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sort merge join sort( a ), Sort( b ); x, y = 0 while !r.empty() && !s.empty(): if ( a[ x ] == b[ y ] ) output += a[ x ] x++; y++ elseif ( a[ x ] < b[ y ] ) x++ else y++ |
Optimized sort merge join algorithm
1) Combine join with the merge phase of sort
2) Sort R and S in M runs of size Mon disk.
3) Merge and join the tuples in one pass.
1) Cost: 3b( r ) + 3b( s )
2) Memory requirement: b( r ) + b( s ) <= m power 2
- because we merge them in one pass
- more efficient but more strict requirement.
S_ID | S_NAME | RATING | S_AGE |
22 | S_Name1 | 7 | 45 |
28 | S_Name2 | 9 | 35 |
31 | S_Name3 | 8 | 55 |
44 | S_Name4 | 5 | 35 |
58 | S_Name5 | 10 | 35 |
S_ID | B_ID | DAY | R_NAME |
28 | 103 | 11-4-96 | R_Name1 |
28 | 103 | 10-3-96 | R_Name2 |
31 | 101 | 9-10-96 | R_Name3 |
31 | 102 | 9-12-96 | R_Name4 |
31 | 101 | 9-11-96 | R_Name5 |
58 | 103 | 10-12-96 | R_Name6 |
1) Cost of the sort is : A( logA )+ ( B logB ) + (A + B)
2) And the scanning of the cost is, ( A + B ), could be ( A*B )
- With ‘35’, ‘100’ or ‘300’ buffer pages, both can be sorted in 2 passes;
- Total cost of the join is = 7500.