table is not reduced at that point, then the count has no significance afterwards. If we want to postpone the reduction, we have to keep an array of counts, i.e. a count for each value of

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Exercises 7 function findIterator(value V ) { /* Returns an iterator for the search on the value V */ Iterator iter (); Set iter .v alue = V ; Set C = root node while ( C is not a leaf node) begin Let i = samllest number such that V <= C . K i if there is no such number i then begin Let P m = last non-null pointer in the node Set C = C . P m ; end else Set C = C . P i ; end /* C is a leaf node */ Let i be the least value such that K i = V if there is such a value i then begin Set iter . index = i ; Set iter . page = C ; Set iter . acti v e = TRUE ; end else if ( V is the greater than the largest value in the leaf) then begin if ( C . P n . K 1 = V ) then begin Set iter . page = C . P n ; Set iter . index = 1; Set iter . acti v e = TRUE ; end else Set iter . acti v e = FALSE ; end else Set iter . acti v e = FALSE ; return ( iter ) } Class Iterator { variables: value V /* The value on which the index is searched */ boolean active /* Stores the current state of the iterator (TRUE or FALSE)*/ int index /* Index of the next matching entry (if active is TRUE) */ PageID page /* Page Number of the next matching entry (if active is TRUE)*/ function next() { if (active) then begin Set retPage = page ; Set retIndex = index ; if ( index + 1 = page . size ) then begin page = page . P n index = 0 end else index = index + 1; if ( page . K index negationslash= V ) then acti v e = FALSE ; return( retPage , retIndex ) end else return null ; } } Figure 11.1 Pseudocode for findIterator and the Iterator class
8 Chapter 11 Indexing and Hashing delete (value K l ) begin j = first i high-order bits of h ( K l ); delete value K l from bucket j ; coalesce (bucket j ); end coalesce (bucket j ) begin i j = bits used in bucket j ; k = any bucket with first ( i j

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• Spring '13
• Dr.Khansari
• hash function, Prime number, Bucket address table

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