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The algorithm in Figure 136 requires that every set of tuples Ss t in main memory; we shall look at extensions of the algorithm to avoid this requirement later in this section Then, the corresponding tuples (if any) of the other relation are read in, and are processed as they are read Figure 137 shows two relations that are sorted on their join attribute a1 It is instructive to go through the steps of the merge join algorithm on the relations shown in the gure Since the relations are in sorted order, tuples with the same value on the join attributes are in consecutive order Thereby, each tuple in the sorted order needs to be read only once, and, as a result, each block is also read only once Since it makes only a single pass through both les, the merge join method is ef cient; the number of block accesses is equal to the sum of the number of blocks in both les, br + bs If either of the input relations r and s is not sorted on the join attributes, they can be sorted rst, and then the merge join algorithm can be used The merge join algorithm can also be easily extended from natural joins to the more general case of equi-joins Suppose the merge join scheme is applied to our example of depositor 1 customer The join attribute here is customer-name Suppose that the relations are already sorted on the join attribute customer-name In this case, the merge join takes a total of 400 + 100 = 500 block accesses Suppose the relations are not sorted, and the memory size is the worst case of three blocks Sorting customer takes 400 (2 log2 (400/3) + 1), or 6800, block transfers, with 400 more transfers to write out the result Similarly, sorting depositor takes 100 (2 log2 (100/3) + 1), or 1300, transfers, with 100 more transfers to write it out Thus, the total cost is 9100 block transfers if the relations are not sorted, and the memory size is just 3 blocks With a memory size of 25 blocks, sorting the relation customer takes a total of just 400 (2 log24 (400/25) + 1) = 1200 block transfers, while sorting depositor takes 300 block transfers Adding the cost of writing out the sorted results and reading them back gives a total cost of 2500 block transfers if the relations are not sorted and the memory size is 25 blocks As mentioned earlier, the merge join algorithm of Figure 136 requires that the set Ss of all tuples with the same value for the join attributes must t in main memory.

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Double-click a transition in the library area to make it appear in this box FIGURE 15-6

d 13

The McGraw Hill Companies, 2001

electric eld Metals such as copper or aluminum, which are good conductors, have very large values of , while a material like glass or wood will have a small value of The inverse of conductivity is resistivity, which is denoted by the Greek symbol Resistivity is the inverse of conductivity = So we could write (23) as E = J (24) 1

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This requirement can usually be met, even if the relation s is large If it cannot be met, a block nested-loop join must be performed between Ss and the tuples in r with the same values for the join attributes The overall cost of the merge join increases as a result It is also possible to perform a variation of the merge join operation on unsorted tuples, if secondary indices exist on both join attributes The algorithm scans the records through the indices, resulting in their being retrieved in sorted order This variation presents a signi cant drawback, however, since records may be scattered throughout the le blocks Hence, each tuple access could involve accessing a disk block, and that is costly To avoid this cost, we can use a hybrid merge join technique, which combines indices with merge join Suppose that one of the relations is sorted; the other is unsorted, but has a secondary B+ -tree index on the join attributes The hybrid merge join algorithm merges the sorted relation with the leaf entries of the secondary B+ tree index The result le contains tuples from the sorted relation and addresses for tuples of the unsorted relation The result le is then sorted on the addresses of tuples of the unsorted relation, allowing ef cient retrieval of the corresponding tuples, in physical storage order, to complete the join Extensions of the technique to handle two unsorted relations are left as an exercise for you

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